Increasing stability for the inverse source problems in electrodynamics

Let $B_{R}=B(0,R)=\{\textbf{x}\in{\mathbb{R}}^{3}|\ |\textbf{x}|<R\}$ for $R>0$. We always assume that the source function $\textbf{F}(\textbf{x},t)$ is required to be real-valued, which implies that $\widehat{\textbf{F}}(-\xi,-\omega)=\overline{\textbf{F}}(\xi,\omega)$ for all $(\xi,\omega)\in{\mathbb{R}}^{3}\times{\mathbb{R}}$. In addition, we also assume that $\nabla\cdot\textbf{{F}}(\textbf{x},t)=0$. It follows from that the source term $\textbf{F}(\textbf{x},t)$ can be decomposed into a sum of radiation and non-radiating parts. The non-radiating part cannot be determined and gives rise to the non-uniqueness issue. By the divergence-free condition of $\textbf{F}(\textbf{x},t)$, we eliminate non-radiating sources in order to ensure the uniqueness of the inverse problem. First we show increasing stability result for the time-dependent inverse problem (IP1). Let $\textbf{F}(\textbf{x},t)=\textbf{f}(\textbf{x})g(t)$, where $\textbf{f}\in{\mathbb{R}}^{3}$ is compactly supported in $B_{R}$ and $g\in{\mathbb{R}}$ is supported in $(0,T_{0})$ for some $T_{0}>0$. It is supposed that $\textbf{f}(\textbf{x})\in H^{1}({\mathbb{R}}^{3})^{3}$ and $g(t)\in H^{4}(0,\infty)$. The one-dimensional Fourier transform of $g$ with respect to the time variable $t$ is defined as

$$\widehat{g}(\omega):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}g(t)e^{-i\omega t}dt=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}g(t)e^{-i\omega t}dt.$$

We suppose there exist a number $b>1$ and a constant $\delta>0$ such that

$\displaystyle|\widehat{g}(\omega)|\geq\delta>0\quad\mbox{for all}\quad\omega\in(0,b).$

(2.4)

Physically, the parameter $b$ in (2.4) is associated with the bandwidth of the temporal signal $g(t)$. The condition (2.4) covers a large class of functions. For example, if $g(t)=e^{-(t-1)^{2}/\eta}\chi(t)$ with some $\eta>0$ and $\chi(t)\in C_{0}^{\infty}({\mathbb{R}})$ such that $\chi(t)=0$ for $t\notin[0,T_{0}]$, the one can always find the parameters $b>0$ and $\delta>0$ such that (2.4) holds true. Since the source function is real valued, (2.4) holds true for all $\omega\in(-b,b)$. We remark that the interval $(0,b)$ in (2.4) can be replaced by $(\omega_{0}-b,\omega_{0}+b)$ for some $\omega_{0}\geq 0$. In this paper we take $\omega_{0}=0$ for simplicity. The condition (2.8) is also similar.

Define a boundary operator

$$T(\textbf{E}\times\nu)=(\nabla\times\textbf{E})\times\nu\quad\mbox{on}\ \partial B_{R}\times(0,T_{0}),$$

(2.5)

where $\nu$ is the unit normal vector on $\partial B_{R}$.

In the following theorem, we establish increasing stability estimate of the $L^{2}$-norm of f about $b$.

Next, we state the stability estimate for the second inverse problem (IP2). Let $n\in(0,b^{2})$, $b>1$. In the following theorem, we establish the increasing stability estimate of $L^{2}$-norm of F about $b$. Let $\textbf{E}=\textbf{E}(\textbf{x},t;n)$ satisfy (1.2) and (1.3).

