Solution of Laplace’s equation between two concentric spheres. We are interested in solving Laplace’s equation ∇2 u = 0 in 3D space in the region between two concentric spheres, Ω. Let us assume that the inner sphere, ΓD , has Dirichlet boundary conditions specified, i.e. u(x) = h(x)

x ∈ ΓD

where h(x) is some known function of position, and the outer sphere, ΓN has Neumann boundary conditions specified, i.e. ∂u = g(x) x ∈ ΓN ∂n where g(x) is some other known function of position. From Green’s Representation Theorem, we have Z Z ∂u(y) ∂U ∗ (x, y) ∗ u(x) = U (x, y) dΓy − u(y) dΓy ∂n ∂n Γ Γ where x ∈ Ω and y ∈ Γ. Here U ∗ (x, y) is the fundamental solution of the 3D Laplace’s equation: U ∗ (x, y) =

1 4π|x − y|

Now break up the this expression for u(x) into integrals over the respective boundaries of the problem: Z Z ∂u(y) ∂u(y) ∗ U (x, y) U ∗ (x, y) u(x) = dΓy + dΓy ∂n ∂n ΓD ΓN Z Z ∂U ∗ (x, y) ∂U ∗ (x, y) dΓy − u(y) dΓy x ∈ Ω − u(y) ∂n ∂n ΓN ΓD Following the notation of Steinbach (2008), define the Dirichlet trace, γ0int as γ0int u(x) =

lim

Ω3˜ x→x∈∂Ω

u(˜ x)

and the Neumann trace γ1int as γ1int u(x) = γ0int

∂u ∂u(˜ x) = lim ∂n Ω3˜x→x∈∂Ω ∂n

Therefore, we can rewrite u(x) in terms of traces to get (c.f. Steinbach, equation (6.1)) Z Z ∗ int u(x) = U (x, y)γ1 u(y)dΓy + U ∗ (x, y)γ1int u(y)dΓy ΓD ΓN Z Z int int ∗ int ∗ − γ0 u(y)γ1,y U (x, y)dΓy − γ0int u(y)γ1,y U (x, y)dΓy x ∈ Ω ΓD

ΓN

Further, taking the normal derivative of u gives Z Z ∂u(x) int int ∗ int ∗ = γ1,x U (x, y)γ1 u(y)dΓy + γ1,x U (x, y)γ1int u(y)dΓy ∂n ΓD ΓN Z  Z  int int int ∗ int int int ∗ − γ1,x γ0 u(y)γ1,y U (x, y)dΓy − γ1,x γ0 u(y)γ1,y U (x, y)dΓy ΓD

ΓN

1

x∈Ω

Now take the Dirichlet trace on ΓD , i.e. for x ∈ ΓD Z Z int(D) ∗ int γ0 u(x) = lim U (˜ x, y)γ1 u(y)dΓy + U ∗ (˜ x, y)γ1int u(y)dΓy Ω3˜ x→x∈ΓD ΓD Γ  Z Z N int int ∗ int int ∗ − γ0 u(y)γ1,y U (˜ x, y)dΓy − γ0 u(y)γ1,y U (˜ x, y)dΓy ΓD

x ∈ ΓD

ΓN

 

1  int(D)  int(D) γ0 u (x) − Kγ0 u (x) = V γ int(D) u (x) + V γ int(N ) u (x) + 2   1  int(N )  int(N ) + γ0 u (x) − Kγ0 u (x) x ∈ ΓD 2 where the superscripts D and N indicate over which boundary the traces are taken. Finally, take the Neumann trace on ΓN , i.e. for x ∈ ΓN Z Z int(N ) int int ∗ int ∗ U (˜ x, y)γ1int u(y)dΓy γ1 u(x) = lim x, y)γ1 u(y)dΓy + γ1,x γ1,x U (˜ Ω3˜ x→x∈ΓN ΓZD  ΓN Z  int int int ∗ int int int ∗ − γ1,x γ0 u(y)γ1,y U (x, y)dΓy − γ1,x γ0 u(y)γ1,y U (x, y)dΓy ΓD

ΓN

    1  int(D)  1  int(N )  int(D) int(N ) = γ1 u (x) + K 0 γ1 u (x) + γ1 u (x) − K 0 γ1 u (x) 2  2    int(N ) int(D) (x) x ∈ ΓN (x) − Dγ0 − Dγ0 where • V is the single layer potential operator: Z (V u)(x) =

U (x, y)u(y)dΓ Γ

• K is the double layer potential operator: Z (Ku)(x) = lim →0

int γ1,y U (x, y)u(y)dΓ

Γ:|y−x|≥

• K 0 is the adjoint double layer potential operator: Z 0 int (K u)(x) = lim γ1,x U (x, y)u(y)dΓ →0

Γ:|y−x|≥

• D is the hypersingular potential operator: Z int (Du)(x) = −γ1,x lim →0

int U (x, y)u(y)dΓ γ1,y

Γ:|y−x|≥

The above equations can be rewritten as follows:       1 1 int(D) int(N ) I +K γ0 u (x)+ − I + K γ0 u (x) 2 2     int(D) int(N ) = V γ1 u (x) + V γ1 u (x) x ∈ ΓD        1 1 int(D) int(N ) I + K0 γ1 u (x)+ − I + K 0 γ1 u (x) 2 2     int(D) int(N ) = Dγ0 u (x) + Dγ0 u (x) x ∈ ΓN

2

int(D)

int(D)

In these equations, γ0 = h(x) are the known Dirichlet values on ΓD , γ1 are the int(N ) int(N ) unknown Neumann values on ΓD , γ0 are the unknown Dirichlet values on ΓN and γ1 = g(x) are the known Neumann values on ΓN . In turn the above equations can be rewritten in a matrix form, with unknown values on the LHS and known values on the RHS.       int(N ) int(D)     1 1 γ u (x) γ u (x) −V V K − 2I  0   = −(K + 2 I)  0   1 1 0 0 int(D) int(N ) D −( 2 I + K ) −D −2I + K γ u (x) γ u (x) 1

1

3