Hölder stability for an inverse medium problem with internal data

Abstract

We are interested in an inverse medium problem with internal data. This problem is originated from multi-waves imaging. We aim in the present work to study the well posedness of the inversion in terms of the boundary conditions. We precisely show that we have actually a stability estimate of Hölder type. For sake of simplicity, we limited our study to the class of Helmholtz equations $\Delta + V$ with bounded potential V.

Appendix A

In this appendix, $\Omega$ is a bounded domain of $\mathbb {R}^n$, $n\ge 2$, with Lipschitz boundary $\Gamma$. Let

$$\begin{aligned} L=\text{ div }(A\nabla \, \cdot )+V, \end{aligned}$$

where $V\in L^\infty (\Omega )$, $A=(a^{ij})$ is a symmetric matrix with coefficients in $W^{1,\infty }(\Omega )$ and there exist $\kappa >0$ and $\Lambda >0$ so that

$$\begin{aligned} A(x)\xi \cdot \xi \ge \kappa |\xi |^2, \quad x\in \Omega , \quad \xi \in \mathbb {R}^n, \end{aligned}$$

(A.1)

and

$$\begin{aligned} \Vert V\Vert _{L^\infty (\Omega )}+\Vert a^{ij}\Vert _{W^{1,\infty }(\Omega )}\le \Lambda ,\quad 1\le i,j\le n. \end{aligned}$$

(A.2)

Recall the following three-ball interpolation inequality, proved in [11] when $V=0$ but still holds for any bounded V (see also [12]).

Theorem A.1

Let $0<k<\ell <m$. There exist $C>0$ and $0<s <1$, only depending on $\Omega$, k, $\ell$, m, $\kappa$ and $\Lambda$, such that

$$\begin{aligned} \Vert v\Vert _{L^2(B(y,\ell r))}\le C\Vert v\Vert _{L^2(B(y,kr))}^s\Vert v\Vert _{L^2(B(y,m r))}^{1-s}, \end{aligned}$$

(A.3)

for all $v\in H^1(\Omega )$ satisfying $Lv=0$ in $\Omega$, $y\in \Omega$ and $0<r< \text{ dist }(y,\Gamma )/m$.

We know from [19, Theorem 2.4.7, page 53] that any Lipschitz domain has the uniform interior cone condition, abbreviated to UICP in the sequel. In particular, there exist $R>0$ and $\theta \in \left]0,\frac{\pi }{2}\right[$ so that, to any $\tilde{x}\in \Gamma$ corresponds $\xi =\xi (\tilde{x})\in \mathbb {S}^{n-1}$ for which

$$\begin{aligned} \mathcal {C}(\tilde{x})=\{x\in \mathbb {R}^n; 0<|x-\tilde{x}|<R, (x-\tilde{x})\cdot \xi >|x-\tilde{x}|\cos \theta \}\subset \Omega . \end{aligned}$$

Define the geometric distance $d_g^D$, on a bounded domain D of $\mathbb {R}^n$, by

$$\begin{aligned} d_g^D(x,y)=\inf \left\{ \ell (\psi ) ; \psi :[0,1]\rightarrow D\;\text{ Lipschitz } \text{ path } \text{ joining } \; x\; \text{ to } \, y\right\} , \end{aligned}$$

where

$$\begin{aligned} \ell (\psi )= \int _0^1|\dot{\psi }(t)|\hbox {d}t \end{aligned}$$

is the length of $\psi$.

Note that, according to Rademacher’s theorem, any Lipschitz continuous function $\psi :[0,1]\rightarrow D$ is almost everywhere differentiable with $|\dot{\psi }(t)|\le k$ a.e. $t\in [0,1]$, where k is the Lipschitz constant of $\psi$.

Lemma A.1

Let D be a bounded Lipschitz domain of $\mathbb {R}^n$. Then $d_g^D\in L^\infty (D \times D )$.

A proof of this lemma can be found in [16].

In the rest of this text

$$\begin{aligned} \mathbf {d}_g=\Vert d_g^\Omega \Vert _{L^\infty (\Omega \times \Omega )}. \end{aligned}$$

Proof

(Proof of Theorem 2.1) In this proof C denotes a generic constant that can depend only on $\Omega$, $v_0$, $V_0$, $\kappa$ and M.

