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### ODE-Linear

Course: MATH 260, Spring 2012
School: UPenn
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Word Count: 1930

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ODEs Second Linear order linear equations Many traditional problems involving ordinary equations arise as second order linear equations au + bu + cu = f , more briey as Lu = f . The problem is, given f , nd u ; often we will want to nd u that satises some auxiliary initial or boundary conditions. Here we have used the notation Lu := a(x)u + b(x)u + c(x)u, (1) so L takes a function u and gives a new function Lu...

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ODEs Second Linear order linear equations Many traditional problems involving ordinary equations arise as second order linear equations au + bu + cu = f , more briey as Lu = f . The problem is, given f , nd u ; often we will want to nd u that satises some auxiliary initial or boundary conditions. Here we have used the notation Lu := a(x)u + b(x)u + c(x)u, (1) so L takes a function u and gives a new function Lu . This operator L is a linear map because it has the two properties L(u) = Lu and L(u + v) = Lu + Lv, where is any constant and u and v are functions. One consequence is that if Lu = 0 and Lv = 0, then L(Au + Bv) = 0 for any constants A and B . The solutiond of Lu = 0 are often called the nullspace or kernel of L . These properties show that the nullspace of a linear map is a linear space. For instance, in the special case where Lu = u + u we know that L cos x = 0 and L sin x = 0. Thus L(A cos x + B sin x) = 0 for any constants A and B . E XAMPLE Well show that any solution of Lu := u + u = 0 has the form u(x) = A cos x + B sin x. This will show that the nullspace of Lu := u + u = 0 has dimension two. First we must pick the constants A and B . Letting x = 0 we see that (if this is to work) A = u(0) . Similarly, taking the derivative we get B = u (0) . Let v(x) := u(0) cos x + u (0) sin x . Our task is to show that u(x) = v(x) . Equivalently, if we let w(x) := u(x) v(x) , we must show that w(x) 0. A key observation motivating us is that by linearity, w + w = 0, and w(0) = 0, w (0) = 0. Introduce the function 1 E (x) = [w 2 + w2 ]. 2 Then since w = w , E (x) = w w + ww = w (w) + ww = 0, so E (x) = constant (in many physical examples, this is conservation of energy). But from w(0) = 0 and w (0) = 0 we nd E (0) = 0. Since E (x) is a sum of squares, the only possibility is w(x) 0, as claimed. This Example generalizes. Assuming a(x) = 0, the nullspace of (1) always has dimension 2. Let (x) and (x) be the solutions of the homogeneous equation Lu = 0 with (0) = 1, (0) = 0, and (0) = 0, (0) = 1, then every solution of the homogeneous equation Lu = 0 has the form u(x) = A(x) + B(x) for some constants A and B . The proof, which we dont give, has two parts. The rst is the existence of the solutions and , the second is their uniqueness. While these proofs are not obvious, they are not killers. 1 For (1) and other linear ordinary and partial differential equations, it is surprising that if one knows the general solution of the homogeneous equation Lu = 0 one can nd an explicit formula for a particular solution of the inhomogeneous equation Lu = f . Based on our experience with the rst order linear inhomogeneous equation u + au = f it is plausable to seek u in the form u(x) = p(x)v(x) where p(x) is chosen cleverly to make the equation for v simple to solve. We do this for Lu := u + u = f (the general case of (1) is then routine). Clearly u = pv + p v and u = pv + 2 p v + p v so Lu = pv + 2 p v + ( p + p)v. This clearly simplies if we pick p as a solution of the homogeneous equation p + p = 0. But we know two solutions of this, cos x and sin x . Which should we use? After some experimentation, Lagrange decided he should use both and instead sought u in the more general form u = pv + qw, (2) where for our example p(x) = cos x and q(x) = sin x . Now he had one equation, f = Lu = pv + 2 p v + qw + 2q w for the two unknowns, v(x) and w(x) so he could impose another condition. After some experimenting he imposed the condition pv + qw = 0, (3) which resulted in the two linear equations: f = Lu = p v + q w 0 = pv + qw , and that is, f = ( sin x)v + (cos x)w 0 = (cos x)v + (sin x)w . and He solved these for v and w : v (x) = sin x f (x) and w (x) = cos x f (x). Thus integrating and using (2), we obtain the simple formula for a particular solution, upart , of the inhomogeneous equation Lu = f : Zx upart (x) = cos x sin s f (s) ds + sin x Zx 0 Zx cos s f (s) ds = 0 sin(x s) f (s) ds (4) 0 To get the general solution of the inhomogeneous equation we simply add the general solution of the homogeneous equation: Zx u(x) = A cos x + B sin x + sin(x s) f (s) ds. 0 In honor of George Green we often write (4) in the symbolic form Zx u(x) = G(x, s) f (s) ds 0 2 (5) and call G(x, s) := sin(x s) Greens function for the equation Lu = f . The point is that (5) can be thought of as writing u = L1 f so we have a conceptually satisfying formula for the inverse operator L1 . Lagranges procedure for nding the formula (4) for a particular solution of the inhomogeneous equation is called variation of parameters. The key step is to seek u in the form (2) with Lp = 0 and Lq = 0. We now carry out the details for the general case of the general second order equation Lu := u + b(x)u + c(x)u = f (x). (6) Note that here the coefcient of u is 1 (if not, then divide by it). Using equation (2) and the condition (3) he found that u = p v + q w, and u = p v + q w + pv + qw. Substitute these into the equation (6) for L . After a computation short that uses Lp = Lq = 0, we get the simple formula Lu = p v + q w . (7) To solve Lu = f we thus need to nd v and w that satisfy this and (3): pv + qw = 0 p v +q w = f. These are two linear equations for v and w . Their solution is v= q f W and w = pf , W where W (x) := pq p q (called the Wronskian of p and q ). Integrating we nd v and w and thus from (2), a particular solution upart upart = p(x) Zx q(s) f (s) 0 W (s) ds + q(x) where G(x, s) := Zx p(s) f (s) 0 W (s) Zx ds = G(x, s) f (s) ds, (8) 0 q(x) p(s) p(x)q(s) W (s) is Greens function for the problem. In the special case of u + u = f done earlier, p(x) = cos x and q(x) = sin x so W (x) = 1 and g(x, s) = sin(x s) , just as in (4). First order linear systems Next consider the rst order system of equations LU := U (x) + A(x)U (x) = F (x), 3 (9) where U and F are vectors with n components and A(x) is an n n matrix. We assume that both A and F depend continuously on x . A typical problem is to seek a solution of (9) that satises some initial condition U (0) = C , where C Rn is a given vector. If A(x) and F (x) are both periodic with period P , another typical problem is to seek a periodic solution U (x) [the simplest scalar example u = 1 has no periodic solutions with any period and shows that answering this question may involve some work]. The homogeneous equation A general theorem, which well not prove (it is not a killer) is Theorem 1 . Given any constant C Rn , the homogeneous equation LU = 0 has a unique solution satisfying U (0) = C . Note that has a unique solution means the same as has one and only one solution. Let e1 := (1, 0, . . . , 0) , e2 := (0, 1, 0, . . . , 0) , . . . , en := (0, 0, . . . , 0, 1) be the standard basis vectors in Rn . It is useful to use the special solutions 1 (x) ,. . . , n (x) that satisfy the homogeneous equation L j (x) = 0 with j (0) = e j and use them to construct the n n matrix (x) whose columns are the vectors 1 (x) ,. . . , n (x) . Then satises (x) + A(x)(x) = 0, and the initial condition (0) = I . This matrix (x) is sometimes called the fundamental solution matrix. E XAMPLE 1 Let U (x) = u1 (x) , u2 (x) A := 0 1 , 1 0 and F (x) = f1 (x) , f2 (x) so the system of equations LU := U + AU = F is u1 u2 = f1 (10) u2 + u1 = f2 . The vectors 1 (x) := cos x , sin x 2 (x) := sin x cos x both satisfy the homogeneous equation L j = 0 with initial conditions 1 (0) = e1 , 2 (0) = e2 , so the fundamental solution matrix is (x) := cos x sin x sin x cos x (11) The inhomogeneous equation Next we show that if you know a fundamental matrix solution (x) for the homogeneous equation, then you can nd a formula for a particular solution of the inhomogeneous equation LU = F , that 4 is, U + AU = F . As before, seek U in the special form U (x) = S(x)V (x) , where S(x) is an n n