2. Direct source problem

Using the Helmholtz decomposition, the source function $\mathbf{f}\in L^2(B_R)^3$ can be decomposed as

\begin{equation*} \mathbf {f} = \mathbf {f}_p +\mathbf {f}_s, \end{equation*}

where $\mathbf{f}_p, \mathbf{f}_s\in L^2(B_R)^3$ with $\nabla \times \mathbf{f}_p = 0$ and $\nabla \cdot \mathbf{f}_s = 0$ . Hence, the solution $\mathbf u$ to the equation (1.1) can be decomposed into the pressure wave $\mathbf u_p$ and shear wave $\mathbf u_s$

(2.1) \begin{align} \mathbf{u}=\mathbf u_p+\mathbf u_s \end{align}

where $\nabla \times \mathbf{u}_p=0$ , $\nabla \cdot \mathbf{u}_s=0$ and

(2.2) \begin{align} \Delta \mathbf u_p+ k_p^2\mathbf u_p=\mathbf{f}_p,\quad \Delta \mathbf u_s+ k_s^2\mathbf u_s=\mathbf{f}_s. \end{align}

Here, $ k_p=\sqrt{\dfrac{\omega ^2 + \textrm{i}\omega \sigma }{\lambda +2\mu }}$ , $ k_s=\sqrt{\dfrac{\omega ^2 + \textrm{i}\omega \sigma }{\mu }}$ are the wavenumbers for the damped pressure and shear waves. Actually, one has

(2.3) \begin{align} \mathbf{u}_p=-\frac{1}{ k_p^2}\nabla \nabla \cdot \mathbf u,\quad \mathbf{u}_s=-\frac{1}{ k_s^2}\nabla \times \nabla \times \mathbf u\quad \text{in}\ \mathbb{R}^3. \end{align}

Motivated by the Helmholtz decomposition (2.1)–(2.2), to investigate the direct scattering problem of the elastic wave we first study the damped Helmholtz equation. Consider the following Helmholtz equation:

(2.4) \begin{align} -\Delta u(x, k) - k^2 u(x, k) - \textrm{i}k\sigma u(x, k) = f(x), \quad x\in \mathbb R^3. \end{align}

Here, we note that $u$ and $f$ are scalar-valued functions. The following theorem concerns its well-posedness.

Theorem 2.1. Given $f\in L^2\left(\mathbb R^3\right)$ with a compact support, there exists a unique exponentially decaying outgoing solution $u\in H^2\left(\mathbb R^3\right)$ to (2.4) for every $k\gt 0$ with the following estimate:

\begin{equation*} |u(x, k)|\leq C(f)e^{-c(k, \sigma )|x|} \end{equation*}

as $|x|\longrightarrow \infty$ . Here, $C(f)$ and $c(k, \sigma )$ are positive constants depending on $f$ and $k, \sigma$ , respectively.

Proof. We define

\begin{align*} u^*(x, k) = \int _{\mathbb R^3} e^{\textrm{i}x\cdot \xi } \frac{1}{|\xi |^2 - k^2 - \textrm{i}k\sigma } \hat{f}(\xi ) \textrm{d}\xi, \quad x\in \mathbb R^3, \end{align*}

where

\begin{equation*} \hat {f}(\xi ) = \frac {1}{(2\pi )^3} \int _{\mathbb R^3} f(x) e^{-\textrm {i}x\cdot \xi } \textrm {d}x. \end{equation*}

This definition is motivated by taking the Fourier transform of $u(x, k)$ formally with respect to the spatial variable $x$ . Then by the Plancherel’s theorem, for each $k\gt 0$ one has that $u^*(\cdot, k)\in H^2\left(\mathbb R^3\right)$ and satisfies equation (2.4).

Since

\begin{equation*} G(x, k) = \int _{\mathbb R^3} e^{\textrm {i}x\cdot \xi } \frac {1}{|\xi |^2 - k^2 - \textrm {i}k\sigma }\textrm {d}\xi = \frac {e^{\textrm {i} \kappa |x|}}{4\pi |x|}, \end{equation*}

where $ \kappa = (k^2 + \textrm{i}k\sigma )^{\frac{1}{2}}$ with $\Im \kappa \gt 0$ , one can rewrite $u^*(x, k)$ as

(2.5) \begin{align} u^*(x, k) = (G*f)(x) = \int _{\mathbb R^3} \frac{e^{\textrm{i} \kappa |x - y|}}{4\pi |x - y|} f(y)\textrm{d}y. \end{align}

Thus, since $f$ has a compact support by (2.5) one has that the solution $u^*(x, k)$ satisfies the estimate

\begin{equation*} |u^*(x, k)|\leq C(f)e^{-c(k, \sigma )|x|}, \end{equation*}

where $C(f)$ and $c(k, \sigma )$ are positive constants depending on $f$ and $k, \sigma$ , respectively. Using direct calculations, one may show that $\nabla u^*$ and $\Delta u^*$ have similar exponential decay estimates.

Now we prove the uniqueness. Assume that $\tilde{u}^*(x, k)$ is another solution to (2.4). Then one has

\begin{align*} ({-}\Delta - k^2 - \textrm{i}k\sigma ) (u^* - \tilde{u}^*) = 0 \end{align*}

and applying Fourier transform to the above equation gives

\begin{equation*} \left(|\xi |^2 - k^2 - \textrm {i}k\sigma \right) \left( \widehat {u^* - \tilde {u}^*}\right) (\xi ) = 0. \end{equation*}

Since for $k\gt 0$ one has that $|\xi |^2 - k^2 - \textrm{i}k\sigma \neq 0$ for all $\xi \in \mathbb R^3$ , the inverse Fourier transform gives $u^* - \tilde{u}^* = 0$ , which proves the uniqueness.

The well-posedness of the direct scattering problem is a direct consequence of the Helmholtz decomposition (2.1)–(2.2) and Theorem 2.1 for the Helmholtz equation.

Theorem 2.2. Let $\mathbf{f}\in L^2\left(\mathbb R^3\right)^3$ with a compact support. Then there exists a unique outgoing solution $\mathbf{u}$ of Schwartz distribution to (1.1) for every $\omega \gt 0$ . Moreover, the solution satisfies

\begin{equation*} |\mathbf {u}(x, \omega )|\leq C(\mathbf {f})e^{-c(\omega, \sigma, \lambda, \mu )|x|} \end{equation*}

as $|x|\to \infty,$ where $C(\mathbf{f})$ and $c(\omega, b, \lambda, \mu )$ are positive constants depending on $\mathbf f$ , and $\omega, \sigma$ and Lamé parameters, respectively.

3. Exact observability bounds for elastic wave equations

In order to bound the unknown source $\mathbf{f}$ by boundary data of $\mathbf{u}$ (see Lemma 4.2), we will prove an observability bound for the initial data of the corresponding time-domain damped elastic wave equation by noting that the solution to the time-harmonic damped elastic wave equation can be connected with the time-domain damped elastic wave equation by Fourier transformation.