Finally, we present the increasing stability for the third inverse problem (IP3). Assume that $\textbf{F}(\textbf{x},t)$ takes the form

$$\textbf{F}(\textbf{x},t)=\textbf{f}(\tilde{x},t)g(x_{3}),\quad\tilde{x}=(x_{1},x_{2})\in{\mathbb{R}}^{2},\quad t\in(0,+\infty),$$

(2.8)

where $\textbf{f}(\tilde{x},t)$ has compact supports such that supp f $\subset$ $\widetilde{B}_{R_{0}}\times(0,T_{0}):=\{(\tilde{x},t)\in{\mathbb{R}}^{2}\times{\mathbb{R}}|\ |\tilde{x}|<R_{0},\ 0<t<T_{0}\}$. Suppose that $g$ is given and supported in $(-R_{0},R_{0})$ for some $0<R_{0}<R/\sqrt{2}$.

The Fourier transform of $g(x_{3})$ of space variable $x_{3}$ is given by

$$\widehat{g}(\xi_{3})=\frac{1}{\sqrt{2\pi}}\int_{{\mathbb{R}}^{3}}g(x_{3})e^{-i\xi_{3}\cdot x_{3}}dx_{3}.$$

(2.9)

We suppose

$\displaystyle|\widehat{g}(\xi_{3})|\geq\delta>0\quad\mbox{for all}\quad\xi_{3}\in(-b,b),$

(2.10)

where $b>1$.

In this section. The electric current density is assumed to take the form

$$\textbf{F}(\textbf{x},t)=\textbf{f}(\textbf{x})g(t)$$

where $g$ is given. Let $\textbf{E}^{inc}$ and $\textbf{H}^{inc}$ be the electric and magnetic plane waves. Explicitly, we have

$$\textbf{E}^{inc}=\textbf{p}e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega t)},\quad\textbf{H}^{inc}=\textbf{q}e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega t)}$$

where $\textbf{d}\in{\mathbb{S}}^{2}$ is a unit vector and p, q are two unit polarization vectors satisfying $\textbf{p}\cdot\textbf{d}=0$, $\textbf{q}=\textbf{d}\times\textbf{p}$.

If $\kappa_{p}^{2}-n\omega^{2}=0$, it is easy to verify that $\textbf{E}^{inc}$ and $\textbf{H}^{inc}$ satisfy the homogeneous Maxwell equations in ${\mathbb{R}}^{3}$:

$$n\partial_{t}^{2}\textbf{E}(\textbf{x},t)+\nabla\times(\nabla\times\textbf{E}(\textbf{x},t))=0$$

and

$$n\partial_{t}^{2}\textbf{H}(\textbf{x},t)+\nabla\times(\nabla\times\textbf{H}(\textbf{x},t))=0.$$

Multiplying the both sides of (1.2) by $\textbf{E}^{inc}$, we obtain

$$\int_{0}^{T}\int_{B_{R}}(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\ \mathrm{d}\textbf{x}\mathrm{d}t=\int_{0}^{T}\int_{B_{R}}\textbf{f}(\textbf{x})g(t)\textbf{E}^{inc}(\textbf{x},t)\ \mathrm{d}\textbf{x}\mathrm{d}t.$$

(3.12)

Integrating by parts, one deduces from the left hand side of (3.12) that

	$\displaystyle\int_{0}^{T}\int_{B_{R}}$	$\displaystyle(n\partial_{t}^{2}\textbf{E}+\nabla\times(\nabla\times\textbf{E})\textbf{E}^{inc}\ \mathrm{d}\textbf{x}\mathrm{d}t$
	$\displaystyle=$	$\displaystyle\int_{B_{R}}n(\partial_{t}\textbf{E}(\textbf{x},t)\textbf{E}^{inc}(\textbf{x},t)-\textbf{E}(\textbf{x},t)\partial_{t}\textbf{E}^{inc}(\textbf{x},t))\big{|}_{0}^{T}\;\mathrm{d}\textbf{x}$
		$\displaystyle\quad+\int_{0}^{T}\int_{\partial B_{R}}\big{(}\nu\times(\nabla\times\textbf{E})\cdot\textbf{E}^{inc}-\nu\times(\nabla\times\textbf{E}^{inc})\cdot\textbf{E}\big{)}ds(\textbf{x})dt.$
	$\displaystyle=$	$\displaystyle-\int_{0}^{T}\int_{\partial B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(\textbf{x})dt.$

