5 4. Properties of Laplace s and Poisson s Equations Chapter 4 Elliptic Equations Contents. Neumann conditions the normal derivative, / = n u is prescribed on the boundar second BP. In this case we have compatibilit conditions i.e. global constraints: E.g., suppose u satisfies u = F on and n u = n u = f on. Then, u d = u d = u n d = d divergence theorem, F d = f d for the problem to be well-defined. 3. Robin conditions a combination of u and its normal derivative such as / + αu is prescribed on the boundar third BP. 4.1 Definitions................................. 49 4. Properties of Laplace s and Poisson s Equations.......... 5 4.3 olving Poisson Equation Using Green s Functions........ 54 4.4 Etensions of Theor:.......................... 68 n 4.1 Definitions Elliptic equations are tpicall associated with stead-state behavior. The archetpal elliptic equation is Laplace s equation and describes u =, stead, irrotational flows, e.g. electrostatic potential in the absence of charge, u + u = in -D, equilibrium temperature distribution in a medium. Because of their phsical origin, elliptic equations tpicall arise as boundar value problems BPs. olving a BP for the general elliptic equation L[u] = n i,j=1 u n + b i + cu = F i j i a ij recall: all the eigenvalues of the matri A = a ij, i, j = 1 n, are non-zero and have the same sign is to find a solution u in some open region of space, with conditions imposed on the boundar of or at infinit. E.g. inviscid flow past a sphere is determined b boundar conditions on the sphere u n = and at infinit u = Const. There are three tpes of boundar conditions for well-posed BPs, 1. Dirichlet condition u takes prescribed values on the boundar first BP. i=1 ometimes we ma have a mied problem, in which u is given on part of and / given on the rest of. If encloses a finite region, we have an interior problem; if, however, is unbounded, we have an eterior problem, and we must impose conditions at infinit. Note that initial conditions are irrelevant for these BPs and the Cauch problem for elliptic equations is not alwas well-posed even if Cauch-Kowaleski theorem states that the solution eist and is unique. As a general rule, it is hard to deal with elliptic equations since the solution is global, affected b all parts of the domain. Hperbolic equations, posed as initial value or Cauch problem, are more localised. From now, we shall deal mainl with the Helmholtz equation u + P u = F, where P and F are functions of, and particularl with the special one if P =, Poisson s equation, or Laplace s equation, if F = too. This is not too severe restriction; recall that an linear elliptic equation can be put into the canonical form n u + = k=1 k and that the lower order derivatives do not alter the overall properties of the solution. 4. Properties of Laplace s and Poisson s Equations Definition: A continuous function satisfing Laplace s equation in an open region, with continuous first and second order derivatives, is called an harmonic function. Functions u 49
Chapter 4 Elliptic Equations 51 5 4. Properties of Laplace s and Poisson s Equations in C with u respectivel u are call subharmonic respectivel superharmonic. 4..1 Mean alue Propert Definition: Let be a point in and let B R denote the open ball having centre and radius R. Let Σ R denote the boundar of B R and let AR be the surface area of Σ R. Then a function u has the mean value propert at a point if u = 1 u d AR Σ R for ever R > such that B R is contained in. If instead u satisfies u = 1 u d, R B R where R is the volume of the open ball B R, we sa that u has the second mean value propert at a point. The two mean value properties are equivalent. so v is harmonic too and equation 4.1 becomes u v Σ r d + u v Σ ε d = u v Σ r r d u v Σ ε r d = u v Σ ε r d = u v Σ r r d i.e 1 ε u d = 1 Σ ε r u d. Σ r ince u is continuous, then as ε the LH converges to 4π u,, z with n = 3, sa, so u = 1 u d. Ar Σ r Recovering the second mean value propert with n = 3, sa is straightforward r ρ u dρ = r3 3 u = 1 4π r Σ ρ u d dρ = 1 4π B r u d. The inverse of this theorem holds too, but is harder to prove. If u has the mean value propert then u is harmonic. Σ R B R R 4.. Maimum-Minimum Principle One of the most important features of elliptic equations is that it is possible to prove theorems concerning the boundedness of the solutions. Theorem: uppose that the subharmonic function u satisfies u = F in, with F > in. Then u, attains his maimum on. Theorem: on. Proof: If u is harmonic in an open region of R n, then u has the mean value propert We need to make use of Green s theorem which sas, v u v d = v u u v d. 4.1 Recall: Appl divergence theorem to the function v u u v to state Green s theorem. ince u is harmonic, it follows from equation 4.1, with v = 1, that d =. Now, take v = 1/r, where r =, and the domain to be B r B ε, < ε < R. Then, in R n, v = 1 r r 1 = r r r Proof: Theorem stated in -D but holds in higher dimensions. uppose for a contradiction that u attains its maimum at an interior point, of. Then at,, since it is a maimum. o, =, =, u and u, u + u, which contradicts F > in. Hence u must attain its maimum on, i.e. if u M on, u < M in. Theorem: The weak Maimum-Minimum Principle for Laplace s equation. uppose that u satisfies u = in a bounded region ; if m u M on, then m u M in.