We will derive an exact observability bound for the initial data $\mathbf{f}$ of the following time-domain damped elastic wave equation

(3.1) \begin{align} \begin{cases} \partial _t^2 \mathbf{U}(x, t) - \Delta ^* \mathbf{U}(x, t) + \sigma \partial _t \mathbf{U}(x, t) = 0, &\quad (x, t)\in \mathbb R^3\times (0, +\infty ),\\[4pt] \mathbf{U}(x, 0) = 0, \quad \partial _t\mathbf{U}(x, 0) = \mathbf{f}(x), &\quad x\in \mathbb R^3, \end{cases} \end{align}

Using Helmholtz decomposition again to the vector-valued initial condition $\mathbf{f}\in L^2(B_R)^3$ , one has

\begin{equation*} \mathbf {f} = \mathbf {f}_p +\mathbf {f}_s, \end{equation*}

where $\mathbf{f}_p, \mathbf{f}_s\in L^2(B_R)^3$ with $\nabla \times \mathbf{f}_p = 0$ and $\nabla \cdot \mathbf{f}_s = 0$ . As a consequence, we decompose the solution $\mathbf U$ to the equation (3.1) into a sum of pressure wave $\mathbf U_p$ and shear wave $\mathbf U_s$ (see, e.g., [Reference Bao, Hu, Kian and Yin4])

(3.2) \begin{align} \mathbf{U}=\mathbf U_p+\mathbf U_s \end{align}

where $\mathbf{U}_p$ and $\mathbf{U}_s$ satisfy $\nabla \times \mathbf{U}_p=0$ , $\nabla \cdot \mathbf{U}_s=0$ and the following damped pressure and shear wave equations

(3.3) \begin{align} \begin{cases} \partial _{tt} \mathbf U_p - c^2_p\Delta \mathbf U_p + \sigma \partial _t \mathbf U_p = 0, \quad &(x, t)\in \mathbb{R}^3\times (0, \infty ),\\[4pt] \mathbf{U}_p(x, 0) = 0, \quad \partial _t \mathbf{U}_p(x, 0) = \mathbf{f}_p(x), \quad &x\in \mathbb R^3, \end{cases} \end{align}

and

(3.4) \begin{align} \begin{cases} \partial _{tt} \mathbf U_s - c^2_s \Delta \mathbf U_s + \sigma \partial _t \mathbf U_s = 0, \quad &(x, t)\in \mathbb{R}^3\times (0, \infty ),\\[4pt] \mathbf{U}_s (x, 0) = 0, \quad \partial _t \mathbf{U}_s (x, 0) = \mathbf{f}_s (x), \quad &x\in \mathbb R^3. \end{cases} \end{align}

Here, $c_p= \sqrt{\lambda +2\mu }$ and $c_s=\sqrt{\mu }$ are the wave speeds for pressure and shear waves, respectively.

The following Carleman estimate is useful in deriving the exact observability bound.

Lemma 3.1. Let $T\gt 2R+1$ , $\varphi (x, t) = |x - a|^2 - \theta ^2\left(t - \frac{T}{2}\right)^2$ where $a\notin \overline{B_R}$ , dist(a, $\partial B_R$ ) = 1 and $\theta = \frac{1}{2}$ . Let $\mathbf{U}$ be a solution to (3.1) with $\mathbf{f}\in H^1(B_R)^3$ , $\text{supp}\mathbf{f}\subset B_R$ . Then we have the following Carleman estimate:

\begin{align*} &\sum _{|\alpha |\leq 1} s^{3 - 2|\alpha |} \int _{B_R \times (0, T)} |\partial _\alpha \mathbf{U}|^2 e^{2s\varphi } \textrm{d}x\textrm{d}t\\ &\lesssim \int _{\partial B_R \times (0, T)} \Big (s^3 |\partial _{tt} \nabla \mathbf{U}|^2 + s^3 |\partial _t\mathbf{U}|^2 \Big ) e^{2s\varphi } \textrm{d}\Gamma (x)\textrm{d}t, \end{align*}

where $\alpha =(\alpha _1, \alpha _2, \alpha _3, \alpha _4)\in \mathbb N^4$ and $\partial _\alpha = \partial _{t}^{\alpha _1}\,\partial _{x_1}^{\alpha _2}\,\partial _{x_2}^{\alpha _3}\,\partial _{x_3}^{\alpha _4}$ .

Proof. Let $v = \nabla \cdot \mathbf{U}$ and $\mathbf{w} = \nabla \times \mathbf{U}$ . Assume $\mathbf{U}$ is the solution to (3.1). By Carleman estimate in [Reference Isakov14] or [Reference Bellassoued and Yamamoto7] one has that

(3.5) \begin{align} &\sum _{|\alpha |\leq 1} s^{3 - 2|\alpha |} \int _{B_R \times (0, T)} \Big (|\partial _\alpha \mathbf{U}|^2 + |\partial _\alpha v|^2 + |\partial _\alpha \mathbf{w}|^2\Big ) e^{2s\varphi } \textrm{d}x\textrm{d}t\notag \\[3pt] &\lesssim \int _{\partial B_R\times (0, T)} s \Big (|\partial _t v|^2 + |\nabla v|^2 + s^2 v^2 \Big )e^{2s\varphi } \textrm{d}\Gamma (x)\textrm{d}t\notag \\[3pt] &\quad + \int _{\partial B_R\times (0, T)} s \Big (|\partial _t \mathbf{w}|^2 + |\nabla \mathbf{w}|^2 + s^2 \mathbf{w}^2 \Big )e^{2s\varphi } \textrm{d}\Gamma (x)\textrm{d}t. \end{align}

Since the Helmholtz decomposition (3.2) gives

\begin{equation*} \Delta v = (c_p^2)^{-1} \partial _{tt}v+(c_p^2)^{-1}\sigma \partial _{t}v, \quad \Delta \mathbf {w} = (c_s^2)^{-1} \partial _{tt}\mathbf {w}+(c_s^2)^{-1} \sigma \partial _{t}\mathbf {w}, \end{equation*}

by the elliptic regularity theory one has

\begin{equation*} \|v\|^2_{H^2(\partial B_R)} \lesssim \|\nabla \mathbf {U}\|_{L^2(\partial B_R)^{3\times 3}}^2+ \|\partial _{t}\nabla \mathbf {U}\|_{L^2(\partial B_R)^{3\times 3}}^2 + \|\partial _{tt}\nabla \mathbf {U}\|_{L^2(\partial B_R)^{3\times 3}}^2 \end{equation*}

and

\begin{equation*} \|\mathbf {w}\|^2_{H^2(\partial B_R)^3} \lesssim \|\nabla \mathbf {U}\|_{L^2(\partial B_R)^{3\times 3}}^2 + \|\partial _{t}\nabla \mathbf {U}\|_{L^2(\partial B_R)^{3\times 3}}^2+ \|\partial _{tt}\nabla \mathbf {U}\|_{L^2(\partial B_R)^{3\times 3}}^2. \end{equation*}

Hence, from (3.5) we have

\begin{align*} &\sum _{|\alpha |\leq 1} s^{3 - 2|\alpha |} \int _{B_R \times (0, T)} |\partial _\alpha \mathbf{U}|^2 \textrm{d}x\textrm{d}t\\ &\lesssim \int _{\partial B_R \times (0, T)} \Big ( s |\partial _{tt}\nabla \mathbf{U}|^2 + s |\partial _{t}\nabla \mathbf{U}|^2 + s |\partial _{t}\mathbf{U}|^2 + s^3 |\nabla \mathbf{U}|^2 + s^3 |\mathbf{U}|^2\Big ) e^{2s\varphi } \textrm{d}\Gamma (x)\textrm{d}t. \end{align*}

Since $\mathbf{f}$ has compact support contained in $B_R$ , one has

\begin{equation*} \|\mathbf {U}\|_{L^2(\partial B_R \times (0, T))^3} \lesssim \|\partial _{t}\mathbf {U}\|_{L^2(\partial B_R \times (0, T))^3} \end{equation*}

and

\begin{equation*} \|\nabla \mathbf {U}\|_{L^2(\partial B_R \times (0, T))^{3\times 3}}, \|\partial _t\nabla \mathbf {U}\|_{L^2(\partial B_R \times (0, T))^{3\times 3}} \lesssim \|\partial _{tt}\nabla \mathbf {U}\|_{L^2(\partial B_R \times (0, T))^{3\times 3}}, \end{equation*}

which completes the proof.

Using Lemma 3.1 and following the arguments in the proof of [[Reference Isakov and Lu15], Theorem 3.1], we obtain the exact observability bound.