Note that, in the last step we have used the fact that $\textbf{E}(\textbf{x},t)=0$ when $|\textbf{x}|<R$ and $t>T_{0}+2R$, which follows straightforwardly from Huygens’ principle (see e.g., [9, Lemma 2.1]). This implies $\textbf{E}(\textbf{x},T)=\partial_{t}\textbf{E}(\textbf{x},T)=0$ for $\textbf{x}\in B_{R}$ and $T>T_{0}+2R$. Hence, the integral over $B_{R}$ on the left hand side of the previous identity vanishes.

$$\int_{B_{R}}\textbf{p}e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\textbf{f}(\textbf{x})d\textbf{x}\int_{0}^{T}g(t)e^{-i\omega t}dt=-\int_{0}^{T}\int_{\partial B_{R}}\big{(}(T(\textbf{E}\times\nu)\cdot\textbf{E}^{inc})+(\textbf{E}\times\nu)\cdot(\nabla\times\textbf{E}^{inc})\big{)}ds(x)dt$$

(3.13)

By the definition of $\textbf{E}^{inc}$, one can check that

$$\nabla\times\textbf{E}^{inc}=-i\kappa_{p}\textbf{d}\times\textbf{p}e^{-i(\kappa_{p}\textbf{x}\cdot\textbf{d}+\omega t)}$$

which gives

$$|\nabla\times\textbf{E}^{inc}|=\sqrt{n}\omega.$$

Using the assumption about $g$ and the fact that f is supported in $B_{R}$, we derive from (3.13) together with the previous two relations that

	$\displaystyle|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})\widehat{g}(\omega)|$	$\displaystyle=$	$\displaystyle\Big{|}(2\pi)^{-2}\int_{0}^{T}\int_{\mathbb{R}^{3}}\textbf{p}\cdot\textbf{f}(\textbf{x})g(t)e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}-i\omega t}\ \mathrm{d}\textbf{x}\mathrm{d}t\Big{|}$
		$\displaystyle=$	$\displaystyle\Big{|}(2\pi)^{-2}\int_{0}^{T}\int_{B_{R}}\textbf{p}\cdot\textbf{f}(\textbf{x})g(t)\textbf{E}^{inc}(\textbf{x},t)\ \mathrm{d}\textbf{x}\mathrm{d}t\Big{|}$
		$\displaystyle\leq$	$\displaystyle C(1+\omega)\epsilon,$

where $\epsilon=\Big{(}\int_{0}^{T}\int_{\partial B_{R}}\big{(}|T(\textbf{E}\times\nu)|^{2}+|\textbf{E}\times\nu|^{2}\big{)}ds(\textbf{x})dt\Big{)}^{\frac{1}{2}}$ represents the measurement data on $\partial B_{R}\times(0,T)$. In view of the assumption (2.4), one obtains for $\omega\in(0,b)$, $b>1$ and $\kappa_{p}^{2}=n\omega^{2}$ that

$$\begin{split}|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|\leq C\frac{\omega\epsilon}{|\hat{g}(\omega)|}&\leq C\,\delta^{-1}(1+\omega)\epsilon\\ &\leq Cb\epsilon,\end{split}$$

(3.14)

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$.