Step 1. Let $y, y_0\in \Omega ^{3\delta }$ and $\psi :[0,1]\rightarrow \Omega$ be a Lipschitz path joining $y_0$ to y so that $\ell (\psi )\le d_g^\Omega (y_0,y)+1$. Let $t_0=0$ and $t_{k+1}=\inf \{t\in [t_k,1]; \psi (t)\not \in B(\psi (t_k),\delta )\}$, $k\ge 0$. We claim that there exists an integer $N\ge 1$ so that $\psi (1)\in B(\psi (t_N),\delta )$. If not, we would have $\psi (1)\not \in B(\psi (t_k),\delta )$ for any $k\ge 0$. As the sequence $(t_k)$ is nondecreasing and bounded from above by 1, it converges to $\hat{t}\le 1$. In particular, there exists an integer $k_0\ge 1$ so that $\psi (t_k)\in B\left( \psi (\hat{t}),\delta /2\right)$, $k\ge k_0$. But this contradicts the fact that $\left| \psi (t_{k+1})-\psi (t_k)\right| =\delta$, $k\ge 0$.

Let us check that $N\le N_0$, where $N_0$ only depends on $\mathbf {d}_g$ and $\delta$. If $\psi =(\psi _1,\ldots ,\psi _n)$, we pick $1\le j\le n$ so that

$$\begin{aligned} \max _{1\le i\le n} \left| \psi _i(t_{k+1})-\psi _i(t_k)\right| =\left| \psi _j(t_{k+1})-\psi _j(t_k)\right| . \end{aligned}$$

Then

$$\begin{aligned} \delta \le n\left| \psi _j (t_{k+1})-\psi _j(t_k)\right| =n\left| \int _{t_k}^{t_{k+1}}\dot{\psi }_j(t)\hbox {d}t\right| \le n\int _{t_k}^{t_{k+1}}|\dot{\psi }(t)|\hbox {d}t . \end{aligned}$$

Consequently, where $t_{N+1}=1$,

$$\begin{aligned} (N+1)\delta \le n\sum _{k=0}^N\int _{t_k}^{t_{k+1}}|\dot{\psi }(t)|\hbox {d}t=n\ell (\psi )\le n(\mathbf {d}_g+1). \end{aligned}$$

Therefore

$$\begin{aligned} N\le N_0=\left[ \frac{n(\mathbf {d}_g+1)}{\delta }\right] . \end{aligned}$$

Here $\left[ n(\mathbf {d}_g+1)/\delta \right]$ is the integer part of $n(\mathbf {d}_g+1)/\delta$.

Let $y_k=\psi (t_k)$, $0\le k\le N$. If $|z-y_{k+1}|<\delta$ then

$$\begin{aligned} |z-y_k|\le |z-y_{k+1}|+|y_{k+1}-y_k|<2\delta . \end{aligned}$$

In other words, $B(y_{k+1},\delta )\subset B(y_k,2\delta )$.

We get from Theorem A.1

$$\begin{aligned} \Vert u\Vert _{L^2(B(y_j,2\delta ))}\le C\Vert u\Vert _{L^2(B(y_j,3\delta ))}^{1-s}\Vert u\Vert _{L^2(B(y_j,\delta ))}^s,\quad 0\le j\le N. \end{aligned}$$

(A.4)

Set $I_j=\Vert u\Vert _{L^2(B(y_j,\delta ))}$, $0\le j\le N$ and $I_{N+1}=\Vert u\Vert _{L^2(B(y,\delta ))}$. Since $B(y_{j+1},\delta )\subset B(y_j,2\delta )$, $1\le j\le N-1$, estimate (A.4) implies

$$\begin{aligned} I_{j+1}\le C M_0^{1-s}I_j^s,\quad 0\le j\le N, \end{aligned}$$

(A.5)

where we set $M_0=\Vert u\Vert _{L^2(\Omega )}$.

Let $C_1=C^{1+s+\ldots +s^{N}}$ and $\beta =s^{N+1}$. Then, by a simple induction argument, estimate (A.5) yields

$$\begin{aligned} I_{N+1}\le C_1M_0^{1-\beta }I_0^\beta . \end{aligned}$$

(A.6)

Without loss of generality, we assume in the sequel that $C\ge 1$ in (A.5). Using that $N\le N_0$, we have

$$\begin{aligned}&\beta \ge \beta _0=s^{N_0+1}\ge se^{-\frac{\kappa }{\delta }}=\psi (\delta ), \, \text{ where } \quad \kappa =n(\mathbf {d}_g+1)|\ln s|,\\&C_1\le C^{\frac{1}{1-s}},\\&\left( \frac{I_0}{M_0}\right) ^\beta \le \left( \frac{I_0}{M_0}\right) ^{\beta _0}. \end{aligned}$$