Theorem 3.2. Assume that the observation time $T$ satisfies $4(2R + 1) \lt T \lt 5(2R + 1)$ . There exists a constant $C$ depending on the domain $B_R$ such that

(3.6) \begin{align} \|\mathbf{f}\|^2_{L^2(B_R)^3} \leq Ce^{C\sigma ^2} \Big (\|\partial _t \mathbf{U}\|^2_{L^2(\partial B_R \times (0, T))^3} + \|\partial _{tt}\nabla \mathbf{U}\|^2_{L^2(\partial B_R \times (0, T))^{3\times 3}}\Big ) \end{align}

for all $\mathbf{U}$ solving (3.1) with $\mathbf{f}\in H^1(B_R)^3$ , $ \text{supp} \mathbf{f}\subset B_R$ .

4. Inverse source problem

In this section, we study the inverse source problem for elastic waves. Denote the resolvent of the elliptic operator $-\Delta ^*$ by

(4.1) \begin{equation} \mathbf{R}_0(\zeta ) = ({-}\Delta ^*- \zeta ^2)^{-1}, \quad \zeta \in \mathbb C. \end{equation}

The following results on analyticity and resolvent estimates of $\mathbf{R}_0(\zeta )$ with respect to complex frequency $\zeta \in \mathbb C$ are useful in the subsequent analysis of the stability estimate. The proof utilises the method for the classical Schrödinger operator in [Reference Dyatlov and Zworski11].

Proposition 4.1. Fix a smooth cut-off function $\rho \in C_0^\infty \left(\mathbb R^3\right)$ . The free resolvent defined by (4.4) is analytic in $\mathbb C$ with respect to $\zeta$ as a family of operators

\begin{equation*} \rho \mathbf {R}_0(\zeta ) \rho \,:\, L^2\left(\mathbb R^3\right)^3 \to L^2\left(\mathbb R^3\right)^3 \end{equation*}

with the following resolvent estimate:

(4.2) \begin{align} \|\rho \mathbf{R}_0(\zeta ) \rho \|_{L^2\left(\mathbb R^3\right)^3 \to H^{j}\left(\mathbb R^3\right)^3} = \mathcal{O}\left(\langle \zeta \rangle ^{j -1} e^{L(\Im \widetilde{k}_s)_-}\right), \quad j = 0, 1, 2, \end{align}

Here, $\langle \zeta \rangle = \sqrt{1 + |\zeta |^2}$ , $\widetilde{k}_s=\zeta \sqrt{\frac{1}{\mu }}$ , $(\Im \widetilde{k}_s)_- = \max \{0, -\Im \widetilde{k}_s\}$ and $L$ satisfies $c_s L\gt \textrm{diamsupp}\rho$ with $\textrm{diamsupp}\, \rho = \textrm{sup}\{|x - y|\,:\, x, y \in \textrm{supp}\rho \}$ .

Proof. Recall the wavenumbers for compressional and shear waves given by

(4.3) \begin{align} \widetilde{k}_p=\zeta \sqrt{\frac{1}{\lambda +2\mu }}, \quad \widetilde{k}_s=\zeta \sqrt{\frac{1}{\mu }}, \end{align}

respectively. Denote the resolvents of the pressure and shear waves in the frequency domain respectively by

\begin{equation*} \mathbf {R}_{p}(\zeta ) = \left({-}\Delta - \widetilde {k}_p^2\right)^{-1}, \quad \mathbf {R}_{s}(\zeta ) = \left({-}\Delta - \widetilde {k}_s^2\right)^{-1}. \end{equation*}

Fix a smooth cut-off function $\rho \in C_0^\infty \left(\mathbb R^3\right)$ and denote $\textrm{diamsupp}\, \rho = \textrm{sup}\{|x - y|\,:\, x, y \in \textrm{supp}\rho \}$ . As in [Reference Dyatlov and Zworski11], for $\zeta \in \mathbb C$ the resolvents of the wave speed for pressure and shear waves can be represented by

\begin{align*} \rho \mathbf{R}_{p}(\zeta ) \rho = \int _0^L e^{\textrm{i}\widetilde{k}_p t} \rho \mathbf W (t) \rho \textrm{d}t, \quad \rho \mathbf{R}_{s}(\zeta ) \rho = \int _0^L e^{\textrm{i}\widetilde{k}_s t} \rho \mathbf W (t) \rho \textrm{d}t, \end{align*}

where $\mathbf W (t) = \frac{\sin t \sqrt{-\Delta }}{ \sqrt{-\Delta } }$ and $L$ satisfies $c_s L\gt \textrm{diamsupp}\rho$ . Consequently, by (3.2), (3.3) and (3.4) we obtain that

(4.4) \begin{align} \rho \mathbf{R}_0(\zeta ) \rho \, \mathbf{f} = \rho \mathbf{R}_{p}(\zeta ) \rho \, \mathbf{f}_p + \rho \mathbf{R}_{p}(\zeta ) \rho \, \mathbf{f}_s. \end{align}

Moreover, since by [[Reference Dyatlov and Zworski11], Theorem 3.1]

\begin{equation*} \|\rho \mathbf {R}_{p}(\zeta ) \rho \|_{L^2\left(\mathbb R^3\right)^3\to L^2\left(\mathbb R^3\right)^3} = \mathcal {O} \left( (1 + |\zeta |)^{-1} e^{L(\Im \widetilde {k}_p)_-} \right) \end{equation*}

and

\begin{equation*} \|\rho \mathbf {R}_{s}(\zeta ) \rho \|_{L^2\left(\mathbb R^3\right)^3\to L^2\left(\mathbb R^3\right)^3} = \mathcal {O} \left( (1 + |\zeta |)^{-1} e^{L(\Im \widetilde {k}_s)_-} \right) \end{equation*}

where $(t)_-: = \max \{0, -t\}$ , we obtain from (4.4) and (4.3) that

(4.5) \begin{align} \|\rho \mathbf{R}_0(\zeta ) \rho \, \mathbf{f}\|_{L^2\left(\mathbb R^3\right)^3} &\lesssim (1 + |\zeta |)^{-1} e^{L(\Im \widetilde{k}_s)_-} \left(\|\mathbf{f}_p\|_{L^2\left(\mathbb R^3\right)^3} + \|\mathbf{f}_s\|_{L^2\left(\mathbb R^3\right)^3}\right)\notag \\ &\lesssim (1 + |\zeta |)^{-1} e^{L(\Im \widetilde{k}_s)_-} \|\mathbf{f}\|_{L^2\left(\mathbb R^3\right)^3}. \end{align}

Now we can derive the analyticity of the free resolvent $\rho \mathbf{R}_0(\zeta ) \rho$ from those of $\rho \mathbf{R}_{p}(\zeta ) \rho$ and $\rho \mathbf{R}_{s}(\zeta ) \rho$ . In what follows, we prove the resolvent estimates (4.2) by standard regularity theory of elliptic equations. The case for $j = 0$ is a consequence of (4.5). For $j = 2$ , taking $\tilde{\rho }\in C^\infty _0\left(\mathbb R^3\right)$ such that $\tilde{\rho } = 1$ near the support of $\rho$ , we have from [[Reference Stein21], $(7.13)$ ] that

\begin{equation*} \|\rho \mathbf {u}\|_{H^{2}\left(\mathbb R^3\right)^3}\leq C\left(\|\tilde {\rho }\mathbf {u}\|_{L^2\left(\mathbb R^3\right)^3} + \|\tilde {\rho } ({-}\Delta ^*) \mathbf {u}\|_{L^2\left(\mathbb R^3\right)^3}\right), \end{equation*}

where $C\gt 0$ is a constant. Let $\mathbf{u} = \mathbf{R}_0(\zeta )(\rho \mathbf{f})$ with $\mathbf{f}\in L^2\left(\mathbb R^3\right)^3$ . One has

\begin{equation*} \|\rho \mathbf {R}_0(\zeta )(\rho \mathbf {f})\|_{H^{2}\left(\mathbb R^3\right)^3}\leq C\left(\|\tilde {\rho }\mathbf {R}_0(\zeta )(\rho \mathbf {f})\|_{L^2\left(\mathbb R^3\right)^3} + \|\tilde {\rho }({-}\Delta ^*)(\mathbf {R}_0(\zeta )(\rho \mathbf {f}))\|_{L^2\left(\mathbb R^3\right)^3}\right). \end{equation*}

