Course Hero - We put you ahead of the curve!

You have requested the below document.

Coursehero >> California >> UCSD >> MATH 110

K. Bruce Driver Math 110, Spring 2004 Notes May 25, 2004 File:110notes.tex Springer Berlin Heidelberg NewYork Hong Kong London Milan Paris Tokyo Contents 1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 The spectral theorem for symmetric matrices . . . . . . . . . . . . . . . 1 1.2 Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . 5 1.2.1 Cylindrical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2.2 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 PDE Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 d Alembert s solution to the 1-dimensional wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Heat Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Other Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 First order linear ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Solving for etA using the Spectral Theorem . . . . . . . . . . . . . . . . . 3.3 Second Order Linear ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 ODE Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 11 15 18 21 23 27 27 32 35 38 2 3 4 Linear Operators and Separation of Variables . . . . . . . . . . . . . . 41 4.1 Introduction to Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 4.2 Application / Separation of variables . . . . . . . . . . . . . . . . . . . . . . . 47 Orthogonal Function Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Generalities about inner products on function spaces . . . . . . . . . 5.2 Convergence of the Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Proof of Theorem 5.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Gibb s Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 53 57 58 70 74 5 4 Contents 5.5 Fourier Series on Other Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . 76 6 7 Boundary value problem examples . . . . . . . . . . . . . . . . . . . . . . . . . 81 Boundary value generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 7.1 Linear Algebra of the Strurm-Liouville Eigenvalue Problem . . . 87 7.2 General Elliptic PDE Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 PDE Applications and Duhamel s Principle . . . . . . . . . . . . . . . . 97 8.1 Interpretation of d Alembert s solution to the 1-d wave equation 97 8.2 Solving 1st - order equations using 2nd - order solutions . . . . . . 97 8.2.1 The Solution to the Heat Equation on R . . . . . . . . . . . . . 99 8.3 Duhamel s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 8.4 Application of Duhamel s principle to 1 - d wave and heat equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Problems In other Coordinates Systems . . . . . . . . . . . . . . . . . . . 115 9.1 A Heat equation in spherical coordinates . . . . . . . . . . . . . . . . . . . 115 9.2 Problems with cylindrical symmetries . . . . . . . . . . . . . . . . . . . . . . 116 9.3 Strum Liouville problem in cylindrical coordinates . . . . . . . . . . . 119 9.4 Bessel Equation and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 9.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Some Complex Variables Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Assigned Homework Problems: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 8 9 A B Page: 4 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 1 Preliminaries 1.1 The spectral theorem for symmetric matrices Let A be a real N N matrix, a11 a12 . . . a1N a21 a22 . . . a2N A= . . , .. . . .. . aN 1 aN 2 . . . aN N f = f1 f2 . . . fN RN (1.1) and be a given vector. As usual we will let ei denote the vector in RN with all entries being zero except for the ith which is taken to be one. We will write N (u, v) := u v = i=1 N 2 ui vi = utr v and u2 = utr u. i i=1 |u| = (u, u) = N Recall that {vi }i=1 RN is said to be an orthonormal basis if (vi , vj ) = ij := 1 if i = j . 0 if i = j (1.2) The following proposition and its in nite dimensional analogue will be the basis for much of this course. 2 1 Preliminaries N N Proposition 1.1. If {vi }i=1 RN satis es Eq. (1.2) then {vi }i=1 is a basis for RN and if u RN we have N u= i=1 (u, vi ) vi . ai vi for some ai R. Then N N (1.3) Proof. Suppose that u = N N i=1 (u, vj ) = i=1 ai vi , vj = i=1 ai (vi , vj ) = i=1 ai ij = aj . In particular if u = 0 we learn that aj = (u, vj ) = 0 and we have shown that N {vi }i=1 is a linearly independent set. Since dim RN = N, it now follows N that {vi }i=1 is a basis for RN and hence every u RN may be written in the N form u = i=1 ai vi . By what we have just proved, we must have ai = (u, vi ) , i.e. Eq. (1.3) is valid. De nition 1.2. A matrix A as in Eq. (1.1) is symmetric A = Atr , i.e. if aij = aji for all i, j. The following characterization of a symmetric matrix will be more useful for our purposes. Lemma 1.3. If A is a real N N matrix then, for all u, v RN , (Au, v) = u, Atr v . Moreover A is symmetric i (Au, v) = (u, Av) for all u, v RN . (1.5) (1.4) Proof. Eq. (1.4) is a consequence of the following matrix manipulations (Au, v) = (Au) v = utr Atr v = u, Atr v which are based on the fact that (AB) = B tr Atr . Hence if A is symmetric, then Eq. (1.5) holds. Conversely, if Eq. (1.5) holds, by taking u = ei and v = ej in Eq. (1.5) we learn that a1i a1j . . aji = . , ej = (Aei , ej ) = (ei , Aej ) = ei , . = aij . . . aN,i aN,j tr tr Corollary 1.4. Suppose that A = Atr and v, w RN are eigenvectors of A with eigenvalues and respectively. If = then v and w are orthogonal, i.e. (v, w) = 0. Page: 2 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 1.1 The spectral theorem for symmetric matrices 3 Proof. If Av = v and Aw = w with = then (v, w) = ( v, w) = (Av, w) = (v, Aw) = (v, w) = (v, w) or equivalently, ( ) (v, w) = 0. Sine = , we must conclude that (v, w) = 0. The following important theorem from linear algebra gives us a method for guaranteeing that a matrix is diagonalizable. Again much of this course is based on an in nite dimensional generalization of this theorem. Theorem 1.5 (Spectral Theorem). If A in Eq. (1.1) is a symmetric matrix, then A has an orthonormal basis of eigenvectors, {v1 , . . . , vN } and the corresponding eigenvalues, { 1 , 2 , . . . , N } are all real. Example 1.6. Suppose that A := then p ( ) = det (A I) = det = 1 2 2 1 2 3 2 3 1 2 2 1 2 3 2 3 2 1 2 , (1.6) 9 4 which we set equal to zero to learn 1 2 or equivalently, 1 2 2 = 9 4 = 3 and hence A has eigenvalues, 2 1 = 1 and 2 = 2. Since A + I := and A 2I = we learn that v1 = v2 = 3 2 3 2 3 2 3 2 1 1 = 00 11 = 00 3 3 2 2 3 3 2 2 1 1 = 1 1 1 2 = 2. 1 date/time: 25-May-2004/11:28 Page: 3 job: 110notes macro: svmono.cls 4 1 Preliminaries Notice that (v1 , v2 ) = 0 as is guaranteed by Corollary 1.4. The normalized 1 eigenvectors are given by 2 1/2 v1 and 2 2 v2 . Consequently if f R2 , we have f = 2 1/2 v1 , f 2 1/2 v1 + 2 1/2 v2 , f 2 1/2 v2 = 1 1 (v1 , f ) v1 + (v2 , f ) v2 . 2 2 (1.7) Remark 1.7. As above, it often happens that naturally we nd a orthogonal N but not orthonormal basis {vi }i=1 for RN , i.e. (vi , vj ) = 0 if i = j but (vi , vi ) = 1. We can still easily expand in terms of these vectors. Indeed, |vi | have 1 N vi i=1 is an orthonormal basis for RN and therefore if f RN we N N f= i=1 f, |vi | 1 vi |vi | 1 vi = i=1 (f, vi ) |vi | 2 vi . Example 1.8. Working as above, one shows the symmetric matrix, 1 7 2 A := 7 1 2 , 2 2 10 has characteristic polynomial given by p ( ) = det (A I) = 3 12 2 36 + 432 = ( 6) ( 12) ( + 6) . (1.8) Thus the eigenvalues of A are given by 1 = 6, 2 = 6 and 3 = 12 and the corresponding eigenvectors are 1 1 1 v1 := 1 6, v2 := 1 6, v3 := 1 12. 0 1 2 Again notice that {v1 , v2 , v3 } is an orthogonal set as is guaranteed by Corollary 1.4. Relative to this basis we have the expansion f = (f, v1 ) = v1 |v1 | 2 + (f, v2 ) v2 |v2 | 2 + (f, v3 ) v3 |v3 | 2 1 1 1 (f, v1 ) v1 + (f, v2 ) v2 + (f, v2 ) v3 . 2 3 6 tr For example if f = (1, 2, 3) , then f= 1 1 v1 + 2v2 + v3 . 2 2 (1.9) Page: 4 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 1.2 Cylindrical and Spherical Coordinates 3 5 Exercise 1.1. Verify that the vectors {vi }i=1 are eigenvectors of A in Eq. (1.8) which have the stated eigenvalues. Hint: you are only asked to verify not solve from scratch. Exercise 1.2. Find eigenvectors {vi }i=1 and corresponding eigenvalues { i }i=1 for the symmetric matrix, 2 1 1 A := 1 2 1 . 1 1 2 Make sure you choose them to be orthogonal. Also express the following vectors, f = (1, 0, 2) tr 3 3 and g = (0, 1, 2) 3 tr and h = ( 1, 1, 0) , tr as linear combinations of the {vi }i=1 that you have found. Exercise 1.3. Suppose that A is a N N symmetric matrix and {vi }i=1 is N a basis of eigenvectors of A with corresponding eigenvalues { i }i=1 . Suppose N f R has been decomposed as N N f= i=1 ai vi . Show: 1. An f = i=1 ai n vi . i 2. More generally, suppose that p ( ) = a0 + a1 + a2 2 + + an n is a polynomial in , then N N p (A) f = i=1 ai p ( i ) vi and in particular p (A) v = p ( ) v is Av = v. 1.2 Cylindrical and Spherical Coordinates Our goal in this section is to work out the Laplacian in cylindrical and spherical coordinates. We will need these results later in the course. Our method is to make use of the following two observations: Page: 5 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 6 1 Preliminaries 3 1. If {ui }i=1 is any orthonormal basis for R3 then 3 3 f g= i=1 ( f, ui ) ( g, ui ) = i=1 ui f ui g and 2. if g has compact support in a region , then by integration by parts f gdV = f gdV. (1.10) The following theorem is a far reaching generalization of Eq. (1.10). Theorem 1.9 (Divergence Theorem). Let Rn be an open bounded region with smooth boundary, n : Rn be the unit outward pointing normal to . If Z C 1 ( , Rn ), then Z(x) n(x)d (x) = Z(x) dx. (1.11) Corollary 1.10 (Integration by parts). Let Rn be an open bounded region with smooth boundary, n : Rn be the unit outward pointing normal to . If Z C 1 ( , Rn ) and f f C 1 ( , R), then f (x) Z(x) dx = f (x) Z(x) dx+ f (x) Z(x) n(x)d (x). (1.12) Also if g C 2 ( , R), then f (x) g(x) dx = f (x) g(x) dx+ f (x) g(x) n(x)d (x). (1.13) Proof. Eq. (1.12) follows by applying Theorem 1.9 with Z replaced by f Z making use of the fact that (f Z) = f Z +f Z. g. Eq. (1.13) follows from Eq. (1.12) by taking Z = 1.2.1 Cylindrical coordinates Recall that cylindrical coordinates, see Figures 1.1, are determined by (x, y, z) = R( , , z) ( cos , sin , z). In these coordinates we have dV = d d dz. Page: 6 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 1.2 Cylindrical and Spherical Coordinates 7 Fig. 1.1. Cylindrical and polar coordinates. Proposition 1.11 (Laplacian in Cylindrical Coordinates). The Laplacian in cylindrical coordinates is given by f = 1 12 2 ( f ) + 2 f + z f. (1.14) Proof. We further observe that R ( , , z) = (cos , sin , 0) R ( , , z) = ( sin , cos , 0) Rz ( , , z) = (0, 0, 1) so that R ( , , z), 1 R ( , , z), Rz ( , , z) is an orthonormal basis for R3 . Therefore, ( f, g) = ( f, R ) ( g, R ) + f, 1 R 1 f g f g f g + + . = 2 z z g, 1 R + ( f, Rz ) ( g, Rz ) If g has compact support in a region , then by integration by parts, Page: 7 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8 1 Preliminaries f gdV = f gdV = = = = 1 f g f g f g + + d d dz 2 z z 12 2 ( f ) + f + z f g d d dz 1 12 2 ( f ) + 2 f + z f g d d dz 1 12 2 ( f ) + 2 f + z f gdV. Since this formula holds for arbitrary g with small support, we conclude that f = 1 12 2 ( f ) + 2 f + z f. 1.2.2 Spherical coordinates We will now work out the Laplacian in spherical coordinates by a similar method. Recall that spherical coordinates, see Figures 1.2, are determined by Fig. 1.2. De ning spherical coordinates of a point in R3 . (x, y, z) = R(r, , ) (r sin cos , r sin sin , r cos ). In this coordinate systems we have dV = r2 sin drd d . See Figure 1.3. Page: 8 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 1.2 Cylindrical and Spherical Coordinates 9 Fig. 1.3. A picture proof that dxdydz = r2 sin drd d , where r2 sin drd d should be viewed as (r sin d )(rd )dr. Proposition 1.12 (Laplacian in spherical coordinates). The Laplacian in spherical coordinates is given by f = 1 1 1 2 r (r2 r f ) + 2 (sin f ) + 2 2 f. r2 r sin r sin (1.15) A simple computation shows (as will be needed later) that 1 12 (r2 r f ) = r (rf ). 2r r r Proof. Since Rr (r, , ) = (sin cos , sin sin , cos ) R (r, , ) = ( r sin sin , r sin cos , 0) R (r, , ) = (r cos cos , r cos sin , r sin ) it is easily veri ed that Rr (r, , ), 1 1 R ( , , z), R ( , , z) r sin r (1.16) is an orthonormal basis for R3 . Therefore ( f, g) = ( f, Rr ) ( g, Rr ) + + 1 ( f, R ) ( g, R ) r2 sin2 1 ( f, R ) ( g, R ) r2 f g 1 f g 1 f g = +2 2 +2 . r r r sin r If g has compact support in a region , then Page: 9 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 10 1 Preliminaries f gdV = f gdV = 1 1 f g f g f g +2 2 + r2 sin drd d r r r2 r sin 1 f g f g f g = + + sin drd d r2 sin r r sin 12 = r r2 r g sin + f + (sin f ) g drd d sin 1 1 1 2 r r2 r g + 2 2 f + 2 = (sin f ) g r2 sin drd d r2 r sin r sin 1 1 1 2 r r2 r g + 2 2 f + 2 = (sin f ) g dV. r2 r sin r sin Since this formula holds for arbitrary g we conclude that f = 1 1 1 2 (r2 r f ) + 2 (sin f ) + 2 2 f. 2r r r sin r sin 1.2.3 Exercises In the following two exercises, I am using the conventions in the Lecture notes and not the book. Exercise 1.4. Compute f where f is given in cylindrical coordinates as: f = 3 cos + z Exercise 1.5. Compute f where f is given in spherical coordinates as: f = r 1 + cos sin . Page: 10 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2 PDE Examples 2.1 The Wave Equation Example 2.1 (Wave Equation for a String). Suppose that we have a stretched string supported at x = 0 and x = L and y = 0. Suppose that the string only undergoes vertical motion (pretty bad assumption). Let u(t, x) and T (t, x) denote the height and tension respectively of the string at (t, x), (x) denote the density in equilibrium and T0 be the equilibrium string tension. Let J = Fig. 2.1. A piece of displace string [x, x + x] [0, L], then PJ (t) := J ut (t, x) (x)dx is the momentum of the piece of string above J. (Notice that (x)dx is the weight of the string above x.) Newton s equations state dPJ (t) = dt utt (t, x) (x)dx = Force on String. J Since the string is to only undergo vertical motion we require 12 2 PDE Examples T (t, x + x) cos( x+ x ) T (t, x) cos( x ) = 0 for all x and therefore that T (t, x) cos( x ) =: H for some constant H, i.e. the horizontal component of the tension is constant. Looking at Figure 2.2, the tension on the piece of string above J = [a, b] at the right endpoint b must Fig. 2.2. Computing the net vertical force due to tension on the part of the string above [a, b]. be given by H (1, ux (t, b)) while the tension at the left endpoint, a, must be given by H ( 1, ux (t, a)) . So the net tension force on the string above J is b H [ux (t, b) ux (t, a)] = H a uxx (t, x) dx. Finally there may be a component due to gravity and air resistance, say b gravity = g a b (x)dx and k(x)ut (t, x)dx. a air resistance = So Newton s equations become b b utt (t, x) (x)dx = a a [Huxx (t, x) g (x) k(x)ut (t, x)] dx. Di erentiating this equation in b at b = x then shows utt (t, x) (x) = Huxx (t, x) g (x) k(x)ut (t, x) or equivalently that utt (t, x) = k(x) H uxx (t, x) g ut (t, x). (x) (x) (2.1) Page: 12 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2.1 The Wave Equation 13 Example 2.2 (Wave equation. for a drum head). Suppose that u (t, x) represents the height at time t of a drum head over a point x being the base of the drum head, see Figure 2.3. As for the string we will make the sim- Fig. 2.3. A deformed membrane attached to a wire base. We are also compute the tension density on a region of the membrane above a region V in the plane. plifying assumption that the membrane only moves vertically or equivalently that the horizontal component of tension/unit-length is a constant value, H. Let V be a test region and consider the membrane which lie above V as in Figure 2.3. Then PV (t) := V ut (t, x) (x)dx is the momentum of the piece of string above V where (x)dx is the weight of the membrane above x. Newton s equations state dPV (t) = dt utt (t, x) (x)dx = Force on membrane. V To nd the vertical force on the membrane above V, let x V, then (n (x) , u (t, x) n (x)) = d |0 (x + sn (x) , u(t, x + sn (x)) ds is a vector orthogonal to the boundary of the region above V and by assumption the tension/unit-length at x is H (n (x) , u (t, x) n (x)) . Thus the vertical component of the force on the membrane above V is given by H V u (t, x) n (x) d (x) = H V u (t, x) dx = H V u (t, x) dx. Page: 13 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 14 2 PDE Examples Finally there may be a component due to gravity and air resistance, say gravity = V g (x)dx and k(x)ut (t, x)dx. V air resistance = So Newton s equations become utt (t, x) (x)dx = V V [H u (t, x) g (x) k(x)ut (t, x)] dx. Since V is arbitrary, this implies (x)utt (t, x) = H u (t, x) g (x) k(x)ut (t, x) or equivalently that utt (t, x) = k(x) H u (t, x) g ut (t, x). (x) (x) (2.2) Example 2.3 (Wave equation for a metal bar). Suppose that have a metal wire which we is going to be deformed and then released. We would like to nd the equation that the displacement u (t, x) of the section of the bar originally at location x must solve, see Figure 2.4 below. Fig. 2.4. The picture represents an elastic bar in its un-deformed state and then in a deformed state. The quantity y (t, x) represents the displacement of the section that was originally at location x in the un-deformed bar. In the above gure y (t, x) < 0 which y (t, x + x) > 0. Page: 14 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2.1 The Wave Equation 15 To do this will write down Newton s equation of motion. First o , the longitudinal force on the left face of the section which was originally between x and x + x is approximately, AE u (t, x + x) u(t, x) , x where E is Young s modulus of elasticity and A is the area of the bar. (The minus represents the fact that we must pull to the left to get the current con guration in the gure.) Letting x 0, we nd the force of the section that was originally at x is given by AEux (t, x) . Now suppose that x is not necessarily small. Then we have the momentum of the region of the bar originally between x and x + x is given by x+ x ut (t, x) (x) dx x where (x) is the linear mass density. Therefore, Mass acceleration = x+ x d x+ x ut (t, x) (x) dx = utt (t, x) (x) dx dt x x = the net force on this section of the bar = AEux (t, x) + AEux (t, x + x) where AEux (t, x) is the force on left end and AEux (t, x + x) is the force on the right end. Hence we have x+ x utt (t, x) (x) dx = AEux (t, x + x) AEux (t, x) x which upon di erentiating in x at x = 0 shows utt (t, x) (x) = AEuxx (t, x) . 