Chapter 4 Elliptic Equations 53 54 4.3 olving Poisson Equation Using Green s Functions Proof: Theorem stated in -D but holds in higher dimensions. Consider the function v = u + ε +, for an ε >. Then v = 4ε > in since + = 4, and using the previous theorem, v M + εr in, where u M on and R is the radius of the circle containing. As this holds for an ε, let ε to obtain u M in, i.e., if u satisfies u = in, then u cannot eceed M, the maimum value of u on. Also, if u is a solution of u =, so is u. Thus, we can appl all of the above to u to get a minimum principle: if u m on, then u m in. This theorem does not sa that harmonic function cannot also attain m and M inside though. We shall now progress into the strong Maimum-Minimum Principle. Theorem: uppose that u has the mean value propert in a bounded region and that u is continuous in =. If u is not constant in then u attains its maimum value on the boundar of, not in the interior of. Proof: ince u is continuous in the closed, bounded domain then it attains its maimum M somewhere in. Our aim is to show that, if u attains its ma. at an interior point of, then u is constant in. uppose u = M and let be some other point of. Join these points with a path covered b a sequence of overlapping balls, B r. 1 Consider the ball with at its center. ince u has the mean value propert then M = u = 1 u d M. Ar Σ r This equalit must hold throughout this statement and u = M throughout the sphere surrounding. ince the balls overlap, there is 1, centre of the net ball such that u 1 = M; the mean value propert implies that u = M in this sphere also. Continuing like this gives u = M. ince is arbitrar, we conclude that u = M throughout, and b continuit throughout. Thus if u is not a constant in it can attain its maimum value onl on the boundar. Corollar: Appling the above theorem to u establishes that if u is non constant it can attain its minimum onl on. Also as a simple corollar, we can state the following theorem. The proof follows immediatel the previous theorem and the weak Maimum-Minimum Principle. Theorem: The strong Maimum-Minimum Principle for Laplace s equation. Let u be harmonic in, i.e. solution of u = in and continuous in, with M and m the maimum and minimum values respectivel of u on the boundar. Then, either m < u < M in or else m = u = M in. Note that it is important that be bounded for the theorem to hold. E.g., consider u, = e sin with = {, < < +, < < }. Then u = and on the boundar of we have u =, so that m = M =. But of course u is not identicall zero in. Corollar: If u = C is constant on, then u = C is constant in. Armed with the above theorems we are in position to prove the uniqueness and the stabilit of the solution of Dirichlet problem for Poisson s equation. Consider the Dirichlet BP u = F in with u = f on and suppose u 1, u two solutions to the problem. Then v = u 1 u satisfies v = u 1 u = in, with v = on. Thus, v in, i.e. u 1 = u ; the solution is unique. To establish the continuous dependence of the solution on the prescribed data i.e. stabilit of the solution let u 