Repeating similar steps, we get

$\displaystyle|\textbf{q}\cdot\hat{\textbf{f}}(\kappa_{p}\textbf{d})|\leq C\,b\,\epsilon\quad\mbox{for all}\ |\kappa_{p}|<\sqrt{n}b.$

(3.15)

On the other hand, Since $\nabla\cdot\textbf{f}=0$, we obtain that

$$-i\kappa_{p}\textbf{d}\cdot\widehat{\textbf{f}}(-\kappa_{p}\textbf{d})=-i\kappa_{p}\int_{B_{R}}\textbf{d}e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\cdot\textbf{f}(\textbf{x})d\textbf{x}=\int_{B_{R}}\nabla e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\cdot\textbf{f}(\textbf{x})d\textbf{x}=\int_{B_{R}}e^{-i\kappa_{p}\textbf{x}\cdot\textbf{d}}\nabla\cdot\textbf{f}(\textbf{x})d\textbf{x}=0,$$

(3.16)

which means $\textbf{d}\cdot\widehat{\textbf{f}}(-\kappa_{p}\textbf{d})=0$. Using the Pythagorean theorem yields

$$|\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}=|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}+|\textbf{q}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}+|\textbf{d}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}=|\textbf{p}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}+|\textbf{q}\cdot\widehat{\textbf{f}}(\kappa_{p}\textbf{d})|^{2}.$$

(3.17)

Let $\xi_{p}=\kappa_{p}\textbf{d}$, we obtain from the Parseval theorem that

$$\|\textbf{f}\|_{L^{2}({\mathbb{R}}^{3})^{3}}^{2}=\|\widehat{\textbf{f}}\|_{L^{2}({\mathbb{R}}^{3})^{3}}^{2}=\int_{{\mathbb{R}}^{3}}|\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}=\int_{{\mathbb{R}}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}+\int_{{\mathbb{R}}^{3}}|\textbf{q}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}.$$

(3.18)

Denote

$$I(s)=\int_{|\xi_{p}|\leq\sqrt{n}s}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}.$$

Since the integrand $I(s)$ is an entire analytic function of $\xi$, the integral $I(s)$ with respect to $\xi_{p}$ can be taken over by any path joining points $0$ and $\sqrt{n}s$ in complex plane. Thus $I(s)$ is an entire analytic function of $s=s_{1}+is_{2}$ $(s_{1},s_{2}\in\mathbb{R})$ and the following estimate holds.

The following Lemma is essential to show the relation between $I(s)$ for $s\in(b,\infty)$ with $I(b).$ Its proof can be found in [13].

Let the sector $S\subset{\mathbb{C}}$ be given as in Lemma 3.2. Now, it follows from Lemma 3.1 that

$$|I(s)e^{-(2R\sqrt{n}+1)s}|\leq CM^{2}\quad\mbox{for all}\quad s\in S,$$

where $C>0$ depends on $n$ and $R$. Recalling from a priori estimate (3.14), we obtain

$$I(s)=\int_{|\xi_{p}|\leq\sqrt{n}s}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}\leq C\,s^{3}\,b^{2}\,\epsilon^{2},\quad s\in(0,b]$$

where $C>0$ depends on $n$, $T_{0}$ and $R$. Hence

$$|I(s)e^{-(2R\sqrt{n}+1)s}|\leq Cb^{2}\,\epsilon^{2},\ \ s\in(0,b].$$

Then applying Lemma 3.2 with $L=b$ to the function $J(s):=\frac{1}{b^{2}}I(s)e^{-(2R\sqrt{n}+1)s}$ , we know that there exists a function $\mu(s)$ satisfying

$$\begin{cases}\mu(s)\geq\frac{1}{2},\ \ &s\in(b,2^{\frac{1}{4}}b),\\ \mu(s)\geq\frac{1}{\pi}((\frac{s}{b})^{4}-1)^{-\frac{1}{2}},\ \ &s\in(2^{\frac{1}{4}}b,\infty)\end{cases}$$

such that

$$|\frac{1}{b^{2}}I(s)e^{-(2R\sqrt{n}+1)s}|\leq CM^{2}\epsilon^{2\mu}\ \ \quad\mbox{for all}\quad s\in(b,\infty),$$

where $C>0$ depends on $n$, $T_{0}$ and $R$.