2.1.1 d Alembert s solution to the 1-dimensional wave equation Here we are going to try to nd solutions to the wave equation, ytt = a2 yxx . Since this equation may be written as 2 2 t a2 x y = 0 (2.3) and 2 2 t a2 x = ( t a x ) ( t + a x ) we are lead to consider the wave equation in the new variables, u = x + at and v = x at. Page: 15 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 16 2 PDE Examples In these variables we have u u + t = t u x = u + x from which it follows that v v = a u a v and t v v = u + v x t a x = 2a v and t + a x = 2a u and hence the wave equation in (u, v) coordinates becomes, 0 = ( t a x ) ( t + a x ) y = 2a v 2a u y = 4a2 yuv , i.e. yuv = 0. Integrating this equation in v shows yu = F (u) and then integrating in u shows y= F (u) du + (v) = (u) + (v) . Thus we have shown if y solves the wave equation then y (t, x) = (x + at) + (x at) for some functions and . Exercise 2.1. Show that if y (t, x) has the form given in Eq. (2.4) with and being twice continuously di erentiable functions, then y solves the wave Eq. (2.3). To get a unique solution to Eq. (2.3) we must introduce some initial conditions. For example, let us further assume that y (0, x) = f (x) and yt (0, x) = 0. This then implies that f (x) = (x) + (x) and 0 = a (x) a (x) , The latter equation shows that (x) = (x) + C and using this in the rst equation implies that f (x) = 2 (x) + C 1 1 (f (x) C) and (x) = (f (x) + C) . 2 2 Thus we have found the solution to be given by (x) = 1 {f (x + at) + f (x at)} . 2 In the homework you are asked to generalize this result to prove the following theorem. y (t, x) = Page: 16 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 (2.4) or 2.1 The Wave Equation 17 Theorem 2.4 (d Alembert s solution). If f (x) is twice continuously differentiable and g (x) is continuously di erentiable for x R, then the unique solution to ytt = a2 yxx with y (0, x) = f (x) and yt (0, x) = g (x) is given by y(t, x) = 1 1 [f (x + at) + f (x at)] + 2 2a x+at (2.5) (2.6) g(s)ds. x at (2.7) Example 2.5. Here we wish to solve for x 0 and t 0, 2 2 t y = x y with y (0, x) = f (x) and y (0, x) = 0 with y (t, 0) = 0. As before we know that y (t, x) = (x + t) + (x t) . We must now implement all of the boundary conditions, f (x) = y (0, x) = (x) + (x) 0 = y (0, x) = (x) (x) and 0 = y (t, 0) = (t) + ( t) . This suggests that we de ne ( t) := (t) for t > 0, and also that (x) = (x) + C f (x) = 2 (x) + C or (x) = 1 (f (x) C) 2 1 (x) = (f (x) + C) . 2 1 [f (x + t) + f (x t)] 2 Thus our answer is given by y (t, x) = where by above, 1 1 (f ( x) C) = ( x) = (x) = (f (x) + C) 2 2 and thus f ( x) := f (x) . 1 [f (x + t) + f (x t)] 2 where f is extend to all of R to be an odd function. y (t, x) = Page: 17 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 Thus we have 18 2 PDE Examples 2.2 Heat Equations Example 2.6 (Heat or Di usion Equation in 1-dimension). Let us consider the temperature in a rod . We will let 1. (x) denote the linear density of the rod 2. c (x) denote the heat capacity of the rod per unit mass at x 3. (x) be the thermal conductivity of the rod at x. By Newton s Law of cooling, the heat ow from left to right in the rod at location x should be approximately equal to (x) (u (x) u (x + )) . (2.8) Notice the appearing in the denominator represents the fact that the thicker the insulation in your house the less heat transfer that you have. Passing to the limit in Eq. (2.8) then gives Fourier s law, namely the heat ow from left to right in the rod at location x is given by (x) u (x) . (2.9) (In the book, it is typically assumed that (x) = , (x) = K and c (x) = are all constant.) 4. u (t, x) be the temperature of the rod at time t and location x. 5. H (t, x) represent heat source at x and time t. For example we may be passing a current through the wire and the resistance of the wire is both spatially and time dependent. Alternatively we may be heating the wire with an external source. Let B = [a, b] be a sub-region of the rod, see Figure 2.5. Then Fig. 2.5. Part of a rod with a test region B = [a, b] being examined. b E (t) = a u (t, x) (x) c (x) dx represents the heat energy in B at time t. Hence Page: 18 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2.2 Heat Equations b 19 E (t) = a ut (t, x) (x) c (x) dx is the rate of change of heat energy in B. This may alternatively be computed as the rate at which heat enters the system which is given by b E (t) = a b H (t, x) dx + ( (b) ux (t, b) (a) ux (t, a)) H (t, x) + a = Hence we conclude that b d ( (x) ux (t, x)) dx. dx b ut (t, x) (x) c (x) dx = a a H (t, x) + d ( (x) ux (t, x)) dx dx for all sub-intervals in the rod and therefore (again just di erentiate in b) that (x) c (x) ut (t, x) = This equation may be written as ut (t, x) = Lu (t, x) + h (t, x) where Lf (x) := 1d p (x) dx d f (x) , dx H (t, x) . p (x) d ( (x) ux (t, x)) + H (t, x) . dx (2.10) (x) p (x) = (x) c (x) and h (t, x) := If we further assume that the rod in not perfectly insulated along its length and the ambient temperature is not constant, we may end up with another terms in computing E (t) of the form b Q (x) [u (t, x) T (x)] dx a and we would then arrive at a heat equation of the form ut (t, x) = Lu (t, x) + h (t, x) where Lf (x) := 1d p (x) dx d 1 f (x) + q (x) f (x) dx p (x) (x) (2.11) for some function q (x) and a modi ed function h (t, x) . Page: 19 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 20 2 PDE Examples Fig. 2.6. A test volume B in centered at x with outward pointing normal, n. Example 2.7 (Heat or Di usion Equation in d - dimensions). Suppose that Rd is a region of space lled with a material, (x) is the density of the material at x and c(x) is the heat capacity. Let u(t, x) denote the temperature at time t [0, ) at the spatial point x . Now suppose that B Rd is a little volume in Rd , B is the boundary of B, and EB (t) is the heat energy contained in the volume B at time t. Then EB (t) = B (x)c(x)u(t, x)dx. So on one hand, EB (t) = B (x)c(x)u(t, x)dx (2.12) while on the other hand, EB (t) = B ( (x) u(t, x) n(x)) d (x), (2.13) where (x) is a d d positive de nite matrix representing the conduction properties of the material, n(x) is the outward pointing normal to B at x B, and d denotes surface measure on B. In order to see that we have the sign correct in (2.13), suppose that x B and u(x) n(x) > 0, then the temperature for points near x outside of B are hotter than those points near x inside of B and hence contribute to a increase in the heat energy inside of B. (If we get the wrong sign, then the resulting equation will have the property that heat ows from cold to hot!) Comparing Eqs. (2.12) to (2.13) after an application of the divergence theorem shows that (x)c(x)u(t, x)dx = B B ( ( ) u(t, ))(x) dx. (2.14) Since this holds for all volumes B , we conclude that the temperature functions should satisfy the following partial di erential equation. Page: 20 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2.3 Other Equations 21 (x)c(x)u(t, x) = or equivalently that u(t, x) = 1 (x)c(x) ( ( ) u(t, ))(x) . (2.15) ( (x) u(t, x)). (2.16) Setting g ij (x) := ij (x)/( (x)c(x)) and d z j (x) := i=1 ( ij (x)/( (x)c(x)))/ xi the above equation may be written as: u(t, x) = Lu(t, x), where (Lf )(x) = i,j (2.17) f (x). xj g ij (x) 2 f (x) + xi xj z j (x) j (2.18) The operator L is a prototypical example of a second order elliptic di erential operator. Example 2.8 (Laplace and Poisson Equations). Laplace s Equation is of the form Lu = 0 and solutions may represent the steady state temperature distribution for the heat equation. Equations like u = appear in electrostatics for example, where u is the electric potential and is the charge distribution. 2.3 Other Equations Example 2.9 (Shrodinger Equation and Quantum Mechanics). i (t, x) = (t, x) + V (x) (t, x) with ( , 0) t 2 2 = 1. Interpretation, | (t, x)| dt = the probability of nding the particle in A at time t. A 2 (Notice similarities to the heat equation.) Example 2.10 (Maxwell Equations in Free Space). E = B t B = E t E = B = 0. Page: 21 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 22 2 PDE Examples Notice that 2E = t2 2 B = t ( E) = E ( E) = E and similarly, B = B so that all the components of the electromagnetic t2 elds satisfy the wave equation. Example 2.11 (Tra c Equation). Consider cars travelling on a straight road with coordinate x R, let u(t, x) denote the density of cars on the road at time t and location x R, and v(t, x) be the velocity of the cars at (t, x). b Then for J = [a, b] R, NJ (t) := a u(t, x)dx is the number of cars in the set J at time t. We must have b u(t, x)dx = NJ (t) = u(t, a)v(t, a) u(t, b)v(t, b) a b = a [u(t, x)v(t, x)] dx. x Since this holds for all intervals [a, b], we must have u(t, x) = [u(t, x)v(t, x)] . x To make life more interesting, we may imagine that v(t, x) = F (u(t, x), ux (t, x)), in which case we get an equation of the form u= G(u, ux ) where G(u, ux ) = u(t, x)F (u(t, x), ux (t, x)). t x A simple model might be that there is a constant maximum speed, vm and maximum density um , and the tra c interpolates linearly between 0 (when u = um ) to vm when (u = 0), i.e. v = vm (1 u/um ) in which case we get u = vm (u(1 u/um )) . t x Example 2.12 (Burger s Equation). Suppose we have a stream of particles travelling on R, each of which has its own constant velocity and let u(t, x) denote the velocity of the particle at x at time t. Let x(t) denote the trajectory of the particle which is at x0 at time t0 . We have C = x(t) = u(t, x(t)). Di erentiating this equation in t at t = t0 implies 0 = [ut (t, x(t)) + ux (t, x(t))x(t)] |t=t0 = ut (t0 , x0 ) + ux (t0 , x0 )u(t0 , x0 ) which leads to Burger s equation 0 = u t + u ux . Page: 22 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2.4 Worked Examples 23 Example 2.13 (Minimal surface Equation). Let D R2 be a bounded region with reasonable boundary, u0 : D R be a given function. We wish to nd the function u : D R such that u = u0 on D and the graph of u, (u) has least area. Recall that the area of (u) is given by A(u) = Area( (u)) = D 1 1 + | u| dx. 2 Assuming u is a minimizer, let v C (D) such that v = 0 on D, then 0= d d 2 1 + | (u + sv)| dx |0 A(u + sv) = |0 ds ds D d 2 = |0 1 + | (u + sv)| dx ds D 1 u v dx = 2 D 1 + | u| 1 = u v dx 2 D 1 + | u| from which it follows that 1 1 + | u| 2 u = 0. Example 2.14 (Navier Stokes). Here u(t, x) denotes the velocity of a uid ad (t, x), p(t, x) is the pressure. The Navier Stokes equations state, u + u u = u p + f with u(0, x) = u0 (x) t u = 0 (incompressibility) (2.19) (2.20) where f are the components of a given external force and u0 is a given divergence free vector eld, is the viscosity constant. The Euler equations are found by taking = 0. Equation (2.19) is Newton s law of motion again, F = ma. See http://www.claymath.org for more information on this Million Dollar problem. 2.4 Worked Examples Example 2.15 (6.3 in book). Slab / wire along [0, c] with surface conductance at the end being H and K being the thermal conductivity along the rod. Assume outside temperature at 0 is 0 and at c is T, Find the steady state temperature of the rod. The relevant heat equation is: Page: 23 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 24 2 PDE Examples ut = Kuxx with Ku (0) = H (u (0) outside temp) = H (u (0) 0) and Ku (c) = H (u (c) outside temp) = H (T u (c)) . Notice the signs used here. Recall that Ku (x) represents the heat movement (or Flux) from right to left in the bar at location x, therefore, Ku (0) = rate of heat leaving bar at 0 while H (u (0) outside temp) = rate of heat leaving bar at 0 do surface conductance Similarly, Ku (c) = rate of heat entering bar at c while H (u (c) T ) = rate of heat leaving bar at 0 do surface conductance and hence Ku (c) = H (u (c) T ) = H (T u (c)) . So in steady state we have ut = 0 and so u (x) = 0 of u (x) = mx + b. Now Km = Ku (0) = Hu (0) = Hb and Km = Ku (c) = H (T u (c)) = H (T mc b) and so if we let h := H/K, then m = hb and m = h (T mc b) = hT hmc hb = hT hmc m from which it follows that (2 + hc) m = hT or m= m T hT and b = = . 2 + hc h 2 + hc Thus the solution is given by u (x) = T [hx + 1] . 2 + hc Example 2.16 (10.1b in book.). Solve uxy = 2x for x, y > 0 and u (0, y) = 0 and u (x, 0) = x2 . Ans., ux = 2xy + C (x) and so Page: 24 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2.4 Worked Examples 25 u (x, y) = x2 y + K (x) + F (y) Since u (x, 0) = x2 = K (x) + F (0) and 0 = u (0, y) = K (0) + F (y) we have x2 = K (x) K (0) and u (x, y) = x2 y + K (x) K (0) = x2 y + x2 = x2 (y + 1) . Page: 25 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 3 Linear ODE 3.1 First order linear ODE We would like to solve the ordinary di erential equation u (t) = Au (t) with u (0) = f. (3.1) (3.2) The method of separation of variables or eigenvector expansions proposed to begin by looking for solutions of the form u (t) = T (t) v to Eq. (3.1). Here v is a xed vector in RN and T (t) is some unknown function of t. Substituting u (t) = T (t) v into Eq. (3.1) gives T (t) v = T (t) Av or equivalently that Av = T (t) v. T (t) Since the left side of this equation is independent of t we must have T (t) = T (t) (3.3) for some R. The solution to Eq. (3.3) is of course T (t) = et T (0) and therefore we have shown the following lemma. Lemma 3.1. If u (t) = T (t) v solves Eq. (3.1), then v is an eigenvector of A and if is the corresponding eigenvalue (i.e. Av = v) then u (t) = e t T (0) v. Conversely if Av = v then u (t) = e t v solves Eq. (3.1). 28 3 Linear ODE Proposition 3.2 (Principle of superposition). If u (t) and v (t) solves Eq. (3.1) then so does u (t) + cv (t) for any c R. Proof. This is a simple consequence of the fact that matrix multiplication and di erentiation are linear operations. In detail, d (u (t) + cv (t)) = u (t) + cv (t) = Au (t) + cAv (t) dt = A (u (t) + cv (t)) . Consequently if Avi = i vi for i = 1, 2, . . . , k, then u (t) = i et i vi . solves Eq. (3.1). Theorem 3.3. Suppose the matrix A is diagonalizable, i.e. there exists a basis N {vi }i=1 for RN consisting of eigenvectors A. Then to any f RN there is a unique solution, u (t) , to Eqs. (3.1) and (3.2). Moreover, if we expand f in N terms of the basis {vi }i=1 as N f= i=1 ai vi , then the unique solution to Eqs. (3.1) and (3.2) is given by N u (t) = i=1 ai et i vi . (3.4) Proof. The fact that Eq. (3.4) solves Eqs. (3.1) and (3.2) follows from the principle of superposition and the fact that et i = 1 when t = 0. Conversely, suppose that u solves Eqs. (3.1) and (3.2), then N u (t) = i=1 ai (t) vi (3.5) for some functions ai (t) with ai (0) = ai . Now on one hand N u (t) = i=1 ai (t) vi while on the other hand N N N u (t) = Au (t) = A i=1 ai (t) vi = i=1 ai (t) Avi = i=1 ai (t) i vi . Page: 28 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 3.1 First order linear ODE 29 Subtracting these two equation shows N N N 0= i=1 N ai (t) vi i=1 ai (t) i vi = i=1 (ai (t) ai (t) i ) vi . Since {vi }i=1 is a basis for RN it follows, for all i, that ai (t) = ai (t) i with ai (0) = ai and therefore, ai (t) = et i ai . Putting this result back into Eq. (3.5) gives Eq. (3.4). De nition 3.4. If f RN we will write etA f for the solution, u (t) , to Eqs. (3.1) and (3.2). Fact 3.5 Eqs. (3.1) and (3.2) have a unique solution independent as to whether A has a basis of eigenvectors or not. Moreover we may compute etA using the matrix power series expansion, etA f = tn n A f. n! n=0 (3.6) Notice that formula in Eq. (3.6) is consistent with our previous results. For example if v RN and Av = v, then An v = n v and therefore, tn n tn n A v= v = et v = etA v. n! n! n=0 n=0 More generally, if u = k i=1 ai vi with Avi = i vi , then k k etA u = i=1 ai et i vi = i=1 ai etAi vi . Remark 3.6. As the notation suggests, it is true that etA esA = e(t+s)A as you are asked to prove in Exercise 3.1 below. However, it is not generally true that e(A+B) = eA eB = eB eA , see Proposition 3.9 below. Example 3.7. Let us nd etA when 110 A = 0 2 2 . 003 Page: 29 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 30 3 Linear ODE The eigenvalues of A are given as the roots of the characteristic polynomial, p ( ) = det (A I) = (1 ) (2 ) (3 ) . These roots are = 1, = 2, and = 3. As usual we nd the corresponding eigenvectors as solutions to the equation (A I) u = 0. The result is, eigenvectors: 1 1 1 v1 := 0 1, v2 := 1 2, v3 := 2 3. 0 0 1 Since 0 1 1 1 = 1 0 0 0 0 1 1 1 0 0 = 2 2 1 + 0 1 1 0 0 and it follows that the columns of etA are given by 1 1 etA 0 = et 0 0 0 0 1 1 1 1 etA 1 = etA 1 etA 0 = e2t 1 et 0 0 0 0 0 0 and 0 1 1 1 etA 0 = etA 2 2etA 1 + etA 0 1 1 0 0 1 1 1 = e3t 2 2e2t 1 + et 0 1 0 0 and therefore, etA Example 3.8. Let 1 7 2 A := 7 1 2 . 