Question

Problem 3 (12 points): Let D be a bounded domain in R" with smooth boundary. Suppose that K(x, y) is a Green's function for the Neumann . For each x E D, the function y H K(x, y) is a smooth harmonic For each x E D, the normal derivative of the function y K(x, y) . For each z e D, the function y K(x,y)-Г(z-y) is smooth near problem. This means the following: function on D(r satisfies (VyK(x, y).v(b))-arefor all y E oD. z. Here, Г denotes the Newton potential. Please show: if u is a smooth solution of Au -f in D with Neumann boundary condition (Vu,g, then u(x)K(x, y) f(y) dy -K(x, y) g(y) +constant.

Answer 1

SDLUTION;-

-;GIVEN DATA;-

'' The Green’s function method for solving Sturm-Liouville problems of the form Lrφ(r) = ∇ · [p(r)∇φ(r)] + q(r)φ(r) = ρ(r) (1) is described in many textbooks.

We list some of the more recent texts that treat Green’s functions in references 1-4, and review the usual textbook treatment below

. For simplicity, we will take q(r) = 0. A motivation for using a Green’s function, G(r, r ′ ), is that it satisfies homogeneous boundary conditions that makes it easier to solve for than the original problem with inhomogeneous boundary conditions.

Application of Green’s theorem Z [φ(r ′ )Lr ′G(r, r ′ )−G(r, r ′ )Lr ′φ(r ′ )]dτ ′ = Z dS′ ·[φ(r ′ )p(r ′ )∇′G(r, r ′ )−G(r, r ′ )p(r ′ )∇′φ(r ′ )] (2) provides the solution to the original problem if G(r, r ′ ) satisfies Lr ′G(r, r ′ ) = δ(r − r ′ ). (3) In that case, Eq (2) leads to φ(r) = Z G(r, r ′ )ρ(r ′ )dτ ′ + Z dS′ · [φ(r ′ )p(r ′ )∇′G(r, r ′ ) − G(r, r ′ )p(r ′ )∇′φ(r ′ )]. (4) The two simplest boundary conditions for which the Green’s function method is applicable are the Dirichlet boundary condition for which the solution φ(r) is given on all bounding surfaces, and the Neumann boundary condition for which its normal derivative ˆn · ∇φ(r) is given.

The Dirichlet Green’s function is generally used for electrostatic problems where the potential is specified on bounding surfaces, while the Neumann Green’s function is useful for finding temperature distributions where the bounding surfaces are heat insulated or have specified heat currents. For the Dirichlet boundary condition, the Green’s function satisfies the homogeneous boundary condition GD(r, r ′ ) = 0, for r ′ on all bounding surfaces. (5) This reduces Eq. (4) to φ(r) = Z GD(r, r ′ )ρ(r ′ )]dτ ′ + Z dS′ · [φ(r ′ )p(r ′ )∇′GD(r, r ′ )], (6) 2 which is the solution to the Dirichlet problem once GD(r, r ′ ) is known. The Dirichlet Green’s function is the solution to Eq. (3) satisfying the homogeneous boundary condition in Eq. (5). Textbooks generally treat the Dirichlet case as above, but do much less with the Green’s function for the Neumann boundary condition, and what is said about the Neumann case often has mistakes of omission and commission. First of all, the Neumann boundary condition for the solution φ(r) must satisfy the constraint Z dS · p(r)∇φ(r) = Z ρdτ, (7) which follows from applying the divergence theorem to Eq. (1). Most texts do not mention this important constraint on the Neumann boundary condition.5 There are cases where the boundary condition is Neumann on some surfaces and Dirichlet on others. In those cases, the normal derivative of φ on the Dirichlet surfaces automatically adjusts to satisfy the constraint surface integral. But for pure Neumann boundary conditions, the normal derivative must satisfy the constraint or no solution exists. In the following sections, we will assume pure Neumann boundary conditions for which constraint equation (7) holds. We treat the one dimensional Neumann Green’s function in Section 2, and then the three dimensional.

