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The Helmholtz Equation Seminar BEM on Wave Scattering Rene Rühr ETH Zürich October 28, 2010

Outline Steklov-Poincare Operator Helmholtz Equation: From the Wave equation to Radiation condition Uniqueness of the Exterior Dirichlet Problem Boundary Integral Operators Representation formula Existence of the Exterior Dirichlet Problem Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 2 / 34

Poincare - Steklov Operator Last time we saw the Calderon projection, enabling us to write the following system of BIE for the Cauchy data γ D u, γ N u ( ) ( γd u 1 = 2 I K ) ( ) 0 V 0 γd u 1 γ N u W 0 2 I + K 0 γ N u (here for for homogeneous case u = 0) where V 0 : H 1 2 (Γ) H 1 2 (Γ) K 0 : H 1 2 (Γ) H 1 2 (Γ) K 0 : H 1 2 (Γ) H 1 2 (Γ) W 0 : H 1 2 (Γ) H 1 2 (Γ) Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 3 / 34

Steklov - Poincare Operator We proved (in 3d) V 0 : H 1 2 (Γ) H 1 2 (Γ) is elliptic invertible! Rewrite first equation: γ N u = V 1 0 ( 1 2 I + K 0)γ D u Define S 0 := V 1 0 ( 1 2 I + K 0) : H 1 2 (Γ) H 1 2 (Γ)...called Steklev-Poincare-Operator S 0 is a Dirichlet-to-Neumann map Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 4 / 34

Steklov - Poincare Operator Second equation of the Calderon projection admits another representation: S 0 = W 0 + ( 1 2 I + K 0)V 1 0 ( 1 2 I + K 0) S 0 admits the same ellipticity estimates as W 0 : Sv, v Γ W 0 v, v Γ H 1 2 (Γ)-elliptic. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 5 / 34

Wave equation reduces to Helmholtz Equation The wave equation is 2 t 2 Ψ c2 Ψ = 0 assume the solution to be time harmonic: Ψ(t, x) = e iwt u(x) then u will satisfy w 2 u c 2 u = 0 or u k 2 u = 0 with k = w c... called Helmholtz equation. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 7 / 34

Wave equation reduces to Helmholtz Equation Fundamental solution of u k 2 u = 0 in R 3 G k (x, y) = 1 e ik x y 4π x y x y also 1 e ik x y 4π x y is a fundamental solution - justify our choice later Just as for the Laplace, in n=2, things are a little bit different, Y 0 the second kind Bessel function of order zero, G k (x, y) = 1 2π Y 0(k x y ) x y whose singularity at x y behaves like the log s as s 0 we study the Exterior Dirichlet Problem (EDP) for a bounded domain Ω R 3 u k 2 u = 0 in Ω c and u = g in Ω = Γ Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 8 / 34

Radiation condition Problem: in general won t be unique... for consider A eikr r Helmholtz equation and B e ikr r spherical waves will both satisfy the want our solution of the EDP also behaves like an outgoing wave! so let s look for a condition only u = A eikr r satisfies. r u iku = o( 1 r 2 ) as r Sommerfeld Radiation Condition! Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 9 / 34

Radiation condition consider u k 2 u = f in R n. u = G k f is a solution.. that satifies, putting x = rω A(ω) eikr r + o(r 2 ) as r right choice of sign! Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 10 / 34

Uniqueness Uniqueness of the EDP! Theorem There is at most one radiating solution u H 1 loc (Ωc ) of where g H 1 2 (Γ). u k 2 u = 0 on Ω c γd c u = g on Γ, Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 12 / 34

Uniqueness need the following Lemma established by Rellich: Lemma Let k > 0 and u a solution of u k 2 u = 0 on B ρ0 (0) c and suppose that then u = 0 on B ρ0 (0) c. lim u(x) 2 dσ = 0 ρ x =ρ Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 13 / 34

Uniqueness Corollary and, with a little work, it follows that Let u H 1 loc (Ωc ) be a solution of the homogeneous exterior Helmholtz equation, u k 2 u = 0 on Ω c and satisfies in addition to the Sommerfeld radiation condition also ( ) I k (γ Nū)(γc c D u) dσ 0, then u = 0 on Ω c. Γ Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 14 / 34