2 2 10 Page: 30 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 et et + e2t et 2e2t + e3t e2t 2e2t + 2e3t . = 0 0 0 e3t 3.1 First order linear ODE 31 Solve the equation u (t) = Au (t) with 1 1 1 u (0) = v1 + 2v2 + 3v3 = 1 + 2 1 + 3 1 0 1 2 i.e. 2 u (0) = f = 0 . 8 Answer. Recall from Example 1.8 that A has eigenvectors 1 1 1 v1 := 1 6, v2 := 1 6, v3 := 1 12. 0 1 2 Therefore, (f, v1 ) |v1 | so that f = v1 + 2v2 + 3v3 . Therefore, u (t) = e 6t v1 + 2e6t v2 + 3e12t v3 solves the ODE. Proposition 3.9. Let A and B be two N N matrices. Then the following are equivalent: 1. 0 = [A, B] := AB BA. 2. etA B = BetA for all t R, 3. etA esB = esB etA for all s, t R. Moreover if [A, B] = 0 then e(A+B) = eA eB and in particular etA esA = e(t+s)A for all s, t R. Proof. If [A, B] = 0, then d tA tA e Be = etA [A, B] e tA = 0 dt and therefore, etA B = BetA for all t R. It now follows that (3.7) 2 = 1, (f, v2 ) |v2 | 2 = 6 (f, v3 ) 18 = 2 and 2 = 6 =3 3 |v3 | Page: 31 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 32 3 Linear ODE d tA sB e e = etA BesB = BetA esB ds d sB tA e e = BesB etA ds and so by uniqueness of solutions to these ODE we conclude etA esB = esB etA for all s, t R. If etA esB = esB etA for all s, t R then AB = d tA dd dd d |0 e B = |0 |0 etA esB = |0 |0 esB etA = |0 BetA = BA. dt dt ds dt ds dt For the last assertion, let T (t) := etA etB , then d T (t) = AetA etB + etA BetB = AetA etB + BetA etB dt = (A + B) T (t) with T (0) = I. So again by uniqueness of solutions, etA etB = T (t) = et(A+B) . 3.2 Solving for etA using the Spectral Theorem Example 3.10. Let A := 1 2 3 2 3 2 1 2 as in Example 1.6 with eigenvectors/eigenvalues given by v1 = and v2 = Recall that f= and hence etA f = 1 (v1 , f ) etA v1 + 2 1 = (v1 , f ) e t v1 + 2 1 (v2 , f ) etA v2 2 1 (v2 , f ) e2t v2 . 2 1 2 = 2. 1 1 1 = 1 1 1 1 (v1 , f ) v1 + (v2 , f ) v2 2 2 (3.8) Taking f = e1 and f = e2 then implies Page: 32 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 3.2 Solving for etA using the Spectral Theorem 33 etA e1 = 1 tA e v1 + 2 1 1 = e t 1 2 1 tA 1 1 e v2 = e t v1 + e2t v2 2 2 2 1 2t 1 +e 1 2 = and etA e2 = 1 e t + e2t . 2 e t e2t 1 tA e v1 2 1 1 = e t 1 2 1 tA 1 1 e v2 = e t v1 e2t v2 2 2 2 1 2t 1 e 1 2 = 1 e t e2t . 2 e t + e2t Thus we may conclude that etA = etA e1 etA e2 = Alternatively, from Eq. (3.8) etA f = and therefore etA = 1 1 tr tr (v1 , f ) e t v1 v1 + e2t v2 v2 2 2 1 1 1 1 1 1 + e2t 1 1 = e t 1 1 2 2 = 1 e t + e2t e t e2t . 2 e t e2t e t + e2t 2 1 1 2 1 1 tr tr (v1 , f ) e t v1 v1 f + e2t v2 v2 f 2 2 1 e t + e2t e t e2t . 2 e t e2t e t + e2t Example 3.11. The matrix A= has eigenvectors/eigenvalues given by v1 := As usual, if f R2 then f= 1 1 (v1 , f ) v1 + (v2 , f ) v2 . 2 2 1 1 1 and v2 := 3 1 1 Page: 33 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 34 3 Linear ODE It then follows that etA f = 1 (v1 , f ) etA v1 + 2 1 = (v1 , f ) e t v1 + 2 1 (v2 , f ) etA v2 2 1 (v2 , f ) e 3t v2 . 2 Taking f = e1 and then f = e2 gives etA e1 = 1 1 (v1 , e1 ) e t v1 + (v2 , e1 ) e 3t v2 2 2 1 1 = e t v1 e 3t v2 2 2 1 e t + e 3t = 2 e t e 3t 1 e t e 3t . 2 e t + e 3t and similarly etA e2 = Therefore, etA = etA e1 etA e2 = 1 e t + e 3t e t e 3t . 2 e t e 3t e t + e 3t Example 3.12. Continuing the notation and using the results of Example 1.8, 1 7 2 A := 7 1 2 2 2 10 with eigenvectors/eigenvalues given by 1 1 1 v1 := 1 6, v2 := 1 6, v3 := 1 12. 1 2 0 and f= 1 1 1 (f, v1 ) v1 + (f, v2 ) v2 + (f, v2 ) v3 . 2 3 6 tr For example if f = (1, 2, 3) , then f= and hence 1 1 v1 + 2v2 + v3 2 2 Page: 34 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 3.3 Second Order Linear ODE 35 etA (1, 2, 3) = tr 1 tA 1 e v1 + 2etA v2 + etA v3 2 2 1 1 = e 6t v1 + 2e6t v2 + e12t v3 . 2 2 A straightforward but tedious computation shows 3e 6t + 2e6t + e12t 3e 6t + 2e6t + e12t 2e6t 2e12t 1 etA = 3e 6t + 2e6t + e12t 3e 6t + 2e6t + e12t 2e6t 2e12t . 6 2e6t 2e12t 2e6t 2e12t 2e6t + 4e12t This can alternatively be done using a computer algebra package, which is what I did. 3.3 Second Order Linear ODE We would like to solve the ordinary di erential equation u (t) = Au (t) with u (0) = f and u (0) = g (3.9) (3.10) for some f, g RN and A a N N matrix. Again we might begin by trying to nd solutions to Eq. (3.9) by considering functions of the form u (t) = T (t) v. In order for u (t) = T (t) v to be a solution we must have T (t) v = T (t) Av and working as above we concluded that there must exists such that T (t) = T (t) and Av = v. The general solution to the equation T (t) = T (t) is T (t) = c (t) T (0) + s (t) T (0) where c (t) := and t cos if 0 cosh t if 0 t sin if < 0 t if = 0 s (t) := sinh t if > 0. Page: 35 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 36 3 Linear ODE Theorem 3.13. Suppose the matrix A is diagonalizable, i.e. there exists a N basis {vi }i=1 for RN consisting of eigenvectors A with corresponding eigenN values { i }i=1 R. Then for any f, g RN there is a unique solution, u (t) , to Eqs. (3.9) and (3.10). Moreover, if we expand f and g in terms of the basis N {vi }i=1 as N N f= i=1 ai vi and g = i=1 bi vi then the unique solution to Eqs. (3.9) and (3.10) is given by N u (t) = i=1 [ai c i (t) + bi s i (t)] vi . (3.11) Proof. It is easy to check that u de ned as in Eq. (3.11) solves Eqs. (3.9) and (3.10) which proves the existence of solutions. The uniqueness of solutions may also be proved similarly to what was done in Theorem 3.3. Indeed, suppose that N u (t) = i=1 i (t) vi then the equation, u = Au, is equivalent to N N N i (t) vi = u (t) = Au (t) = i=1 N i=1 i (t) Avi = i=1 i (t) i vi and since {vi }i=1 is a basis for RN we must have i (t) = i i (t) for all i. Moreover, N N (3.12) ai vi = f = u (0) = i=1 N i=1 N i (0) vi and bi vi = g = u (0) = i=1 i=1 i (0) vi implies that i (0) = ai and i (0) = bi for all i. (3.13) This completes the proof, since the unique solution to Eqs. (3.12) and (3.13) is given by i (t) = ai c i (t) + bi s i (t) . Page: 36 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 3.3 Second Order Linear ODE 37 Notation 3.14 From now on, let us agree that cos t := cosh t if 0 sin t sinh t if > 0 and := sin t := t if = 0. With the above notation it is natural to write the general solution Eqs. (3.9) and (3.10) as sin At u (t) = cos At f + g A with the understanding that At 1. cos At and sin A are linear (i.e. matrices) and 2. if Av = v then cos At v := cos t v and sin At sin t v. v := A Example 3.15. Continuing the notation and using the results of Example 1.8, 1 7 2 A := 7 1 2 2 2 10 with eigenvectors/eigenvalues given by 1 1 1 v1 := 1 6, v2 := 1 6, v3 := 1 12. 1 2 0 We will solve, u (t) = Au (t) with u (0) = f = (1, 2, 3) As above we have f= 1 1 1 (f, v1 ) v1 + (f, v2 ) v2 + (f, v2 ) v3 2 3 6 1 1 = v1 + 2v2 + v3 2 2 tr and u (0) = g = (1, 1, 1) . tr Page: 37 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 38 3 Linear ODE and 1 2 1 g = 0v1 + v2 + (f, v2 ) v3 = (2v2 + v3 ) . 3 6 3 1 1 6t v1 + 2 cosh 6t v2 + cosh 12t v3 cos 2 2 1 g= 3 sinh 6t sinh 12t 2 v2 + v3 6 12 Therefore, cos and sin At f = At A and the solution is given by u (t) = cos = At f + sin At g A 1 cos 6t v1 2 2 sinh 6t v2 6t + + 2 cosh 3 6 1 sinh 12t 1 cosh 12t + + v3 . 2 3 12 3.4 ODE Exercises Exercise 3.1. Here you are asked to give another proof of Eq. (3.7). Let A be an N N, matrix, f RN and s, t R. Show etA esA f = e(t+s)A f. Outline: Let u (t) := etA esA f and v (t) = e(t+s)A f and show both u and v solve the di erential equation, w (t) = Aw (t) with w (0) = esA f and then use uniqueness of solutions of this equation (see Fact 3.5) to conclude that u (t) = v (t) . Exercise 3.2. Let A= Show etA = using the following three methods. Page: 38 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 01 1 0 . cos t sin t sin t cos t 3.4 ODE Exercises 39 1. Showing d dt and cos t sin t sin t cos t and 2. by explicitly summing the series etA = 2 cos t sin t sin t cos t =A cos t sin t sin t cos t 10 01 =I= t=0 tn n A. n! n=0 d 3. Show dt2 etA = etA and then solve this equation using etA |t=0 = I and d tA = A. dt |0 e Exercise 3.3. Combine Exercises 3.1 and 3.2 to give a proof of the trigonometric identities: cos(s + t) = cos s cos t sin s sin t and sin (s + t) = cos s sin t + cos t sin s. Exercise 3.4. Let a, b, c R and 0ab A = 0 0 c . 000 Show etA 1 at bt + 1 act2 2 ct = 0 1 00 1 (3.14) (3.15) by summing the matrix power series. Also nd et( I+A) where R and I is the 3 3 identity matrix. Exercise 3.5. Let 2 1 1 A := 1 2 1 1 1 2 f = (1, 0, 2) tr and and g = (0, 1, 2) . tr be as in Exercise 1.2. Solve the following equations u (t) = Au (t) with u (0) = f and u (t) = Au (t) with u (0) = f and u (0) = g. Page: 39 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 40 3 Linear ODE Write your solutions in the form 3 u (t) = i=1 ai (t) vi where the functions ai are to be determined. Hint: Recall from Exercise 1.2 (you should have shown) that 1 1 1 v1 = 1 0, v2 = 0 3, v3 = 2 3. 1 1 1 is an orthogonal basis of eigenvectors (with corresponding eigenvalues) for A. Page: 40 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 4 Linear Operators and Separation of Variables De nition 4.1. A linear combination of the vectors {vi }i=1 R3 (or n {vi }i=1 V with V being any vector space) is a vector of the form, c1 v1 + c2 v2 + + cn vn , with {ci }i=1 being real (or complex) constants. We are going to be interested in the case that the vector space V consists of a class of functions on some domain, Rn . If u1 and u2 are functions on Rn , and c1 , c2 R we write (c1 u1 + c2 u2 ) =: u for the function, u : R such that u (x) = c1 u1 (x) + c2 u2 (x) for all x . For example we may consider, u1 + u2 , u1 + 3u2 , and 0 = 0u1 + 0u2 . De nition 4.2. A linear space of functions, V, is a class of functions with common domain so that if u1 , u2 are in the class then so is c1 u1 + c2 u2 for all c1 , c2 R, i.e. the space of functions V is closed under taking linear combinations. Example 4.3. D = {f : R R : f is di erentiable on R} or C = {f : R R : f is continuous on R}. Consider operator, L : D {all functions on R} de ned by Lf = f . This operator is linear, namely, we have L(c1 f1 + c2 f2 ) = (c1 f1 + c2 f2 ) = c1 f1 + c2 f2 = c1 L(f1 ) + c2 L(f2 ). It is interesting to note that L does not map D to C. For example, let n n 42 4 Linear Operators and Separation of Variables f (x) = then f (x) = so that D however f C. / 1 x2 sin x 0 x=0 x=0 x=0 x=0 1 1 2x sin x cos x x2 sin lim x x 0 1 x =0 De nition 4.4. A Linear operator, is a mapping, L, of one linear space of functions to another such that L(c1 u1 + c2 u2 ) = c1 L(u1 ) + c2 L(u2 ) for all u1 , u2 in the domain function space and c1 , c2 R. An induction argument shows the linearity condition in De nition 4.4 implies n N L i=1 ci ui = i=1 ci L(ui ) for all ui in the domain function space and ci R. Example 4.5. Let be some open subset of R2 , for example = R2 or = x R2 |x2 + x2 < 5 and let D denote those functions u : R such 1 2 that u and all of its partial derivatives up to order two exist and are continuous. (In the future we denote this class of functions by C 2 ( ) .) Then the following are example of linear operators taking D to the class of continuous functions on : 1. Lu = u x2 2u 2. Lu = x y 3. (Lu) (x, y) = x u (x, y) + y u (x, y) . y x Whereas, the following operator is an example of a non-linear operator; Lu = x u + u2 . To see this operator is not linear, notice that L(u1 + u2 ) = u1 + u2 + (u1 + u2 )2 x x = u1 + u2 + u2 + u2 = Lu1 + Lu2 . 1 2 x x and u2 = 1 x+5 2 For example, let u1 = 1 x (also see Exercise 13.9), then L(u1 ) = 1 1 + 2 =0 x2 x 1 1 L(u2 ) = + =0 (x + 5)2 (x + 5)2 Page: 42 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 4 Linear Operators and Separation of Variables 43 while L(u1 + u2 ) = L = 1 1 + x x+5 1 1 + x x+5 1 1 2 2 + + = = 0. 2 2 x (x + 5) x + 5) x(x + 5) 2 1 1 + x2 (x + 5)2 1 1 = 2 + x (x + 5)2 De nition 4.6. If L, M are two linear operators on the same class of functions we de ne L + M (L + M )(u) := Lu + M u. If M is a linear operator on the range-space of L we also de ne LM by ((LM )u) = L(M u). These new operators are still linear, for example, (LM )(c1 u1 + c2 u2 ) = L(c1 M (u1 ) + c2 M (u2 )) = c1 L(M (u1 )) + c2 L(M (u2 ))) = c1 (LM )(u1 ) + c2 (LM )(c2 ). However, it is in general not true (see Exercise 13.2) that LM = M L, in fact LM may be de ned while M L is not de ned. For example, let Lu = x2 u and M u = x u then taking u = exy we nd LM (u) = L while xy e x = L(y exy ) = x2 yexy 2 xy (x e ) = 2xexy + yx2 exy = LM (u) . x In general, in this class we will be interested in linear di erential operators of the form M (Lu) = Lu = Auxx + Buxy + Cuyy + Dux + Euy + F u where A, B, C, D, E, F are functions of x and y. The homogeneous partial di erential equations, Lu = 0 is shorthand notation for u solving the equation, Auxx + Buxy + Cuyy + Dux + Euy + F u = 0. Lemma 4.7 (Principle of Superposition). If L is a linear di erential operator as above and u1 and u2 solve the homogeneous partial di erential equations, Lu1 = 0 and Lu2 = 0, then any linear combination, c1 u1 + c2 u2 also satis es the same equation, namely, L (c1 u1 + c2 u2 ) = 0. Page: 43 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 44 4 Linear Operators and Separation of Variables Example 4.8 (Homogeneous Wave Equation ). The wave equation utt = 2 2 a2 uxx , is equivalent to writing Lu = 0 where L = t2 a2 x2 . So if u1 , . . . , uM are solutions to the wave equation, (i.e., Lun = 0), then any linear combination, c1 u1 + + cn uM , is another solution as well. (See Exercises 13.4, 13.6, 13.8 for more on this and the issue of boundary conditions.) To be more explicit let us notice that 1. Show that un (t, x) := sin(nx) sin(ant) for n N all solve the equation, Lu = 0. Indeed, Lun = 2 2 [sin(nx) sin(ant)] a2 2 [sin(nx) sin(ant)] t2 x = (sin(nx)an cos(ant)) a2 (n cos(nx) sin(ant)) t x = a2 n2 sin(nt) sin(ant) a2 [ n2 sin(nx) sin(ant)] = 0. 2. So by the superposition principle, N u(x, y) = n=1 cn sin(nx) sin(ant) with c1 , . . . , cM R also satis es the wave equation. (Later we will allow for in nite linear combinations and we will then choose the constants, ci , so that certain boundary conditions are satis ed.) 4.1 Introduction to Fourier Series In this section, I would like to explain how certain functions like sin nx and cos nx are going to appear in our study of partial di erential equations. Supd2 pose L is the di erential operator, L = dx2 , acting on functions on = [a, b] . De ne the inner product, (f, g) := =[a,b] f (x) g (x) dx, for functions f, g : R. Two integration by parts now shows, (Lf, g) = f (x) g (x) dx = f (x) g (x) dx + f (x) g (x) |b a = f (x) g (x) dx + [f (x) g (x) f (x) g (x)] |b a = (f, Lg) + [f (x) g (x) f (x) g (x)] |b . a Page: 44 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 4.1 Introduction to Fourier Series 45 There are now a number of boundary conditions that may be imposed on f and g so that boundary terms in the previous equation are zero. For example we may assume f, g Dper where Dper denotes those twice continuously di erentiable functions such that f (b) = f (a) and f (b) = f (a) . Or we might assume f, g DDirichlet or f, g DNeumann where DDirichlet (DNeumann ) consists of those twice continuously di erentiable functions such that f (a) = 0 = f (b) (f (a) = 0 = f (b)). In any of these cases, we will have (Lf, g) = (f, Lg) and so in analogy with the Spectral Theorem 1.5 we should expect that L has an orthonormal basis of eigenvectors. Let us nd these eigenvectors in a few examples. Before doing this it is useful to record a few integrals. Lemma 4.9. Let n be a positive integer, then sin2 nxdx = 0 0 cos2 nxdx = , 2 (4.1) (4.2) sin2 nxdx = cos2 nxdx = , and sin nx cos nxdx = 0. (4.3) Proof. Recall that cos 2 = cos2 sin2 = 1 2 sin2 = 2 cos2 1. Therefore, taking = nx and integrating we nd, 0= = 0 1 sin 2nx| = 0 2n cos 2nxdx 0 1 2 sin2 nx dx = 0 2 cos2 nx 1 dx cos2 nxdx 0 0 = 2 0 sin nxdx = 2 2 which gives Eq. (4.1). Similarly, replacing valid as well. Finally, by above shows Eq. (4.2) is sin nx cos nxdx = 1 sin2 nx| = 0 0 = 0. 2n Example 4.10 (Fourier Series). Let a = and b = , L = so that d2 dx2 and D = Dper Page: 45 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 46 4 Linear Operators and Separation of Variables (f, g) = f (x) g (x) dx (4.4) and (Lf, g) = (f , g ) = f (x) g (x) dx = (f, Lg). Thus if Lf = f with f D we have (f, f ) = (Lf, f ) = [f (x)] dx 0 2 from which it follows that 0. So we need only look for negative eigenvalues. If = 0 the eigenvalue equation becomes f = 0 and hence f (x) = Ax + B. We will only have f D if A = 0 and therefore let f0 = 1. We may now suppose that = 2 < 0 in which case the eigenvalue equation becomes f = 2 f which has f (x) = A cos x + B sin x as the general solution. We still must enforce the boundary values. For example f ( ) = f ( ) implies A cos ( ) + B sin ( ) = A cos + B sin or B sin = 0. Similarly, f ( ) = f ( ) implies A sin ( ) + B cos ( ) = A sin + B cos that A sin = 0. Hence we either have A = B = 0 (in which case f 0 which is not allowed) or sin = 0 from which it follows that = n Z. Hence we have := {cos nx, sin nx : n N} {1} as our possible eigenvectors. The eigenvalue associated to cos nx and sin nx is n = n2 . By Lemma 4.9, (cos nx, sin nx) = 0, so that is an orthogonal set and moreover, (cos nx, cos nx) = (sin nx, sin nx) = and (1, 1) = 2 . Thus we expect that any reasonable function f on [ , ] may be written as f (x) = 1 1 (f, 1) 1 + 2 [(f, cos n ( )) cos nx + (f, sin n ( )) sin nx] . (4.5) n=1 See Theorem 5.8, Theorem 5.17 and Theorem 7.2 below for more details on this point. Page: 46 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 4.2 Application / Separation of variables 47 Example 4.11 (Fourier Sine Series / Dirichlet boundary conditions). Suppose d2 a = 0 and b = , L = dx2 and D = DDirichlet so if f, g D then f (0) = 0 = f ( ) . We now take (f, g) = 0 f (x) g (x) dx and working as in the previous example we nd un (x) = sin nx with n = n2 for n N. By Lemma 4.9, (sin n ( ) , sin n ( )) = and so by Theorem 7.2, f (x) = 2 2 (f, sin n ( )) sin nx n=1 for any reasonable function f on [0, ] . Example 4.12 (Fourier Cosine Series / Neumann boundary conditions.). Supd2 pose a = 0 and b = , L = dx2 and D = DNeumann so if f, g D then f (0) = 0 = f ( ) . Again we take (f, g) = 0 f (x) g (x) dx and we nd the eigenfunctions and eigenvalues to be un (x) = cos nx with n = n2 for n N {0} . By Lemma 4.9, (cos n ( ) , cos n ( )) = for n N 2 and (1, 1) = , Thus Theorem 7.2 asserts that any reasonable function f on [0, ] may be written as 2 1 f (x) = (f, 1) 1 + (f, cos n ( )) cos nx. n=1 4.2 Application / Separation of variables Example 4.13. Use separation of variables to solve the heat equation, Page: 47 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 48 4 Linear Operators and Separation of Variables ut (t, x) = uxx (t, x) with u (t, 0) = u (t, 5) = 0 and u (0, x) = f (x) . (4.6) The technique is to rst ignore the nonhomogeneous condition u (0, x) = f (x) and look for any solutions to the Eq. (4.6) of the form u (t, x) = T (t) X (x) . From this we get, T (t) X (x) = T (t) X (x) or equivalently that T (t) X (x) = = T (t) X (x) where is a constant. Thus we require that X (x) = X (x) with X (0) = 0 and X (5) = 0. The solutions to this Sturm-Liouville problem are given by Xn (x) = sin This then forces Tn (t) = e t( n n x with = n = 5 5 n 5 2 . ) . Thus we nd that n 5 2 un (t, x) = e t( ) sin n x 5 2 are all solutions to Eq. (4.6). We then look for a general solution to our problem in the form u (t, x) = n=1 bn un (t, x) where we wish to choose the constants, bn such that f (x) = u (0, x) = n=1 bn un (0, x) = n=1 bn sin n x . 