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Harmonic-lab2 - H HARMONIC FUNCTIONS This chapter is...

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Unformatted text preview: H HARMONIC FUNCTIONS This chapter is devoted to the Laplace equation. We introduceOtwo of its important properties, the maximum pr1n01p1e and the rotatlonal invarianceCi Then we solve the equation in series form 1n rectangles, c1rcles, and relate shapes. The case of a circle leads to the beautlful P01sson formula. 6. 1 LAPLACE’S EQUATION If a diffusion or wave process is stationary (independent of time), then u, E 0 and ut, E 0. Therefore, both the diffusion and the wave equatlons reduce to the Laplace equation: uxx = 0 in one dimension V - Vu = Au = um + uyy = 0 in two dimensions V - Vu = Au = uxx + uyy + uzz = 0 in three dimensions ' ' ' ' tion. A solution of the Laplace equation is called a harmonic func _ . In one dimension, we have simply uxx = 0, so the only harmonic functions in one dimension are u(x) = A + Bx. But this is so s1mple that it hardly gives us a clue to what happens in higher dimensions. . The inhomogeneous version of Laplace’s equatlon M = f (I) with f a given function, is called Poisson’s equation. . f Besides stationary diffusions and waves, some other instances 0 I .nnlnm’g and Poisson’s equations include the following. 6.1 LAPLACE'S EQUATION 153 Electrostatics. From Maxwell’s equations, one has curl E = 0 and div E 2 471,0, where ,o is the charge density. The first equation implies E = — grad d) for a scalar function (15 (called the electric potential). Therefore, A¢ = div(grad (/5) = —div E = —4np, which is Poisson’s equation (with f = —47tp). Steady fluid flow. Assume that the flow is irrotational (no eddies) so that curl v = 0, where v = v(x, y, z) is the velocity at the position (x, y, z), assumed independent of time. Assume that the fluid is incompressible (e.g., water) and that there are no sources or sinks. Then div v = 0. Hence v = — grad 45 for some q) (called the velocity potential) and Arp = —divv = 0, which is Laplace’s equation. Analytic functions of a complex variable. Write z = x + iy and f(z) = u(Z) + iv(z) = We + iy) + iv(x + iy), where u and v are real—valued functions. An analytic function is one that is expressible. as a power series in z. This means that the powers are not xmy” but 2” = (x + iy)". Thus f(z) = Zanz" n=0 (an ’complex constants). That is, 00 u<x + iy) + W + iy) = 2an + iy)". n=0 Formal differentiation of this series shows that 8u 3v Bu av —— = — and — 2 ~— Bx 8y 8y 8x (see Exercise 1). These are the Cauchy—Riemann equations. If we differ- entiate them, we find that ”xx = vyx = vxy = _uyyy so that Au 2 0. Similarly Av = 0, where A is the two-dimensional laplacian. Thus the real and imaginary parts of an analytic function are harmonic. Brownian motion. Imagine brownian motion in a containerD. This means that particles inside D move completely randomly until they hit the bound- ary, when they stop. Divide the boundary arbitrarily into two pieces, C1 and C2 (see Figure 1). Let u(x, y, z) be the probability that a particle that begins at the point (x, y, z) stops at some point of C1. Then it can be deduced that ‘ Au = 0 in D u=10nC1 u=00nC2. Thus u is the solution of a Dirichlet problem. 154 CHAPTER 6 HARMONIC FUNCTIONS Figure 1 As we discussed in Section 1.4, the basic mathematical problem is. to solve Laplace’s or Poisson’s equation in a given domain D w1th a conditlon on bdy D: In one dimension the only connected domain is an interval {a 5 x 5 b}. We will see that what is interesting about the two— and three-dimens1onal cases is the geometry. MAXIMUM PRINCIPLE We begin our analysis with the maximum principle, which is easier for Laplace’s equation than for the diffusion equation. By an open set we mean a set that includes none ofiits boundary points (see Section A.1). Maximum Principle. Let D be a connected bounded open set (in ei- ther two— or three—dimensional space). Let eitiEr u(x, y) or u(x, y, z) be a harmonic function in D that is continuous on D = D U (bdy D). Then the maximum and the minimum values of u are attained 0n bdy D and nowhere inside (unless u E constant). In other words, a harmonic function is its biggest somewhere on the boundary and its smallest somewhere else on the boundary. To understand the maximum principle, let us use the vector shorthand x = (x, y) in two dimensions or x = (x, y, z) in three dimensions. Also, 1th . . . 