Boundary element solutionof a parabolic equation

based on the Laplace transform

Thorsten Hohage, University of Göttingen, Germany

and

Javier Sayas, University of Zaragoza, Spain

Third International Conference on Boundary Integral Methods:Theory and Applications

Reading, UK, September 14-18, 2004

. – p.1/21

overview

1. problem formulation

2. Laplace transform method

3. boundary integral equations

4. convergence analysis and numerical results

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problem description

Ω−

Ω+

measurements / heat sources

F. Garrido and A. Salazar. Thermal wave scattering by spheres. J. Appl. Phys,95:140–149, 2004.

A. Ocariz, A. Sánchez-Lavega, and A. Salazar. Photothermal study of subsurfacecylindrical structures. 2. experimental results. J. Appl. Phys, 81:7561–7566, 1997.

J. M. Terrón, A. Salazar, and A. Sánchez-Lavega. General solution for the thermalwave scattering in fiber composites. J. Appl. Phys, 91:1087–1098, 2002.

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experimental setup

Grupo de Fototermia, Univ. Pais Vasco (Spain)

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mathematical formulation

differential equation: Ω+ = exterior domain, Ω− = interior domainκ± = conductivity/(density · specific heat) in Ω±

∂tu+ = κ+∆u+, in Ω+ × (0,∞),

∂tu− = κ−∆u−, in Ω− × (0,∞)

transmission conditions: uhom = solution without inclusionf proportional to corrosion, unknown of inverse problem!α = interior conductivity/ exterior conductivity

u+ + uhom = u− + f∂νu− on Γ × (0,∞)

∂νu+ + ∂νuhom = α∂νu− on Γ × (0,∞)

boundary condition at top barrier H : ∂νu+|H = 0.

initial condition: u+( · , 0) = 0, u−( · , 0) = 0.

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2. Laplace transform method

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stable inversion of Laplace transform

The Laplace transform of a continuous, bounded mappingu : [0,∞) → X taking values in a Banach space X is

U(s) =

∫ ∞

0e−stu(t) dt, Re s > 0.

assumption: U has a holomorphic extension U : C \ Σδ,r → X tothe complement of Σδ,r := s ∈ C : | arg(−s)| ≤ δ or |s| ≤ r,δ ∈ (0, π

2 ), r > 0, and there exists M > 0 such that

‖U(s)‖ ≤ M

|s| for all s ∈ C \ Σδ,r.

stable inversion formula: Γ = path connecting −i∞ to i∞

u(t) =1

2πi

∫

ΓetsU(s) ds, t > 0.

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inversion contours

−1.5 −1 −0.5 0 0.5−3

−2

−1

0

1

2

3

Re(s)

Im(s

)Lopez−F./PalenciaTalbot

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numerical inversion of Laplace transform

quadrature: For t in a compact interval [t0, T ] ⊂ (0,∞)

1

2πi

∫

ΓetsU(s) ds ≈

N∑

j=−N

γ(N)j es

(N)j tU(x, s

(N)j ) =: uN (t)

Talbot, 1979:

‖uN (t) − u(t)‖ ≤ CM exp(

−c√N

)

, t ∈ [t0, T ] ???

Sheen, Sloan & Thomée, 1999/2003: (applications to FEM)If Ml is a bound on U (l)(s), l ∈ N, then

‖uN (t) − u(t)‖ ≤ ClMlN−l, t ∈ [t0, T ].

López-Fernández & Palencia, 2004:

‖uN (t) − u(t)‖ ≤ CM exp (−cN/ lnN) , t ∈ [t0, T ]. – p.9/21

other boundary integral methods for heat equation

boundary integral formulation using fundamental solution tothe heat equation as kernel.see Costabel ’90, Hsiao ’94, Kress ’89, ...

Lubich & Schneider ’92: operational quadrature methods tosolve the Volterrà–Fredholm integral equations ofconvolution type sequence of boundary integralequations involving Laplace transformed kernels.see also Chapko & Kress ’97

advantages of Laplace transform method:

exponential instead of polynomial order of convergence.

Laplace transformed problems can be solved in parallel.

no need to store previous time-steps.

disadvantage:

convergence only on compact intervals [t0, T ] ⊂ (0,∞).Estimates deteriorate as T/t0 → ∞.

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3. boundary integral equations

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Laplace transformed transmission problem

differential equation:

κ+∆U+ − sU+ = 0 in Ω+,

κ−∆U− − sU− = 0 in Ω−

“radiation condition”:

U+(s, ·) ∈ H2(Ω+)

transmission/boundary conditions:

U+ + Uhom = U− + f∂νU− on Γ,

∂νU+ + ∂νUhom = α∂νU− on Γ,

∂νU+ = 0 on H.