1 and u satisf u {1,} = F in with u {1,} = f {1,} on, with ma f 1 f = ε. Then v = u 1 u is harmonic with v = f 1 f on. As before, v must have its maimum and minimum values on ; hence u 1 u ε in. o, the solution is stable small changes in the boundar data lead to small changes in the solution. We ma use the Maimum-Minimum Principle to put bounds on the solution of an equation without solving it. The strong Maimum-Minimum Principle ma be etended to more general linear elliptic equations n u n L[u] = a ij + b i + cu = F, i j i i,j=1 and, as for Poisson s equation it is possible then to prove that the solution to the Dirichlet BP is unique and stable. 4.3 olving Poisson Equation Using Green s Functions We shall develop a formal representation for solutions to boundar value problems for Poisson s equation. i=1 the
Chapter 4 Elliptic Equations 55 56 4.3 olving Poisson Equation Using Green s Functions 4.3.1 Definition of Green s Functions Consider a general linear PDE in the form n Lu = F in, where L is a linear self-adjoint differential operator, u is the unknown and F is the known homogeneous term. Recall: L is self-adjoint if L = L, where L is defined b v Lu = L v u and where v u = vwud w is the weight function. The solution to the equation can be written formall u = L 1 F, where L 1, the inverse of L, is some integral operator. We can epect to have LL 1 = LL 1 = I, identit. We define the inverse L 1 using a Green s function: let u = L 1 F = G, ξf ξdξ, 4. where G, ξ is the Green s function associated with L G is the kernel. Note that G depends on both the independent variables and the new independent variables ξ, over which we integrate. Recall the Dirac δ-function more precisel distribution or generalised function δ which has the properties, R n δ d = 1 and Now, appling L to equation 4. we get hence, the Green s function G, ξ satisfies R n δ ξ hξ dξ = h. Lu = F = LG, ξf ξ dξ; u = G, ξ F ξ dξ with L G, ξ = δ ξ and, ξ. 4.3. Green s function for Laplace Operator Consider Poisson s equation in the open bounded region with boundar, u = F in. 4.3 Then, Green s theorem n is normal to outward from, which states u v v u d = u v v d, for an functions u and v, with h/ = n h, becomes u v d = vf d + u v v d; so, if we choose v v, ξ, singular at = ξ, such that v = δ ξ, then u is solution of the equation uξ = vf d u v v d 4.4 which is an integral equation since u appears in the integrand. To address this we consider another function, w w, ξ, regular at = ξ, such that w = in. Hence, appl Green s theorem to the function u and w u w w d = u w w u d = wf d. Combining this equation with equation 4.4 we find uξ = v + wf d u v + w v + w d, so, if we consider the fundamental solution of Laplace s equation, G = v + w, such that G = δ ξ in, uξ = GF d u G d. 4.5 Note that if, F, f and the solution u are sufficientl well-behaved at infinit this integral equation is also valid for unbounded regions i.e. for eterior BP for Poisson s equation. The wa to remove u or / from the RH of the above equation depends on the choice of boundar conditions.