1 2 radial coordinate IS written as |x| = (x2 + yz) / or |x| = (x2 + y2 + z2) . The maximum principle asserts that there exist pomts XM and xm on bdy D such that u(xm) : u(X) s u(xM) (2) 6.1 LAPLACE’S EQUATION 155 Figure 2 for all x e D (see Figure 2). Also, there are no points inside D with this property (unless u E constant). There could be several such points on the boundary. The idea of the maximum principle is as follows, in two dimen- sions, say. At a maximum point inside D, if there were one, we’d have uxx 5 0 and an S 0. (This is the second derivative test of calculus.) So uxx + uyy E 0. At most maximum points, uxx < 0 and uyy < 0. So we’d get a contradiction to Laplace’s equation. However, since it is possible that uxx = 0 = uyy at a maximum point, we have to work a little harder to get a proof. " Here we go. Let 6 > 0. Let v(x) = u(x) + elxlz. Then, still in two dimen- sions, say, Av=Au+eA(x2+y2)=0+4e>0 inD. But Av = Um + vyy 5 0 at an interior maximum point, by the second deriva— tive test in calculus! Therefore, v(x) has no interior maximum in D. Now v(x), being a continuous function, has to have a maximum some- where in the closure D = D U bdy D. Say that the maximum of v(x) is attained at x0 6 bdy D. Then, for all X e D, ”(X) S v(X) S v(Xo) = u(Xo) + elXol2 5 gay + el2, y where l is the greatest distance from bdy D to the origin. Since this is true for any 6 > 0, we have < f . u(x) _ trriryaxu or all x E D (3) Now this m_aximum is attained at some point XM e bdy D. So u(x) S u(XM) for all x e D, which is the desired conclusion. The existence of a minimum point xm is similarly demonstrated. (The absence of such points inside D will be proved by a different method in Section 6.3.) UNIQUENESS OF THE DIRICHLET PROBLEM To prove the uniqueness, suppose that Au=f inD Av=f inD u=h onbdyD u=h onbdyD. 156 CHAPTER 6 HARMONIC FUNCTIONS We want to show that u E v in D. So we simply subtract equations and let w = a — 1). Then Aw = 0 in D and w = 0 on bdy D. By the maximum principle 0 = w(xm) S w(x) 5 w(XM) = 0 for all x E D. Therefore, both the maximum and minimum of w(x) are zero. This means that W E 0 and u E v. INVARIANCE IN TWO DIMENSIONS The Laplace equation is invariant under all rigid motions. A rigid motion in the plane consists of translations and rotations. A translation in the plane is a transformation x’=x+a y’=y+b. Invariance under translations means simply that an + an = ax/x/ + uy/yu A rotation in the plane through the angle or is given by x’ =xcosa +ysina (4) y’ = —x sina +ycosoi. By the chain rule we calculate ax = ax, COSOl — uy/ sina My = my sina + uy/ COSOl uxx = (ux/ cosoz — My! sin oz)x« COSOl — (ax, cosa — uyz sin 00y: SlIlOl uyy = (ax, sina + uy/ cos 00x, smog +(uxzs1noz + uy/ cos 00y, cosa. Adding, we have “xx + “w = (ux’x' + uy/y«)(cos2a + $11120!) + “x’y’ ' (0) = “x’x’ + Myry/, This proves the invariance of the Laplace operator. In engineering the laplacian A is a model for isotropic physical situations, in which there is no preferred direction. The rotational invariance suggests that the two—dimensional laplacian should take a particularly simple form in polar coordinates. The transforma- tion x=rcos6 y=rsin6 6.1 LAPLACE'S EQUATION 157 has the jacobian matrix 8x 8y 9: 8r 8r = cos6 sin6 812): —rsin6 rcos6 86 86 with the inverse matrix 8r 86 —sin6 — — cos 6 $4 = 8x 8x _ r 8r 86 _ , 9 cos 6 ~ 8y 8y s1n r (Beware, however, that 8r / 8x 75 (8x / 8r)'1.) So by the chain rule we have 8 _ cos 6 8 sin6 8 8x _ 8r r 86’ 8 . 8 cos6 8 — = sm 6— + —. 