5 Letting 5 (f, g) = 0 f (x) g (x) dx we have sin and therefore, bn = m x 5 n x , sin = mn 5 5 2 5 f, sin n x 2 5 = 5 sin n x , sin n x 5 5 f (x) sin 0 n x dx. 5 Page: 48 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 4.2 Application / Separation of variables 49 Example 4.14. Use separation of variables to solve Laplace s equation, uxx (x, y) + uyy (x, y) = 0 for 0 x 5 and 0 y 2 with boundary conditions, u (0, y) = u (5, y) = 0, u (x, 0) = 0 and u (x, 2) = f (x) . To do this we will work as above and begin by ignoring the non-homogeneous boundary condition and solve the rest by separation of variables. So we write u (x, y) = X (x) Y (x) and require that Y X + = 0 with X (0) = X (5) = 0 = Y (0) . X Y As before we must have X = X and we know the solutions are given by Xn (x) = sin It them implies that Y (y) = From this we concluded that Yn (y) = sinh and we nd that n y 5 n 5 2 n x n with = n = 5 5 2 . Y (y) with Y (0) = 0. n y n x sinh 5 5 in this case. So working as above we try to nd a solution of the form un (x, y) = sin u (x, y) = n=1 bn un (x, y) . All the boundary conditions are now satis ed except for f (x) = u (x, 2) = n=1 bn un (x, 2) = n=1 bn sinh 2n n x sin . 5 5 By the same logic as above we must have bn sinh and thus that Page: 49 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 2n 2 = 5 5 5 f (x) sin 0 n x dx 5 50 4 Linear Operators and Separation of Variables u (x, y) = n=1 bn sinh 5 n x 2n sin 5 5 n x dx. 5 with bn = 2 5 sinh 2n 5 0 f (x) sin Example 4.15. Solve the wave equation, for 0 x 5 and t R, utt (t, x) = ux x (t, x) with u (t, 0) = u (t, 5) = 0 and u (0, x) = f (x) and ut (0, x) = 0. We could go through separation of variables here to answer this question, but this is getting tedious. I will just write down the answer as u (t, x) = cos As we have seen, f (x) = n=1 2 x t f (x) . bn sin 5 n x 5 n x dx 5 n x 5 with bn = and hence 2 5 f (x) sin 0 u (t, x) = n=1 bn cos bn cos n=1 2 x t sin = n t 5 sin n x . 5 It is interesting to notice that since sin (A + B) = cos A sin B + sin A cos B we have sin (A + B) + sin (A B) = 2 sin A cos B. Thus we may write, sin and thus u (t, x) = = 1 bn sin 2 n=1 n x cos 5 n t 5 = 1 sin 2 n (x + t) 5 + sin n (x t) 5 n (x + t) 5 + sin n (x t) 5 1 [F (x + t) + F (x t)] 2 Page: 50 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 4.2 Application / Separation of variables 51 where F (x) = n=1 bn sin n x 5 = the 5 periodic extensions of f (x) , i.e. if x = 5n + y with 0 < y < 5 and n Z, then F (x) = f (y) . Page: 51 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5 Orthogonal Function Expansions 5.1 Generalities about inner products on function spaces Let be a region in Rd (most of the time d will be one for us) and p : (0, ) be a positive function. For functions f, g : R de ne (f, g) := f (x) g (x) p (x) dx. This is an example of a inner product, i.e. something that behaves like the dot product on RN . For example we still have the following properties: (f1 + cf2 , g) = (f1 , g) + c (f2 , g) (f, g) = (g, f ) f 2 := (f, f ) = 0 implies f = 0. The following computation will be used frequently in this class: f +g 2 = (f + g, f + g) = f =f 2 2 +g 2 + (f, g) + (g, f ) (5.1) +g 2 + 2(f, g). De nition 5.1. Two functions f, g : R are orthogonal and we write n f g i (f, g) = 0. More generally, a collection of functions, { i }i=1 , is an orthogonal set if i j (i.e. ( i , j ) = 0) for i = j. If we further have n i = 1 then we say { i }i=1 is an orthonormal set. Exercise 5.1. Put in some exercise on orthogonal sets from the book here. Theorem 5.2 (Schwarz Inequality). For all f, g : R, |(f, g)| f g and equality holds i f and g are linearly dependent. 54 5 Orthogonal Function Expansions Proof. If g = 0, the result holds trivially. So assume that g = 0 and 2 observe; if f = g for some C, then (f, g) = g and hence |(f, g)| = | | g 2 =f g. Now suppose that f H is arbitrary, let h := f g 2 (f, g)g. (So z is the orthogonal projection of f onto g, see Figure 5.1.) Then Fig. 5.1. The picture behind the proof of the Schwarz inequality. 0 h 2 = f (f, g) g =f g2 (f, g)2 = f 2 g2 2 2 2 + (f, g)2 g g4 2 2 f, (f, g) g g2 from which it follows that 0 g equivalently i f = g 2 (f, g)g. f 2 (f, g)2 with equality i h = 0 or Corollary 5.3 (Triangle inequality). Let f, g : R be functions and a R, then f +g f + g af = |a| f . Proof. f +g 2 and (5.2) (5.3) =f f 2 2 +g +g 2 2 + 2(f, g) +2 f g = ( f + g )2 . Taking the square root of this inequality shows Eq. (5.2) holds. Taking the square root of the identity, af 2 = |a| |f (x)| dx = |a| 2 2 2 |f (x)| dx = |a| 2 2 f 2 , proves Eq. (5.3). Page: 54 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.1 Generalities about inner products on function spaces n 55 Proposition 5.4 (Pythagorean s Theorem). Suppose that { i }i=1 is an orthogonal set, then n 2 n i i=1 = i=1 i 2 . (5.4) Proof. Let s := s and n i=1 2 i , then n n = (s, s) = i=1 i , s = i=1 ( i , s) ( i , s) = i , n j = n ( i , j ) = ( i , i ) = i j=1 2 . j=1 The last two equations proves Eq. (5.4). Theorem 5.5 (Best Approximation Theorem). Suppose { i }i=1 is an orthonormal set and ai R, then n 2 n 2 n n f i=1 ai i = f i=1 (f, i ) i + i=1 |(f, i ) ai | 2 (5.5) n i=1 and therefore the best approximation to f by functions of the form occurs when ai = (f, i ) . Proof. The function (vector), n ai i h := f i=1 (f, i ) i , is orthogonal to { i }i=1 since n n n (h, j ) = f i=1 (f, i ) i , j n = (f, j ) i=1 (f, i ) ( i , j ) = (f, j ) i=1 (f, i ) ij = (f, j ) (f, j ) = 0. Since n n n f i=1 ai i = f i=1 n (f, i ) i + i=1 [(f, i ) ai ] i =h+ i=1 [(f, i ) ai ] i , Page: 55 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 56 5 Orthogonal Function Expansions it follows by Pythagorean s Theorem, Proposition 5.4, that n 2 n f i=1 ai i =h =h 2 + i=1 n [(f, i ) ai ] i |(f, i ) ai | i=1 2 2 2 2 + n n = f i=1 (f, i ) i + i=1 |(f, i ) ai | . 2 De nition 5.6. Let f be a function such that |f (x)| p (x) dx < and { i }i=1 be an orthonormal set, we will write f i=1 (f, i ) i to mean n n 2 2 lim f (x) i=1 (f, i ) i (x) p (x) dx n 2 = lim n f i=1 (f, i ) i = 0. We say { i }i=1 is complete (or closed in the book s terminology) if f 2 < . i=1 (f, i ) i whenever f Corollary 5.7 (Bessel s (In)equality ). Suppose { i }i=1 is an orthonormal set, then n n |(f, i )|2 f i=1 n 2 for all f, (5.6) Moreover we get equality i f = i=1 (f, i ) i . These statements remain true even when n = provided we interpret, f = i=1 (f, i ) i to mean f i=1 (f, i ) i . So we have f i=1 (f, i ) i i Pythagorean s theorem holds, i.e. i |(f, i )|2 = f i=1 2 . Proof. Taking ai = 0 in Eq. (5.5) shows n 2 n f and hence that n 2 = f i=1 (f, i ) i + i=1 |(f, i )| 2 n 2 |(f, i )| = f i=1 2 2 f i=1 (f, i ) i f 2 Page: 56 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.2 Convergence of the Fourier Series 57 with equality i f = shows, 2 n i=1 (f, i ) i . Letting n in the previous equation n 2 |(f, i )| = f i=1 2 lim n f i=1 (f, i ) i f 2 with equality i n n 2 lim f i=1 (f, i ) i = 0, i.e. i f i=1 (f, i ) i . 5.2 Convergence of the Fourier Series For this section it will be convenient to de ne (f, g) = f (y) g (y) 1 dy Recall from Example 4.10, if f : R R is reasonable 2 - periodic function (i.e. f (x + 2 ) = f (x) for all x R), we expect by analogy with the nite dimensional spectral theorem that f (x) = where an := (f, cos n ( )) = and bn := (f, sin n ( )) = 1 1 a0 + [an cos nx + bn sin nx] 2 n=1 (5.7) 1 f (y) cos ny dy for n = 0, 1, 2, . . . f (y) sin ny dy for n = 1, 2, . . . . The following theorem gives a precise version of this statement. Theorem 5.8 (Fourier Convergence Theorem). Let f : R R be a 2 - periodic function which is piecewise continuous on ( , ). Then at points x X where f (x ) exist we have f (x+) + f (x ) 1 a0 + [an cos nx + bn sin nx] = . 2 2 n=1 Page: 57 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 58 5 Orthogonal Function Expansions Fact 5.9 If f : [ , ] R is any function such that we may still de ne N |f (x)| dx < , 2 fN (x) = 1 a0 + [an cos nx + bn sin nx] 2 n=1 with an and bn as above. With this de nition we will always have, 1 lim N i.e. that f (x) |f (x) fN (x)| dx = lim 2 N f fN 2 = 0, 1 a0 + [an cos nx + bn sin nx] . 2 n=1 5.3 Examples Remark 5.10. We will use the following identities repeatedly. sin (A + B) = cos A sin B + sin A cos B, cos (A + B) = cos A cos B sin A sin B, 1 sin A cos B = (sin (A + B) + sin (A B)) 2 1 cos A cos B = (cos (A + B) + cos (A B)) 2 1 sin A sin B = (cos (A B) cos (A + B)) . 2 Example 5.11. Suppose f (x) = 1 if 0 < x < , 1 if < 0 < x (5.8) (5.9) (5.10) (5.11) (5.12) then an := (f, cos n ( )) = 0 because f is odd while bn = (f, sin n ( )) = 2 0 sin ny dy = 2 = (1 cos n ) = n Thus we conclude that f (x) n odd 2 cos ny| 0 n 0 if n is even 4 n if n is odd. 4 4 sin nx = sin (2n 1) x. n (2n 1) n=1 Page: 58 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.3 Examples 59 The series converges for every x R and x ( , ) \ {0} it converges to f (x) and at x = 0 it converges to 0. By the way, by Bessel s equality we have n=1 4 (2n 1) 2 =f 1 = 2 = 1 |f (x)| dx 2 1dx = 2 and hence we conclude that 1+ 1 1 1 2 1 + + + = = . 2 9 25 49 8 n=1 (2n 1) N 4 n=1:n odd n Here are some related graphs wherein fN (x) := sin nx. Page: 59 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 60 5 Orthogonal Function Expansions Fig. 5.2. A plot of f5 (x) = f (x) . 4 sin x + 1 3 sin 3x + 1 5 sin 5x . which is approximating Fig. 5.3. A plot of f11 (x) . Page: 60 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.3 Examples 61 Fig. 5.4. A plot of f19 (x) . These last few picture illustrate what is known as the Gibb s e ect; namely the convergence is not uniform. There is always the pesky little bump appearing near 0 (and ). 4 Fig. 5.5. A plot (at t = 1/2) of e t sin x + 1 e 9t sin 3x + 1 e 25t sin 5x which is 3 5 approximating the solution to the heat equation with periodic boundary condtions and with initial condition, u (0, x) = f (x) . The following lemma will be useful in simplifying the computations in some of the examples below. Lemma 5.12. Suppose f : [a, b] R is a continuous function and p (x) = n k k=0 pk x is a polynomial, then f pdx = pF1 p F2 + p F3 + ( 1) p(n) Fn+1 + C where F1 = f dx and Fk+1 = Fk dx. n Page: 61 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 62 5 Orthogonal Function Expansions Proof. This is simply a matter of repeated integration by parts. Explicitly, f pdx = F1 pdx = F1 p F1 p dx + C. F1 p dx = F2 p and hence f pdx = F2 p dx + C F1 pdx = F1 p F2 p + F2 p dx + C F3 p(3) dx + C, = F1 p F2 p + F2 p dx etc. Using this fact we may now compute the Fourier series of a number of functions. Example 5.13. Let (f, g) = 1. (x, 1) = 0, (x, cos nx) = 0 by symmetry and (x, sin nx) = 1 1 1 x sin nxdx = x cos nx 2 sin nx n n 1 2 n+1 = [ 2 cos n ] = ( 1) n n 1 f (x) g (x) dx, then wherein we have used F1 = F2 = Thus we expect that x= 2 n+1 ( 1) sin nx = 2 n n=1 sin nxdx = 1 n 1 cos nx and n 1 cos nxdx = 2 sin nx. n 1 1 1 1 sin x sin 2x + sin 3x sin 4x + . . . 1 2 3 4 . Page: 62 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.3 Examples 63 Plot of x and 2 2 2 2 sin x sin 2x + sin 3x sin 4x 1 2 3 4 . 2. f (x) = x expansion. x2 , sin nx = 0 by symmetry and 1 2 x2 , cos nx = x cos nxdx 1 sin nx 1 sin nx + 2x 2 cos nx 2 3 n n n 1 1 1 n = 2 2 [cos n + cos n ( )] = 4 2 ( 1) n n = x2 wherein we have used F1 = F2 = F3 = We also have x2 , 1 = 1 cos nxdx = 1 sin nx, n 1 1 sin nxdx = 2 cos nx and n n 1 1 2 cos nxdx = 3 sin nx. n n 2 x2 dx = x3 2 2 | = 3 3 Thus we expect from Eq. (4.5) that Page: 63 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 64 5 Orthogonal Function Expansions 12 x= x ,1 1 + 2 n=1 2 x2 , cos n ( ) cos nx + x2 , sin n ( ) sin nx = = = 1 1 2 2 n 1+ 4 2 ( 1) cos nx 23 n n=1 2 ( 1) cos nx +4 3 n2 n=1 1 2 1 + 4 cos x + 2 cos 2x 2 cos 3x + . . . . . 3 2 3 n Plot of x2 (in red) and 1 1 2 + 4 cos x + 2 cos 2x 2 cos 3x + . . . 3 2 3 . 3. Integrating the equation, x2 = one expects, ( 1) 2 +4 cos nx, 3 n2 n=1 n Page: 64 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.3 Examples 65 x3 2 ( 1) sin nx = x+4 3 3 n2 n n=1 = 2 ( 1) 2 3 n n=1 n+1 n sin nx 4 4 2 ( 1) n3 n=1 sin nx n+1 sin nx = n=1 2 2 ( 1) 3 n ( 1) n+1 n+1 ( 1) n3 n+1 =2 n=1 ( n) 6 sin nx 3n3 and hence that x3 = 2 n=1 ( 1) n+1 ( n) 6 sin nx n3 2 also see top of page 77 of the book 4. The expansions for functions of the form f (x) = ax + bx2 + cx3 are now easily found. For example, if x2 x = 2 ( 1) 2 n+1 +4 cos nx ( 1) sin nx. 2 3 n n n=1 n=1 n The plots of this function and the following approximation, S2 (x) = 2 +4 3 1 1 cos x + 2 cos 2x 2 12 2 1 1 sin x sin 2x , 1 2 S3 (x) = 1 2 1 1 +4 cos x + 2 cos 2x + 2 cos 3x 3 12 2 3 1 1 1 sin x sin 2x + sin 3x 2 1 2 3 and S4 (x) = 2 1 1 +4 cos x + + 2 cos 4x 2 3 1 4 1 1 1 1 2 sin x sin 2x + sin 3x sin 4x 1 2 3 4 are given in Figure 5.6 below. Example 5.14. Suppose that f (x) is a function de ned for 0 x . Suppose we extend f to be an odd function by setting Page: 65 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 66 5 Orthogonal Function Expansions Fig. 5.6. Plot of x2 x and the approximats, S2 , S3 , and S4 . Notice that the approximations are not doing to well at the end points. This is because they are convergeing to 2 at x = and x = . Fig. 5.7. The function f with its extension to an odd function on [ , ]. F (x) := f (x) if 0 < x < f ( x) if < x < 0, see Figure 5.7. Then computing the Fourier series of F, we learn an := (F, cos n ( )) = and bn := (F, sin n ( )) = From this we learn that 1 1 F (y) cos ny dy = 0 for n = 0, 1, 2, . . . F (y) sin ny dy = 2 f (y) sin ny dy. f (x) = n=1 2 f (y) sin ny dy sin nx. Page: 66 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.3 Examples 67 Notice that {sin ny : n N} form an orthonormal set relative to the inner product, 2 f (x) g (x) dx. (f, g) := 0 As an explicit example, let us consider the sin - series expansion of cos x for 0 < x < . For this we have bn = 2 1 = = = 1 1 cos y sin ny dy 0 (sin ((n + 1) y) + sin ((n 1) y)) dy 0 cos ((n + 1) y) cos ((n 1) y) + n+1 n 1 n+1 n 1 0 ( 1) ( 1) 1 1 + n+1 n 1 n+1 n 1 = 1n-even 2 2n n2 1 = 1n-even 1 2 1 + n+1 n 1 4n = 1n-even n2 1 and we conclude that cos x = 4 4 n sin nx n2 1 n=2,4,6,... 2n sin 2nx. 4n2 1 n=1 = 4 Here is a plot of cos x along with 2 22 1 sin 2x + 12 122 1 4 42 1 sin 4x + 6 62 1 sin 6x and 4 2 22 1 sin 2x + + sin 12x Page: 67 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 68 5 Orthogonal Function Expansions Example 5.15. Suppose that f (x) is a function de ned for 0 x . Suppose we extend f to be an even function by setting F (x) := f (x) if 0 < x < f ( x) if < x < 0, see Figure 5.8. Then computing the Fourier series of F, we learn Fig. 5.8. The function f with its extension to an even function on [ , ]. an := (F, cos n ( )) = and bn := (F, sin n ( )) = From this we learn that 1 1 F (y) cos ny dy = 2 f (y) cos ny dy 0 F (y) sin ny dy = 0. f (x) = where an := Notice that {cos ny : n N} inner product, (f, g) := a0 + an cos nx 2 n=1 2 f (y) cos ny dy. 0 1 2 form an orthonormal set relative to 2 the f (x) g (x) dx. 0 As an explicit example, let us consider the cos - series expansion of sin x for 0 < x < . For this we have Page: 68 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.3 Examples 69 an := 2 sin y cos ny dy 0 1 (sin ((n + 1) y) + sin ((1 n) y)) dy = 0 1 cos ((n + 1) y) cos ((1 n) y) = + n+1 1 n 0 1 1 1 = 2 1n-even + 1+n 1 n 4 4 1 1 = 1n-even 2 = 1n-even . 1 n2 n 1 So we conclude that sin x = 4 2 2 4 2 n2 1 cos nx 1 n=2,4,6,... = = 1 cos 2nx 4n2 1 n=1 2 2 2 cos 2x 2 cos 4x 2 cos 6x . . . 22 1 4 1 6 1 . 1 Plot of sin x and the function 2 1 2 2 cos 2x 2 cos 4x 22 1 4 1 which consists of the rst 3 terms in the cosine expansion of sin x. Page: 69 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 70 5 Orthogonal Function Expansions 5.4 Proof of Theorem 5.8 Before giving the proof of Theorem 5.8, we will need the following simple consequence of Bessel s inequality in Corollary 5.7. Lemma 5.16. Suppose f is a continuous function on [ , ] or more gener 2 ally any function such that |f (x)| dx < and an and bn are given as above, then 1 12 2 2 a+ |an | + |bn | 2 0 n=1 Moreover we also have N |f (x)| dx < . 