Chapter 4 Elliptic Equations 57 58 4.3 olving Poisson Equation Using Green s Functions Dirichlet Boundar Conditions Here, the solution to equation 4.3 satisfies the condition u = f on. o, we choose w such that w = v on, i.e. G = on, in order to eliminate / form the RH of equation 4.5. Then, the solution of the Dirichlet BP for Poisson s equation is u = F in with u = f on uξ = GF d f d, where G = v+w w regular at = ξ with v = δ ξ and w = in and v+w = on. o, the Green s function G is solution of the Dirichlet BP Neumann Boundar Conditions G = δ ξ in, with G = on. Here, the solution to equation 4.3 satisfies the condition / = f on. o, we choose w such that w/ = v/ on, i.e. / = on, in order to eliminate u from the RH of equation 4.5. However, the Neumann BP with G = δ ξ in, = on, which does not satisf a compatibilit equation, has no solution. Recall that the Neumann BP u = F in, with / = f on, is ill-posed if F d fd. We need to alter the Green s function a little to satisf the compatibilit equation; put G = δ + C, where C is a constant, then the compatibilit equation for the Neumann BP for G is δ + C d = d = C = 1, where is the volume of. Now, appling Green s theorem to G and u: G u u G d = G u d we get uξ = GF d + Gf d + 1 u d. }{{} ū This shows that, whereas the solution of Poisson s equation with Dirichlet boundar conditions is unique, the solution of the Neumann problem is unique up to an additive constant ū which is the mean value of u over. Thus, the solution of the Neumann BP for Poisson s equation u = F in with = f on is uξ = ū GF d + Gf d, where G = v + w w regular at = ξ with v = δ ξ, w = 1/ in and w/ = v/ on. o, the Green s function G is solution of the Neumann BP with Robin Boundar Conditions G = δ ξ + 1 in, = on. Here, the solution to equation 4.3 satisfies the condition / + αu = f on. o, we choose w such that w/ + αw = v/ αv on, i.e. / + αg = on. Then, u G d = u + Gαu f d = Gf d. Hence, the solution of the Robin BP for Poisson s equation u = F in with + αu = f on is uξ = GF d + Gf d, where G = v + w w regular at = ξ with v = δ ξ and w = in and w/ + αw = v/ αv on. o, the Green s function G is solution of the Robin BP with mmetr of Green s Functions G = δ ξ in, + αg = on. The Green s function is smmetric i.e., G, ξ = Gξ,. To show this, consider two Green s functions, G 1 G, ξ 1 and G G, ξ, and appl Green s theorem to these, G1 G G G 1 d = G 1 G 1 d. Now, since, G 1 and G are b definition Green s functions, G 1 = G = on for Dirichlet boundar conditions, 1 / = / = on for Neumann boundar conditions or G 1 / = G 1 / on for Robin boundar conditions, so in an case the right-hand side is equal to zero. Also, G 1 = δ ξ 1, G = δ ξ and the equation becomes G, ξ 1 δ ξ d = G, ξ δ ξ 1 d, Gξ, ξ 1 = Gξ 1, ξ. Nevertheless, note that for Neumann BPs, the term 1/ which provides the additive constant to the solution to Poisson s equation breaks the smmetr of G.
Chapter 4 Elliptic Equations 59 6 4.3 olving Poisson Equation Using Green s Functions Eample: Consider the -dimensional Dirichlet problem for Laplace s equation, u = in, with u = f on boundar of. ince u is harmonic in i.e. u = and u = f on, then Green s theorem gives u v d = f v v d. Note that we have no information about / on or u in. uppose we choose, v = 1 4π ln ξ + η, then v = on for all points ecept P = ξ, = η, where it is undefined. To eliminate this singularit, we cut this point P out i.e, surround P b a small circle of radius ε = ξ + η and denote the circle b Σ, whose parametric form in polar coordinates is Σ : { ξ = ε cos θ, η = ε sin θ with ε > and θ, }. Hence, v = 1/ ln ε and dv/dε = 1/ε and appling Green s theorem to u and v in this new region with boundaries and Σ, we get f v v d + u v v d =. 