matrix. The goal is to choose a clever S so the resulting differential equation for V (x) is simple. Clearly LU = SV + (S + AS)V. This evidently simplies dramatically if S + AS = 0, so we let S(x) = (x) be the fundamental matrix solution of the homogeneous equation L = 0. Because (0) = I , we know that S(x) is invertible, at lease for x near 0 (In fact, it is invertible for all x . We leave that for you). The equation LU = F is thus SV = F so V (x) = S1 (x)F (x) . Integrating this we can obtain the desired particular solution, Upart of LU = F . Since we just want a particular solution, we can let Upart (0) = 0, which implies V (0) = 0. Thus the desired formula is: Zx Upart (x) =S(x)V (x) = S(x) V (0) + Zx = S(x)S1 (s)F (s) ds = 0 S1 (s)F (s) ds (12) G(x, t )F (s) ds, (13) 0 Zx 0 where G(x, t ) := S(x)S1 (s) is Greens function for this problem. E XAMPLE 1 ( CONTINUED ) We are now in a position to write a formula for a particular solution of LU = F for Example 1 above. Then (11) is the fundamental matrix solution for the homogeneous equation, S(x) = (x) . Since this happens to be an orthogonal matrix, its inverse is just the transpose. Consequently G(x, s) = (x)1 (s) = = cos x sin x sin x cos x cos s sin s sin s cos s cos(x s) sin(x s) . sin(x s) cos(x s) (14) (15) Consequently Zx Upart (x) = 0 cos(x s) sin(x s) sin(x s) cos(x s) f1 (s) d s, f2 (s) (16) E XAMPLE 2 We can write any second order equation u + bu + cu = f as a rst order system by letting u1 (x) = u(x) and u2 (x) = u (x) . Then, using the differential equation, u1 = u2 and u2 = u = bu2 cu1 + f . that is, u1 (x) u2 (x) + 0 1 c b u1 (x) = u2 (x) 0 f (x) In the special case of u + u = f we have b = 0 and c = 1 so the previous equation becomes u1 (x) u2 (x) + 0 1 1 0 u1 (x) = u2 (x) 5 0 , f (x) which is exactly (10) with f1 = 0 and f2 = f . Now (16) gives a formula for a particular solution of this inhomogeneous equation. It is Zx Upart (x) = 0 Zx = 0 cos(x s) sin(x s) sin(x s) cos(x s) 0 ds f (s) sin(x s) f (s) ds cos(x s) Since u1 (x) = u(x) , this formula is exactly the same as (4) found earlier. 6
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UPenn - MATH - 425
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UPenn - MATH - 425
UPenn - MATH - 425
UPenn - MATH - 425
First-order ordinary differential equations Before we get involved trying to understand partial differential equations (PDEs), we'll review the highlights of the theory of ordinary differential equations (ODEs). We'll do this in such a way that we can beg
UPenn - MATH - 425
Linear ODEsSecond order linear equationsMany traditional problems involving ordinary equations arise as second order linear equationsau + bu + cu = f ,more briey asLu = f .The problem is, given f , nd u ; often we will want to nd u that satises some
UPenn - MATH - 425
Math 425 Midterm 1Dr. DeTurck February 8, 20071. Solve ux + yuy + u = 0, u(0, y ) = y . In what domain in the plane is your solution valid? As usual, we construct the graph of the solution by propagating the initial data o the line in the xy -plane wher
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UPenn - MATH - 425
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Exam 1Math 508October 12, 2006Jerry L. Kazdan12:00 1:20Directions This exam has three parts, Part A has 4 problems asking for Examples (20 points, 5points each), Part B asks you to describe some sets (20 points), Part C has 4 traditional problems(6
UPenn - MATH - 508
SignaturePrinted NameExam 2Math 508December 8, 2006Jerry L. Kazdan12:00 1:20Directions This exam has two parts, Part A has 3 shorter problems (8 points each, so 24 points),Part B has 5 traditional problems (15 points each, so 75 points).Closed bo
UPenn - MATH - 508
Exam 2Math 508December 8, 2006Jerry L. Kazdan12:00 1:20Directions This exam has two parts, Part A has 3 shorter problems (8 points each, so 24 points),Part B has 5 traditional problems (15 points each, so 75 points).Closed book, no calculators but
UPenn - MATH - 508
Exam 1Math 508October 16, 2008Jerry L. Kazdan10:30 11:50Directions This exam has three parts, Part A has 4 problems asking for Examples (20 points, 5points each), Part B asks you to describe some sets (20 points), Part C has 4 traditional problems(