PROBLEM SET 6. E [w] = 1 2 D. is the smallest for w = u, where u is the solution of the Neumann problem. u = 0 in D u = h (x) on D,

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1 PROBLEM SET 6 UE ATE: - APR 25 Chap 7, 9. Questions are either directl from the text or a small variation of a problem in the text. Collaboration is oka, but final submission must be written individuall. Mention all collaborators on our submission. The terms in the bracket indicate the problem number from the text. Section 7. Prob 5, Pg 84 Prove irichlet s Principle for Neumann boundar condition. It asserts that among all real-valued functions w x on, the quantit E [w = w 2, dx hw ds, 2 is the smallest for w = u, where u is the solution of the Neumann problem where h satisfies the constraint u = in = h x on, h x ds =. Note that there are no restrictions on w as opposed to the irichlet principle for irichlet boundar conditions, the function h x appears in the energ and the energ does not change if ou add a constant to w. Comment on the last bit in context of solutions to the Neumann problem for Laplace s equation. Let w be the minimizer of the energ above, then given an function v C 2, consider f v ɛ = E [w + ɛv = 2 Since w is a minimizer, f =. Using our standard calculation f ɛ = v w dx + ɛ f = On using integration b parts for the first term, we get = f = The above identit holds for all functions v and hence we must have w + ɛv 2 dx h w + ɛv v 2 hv ds v w dx hv ds v w dx v h w w = w = h x x Now suppose u satisfies the above Neumnn boundar value problem. Then

2 PROBLEM SET 6 2 E [u = E [w + u w = E [w + E [u w + w u w dx = E [w + u w 2 dx u w h ds + 2 = E [w + u w 2 dx + u w w dx 2 = E [w + u w 2 dx. 2 w u w dx Thus, E [u E [w. The fact that the energ does not change b adding constants is consistent with the fact that we can onl compute solutions of the Neumann equation up to a constant. Section Prob, Pg 87 erive the representation formula for hamronic functions in two dimensions u x = [ u x log x x log x x ds Without loss of generalit, let x =. Let v x = log x and appl Green s second identit to the domain \ B ɛ. Then = \B ɛ u v v u dx = \B ɛ u v v The boundar again separates into two parts, and B ɛ where the normal to B ɛ is inward facing. \B ɛ u v v ds = B ɛ u x log x + log x ds+ u x log x + log x On B ɛ, n = r and x = ɛr, and ds = ɛdθ where r is the unit vector in the outward radial direction. u x log x ds = B ɛ = lim u x log x ds = u, ɛ B ɛ where the last equalit follows from the continuit of u. Similarl, u x u ɛ, θ dθ x n x 2 ɛdθ

3 PROBLEM SET 6 3 log x ds = log x ɛdθ B ɛ lim ɛ B ɛ B ɛ = log ɛ ɛdθ log x ds = log ɛ ɛdθ log ɛ ɛ log x ds = Taking the limit ɛ in Equation, we get u x log x + log x ds + u =, which proves the result. Section Prob, Pg 9 Show that the Green s function is unique. Hint: Take the difference of two of them Let G x, = 4π x + H x, and G 2 x, = 4π x + H 2 x, be two Green s functions. Then, their difference G G 2 = H x, H 2 x, is a harmonic in the x variable for ever and moreover, for a fixed, H H 2 = on the boundar since G = G 2 = on the boundar as a function of x. Thus, for a fixed, the difference of two Green s functions is a harmonic function in x and on the boundar. B uniqueness of the interior irichlet problem, we conclude that the difference must be identicall zero in the interior. Section Prob 7,8 Pg 96 a If u x, = f x is a harmonic function, solve the OE satisfied b f. b Show that r u, where r = x c Suppose v x, is an { > } such that r v. Show that v x, is a function of the quotient x. d Find the boundar values lim u x, = h x e Find the harmonic function in the half plane { > } with boundar data h x = for x > and h x = for x <. f Find the harmonic function in the half plane { > } with boundar data h x = for x > a and h x = for x < a. a b x = r cos θ, = r sin θ c r u = d dx f f s = A arctan s + B = f x = x x r + d x r f x d f x r v x, = x r xv + r v = r x 2 r Use method of characteristics to conclude that d v = f x. h x = { 2 πa + B x > 2 πa + B x <

4 PROBLEM SET 6 4 e f u x, = 2 + π arctan x u x, = 2 + x a π arctan 5 Prob 7, Pg 97 a Find the Green s function for the quadrant b Use the answer in part a to solve the irichlet problem Q = {x, : x >, > }. u = in Q u, = g > u x, = h x x >. Use method of images to place appropriate charges in each quadrant. a b G x, x = x log x log u x, = x x log x log + x x + x [ xg η η 2 + x 2 + η 2 + x 2 [ h ξ x ξ x + ξ Prob 2, Pg 98 The Neumann function N x, for a domain is defined exactl like the Green s function with the following conditions: N x, = + H x, 4π x where H x, is a harmonic function of x for each fixed, and = c x for a suitable constant c. a Show that c = A where A is the area of the boundar. b State and prove the analog of Theorem 7.3., expressing the solution of the Neumann problem in terms of the Neumann function. a Without loss of generalit, we ma assume that =. = N x, dv = ds \B ɛ = \B ɛ = ca. ds + B ɛ Thus, c = A. b u x = N x, ds + u A 7 Prob 22: Pg 98 Solve the Neumann problem in the half plane: u = in { > }, x, = h x dη π + dξ π. ds

5 PROBLEM SET 6 5 and u x, is bounded at. u x, = C + h x ξ log 2 + ξ 2 dξ, boundedness follows from the fact that h x ξ dξ =. Section 9. 8 Prob, Pg 233 Find all three-dimensional plane waves: i.e., all the solutions of the wave equation of the form u x, t = f k x ct where k is a fixed vector and f is a function of one variable Either k = and an arbitar f works or 9 Prob 8, Pg 234 Consider the equation where m >, known as the Klein-Gordon equation. a What is the energ? Show that it is a constant. b Prove the causalit principle for this equation. a E t = 2 u = a + b k x ct tt u c 2 u + m 2 u =, R 2 u t 2 + c 2 u 2 + m 2 u 2 dx d dt E t = ut u tt + c 2 u u t + m 2 uu t dx R 2 = ut u tt c 2 u t u + m 2 uu t dx integration b parts R 2 = b Proof of causalit follows in exactl the same fashion, since t 2 u t c2 u m2 u 2 c 2 u t u = u tt c 2 u + m 2 u u t