Solving Ill-Posed Cauchy Problems in Three Space Dimensions using Krylov Methods

Transcription

1 Solving Ill-Posed Cauchy Problems in Three Space Dimensions using Krylov Methods Lars Eldén Department of Mathematics Linköping University, Sweden Joint work with Valeria Simoncini February 21 Lars Eldén (Linköping Univ.) Krylov Method February 21 1 / 59

2 Motivating example: Ilmenite iron melting furnace Thermocouple Electrode Level K to level D thermocouples with level K highest Center Under electrodes Between electrodes The furnace material properties are temperature dependent. Problem: find the inner shape of the furnace. Nonlinear, and (rather) complex geometry PhD thesis: I M Skaar, Monitoring the Lining of a Melting Furnace, NTNU, Trondheim, 21 Lars Eldén (Linköping Univ.) Krylov Method February 21 2 / 59

3 Inverse Heat Conduction Problem Steady state heat conduction problem: The upper boundary is unavailable for measurements 3D problem! See also Egger et al., Inverse Problems, 29 Lars Eldén (Linköping Univ.) Krylov Method February 21 3 / 59

4 Outline 1 Cauchy Problem Regularization Error Estimate 2 Rational Krylov method Krylov/Regularization 3 Numerical Examples Example 1 Animation Example 2 4 Conclusions Future work Lars Eldén (Linköping Univ.) Krylov Method February 21 4 / 59

5 Ill-Posed Cauchy Problem Ω: connected in R 2 with smooth boundary Ω L: linear, self-adjoint, positive definite elliptic in Ω. u zz Lu =, (x, y) Ω, z [, z 1 ], u(x, y, z) =, (x, y) Ω, z [, z 1 ], u(x, y, ) = g(x, y), (x, y) Ω, u z (x, y, ) =, (x, y) Ω. Sought: f(x, y) = u(x, y, z 1 ), (x, y) Ω. Formal solution u(x, y, z) = cosh(z L)g BUT: L is a pos. def. unbounded operator ILL-POSED! Lars Eldén (Linköping Univ.) Krylov Method February 21 5 / 59

6 Standard Iterative Procedure Guess f (1) 1 for k = 1, 2,... until convergence 1 Solve u zz Lu =, (x, y) Ω, z [, z 1 ], u(x, y, z) =, (x, y) Ω, z [, z 1 ], u(x, y, z 1 ) = f (k), (x, y) Ω, u z (x, y, ) =, (x, y) Ω. giving u (k) 2 Evaluate g(, ) u (k) (,, ) and adjust f (k) f (k+1) In every iteration: Solve a 3D well-posed problem Often slow convergence. Lars Eldén (Linköping Univ.) Krylov Method February 21 6 / 59

7 Other possible methods? Tikhonov regularization? Impossible, because we do not know the integral operator for equations with variable coefficients and/or complicated geometry. Replace unbounded L by a bounded approximation? Possible in connection with finite difference approximation, but more difficult with finite elements. BUT: Krylov method! Lars Eldén (Linköping Univ.) Krylov Method February 21 7 / 59

8 Regularization Formal solution u(x, y, z) = cosh(z L)g BUT: L is a pos. def. unbounded operator ILL-POSED! High frequency perturbations in g are blown up Regularization Replace unbounded operator by a bounded one! Lars Eldén (Linköping Univ.) Krylov Method February 21 8 / 59

9 Cut Off High Frequencies Eigenvalues of L: λ 2 j, j = 1, 2,... and λ j + as j General approach: Compute the k eigenvalues of smallest modulus: LX k = X k D k, where X k holds orthonormal eigenvectors Approximate by projection cosh(z L)g cosh(z L)X k Xk g = X k cosh(z D k )Xk g Lars Eldén (Linköping Univ.) Krylov Method February 21 9 / 59

10 Eigenvalues of L L is large and sparse (of the order , say) Compute the smallest eigenvalues Operate with L 1 Solve many standard 2D elliptic problems Lw = v Lars Eldén (Linköping Univ.) Krylov Method February 21 1 / 59

11 Error Estimate L 2 (Ω) setting, u is an exact solution v is an approximate solution with perturbed data g m Theorem Assume that u(,, 1) M and that the data perturbation satisfies g g m ǫ. Then if v is computed by projection using the eigenvalues satisfying λ j λ c, where then λ c = (1/z 1 ) log(m/ǫ) u(,, z) v(,, z) 3ǫ 1 z/z 1 M z/z 1, z z 1. Optimal error bound Lars Eldén (Linköping Univ.) Krylov Method February / 59