2 (5.13) lim f (y) sin N+ 1 2 y dy = 0. (5.14) 1 Proof. Since 2 , cos nx, sin nx : n N forms an orthonormal set, it follows from Corollary 5.7 that 12 2 2 a+ |an | + |bn | 2 0 n=1 = 1 f, 2 2 + n=1 (f, cos n ( )) + (f, sin n ( )) |f (x)| dx < . 2 2 2 1 (f, f ) = In particular this implies that n lim f (y) cos ny dy = lim bn = 0 and n n lim f (y) sin ny dy = lim an = 0. n Since sin N+ 1 2 y = cos 1 y sin (N y) + sin 2 1 y cos (N y) , 2 Eq. (5.14) now follows from the previous limit formulas with f replaced by 1 f (y) cos 2 y and f (y) sin 1 y respectively. 2 Proof of the Fourier Convergence Theorem 5.8.. To concentrate on the basic ideas of the argument, I am only going to give the proof under the additional assumption that f is continuously di erentiable. The full proof may be found in the book. Page: 70 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.4 Proof of Theorem 5.8 71 Let fN (x) = 1 a0 + [an cos nx + bn sin nx] . 2 n=1 N (5.15) We begin by deriving a more tractable form for the function fN (x) . To do this notice that an cos nx + bn sin nx 1 1 = 1 = = f (y) cos ny dy cos nx + 1 f (y) sin ny dy sin nx f (y) (cos ny cos nx + sin ny sin nx) dy f (y) cos n(x y)dy wherein the last equality we have used Eq. (3.14) with t = nx and s = ny. Using this observation in Eq. (5.15) shows fN (x) = = where DN ( ) = 1 2 1 f (y) dy + 1 N f (y) cos n(x y)dy n=1 f (y) DN (x y)dy, (5.16) 1 sin N + 2 1 + cos n = , 2 n=1 2 sin 1 2 N (5.17) the second equality being Problem 14 of Section 32 of the book. (Also see Remark 5.18 below.) To see what DN ( ) looks like, see Figure 5.9. Making the change of variables, z = x y in Eq. (5.16) and using the fact that f and DN are 2 - periodic, we have fN (x) = = 1 1 f (y) DN (x y)dy = +x 1 1 x f (x z) DN (z)dz x+ f (x z) DN (z)dz = x f (x z) DN (z)dz. Also notice that it follows from Eq. (5.17) that 1 DN (y)dy = (1, DN ) = 1, 1 + cos n ( ) 2 n=1 N = 1 (1, 1) = 1. 2 Page: 71 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 72 5 Orthogonal Function Expansions Fig. 5.9. This is a plot D1 and D10 . Hence we may write fN (x) f (x) = 1 1 = 1 [f (x z) f (x)] DN (z)dz f (x z) f (x) sin 1 2 sin 2 z g (z) sin N+ 1 2 N+ 1 2 z dz (5.18) = where z dz g (z) := f (x z) f (x) 2 sin 1 z 2 f (z) if z = 0 if z = 0. Notice that, by l Hopital s rule, g is continuous and hence 1 |g (z)| dz < 2 and so we may let N in Eq. (5.18) with the aid of Lemma 5.16 to nd lim fN (x) f (x) = 1 lim N N g (z) sin N+ 1 2 z dz = 0. The following strengthens the convergence of the sum in Eq. (5.7) when : R R is piecewise C 1 . Theorem 5.17 (Uniform Convergence of Fourier Series). If f : R R is piecewise C 1 the convergence in Eq. (5.7) is uniform. To be explicit, if we let fN (x) be as in Eq. (5.15), i.e. Page: 72 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.4 Proof of Theorem 5.8 73 fN (x) = then 1 1 a0 + 2 N [an cos nx + bn sin nx] , n=1 max |f (x) fN (x)| 0 as N . x (5.19) Proof. We have max |f (x) fN (x)| = x 1 max x 1 max x 1 [an cos nx + bn sin nx] n=N +1 |an cos nx + bn sin nx| n=N +1 |an | + n=N +1 1 |bn | . n=N +1 (5.20) By the Cauchy-Schwarz inequality, |an | = n=1 n=1 n |an | 1 n (n |an |) n=1 2 1 n2 n=1 . (5.21) Finally, by an integration by parts (where the boundary terms vanish using the 2 - periodicity of the function f ) we nd nan = 1 f (y) n cos ny dy = 1 f (y) d cos ny dy dy 1 = f (y) cos ny dy = (f , cos (n ( ))) Therefore by Bessel s inequality, with f replaced by f , it follows that 2 n=1 (n |an |) < and so by Eq. (5.21), n=1 |an | < . Similarly we may also show that n=1 |bn | < and hence it Eq. (5.19) follows from Eq. (5.20). Remark 5.18 (The Dirichlet kernel for those who know complex variables). Recall Euler s formula which states, ei = cos + i sin . With this notation we have DN ( ) = 1 + cos n 2 n=1 1 1 in 1 + e e in = 2 n=1 2 2 N N N = ein . n= N Page: 73 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 74 5 Orthogonal Function Expansions Letting = ei /2 , we have N 2DN ( ) = n= N 2n = 2N +1 (2N +1) 2(N +1) 2N = 2 1 1 = and therefore 1 1 2i sin(N + 2 ) sin(N + 2 ) = . 1 2i sin 1 sin 2 2 N DN ( ) := 1 2 ein = n= N 1 sin(N + 2 ) , 1 2 sin 2 (5.22) with the understanding that the right side of this equation is N + 1 whenever 2 2 Z. 5.4.1 Gibb s Phenomenon Problem 38.7: This problems deals with the Fourier expansion of the function f (x) = (1 <x<0 + 10<x< ) . 2 The expansion is given by sN (x) = 2 =2 sin((2n 1)x) 2n 1 n=1 sin x sin 3x sin 5x sin 7x sin 9x sin 11x + + + + + ... 1 3 5 7 9 11 N whose plots are The goal of this problem is to show how that convergence is not uniform which is referred to as Gibb s phenomenon in this setting. (a) We have 2 sin x cos((2n 1)x) = sin 2nx sin((2n 2)x). So summing both sides of the above equation on n, N N 2 sin x n=1 cos((2n 1)x) = n=1 [sin 2nx sin((2n 2)x)] = sin 2x + sin 4x + + sin((2N 2)x) + sin 2N x sin 2x sin 4x sin((2N 2)x) = sin 2N x. Thus, N 2 n=1 cos((2n 1)x) = sin 2N x sin x (5.23) Page: 74 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.4 Proof of Theorem 5.8 75 Fig. 5.10. Plot of s2 . Fig. 5.11. A plot of S6 . for x = k , k an integer. Since for 0 < x < , sN (x) = 2 we may then write N sin((2n 1)x) , 2n 1 n=1 N sN (x) = 2 n=1 cos(2n 1)x) = sin 2N x . sin x (b) To nd local extrema in 0 < x < , we of course must nd the points in the interval at which sN (x) = 0. This occurs when sin 2N x = 0, at 2N x = 0, , 2 , . . .. So the extrema of sN (x) on the interval 0 < x < Page: 75 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 76 5 Orthogonal Function Expansions 2 3 (2N 1) , , ,..., . 2N 2N 2N 2N Thus the rst extremum on the relevant interval occurs at x = x= sN (x) = sN ( (sin x)(2N cos 2N x) (sin 2N x)(cos x) sin2 x 2N cos 2N )= = 2N sin 2N sin 2N < 0, occur when 2N and since we know a relative maximum occurs at this point. (c & d) Integrating Eq. (5.23) implies sN ( / (2N )) = 2 sin((2n 1) / (2N )) 2n 1 n=1 sin((2n 1)x) /(2N ) |0 2n 1 n=1 N /(2N ) N N =2 =2 n=1 0 /(2N ) cos((2n 1)x)dx sin 2N x dx sin x sin 2N x dx + x /(2N ) 0 /(2N ) 0 = 0 /(2N ) = 0 (x sin x) sin 2N xdx x sin x (5.24) = 0 sin x dx + x (x sin x) sin 2N xdx x sin x wherein we have used 1 1 (x sin x) = . sin x x x sin x Because (x sin x) (x sin x) sin 2N x x sin x x sin x is bounded near 0, we may let N in Eq. (5.24) to nd, N lim sN 2N = 0 sin x dx =: 1.8519. = x 5.5 Fourier Series on Other Intervals c Suppose f (x) is de ned for c < x < c. By setting F (y) := f y , we get a function de ned for < y < . This function may be expanded into a Fourier series as Page: 76 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.5 Fourier Series on Other Intervals 77 c 1 f y = F (y) = a0 + [an cos ny + bn sin ny] 2 n=1 where an := 1 1 bn := c y cos ny dy and c f y sin ny dy. f c Making the change of variables, x = y (or y = x) in the above equations c gives 1 f (x) = a0 + an cos n x + bn sin n x 2 c c n=1 where an := 1 c 1 bn := c f (x) cos n x dx c c c f (x) sin n x dx. c c c The convergence properties of these sum are the same as those for the Fourier series on ( c, c). Similarly if f (x) is de ned on 0 < x < c then we have the sine and cosine series expansions f (x) = f (x) = where now an := 2 c 2 bn := c f (x) cos n x dx c 0 c f (x) sin n x dx. c 0 c bn sin n x for 0 < x < c, and c n=1 1 a0 + an cos n x for 0 < x < c, 2 c n=1 Example 5.19. We have seen y= By letting y = cx 2 n+1 ( 1) sin ny for < y < . n n=1 in the above formula, we conclude that 2 n+1 x= ( 1) sin n x for c < x < c, c n c n=1 Page: 77 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 78 5 Orthogonal Function Expansions i.e. x= 2c n+1 ( 1) sin n x for c < x < c. n c n=1 Suppose that we want the cosine expansion of x on the interval, 0 < x < c. In this case we will have x= with an = 2c c d 2c x cos n x dx = x sin n x dx c0 c c0 n dx c c c c 2 c 2 sin n x|0 sin n x dx =x c n c c n 0 c 2 2 2c 2c n = cos n x|c = (( 1) 1) c n c0 c n 2 c c 1 a0 + an cos n x 2 c n=1 for n = 0 and a0 = Thus we have x= 0 x dx = x2 c | =c c0 c 4c 2 2 1 cos n x. n2 c n=1,3,5,... For example if c = 1, we have x= 1 4 2 2 cos x + 1 1 cos 3 x + 2 cos 5 x + . . . 2 3 5 , see Figure 5.19 below. A plot of x and its Fourier cosine series approximation, 1 4 1 1 2 2 cos x + 32 cos 3 x + 52 cos 5 x . Page: 78 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 5.5 Fourier Series on Other Intervals 79 This series is convergent for all 0 x c. Taking x = 0 in this expansion we may conclude that 0= or that 1+ 1 1 + 2 + ... 32 5 = 1 4 2 2 1+ 1 1 + 2 + ... 2 3 5 2 8 and taking x = 1 in this expansion we nd 1= which again gives 1+ 1 1 + 2 + ... 32 5 = 2 . 8 1 4 2 2 1 1 1 cos 3 x 2 cos 5 x + . . . 32 5 Page: 79 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 6 Boundary value problem examples We will begin in this chapter by solving three problems related to the operator d2 L = dx2 on [0, ] with Neumann boundary conditions and the function f (x) = x ( x) = x x2 Recall that the eigenfucntions/eigenvalues to this problem are Xn (x) = cos nx with n = n2 for n N {0} . Also we have and we have f (x) where a0 + an Xn (x) 2 n=1 82 6 Boundary value problem examples an = 2 f (x) Xn (x) dx = 0 2 x ( x) cos nxdx 0 d2 2 x ( x) 2 cos nxdx n2 0 dx 2 d d = 2 x ( x) cos nx | ( 2x) cos nxdx 0 n dx dx 0 d 2 ( 2x) = cos nxdx 2 n dx 0 2 [( 2x) cos nx] | ( 2) cos nxdx = 0 n2 0 2 sin nx = [( 2x) cos nx] | + 2 | 0 n2 n0 2 2 n n = 2 [( 1) + 1] = 2 [( 1) + 1] n n = and a0 = Therefore we have x ( x) = = 2 x ( x) dx = 0 12 . 3 1 2 n 2 [( 1) + 1] cos nx 6 n2 n=1 2 4 6 1 1 cos 2x + 2 cos 4x + . . . 2 2 4 Plot of 2 6 4 1 22 cos 2x + 1 42 cos 4x which is approximating x ( x) . Example 6.1. utt = uxx with ux (t, 0) = 0 = ux (t, ) and u(0, x) = 0 and ut (0, x) = x x2 . Page: 82 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 6 Boundary value problem examples 83 The answer is given by sin Lt sin Lt u (t, x) = f (x) = L L = = = Example 6.2. a0 sin nt t+ an Xn (x) 2 n n=1 a0 X0 (x) + an Xn (x) 2 n=1 2 1 n [( 1) + 1] cos nx sin nt t 2 6 n3 n=1 2 t 4 6 1 1 cos 2x sin 2t + 3 cos 4x sin 4t + . . . 3 2 4 ut = uxx with ux (t, 0) = 0 = ux (t, ) and u (0, x) = x x2 . Answer is given by u (t, x) = etL f (x) = etL a0 an Xn (x) X0 (x) + 2 n=1 = a0 tL e X0 (x) + an etL Xn (x) 2 n=1 2 a0 + an e tn Xn (x) 2 n=1 = 2 2 1 n = 2 [( 1) + 1] e tn cos nx 6 n3 n=1 = Notice that 2 4 6 2 1 t22 1 e cos 2x + 2 e t4 cos 4x + . . . 22 4 2 . 6 x ( x) dx = 0 13 6 because heat is conserved with Neumann boundary conditions. Example 6.3. Here we take 0 < x < and 0 < y < 3 and we wish to solve uxx + uyy = 0, ux (0, y) = 0 = ux ( , y) , u (x, 3) = 0 and u(x, 0) = x ( x) . Page: 83 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 84 6 Boundary value problem examples In this case we use the separation of variables mantra and look for Y (y) such that Y (3) = 0 and so that un (x, y) = Xn (x) Y (y) solves Laplace s equation. To do this we have 0 = un = n2 Xn (x) Y (y) + Xn (x) Y (y) from which we must demand that Y (y) n2 Y (y) = 0 so that Y (y) = Aeny + Be ny and then requiring 0 = Y (3) = Ae3n + Be 3n shows that B = Ae6n and thus that Y (y) = Aeny Ae6n e ny . 1 Taking A = 2 e 3n gives Y (y) = sinh n (3 y) . For n = 0 we have Y (y) = 0, so that Y (y) = Ay + B and to match the boundary condition we should take Y (y) = 3 y. Thus we look for a solution of the form u (x, y) = with f (x) = 1 2 n 2 [( 1) + 1] cos nx 6 n2 n=1 a0 (3 y) + an cos nx sinh n (3 y) 2 n=1 a0 an cos nx sinh n3 = u (x, 0) = 3 + 2 n=1 and so a0 2 2 3= or a0 = and 2 6 9 1 1 n n an sinh n3 = 2 2 [( 1) + 1] or an = 2 2 [( 1) + 1] . n n sinh n3 Thus the solution is given by u (x, y) = 2 1 sinh n (3 y) n (3 y) 2 . [( 1) + 1] cos nx 2 18 n sinh 3n n=1 Page: 84 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 6 Boundary value problem examples 85 Example 6.4. Suppose now 0 x and 0 y 1 and we wish to solve Laplace s equation, u (x, y) := uxx (x, y) + uyy (x, y) = 0 with u (0, y) = u ( , y) = 0, u (x, 1) = 0 and u (x, 0) = f (x) . Solution: By separation of variable, one nds the functions un (x, y) = sin nx sinh n (1 y) solve Eqs. (6.1) and (6.2) for all n N. By the superposition principle, (6.1) (6.2) (6.3) u (x, y) = n=1 bn un (x, y) = n=1 bn sin nx sinh n (1 y) (6.4) will also solve Eqs. (6.1) and (6.2). Finally we wish to chose the bn so that f (x) = n=1 bn sin nx sinh n (1 y) |y=0 = n=1 bn sin nx sinh n. But from the theory of the Fourier sine series, we know that bn sinh n = and therefore we must choose bn = 2 sinh n 2 f (x) sin nxdx 0 f (y) sin nydy. 0 We thus nd the solution to our problem is u (x, y) = n=1 2 sinh n f (y) sin nydy sin nx sinh n (1 y) . 0 For example if f (x) = sin 2x sin 5x, then we want sin 2x sin 5x = n=1 bn sin nx sinh n and hence that b2 sinh 2 = 1 and b5 sinh 5 = with all of the other bn = 0. Thus we nd u (x, y) = 1 sin 2x sinh 2 (1 y) sin 5x sinh 5 (1 y) . sinh 2 sinh 5 Page: 85 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 86 6 Boundary value problem examples Example 6.5 (Example 39.2 on p. 122). Here we wish to solve, ut = kuxx with u (t, 0) = 0, u (t, ) = u0 and u(x, 0) = 0. Notice that the boundary condition is not homogeneous. To take care of this we begin by nding the equilibrium solution, (x) to the problem. Namely, (x) = 0 with (0) = 0 and ( ) = u0 . The answer is of course, (x) = x u0 . We now write u (t, x) = U (t, x) + (x) and notice that U (t, x) = u (t, x) = kuxx = k (Uxx (t, x) + (x)) = kUxx (t, x) and U now has the boundary conditions, x U (t, 0) = 0, U (t, ) = 0 and U (x, 0) = (x) = u0 . Hence the solution to our problem now is U (t, x) = ektL (x) = Recalling that x= 2 n+1 ( 1) sin nx = 2 n n=1 u0 ktL e x. 1 1 1 1 sin x sin 2x + sin 3x sin 4x + . . . 1 2 3 4 it now follows that U (t, x) = ektL (x) = = u0 u0 2 n+1 ktL ( 1) e sin nx n n=1 2 n+1 ktn2 ( 1) e sin nx n n=1 and the full solution is given by u (t, x) = x u0 u0 2 n+1 ktn2 ( 1) e sin nx n n=1 n = u0 ( 1) ktn2 x+2 e sin nx . n n=1 Page: 86 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 7 Boundary value generalities 7.1 Linear Algebra of the Strurm-Liouville Eigenvalue Problem Suppose L is a di erential operator on functions on = [a, b] of the form Lf (x) := 1d p (x) dx (x) 1 d f (x) + q (x) f (x) . dx p (x) (7.1) De ne the inner product, (f, g) := =[a,b] f (x) g (x) p (x) dx, for functions f, g : R. With this notation we nd (Lf, g) = d f (x) + q (x) f (x) g (x) dx dx d d = (x) f (x) g (x) + q (x) f (x) g (x) dx + (x) f (x) g (x) |b a dx dx d d (x) g (x) + q (x) g (x) dx = f (x) dx dx (x) b d dx + [ (x) (f (x) g (x) f (x) g (x))]|a . Hence we see that (Lf, g) (f, Lg) = [ (x) (f (x) g (x) f (x) g (x))]|a . Let (a1 , a2 ) and (b1 , b2 ) be two non-zero vectors in R2 and de ne Bf (a) = a1 f (a) + a2 f (a) and Bf (b) = b1 f (b) + b2 f (b) . b (7.2) (7.3) 88 7 Boundary value generalities In the sequel we will be interested on imposing the boundary conditions on Bf (a) = 0 = Bf (b) . If we assume that f and g satisfy these boundary conditions in Eq. (7.3) then it follows that (f (a) , f (a)) and (g (a) , g (a)) line on the same line and therefore W (f, g) (a) := f (a) g (a) f (a) g (a) = det g (a) g (a) = 0. f (a) f (a) (W (f, g) is called the Wronskian of f and g.) Similar reasoning shows f (b) g (b) f (b) g (b) = det g (b) g (b) =0 f (b) f (b) and therefore it follows from Eq. (7.3) that L satis es the symmetry condition, (Lf, g) = (f, Lg) if Bf = 0 = Bg on = {a, b} . It also worth noting that if (a) = (b) and f and g satisfy periodic boundary conditions; i.e. f (a) = f (b) and f (a) = f (a) then we still have (Lf, g) = (f, Lg) . As we will see we are going to be interested in the following eigenvalue problem, namely we will be interested in nding solutions to the eigenfunction equation, Lf = f with Bf = 0 on . This may be rewritten as d d (x) f (x) + [q (x) + p (x)] f (x) = 0 with Bf = 0 on . dx dx This is the general form of the Sturm-Liouville eigenvalue problem as in Chapter 6 of the book. There (x) = r (x) . Let DB denote those functions f : R with are twice continuously di erentiable and satisfy the boundary conditions, Bf = 0 on and if (a) = (b) let Dper denote those functions f : R with are twice continuously di erentiable and satisfy periodic boundary conditions in Eq. (7.4). The next result is the formal analogue of Corollary 1.4. Proposition 7.1. If f, g DB or (a) = (b) and f, g Dper such that Lf = f and Lg = g with = , then (f, g) = 0. Proof. The same proof as that given for Corollary 1.4 work here without change. The next theorem is an analogue of the spectral theorem for the operator L. Page: 88 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 (7.4) 7.1 Linear Algebra of the Strurm-Liouville Eigenvalue Problem 89 Theorem 7.2 (Strurm-Liouville Spectral Theorem). Let L be as above, assume > 0 on [a, b] and let D = DB or D = Dper (in which case we assume additionally that (b) = (a)), then there exists un D and n R such that: 1. Lun = n un for all n, 2. the eigenvalues are increasing, i.e. 1 2 3 . . . , 3. limn n = (in fact # {n : n a} a1/2 or equivalently n n2 ). 4. Every nice function f on [a, b] may be expanded as f (x) = n=1 (f, un ) un (x) b = n=1 a f (x) un (x) p (x) dx un (x) . Example 7.3 (Problem 53.2 on p. 173). Consider the eigenfunction problem, X + X = 0 with X (0) = 0 and hX (1) + X (1) = 0 with h > 0. Notice again that 1 (X, X) = (X , X) = 0 1 X (x) X (x) dx = 0 X (x) X (x) dx + X (x) X (x) |1 0 2 = (X , X ) hX (1) (7.5) from which it follows that 0. Moreover if = 0, then X = 0 and hence X is constant. But since hX (1) + X (1) = 0 it follows that X = 0, hence we must have > 0. Therefore, X (x) = sin x for some . Implementing the last boundary condition then gives, h sin + cos = 0 or equivalently that tan = /h. Let n denote the solutions to this equation, see Figure 7.3 below. Page: 89 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 90 7 Boundary value generalities The x coordinates of the points of intersection of the line with the curves gives the values of when h = 1. So we have 2 Xn (x) = Cn sin n x where tan n = n /h with n > 0 and n = n . Let us now choose Cn so that (Xn , Xn ) = 1. To this end we have 1 2 (Xn , Xn ) = Cn 0 [sin n x] dx. 2 Now recall that cos 2 = 1 2 sin2 , and therefore, sin2 = Using this above gives 1 0 1 cos 2 . 