These estimates in (A.6) gives

$$\begin{aligned} \frac{I_{N+1}}{M_0}\le C\left( \frac{I_0}{M_0}\right) ^{\psi (\delta )}. \end{aligned}$$

In other words,

$$\begin{aligned} \frac{\Vert u\Vert _{L^2(B(y,\delta ))}}{\Vert u\Vert _{L^2(\Omega )}}\le C\left( \frac{\Vert u\Vert _{L^2(B(y_0,\delta ))}}{\Vert u\Vert _{L^2(\Omega )}}\right) ^{\psi (\delta )}. \end{aligned}$$

Applying Young’s inequality, we get from this inequality

$$\begin{aligned} \Vert u\Vert _{L^2(B(y,\delta ))}\le C\left( \epsilon ^{\frac{1}{1-\psi (\delta )}}\Vert u\Vert _{L^2(\Omega )}+\epsilon ^{-\frac{1}{\psi (\delta )}}\Vert u\Vert _{L^2(B(y_0,\delta ))}\right) , \end{aligned}$$

(A.7)

$\epsilon >0$, $y,y_0\in \Omega ^{3\delta }$.

Step 2. Fix $\tilde{x}\in \Gamma$ so that $|u(\tilde{x})|=\Vert u\Vert _{L^\infty (\Gamma )}$. Let $\xi =\xi (\tilde{x})$ be as in the definition of the UICP. Let $x_0=\tilde{x}+\delta \xi$, $\delta \le R/2$, $d_0=|x_0-\widetilde{x}|=\delta$ and $\rho _0=d_0\sin \theta /3$. Note that $B(x_0,3\rho _0)\subset \mathcal {C}(\tilde{x})$.

By induction in k, we construct a sequence of balls $(B(x_k, 3\rho _k))$, contained in $\mathcal {C}(\tilde{x})$, as follows

$$\begin{aligned} \left\{ \begin{array}{ll} x_{k+1}=x_k-\alpha _k \xi ,\\ \rho _{k+1}=\mu \rho _k,\\ d_{k+1}=\mu d_k, \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} d_k=|x_k-\tilde{x}|, \quad \rho _k=\vartheta d_k, \quad \alpha _k=(1-\mu )d_k , \end{aligned}$$

with

$$\begin{aligned} \vartheta =\frac{\sin \theta }{3},\quad \mu =\frac{3-2\sin \theta }{3-\sin \theta } . \end{aligned}$$

Note that this construction guarantees that, for each k, $B(x_k,3\rho _k)\subset \mathcal {C}(\tilde{x})$ and

$$\begin{aligned} B(x_{k+1},\rho _{k+1})\subset B(x_k,2\rho _k). \end{aligned}$$

(A.8)

We get, by applying Theorem A.1, that there exist $C>0$ and $0<s<1$, only depending on $\Omega$, $v_0$ and $V_0$, so that

$$\begin{aligned} \Vert u\Vert _{L^2(B(x_k,2\rho _k))}&\le C\Vert u\Vert _{L^2(B(x_k,3\rho _k))}^{1-s}\Vert u\Vert _{L^2(B(x_k,\rho _k))}^s \\&\le CM^{1-s}\Vert u\Vert _{L^2(B(x_k,\rho _k))}^s.\nonumber \end{aligned}$$

(A.9)

In light of (A.8), (A.9) gives

$$\begin{aligned} \Vert u\Vert _{L^2(B(x_{k+1},\rho _{k+1}))}\le CM^{1-s}\Vert u\Vert _{L^2(B(x_k,\rho _k))}^s. \end{aligned}$$

(A.10)

Let $J_k=\Vert u\Vert _{L^2(B(x_k,\rho _k))}$, $k\ge 0$. Then (A.10) is rewritten as follows:

$$\begin{aligned} J_{k+1}\le CM^{1-s}J_k^s. \end{aligned}$$

An induction in k yields

$$\begin{aligned} J_k\le C^{1+s +\cdots +s^{k-1}}M^{(1-s)(1+s +\cdots +s^{k-1})}J_0^{s^k}. \end{aligned}$$

That is

$$\begin{aligned} J_k\le \left[ C^{\frac{1}{1-s}}M\right] ^{1-s^k}J_0^{s^k}. \end{aligned}$$

(A.11)