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ansatz

ansatz:

U+(s, x) =

∫

ΓΦ√

−s/κ+(x, y)ψ+(y) ds(y), x ∈ Ω+

U−(s, x) =

∫

ΓΦ√

−s/κ−

(x, y)ψ−(y) ds(y), x ∈ Ω−

fundamental solution of ∆ + ρ2 in Rd:

Φρ(x, y) :=

ı

4H

(1)0 (ρ|x− y|), when d = 2,

exp(ı ρ |x− y|)4π |x− y| , when d = 3,

fundamental solution of ∆ + ρ2 in half-space: (d = 2)

Φρ((x1, x2), y) := Φρ((x1, x2), y) + Φρ((x1,−x2), y). – p.13/21

boundary integral equations

(12ψ− + J√

−s/κ−

ψ−)f + V√−s/κ−

ψ− −V√−s/κ+

ψ+ = Uhom(s, ·)α

(

12ψ− + J√

−s/κ−

ψ−

)

12ψ+ − J√

−s/κ+ψ+ = ∂νUhom(s, ·)

potential operators:

(Vρψ)(x) :=

∫

ΓΦρ(x, y)ψ(y) ds(y), x ∈ Γ,

(Jρψ)(x) :=

∫

Γ

∂Φρ(x, y)

∂ν(x)ψ(y) ds(y), x ∈ Γ.

assumption: f(x) ≥ f0 > 0 for all x ∈ Γ

uniqueness of transmission problem:

reflection argument (Rapun & Sayas, submitted)

+ uniqueness of transmission problem in Rd (Kress & Roach, ’78)

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self-adjointness

inner product on L2(Ω− × Ω+):〈u, v〉 :=

∫

Ω−

(α/κ−)u− v−dx+∫

Ω+(1/κ+)u+ v+dx

operator:

Au :=

κ−∆u, in Ω−,

κ+∆u, in Ω+

domain of A : D(A) → L2(Ω− × Ω+):

D(A) :=

u ∈ H2(Ω− ∪ Ω+) :u|Γ,− + f ∂νu|Γ,− − u|Γ,+ = 0,α ∂νu|Γ,− − ∂νu|Γ,+ = 0,∂νu|H = 0

.

Theorem: A is self-adjoint, and σ(A) ⊂ (−∞, 0].

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4. convergence analysis and

numerical results

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bounds on boundary integral operators explicit in s

Theorem: Let W (s) : Hν(Γ) ×Hν(Γ) → Hν(Γ) ×Hν(Γ) denotethe matrix of integral operators. Then for all k ∈ −1, 0, 1, . . . there exists a constant C > 0 such that

‖W (s)−1‖k+1/2 ≤ C(1 + |s|)2⌊(k+1)/2⌋+4

dist (s, (−∞, 0])2for all s ∈ C\(−∞, 0]

and

‖W (s)‖±1/2 ≤ C(1 + |s|)2

dist (s, (−∞, 0])for all s ∈ C \ (−∞, 0].

proof. Using standard bounds on resolvent of A . . .

alternative approach by Lubich and Schneider ’92 forV : H−1/2(Γ) → H1/2(Γ) using estimates of kernel andparameter-dependent pseudodifferential calculus.

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convergence result

space discretization: We consider a Galerkin method withcontinuous, piecewise polynomial functions of degree ≤ k on aquasi-uniform triangulation of Γ with mesh size h.

Theorem: Let uN,h be the numerical approximation of u meshsize h and 2N + 1 quadrature points for the Laplace inversion.Then there exists constants c, C1, C2 > 0 for each compact timeinterval [t0, T ] ⊂ (0,∞) such that

‖u(t) − uN,h(t)‖H1(Ω−∪Ω+) ≤ C1ecN/ ln N + C2h

k+3/2 for all t ∈ [t0, T ],

and on the top barrier

‖u(t) − uN,h(t)‖L∞(H) ≤ C1ecN/ ln N + C2h

2k+2 for all t ∈ [t0, T ].

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numerical tests

source

numerical experiments by Maria-Luisa Rapun

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convergence as N → ∞

2 2.5 3 3.5 4

2.2

2.4

2.6

2.8

3

3.2

log(N)

log(

−lo

g(E

))

N rel. error5 3.18 e-410 1.37 e-415 1.16 e-520 1.44 e-625 2.55 e-730 1.94 e-835 5.46 e-940 2.93 e-1045 1.18 e-1050 5.32 e-1255 2.98 e-1260 2.46 e-13

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time dependence of the error

0.01 0.1 1 10 100

−15

−10

−5

0

5

10

15

20

25

t

log(

E)

Total error

λ=10λ=1λ=0.1

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