Using Green s functions with inhomogeneous BCs
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1 Using Green s functions with inhomogeneous BCs
2 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!
3 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.
4 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.
5 Using Green s formula for inhomogeneous boundary conditions Want to solve Lu(x) = f (x) + inhomogeneous boundary conditions.
6 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.
7 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
8 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 Collapsing the integral involving the δ function, u(x 0 ) = G(x, x 0 )f (x)dx + boundary terms
9 Green s formula for Laplacian Want to solve u = f in, u = h on,
10 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.
11 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.
12 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
13 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 Simplifies to u(x 0 ) = G(x, x 0 )f (x)dx + h(x) x G(x, x 0 ) ˆn(x) dx,
14 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 ) ).
15 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 ) = r 2 0 u = 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 ) u(a, θ) = h(θ) h(x) xg(x, x 0 ) ˆn(x) dx. ).
16 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 a4 2rr 0 a 2 cos(θ θ 0 ) ( a 2π 1 (r/a) 2 r 2 + a 2 2ar cos(θ θ 0 ) ). ),
17 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 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. ),
18 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}.
19 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).
20 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.
21 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 ) (x x0 ) 2 + (y y 0 ) 2 + (z + z 0 ) 2
22 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 ) (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,
23 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 ) (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.
24 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 ) (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. Since ˆn is directed outward, u(x) ˆn(x) = u z (x, y, 0), and u(x 0, y 0, z 0 ) = 1 2π h(x, y) (x x 0 ) 2 + (y y 0 ) 2 + z0 2 dxdy.
25 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"
26 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.
27 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).
28 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.
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