8y 8r r 86 These operators are squared to give a2 a sin6 a 2 — = cos 6— — —— 8162 8r r 86 82 sin6 cos6 82 2 = 9— _ 2 _ COS w < r ) 8r86 sin26 32 +25in6cos6 a +sin26 a r2 862 r2 86 r 8r 32 ( a cosea)2 = s1n6—+ — + 8—322 8r r 86 82 ' 2 =sin26—+2 sm6cos6 8 8r2 r 8r86 +cos26 82 25in6cos6 8 +00326 8 r2 862 r2 86 r 8r' (The last two terms come from differentiation of the coefficients.) Adding these operators, we get (10 and behold!) A_a_2 az_az+1a+132 75) 2—3.362 8y2—8r2 r8r r2862. ( It is also natural to look for special harmonic fimctions that themselves nrp rntntirmnllv immrinnt Tn furn flimnnoinno Hm'n mnnnn Mane “m "N. «AL... 158 CHAPTER 6 HARMONIC FUNCTIONS coordinates (r, 6) and look for solutions depending only on r. Thus by (5) 1 0=uxx+uyy=urr+;ur if it does not depend on 9. This ordinary differential equation is easy to solve: (rur), = 0, ru, = (:1, u = cllogr + oz. The function log r will play a central role later. INVARIANCE IN THREE DIMENSIONS The three—dimensional laplacian is invariant under all rigid motions in space. To demonstrate its rotational invariance we repeat the preceding proof using vector-matrix notation. Any rotation in three dimensions is given by x’ = Bx, where B is an orthogonal matrix (’33 = B’B = I). The laplacian is Au = 219:1 14,-,- = 22].:1 8,-J-uij where the subscripts on u denote partial derivatives. Therefore, Au = Z (Z bkiaijblj) 14qu = 25k] uk’l’ Lj kJ k,l = E uk/k/ k because the new coefficient matrix is Zbkisijblj =Zbkibli = (B’B)k, = 31d- tlj i So in the primed coordinates Au takes the usual form Au 2 ux/x/ + uy/y/ + “111/, For the three-dimensional laplacian 32 32 32 A = —— —— — 3 8x2 + 8322 + 8Z2 it is natural to use spherical coordinates (r, 0, ¢) (see Figure 3). We’ll use the notation ”mam—22‘ um x=scos¢ z=rcos0 y=ssin¢ s=rsin6. (Watch out: In some calculus books the letters (1) and 6 are switched.) The calculation, which is a little tricky, is organized as follows. The chain of 6.1 LAPLACE’S EQUATION 159 Figure 3 variables is_(x, y, z) —> (s, ¢, z) —> (r, 9, (15).Bythetwo-dimensionalLaplace calculation, we have both 1 1 I"zz + “ss = urr + ‘ur + —2u00 r r and 1 1 uxx + an = a“ + Eu; + S—2u¢¢. We add these two equations, and cancel ass, to get A3 = uxx + uyy + an 1 1 1 1 = urr + ;ur + r—Zuoe + EMS + gums. In the last term we substitute s2 = rzsinze and in the next-to-last term _ 3“ ar 36 29¢ us—g=ur5;+u05;+u¢g s cos0 =urg—+u9~ +u¢-0. r r This leaves us with 1 1 [Meg + (COt (9)149 + . “(1505], (6) 1n20 82 2 8 l 3 sin 6 8 + 1 32 _ ar2 r ar rzsin e 30 as rzsinze a¢2' Finally, let’s look for the special harmonic functions in three dimensions which don’t change under rotations, that is, which depend only on r. By (7) i (7) 160 CHAPTER 6 HARMONIC FUNCTIONS they satisfy the ODE 2 0 = A314 = u,, + —ur. r 2 860214,), = 0. It has the solutionsr u. = c1. That is, u = ~—clr_1 + C2.This important harmonic function 1 ; = (x2 + y2 +z2)_1/2 is the analog of the special two—dimensional function log(x2 + y2)1/ 2 found before. Strictly speaking, neither function is finite at the origin. In electrostat- ics the function u(x) = r‘1 turns out to be the electrostatic potential when a unit charge is placed at the origin. For further discussion, see Section 12.2. EXERCISES 1. Show that a function which is a power series in the complex variable x + iy must satisfy the Cauchy—Riemann equations and therefore Laplace’s equation. 2. Find the solutions that depend only on r of the equation um + uyy + uzz = kzu, where k is a positive constant. (Hint: Substitute u = v / r.) 3. Find the solutions that depend only on r of the equation uxx + uyy = kzu, where k is a positive constant. (Hint: Look up Bessel’s differential equation in [MF] or in Section 10.5.) 4. Solve um + uyy + uzz = Oin the spherical shellO < a < r < b with the boundary conditions u = A on r = a and u = B on r = b, where A and B are constants. (Hint: Look for a solution depending only on r.) 5. Solve uxx + uyy = 1 in r < a with u(x, y) vanishing on r = a. 6. Solve uxx + uyy = 1 in the annulus a < r < b with u(x, y) vanishing on both parts of the boundary r = a and r = b. 7. Solve um + uyy + uzz = 1 in the spherical shell a < r < b with u(x, y, z) vanishing on both the inner and outer boundaries. 