2 [sin n x] dx = 2 1 2 1 = 2 1 = 2 1 (1 cos 2 n x) dx 0 sin 2 n x 1 | 2 n 0 sin 2 n 1 1 = 2 n 2 1 1 sin n cos n n . Since tan n = n /h, we have sin n = and therefore, 1 0 n cos n h (7.6) [sin n x] dx = 2 1 2 1+ cos2 n h = 1 h + cos2 n . 2h So we need to choose Cn so that Page: 90 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 7.2 General Elliptic PDE Theory 2 Cn 91 h + cos2 n = 1, 2h i.e. 2h h + cos2 n and hence the nal answer is given by Cn = Xn (x) = 2h sin n x where tan n = n /h. h + cos2 n 1 0 1 0 2 Here n > 0 and n = n . Alternative way to compute 1 2 n 0 2 [sin n x] dx. From Eq. (7.5), [ n cos n x] dx h sin2 n 2 2 [sin n x] dx = and therefore 1 0 [sin n x] dx = 0 1 2 1 [cos n x] dx + 2 h sin2 n 2 n h sin2 n . 2 n = 0 1 sin2 n x dx + From this equation and Eq. (7.6) it follows that 1 2 0 [sin n x] dx = 1 + 2 h h n sin2 n = 1 + 2 cos n 2 n n h h + cos2 n cos2 n = =1+ h h [sin n x] dx = 2 2 and therefore that 0 1 h + cos2 n 2h as before. 7.2 General Elliptic PDE Theory In this section we will state the generalization of theorem 7.2 to higher dimensional situations. Let be a nice open bounded region in Rd (typically we have in mind d = 1, 2, 3 here). Suppose that i,j (x) and p (x) be smooth functions on such that matrix (x) 11 (x) 12 (x) . . . 1d (x) 21 (x) 22 (x) . . . 2d (x) (x) := . . .. . . . . . d1 (x) d2 (x) . . . dd (x) Page: 91 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 92 7 Boundary value generalities Fig. 7.1. A region represnting some material of some region in space. is positive de nite, i.e. tr = and v v > 0 for all v Rd . We now form the inner product (u, v) := u (x) v (x) p (x) dV for function u, v on and let Lu := 1 i ( ij j u) + u. p i,j=1 d (7.7) Example 7.4. Look at the formula for the Laplacian in polar, cylindrical, and spherical coordinates to nd natural operators written in this form. (There are some singularities involved here which are arti ces of these coordinates system. We will have to deal with them later.) The general form of the heat equation also produced such operators L as in Eq. (7.7) above. By the divergence Theorem 1.9, d 1 (Lu, v) = i ( ij j u) + u vpdV i,j=1 p d = i,j=1 i ( ij j u) v + uvp dV d = ij j u i v + uvp dV + d i,j=1 ( ij j u) ni vd i,j=1 = [ u v + uv] p dV + ( u n) v d where n (x) is the outward pointing normal and d is the surface measure on . By interchanging the roles of u and v in the above formula, it follows that Page: 92 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 7.2 General Elliptic PDE Theory 93 (u, Lv) = [ u v + uv] dV + ( v n) u d ( v n) u d = [ u v + uv] dV + and therefore, (Lu, v) (u, Lv) = [( u n) v ( v n) u] d . (7.8) Let us now further suppose is a given function on and u and v satisfy the boundary conditions, Bu (x) := (x) u (x) n (x) + (x) u (x) = 0 where we allow for (x) = by which we mean u (x) = 0 at such points. Then using these boundary condition in Eq. (7.8) shows that [( u n) v ( v n) u] = vu uv = 0 on and hence we have (Lu, v) = (u, Lv) whenever Bu = Bv = 0. The following is an analogue of the spectral theorem for matrices in this context. Theorem 7.5. Keeping the above set up, there exists an orthonormal set {un }n=1 of eigenvectors for ( L, B) , i.e. Bun = 0 on and Lun = n un . Moreover these may be chosen so that: 1. the eigenvalues are increasing, i.e. 1 2 3 . . . , 2. limn n = (in fact # {n : n a} ad/2 or equivalently n n2/d ). 3. Every nice function f on may be expanded as set f (x) = n=1 (f, un ) un (x) f (x) un (x) p (x) dV un (x) . n=1 = Example 7.6. Let us nd the eigenfunctions u : = [0, a] [0, b] R such that u = u and u = 0 on . Page: 93 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 94 7 Boundary value generalities As usual, let us try to nd solutions of the form u (x, y) = X (x) Y (y) . In order for this to work we must have = and therefore, X (x) Y (y) = k and = k with X (x) Y (y) X (0) = 0 = X (a) and Y (0) = 0 = Y (b) for some constant k. The solutions to these one-dimensional Strum-Liouville problems are already known, namely we must have X (x) = sin m x and Y (y) = sin n y a b for some integers m, n. The functions, um,n (x, y) := sin m x sin n y a b satisfy um,n = m,n um,n with um,n = 0 on where m,n := 2 m a 2 u u X (x) Y (y) = = + u u X (x) Y (y) + n b 2 . All of these functions are orthogonal and in fact normalized if we de ne (f, g) := and we may expand f as 22 ab a b dx 0 0 dyf (x, y) g (x, y) f= m,n=1 Bm,n um,n where Bm,n = (f, um,n ) = 22 ab a b dx 0 0 dyf (x, y) um,n (x, y) . To see this is correct, notice that f (x, y) = n=1 2 b b 0 f (x, y ) sin n y d sin n y y b b (7.9) Page: 94 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 7.2 General Elliptic PDE Theory 95 and f (x, y ) = m=1 2 a a 0 f ( , y ) sin m xd sin m x. x x a a (7.10) Formally inserting Eq. (7.10) into Eq. 7.9) shows f (x, y) = n=1 2 b b d y 0 m=1 2 a a 0 a 0 f ( , y ) sin m xd sin m x sin n y sin n y x x a a b b = n=1 m=1 2 b b d y 0 2 a d f ( , y ) sin m x sin n y sin m x sin n y x x a b a b = m,n=1 Bm,n um,n (x, y) . It also interesting to observe that if u = u with u = 0 on and we let vm (y) := then vm (y) = = 2 a 2 a 2a u (x, y) sin m xdx = u (x, y) sin m xdx a a0 a 0 a 2 a2 2 x u (x, y) sin m xdx + y u (x, y) sin m xdx a a0 a 0 22 a 22 a m u (x, y) sin m xdx + y u (x, y) sin m xdx a a0 a a a 0 2 2 vm (y) + y vm (y) m a a 2 a a 0 u (x, y) sin m xdx, a = = and hence 2 y vm (y) = + m a 2 vm (y) with vm (0) = vm (b) = 0. 2 Since vm (y) is an eigenfunction of y with Dirichlet boundary conditions, we must have vm (y) = Cm sin nm b y for some constant Cm and some integer nm such that nm 2 m2 + 2 = 2 , a b i.e. we must have m2 nm 2 = 2 + . a b From all of this it follows that Page: 95 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 96 7 Boundary value generalities u (x, y) = = vm (y) sin m x a m=1 Cm sin nm y sin m x, b a m=1 with only a nite number of non-zero terms in the above sum. This shows that the functions {um,n : m, n N} forms a basis for all of the eigenfunctions of with Dirichlet boundary conditions on . Page: 96 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8 PDE Applications and Duhamel s Principle 8.1 Interpretation of d Alembert s solution to the 1-d wave equation Example 8.1. We may use d Alembert s solution to the wave equation to for2 sin t a2 x 2 2 2 . To see what mally work out the meaning of cos t a2 x and a x we should get, let A = y (t, x) = cos t 2 2 a2 x and A = sin t 2 a2 x 2 a2 x 2 a2 x then g (x) for < x < 2 a2 x f (x) + should solve Eqs. (2.5) and (2.6). By comparing this with Eq. (2.7), d Alembert s solution which I recall here, y(t, x) = we conclude that cos t sin t 2 a2 x f (x) = 2 a2 x 2 a2 x 1 1 [f (x + at) + f (x at)] + 2 2a x+at g(s)ds, x at 1 [f (x + at) + f (x at)] and 2 1 2a x+at (8.1) (8.2) g(x) = g(s)ds. x at We will use these results and the results of the next section to allow of to solve the forced wave equation. 8.2 Solving 1st - order equations using 2nd - order solutions Lemma 8.2 (A Key Fourier Transform Formula). For all R and t > 0, 98 8 PDE Applications and Duhamel s Principle 2 e 4t s cos ( s) ds = e t . 4 t 1 2 (8.3) Proof. Fix t > 0 and let g ( ) := e 4t s cos ( s) ds. 4 t 1 2 Then g ( ) = e 4t s s sin ( s) ds 4 t d e 4t s sin ( s) ds ds 4 t d e 4t s sin ( s) ds 4 t ds e 4t s cos ( s) ds = 2t g ( ) . 4 t 1 2 1 2 1 2 1 2 = 2t = 2t = 2t Solving this ODE for g gives, g ( ) = e t g (0) . This completes the proof since g (0) = 1 as we now show. Letting s = 2 tx, g 2 (0) = e 4t s ds 4 t 1 x2 4 1 2 2 = e 4 x dx 4 1 2 2 = e 4 e dx 1 4 r2 e 1 4 y2 4 dy = R2 2 2 1 e 4 (x +y ) dxdy 4 1 2 2 0 = 0 4 rdrd = 1 2 0 e 4 r rdr = e 4 r | = 1. 0 1 2 wherein the fth equality we have gone to polar coordinates. Theorem 8.3 (Solving for etA via cos At ). Suppose A is a N N symmetric matrix with all non-positive eigenvalues. Then 12 e 4t s cos As ds for all t > 0. (8.4) 4 t Proof. Formally we are taking = A in Eq. (8.3). To rigorously N prove Eq. (8.4), let {vi }i=1 be an orthonormal basis of eigenvectors for A. By assumption we may write eigenvalue for vi as 2 , i.e. Avi = 2 vi . i i Then etA = Page: 98 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.2 Solving 1st - order equations using 2nd - order solutions 12 e 4t s cos As vi ds = 4 t 99 t 2 i e 4t s cos ( i s) vi ds 4 t vi = etA vi 1 2 =e and the result follows since both sides of Eq. (8.4) are linear. 8.2.1 The Solution to the Heat Equation on R Example 8.4 (Heat Equation). Let us try to formally use Theorem 8.3 to solve the heat equation, 2 ut (t, x) = x u (t, x) with u (0, x) = f (x) . According to theorem 8.3 and Example 8.1 with a = 1, the solution should be given by u (t, x) = et x f (x) = 1 2 2 e 4t s cos 4 t 1 2 2 x s f (x) ds = e 4t s 1 [f (x + s) + f (x s)] ds 4 t 2 1 2 = e 4t s ds = f (x s) 4 t f (y) e 4t (x y) dy. 4 t 1 2 Exercise 8.1. Suppose f is a bounded continuous function, show u (t, x) := f (y) e 4t (x y) dy = 4 t 1 2 f (y)p (t, x y) dy (8.5) 2 solves the heat equation, ut (t, x) = x u (t, x) for t > 0 where e 4t x p (t, x) := . 4 t 1 2 (8.6) Hint: rst show p (t, x) solves the heat equation for t > 0. Then check u solves the heat equation by di erentiating past the integral, which you should assume to be valid here. It is a fact that we will discuss later that lim u (t, x) = f (x) . t 0 (8.7) This is based on the idea that p (t, x) is approximating a function, see Figure 8.1 We will abbreviate all this by the suggestive formula, et x f (x) = 2 f (y)p (t, x y) dy. (8.8) Page: 99 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 100 8 PDE Applications and Duhamel s Principle 1 1 Fig. 8.1. Plots of x p (t, x) for t = 2, t = 1 , t = 32 and t = 64 . Notice that p (t, x) 4 is being more and more concentrated near x = 0 as t 0 while always keeping the total area under x p (t, x) equal to one. 8.3 Duhamel s Principle Theorem 8.5 (Duhamel s Principle I). Suppose A is an N N, f RN and h (t) RN be given. Then the ordinary di erential equation, u (t) = Au (t) + h (t) with u (0) = f has a unique solution given by t (8.9) (8.10) u (t) = etA f + 0 e(t )A h ( ) d . (8.11) In words, u (t) is constructed by adding together the solutions to a bunch of initial value problems where h 0. Namely, etA f is the solution to u (t) = Au (t) with u (0) = f while e(t )A h ( ) is the solution to u (t) = Au (t) with u ( ) = h ( ) . Hence we have u (t) = + 0 solution at time t to u (t) = Au (t) & u (0) = f t solution at time t to u (t) = Au (t) & u (0) = h ( ) d . Page: 100 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.3 Duhamel s Principle 101 or equivalently that u (t) = + 0 solution at time t to u (t) = Au (t) & u (0) = f t solution at time t to u (t) = Au (t) & u ( ) = h ( ) d . Proof. Suppose u solves Eq. (8.9), then by the product rule d tA e u (t) = Ae tA u (t) + e tA u (t) dt = e tA Au (t) + e tA (Au (t) + h (t)) = e tA h (t) . Integrating this equation then shows t t e tA u (t) = u (0) + 0 e A h ( ) d = f + 0 e A h ( ) d . Multiplying this equation by etA on the left shows that if u exists it must be given by Eq. (8.11). To prove existence, let u now be de ned by Eq. (8.11) and notice that we may write it as t u (t) = etA f + 0 e A h ( ) d . Thus u (0) = f and, by the product rule and the fundamental theorem of calculus, t u (t) = AetA f + 0 e A h ( ) d + etA e tA h (t) = Au (t) + h (t) . Example 8.6. Continuing the notation and using the results of Example 1.8, 1 7 2 A := 7 1 2 2 2 10 with eigenvectors/eigenvalues given by 1 1 1 v1 := 1 6, v2 := 1 6, v3 := 1 12. 0 1 2 We wish to solve, Page: 101 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 102 8 PDE Applications and Duhamel s Principle u (t) = Au (t) + tf with u (0) = 0 where f = (1, 2, 3) = The solution is t tr 1 1 v1 + 2v2 + v3 2 2 u (t) = 0 t e(t )A f d 0 = 1 6(t ) 1 e v1 + 2e6(t ) v2 + e12(t ) v3 d . 2 2 To work this out we notice that t e d = 0 d d t e d = 0 d e =t | d =0 d e t 1 = d 1 = 2 1 (t + 1) e t and therefore, t e (t ) d = e t 0 1 1 (t + 1) e t 2 = Hence the answer is given by u (t) = 1 2 62 1 t e (1 + t ) . 2 2 e 6t 1 + 6t v1 + 62 e6t 1 6t v2 1 12t + 2 122 e 1 12t v3 . Theorem 8.7 (Duhamel s Principle II). Suppose A is an N N, f, g RN and h (t) RN be given. Then the ordinary di erential equation, u (t) = Au (t) + h (t) with u (0) = f and u (0) = g has a unique solution given by u (t) = cos sin At g+ At f + A t 0 (8.12) (8.13) sin A (t ) h ( ) d . A (8.14) Again, in words, u (t) is constructed by adding the solutions to a bunch of initial value problems where h 0. Namely, sin At cos At f + g A Page: 102 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.3 Duhamel s Principle 103 is the solution to u (t) = Au (t) with u (0) = f and u (0) = g while sin A(t ) h ( ) A is the solution to u (t) = Au (t) with u ( ) = 0 and u ( ) = h ( ) so solution at time t to u (t) = u (t) = Au (t) with u (0) = f & u (0) = g solution at time t to t u (t) = Au (t) with d . + 0 u ( ) = 0 & u ( ) = h ( ) Proof. The best way to understand this theorem is to reduce it to the rst version of Duhamel s principle in Theorem 8.5. To this end, let v (t) = u (t) , u (t) N N then the pair R R solves the equation, v (t) d dt u (t) v (t) = = = Let B := 0I A0 u (t) v (t) = v (t) u (t) v (t) Au (t) + h (t) 0I A0 u (t) v (t) + 0 h (t) . a 2N 2N matrix. Then by Theorem 8.5, we have u (t) v (t) = etB f g t + 0 e(t )B 0 h ( ) d . (8.15) When h 0, we know that u (t) = cos from which we deduce e tB At f + sin At g A sin At g A f g = cos At f + . (8.16) Page: 103 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 104 8 PDE Applications and Duhamel s Principle The second component of etB f is easily found as well (just di erentiate g the rst component) but we will not need it. Because of Eq. (8.16), e (t )B 0 h ( ) sin = A(t ) h ( ) A . (8.17) Hence taking the rst component of Eq. (8.15), using Eqs. (8.16) and (8.17), gives Eq. (8.14). We may also directly check that Eq. (8.14) works. Indeed, for simplicity assume f = 0 = g so that t sin A (t ) h ( ) d . u (t) = A 0 Then u (0) = 0, u (t) = sin A (t ) h ( ) | =t + A t = cos A (t ) h ( ) d 0 t 0 d sin A (t ) h ( ) d dt A (so u (0) = 0) and similarly, u (t) = cos A (t ) h ( ) | =t + 0 t t d cos A (t ) h ( ) d dt = h (t) + A sin A (t ) h ( ) d 0 t 2 sin A (t ) A h ( ) d = h (t) A 0 = h (t) + Au (t) as desired. Example 8.8. Continuing the notation and using the results of Example 1.8, 1 7 2 A := 7 1 2 2 2 10 with eigenvectors/eigenvalues given by 1 1 1 v1 := 1 6, v2 := 1 6, v3 := 1 12. 0 1 2 Page: 104 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.3 Duhamel s Principle 105 We wish to solve, u (t) = Au (t) + h with u (0) = 0 = u (0) where h = (1, 2, 3) = The solution is given by sin A (t ) hd u (t) = A 0 t sin A (t ) 1 1 = v1 + 2v2 + v3 d 2 2 A 0 6(t sin 6(t ) 1 t v1 + sinh 6 ) 2v2 2 6 d . = 12(t + sinh 12 ) 1 v3 0 2 t tr 1 1 v1 + 2v2 + v3 . 2 2 Now 0 t cos a (t ) =t 1 cos at sin a (t ) d = | =0 = a a2 a2 and similarly, t 0 sinh a (t ) cosh a (t ) =t 1 cosh at d = | =0 = a a2 a2 6t 1 cosh 6t 1 cosh 12t 1 1 v1 + 2v2 + v3 . 2 6 12 2 so that u (t) = 1 cos 6 Let us do a quick check that this solution is correct. For example let us check that 1 cos 6t u (t) = v1 6 solves u (t) = Au (t) + v1 . This is the case since, u (t) Au (t) = cos 6t v1 Au (t) 1 cos 6t = cos 6t v1 Av1 6 = cos 6t v1 + 1 cos 6t v1 = v1 as desired. Exercise 8.2. Let 2 1 1 A := 1 2 1 1 1 2 Page: 105 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 106 8 PDE Applications and Duhamel s Principle tr and h = ( 1, 1, 0) . Solve the following equations for u, u (t) = Au (t) + h with u (0) = 0 = (0, 0, 0) and u (t) = Au (t) + h with u (0) = 0 and u (0) = 0. Write your solutions in the form 3 tr u (t) = i=1 ai (t) vi where the functions ai are to be determined. 8.4 Application of Duhamel s principle to 1 - d wave and heat equations Using the formulas for cos t 2 x and sin t 2 2x x in Eqs. (8.1) and (8.2) respectively we are now in a position to formally apply Duhamel s principle in order to solve the forced wave equation; utt = uxx + h with u(0, ) = f and ut (0, ) = g. (8.18) Theorem 8.9. If f C 2 (R, R) and g C 1 (R, R), and h C(R2 , R) such that hx exists and hx C(R2 , R), then Eq. (8.18) has a unique solution u(t, x) given by Eq. (8.19). 2 Proof. By a formal application of Theorem 8.5 with A = x suggest that u(t, ) = cos(t 2 x )f + sin(t 2 x ) 2 x t g+ 0 sin((t ) 2 x 2 x ) h( , )d . Moreover using the formulas in Eqs. (8.1) and (8.2) then implies u(t, x) = 1 1 [f (x + t) + f (x t)]+ 2 2 t g(x+s)ds+ t 1 2 t x+t d 0 x t+ dy h( , y). (8.19) To verify that u de ned in Eq. (8.19) satis es Eq. (8.18) it su ces (by what we have already done) to assume f = g = 0 so that u(t, x) = Now 1 2 t x+t d 0 x t+ dy h( , y). Page: 106 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.4 Application of Duhamel s principle to 1 - d wave and heat equations 107 1 2 1 utt = 2 1 ux (t, x) = 2 1 uxx (t, x) = 2 ut = t [h( , x + t ) + h( , x t + )] d , 0 t [hx ( , x + t ) hx ( , x t + )] d + h(t, x) 0 t d [h( , x + t ) h( , x t + )] and 0 t d [hx ( , x + t ) hx ( , x t + )] 0 so that utt uxx = h and u(0, x) = ut (0, x) = 0. The only thing left to prove is the uniqueness assertion. For this suppose that v is another solution, then (u v) solves the wave equation (8.18) with f = g = 0 and hence by the uniqueness assertion in Theorem 2.4 (with a = 1), u v 0. Similarly we may solve the forced heat equation as well. Theorem 8.10 (The Forced Heat Equation). Suppose that f : R R and h : R2 R are bounded continuous functions, then the function t u (t, x) = p (t, x y) f (y) dy + 0 d p (t , x y) h ( , y) dy (8.20) solves the heat equation ut = uxx + h with lim u(t, x) = f (x) . t 0 2 Proof. Formally applying Theorem 8.5 with A = x suggests that (8.21) u (t, x) = et x f (x) + 0 2 t e(t ) x h ( , ) (x) d . 