Since

\begin{align*} \|\tilde{\rho }({-}\Delta ^*)(\mathbf{R}_0(\zeta )(\rho \mathbf{f}))\|_{L^2\left(\mathbb R^3\right)^3}&=\| \rho \mathbf{f} + \tilde{\rho }|\zeta |^{2}\mathbf{R}_0(\zeta )(\rho \mathbf{f})\|_{L^2\left(\mathbb R^3\right)^3}\\ & \lesssim \langle \zeta \rangle ^{2} e^{L(\Im \widetilde{k}_s)_-}\|\mathbf{f}\|_{L^2\left(\mathbb R^3\right)^3}, \end{align*}

we get

\begin{equation*} \|\rho \mathbf {R}_0(\zeta )\rho \|_{L^2\left(\mathbb R^3\right)^3\rightarrow H^{2}\left(\mathbb R^3\right)^3}\lesssim \langle \zeta \rangle ^{2} e^{L(\Im \widetilde {k}_s)_-}. \end{equation*}

Finally, the case of $j = 1$ can be obtained from the interpolation between $j=0$ and $j = 2$ , which completes the proof.

Define a vector-valued real function space

\begin{equation*} \mathcal {C}_Q = \left\{\mathbf {f} \in H^{n}(B_R)^3\,:\, n\geq 5, \|\mathbf {f}\|_{H^{n}(B_R)^3}\leq Q, \text {supp} \mathbf {f}\subset B_R, \mathbf {f}: B_R \rightarrow \mathbb R^3 \right\}. \end{equation*}

Due to the exact observability bound (3.6), we introduce the boundary measurement of the elastic wave

(4.6) \begin{align} \|\mathbf{u}(x, \omega )\|^2_{\partial B_R}&\,:\!=\, \int _{\partial B_R}\Big ( \omega ^2 |\mathbf{u}(x, \omega )|^2 + \omega ^4|\nabla \mathbf{u}(x, \omega )|^2\Big ) \textrm{d}\Gamma (x). \end{align}

In the following lemma, we bound the unknown source by the boundary measurements (4.6).

Lemma 4.2. There holds

\begin{align*} \|\mathbf{f}\|_{L^2(B_R)^3}^2&\lesssim e^{C\sigma ^2} \int _{0}^{+\infty } \|\mathbf{u}(x, \omega )\|^2_{\partial B_R} \textrm{d}\omega, \end{align*}

where $\mathbf{u}(x, \omega )$ is the solution to the direct scattering problem (1.1) with $\mathbf{f}\in \mathcal C_Q$ .

Remark 4.3. The proof of Lemma 4.2 depends on the exact observability inequality (3.6) and Fourier transform. Intuitively, by letting $\mathbf{U}(x,t)=0$ , $t\lt 0$ , we can see $\mathbf{u}(x,\omega )=\int _{-\infty }^{+\infty } \mathbf{U}(x,t)e^{\textrm{i}\omega t}dt=\int _{0}^{+\infty } \mathbf{U}(x,t)e^{\textrm{i}\omega t}dt$ . By Theorem 3.6 and Plancherel’s theorem, we have

(4.7) \begin{align} \|\mathbf{f}\|^2_{L^2(B_R)^3} &\leq Ce^{C\sigma ^2} \Big (\|\partial _t \mathbf{U}\|^2_{L^2(\partial B_R \times (0, T))^3} + \|\partial _{tt}\nabla \mathbf{U}\|^2_{L^2(\partial B_R \times (0, T))^{3\times 3}}\Big )\notag \\ &\leq Ce^{C\sigma ^2} \Big (\|\partial _t \mathbf{U}\|^2_{L^2(\partial B_R \times (0, +\infty ))^3} + \|\partial _{tt}\nabla \mathbf{U}\|^2_{L^2(\partial B_R \times (0, +\infty ))^{3\times 3}}\Big )\notag \\ & =\frac{C}{2\pi }e^{C\sigma ^2}\int _{0}^{+\infty } \|\mathbf{u}(x, \omega )\|^2_{\partial B_R} \textrm{d}\omega. \end{align}

Thus, we formally proved Lemma 4.2. To justify the Fourier transform rigorously, one can obtain the decay estimates of the solution $\mathbf{U}$ to the time-domain initial-valued elastic wave (3.1) by combining the Helmholtz decomposition (3.2) and the decay estimates for acoustic wave equations proved in Theorem 5.1 assuming that the source $\mathbf{f}$ has sufficient regularity ( $\mathbf{f}\in H^n, n\geq 5$ ). As the proof follows the arguments in [[Reference Li, Yao and Zhao17], Lemma 3.1] in a straightforward way, we omit it for brevity.

Define

\begin{align*} I_0(s) &= \int _0^s \omega ^2 \int _{\partial B_R} \mathbf{u}(x, \omega )\cdot \mathbf{u}(x, -\omega ) \textrm{d}\Gamma (x) \textrm{d}\omega, \\ I_1(s) &= \int _0^s \omega ^4\int _{\partial B_R} \nabla \mathbf{u}(x, \omega )\,:\, \nabla \mathbf{u}(x, -\omega ) \textrm{d}\Gamma (x) \textrm{d}\omega, \end{align*}

where $:$ denotes the double dot product which is defined by $A\,:\,B=\sum \limits _{i,j=1}^3 a_{ij}b_{ij}$ for $A=(a_{ij})_{i,j=1}^3$ , $B=(b_{ij})_{i,j=1}^3$ . Since the integrands are entire analytic functions of $\omega$ , the integrals in $I_0(s)$ and $I_1(s)$ with respect to $\omega$ can be taken over any path joining points $0$ and $s$ of the complex plane. Consequently, $I_0(s)$ and $I_1(s)$ are entire analytic functions of $s=s_1+\textrm{i}s_2$ .

The following lemma gives estimates of $I_0(s)$ and $I_1(s)$ . The proof employs the resolvent estimates in Proposition 4.1.

Lemma 4.4. Denote $S=\{z=x+\textrm{i}y\in \mathbb{C}\,: -\frac{\pi }{4}\lt \textrm{arg} z\lt \frac{\pi }{4}\}$ . Let $s=s_1+\textrm{i}s_2\in S$ . The following estimates hold:

\begin{align*} |I_0(s)| \lesssim (1 + |s|)^3 e^{C(4s_1 + \sigma )} \|\mathbf{f}\|^2_{L^2(B_R)^3},\\ |I_1(s)| \lesssim (1 + |s|)^5 e^{C(4s_1 + \sigma )} \|\mathbf{f}\|^2_{L^2(B_R)^3}.\\ \end{align*}

Proof. We first show that $ \kappa (\omega ) = \sqrt{\omega ^2 + \textrm{i}\omega \sigma }$ is analytic for $\omega \in S$ . When $\omega = \omega _1 + \textrm{i}\omega _2 \in S$ the image of $\omega ^2 + \textrm{i}\sigma \omega$ satisfies

\begin{equation*} \{\omega ^2 + \textrm {i}\sigma \omega | \omega \in S\} \cap \textrm {i}({-}\infty, 0] = \phi. \end{equation*}

In fact, assume that $\omega \in S$ . Since

\begin{equation*} \omega ^2 + \textrm {i}\sigma \omega = (\omega _1^2 - \omega _2^2 - \sigma \omega _2) + \textrm {i}(2\omega _1\omega _2 + \sigma \omega _1), \end{equation*}

if $\omega ^2 + \textrm{i}\sigma \omega \in \textrm{i}({-}\infty, 0]$ , then we have

\begin{equation*} \omega _1^2 - \omega _2^2 - \sigma \omega _2 = 0, \quad 2\omega _1\omega _2 + \sigma \omega _1\leq 0, \end{equation*}

which gives $\omega _2 \leq -\frac{\sigma }{2}$ . Then, we have

\begin{equation*} \omega _1^2 = \omega _2^2 + \sigma \omega _2 \leq \omega _2^2 - \frac {\sigma ^2}{2} \lt \omega _2^2, \end{equation*}

which is in contradiction with the assumption that $\omega \in S$ where $\omega _1\geq |\omega _2|$ . Therefore, by choosing the branch cut of $\sqrt{z}$ to be $z\in \mathbb C\backslash \textrm{i}({-}\infty, 0]$ we obtain that $ \kappa (\omega ) = \sqrt{\omega ^2 + \textrm{i}\omega b}$ is analytic for $\omega \in S$ .