8. Solve uxx + uyy + uzz = 1 in the spherical shell a < r < b with u = 0 on r = a and Bu/Br = 0 on r = b. Then let a —> 0 in your answer and interpret the result. 9. A spherical shell with inner radius 1 and outer radius 2 has a steady-state temperature distribution. Its inner boundary is held at 100°C. Its outer boundary satisfies flu/8r = —y < 0, where y is a constant. (a) Find the temperature. (Him: The temperature depends only on the radius.) (b) What are the hottest and coldest temperatures? (c) Can you choose 7/ so that the temperature on its outer boundary is 20°C? 6.2 RECTANGLES AND CUBES 161 10. Prove the uniqueness of the Dirichlet problem Au 2 f in D, u = g on bdy D by the energy method. That is, after subtracting two solutions w = u # v, multiply the Laplace equation for w by w itself and use the divergence theorem. 11. Show that there is no solution of 3 Au=f inD, a—u=g onbdyD n in three dimensions, unless fl fdxdydz: f/gdS. bdy(D) (Hint: Integrate the equation.) Also show the analogue in one and two dimensions. 12. Check the‘validity of the maximum principle for the harmonic func- tion(1— X2 — y2)/(1— 2x +x2 + y2) in the disk D = {x2 + y2 :1}. Explain. 13. A function u(x) is subharmonic in D if Au 3 0 in D. Prove that its maximum value is attained on bdy D. [Note that this is not true for the minimum value.] 6.2 RECTANGLES AND CUBES Special geometries can be solved by separating the variables. The general procedure is the same as in Chapter 4. (i) Look for separated solutions of the PDE. (ii) Put in the homogeneous boundary conditions to get the eigenvalues. This is the step that requires the special geometry. (iii) Sum the series. (iv) Put in the inhomogeneous initial or boundary conditions. It is important to do it in this order: homogeneous BC first, inhomogeneous BC last. We begin with Azu = uxx+uyy = 0 in D, ‘ . (1) whereD is the rectangle {0 < x < a, 0 < y < b} on each of whose sides one of the standard boundary conditions is prescribed (inhomogeneous Dirichlet, Neumann, or Robin). _7_ GREEN’S IDENTITIES AND GREEN’S FUNCTIONS The Green’s identities for the laplacian lead directly to the maximum principle and to Dirichlet’s principle about minimizing the energy. The Green’s function is a kind of universal solution for harmonic functions in a domain. All other harmonic functions can be expressed in terms of it. Combined with the method of reflection, the Green’s function leads in a very direct way to the solution of boundary problems in special geometries. George Green was interested in the new phenomena of electricity and magnetism in the early 19th century. 7.1 GREEN’S FIRST IDENTITY NOTATION In this chapter the divergence theorem and vector notation will be used ex— tensively. Recall the notation (in three dimensions) grad f = V f = the vector (fx, fy, fz) 3F divF=V.F=E+a—Fz—+—3, 8x 8y Bz where F = (F1, F2, F3) is a vector field. Also, Au =divgradu =V-Vu =um+uyy+uzz 2 _ 2 _ 2 |Vu| — lgrad ul — ux + u; + (43. Watch out which way you draw the triangle: in physics texts one often finds the laplacian V - V written as V2, but we write it as A. 7.1 GREEN'S FIRST IDENTITY 179 We will write almost everything in this chapter for the three—dimensional case. (However, using two dimensions is okay, too, even n dimensions.) Thus wewrlte fff"dx=///Ndxdydz D D if D is a three-dimensional region (a solid), and ff...dS=//...dS, bdy D S where S =. bdy D is the bounding surface for the solid region D. Here dS indicates the usual surface integral, as in the calculus. Our basic tool in this chapter will be the divergence theorem: (1) where F is any vector function, D is a bounded solid region, and n is the unit outer normal on bdy D (see Figure 1) (see Section A.3). GREEN'S FIRST IDENTITY We start from the product rule (vux)x = vxux + vuxx and the same with y and z derivatives. Summing, this