2 In light of Eq. (8.8), this equation then gives rise to Eq. (8.20). It is of course possible to directly check that Eq. (8.20) solves Eq. (8.21), however I will not stop to do it here. Example 8.11 (Problem 40.4 on p. 128). Solve the following problem on 0 < x < , ut = kuxx with u (0, x) = 0 and u (t, 0) = 0 and u (t, ) = F (t) . where F (0) = 0. To do this, let (t, x) = We then have x F (t) and then let U (t, x) = u (t, x) (t, x) . Page: 107 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 108 8 PDE Applications and Duhamel s Principle x 2 2 t k x U = t k x [u (t, x) (t, x)] = F (t) with U (0, x) = (0, x) = 0 Hence the solution to the problem is U (t, x) = x e(t )L F ( ) d 0 t x d x = e(t )L F ( ) |t + F ( ) e(t )L d 0 d 0 t x x F ( ) Le(t )L d = F (t) 0 t and therefore, u (t, x) = x x F (t) F (t) t 0 t 0 x F ( ) Le(t )L d = Recalling that x F ( ) Le(t )L d . 2 n+1 ( 1) sin nx n n=1 x= we have Le(t )L x = = we get u (t, x) = = 2 n+1 ( 1) Le(t )L sin nx n n=1 2 2 n+1 ( 1) kn2 e (t )kn sin nx n n=1 2 n ( 1) kn2 n n=1 2k t 0 e (t )kn F ( ) d e (t )kn F ( ) d 2 2 sin nx ( 1) n=1 n+1 0 t n sin nx. Example 8.12 (Problem 40.4 on p. 128). Solve the following problem on 0 < x < , ut = kuxx with u (0, x) = 0 and u (t, 0) = 0 and u (t, ) = F (t) . where F (0) = 0. Page: 108 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.4 Application of Duhamel s principle to 1 - d wave and heat equations 109 To do this, let (t, x) = We then have x F (t) and then let U (t, x) = u (t, x) (t, x) . x 2 2 t k x U = t k x [u (t, x) (t, x)] = F (t) with U (0, x) = (0, x) = 0 U (t, 0) = 0 = U (t, ) . Hence the solution to the problem is U (t, x) = x e(t )L F ( ) d 0 t x d x = e(t )L F ( ) |t + F ( ) e(t )L d 0 d 0 t x x = F (t) F ( ) Le(t )L d 0 t and therefore, u (t, x) = x x F (t) F (t) t 0 t 0 x F ( ) Le(t )L d = Recalling that x F ( ) Le(t )L d . 2 n+1 ( 1) sin nx n n=1 x= we have Le(t )L x = = we get u (t, x) = = 2 n+1 ( 1) Le(t )L sin nx n n=1 2 2 n+1 ( 1) kn2 e (t )kn sin nx n n=1 2 n ( 1) kn2 n n=1 2k t 0 e (t )kn F ( ) d e (t )kn F ( ) d 2 2 sin nx ( 1) n=1 n+1 0 t n sin nx. Example 8.13 (Forced Wave Equation problem). Consider the forced wave equation, Page: 109 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 110 8 PDE Applications and Duhamel s Principle x with u (t, 0) = ux (t, ) = 0 2 u (0, x) = f (x) and ux (0, x) = g (x) . utt = uxx + sin d Letting L = dx2 acting on functions satisfying h (0) = 0 = h ( ) , the formal solution to our problems is given by sin t L u (t, x) = cos t L f (x) + g (x) L t sin (t ) L x d . sin + 2 L 0 2 In order to work this out we need to recall (see Problem 53.1 on p. 173) that the eigenfunctions n (x) = sin n x where n = 2 and L n = n n and 2n 1 for n N 2 sin2 n xdx = 0 0 1 sin 2 n x 1 cos 2 n x dx = |= . 2 2 2 2 n 0 2 Thus we have f (x) = n=1 fn sin n x and g (x) = n=1 gn sin n x where fn = 2 f (x) sin n xdx and gn = 0 2 g (x) sin n xdx. 0 The solution to our problem is thus sin t L u (t, x) = fn cos t L sin n x + gn sin n x L n=1 n=1 t + 0 sin ((t ) /2) x sin d 1/2 2 = n=1 fn cos ( n t) sin n x + n=1 gn sin ( n t) sin n x n 4 cos ((t ) /2) |t 0 x sin 2 = n=1 fn cos ( n t) sin n x + n=1 gn sin ( n t) t x sin n x + 4 1 cos sin . n 2 2 Page: 110 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.4 Application of Duhamel s principle to 1 - d wave and heat equations 111 t x sin 2 2 does satisfy the correct equation. It is easily seen that v (0, x) = v (0, x) = 0 and v (t, x) = 4 1 cos vtt vxx = 4 1 t x1 t x cos sin + 1 cos sin 4 2 24 2 2 x = sin 2 Let us check that as required. Example 8.14 (Ressonance in the Wave Equation). Consider now the forced equation with a time dependent forcing, namely x t sin with u (t, 0) = ux (t, ) = 0 2 2 u (0, x) = 0 and ux (0, x) = 0. utt = uxx + sin The solution is given by t u (t, x) = 0 t = 0 sin (t ) L x sin sin d 2 2 L sin (t ) 1 x 2 d sin sin 1 2 2 2 t 1 x sin d sin 2 2 2 0 1 1 1 x = 2 sin t t cos t sin 2 2 2 2 1 1 x = 2 sin t t cos t sin 2 2 2 =2 sin (t ) wherein we have computed this integral using the identity, sin A sin B = 1 (cos (A B) cos (A + B)) 2 t with A = (t ) 1 and B = /2 to nd 2 t 2 0 sin t 2 sin d = 2 = cos 0 t 2 t 2 sin t 2 t |t t cos 0 2 cos d t t t cos . 2 2 This is an example the phenomenon of resonance, namely, the amplitude of u (t, x) can be arbitrarily large for most values of large t. = 2 sin Page: 111 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 112 8 PDE Applications and Duhamel s Principle Proposition 8.15. The solution to utt = kuxx with u (0, x) = 0 = ut (0, x) and u (t, 0) = 0, and u (t, ) = F (t) , where 0 < x < and (0) = 0 is given by u (t, x) = 1 ( 1) n=1 n+1 0 t sin ((t ) n) F ( ) d sin nx. Moreover if F ( ) = sin , then u (t, x) = 1 1 1 sin t t cos t sin nx + (Terms remaining bounded) . 2 2 Notice that t cos t can be arbitrarily large, another manifestation of resonance. Proof. To prove this, let (t, x) = (t, x) . We then have x F (t) and then let U (t, x) = u (t, x) x 2 2 2 2 t k x U = t k x [u (t, x) (t, x)] = F (t) with U (0, x) = (0, x) = 0 x Ut (0, x) = F (0) and U (t, 0) = 0 = U (t, ) . Hence the solution to the problem is t (0) sin L x U (t, x) = F L sin (t ) L x F ( ) d . L t 0 Integrating the last term by parts shows, t 0 =t sin (t ) L x sin (t ) L x F ( ) d = F ( ) L L =0 t d sin (t ) L x F ( ) d d L 0 sin t L x F (0) = L t x + cos (t ) L F ( ) d . 0 Combining the last two equations then implies that t U (t, x) = 0 x cos (t ) L F ( ) d . Page: 112 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 8.4 Application of Duhamel s principle to 1 - d wave and heat equations 113 We now integrate by parts one more time to nd, t =t x U (t, x) = cos (t ) L F ( ) + =0 0 t x = F (t) + L sin (t ) L 0 d x cos (t ) L F ( ) d d x F ( ) d . and therefore, u (t, x) = t x x x F (t) F (t) + L sin (t ) L F ( ) d 0 t x L sin (t ) L = F ( ) d 0 t = 0 1 L sin (t ) L t 2 n+1 ( 1) sin nx F ( ) d n n=1 = = 1 1 1 n+1 ( 1) n n=1 n sin ((t ) n) F ( ) d (sin nx) 0 t ( 1) n=1 n+1 0 sin ((t ) n) F ( ) d sin nx. To be more explicit, let us suppose that F ( ) = sin , in this case t sin ((t ) n) sin d = 0 1 2 t (cos ((n + 1) nt) cos (nt + (1 n) )) d 0 =t =0 = 1 sin ((n + 1) nt) sin (nt + (1 n) ) 2 n+1 (1 n) for n = 1 and for n = 1 we have t sin ((t )) sin d = 0 1 1 sin t t cos t. 2 2 Therefore u (t, x) = 1 1 1 sin t t cos t sin x + 2 2 (Terms remaining bounded) . Page: 113 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9 Problems In other Coordinates Systems 9.1 A Heat equation in spherical coordinates Example 9.1 (Example 40.3 on p. 126 (Spherical Coordinates)). Let x R3 , |x| = x x and = x R3 : |x| a . We wish to solve ut (t, x) = u (t, x) for x with u (t, x) = 0 if |x| = a and u (0, x) = f (|x|) for x We will look for a solution which only depends on r = |x| , where r is the radial variable in spherical coordinates. Recalling from Eqs. (1.15) and (1.16) that 12 1 1 2 g = r (rg) + 2 (sin g) + 2 2 g r r sin r sin our equations becomes ut (t, r) = 12 1 r (ru (t, r)) = (ru)rr (t, r) with r r u (t, a) = 0 and u (0, r) = f (r) for 0 r a. Hence if we let v (t, r) := ru (t, r) , we nd that vt (t, r) = vrr (t, r) with v (t, a) = 0 = v (t, 0) and v (0, r) = rf (r) for 0 r a. We can now solve this equation for v by the standard means, namely, rf (r) = with bn := 2 a a bn sin n r a rf (r) sin n r dr. 0 116 9 Problems In other Coordinates Systems Fig. 9.1. The region . Then we have, with L = 2 r 2 with Dirichlet boundary conditions, that bn sin n bn etL sin n sin n r a r a v (t, r) = etL = and so u (t, r) = 1 r r = a n2 2 bn exp t 2 a bn exp t n2 2 a2 r sin n . a 9.2 Problems with cylindrical symmetries Example 9.2 (Example 43.1 on p. 137 (Harmonic functions in cylindrical coordinates)). In this example, denotes those points which may be written in polar coordinates ( , ) with 1 < < b and 0 < < , see Figure 9.1 below. We wish to solve u = 0 with u ( , 0) = 0 = u ( , ) for 1 < < b and u (1, ) = f ( ) and u (b, ) = g ( ) for 0 < < . Recalling that in polar coordinates is given by u = 1 12 ( u) + 2 u (9.4) (9.1) (9.2) (9.3) we look for some solutions to Eq. (9.1) and Eq. (9.2) of the form u ( , ) = R ( ) ( ) . To short cut things, from experience we know that ( ) = sin n for some n N and hence un ( , ) = R ( ) sin n . Plugging this into Eq. (9.1) then implies Page: 116 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.2 Problems with cylindrical symmetries 117 0= 1 n2 ( R) 2 R. This is an Euler type equation which may be solved by looking for solution of the form R ( ) = . This happens i 0 = 2 2 n2 2 and hence we must have 2 = n2 for = n. Thus we nd un ( , ) = An n + Bn n sin n . So we now look for solutions of the form u ( , ) = n=1 An n + Bn n sin n and we wish to choose An and Bn so that Eq. (9.3) is satis ed. If we write f ( ) = n=1 f (n) sin n and g ( ) = n=1 g (n) sin n with 2 f (n) = 2 g (n) = then we must have f ( ) sin n d and 0 g ( ) sin n d 0 f (n) sin n = n=1 n=1 (An + Bn ) sin n and An bn + Bn b n sin n n=1 g (n) sin n = n=1 which implies that An + Bn = f (n) An bn + Bn b n = g (n) . Solving these equations gives Bn bn b n = bn f (n) g (n) An b n bn = b n f (n) g (n) or Page: 117 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 118 9 Problems In other Coordinates Systems Fig. 9.2. The unit disk in R2 . Bn = bn f (n) g (n) n b n b b n f (n) g (n) g (n) b n f (n) An = = n bn n b n b b and we nd u ( , ) = n=1 g (n) b n f (n) n f (n) bn g (n) n + n b n n b n b b g (n) n n + f (n) b n sin n b n = 1 n b n b n=1 sin n . Example 9.3 ( Laplace s equation on the disk). We wish to solve Laplace s equation, u = 0 on with u = f on where = (x, y) R2 : x2 + y 2 < 1 is the unit disk as in the Figure below. The usual separation of variables arguments shows that un ( , ) = (A cos n + B sin n ) a n + b n for n = 1, 2, 3, . . . 1 and for n = 0 we have 0 = ( R) , which implies R = a and R = a ln + b so that u0 ( , ) = 1 (a ln + b) all solve u = 0. However, we want our solution to be continuous and hence bounded near (0, 0) which requires us to take a = 0. Thus we look for a solution of the form u ( , ) = A0 + (An cos n + Bn sin n ) n . 2 n=1 Page: 118 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.3 Strum Liouville problem in cylindrical coordinates 119 To satisfy the boundary condition we must require, f ( ) = i.e. An = A0 + (An cos n + Bn sin n ) , 2 n=1 1 1 f ( ) cos n d and Bn = f ( ) sin n d . You are asked to show in Problem 48.7 that the solution may be written as u ( , ) = 1 2 1 2 f ( ) d for < 1. 1 2 cos ( ) + 2 This is called the Poisson integral formula. 9.3 Strum Liouville problem in cylindrical coordinates In this section we will consider problems with symmetry about the z - axis. We will, for simplicity, restrict ourselves the region, = (x, y) R2 : x2 + y 2 c2 R2 and consider problems with associated to Dirichlet boundary conditions. That is we wish to solve the heat and wave equations, ut = u with u = 0 on and utt = u with u = 0 on . Following the strategy used throughout this course we rst will look for solutions to the eigenfunction problem, = on with = 0 on . As above we will let (f, g) := (9.5) f (x, y) g (x, y) dxdy. (9.6) Let us recall that if is a solution to Eq. (9.5), then ( , ) = ( , ) = ( , ) = | | dxdy 2 2 and hence it follows that 0. Moreover, if = 0, then | | dxdy = 0 which implies = 0. This condition along with the Dirichlet boundary conditions implies that 0. Page: 119 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 120 9 Problems In other Coordinates Systems Conclusion: When looking for solutions to Eq. (9.5) we may restrict our attention to the case where > 0. Because of the symmetry of the domain, we will view as a function of polar coordinate, ( , ) in which case Eq. (9.5) may be written as 12 1 ( ) + 2 = with (c, ) = 0, see Eq. (1.14). Moreover in polar coordinates, Eq. (9.6) becomes 2 c (9.7) (f, g) := 0 0 f ( , ) g ( , ) d d . (9.8) The structure of Eq. (9.7), suggests we look for solutions of the form, ( , ) = R ( ) ( ) . Plugging this into Eq. (9.7) implies 1 R2 ( R) + 2 = R and then multiplying by 2 R gives 2 ( R) + 2 = = k R for some constant k. We begin by recalling the solutions to 2 = k with (0) = (2 ) and (0) = (2 ) are of the form n ( ) = An cos n + Bn sin n for some n = 0, 1, 2, . . . and constants A and B. For such a n we have k = n2 . Fixing such an n, we must now solve ( R) + 2 = n2 with R (c) = 0. R Multiplying this equation through by R, we get ( R) + 2 n2 R = 0 with R (c) = 0 or equivalently that 2 R + R + 2 n2 R = 0 with R (c) = 0. 9.4 Bessel Equation and Functions Before proceeding, we will need to understand better the solutions to the ordinary di erential equation, ( R) + 2 n2 R = 0 with > 0. (9.9) Page: 120 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.4 Bessel Equation and Functions 121 Lemma 9.4. If R solves Eq. (9.9), x := and x , y (x) := R ( ) = R then y solves Bessel s equation of order n, 0=x d dx x d y + x2 n2 y dx (9.10) = x2 y (x) + xy (x) + x2 n2 y (x) . Alternatively put, if y solves Bessel s equation of order n, then R ( ) := y will solve Eq. (9.9). Proof. By the chain rule, d dx d x d d = = =x d d dx dx dx and hence in this new variable, Eq. (9.9) becomes 0=x =x d dx d dx x x d y+ dx x 2 n2 y d y + x2 n2 y dx and hence we have eliminate the from the equation. The previous results show that if y solves Bessel s equation of order n and c = 0, then y u ( , ) = (An cos n + Bn sin n ) y satis es u = u with u = 0 on . (9.12) To understand this better we need to understand the solution to Eq. (9.10) better. We will do this in the next few results. Theorem 9.5. Every solution to Bessel s Eq. (9.10) of order n which is bounded near x = 0 is a multiple the Bessel function of the rst kind of order n de ned by (9.11) Jn (x) := k=0 ( 1) k! (n + k)! k x 2 n+2k . (9.13) Page: 121 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 122 9 Problems In other Coordinates Systems Fig. 9.3. A plot of J0 , J3 and J10 . Notice that J0 (0) = 1 whereas Jn (0) = 0 for all n 1. Proof. This solution can be found by the usual power series method for solving such equations, namely plugging in the series expansion of the form y (x) = k=0 ak xk+r into Eq. (9.10) and then nding relations that r and the coe cients ak have to satisfy. This method produces the solution in Eq. (9.13) above. Moreover, it can be veri ed by straightforward computation that Jn (x) does indeed solve Eq. (9.10). To nd more solutions one may use the method of reduction of order to show that the general solution to Eq. (9.13) is of the form y (x) = A 11 dx + B 2 Jn (x) x for some constants A and B. Since Jn (x) xn for x near zero we learn that u (x) 1 x2n+1 dx x 2n for x near 0 if n 1, u (x) ln |x| for x near 0 if n 1 and therefore, we nd u (x) Jn (x) x n for x near zero if n 1 or ln |x| J0 (x) if n = 0. At any rate, Jn (x) is the only solution to Bessel s equation which is bounded near x = 0. Theorem 9.6. The functions Jn (x) behave as shown in Figure 9.3. More precisely, there are constants cn and n such that Jn (x) cn x 1/2 sin(x + n ) as x . (9.14) Page: 122 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.4 Bessel Equation and Functions 123 In particular each Bessel function Jn has an in nite number of zeros 0 < n,1 < n,2 < . . . n,j < . . . and limj n,j = . (By convention, when n = 0 we let 0,1 denote the rst j zero of J0 which is positive.) Proof. We give the idea for the proof without giving the full proof which would take us to far a eld. We begin by writing a solution to Bessel s Eq. (9.13) in the form y (x) = xr u (x) where r is to be determined so that u solves a second order di erential equation where u (x) does not appear. Since y = rxr 1 u + xr u and y = r (r 1) xr 2 u + 2rxr 1 u + xr u we have 0 = x2 y + xy + x2 n2 y = r (r 1) xr u + 2rxr+1 u + xr+2 u + rxr u + xr+1 u + x2 n2 xr u = xr+2 u + (2r + 1) xr+1 u + r2 + x2 n2 xr u. So we choose r = 1 and then divide the above result by xr+2 to learn u 2 solves, 1/4 n2 u + 1+ u = 0. x2 For x large this equation looks like u + u = 0 (the spring constant, 1 + 1 2 x 2 1 as x ), and thus we expect u(x) c sin(x + ) for 4 n x >> 1. Thus we are lead to expect that Jn (x) = x 1/2 u (x) should behave as described in Eq. (9.14). Corollary 9.7 (Eigenfunctions). The functions, un,j ( , ) := (An cos n + Bn sin n ) Jn n,j satisfy un,j = 2 n,j un,j with un,j = 0 on . c2 , c (9.15) (9.16) Proof. This is a consequence of Eqs. (9.11), (9.12), Lemma 9.4, Theorem 9.5 and Theorem 9.6. Theorem 9.8 (Recurrence relations for Bessel functions). Let Jn (x) be Bessel function of the rst kind of order n in Eq. (9.13), then: Page: 123 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 124 9 Problems In other Coordinates Systems d x n Jn (x) = x n Jn+1 (x) or equivalently dx xJn (x) = nJn (x) xJn+1 (x) , for n = 0, 1, 2, 3, . . . , dn [x Jn (x)] = xn Jn 1 (x) or equivalently dx xJn (x) = nJn (x) + xJn 1 (x) , for n = 1, 2, 3, . . . , and xJn+1 (x) = 2nJn (x) xJn 1 (x) for n = 1, 2, 3, . . . In particular from Eq. (9.18) with n = 0, J1 (x) = J0 (x) (9.17) (9.18) (9.19) (9.20) (9.21) (9.22) and form Eq. (9.20) with n = 1, xJ1 (x) = J1 (x) + xJ0 (x) which may be written as d (xJ1 (x)) . (9.23) xJ0 (x) = dx Proof. Computing d d x n Jn (x) = 2 n dx dx k=0 ( 1) k! (n + k)! k k x 2 2k = 2 n k=1 ( 1) k k! (n + k)! k x 2 2k 1 = 2 n k=1 ( 1) (k 1)! (n + k)! ( 1) k! (n + k + 1)! ( 1) k! (n + k + 1)! k k+1 x 2 x 2 2k 1 = 2 n = x = x This relation also implies Jn+1 (x) = xn 2k+1 k=0 n k=0 n x 2 2k+1+n Jn+1 (x) . d x n Jn (x) = nx 1 Jn (x) + Jn (x) dx which his equivalent to Eq. (9.18). Eq. (9.20) is proved similarly, see Problems 69.1 from Brown and Churchill. Subtracting Eq. (9.20) from Eq. (9.18) gives Eq. (9.21). Page: 124 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.4 Bessel Equation and Functions 125 Theorem 9.9 (Bessel function normalizations). Let n,j denote the j th zero of Jn (x) . Then 1 2 Jn ( n,j ) d = 0 1 1 2 2 (Jn ( n,j )) = (Jn+1 ( n,j )) 2 2 (9.24) More generally, if n,j ( ) = Jn ( n,j ) where n,j = n,j /c is the j th zero of the function Jn (c ) , then c 2 ( ) d = n,j 0 c2 c2 2 2 (Jn ( n,j c)) = (Jn+1 ( n,j )) . 