Uniqueness Proof: Let s bound our domain: Ω c ρ = Ω c B ρ Green s 1st Ω c ρ ū u k 2 ūu dx = ( k 2 )u, u L 2 (Ω c ρ) + γ N u, γ D u L 2 ( Ω c ρ ) first term vanishes by assumption. Multiply with k and take the imaginary part ( I(k) u 2 k 2 u 2 dx = I k ū ) ( B ρ ν u dσ I k ū ) Γ ν u dσ Ω c ρ Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 15 / 34

Uniqueness last integral can be rewritten since u ρ iku 2 = u ρ I(k) u 2 + k 2 u 2 dx + 1 Ω c 2 ρ + u B ρ ρ iku I(k) > 0 Ω c ρ u 2 dx 0 I(k) = 0 Rellich applies 2 ( + k 2 u 2 + 2I k ū ) ν u B ρ 2 u 2 ρ + k 2 u 2 dσ = ( dσ I k ū ) ν u dσ. Γ Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 16 / 34

Integral operators and Boundary Integral Equations As for the Laplace, we define the boundary integral operators as (V k w)(x) = G k (x, y)w(y)ds y Γ (K k v)(x) = G k (x, y)v(y)ds y Γ n y (K k v)(x) = G k (x, y)v(y)ds y Γ n x (W k v)(x) = G k (x, y)v(y)ds y n x n y Γ Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 18 / 34

Integral operators and Boundary Integral Equations with the same properties as for the Laplace Theorem For a bounded Lipschitz domain, the boundary integral operators are continous. Remark V k : H 1 2 (Γ) H 1 2 (Γ) K k : H 1 2 (Γ) H 1 2 (Γ) K k : H 1 2 (Γ) H 1 2 (Γ) W k : H 1 2 (Γ) H 1 2 (Γ) One has the equalities W k = K k + 1 2 Id and V k = K k + 1 2 Id. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 19 / 34

Integral operators and Boundary Integral Equations Lemma well... not exactly the same properties: V k no longer elliptic! but we still have V k : H 1 2 (Γ) H 1 2 (Γ) is coercive, i.e. there exists a compact operator C : H 1 2 (Γ) H 1 2 (Γ) such that the Gardings inequality V k w, w L 2 (Γ) + Cw, w L 2 (Γ) const w 2 H 1 2 (Γ) w H 1 2 (Γ) holds. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 20 / 34

Integral operators and Boundary Integral Equations Theorem the single and double layer potential give us radiating solution: If φ H 1 2 (Γ) then for u = Ψ SL φ satisfies the Sommerfeld radiation condition and ( k 2 )u = 0 in R n \ Γ Analogously for u = Ψ DL φ if φ H 1 2 (Γ). Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 21 / 34

Integral operators and Boundary Integral Equations suppose u H 1 loc (Ωc ) with ( k 2 )u = 0 in Ω Ω c satisfies u = Ψ SL [γ N u] + Ψ DL [γ D u] in Ω Ω c Integral representation formula take traces to get the exterior Calderon projection ( ) ( γ c D u 1 γn c u = 2 I + K ) ( ) k V k γ c D u 1 W k 2 I K k γn c u different signs w.r.t to the interior Calderon Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 22 / 34

Integral operators and Boundary Integral Equations boundary integral equations to the exterior Dirichlet problem: Direct Method - integral equality of the first kind: Let g H 1 2 (Γ). Find ψ H 1 2 (Γ) such that V k ψ, ϕ L 2 (Γ) = 1 2 g, ϕ L 2 (Γ) + K k g, ϕ L 2 (Γ) ϕ H 1 2 (Γ) Direct Method - integral equality of the second kind: Let g H 1 2 (Γ). Find ψ H 1 2 (Γ) such that 1 2 ψ, ϕ L 2 (Γ) + K kψ, ϕ L2 (Γ) = W k g, ϕ L2 (Γ) ϕ H 1 2 (Γ) Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 23 / 34