12 Eigenvalue method Is it necessary to compute the eigenvalues and eigenvectors accurately? Do we need all the information that we get in the eigenvalues? Can we take advantage of the fact that we want to compute an approximation of cosh(z L)g for this particular vector? Lars Eldén (Linköping Univ.) Krylov Method February / 59

13 Eigenvalue method Is it necessary to compute the eigenvalues and eigenvectors accurately? NO! Do we need all the information that we get in the eigenvalues? NO! Can we take advantage of the fact that we want to compute an approximation of cosh(z L)g for this particular vector? YES! Lars Eldén (Linköping Univ.) Krylov Method February / 59

14 Lanczos tridiagonalization Choose q 1 and iterate L 1 q k = q k 1 β k 1 + q k α k + q k+1 β k, k = 1, 2,..., with α k = qk L 1 q k and β k = qk+1 L 1 q k ; One matrix-vector multiply L 1 q k per step One standard 2D elliptic solve (black box) per step Lw = q k Lars Eldén (Linköping Univ.) Krylov Method February / 59

15 Lanczos properties Initial convergence influenced by starting vector v 1. Choose q 1 = 1/β g m, β = g m Faster for largest eigenvalues of L 1 Fast convergence for some of the smallest eigenvalues of L Optional: To get faster convergence for eigenvalues in [,λ c ] operate with (L τi) 1, where τ = λ c /2 Lars Eldén (Linköping Univ.) Krylov Method February / 59

16 Lanczos reduction L 1 Q k = Q k T k + β k+1 q k+1 e k+1 Q kt k Approximation cosh(z L)g Q k cosh(zt 1/2 k )Qk g Problem: We cannot prevent T k from approximating large eigenvalues! Solution: Regularize T k : cut off large eigenvalues 1 1 Krylov+regularization: O Leary & Simmons (1981), Björck, Grimme & Van Dooren (1994) Lars Eldén (Linköping Univ.) Krylov Method February / 59

17 Projected and Truncated Approximation Let ((θ (k) j ) 2, y (k) j ), j = 1,...,k be the eigenpairs of T 1 k Define F(z,λ) = cosh(zλ 1/2 ) and S k = T 1 k Truncated approximation: v k (z) = Q k F(z, S c k )Q k g m := Q k θ (k) j λ c y (k) j cosh(zθ (k) j )(y (k) j ) e 1 g m. Lars Eldén (Linköping Univ.) Krylov Method February / 59

18 Error Estimate Recall F(z,λ) = cosh(zλ 1/2 ) and S k = T 1 k Theorem Let u be the exact solution and v k (z) = Q k F(z, Sk c )Q k g m Under the same hypotheses as earlier, u(z) v k (z) 3ǫ 1 z/z 1 M z/z [F(z, L c ) Q k F(z, Sk c )Q k ]g. Lars Eldén (Linköping Univ.) Krylov Method February / 59

19 Krylov/Regularization, First version Starting vector q 1 = 1/β g m for k = 2, 3,... until stable [Q k, T k ] = krylovstep(l 1, Q k 1, T k 1 ) Compute v k (z) = Q k F(z, S c k )Q k g m end Check residual Kv k g m < ǫ Kv k is the solution of the 3D problem with u = v k at the upper boundary and u z = at the lower. Expensive! Lars Eldén (Linköping Univ.) Krylov Method February / 59

20 Residual: Kv k g m < ǫ Solve 3D problem (denote solution u k ) u zz Lu =, (x, y) Ω, z [, z 1 ], u(x, y, z) =, (x, y) Ω, z [, z 1 ], u(x, y, 1) = v k (x, y), (x, y) Ω, u z (x, y, ) =, (x, y) Ω. Well-posed but expensive! Kv k = u k (x, y, ) We only want to compute this when we are sure that Kv k g m < ǫ Lars Eldén (Linköping Univ.) Krylov Method February 21 2 / 59

21 Krylov/Regularization: stable Starting vector q 1 = 1/β g m for k = 1, 2,... until stable [Q k, T k ] = krylovstep(l 1, Q k 1, T k 1 ) Compute v k (z) = Q k F(z, S c k )Q k g m end Check residual Kv k g m < ǫ How can we quantify stable? Lars Eldén (Linköping Univ.) Krylov Method February / 59

22 Krylov/Regularization Recall F(z,λ) = cosh(zλ 1/2 ) and S k = T 1 k Cheap (2D) approximate residual: r (k+p) k Starting vector q 1 = 1/β g m for k = 1, 2,... maxit [Q k, T k ] = krylovstep(l 1, Q k 1, T k 1 ) Compute w k (z) = F(z, S k c)q k g m if r (k 1+p) < tol then end r (k+p) k k 1 / r (k+p) k if Kv k g m < ǫ then stop iterating endif Compute v k = Q k w k Lars Eldén (Linköping Univ.) Krylov Method February / 59