Applying Young’s inequality we obtain, for any $\epsilon >0$,

$$\begin{aligned} J_k&\le (1-s^k)\epsilon ^{\frac{1}{1-s^k}}C^{\frac{1}{1-s}}M+s^k \epsilon ^{-\frac{1}{s^k}}J_0 \nonumber \\&\le \epsilon ^{\frac{1}{1-s^k}}C^{\frac{1}{1-s}}M+\epsilon ^{- \frac{1}{s^k}}J_0\nonumber \\&\le C\epsilon ^{\frac{1}{1-s^k}}M+\epsilon ^{-\frac{1}{s^k}}J_0. \end{aligned}$$

(A.12)

Now, since $u\in C^{0,1/2}(\overline{\Omega })$,

$$\begin{aligned} |u(\tilde{x})|\le [u]_{1/2} |\tilde{x}-x|^{1/2} +|u(x)|,\quad x\in B(x_k,\rho _k). \end{aligned}$$

Hence

$$\begin{aligned} |\mathbb {S}^{n-1}|\rho _k^n|u(\tilde{x})|^2\le 2[u]_{1/2}^2\int _{B(x_k,\rho _k)} |\tilde{x}-x|\hbox {d}x+ 2\int _{B(x_k,\rho _k)} |u(x)|^2\hbox {d}x. \end{aligned}$$

Or equivalently

$$\begin{aligned} |u(\tilde{x})|^2\le 2|\mathbb {S}^{n-1}|^{-1}\rho _k^{-n}\left( [u]_{1/2}^2\int _{B(x_k,\rho _k)} |\tilde{x}-x|\hbox {d}x+ \int _{B(x_k,\rho _k)} |u(x)|^2\hbox {d}x\right) . \end{aligned}$$

A simple computation shows that $d_k=\mu ^kd_0$. Then

$$\begin{aligned} |\tilde{x}-x|\le |\tilde{x}-x_k|+|x_k-x|\le d_k+\rho _k=(1+\vartheta )d_k=(1+\vartheta )\mu ^kd_0. \end{aligned}$$

Therefore,

$$\begin{aligned} |u(\tilde{x})|^2\le 2\left( M^2(1+\vartheta )^{1/2} d_0^{1/2}\mu ^{k}+ |\mathbb {S}^{n-1}|^{-1}(\vartheta d_0)^{-n}\mu ^{-nk}\Vert u\Vert _{L^2(B(x_k,\rho _k))}^2\right) \end{aligned}$$

implying, when $d_0 (=\delta ) \le 1$,

$$\begin{aligned} |u(\tilde{x})|\le C\left( M\mu ^{k/2}+\mu ^{-nk/2}\delta ^{-n/2}J_k\right) . \end{aligned}$$

(A.13)

Inequalities (A.12) and (A.13) give

$$\begin{aligned} |u(\tilde{x})|\le C\left( \mu ^{k/2}M+\mu ^{-nk/}\epsilon ^{1/(1-s^k)}\delta ^{-n/2}M+\mu ^{-nk/2}\epsilon ^{-1/s^k}\delta ^{-n/2}J_0\right) . \end{aligned}$$

(A.14)

We get, by choosing $\epsilon =\mu ^{(1-s^k)(n+1)k/2}$ in (A.14),

$$\begin{aligned} |u(\tilde{x})|\le C\left( \mu ^{k/2}M+ \mu ^{k/2}\delta ^{-n/2}M+ \mu ^{-(n+1)k/(2s^k)+k/2}\delta ^{-n/2}J_0\right) . \end{aligned}$$

Hence

$$\begin{aligned} |u(\tilde{x})|\le C\delta ^{-n/2}\left( \mu ^{k/2}M+ \mu ^{-(n+1)k/(2s^k)}J_0\right) , \end{aligned}$$

(A.15)

by using $\delta \le \text{ diam }(\Omega )$.

Let $t>0$ and k be the integer so that $k\le t<k+1$. It follows from (A.15)

$$\begin{aligned} |u(\tilde{x})|\le C\delta ^{-n/2}\left( \mu ^{t/2}M+\mu ^{-t(n+1)/(2s^t)}J_0\right) . \end{aligned}$$

(A.16)

Let $p =(n+1)/2+|\ln s|$. Then (A.16) yields

$$\begin{aligned} |u(\tilde{x})|\le C\delta ^{-n/2}\left( \mu ^{t/2}M+\mu ^{-e^{p t}}J_0\right) . \end{aligned}$$

(A.17)

Putting $e^{p t}=1/\epsilon$, $0<\epsilon <1$, we get from (A.17)

$$\begin{aligned} |u(\tilde{x})|\le C\delta ^{-n/2}\left( \epsilon ^{\beta }M+e^{|\ln \mu |/\epsilon } J_0\right) , \end{aligned}$$

(A.18)

where $\beta = |\ln \mu |/(2p)$.