Let $\kappa (\omega ) = \kappa _1(\omega ) + \textrm{i}\kappa _2(\omega )$ . A direct calculation gives $| \kappa _2(\omega )|^2 \leq (1 + \sqrt{2})\Big (\omega _1 + \frac{\sigma }{4}\Big )^2$ for $\omega \in S$ . Consequently, by the resolvent estimates in Proposition 4.1 we have that for $j = 0, 1, 2$

\begin{align*} \|\mathbf{u}(x, \omega )\|_{H^j(B_R)^3} &= \|\mathbf{R}_0(\kappa (\omega )) \mathbf{f}\|_{H^j(B_R)^3} \\ &\lesssim e^{C(\Im \omega )_-} (1 + | \omega |)^{j - 1} \|\mathbf{f}\|_{L^2(B_R)^3} \\ &\lesssim (1 + | \omega |)^{j - 1} e^{C (\omega _1 + \frac{\sigma }{4})} \|\mathbf{f}\|_{L^2(B_R)^3} \\ &\lesssim (1 + |\omega |)^{j - 1} e^{C (4\omega _1 + \sigma )} \|\mathbf{f}\|_{L^2(B_R)^3}. \end{align*}

Then letting $\omega = s t, t\in [0, 1]$ , we obtain that

\begin{align*} |I_0(s)| &\lesssim |s|^3 \int _0^1 t^2 \left(\int _{\partial B_R} |\mathbf{u}(x, st)|^2 \textrm{d}\Gamma (x)\right)^{1/2} \left(\int _{\partial B_R} |\mathbf{u}(x, -st)|^2 \textrm{d}\Gamma (x)\right)^{1/2} \textrm{d}t\\ &\lesssim (1 + |s|)^3 \int _0^1 t^2 \|\mathbf{u}(x, st)\|_{H^1(B_R)^3} \|\mathbf{u}(x, -st)\|_{H^1(B_R)^3} \textrm{d}t \\ &\lesssim (1 + |s|)^3 e^{C(4s_1 + \sigma )} \|\mathbf{f}\|^2_{L^2(B_R)^3}. \end{align*}

For $I_1(s)$ repeating the above arguments for $I_0(s)$ , we have that

\begin{align*} |I_1(s)| &\lesssim |s|^5 \int _0^1 t^4 \left(\int _{\partial B_R} |\nabla \mathbf{u}(x, st)|^2 \textrm{d}\Gamma (x)\right)^{1/2} \left(\int _{\partial B_R} |\nabla \mathbf{u}(x, -st)|^2 \textrm{d}\Gamma (x)\right)^{1/2} \textrm{d}t\\ &\lesssim |s|^5 \int _0^1 t^4 \|\mathbf{u}(x, st)\|_{H^2(B_R)^3} \|\mathbf{u}(x, -st)\|_{H^2(B_R)^3} \textrm{d}t \\ &\lesssim (1 + |s|)^5 e^{C(4s_1 + \sigma )} \|\mathbf{f}\|^2_{L^2(B_R)^3}. \end{align*}

The proof is completed.

The following lemma proved in [Reference Bao, Li and Zhao6] provides an estimate of the high frequency tail of $\|\mathbf{u}(x, \omega )\|^2_{\partial B_R}$ .

Lemma 4.5. Let $\mathbf{f}\in \mathcal C_Q$ . Then the following estimate holds:

\begin{align*} &\int _s^\infty \|\mathbf{u}(x, \omega )\|^2_{\partial B_R}\textrm{d}\omega \lesssim \frac{1}{s^{2n-3}}\|\mathbf{f}\|^2_{H^n(B_R)^3}. \end{align*}

The following lemma on analytic continuation is proved in [Reference Cheng, Isakov and Lu8] which will be useful in the subsequent analysis.

Lemma 4.6. Let $J(z)$ be analytic in $S=\{z=x+\textrm{i}y\in \mathbb{C}\,: -\frac{\pi }{4}\lt \textrm{arg} z\lt \frac{\pi }{4}\}$ and continuous in $\bar{S}$ satisfying

\begin{equation*} \begin {cases} |J(z)|\leq \varepsilon, & z\in (0, K],\\[4pt] |J(z)|\leq M, & z\in S,\\[4pt] |J(0)|=0. \end {cases} \end{equation*}

Then there exits a function $\beta (z)$ satisfying

\begin{equation*} \begin {cases} \beta (z)\geq \frac {1}{2}, & z\in \left(K, 2^{\frac {1}{4}}K\right),\\[4pt] \beta (z)\geq \frac {1}{\pi }\left(\left(\frac {z}{K}\right)^4-1\right)^{-\frac {1}{2}}, & z\in \left(2^{\frac {1}{4}}K, \infty \right) \end {cases} \end{equation*}

such that

\begin{equation*} |J(z)|\leq M\varepsilon ^{\beta (z)}\quad \forall \, z\in (K, \infty ). \end{equation*}

The following lemma is a direct consequence of Lemmas 4.4 and 4.6.

Lemma 4.7. Let $f\in \mathcal C_Q$ . Then there exists a function $\beta (s)$ satisfying

(4.8) \begin{equation} \begin{cases} \beta (s)\geq \frac{1}{2}, \quad & s\in \left(K, 2^{\frac{1}{4}}K\right),\\[4pt] \beta (s)\geq \frac{1}{\pi }\left(\left(\frac{s}{K}\right)^4-1\right)^{-\frac{1}{2}}, \quad & s\in \left(2^{\frac{1}{4}}K, \infty \right), \end{cases} \end{equation}

such that

\begin{equation*} |I_0(s)+I_1(s)|\lesssim Q^2 e^{(C+1)s}\epsilon ^{2\beta (s)},\quad \forall s\in (K, \infty ), \end{equation*}

where

\begin{equation*} \epsilon ^2 = \int _0^K \|\mathbf {u}(x, \omega )\|^2_{\partial B_R} \textrm {d}\omega. \end{equation*}

Proof. It follows from Lemma 4.4 that there exists $C\gt 0$ such that

\begin{equation*} |\!\left (I_0(s)+I_1(s)\right )e^{-(C+1)|s|}|\lesssim Q^2,\quad \forall s\in S. \end{equation*}

Moreover, we have

\begin{equation*} |\!\left (I_0(s)+I_1(s)\right )e^{-(C+1)|s|}|\leq \epsilon ^2,\quad s\in [0, K]. \end{equation*}

A direct application of Lemma 4.6 shows that there exists a function $\beta (s)$ satisfying (4.8) such that

\begin{equation*} |\!\left (I_0(s)+I_1(s)\right )e^{-(C+1)s}|\lesssim Q^2\epsilon ^{2\beta },\quad \forall s\in (K, \infty ), \end{equation*}

which completes the proof.

In the following theorem, we present an increasing stability estimate for the inverse source problem following the arguments in [Reference Li and Yuan19].