Integral operators and Boundary Integral Equations Indirect Method - using the single layer potential: Let g H 1 2 (Γ). Find ψ H 1 2 (Γ) such that V k ψ, ϕ L2 (Γ) = g, ϕ L2 (Γ) ϕ H 1 2 (Γ) Indirect Method - using the double layer potential: Let g H 1 2 (Γ). Find ψ H 1 2 (Γ) such that 1 2 ψ, ϕ L 2 (Γ) + K k ψ, ϕ L2 (Γ) = g, ϕ L2 (Γ) ϕ H 1 2 (Γ) Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 24 / 34

Representation Formula for the exterior Calderon operator we assumed Greens representation formula.. does it hold? Theorem yes - thanks to the radiating property! Let g H 1 2 (Γ). And suppose u H 1 loc (Ωc ) is a radiating solution of u k 2 u = 0 on Ω c then u has the integral representation γd c u = g on Γ, u = Ψ DL g Ψ SL (γ c N u). Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 26 / 34

Representation Formula Proof idea: Ω c ρ = Ω c B ρ Green u(x) = +Ψ SL (γn c u)(x) Ψ DL(γD c u)(x) G(x, y)γ N u(y) dσ + γ N G(x, y)γ D u(y) dσ B ρ B ρ Sommerfeld radiation condition two last terms vanish as ρ 0 Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 27 / 34

Existence Theorem we re ready to prove the existence!..by showing that we can solve the BIE. so we need Let g H 1 2 (Γ). And suppose u H 1 loc (Ωc ) is a radiating solution of u k 2 u = 0 on Ω c γ c Du = g on Γ, then φ = γ c N H 1 2 (Γ) is a solution of the boundary integral equation and u has the integral representation V k φ = ( Id /2 + K)g on Γ, (1) u = Ψ DL g Ψ SL φ. (2) Conversely, if φ H 1 2 (Γ) is a solution (1), then formula (2) defines a solution u H 1 loc (Ωc ) of the exterior Dirichlet problem. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 29 / 34

Existence Theorem the wave number k has influence on the BIE, namely V k is injective on H 1 2 (Γ) iff k 2 is not an eigenvalue of of the interior Dirichlet problem, i.e. u = k 2 u on Ω, γ D u = 0 u = 0 on Ω The kernel of V k is given by ker(v k ) = span [ γ N v : v = k 2 v on Ω and γ D v = 0 on Γ ] Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 30 / 34

Existence Theorem to finalize, we also need the Fredholm Alternative Let A B(X, Y ) coercive. If Au = 0 only allows the trivial solution u = 0, then Au = f is uniquely solvable for all f Y. Else Au = f is solvable iff v, f = 0 for all v Y : A v = 0. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 31 / 34

Existence so at last Theorem Let Ω be a bounded Lipschitz domain with boundary Γ. Then for every g H 1 2 (Γ) the exterior Dirchlet problem u k 2 u = 0 on Ω c γ c D u = g on Γ has a unique solution u H 1 loc (Ωc ) that satisfies the Sommerfeld radiation condition. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 32 / 34

Existence need to solve V k φ = ( Id /2 + K k )g on Γ V k coercive, so apply Fredholm! k 2 not an eigenvalue Fredholm V k invertible Else: need to check solvability condition w, ( Id /2 + K k )g L 2 (Γ) = 0 for all w ker(v k ) What s ker(vk )? Can modifiy thm about ker(v k)! [ ] ker(v ) = span γ N v : v = k 2 v on Ω, γ D v = 0 on Γ Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 33 / 34

Existence so take such a γ N v and apply Green s Identity γ N v, ( 1 2 I + K k)g L 2 (Γ) = γ N v, γ D (Ψ DL g) L 2 (Γ) = γ D v, γ N (Ψ DL g) L 2 (Γ) ( k 2 )v, Ψ DL g L 2 (Ω) + v, ( k 2 )Ψ DL g L 2 (Ω) = 0...since γ D v = 0 on Γ, ( k 2 )v = 0 and ( k 2 )Ψ DL g = 0 on Ω. Rene Rühr (ETH Zürich) The Helmholtz Equation October 28, 2010 34 / 34