Step 3. A combination of (A.7) and (A.18) entails, with $0<\epsilon <1$ and $\epsilon _1>0$,

$$\begin{aligned} |u(\tilde{x})|\le C\delta ^{-n/2}\left( \epsilon ^{\beta }M+e^{|\ln \mu |/\epsilon } \left( \epsilon _1^{1/(1-\psi (\delta ))}M+\epsilon _1^{-1/\psi (\delta )}\Vert u\Vert _{L^2(B(y_0,\delta ))}\right) \right) . \end{aligned}$$

Hence, where $\ell =n/2$ and $\rho =|\ln \mu |$,

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Gamma )}\le C\delta ^{-\ell }\left( \epsilon ^{\beta }M+e^{\rho /\epsilon } \left( \epsilon _1^{1/(1-\psi (\delta ))}M+\epsilon _1^{-1/\psi (\delta )}\Vert u\Vert _{L^2(B(y_0,\delta ))}\right) \right) . \end{aligned}$$

In this inequality, we take $\epsilon _1=\epsilon ^{\beta (1-\psi (\delta ))}e^{-\rho (1-\psi (\delta ))/\epsilon }$. Using that

$$\begin{aligned} \epsilon _1^{-1/\psi (\delta )}\le e^{(\rho +\beta )(1-\psi (\delta ))/(\epsilon \psi (\delta ))}, \end{aligned}$$

we obtain in a straightforward manner

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Gamma )}\le C\delta ^{-\ell }\left( \epsilon ^{\beta }M+e^{(\rho +\beta )(1-\psi (\delta ))/(\epsilon \psi (\delta ))}\Vert u\Vert _{L^2(B(y_0,\delta ))}\right) . \end{aligned}$$

If $\phi (\delta )=\rho +(\rho +\beta )(1-\psi (\delta ))/\psi (\delta )$ then we can rewrite the previous estimate as follows:

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Gamma )}\le C\delta ^{-\ell }\left( \epsilon ^{\beta }M+e^{\phi (\delta )/\epsilon }\Vert u\Vert _{L^2(B(y_0,\delta ))}\right) , \end{aligned}$$

or equivalently

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Gamma )}\le C\delta ^{-\ell }\left( t^{-\beta }M+e^{t\phi (\delta )}\Vert u\Vert _{L^2(B(y_0,\delta ))}\right) , \quad t>1. \end{aligned}$$

(A.19)

If $M/\Vert u\Vert _{L^2(B(y_0,\delta ))}> e^{\phi (\delta )}$, we find $t>1$ so that $M/\Vert u\Vert _{L^2(B(y_0,\delta ))}=t^\beta e^{t\phi (\delta )}$. The estimate (A.19) with that t yields

$$\begin{aligned} \Vert u\Vert _{L^\infty (\Gamma )}\le C\delta ^{-\ell }M\left( \frac{1}{(\beta +\phi (\delta )}\ln \left( \frac{M}{\Vert u\Vert _{L^2(B(y_0,\delta ))}} \right) \right) ^{-\beta } . \end{aligned}$$

In light of the inequality $\Vert u\Vert _{L^\infty (\Gamma )}\ge \eta$, this estimate implies

$$\begin{aligned} \eta \le C\delta ^{-\ell }M\left( \frac{1}{\beta +\phi (\delta )}\ln \left( \frac{M}{\Vert u\Vert _{L^2(B(y_0,\delta ))}} \right) \right) ^{-\beta } . \end{aligned}$$

This inequality is equivalent to the following one

$$\begin{aligned} Me^{-C(\beta +\phi (\delta ))\delta ^{-\ell /\beta }\left( M/\eta \right) ^{1/\beta }}\le \Vert u\Vert _{L^2(B(y_0,\delta ))}. \end{aligned}$$

(A.20)

Otherwise,

$$\begin{aligned} Me^{-\phi (\delta )}\le \Vert u\Vert _{L^2(B(y_0,\delta ))}. \end{aligned}$$

(A.21)

We derive from (A.20) and (A.21) that, there exist $C>0$ and $\delta ^*$ so that

$$\begin{aligned} e^{-e^{c/\delta }}\le \Vert u\Vert _{L^2(B(y_0,\delta ))}, \quad 0< \delta \le \delta ^*. \end{aligned}$$

Obviously, a similar estimate holds for $\delta \ge \delta ^*$. $\square$