Partial Differential Equations in Biology The boundary element method. March 26, 2013

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1 The boundary element method March 26, 203

2 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet problem]). Given ξ R d K(, ξ) denotes the fundamental solution of the Laplace operator: K = δ ξ. T (, ξ) denotes the normal derivative of K(, ξ), defined on D = Γ (n is the unit outward normal vector on Γ): T (x, ξ) = x K(x, ξ) n(x).

3 Fundamental solutions If D R 2 K(x, ξ) = 2π log x ξ If D R 3 K(x, ξ) = 4π T (x, ξ) = x ξ 2π n(x). x ξ 2 x ξ T (x, ξ) = x ξ 4π n(x). x ξ 3

4 Basic integral equations Internal points: ξ D u(ξ) + Γ u(x)t (x, ξ) ds(x) u Γ n (x)k(x, ξ) ds(x) = () D f (x)k(x, ξ) dx. Boundary points: ξ Γ (regular boundary) 2 u(ξ) + Γ u(x)t (x, ξ) ds(x) u Γ n (x)k(x, ξ) ds(x) = (2) D f (x)k(x, ξ) dx.

5 Polyhedral domain and piecewise-polynomial functions The domain D is approximated by a polyhedral domain D h (union of triangles or tetrahedra; here h is the maximum of their diameters). The boundary D = Γ is therefore approximated by D h (union of M segments or triangles S k ). The approximate solution is piecewise-polynomial (on D h, if the problem to approximate is defined in D, or on D h, if the problem to approximate is defined in Γ). In our case (approximation of the integral equation (2), defined on Γ, by piecewise-constant functions on D h ): the unknowns are the values at the mid-point (or baricenter) ξ j of S j, j =,..., M.

6 The approximate equations: collocation method Let u h the approximation of u on the boundary and q h the approximation of u n on the boundary (clearly, u h and q h are defined on D h, while u and u n are defined on Γ). They are identified by their constant values α k and β k on the element S k. We can start considering an approximate form of (2), valid for ξ D h : 2 u h(ξ) + D h u h (x)t (x, ξ) ds(x) D h q h (x)k(x, ξ) ds(x) = D f (x)k(x, ξ) dx. (3)

7 The approximate equations: collocation method (cont d) Dirichlet problem. We know u = ϕ on the boundary Γ, therefore we can construct a suitable approximation ϕ h on the boundary D h. We look for q h, the approximation of u n. For j =,..., M 2 ϕ h(ξ j ) + D h ϕ h (x)t (x, ξ j ) ds(x) D h q h (x)k(x, ξ j ) ds(x) = D f (x)k(x, ξ j) dx. (4) We can write D h q h (x)k(x, ξ j ) ds(x) = M k= = M k= β k S k q h (x)k(x, ξ j ) ds(x) S k K(x, ξ j ) ds(x).

8 The approximate equations: collocation method (cont d) We have thus obtained the linear system A Dir β = b Dir, where A Dir jk = K(x, ξ j ) ds(x), S k bj Dir = 2 ϕ h(ξ j ) + ϕ h (x)t (x, ξ j ) ds(x) D h The matrix A Dir is not symmetric. D f (x)k(x, ξ j ) dx.

9 The approximate equations: collocation method (cont d) Neumann problem. We know u n = g on the boundary Γ, therefore we can construct a suitable approximation g h on the boundary D h. We look for u h, the approximation of u. For j =,..., M 2 u h(ξ j ) + D h u h (x)t (x, ξ j ) ds(x) D h g h (x)k(x, ξ j ) ds(x) = D f (x)k(x, ξ j) dx. (5) We can write D h u h (x)t (x, ξ j ) ds(x) = M k= = M k= α k S k u h (x)t (x, ξ j ) ds(x) S k T (x, ξ j ) ds(x).

10 The approximate equations: collocation method (cont d) We have thus obtained the linear system where A Neu α = b Neu, A Neu jk = 2 δ jk + T (x, ξ j ) ds(x), S k bj Neu = g h (x)k(x, ξ j ) ds(x) + f (x)k(x, ξ j ) dx. D h D The matrix A Neu is not symmetric. [The matrix A Neu is singular, the kernel being given by constant vectors c(,,...,, ): a technical problem that we do not look at in depth.]

11 The approximate equations: Galerkin method Another set of approximate equations can be obtained by projecting equation (3) on the subspace of piecewise-constant functions on D h. This is an example of Galerkin method. Let us define by {ψ j }, j =,..., M, a set of basis functions of the vector space given by the piecewise-constant functions on D h. We can write u h (ξ) = M k= α kψ k (ξ), q h (ξ) = M k= β kψ k (ξ). [The simplest choice is given by ψ j equal to in S j, and equal to 0 in all the other S l for l j.]

12 The approximate equations: Galerkin method (cont d) Multiplying equation (3) by ψ j and integrating on D h we find for each j =,..., M: 2 D h u h (ξ)ψ j (ξ)ds(ξ) + ( ) D h D h u h (x)t (x, ξ) ds(x) ψ j (ξ)ds(ξ) ( ) (6) D h D h q h (x)k(x, ξ) ds(x) ψ j (ξ)ds(ξ) = D h ( D f (x)k(x, ξ) dx )ψ j (ξ)ds(ξ).