leads to the identity V - (vVu) = er- Vu + vAu. Figure 1 180 CHAPTER 7 GREEN'S IDENTITIES AND GREEN’S FUNCTIONS Then we integrate and use the divergence theorem on the left side to get f/U%dS=///VUVHdX‘i‘f/fUAMdX, (G1) D D bdy D where au/Bn = n - Vu is the directional derivative in the outward normal direction. This is Green’s first identity. It is valid for any solid region D and any pair of functions u and v. For example, we could take 1) E 1 to get f/ggdsz fifth... (2) bdy D As an immediate application of (2), consider the Neumann problem in any domain D. That is, Au = f(x) in D 8 (3) —“ = h(x) on bdy D. an ffhdS=/£ fdx. .. <4) bdy D By (2) we have It follows that the data (f and h) are not arbitrary but are required to satisfy condition (4). Otherwise, there is no solution. In that sense the Neumann problem (3) is not completely well—posed. On the other hand, one can show that if (4) is satisfied, then (3) does have a solution—so the situation is not too bad. What about uniqueness in problem (3)? Well, you could add any constant to any solution of (3) and still get a solution. So problem (3) lacks uniqueness as well as existence. MEAN VALUE PROPERTY In three dimensions the mean value property states that the average value of any harmonic function over any sphere equals its value at the center. To prove this statement, let D be a ball, {|X| < a}, say; that is, {x2 + y2 + Z2 < a2}. Then bdy D is the sphere (surface) {lxl = a}. Let Au = 0 in any region that contains D and bdy D. For a sphere, n points directly away from the origin, so that " Bu x x y z Bu ——=n-Vu=—‘Vu=—u +—u +-u an. r rx ry rZ 8r II I 7.1 GREEN'S FIRST IDENTITY 181 where r = (x2 + y2 + z2)1/2 = [XI is the spherical coordinate, the distance of ‘ the point (x, y, z) from the center 0 of the sphere. Therefore, (2) becomes Bu f/EdS=O. (5) bdy D Let’s write this integral in spherical coordinates, (r, 6, 45). Explicitly, (5) takes the form 271 n f / ur(a, e, ¢)a2sin0d6 d¢ = 0 0 0 since r = a on bdy D. We divide this by the constant 47m2 (the area of bdy D). This result is valid for all a > 0, so that we can think of a as a variable and call it r. Then we pull B/Br outside the integral (see Section A.3), obtaining a 1 27f 7t ' ELIE/0 f0 ”(ra9,¢)81n0d6d¢:|=0_ 1 221 If E/ / u(r,6,¢) sin6d9 d¢ 0 0 is independent of r. This expression is precisely the average value of u on the sphere {lxl = r}. In particular, if we let r —> 0, we get Thus 1 271 Jr . 5/0 /0 u(0) Sln9 d6 d¢ = u(0). That is, (6) This proves the mean value property in three dimensions. Actually, the same idea works in n dimensions. For n = 2 recall that we found another proof in Section 6.3 by a completely different method. MAXIMUM PRINCIPLE Exactly as in two dimensions in Section 6.3, we deduce from the mean value property the maximum principle. If D is any solid region, a nonconstant harmonic function in D cannot take its maximum value inside D, but only on bdy D. It can also be shown that the outward normal derivative flu/8n is strictly positive at a maximum point: {in/an > 0 there. The last assertion is called the Hopf maximum principle. For a proof, see [PW]. 182 CHAPTER 7 GREEN’S IDENTITIES AND GREEN’S FUNCTIONS UNIQUENESS OF DIRICHLET'S PROBLEM We gave one proof in Section 6.1 using the maximum principle. Now we give another proof by the energy method. If we have two harmonic functions u1 and uz with the same boundary data, then their difference u = u] — M is harmonic and has zero boundary data. We go back to (G1) and substitute 1) = u. Since u is harmonic, we have Au = 0 and f/u?—udS=// |Vu|2dx. (7) (in D bdy D Since u = 0 on bdy D, the left side of (7) vanishes. Therefore, flD IVu |2dx = 0. By the first vanishing theorem in Section A.1, it follows that |Vu|2 E Oin D. Now a function with vanishing gradient must be a constant (provided that D is connected). So u(x) E C throughout D. But a vanishes somewhere (on bdy D), so C must be 0. Thus u(x) E O in D. This p...
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