Theorem 4.8. Let $\mathbf{u}(x, \omega )$ be the outgoing solution of the scattering problem (1.1) corresponding to the source $\mathbf{f} \in \mathcal{C}_Q$ . Then for $\epsilon$ sufficiently small the following estimate holds:

(4.9) \begin{align} \|\mathbf{f}\|^2_{L^2(B_R)^3} \lesssim e^{C\sigma ^2} \left(\epsilon ^2 + \frac{Q^2}{\Big (K^{\frac{2}{3}} |\ln \epsilon |^{\frac{1}{4}}\Big )^{2n - 3}}\right) \end{align}

where

\begin{equation*} \epsilon ^2 = \int _0^K \|\mathbf {u}(x, \omega )\|^2_{\partial B_R} \textrm {d}\omega. \end{equation*}

Proof. We can assume that $\epsilon \lt e^{-1}$ , otherwise the estimate is obvious. Let

\begin{equation*} s=\begin {cases} \dfrac {1}{((C+3)\pi )^{\frac {1}{3}}}K^{\frac {2}{3}}|\ln \epsilon |^{\frac {1}{4}},& 2^{\frac {1}{4}} ((C+3)\pi )^{\frac {1}{3}}K^{\frac {1}{3}}\lt |\ln \epsilon |^{\frac {1}{4}},\\[11pt] K, &|\ln \epsilon |^{\frac {1}{4}}\leq 2^{\frac {1}{4}}((C+3)\pi )^{\frac {1}{3}}K^{\frac {1}{3}}. \end {cases} \end{equation*}

If $2^{\frac{1}{4}}((C+3)\pi )^{\frac{1}{3}}K^{\frac{1}{3}} \lt |\ln \epsilon |^{\frac{1}{4}}$ , then we have from Lemma 4.7 that

\begin{align*} |I_0(s)+I_1(s)|&\lesssim Q^2 e^{(C+3)s} e^{-\frac{2|\ln \epsilon |}{\pi }\left(\left(\frac{s}{K}\right)^4-1\right)^{-\frac{1}{2}}}\cr &\lesssim Q^2 e^{\frac{(C+3)}{((C+3)\pi )^{\frac{1}{3}}}K^{\frac{2}{3}}|\ln \epsilon |^{\frac{1}{4}}-\frac{ 2|\ln \epsilon |}{\pi }(\frac{K}{s})^2}\\ &=M^2 e^{-2\left (\frac{(C+3)^2}{\pi }\right )^{\frac{1}{3}}K^{\frac{2}{3}} |\ln \epsilon |^{\frac{1}{2}}\left (1-\frac{1}{2} |\ln \epsilon |^{-\frac{1}{4}}\right )}. \end{align*}

Noting

\begin{equation*} \frac {1}{2} |\ln \epsilon |^{-\frac {1}{4}}\lt \frac {1}{2},\quad \left (\frac {(C+3)^2}{\pi }\right )^{\frac {1}{3}}\gt 1, \end{equation*}

we have

\begin{equation*} |I_0(s)+I_1(s)| \lesssim Q^2 e^{-K^{\frac {2}{3}}|\ln \epsilon |^{\frac {1}{2}}}. \end{equation*}

Using the elementary inequality

(4.10) \begin{equation} e^{-x}\leq \frac{(6n-9)!}{x^{3(2n - 3)}}, \quad x\gt 0, \end{equation}

we get

(4.11) \begin{equation} |I_0(s)+I_1(s)|\lesssim \frac{Q^2}{\left (\dfrac{K^2|\ln \epsilon |^{\frac{3}{2} }}{(6n-9)^3}\right )^{2n - 3}}. \end{equation}

If $|\ln \epsilon |^{\frac{1}{4}}\leq 2^{\frac{1}{4}}((C+3)\pi )^{\frac{1}{3}}K^{\frac{1}{3}}$ , then $s=K$ . We have that

\begin{equation*} |I_0(s)+I_1(s)|\leq \epsilon ^2. \end{equation*}

Here, we have used the fact that

\begin{equation*} I_0(s)+I_1(s)=\int _0^s \| \mathbf {u}(x, \omega )\|^2_{\partial B_R}\textrm {d}\omega,\quad s\gt 0. \end{equation*}

Hence, we obtain from Lemma 4.5 and (4.11) that

\begin{align*} &\int _0^\infty \|\mathbf{u}(x, \omega )\|^2_{\partial B_R}\textrm{d}\omega \\ &\leq I_0(s)+I_1(s)+\int _s^\infty \| \mathbf{u}(x, \omega )\|^2_{\partial B_R}\textrm{d}\omega \\ &\lesssim \epsilon ^2+\frac{Q^2}{\left (\frac{ K^2|\ln \epsilon |^{\dfrac{3}{2}}}{(6n-9)^3} \right )^{2n - 3}}+\frac{Q^2}{\left (2^{ -\frac{1}{4}}((C+3)\pi )^{-\frac{1}{3}}K^{\frac{2}{3}} |\ln \epsilon |^{\frac{1}{4}}\right )^{2n - 3}}. \end{align*}

By Lemma 4.2, we have

\begin{equation*} \|\mathbf {f}\|^2_{L^2(B_R)}\lesssim e^{C\sigma ^2} \left(\epsilon ^2 +\frac {Q^2}{\left (K^2|\ln \epsilon |^{\frac {3}{2}} \right )^{2n - 3}}+\frac {Q^2}{\left (K^{\frac {2} {3}}|\ln \epsilon |^{\frac {1}{4}}\right )^{2n - 3}}\right). \end{equation*}

Since $K^{\frac{2}{3}}|\ln \epsilon |^{\frac{1}{4}}\leq K^2 |\ln \epsilon |^{\frac{3}{2}}$ when $K\gt 1$ and $|\ln \epsilon |\gt 1$ , we finish the proof and obtain the stability estimate (4.9).

The stability estimate (4.9) consists of two parts: the first part is the Lipschitz type of data discrepancy and the second part is the high frequency tail of the source function. As the upper bound $K$ of the frequency increases, the stability estimate (4.9) tends to a Lipschitz-type stability which suggests that the inverse source problem becomes more stable when data of higher frequency are used. It also shows that the stability deteriorates if the attenuation $\sigma$ becomes larger.

5. Useful decay estimates for acoustic waves

To justify the Fourier transform in Remark 4.3, we need to prove some decay estimates for the solution $\mathbf{U}$ to the time-domain initial-valued elastic wave (3.1). As mentioned in Remark 4.3, in order to obtain the decay estimates of the solution to the time-domain initial-valued elastic wave (3.1), we just need to prove decay estimates for the dumped acoustic wave equation since we have the Helmholtz decomposition (3.2). In this section, we will prove some decay estimates which can guarantee that the Fourier transform for the solution of the dumped acoustic wave equation is well defined (see Remarks 5.2 and 5.3).

We consider the following initial-valued damped acoustic wave equation in $\mathbb R^3$

(5.1) \begin{align} \begin{cases} \partial _t^2 U(x, t) - \Delta U(x, t) + \sigma \partial _t U(x, t) = 0, &\quad (x, t)\in \mathbb R^3\times (0, +\infty ),\\[5pt] U(x, 0) = 0, \quad \partial _tU(x, 0) = f(x), &\quad x\in \mathbb R^3, \end{cases} \end{align}

where $f(x)\in L^1\left(\mathbb R^3\right)\cap H^s\left(\mathbb R^3\right)$ . The regularity assumption $H^s\left(\mathbb R^3\right)$ for $f(x)$ will be specified later. Since the equation in (5.1) has constant coefficients, the decay estimates of the solution $U(x, t)$ can be derived by the standard Fourier transform.