2 2 (9.25) When n = 0, this may be written as c 2 J0 ( n,j ) ( ) d = 0 c2 c2 2 2 (J0 ( n,j c)) = J1 ( n,j c) . 2 2 (9.26) Proof. To prove Eq. (9.24) we begin by recalling that R ( ) := Jn ( ) where = n,j , solves the di erential equation, n2 1 ( R) + 2 2 or equivalently that ( R) + 2 Therefore, 1 R=0 n2 R = 0. 0= 0 ( R) + 2 1 2 1 0 1 2 n2 R R d = ( R) + 2 2 n2 R2 d 2 2 n2 R2 d 0 1 1 1 2 = ( R) |1 + 0 2 2 = = 1 1 2 ( R) + 2 2 n2 R2 |1 0 2 2 1 2 ( R) + 2 2 n2 R2 |1 2 0 2 2 2 n2 R2 d 0 1 R2 d 0 and we have shown, 1 R2 d = 0 1 2 ( R) + 2 2 n2 R2 |1 . 0 2 2 Thus we have shown, Page: 125 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 126 1 0 9 Problems In other Coordinates Systems 2 Jn ( n,j ) d = 1 2 2 ( n,j Jn ( n,j )) + 2 2 n2 Jn ( n,j ) |1 0 2 2 n,j 1 1 2 2 2 = ( n,j Jn ( n,j )) + 2 n2 Jn ( n,j 1) = (Jn ( n,j )) . 2 2 n,j 2 To get the second equality in Eq. (9.24), set x = n,j in recursion relation, xJn (x) = nJn (x) xJn+1 (x) , to nd n,j Jn ( n,j ) = nJn ( n,j ) n,j Jn+1 ( n,j ) = n,j Jn+1 ( n,j ) , which implies that [Jn ( n,j )] = [Jn+1 ( n,j )] . More generally, if n,j ( ) = Jn n,j , then letting x = c nd c c 2 2 c or = cx we 2 ( ) d = n,j 0 0 2 Jn n,j 1 2 0 d c c2 2 c2 2 (Jn ( n,j )) == Jn+1 ( n,j ) . 2 2 =c 2 Jn ( n,j x) xdx = Notice that n,j = n,j /c is the j th zero of the function Jn (c ) . With this notation we have n,j ( ) = Jn ( n,j ) and c 2 ( ) d = n,j 0 c2 2 c2 2 (Jn ( n,j c)) = Jn+1 ( n,j ) . 2 2 Corollary 9.10 (Eigenfunction Expansions). Any reasonable function , F ( , ) , may be expressed as F ( , ) = 1 2 A0,j J0 0,j j=1 c c (9.27) + n=1 j=1 (An cos n + Bn sin n ) Jn n,j where An,j := and Bn,j := 2 2 c2 Jn+1 c 2 2 c2 Jn+1 c ( n,j ) F ( , ) Jn n,j =0 = cos n d d c (9.28) ( n,j ) F ( , ) Jn n,j =0 = sin n d d . c (9.29) Page: 126 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.4 Bessel Equation and Functions 127 Proof. The existence of an expansion in Eq. (9.27) is to be expected from our previous experience with expansions related to symmetric eigenvalue problems. (We will have to take this one on faith.) Given such an expansion exists, the usual orthogonality arguments shows that the coe cients must be given as in Eqs. (9.28) and (9.29). Corollary 9.11 (Heat and Wave Eq. Solutions). The solution to the heat equation, ut = u with u = 0 on and u (t, , ) = F ( , ) is given by u (t, , ) = 1 2 A0,j e t 0,j J0 0,j e t n,j Jn n,j 2 2 j=1 c + n=1 j=1 (An cos n + Bn sin n ) . c The solution to the wave equation, utt = u with u = 0 on and u (t, , ) = F ( , ) and ut (t, , ) = 0 is given by u (t, , ) = 1 2 A0,j cos ( 0,j t) J0 0,j cos ( n,j t) Jn n,j j=1 c + n=1 j=1 (An cos n + Bn sin n ) . c The following result will enable us to work out some explicit examples. Theorem 9.12 (Reduction formula, Problem 69.3). x sn Jn 1 (s) ds = xn Jn (x) for n N 0 x (9.30) (9.31) 2 0 x sJ0 (s) ds = xJ1 (x) 0 x 0 sn J0 (s) ds = xn J1 (x) + (n 1) xn 1 J0 (x) (n 1) sn 2 J0 (s) ds (9.32) The scaled version of Eq. (9.32) is Page: 127 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 128 9 Problems In other Coordinates Systems x tn J0 ( t) dt = 0 xn n 1 n 1 x J0 ( x) J1 ( x) + 2 (n 1) 2 2 0 x tn 2 J0 ( t) dt. (9.33) Proof. Equation (9.30) is an easy consequence of the fundamental theorem of calculus and Eq. (9.19). The proof of Eq. (9.32) is Problem 69.3, see Solution ??. Making the change of variables, s = t, we nd x x tn J0 ( t) dt = 0 0 ds sn J (s) = (n+1) n0 n x sn J0 (s) ds 0 n 1 ( x) J1 ( x) + (n 1) ( x) J0 ( x) 2 x (n 1) 0 sn 2 J0 (s) ds n 1 n 1 xn x J0 ( x) = J1 ( x) + 2 2 (n 1) n 1 x n 2 t J0 ( t) dt n+1 0 = (n+1) = xn n 1 n 1 (n 1) J1 ( x) + x J0 ( x) 2 2 2 0 x tn 2 J0 ( t) dt. 9.5 Examples Example 9.13. We want to expand 1 in terms of the functions j (x) = J0 ( j x) where j is the j th root of J0 (x) . To this end we have 1= j=1 (1, j ) j = ( j , j ) 1 12 J j=1 2 1 (1, j ) j (x) ( j ) where (f, g) := 0 f (x) g (x) xdx. Now, using Eq. (9.23) shows 1 1 (1, j ) = 0 x j (x) dx = 0 j xJ0 ( j x) dx 1 2 j j 0 = Hence we nd 1 2 j sJ0 (s) ds = 0 d 1 [sJ1 (s)] ds = J1 ( j ) . ds j Page: 128 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.5 Examples 129 1= j=1 (1, j ) j = ( j , j ) J1 ( j ) j 12 J ( j ) j=1 2 1 (x) =2 j=1 1 J0 ( j x) . j J1 ( j ) Solution to Exercise (76.1). We want to expand x in terms of the functions j (x) = J0 ( j x) where j is the j th root of J0 (x) . To this end we have (x, j ) (x, j ) j (x) x= j = 12 ( j , j ) J ( j ) j=1 j=1 2 1 where (f, g) := 0 1 f (x) g (x) xdx. Using the reduction formula in Eq. (9.33) we with x = 1 and t replaced by x, 1 xn J0 ( j x) dx = 0 1 (n 1) n 1 J1 ( j ) + 2 J0 ( j ) 2 j j j 2 0 1 xn 2 J0 ( j x) dx, with n = 2, we nd 1 1 (x, j ) = 0 x2 j (x) dx = 0 x2 J0 ( j x) dx 1 1 1 1 = J1 ( j ) + 2 J0 ( j ) 2 j j j = Therefore we nd, J0 ( j x) dx 0 1 1 J1 ( j ) 2 j j 1 J0 ( j x) dx. 0 x= j=1 (x, j ) j = ( j , j ) 1 1 j J1 ( j ) 1 2 j 1 0 J0 ( j x) dx j (x) j=1 1 12 2 J1 ( j ) J0 ( j x) j J1 ( j ) J0 ( j x) . j J1 ( j ) =2 j=1 1 j J1 ( j ) 2 j J1 J0 ( j x) dx 0 j =2 j=1 1 1 ( j ) J0 (s) ds 0 Solution to Exercise (76.3). We want to expand f (x) = 1 if 0 < x < 1 0 if 1 < x < 2. Page: 129 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 130 9 Problems In other Coordinates Systems We will begin by rst expanding f (2x) instead, and for this we have f (2x) = j=1 (f (2x) , j ) j = ( j , j ) j=1 (f (2x) , j ) j (x) 12 2 J1 ( j ) =2 j=1 (f (2x) , j ) j (x) 2 J1 ( j ) where, using Eq. (9.31), we have, 1 1/2 (f (2x) , j ) = 0 f (2x) J0 ( j x) xdx = 0 j /2 0 2 J0 (s) sds = j J0 ( j x) xdx j j J1 2 2 2 = j j 1 . = J1 2 j 2 Therefore, f (2x) = 2 j 1 2 j J1 2 2 J1 ( j ) j=1 J1 2j J0 J 2 ( j ) j=1 j 1 J0 ( j x) ( j x) . = Replacing x by x/2 shows f (x) = j=1 J1 2j J0 ( j x/2) 2 j J1 ( j ) and then letting j = j /2, shows f (x) = where J0 (2 j ) = 0 for each j. Solution to Exercise (77.2). In this problem we are asked to solve u = 0 on := {( , , z) : 0 , z 1} with u (1, , z) = 0, uz ( , , 0) = 0 and u ( , , 1) = 1. Given the symmetry of this problem, the solution we are looking for should only depend on ( , z) and we should begin by looking for solutions of the form 1 2 j=1 J1 ( j ) 2 (2 ) J0 ( j x) j J1 j Page: 130 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 9.5 Examples 131 u ( , z) = J0 ( 0,j ) Z (z) . Since 1 2 ( J0 ( 0,j )) = 0,j J0 ( 0,j ) 12 1 2 ( u) + 2 u + z u 2 = 0,j J0 ( 0,j ) Z (z) + J0 ( 0,j ) Z (z) 2 Z (z) = 0,j Z (z) . we have 0 = u = and we conclude that The solutions to this equation are given by Z (z) = A cosh 0,j z + B sinh 0,j z. Because of the boundary condition, uz ( , , 0) = 0, we must require B = 0 and we will look for a solution of the form u ( , z) = j=1 Aj J0 ( 0,j ) cosh 0,j z. To x the constants Aj we must require that 1= j=1 Aj J0 ( 0,j ) cosh 0,j . Recalling from Example 9.13 that 1=2 j=1 1 J0 ( j ) j J1 ( j ) 1 j J1 ( j ) we must require that Aj cosh 0,j = 2 and hence our solution is given by u ( , z) = 2 j=1 J0 ( 0,j ) cosh 0,j z . j J1 ( j ) cosh 0,j Solution to Exercise (77.7). Here we are to solve, ut = u bu = ( b) u on := {(t, , ) : 0 1 and t > 0} with u (t, 1, ) = 0 and u (0, , ) = 1. Page: 131 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 132 9 Problems In other Coordinates Systems In this case we have u (t, ) = et( b) 1 = e tb et 1 1 = e tb et 2 J0 ( j ) j J1 ( j ) j=1 = 2e tb j=1 J0 ( j ) t 2 e 0,j . j J1 ( j ) Page: 132 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 A Some Complex Variables Facts Here we suppose w (t) = c (t) + id (t) where c (t) and d (t) are two real valued functions of t. An important example of such a complex valued function is found in the next de nition. De nition A.1 (Euler s Formula). For t R let eit := cos t + i sin t and for z = x + iy let ez := ex eiy = ex (cos y + i sin y) . (A.2) (A.1) Notice that any complex number, z = x + iy, may be written as z = rei where (r, ) are the polar coordinates of the point (x, y) R2 . De nition A.2. If c (t) and d (t) are di erentiable, then we de ne w (t) := c (t) + id (t) and w (t) dt := c (t) dt + i d (t) dt Example A.3. If w (t) = et + i sin t, then w (t) = et i cos t and /2 /2 w (t) dt = 0 0 et + i sin t dt = e 2 1 + i. 1 Example A.4. Suppose w (t) = eit , then d it d e= (cos t + i sin t) = sin t + i cos t dt dt = i (cos t + i sin t) = ieit 134 A Some Complex Variables Facts and b b b b eit dt = a a (cos t + i sin t) dt = a cos tdt + i a sin tdt = (sin t i cos t) |b = a eit b |. ia Example A.6 below, generalizes this result. Theorem A.5 (These de nitions work just as in real variables). If z (t) = a (t) + ib (t) and w (t) = c (t) + id(t) and = u + iv C then 1. 2. 3. 4. 5. d dt (w (t) + z (t)) = w (t) + z (t) d dt [w (t) z (t)] = w z + wz [w (t) + z (t)] dt = w (t) dt + z (t) dt w(t)dt = w( ) w( ) In particular if w = 0 then w is constant. w(t)z(t)dt = w(t)z(t)dt + w (t) z (t) | . 6. w (t) dt |w (t)| dt. Proof. 1. and 4. are easy. 2. d d d [wz] = (ac bd) + i (bc + ad) dt dt dt = (ac bd) + i(bc + ad) + i(bc + ad) + (ac bd) = wz + wz. 3. The only interesting thing to check is that z (t) dt = z (t) dt. Again we simply write out the real and imaginary parts: z (t) dt = (u + iv) (a (t) + ib (t)) dt (ua(t) vb(t) + i [ub(t) + va(t)]) dt = = (ua(t) vb(t)) dt + i [ub(t) + va(t)] dt Page: 134 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 A Some Complex Variables Facts 135 while z (t) dt = (u + iv) [a (t) + ib (t)] dt = (u + iv) a (t) dt + i b (t) dt [ub(t) + va(t)] dt. = (ua(t) vb(t)) dt + i Shorter Alternative: Just check it for = i, this is the only new thing over the real variable theory. 5. w (t) z (t) | = d [w (t) z (t)] dt = dt w(t)z(t)dt + w(t)z(t)dt. 6. (Skip this one!) Let 0 and R be chosen so that w (t) dt = ei , then w (t) dt = = e i w (t) dt = e i w (t) dt Re e i w (t) dt = Re e i w (t) dt e i w (t) dt = |w (t)| dt. Example A.6. Suppose z = x + iy, then ezt := ext eiyt := ext cos yt + iext sin yt and so, again by de nition, d zt d xt e= e cos yt + iext sin yt dt dt = ext (x cos yt y sin yt) + iext (x sin yt + y cos yt) = (x + iy) ext cos yt + iext sin yt = zezt . A better proof. By the product rule and Example A.4, d tx ity d tz e= ee = xetx eity + etx iyeity = zetx eity = zetz . dt dt Page: 135 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 136 A Some Complex Variables Facts Using this fact and item 4. of Theorem A.5 we may conclude, b ezt dt = a ezt b |. za If we write out what this means by comparing the real and imaginary parts of both sides we nd b b b ezt dt = a a ext cos (yt) dt + i a ext sin (yt) dt while ezt ext cos yt + iext sin yt x iy = z x + iy x iy ext =2 [x cos yt + y sin yt + i (x sin yt y cos yt)] x + y2 from which we may conclude that b ext cos (yt) dt = a b ext [x cos yt + y sin yt] x2 + y 2 b and a b ext sin (yt) dt = a ext [x sin yt y cos yt] . x2 + y 2 a Theorem A.7 (Addition formula for ez ). The function ez de ned by Eq. (A.2) satis es Proposition A.8. 1. e z = 2. ew+z = ew ez . 1 ez and Proof. By the previous example we know d tz e = zetz with e0z = e0+i0 = 1. dt Similarly, using the chain rule or by direct computation, one shows d tz e = zetz with e0z = e0+i0 = 1. dt 1. By the product rule, d tz tz e e = ze tz etz + e tz zetz = 0 dt and therefore e tz etz is independent of t and hence e tz etz = e 0z e0z = 1. Taking t = 1 proves 1. 2. Again by the product rule shows Page: 136 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 A Some Complex Variables Facts 137 d t(w+z) tw tz (w + z) e t(w+z) etw etz e = ee t(w+z) +e wetw etz + e t(w+z) etw zetz dt =0 and so e t(w+z) etw etz = e t(w+z) etw etz |t=0 = 1. Taking t = 1 then shows e (w+z) ew ez = 1 and then using Item 1. we get item 2. Corollary A.9 (Addition formulas cos and sin). For , R we have cos ( + ) = cos cos sin sin sin ( + ) = cos sin + cos sin . Proof. These follow by comparing the real and imaginary parts of the identity ei ei = ei( + ) = cos ( + ) + i sin ( + ) while ei ei = (cos + i sin ) (cos + i sin ) = cos cos sin sin + i (cos sin + cos sin ) . Page: 137 job: 110notes macro: svmono.cls date/time: 25-May-2004/11:28 B Assigned Homework Problems: 1. Exercises from Lecture notes: 1.1, 1.2, 1.3, 2.1 Exercises from the Book: 6.1, 6.2, 7.1, 10.1, 10.4, 10.5 2. Exercises from Lecture notes: 3.1, 3.2, 3.3, 3.4 Exercises from the Book: 6.4, 6.5, 8.1, 10.6, 10.7 6.4-6.5, 8.1, 8.4, 10.6, 10.7 3. Exercises from Lecture notes: 3.5, Exercises from the Book: 13.2, 13.4, 13.6, 13.7, 13.9 4. Exercises from the Book: 15.1, 15.2, 15.4a, 17.3, 17.4, 17.6, 22.2 5. Exercises from the Book: 22.3, 22.4, 24.3, 24.4, 24.5, 24.6, 28.1, 28.2 6. Exercises from the Book: 28.4, 28.5 32.1, 32.2, 32.3, 32.5, 32.6a, (32.14 optional), 33.2, 33.5 7. Exercises from the Book: 33.15, 40.1, 40.2, 40.3, 43.1, 43.4, 53.1 8. Exercises from Lecture notes: 8.2. From book: 40.6, 40.7, 41.1, 41.2, 41.8, 43.2, 43.5, 45.7a (Read b only) 9. Exercises from the Book: 45.2, 45.5, 48.1, 53.7, 69.1, 69.3, 76.4

Find millions of documents here - Study Guides, Homework Solutions, Papers, Exam Answer Keys and more. Course Hero has millions of course related materials that will enable you to learn better, faster and get an A in all your courses.
Below is a small sample set of documents:

review.pdf
Path: UCSD >> MATH >> 110 Winter, 2008

110notes2p.pdf
Path: UCSD >> MATH >> 110 Winter, 2008
IMM060127.pdf
Path: UCSD >> MATH >> 111a Fall, 2008
Study_Guide_MidTerm1.pdf
Path: UCSD >> MATH >> 120a Fall, 2008
130hw.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw1soln.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw2soln.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw5soln.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw5.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
page1.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw3soln.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw6soln.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw0soln.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l24w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l22w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l27w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l1w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l21w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l10w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l2w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l7w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l11w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l18w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l19w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l26w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
midw03ps.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l8w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l23w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
l9w03.pdf
Path: UCSD >> MATH >> 130a Fall, 2008
hw7soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw6.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw2soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw1soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
midterm.1.math130b.spring.2006.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
homework.8.math130b.spring.2006.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
homework.5.math130b.spring.2006.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
mids03ps.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
l32w03.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
l31w03.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
problems11.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
l34w03.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw5soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
digraphs.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
paper.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
disc_matrix.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
pos_norm.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
97_04_euclidean_ramsey.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw5.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
10Aevals.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
jumping.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
m1soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
generalizing.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
anti-covers.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
cmj105-106.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
page2.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
140hypercubecycles.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw4soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
P-B-notes.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
tangent_lines.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw4.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
midterm2.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hats.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
IPCO05_Spielman_LectureNotes.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
page1.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
review.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw3soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw6soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
thesis.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
hw8soln.pdf
Path: UCSD >> MATH >> 130b Winter, 2008
FoxLinThibault.pdf
Path: UCSD >> MATH >> 130b Winter, 2008