13 The approximate equations: Galerkin method (cont d) Dirichlet problem. We know u = ϕ on the boundary Γ, therefore we can construct a suitable approximation ϕ h on the boundary D h. We look for q h, the approximation of u n. For j =,..., M we have 2 D h ϕ h (ξ)ψ j (ξ)ds(ξ) + ( ) D h D h ϕ h (x)t (x, ξ) ds(x) ψ j (ξ)ds(ξ) ( ) D h D h q h (x)k(x, ξ) ds(x) ψ j (ξ)ds(ξ) = D h ( D f (x)k(x, ξ) dx )ψ j (ξ)ds(ξ). (7) We can write ( ) D h D h q h (x)k(x, ξ) ds(x) ψ j (ξ)ds(ξ) = M k= β k D h D h K(x, ξ)ψ k (x)ψ j (ξ) ds(x)ds(ξ).

14 The approximate equations: Galerkin method (cont d) We have thus obtained the linear system  Dir β = b Dir, where  Dir jk = K(x, ξ)ψ k (x)ψ j (ξ) ds(x)ds(ξ), D h D h b j Dir = 2 D h ϕ h (ξ)ψ j (ξ)ds(ξ) + ( ) D h D h ϕ h (x)t (x, ξ) ds(x) ψ j (ξ)ds(ξ) D h ( D f (x)k(x, ξ) dx )ψ j (ξ)ds(ξ). Since K(x, ξ) = K(ξ, x), the matrix ÂDir is clearly symmetric.

15 The approximate equations: Galerkin method (cont d) Neumann problem. We know u n = g on the boundary Γ, therefore we can construct a suitable approximation g h on the boundary D h. We look for u h, the approximation of u. For j =,..., M we have 2 D h u h (ξ)ψ j (ξ)ds(ξ) + ( ) D h D h u h (x)t (x, ξ) ds(x) ψ j (ξ)ds(ξ) ( ) D h D h g h (x)k(x, ξ) ds(x) ψ j (ξ)ds(ξ) (8) = D h ( D f (x)k(x, ξ) dx )ψ j (ξ)ds(ξ).

16 The approximate equations: Galerkin method (cont d) We can write D h u h (ξ)ψ j (ξ)ds(ξ) = M α k ψ k (ξ)ψ j (ξ)ds(ξ), D h k= ( ) D h D h u h (x)t (x, ξ) ds(x) ψ j (ξ)ds(ξ) = M k= α k D h D h T (x, ξ)ψ k (x)ψ j (ξ) ds(x)ds(ξ).

17 The approximate equations: Galerkin method (cont d) We have thus obtained the linear system where  Neu α = b Neu,  Neu jk = 2 D h ψ k (ξ)ψ j (ξ)ds(ξ) b Neu j + D h D h T (x, ξ)ψ k (x)ψ j (ξ) ds(x)ds(ξ), = ( ) D h D h g h (x)k(x, ξ) ds(x) ψ j (ξ)ds(ξ) Since T (x, ξ) = + D h ( D f (x)k(x, ξ) dx )ψ j (ξ)ds(ξ). { 2π 4π x ξ n(x) x ξ 2 for d=2 x ξ n(x) x ξ 3 for d=3, we see that T (x, ξ) T (ξ, x), and therefore the matrix ÂNeu is not symmetric [moreover, as in the collocation case, it is singular...].

18 The approximate equations: Galerkin method (cont d) A symmetric Galerkin formulation for the Neumann problem can be derived by using a different integral equation, that can be proved to hold for ξ Γ: u 2 n (ξ) + n ξ u(x)t (x, ξ) ds(x) Γ n ξ Γ u n (x)k(x, ξ) ds(x) = n ξ f (x)k(x, ξ) dx. D (9) Its approximate form for ξ D h, in terms of u h and q h, is 2 q h(ξ) + n ξ D h u h (x)t (x, ξ) ds(x) n ξ D h q h (x)k(x, ξ) ds(x) = n ξ f (x)k(x, ξ) dx. D (0)

19 The approximate equations: Galerkin method (cont d) Multiplying equation (0) by ψ j and integrating on D h we find for each j =,..., M: 2 D h q h (ξ)ψ j (ξ)ds(ξ) + ( ) D h n ξ D h u h (x)t (x, ξ) ds(x) ψ j (ξ)ds(ξ) ( ) () D h n ξ D h q h (x)k(x, ξ) ds(x) ψ j (ξ)ds(ξ) = ( D h n ξ )ψ D f (x)k(x, ξ) dx j (ξ)ds(ξ). We can write ( ) D h n ξ D h u h (x)t (x, ξ) ds(x) ψ j (ξ)ds(ξ) = M k= α k D h n ξ T (x, ξ)ψ k (x)ψ j (ξ) ds(x)ds(ξ). D h

20 The approximate equations: Galerkin method (cont d) We have thus obtained the linear system à Neu α = b Neu, where à Neu jk = D h D h n ξ T (x, ξ)ψ k (x)ψ j (ξ) ds(x)ds(ξ), b j Neu = 2 D h q h (ξ)ψ j (ξ)ds(ξ) + ( ) D h n ξ D h q h (x)k(x, ξ) ds(x) ψ j (ξ)ds(ξ) + ( D h n ξ )ψ D f (x)k(x, ξ) dx j (ξ)ds(ξ). n ξ Since n ξ T (x, ξ) = n x K(x, ξ) is symmetric in x and ξ, we see that the matrix à is symmetric.

21 The approximate equations: Galerkin method (cont d) Let us remark that the construction of the matrices ÂDir, ÂNeu and à Neu only requires that we have a basis ψ k for the space where we look for the approximate solution: namely, we can repeat the same construction for piecewise-linear functions, or piecewise-polynomial functions, once we have a basis for that space of functions. In the particular case of piecewise-constant functions, we can compute the entries of these matrices in a more explicit way: for instance, since ψ k = in S k and ψ k = 0 outside S k, we have  Dir jk = D h D h K(x, ξ)ψ k (x)ψ j (ξ) ds(x)ds(ξ) = S k S j K(x, ξ) ds(x)ds(ξ).