Applying the Fourier transform to the solution $U(x, t)$ to (5.1) with respect to the spatial variable $x$ , we obtain that

\begin{align*} U(x, t) = \mathcal{F}^{-1} (m_\sigma (t, \xi ) \hat{f}(\xi ))(x), \end{align*}

where $\mathcal{F}^{-1}$ denotes the inverse Fourier transform, the multiplies $m_\sigma (t, \xi )$ takes the form

(5.2) \begin{align} m_\sigma (t, \xi ) = \frac{e^{-\frac{\sigma }{2}t}}{\sqrt{\sigma ^2 - 4|\xi |^2}} \Big (e^{\frac{1}{2}t \sqrt{\sigma ^2 - 4|\xi |^2}} - e^{-\frac{1}{2}t \sqrt{\sigma ^2 - 4|\xi |^2}}\Big ), \end{align}

and $\hat{f}(\xi )$ is the Fourier transform of $f$ defined as follows:

\begin{equation*} \hat {f}(\xi ) = \frac {1}{(2\pi )^3} \int _{\mathbb R^3} e^{-\textrm {i}x\cdot \xi } f(x) \textrm {d}x. \end{equation*}

Assume that $\sqrt{\sigma ^2 - 4|\xi |^2} = \textrm{i} \sqrt{4|\xi |^2 - \sigma ^2}$ when $|\xi |^2\gt \frac{\sigma ^2}{4}$ . Then (5.2) becomes

\begin{align*} m_\sigma (t, \xi ) = \begin{cases} 2e^{-\frac{\sigma }{2}t} \frac{\sinh \left(\frac{t}{2}\sqrt{\sigma ^2 - 4|\xi |^2 }\right)}{\sqrt{\sigma ^2 - 4|\xi |^2}},&\quad |\xi |^2\lt \frac{\sigma ^2}{4},\\[10pt] e^{-\frac{\sigma }{2}t} \frac{\sin \left(\frac{t}{2}\sqrt{4|\xi |^2 - \sigma ^2}\right)}{\sqrt{4|\xi |^2 - \sigma ^2}},&\quad |\xi |^2\gt \frac{\sigma ^2}{4}. \end{cases} \end{align*}

Notice from the representation of the multiplies $m_\sigma (t, \xi )$ above that the solution $U(x, t)$ behaves as a “parabolic type” of $e^{-t\Delta } f$ in the low frequency, while for the high frequency part it behaves like a “dispersive type” of $e^{\textrm{i}t\Delta } f$ .

Theorem 5.1. Let $U(x, t)$ be the solution of (5.1). Then $U(x, t)$ satisfies the decay estimate

(5.3) \begin{align} \textrm{sup}_{x\in \mathbb R^3} |\partial ^\alpha _x\partial _t^j U(x, t)| \lesssim (1 + t)^{-\frac{3 + |\alpha |}{2}} \|f\|_{L^1\left(\mathbb R^3\right)} + e^{-ct} \|f\|_{H^s\left(\mathbb R^3\right)}, \end{align}

where $j\in \mathbb N$ , $\alpha$ is a multi-index vector in $\mathbb N^3$ such that $\partial _x^\alpha = \partial _{x_1}^{\alpha _1}\,\partial _{x_2}^{\alpha _2}\,\partial _{x_3}^{\alpha _3}$ , $s\gt j + |\alpha | + \frac{1}{2}$ and $c\gt 0$ is some positive constant. In particular, for $|\alpha | = j = 0$ , the following estimate holds:

(5.4) \begin{align} \textrm{sup}_{x\in \mathbb R^3} |U(x, t)|\lesssim (1 + t)^{-\frac{3}{2}} \left(\|f\|_{L^1\left(\mathbb R^3\right)} + \|f\|_{H^1\left(\mathbb R^3\right)}\right). \end{align}

Remark 5.2. The estimate (5.4) provides a time decay of the order $O((1 + t)^{-\frac{3}{2}})$ for $U(x, t)$ uniformly for all $x\in \mathbb R^3$ , which gives

\begin{align*} \sup _{x\in \mathbb R^3} \int _0^\infty |U(x, t)|^2 \textrm{d}t \lesssim \int _0^\infty (1 + t)^{-3}\textrm{d}t \lt +\infty. \end{align*}

Hence, letting $U(x, t) = 0$ when $t\lt 0$ then $U(x, t)$ has a Fourier transform $\hat{U}(x, k) \in L^2(\mathbb R)$ for each $x\in \mathbb R^3$ . Moreover, the following Plancherel equality holds:

\begin{align*} \int _0^{+\infty } |U(x, t)|^2\textrm{d}t = \int _{-\infty }^{+\infty } |\hat{U}(x, k)|^2 \textrm{d}k. \end{align*}

Remark 5.3. To study the inverse source problem, it suffices to assume that $f\in H^s\left(\mathbb R^3\right), s\geq 5$ . In this case, it follows from the above theorem that both $\partial _t U(x, t)$ and $\partial _{tt}\nabla U(x, t)$ are continuous functions. Moreover, we have from (5.3) that the following estimate holds:

\begin{align*} \textrm{sup}_{x\in \mathbb R^3} |\partial _t U(x, t)|& \lesssim (1 + t)^{-\frac{3}{2}} \|f\|_{L^1\left(\mathbb R^3\right)} + e^{-ct} \|f\|_{H^s\left(\mathbb R^3\right)},\\ \textrm{sup}_{x\in \mathbb R^3} |\partial _{tt}\nabla U(x, t)|& \lesssim (1 + t)^{-2} \|f\|_{L^1\left(\mathbb R^3\right)} + e^{-ct} \|f\|_{H^s\left(\mathbb R^3\right)}. \end{align*}

Proof. Without loss of generality, we may assume that $\sigma = 1$ , and then

\begin{align*} m(t, \xi )\triangleq m_\sigma (t, \xi ) = \frac{e^{-\frac{1}{2}t}}{\sqrt{1 - 4|\xi |^2}} \Big (e^{\frac{1}{2}t \sqrt{1 - 4|\xi |^2}} - e^{-\frac{1}{2}t \sqrt{1 - 4|\xi |^2}}\Big ). \end{align*}

First we prove (5.3) for $j = 0$ . Choose $\chi \in C_0^\infty \left(\mathbb R^3\right)$ such that $\textrm{supp}\chi \subset B(0, \frac{1}{4})$ and $\chi (\xi ) = 1$ for $|\xi |\leq \frac{1}{16}$ . Let

\begin{align*} U(x, t) &= \mathcal{F}^{-1} (m(t, \xi )\chi (\xi )\hat{f}) + \mathcal{F}^{-1}(m(t, \xi )(1 - \chi (\xi ))\hat{f})\\ &\,:\!=\, U_1(x, t) + U_2(x, t). \end{align*}

For $U_1(x, t)$ , since $\sqrt{1 - 4|\xi |^2}\leq 1 - 2|\xi |^2$ when $0\leq |\xi | \leq \frac{1}{4}$ , we have for $|\xi |\leq \frac{1}{4}$ that

\begin{equation*} \frac {1}{\sqrt {1 - 4|\xi |^2}} e^{-\frac {t}{2}\left(1 \pm \sqrt {1 - 4|\xi |^2}\right)}\leq 2e^{-t|\xi |^2}, \quad t\geq 0. \end{equation*}

For each $x\in \mathbb R^3$ , we have

\begin{equation*} \partial ^\alpha _x U_1(x, t) = \int _{\mathbb R^3} e^{\textrm {i}x\cdot \xi } (\textrm {i}\xi )^\alpha m(t, \xi ) \chi (\xi ) \hat {f}(\xi ) \textrm {d}\xi, \end{equation*}

which gives

\begin{align*} \sup _{x\in \mathbb R^3} |\partial _x^\alpha U_1(x, t)| \leq \int _{|\xi |\leq \frac{1}{4}} |\xi |^\alpha e^{-t|\xi |^2} |\hat{f}(\xi )|\textrm{d}\xi \lesssim \|\hat{f}\|_{L^\infty \left(\mathbb R^3\right)} \int _{|\xi |\leq \frac{1}{4}} |\xi |^\alpha e^{-t|\xi |^2}\textrm{d}\xi. \end{align*}

