A backward uniqueness result for the wave equation with absorbing boundary conditions

We consider the wave equation $u_{tt}=\Delta u$ on a bounded domain $\Omega\subset{\mathbb R}^n$, $n>1$, with smooth boundary of positive mean 
curvature. On the boundary, we impose 
the absorbing boundary condition ${\partial u\over\partial\nu}+u_t=0$. We prove uniqueness of solutions backward in time.


1.
Introduction. Well-posed initial-value problems for partial differential equations need not satisfy backward uniqueness, for instance, for the problem posed on the unit interval, all solutions become equal to zero in finite time. It is therefore of interest to characterize problems for which backward uniqueness holds. The property of backward uniqueness, aside from being of interest in its own right, plays a role in a number of results on unique continuation and control. For instance, approximate controllability follows from null controllability if backward uniqueness holds for an adjoint problem. Also, for many systems, results on approximate controllability can be deduced from backward uniqueness in conjunction with a characterisitics argument. A number of methods exist for proving backward uniqueness. Backward uniqueness is obvious for C 0 -groups and analytic semigroups. Other approaches include the logarithmic convexity method and Carleman estimates. In this work, we shall use the method introduced in [4], which is based on the Phragmen-Lindelöf theorem. This method was applied in [4] to thermoelastic plates. It has since been used for a number of other systems which also involve a coupling of hyperbolic and parabolic equations [1,2,3]. These examples are in a sense perturbations of a system (the uncoupled hyperbolic and parabolic equations) for which backward uniqueness is easy, and the task is to show that the perturbation imposed by coupling preserves this property. In a recent paper by the author [5], it was shown that the method can also be applied to certain problems where the "unperturbed" problem does not have backward uniqueness. The examples discussed in [5] included a linearized problem for one-dimensional compressible flow and the damped wave equation on an interval with an absorbing boundary condition. Note that, with this boundary condition, backward uniqueness does not hold for the undamped wave equation. The damping term, although it is a lower order perturbation, is responsible for the uniqueness property.
The abstract result for backward uniqueness, as proved in [5], is the following.
Theorem 1. Let A be the infinitesimal generator of a C 0 -semigroup on a Banach space X. Suppose that there exists θ with π/2 < θ < 3π/2 and R such that for λ = r exp(iθ) and r > R we have the resolvent estimate where ρ < 1. Then exp(At) is injective for every t > 0.
In this paper, we consider the wave equation in more than one space dimension. In this case, it turns out that backward uniqueness actually holds without damping. This is because "absorbing" boundary conditions are actually not exactly absorbing: A radially propagating wave is not exactly of the form u(r−t) and therefore does not satisfy a boundary condition u r + u t = 0 on the boundary of a ball. The curvature of the boundary will play the same role as damping plays in one dimension. Let Ω be a bounded domain in R n with a smooth boundary of positive mean curvature. We consider the following initial boundary value problem: Here ∂u/∂ν denotes the derivative in the direction of the outer normal. Our goal is the following result: Theorem 2. Let u be a solution of (3). If, for any time T > 0, u(·, T ) = u t (·, T ) = 0, then u = 0.
In the usual fashion, we associate our problem with a C 0 -semigroup on H 1 (Ω) × L 2 (Ω). Let u t = v, and define In order to apply Theorem 1, we need to consider the resolvent problem (A − λI)(u, v) = (f, g). We can combine this problem into the single equation and boundary condition In view of Theorem 1, our result follows from the following.
Theorem 3. Let θ be any angle in (π/2, π). For λ = r exp(iθ) and r sufficiently large, a unique solution of (5) exists and satisfies a bound of the form 2. Iterative construction. We shall construct the solution of the resolvent problem (5) in the form of an infinite series u = ∞ n=0ũ n . At each step of the iteration, we start from a problem ∆u n − λ 2 u n = g n + λf n , ∂u n ∂ν + λu n = −f n on ∂Ω.
We then find an approximate solutionũ n for u n and end up with a new problem for the residual u n+1 = u n −ũ n . The convergence of the iteration follows from estimates of the form and We now describe the construction ofũ n . For convenience, we suppress the index n, i.e. f shall stand for f n etc. Let G denote a smooth bounded domain containing Ω in its interior. Let E denote an extension operator which is bounded from If λ is on a ray in the complex plane other than the imaginary axis, it is easy to Next, we set u = v + w, so that the problem for w becomes From the estimate (12), it follows that The idea in finding an approximation for w is that, for large |λ|, w will be essentially confined to a small neighborhood of the boundary ∂Ω. We shall need a cut-off function ψ which is smooth, taking values between 0 and 1, having support in a sufficiently small neighborhood of ∂Ω, and equal to 1 in an even smaller neighborhood of ∂Ω. Below we shall construct an operator L which in a sense approximates the Laplacian in a neighborhood of ∂Ω. We approximate w byw wherew satisfies on the support of ψ and ∂w ∂ν on ∂Ω. Finally, we setũ = v + ψw. For the next iterate, we then have f n+1 = 0, and To complete our argument, we need to show an estimate of the form