23 Test example 1: Laplace equation Ω: unit square u zz + u =, (x, y, z) Ω [,.1], u(x, y, z) =, (x, y, z) Ω [,.1], u(x, y, ) = g(x, y), (x, y) Ω, u z (x, y, ) =, (x, y) Ω. Determine the values at the upper boundary, f(x, y) = u(x, y,.1), (x, y) Ω. Data perturbation: g g m / g eigenvalues are smaller than the tolerance eigs performs approximately 3 2D elliptic solves. Lars Eldén (Linköping Univ.) Krylov Method February / 59

24 Solution and Exact Data Lars Eldén (Linköping Univ.) Krylov Method February / 59

25 Convergence History (.8λ c ) 1 1 True error Residual Res k+2 ε/ g m step Lars Eldén (Linköping Univ.) Krylov Method February / 59

26 Solutions as function of iteration index.45.4 k= 3 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

27 Solutions as function of iteration index.45.4 k= 4 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

28 Solutions as function of iteration index.6.5 k= 5 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

29 Solutions as function of iteration index.45.4 k= 6 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

30 Solutions as function of iteration index.45.4 k= 7 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February 21 3 / 59

31 Solutions as function of iteration index.45.4 k= 8 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

32 Solutions as function of iteration index.6.5 k= 9 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

33 Solutions as function of iteration index.45.4 k= 1 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

34 Solutions as function of iteration index.45.4 k= 11 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

35 Solutions as function of iteration index.45.4 k= 12 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

36 Solutions as function of iteration index.45.4 k= 13 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

37 Solutions as function of iteration index.45.4 k= 14 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

38 Solutions as function of iteration index.45.4 k= 15 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

39 Solutions as function of iteration index.45.4 k= 16 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

40 Solutions as function of iteration index.45.4 k= 17 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February 21 4 / 59

41 Solutions as function of iteration index.45.4 k= 18 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

42 Solutions as function of iteration index.45.4 k= 19 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

43 Solutions as function of iteration index.45.4 k= 2 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

44 Solutions as function of iteration index.45.4 k= 21 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

45 Solutions as function of iteration index.45.4 k= 22 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

46 Solutions as function of iteration index.45.4 k= 23 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

47 Solutions as function of iteration index.45.4 k= 24 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

48 Stopping criterion satisfied here.45.4 k= 25 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

49 Solutions as function of iteration index.45.4 k= 26 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

50 Solutions as function of iteration index.45.4 k= 27 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February 21 5 / 59

51 Solutions as function of iteration index.45.4 k= 28 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

52 Solutions as function of iteration index.45.4 k= 29 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

53 Solutions as function of iteration index.45.4 k= 3 exact solution approx. sol Lars Eldén (Linköping Univ.) Krylov Method February / 59

54 Example 2 Finite element discretization of L with variable coefficients on the ellipse Ω = {(x, y, z) x 2 + y 2 /4 1, z z 1 =.6}. The stiffness matrix has dimension 865 Data perturbation: 3% Lars Eldén (Linköping Univ.) Krylov Method February / 59

55 Solution and Exact Data Lars Eldén (Linköping Univ.) Krylov Method February / 59

56 Perturbed Data Lars Eldén (Linköping Univ.) Krylov Method February / 59

57 Convergence history (.6λ c ). Solution after 9 steps 1 True error Residual Res k+2 ε/ g m step Lars Eldén (Linköping Univ.) Krylov Method February / 59

58 Conclusions 3D Cauchy problem: complex 2D geometry + cylinder in z Krylov method + black box 2D elliptic solver Stability theory = recipe for cut-off level Exponential of small matrix is computed (cheap) Safe-guarded stopping criterion: Approximate residual (cheap) + True residual (rather expensive) Much fewer 2D elliptic solves than eigenvalue computation: 98 eigenvalues were smaller than the tolerance. MATLAB s eigs: 3 2D solves Krylov: 18 solves Highly accurate eigenvalues are not needed The data influence the basis (projection) vectors Lars Eldén (Linköping Univ.) Krylov Method February / 59

59 Extensions Variable coefficients in z? Other Cauchy problems: parabolic: Zohreh Ranjbar s thesis Helmholtz, transient electromagnetics? Paper: Inverse Problems 29 Lars Eldén (Linköping Univ.) Krylov Method February / 59