Since

\begin{align*} \int _{|\xi |\leq \frac{1}{4}} |\xi |^\alpha e^{-t|\xi |^2}\textrm{d}\xi \lesssim \left\{\begin{array}{ll} C, &\quad 0\leq t \leq 1,\\[5pt] t^{-\frac{3 + |\alpha |}{2}}, &\quad t\geq 1, \end{array}\right. \end{align*}

and $\|\hat{f}\|_{L^\infty \left(\mathbb R^3\right)} \lesssim \|f\|_{L^1\left(\mathbb R^3\right)}$ , we obtain

(5.5) \begin{align} \sup _{x\in \mathbb R^3}|\partial _x^\alpha U_1(x, t)| \lesssim (1 + t)^{-\frac{3 + |\alpha |}{2}} |f\|_{L^1\left(\mathbb R^3\right)} \quad \forall \alpha \in \mathbb N^3. \end{align}

To estimate $U_2(x, t)$ , noting

\begin{equation*} (1 - \Delta )^{\frac {p}{2}} U_2(x, t) = \int _{\mathbb R^3} e^{\textrm {i}x\cdot \xi } (1 + |\xi |^2)^{\frac {p}{2}} m(t, \xi ) (1 - \chi (\xi )) \hat {f}(\xi ) \textrm {d}\xi, \end{equation*}

we have from Plancherel’s theorem that

(5.6) \begin{align} \int _{\mathbb R^3} |(1 - \Delta )^{\frac{p}{2}} U_2(x, t)|^2 \textrm{d}x = \int _{\mathbb R^3} (1 + |\xi |^2)^p |m(t, \xi ) (1 - \chi (\xi )) \hat{f}(\xi )|^2 \textrm{d}\xi. \end{align}

There exists a positive constant $c$ such that

\begin{align*} |m(t, \xi )|\leq \begin{cases} te^{-\frac{t}{2}(1 - \sqrt{1 - 4|\xi |^2})} \Big |\frac{1 - e^{-t\sqrt{1 - 4|\xi |^2}}}{t\sqrt{1 - 4|\xi |^2}}\Big |\lesssim e^{-ct}, &\quad \frac{1}{16}\lt |\xi |\leq \frac{1}{2},\\[9pt] \frac{1}{2} t e^{-\frac{t}{2}} \frac{\sin \frac{t}{2}\sqrt{4|\xi |^2 - 1}}{\frac{t}{2}\sqrt{4|\xi |^2 - 1}}\lesssim e^{-ct}, &\quad \frac{1}{2}\lt |\xi |\leq 1,\\[9pt] \frac{e^{-\frac{t}{2}}}{\sqrt{4|\xi |^2 - 1}} |\sin \frac{t}{2}\sqrt{4|\xi |^2 - 1}|\leq \frac{e^{-ct}}{\sqrt{4|\xi |^2 - 1}}, &\quad |\xi |\gt 1. \end{cases} \end{align*}

Hence, when $|\xi |\geq \frac{1}{16}$ we have

\begin{equation*} |(1 + |\xi |^2)^{1/2}m(t, \xi )| \lesssim e^{-ct}. \end{equation*}

It follows from (5.6) that

\begin{align*} \|U_2(x, t)\|^2_{H^p\left(\mathbb R^3\right)} &\leq \int _{|\xi |\geq \frac{1}{16}} | (1 + |\xi |^2)^{\frac{p}{2}} m(t, \xi ) \hat{f}(\xi ) |^2 \textrm{d}\xi \\ &\leq e^{-2ct} \int _{\mathbb R^3} | (1 + |\xi |^2)^{-\frac{1}{2} + \frac{p}{2}} \hat{f}(\xi ) |^2 \textrm{d}\xi = e^{-2ct} \|f\|^2_{H^{p-1}\left(\mathbb R^3\right)}. \end{align*}

On the other hand, by Sobolev’s theorem, we have for $p\gt \frac{3}{2}$ that

\begin{equation*} \sup _{x\in \mathbb R^3} |U_2(x, t)| \leq \|U_2(\cdot, t)\|_{H^{p}\left(\mathbb R^3\right)} \lesssim e^{-ct}\|f\|_{H^{p-1}\left(\mathbb R^3\right)}. \end{equation*}

More generally, for any $\alpha \in \mathbb N^3$ it holds that

\begin{equation*} (1 - \Delta )^{\frac {p}{2}}\partial _x^\alpha U_2(x, t) = \mathcal {F}^{-1} ((1 + |\xi |^2)^{\frac {p}{2}} m(t, \xi ) (1 - \chi (\xi ))\widehat {\partial ^\alpha _x f}), \end{equation*}

which leads to

(5.7) \begin{align} \sup _{x\in \mathbb R^3} |\partial _x^\alpha U_2(x, t)| \lesssim e^{-ct} \|\partial _x^\alpha f\|_{H^{p-1}\left(\mathbb R^3\right)}\lesssim e^{-ct} \|f\|_{H^{s}\left(\mathbb R^3\right)}. \end{align}

Here $s = p - 1 + |\alpha |\gt |\alpha | + \frac{1}{2}$ by choosing $p \gt \frac{3}{2}$ . Combining the estimate (5.5) with (5.7) yields (5.3) for $j = 0.$

Next we consider the general case with $j\neq 0$ . Noting

\begin{equation*} \partial _t^j U(x, t) = \int _{\mathbb R^3} e^{\textrm {i}x\cdot \xi } \partial _t^j m(t, \xi ) \hat {f}(\xi ) \textrm {d}\xi, \end{equation*}

we obtain from direct calculations that

\begin{align*} \partial _t^j m(t, \xi ) &= \partial _t^j \left(\frac{e^{-\frac{1}{2}t}}{\sqrt{1 - 4|\xi |^2}} \Big (e^{\frac{1}{2}t \sqrt{1 - 4|\xi |^2}} - e^{-\frac{1}{2}t \sqrt{1 - 4|\xi |^2}}\Big )\right)\\ &= \sum _{l=0}^j\left (\begin{array}{c} l\\ j \end{array}\right )({-}1)^{j-l} 2^{-j} \left(\sqrt{1 - 4|\xi |^2}\right)^{l-1} e^{-\frac{t}{2}} \Big (e^{\frac{1}{2}t \sqrt{1 - 4|\xi |^2}} + ({-}1)^{l+1}e^{-\frac{1}{2}t \sqrt{1 - 4|\xi |^2}}\Big )\\ &\,:\!=\, \sum _{l=0}^j m_l(t, \xi ), \end{align*}

where $\left (\begin{array}{c} l\\ j \end{array}\right )=\frac{j!}{(j-l)!l!}$ . Hence, we can write $\partial _t^j U(x, t)$ as

(5.8) \begin{align} \partial _t^j U(x, t) = \sum _{l=0}^j \int _{\mathbb R^3} e^{\textrm{i}x\cdot \xi } m_l(t, \xi ) \hat{f}(\xi )\textrm{d}\xi \,:\!=\, \sum _{l=0}^j W_l(x, t). \end{align}

For each $0\leq l \leq j, \, j\neq 0$ , using similar arguments for the case $j = 0$ we obtain

(5.9) \begin{align} \sup _{x\in \mathbb R^3} |\partial _x^\alpha W_l(x, t)| \leq (1 + t)^{-\frac{3 + |\alpha |}{2}}\|f\|_{L^1\left(\mathbb R^3\right)} + e^{-ct}\|f\|_{H^{s}\left(\mathbb R^3\right)} \end{align}

for $s\gt l + |\alpha | - \frac{1}{2}$ . Combining (5.8) and (5.9), we obtain the general estimate (5.3).