4.6 since u = v = for all point in. B transforming to polar coordinates, d = εdθ and / = / ε unit normal is in the direction ε onto Σ; then v ε ln ε d = dθ as ε, ε and also u v d = Σ Σ u v ε ε dθ = 1 Σ ε ξ Σ u ε 1 ε dθ = 1 u dθ uξ, η as ε, and so, in the limit ε, equation 4.6 gives uξ, η = v f v d, where v = 1 4π ln ξ + η. now, consider w, such that w = in but with w regular at = ξ, = η, and with w = v on. Then Green s theorem gives u w w u d = u w w d f w + v d = since u = w = in and w = v on. Then, subtract this equation from equation above to get uξ, η = v f v d f w + v d = f v + w d. etting G, ; ξ, η = v + w, then uch a function G then has the properties, 4.3.3 Free pace Green s Function We seek a Green s function G such that, uξ, η = f d. G = δ ξ in, with G = on. G, ξ = v, ξ + w, ξ where v = δ ξ in. How do we find the free space Green s function v defined such that v = δ ξ in? Note that it does not depend on the form of the boundar. The function v is a source term and for Laplace s equation is the potential due to a point source at the point = ξ. As an illustration of the method, we can derive that, in two dimensions, v = 1 4π ln ξ + η, as we have alread seen. We move to polar coordinate around ξ, η, ξ = r cos θ & η = r sin θ, and look for a solution of Laplace s equation which is independent of θ and which is singular as r. η C r D r ξ r θ
Chapter 4 Elliptic Equations 61 6 4.3 olving Poisson Equation Using Green s Functions Laplace s equation in polar coordinates is 1 r v = v r r r r + 1 v r r = which has solution v = B ln r + A with A and B constant. Put A = and, to determine the constant B, appl Green s theorem to v and 1 in a small disc D r with boundar C r, of radius r around the origin η, ξ, C r so we choose B to make v d = v d = δ ξ d = 1, D r D r C r v d = 1. Now, in polar coordinates, v/ = v/ r = B/r and d = rdθ going around circle C r. o, B r rdθ = B dθ = 1 B = 1. Hence, v = 1 ln r = 1 4π ln r = 1 4π ln ξ + η. We do not use the boundar condition in finding v. imilar but more complicated methods lead to the free-space Green s function v for the Laplace equation in n dimensions. In particular, 1 ξ, n = 1, v, ξ = 1 4π ln ξ, n =, 1 na n 1 ξ n, n 3, Having found the free space Green s function v which does not depend on the boundar conditions, and so is the same for all problems we still need to find the function w, solution of Laplace s equation and regular in = ξ, which fies the boundar conditions v does not satisfies the boundar conditions required for G b itself. o, we look for the function which satisfies w = or 1/ in, ensuring w is regular at ξ, η, with w = v i.e. G = on for Dirichlet boundar conditions, w or = v i.e. = on for Neumann boundar conditions. To obtain such a function we superpose functions with singularities at the image points of ξ, η. This ma be regarded as adding appropriate point sources and sinks to satisf the boundar conditions. Note also that, since G and v are smmetric then w must be smmetric too i.e. w, ξ = wξ,. Eample 1 uppose we wish to solve the Dirichlet BP for Laplace s equation u = u + u = in > with u = f on =. We know that in -D the free space function is If we superpose to v the function v = 1 4π ln ξ + η. w = + 1 4π ln ξ + + η, solution of w = in and regular at = ξ, = η, then G,, ξ, η = v + w = 1 ξ 4π ln + η ξ + + η. where and ξ are distinct points and A n 1 denotes the area of the unit n-sphere. We shall restrict ourselves to two dimensions for this course. Note that Poisson s equation, u = F, is solved in unbounded R n b u = v, ξ F ξ dξ R n where from equation 4. the free space Green s function v, defined above, serves as Green s function for the differential operator when no boundaries are present. ξ, η + ξ, η = ξ = η = η G = v + w v w 4.3.4 Method of Images In order to solve BPs for Poisson s equation, such as u = F in an open region with some conditions on the boundar, we seek a Green s function G such that, in G, ξ = v, ξ + w, ξ where v = δ ξ and w = or 1/. Note that, setting = in this gives, G,, ξ, η = 1 4π ln ξ + η ξ + η =, as required.
