On Hyperbolic Initial-Boundary Value Problems with a Strictly Dissipative Boundary Condition

Abstract

The well-posedness of the initial-boundary value problems for symmetric hyperbolic systems with strictly dissipative boundary conditions is proved. The regularity assumptions on the coefficients of the differential operator and the boundary condition as well as the boundary itself are quite minimal. Characterizations of strictly dissipative boundary operators are given and the example of Maxwell’s equations is discussed.

Appendix

The singular value decomposition of B is B = UTV ^H where $V\in L^\infty (\Sigma ,\mathbb {C}^{N\times N})$ and $U \in L^\infty (\Sigma ,\mathbb {C}^{p\times p})$ are unitary matrices and $T \in L^\infty (\Sigma , \mathbb {R}^{p\times N})$ is of the form

since B is strictly dissipative. The singular values $s_1,\ldots ,s_{N_-}$ are uniformly positive a.e. (t, x) ∈ Σ. Define now E = −UT ^# V ^H A where

The unitary matrix V = [v ₁ … v _N] is structured as follows. The first N ₋ columns span N(B)^⊥ and the last N − N ₊ − N ₋ columns span N(A). This can be done since N(A) ⊂ N(B). Introduce two N ₊ × N matrices G and F by

and observe that N(G) = N(B)^⊥⊕ N(A). One computes

and

since the first N ₋ + N ₋ columns of V  are an orthonormal basis of N(A)^⊥ = Im A. This proves the first statement.

For the second statement we start by proving N(F)^⊥ = AN(B) and N(F) = [AN(B)]^⊥. Let y ∈ N(F). Then GAy = 0, that is Ay ∈ N(G) and hence, 〈y, Az〉 = 〈Ay, z〉 = 0 for all z ∈ N(B). This shows N(F) ⊂ [AN(B)]^⊥. Now let y ∈ [AN(B)]^⊥. Then 0 = 〈y, Az〉 for all z ∈ N(B), which implies Ay ∈ N(G). But this means y ∈ N(F). Note that $\dim N(F)^\perp = N_+$.

Let z ∈ N(B) ∩ N(F). Then 〈Az, z〉 = 0 and by the strictly dissipativity of B we have 〈Az, z〉≳|Az|². Hence Az = 0 and we have established N(A) = N(B) ∩ N(F). Let now $w_1,w_2,..,w_{N_-},v_{N_+ + N_-+1},\ldots ,v_N$ be a basis for N(F). Then $w_1,w_2,..,w_{N_-},v_{N_-+1},\ldots ,v_N$ is a basis for $\mathbb {C}^N$ and the square matrix W, whose columns are the vectors of this basis, block-diagonalizes A, that is

By the strict dissipativity of B the matrix A ₊ is uniformly positive definite. Hence, since the matrix A is assumed to have constant signature independent of (t, x) ∈ Σ almost everywhere, we infer that A ₋ must be uniformly negative definite. Thus $\langle Az,z \rangle \lesssim -|Az|{ }^2$ for all z ∈ N(F).

Finally, observe that F ^H F = AG ^H GA where the matrix G ^H G is the orthogonal projection onto N(B) ⊖ N(A). By the strict dissipativity A is uniformly positive definite on N(B) ⊖ N(A). Thus, the matrix F ^H F has N ₊ positive eigenvalues bounded away from zero, a.e. (t, x) ∈ Σ.