  '' We first review how the Dirac delta function arises when a function f(x), defined in the finite range 0 ≤ x ≤ L, is expanded in orthonormal Dirichlet eigenfunctions un(x) of a Sturm-Liouville operator Lx. The eigenfunctions satisfy the differential equation Lxun(x) = d dx " p(x) dun dx # = λnun(x). (8) with homogeneous Dirichlet boundary conditions un(0) = 0, un(L) = 0. (9) The function f(x) can be expanded in the Dirichlet eigenfunctions as f(x) = X∞ n=1 bnun(x), (10) 3 with the expansion coefficients bn given by bn = Z L 0 u ∗ n (x)f(x)dx. (11) If Eq. (11) for the expansion coefficients is substituted into Eq. (10) for f(x), and the sum executed before the integral, we get f(x) = Z L 0 dx′ X∞ n=1 u ∗ n (x ′ )un(x)f(x ′ ). (12) From the definition of the Dirac delta function by its sifting property, f(x) = Z L 0 dx′ δ(x − x ′ )f(x ′ ) (13) for x and x ′ in the range [0, L], we see that the delta function can be represented by a sum over Dirichlet eigenfunctions as δ(x − x ′ ) = X∞ n=1 u ∗ n (x ′ )un(x). (14) For example, for the simple case Lxun = u ′′ n = λnun, un = q 2/L sin(nπx/L), λn = − nπ L 2 , (15) the delta function is represented by δ(x − x ′ ) = (2/L) X∞ n=1 sin(nπx′ /L) sin(nπx/L) (16) A Dirichlet Green’s function that satisfies the differential equation Lx′GD(x, x′ ) = δ(x − x ′ ), (17) and satisfies the homogeneous Dirichlet boundary conditions in the variable x ′ can be formed from the Dirichlet eigenfunctions as GD(x, x′ ) = X∞ n=1 u ∗ n (x ′ )un(x) λn . (18) Acting on this Green’s function with the Sturm Liouville operator Lx′ removes the denominator in Eq. (18) leaving the delta function of Eq. (17), showing that this is the appropriate Dirichlet Green’s function. The above straightforward derivation for Dirichlet boundary conditions is given in most texts, but a corresponding derivation for Neumann boundary conditions is generally absent. 4 Neumann eigenfunctions, vn(x), satisfy the same differential equation (8) as the Dirichlet eigenfunctions, but have the boundary conditions v ′ n (0) = 0, v′ n (L) = 0. (19) The expansion in Neumann eigenfunctions has a constant eigenfunction corresponding to a zero eigenvalue, so the expansion is given by f(x) = a0 + X∞ n=1 anvn(x), (20) The Neumann expansion coefficients an are given by the integrals an = Z L 0 v ∗ n (x)f(x)dx, n ≥ 1, (21) and a0 = 1 L Z L 0 f(x)dx =< f >, (22) where < f > represents the average value of the function f(x) over the interval [0, L]. Now putting Eqs. (21) and (22) into the expansion Eq. (20) results in f(x) = Z L 0 dx′ " 1 L + X∞ n=1 v ∗ n (x ′ )vn(x) # f(x ′ ), (23) so the representation of the delta function in terms of Neumann eigenfunctions is δ(x − x ′ ) = 1 L + X∞ n=1 v ∗ n (x ′ )vn(x). (24) The additional constant term 1/L is not generally recognized in textbooks. For the simple case of Lxvn = v ′′ n = λnv with the eigenfunctions satisfying the homogeneous Neumann boundary conditions of Eq. (19) the delta function is represented by δ(x − x ′ ) = 1 L + 2 L X∞ n=1 cos(nπx′ /L) cos(nπx/L). (25) A Neumann’s Green function can be formed using Neumann eigenfunctions of the operator Lx as the sum GN (x, x′ ) = X∞ n=1 v ∗ n (x ′ )vn(x) λn . (26) This Green’s function satisfies the homogeneous Neumann boundary conditions ∂x′GN (x, x′ )|(x′=0) = 0, ∂x′GN (x, x′ )|(x′=L) = 0. (27) 5 However, because of the constant term 1/L in Eq. (24), the operation on GN (x, x′ ) by Lx′ is Lx′GN (x, x′ ) = δ(x − x ′ ) − 1/L. (28) Thus, the Neumann Green’s function satisfies a different differential equation than the Dirichlet Green’s function. We now use the Green’s function GN (x, x′ ) to find the solution of the differential equation Lxf(x) = d dx " p(x) df dx# = ρ(x), (29) with the inhomogeneous Neumann boundary conditions f ′ (0) = f ′ 0 , f′ (L) = f ′ L . (30) The boundary values must satisfy the constraint p(L)f ′ L − p(0)f ′ 0 = Z L 0 ρ(x)dx, (31) which follows from a first integral of Eq. (29). The Neumann Green’s function must also satisfy this constraint, which it does because the right hand side of Eq. (28) integrates to zero. Green’s theorem in one dimension (or integration by parts) for this differential equation leads to Z L 0 [f(x ′ )Lx′GN (x, x′ ) − GN (x, x′ )Lx′f(x ′ )]dx′ = −GN (x, L)p(L)f ′ L + GN (x, 0)p(0)f ′ 0 . (32) We have used the boundary conditions (27) to eliminate terms containing ∂x′G(x, x′ ) at the endpoints. Then, using Eq. (28), we get f(x) =< f > + Z L 0 GN (x, x′ )ρ(x ′ )dx′ − GN (x, L)p(L)f ′ L + GN (x, 0)p(0)f ′ 0 , (33) which constitutes the solution to the Sturm-Liouville problem for Neumann boundary conditions. The constant < f >, the average value of f(x), arises when the term 1/L in Eq. (28) is substituted into Eq. (32). Actually, any constant can be added to the solution f(x) since the solution of the Neumann problem is only unique up to an additive constant. The constant term in Eq. (33) will always be the average value of the solution because the variable terms have zero average value.