MICHAEL RENARDY
where γ → 0 as |λ| → ∞. We note that, from (17), we find 3. Construction of a boundary layer approximation. To construct the operator L, we make a change of coordinates near ∂Ω. Let ξ denote the distance of a point from ∂Ω, and let η be a surface coordinate identifying the nearest point on ∂Ω. In these coordinates, the domain Ω corresponds to ξ > 0, and we have where κ denotes the sum of principal curvatures of the boundary, ∆ S is the surface Laplacian on ∂Ω, and Q is a differential operator which is first order in ξ, and second order in η. We now set We verify that where R is a differential operator of first order in η. In summary, we have As a consequence, we have We now consider the solution of for ξ > 0, subject to the boundary condition for ψ = 0. We substitutew = exp(κ(η)ξ/2)ŵ. Consequently, we obtain the equation with the boundary condition Let µ n denote the eigenvalues of −∆ S , and let φ n denote the corresponding eigenfunctions, normalized in L 2 (∂Ω). We can respresentŵ aŝ w = ∞ n=1 a n exp(− λ 2 + µ n ξ)φ n (η). (29) Here the branch of the square root is the one taking values in the right half plane. The boundary condition becomes Let α n = |λ| + √ µ n and q n = λ 2 + µ n + λ. Up to equivalence of norms, we have the following: We fix integers m and M (depending on λ) such that m = β 1 |λ| M = β 2 |λ| 2 , with β 1 , β 2 chosen sufficiently small but independent of λ so that in particular Re q n < 1 4 min ∂Ω κ for µ n < m. We divide the natural numbers into three subsets: Recall that |λ| is assumed large, and π/2 < arg λ < π. We note the following facts about q n : 1. On I 1 , |q n | is small of order m/|λ|. 2. On I 2 , the leading contribution to q n is −µ n /(2λ). In particular, this implies that real and imaginary parts of q n have the same order of magnitude. 3. On I 3 , the real part of q n is positive and dominates over the imaginary part.
Due to the positivity of mean curvature, there are constants such that Moreover, we note for the following that, for some constant, Below, we also need to consider where we set h k = I k a n φ n .
As before, we have For the remaining term, however, we first integrate by parts: Here we have used that µ n ≤ m on I 1 and µ n ≥ M on I 3 .
We now multiply (30) by h(η), integrate over ∂Ω and isolate the term − ∂Ω κ|h| 2 /2 dη on the left side. The result is an estimate of the form Next, we write the equation (30) in the form ∞ n=1 q n a n φ n = ∞ n=1 b n φ n + κ 2 ∞ n=1 a n φ n .
We multiply by −∆ S h(η), integrate, and then take the imaginary part. Since the imaginarty part of q n is always positive, we find I2 µ n |a n | 2 (Im q n ) ≤ Im This results in a bound of the form Here we are free to choose K as large as we wish. Finally, we multiply (41) by −∆ S h 3 (η) and integrate, resulting in Taking the real part, we find a bound of the form We now note the assumptions on the size of µ n on the subsets I k . If we choose K suitably large, we can combine the inequalities (40), (43) and (45) to obtain The rest now follows from this bound.

Remarks.
1. The argument given above proves that the resolvent is bounded for large |λ|.
Hence, in contrast to the application to linearized compressible flow in [5], the full force of the abstract result proved there is not needed; the theorem from [4] is sufficient. However, I note that the norm of the resolvent does not decay at a rate of |λ| −1 . 2. The argument above works just as well if a component of the boundary has uniformly negative rather than positive mean curvature. Thus the result can easily be extended to domains with holes or exterior domains. 3. The argument can easily be modified to cover a damped wave equation u tt + γu t = ∆u. In that case the role of κ in the analysis above is taken over by κ + γ. Thus the theorem holds under the assumption that κ + γ is either uniformly positive or uniformly negative on each component of the boundary. 4. In [5], Theorem 1 is also applied to linearized one-dimensional compressible flow. In a way, this problem is more interesting than the wave equation, since the resolvent along a ray in the left half plane is actually not bounded, but grows like exp(C|λ| 1/2 ). Unfortunately, this result does not seem to extend to higher dimensions. A case amenable to analysis is that of a parallel strip, where separation of variables can be done. In this case, backward uniqueness holds, since each transverse wave number can be analyzed separately, and that analysis is analogous to the one-dimensional case. However, I have done a formal analysis of the eigenvalues which shows that we cannot expect any sector in the left half plane to be free from eigenvalues. Thus a direct application of Theorem 1 to the two-dimensional problem is not possible.