Chapter 4 Elliptic Equations 63 64 4.3 olving Poisson Equation Using Green s Functions The solution is then given b uξ, η = f d. Now, we want / for the boundar =, which is Thus, = = 1 η = π ξ + η eercise, check this. uξ, η = η π + and we can relabel to get in the original variables u, = π + f ξ + η d, fλ λ + dλ. Eample 3 Consider the Neumann BP for Laplace s equation in the upper half-plane, u = u + u = in > with = = f on =. ξ, η ξ, η = ξ = η G = v + w = η v w Eample Find Green s function for the Dirichlet BP u = u + u = F in the quadrant >, >. We use the same technique but now we have three images. Then, the Green s function G is o, ξ, η + ξ, η ξ, η + ξ, η G,, ξ, η = 1 4π ln ξ + η + 1 4π ln ξ + + η 1 4π ln + ξ + + η + 1 4π ln + ξ + η. [ G,, ξ, η = 1 ξ 4π ln + η + ξ + + η ] ξ + + η + ξ + η, and again we can check that G,, ξ, η = G,, ξ, η = as required for Dirichlet BP. Add an image to make / = on the boundar: Note that, G,, ξ, η = 1 4π ln ξ + η 1 4π ln ξ + + η. = 1 4π and as required for Neumann BP, = = 1 = 4π η ξ + η + + η ξ + + η Then, since G,, ξ, η = 1/ ln ξ + η, uξ, η = 1 i.e. u, = 1 + η ξ + η + η ξ + η + f ln ξ + η d, fλ ln λ + dλ,, =. Remind that all the theor on Green s function has been developed in the case when the equation is given in a bounded open domain. In an infinite domain i.e. for eternal problems we have to be a bit careful since we have not given conditions on G and / at infinit. For instance, we can think of the boundar of the upper half-plane as a semi-circle with R +. R 1 +R
Chapter 4 Elliptic Equations 65 66 4.3 olving Poisson Equation Using Green s Functions Green s theorem in the half-disc, for u and G, is G u u G d = G u d. plit into 1, the portion along the -ais and, the semi-circular arc. Then, in the above equation we have to consider the behaviour of the integrals 1 G π d G R dθ and u π R d u R R dθ as R +. Green s function G is Oln R on, so from integral 1 we need / R to fall off sufficientl rapidl with the distance: faster than 1/R ln R i.e. u must fall off faster than lnlnr. In integral, / R = O1/R on provides a more stringent constraint since u must fall off more rapidl that O1 at large R. If both integrals over vanish as R + then we recover the previousl stated results on Green s function. Eample 4 olve the Dirichlet problem for Laplace s equation in a disc of radius a, u = 1 r + 1 u r r r r = in r < a with u = fθ on r = a. θ, Consider image of point P at inverse point Q with ρ q = a i.e. OP OQ = a. r θ P ρ ξ, η φ P = ρ cos φ, ρ sin φ, Q = q cos φ, q sin φ, + Q G,, ξ, η = 1 4π ln ξ + η + 1 4π ln a ρ cos φ + a sin φ + h,, ξ, η with ξ + η = ρ. ρ We need to consider the function h,, ξ, η to make G smmetric and zero on the boundar. We can epress this in polar coordinates, = r cos θ, = r sin θ, Gr, θ, ρ, φ = 1 r cos θ a 4π ln /ρ cos φ + r sin θ a /ρ sin φ r cos θ ρ cos φ + r sin θ ρ sin φ + h, = 1 r 4π ln + a 4 /ρ a r/ρ cosθ φ r + ρ + h. rρ cosθ φ Choose h such that G = on r = a, G = 1 a 4π ln + a 4 /ρ a 3 /ρ cosθ φ a + ρ aρ cosθ φ = 1 a 4π ln ρ + a aρ cosθ φ ρ ρ + a aρ cosθ φ Note that, + h, + h = h = 1 4π ln wr, θ, ρ, φ = 1 r 4π ln + a4 ρ r a cosθ φ + 1 ρ ρ 4π ln a = 1 a 4π ln + r ρ a rρ cosθ φ is smmetric, regular and solution of w = in. o, Gr, θ, ρ, φ = v + w = 1 a 4π ln + r ρ /a rρ cosθ φ r + ρ, rρ cosθ φ G is smmetric and zero on the boundar. This enable us to get the result for Dirichlet problem for a circle, uρ, φ = fθ r a dθ, where so Then ρ r = 1 rρ /a ρ cosθ φ 4π a + r ρ /a rρ cosθ φ r ρ cosθ φ r + ρ, rρ cosθ φ = r = 1 and relabelling, a. ρ /a ρ cosθ φ a + ρ aρ cosθ φ a ρ cosθ φ a + ρ, aρ cosθ φ = 1 ρ a a a + ρ aρ cosθ φ. uρ, φ = 1 a ρ a + ρ fθ dθ, aρ cosθ φ ur, θ = a r fφ a + r a r cosθ φ dφ. Note that, from the integral form of ur, θ above, we can recover the Mean alue Theorem. If we put r = centre of the circle then, u = 1 fφ dφ, i.e. the average of an harmonic function of two variables over a circle is equal to its value at the centre.
