Using Green s functions with inhomogeneous BCs

Using Green s functions with inhomogeneous BCs Surprise: Although Green s functions satisfy homogeneous boundary conditions, they can be used for problems with inhomogeneous BCs!

Using Green s functions with inhomogeneous BCs Surprise: Although Green s functions satisfy homogeneous boundary conditions, they can be used for problems with inhomogeneous BCs! For self adjoint L and u, v with homogeneous boundary conditions it follows that (Lv)u dx (Lu)v dx = 0. But if u, v don t satisfy homogeneous boundary conditions, get (Lv)u dx (Lu)v dx = boundary terms involving u and v. This is called the Green s formula, which depends on L and.

Using Green s formula for inhomogeneous boundary conditions Want to solve Lu(x) = f (x) + inhomogeneous boundary conditions.

Using Green s formula for inhomogeneous boundary conditions Want to solve Lu(x) = f (x) + inhomogeneous boundary conditions. Set v(x) = G(x, x 0 ) in Green s formula (LG)u dx (Lu)G dx = boundary terms. Since LG = δ(x x 0 ) and Lu = f, we have δ(x x 0 )u(x)dx f (x)g(x, x 0 )dx = boundary terms

Green s formula for Laplacian Want to solve u = f in, u = h on,

Green s formula for Laplacian Want to solve u = f in, u = h on, For dimensions 2, the Green s formula is just Green s identity u v v u dx = u v ˆn v u ˆn dx.

Green s formula for Laplacian Want to solve u = f in, u = h on, For dimensions 2, the Green s formula is just Green s identity u v v u dx = u v ˆn v u ˆn dx. Let G solve G = δ(x x 0 ) and G = 0 on boundary. Substituting v(x) = G(x, x 0 ) into Green s formula, u(x)δ(x x 0 ) G(x, x 0 )f (x) dx = u(x) xg(x, x 0 ) ˆn(x) G(x, x 0 ) u(x) ˆn(x) dx

Example: the Poisson integral formula revisited In the case that is a disk of radius a, Green s function is G(r, θ; r 0, θ 0 ) = 1 ( a 2 4π ln r 2 0 r 2 + r 2 0 2rr 0 cos(θ θ 0 ) r 2 + a 4 /r 2 0 2ra2 /r 0 cos(θ θ 0 ) ).

Example: the Poisson integral formula revisited In the case that is a disk of radius a, Green s function is G(r, θ; r 0, θ 0 ) = 1 ( a 2 4π ln The boundary value problem has a solution u(x 0 ) = Need normal derivative of G r 2 0 u = 0, xg(x, x 0 ) ˆn(x) = G r (r, θ; r 0, θ 0 ) = 1 4π which at r = a is r 2 + r 2 0 2rr 0 cos(θ θ 0 ) r 2 + a 4 /r 2 0 2ra2 /r 0 cos(θ θ 0 ) u(a, θ) = h(θ) h(x) xg(x, x 0 ) ˆn(x) dx. ). ( 2r0 2r cos(θ θ 0 ) r 2 + r 2 0 2rr 0 cos(θ θ 0 ) 2r 0 r 2 2ra 2 cos(θ θ 0 ) r 2 r 2 0 + a4 2rr 0 a 2 cos(θ θ 0 ) ( a 2π 1 (r/a) 2 r 2 + a 2 2ar cos(θ θ 0 ) ). ),

Example: the Poisson integral formula revisited In the case that is a disk of radius a, Green s function is G(r, θ; r 0, θ 0 ) = 1 ( a 2 4π ln The boundary value problem has a solution u(x 0 ) = Need normal derivative of G r 2 0 u = 0, xg(x, x 0 ) ˆn(x) = G r (r, θ; r 0, θ 0 ) = 1 4π r 2 + r 2 0 2rr 0 cos(θ θ 0 ) r 2 + a 4 /r 2 0 2ra2 /r 0 cos(θ θ 0 ) u(a, θ) = h(θ) h(x) xg(x, x 0 ) ˆn(x) dx. ). ( 2r0 2r cos(θ θ 0 ) r 2 + r 2 0 2rr 0 cos(θ θ 0 ) 2r 0 r 2 2ra 2 cos(θ θ 0 ) r 2 r 2 0 + a4 2rr 0 a 2 cos(θ θ 0 ) which at r = a is ( ) a 1 (r/a) 2. 2π r 2 + a 2 2ar cos(θ θ 0 ) Parameterize boundary integral using θ and dx = a dθ, u(r 0, θ 0 ) = 1 2π 2π 0 (a 2 r 2 0 )h(θ) a 2 + r 2 0 2ar 0 cos(θ θ 0 ) dθ. ),

Neumann boundary conditions Want to solve u = 0, lim u(x, y, z) = 0, z u z (x, y, 0) = h(x, y), in upper half space {(x, y, z) z > 0}.

Neumann boundary conditions Want to solve u = 0, lim u(x, y, z) = 0, z u z (x, y, 0) = h(x, y), in upper half space {(x, y, z) z > 0}. Green s formula u v v u dx = u v ˆn v u ˆn dx. has both Dirichlet and Neumann boundary terms in u, but only know u(x, y, 0) ˆn = u z (x, y, 0). To make v ˆn = G ˆn vanish on boundary, need Green s function to respect boundary condition principle": The Green s function must have the same type of boundary conditions as the problem to be solved, and they must be homogeneous.

Neumann boundary conditions,cont. Method of images prescribes even reflection so G z = 0 when z = 0: G(x, y, z; x 0, y 0, z 0 ) = G 3 (x, y, z; x 0, y 0, z 0 ) + G 3 (x, y, z; x 0, y 0, z 0 ) ( ) = 1 1 4π (x x0 ) 2 + (y y 0 ) 2 + (z z 0 ) + 1 2 (x x0 ) 2 + (y y 0 ) 2 + (z + z 0 ) 2 Substituting v(x) = G(x, x 0 ) into Green s formula and collapsing the δ-function integral, u(x 0 ) = G(x; x 0 ) u(x) ˆn(x) dx, Boundary is both xy-plane and the effective boundary at infinity, but integrand vanishes on the latter.

Symmetry (reciprocity) of the Green s function If L is self-adjoint, might expect that its inverse to also be: G(x, x 0 ) = G(x 0, x), Reciprocity"

Symmetry (reciprocity) of the Green s function If L is self-adjoint, might expect that its inverse to also be: G(x, x 0 ) = G(x 0, x), Reciprocity" Proof: Insert v(x) = G(x, x 1 ), u(x) = G(x, x 2 ), Lv = δ(x x 1 ) and Lu = δ(x x 2 ) into Green s formula: δ(x x 1 )G(x, x 2 )dx G(x, x 1 )δ(x x 2 )dx = 0 which simplifies to G(x 1, x 2 ) G(x 2, x 1 ) = 0. Also: can interchange arguments of partial derivatives, e.g. xg(x, x 0 ) = lim h 0 (G(x + h, x 0 ) G(x, x 0 ))/h = lim h 0 (G(x 0, x + h) G(x 0, x))/h = x0 G(x 0, x). For example, representation formula u(x 0 ) = G(x, x 0 )f (x)dx + h(x) xg(x, x 0 ) ˆn(x) dx, can be rewritten by exchanging the notation for x and x 0 and using reciprocity, u(x) = G(x, x 0 )f (x 0 )dx 0 + h(x 0 ) x0 G(x, x 0 ) ˆn(x 0 ) dx 0.