Chapter 4 Elliptic Equations 67 Furthermore we ma introduce more subtle inequalities within the class of positive harmonic functions u. ince 1 cosθ φ 1 then a r a ar cosθ φ+r a+r. Thus, the kernel of the integrand in the integral form of the solution ur, θ can be bounded 1 a r a + r 1 a r a r a ar cosθ φ + r 1 a + r a r. For positive harmonic functions u, we ma use these inequalities to bound the solution of Dirichlet problem for Laplace s equation in a disc 1 a r a + r fθ dθ ur, θ 1 a + r a r i.e. using the Mean alue Theorem we obtain Harnack s inequalities Eample 5 a r a + r u ur, θ a + r a r u. Interior Neumann problem for Laplace s equation in a disc, Here, we need r r u = 1 r r = fθ on r = a. + 1 u r = in r < a, θ fθ dθ, G = δ ξδ η + 1 with r =, where = πa is the surface area of the disc. In order to deal with this term we solve the equation κr = 1 r κ = 1 r r r πa κr = r 4πa + c 1 ln r + c, and take the particular solution with c 1 = c =. Then, add in source at inverse point and an arbitrar function h to fi the smmetr and boundar condition of G Gr, θ, ρ, φ = 1 4π ln r + ρ rρ cosθ φ 1 [ a a 4π ln ρ + ρ r ] a rρ cosθ φ + r 4πa + h. o, r = 1 r ρ cosθ φ 4π r + ρ rρ cosθ φ 1 r a /ρ cosθ φ 4π r + a 4 /ρ a r/ρ cosθ φ + r a + h r, r = 1 a ρ cosθ φ a + ρ aρ cosθ φ + a a /ρ cosθ φ a + a 4 /ρ a 3 + 1 /ρ cosθ φ a + h r, = 1 a ρ cosθ φ + ρ /a ρ cosθ φ ρ + a + 1 aρ cosθ φ a + h r, 68 4.4 Etensions of Theor: r = 1 a + 1 a + h r and h r = implies = on the boundar. r Then, put h 1/ lna/ρ ; so, Gr, θ, ρ, φ = 1 [ r 4π ln + ρ rρ cosθ φ a + ρ r ] a rρ cosθ φ + r 4π a. On r = a, Then, G = 1 [ a 4π ln + ρ aρ cosθ φ ] + 1 4π, = 1 [ ln a + ρ aρ cosθ φ 1 ]. uρ, φ = ū + = ū a fθ G a dθ, [ ln a + ρ aρ cosθ φ 1 ] fθ dθ. Now, recall the Neumann problem compatibilit condition, Indeed, o the term involving Eercise: or u d = d uρ, φ = ū a fθ dθ =. from divergence theorem fθdθ in the solution uρ, φ vanishes; hence ur, θ = ū a ln a + ρ aρ cosθ φ fθ dθ, ln a + r ar cosθ φ fφ dφ. Eterior Neumann problem for Laplace s equation in a disc, ur, θ = a 4.4 Etensions of Theor: ln a + r ar cosθ φ fφ dφ. fθ dθ =. Alternative to the method of images to determine the Green s function G: a eigenfunction method when G is epended on the basis of the eigenfunction of the Laplacian operator; conformal mapping of the comple plane for solving -D problems. Green s function for more general operators.