Now we show the proof of Theorem 2.1. We assume that $\epsilon<e^{-1}$, otherwise the estimate is obvious. Let

$$s=\begin{cases}\frac{1}{((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{4}},\qquad&\mbox{if}\quad 2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}<|\ln\epsilon|^{\frac{1}{4}},\\ b,\qquad&\mbox{if}\quad|\ln\epsilon|^{\frac{1}{4}}\leq 2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}.\end{cases}$$

Case (i): $2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}<|\ln\epsilon|^{\frac{1}{4}}$. Then we have

	$\displaystyle|I(s)|$	$\displaystyle\leq$	$\displaystyle CM^{2}b^{2}\epsilon^{2\mu}e^{(2R\sqrt{n}+1)s}$
		$\displaystyle=$	$\displaystyle CM^{2}b^{2}e^{(2R\sqrt{n}+1)s-2\mu(s)|\ln\epsilon|}$
		$\displaystyle\leq$	$\displaystyle CM^{2}b^{2}e^{\frac{(2R\sqrt{n}+3)}{((2R+3)\pi)^{\frac{1}{3}}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{4}}-\frac{2|\ln\epsilon|}{\pi}(\frac{b}{s})^{2}}$
		$\displaystyle=$	$\displaystyle CM^{2}b^{2}e^{-2\big{(}\frac{(2R\sqrt{n}+3)^{2}}{\pi}\big{)}^{\frac{1}{3}}b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{2}}(1-\frac{1}{2}|\ln\epsilon|^{-\frac{1}{4}})},$

where $C>0$ depends on $n$, $T_{0}$ and $R$. Noting that $|\ln\epsilon|^{-\frac{1}{4}}<1$ and $\big{(}\frac{(2R\sqrt{n}+3)^{2}}{\pi}\big{)}^{\frac{1}{3}}>1$, we obtain

$$|I(s)|\leq CM^{2}b^{2}e^{-b^{\frac{2}{3}}|\ln\epsilon|^{\frac{1}{2}}}.$$

Using the inequality $e^{-t}\leq\frac{6!}{t^{6}}$ for $t>0$, we get

$$|I(s)|\leq C\frac{M^{2}}{b^{2}|\ln\epsilon|^{3}}.$$

(3.20)

Since $b^{2}|\ln\epsilon|^{3}\geq b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}$ when $b>1$ and $|\ln\epsilon|>1$. Hence

$$\begin{split}I(s)+\int_{|\xi_{p}|>\sqrt{n}s}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}\,\mathrm{d}\xi_{p}&\leq I(s)+\frac{M^{2}}{n^{2}s^{2}}\\ &\leq C(\frac{M^{2}}{b^{4}|\ln\epsilon|^{3}}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}})\\ &\leq\frac{CM^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}},\end{split}$$

(3.21)

where $C>0$ depends on $n$, $T_{0}$ and $R$.

Case (ii): $|\ln\epsilon|^{\frac{1}{4}}\leq 2^{\frac{1}{4}}((2R\sqrt{n}+3)\pi)^{\frac{1}{3}}b^{\frac{1}{3}}$. In this case we have $s=b$ by the choice of $s$ and $|I(b)|\leq C\,b^{5}\epsilon^{2}$. Hence,

$$\begin{split}I(b)+\int_{|\xi_{p}|>\sqrt{n}b}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}\,\mathrm{d}\xi_{p}\leq C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{2}})\\ \leq C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}),\end{split}$$

(3.22)

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$. Combining (3.21) and (3.22), we finally obtain

$$\int_{{\mathbb{R}}^{3}}|\textbf{p}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}\leq C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}).$$

Repeating similar steps by multiplying $\textbf{H}^{inc}$ on both sides of the equation (1.2), we get

$$\int_{{\mathbb{R}}^{3}}|\textbf{q}\cdot\widehat{\textbf{f}}(\xi_{p})|^{2}d\xi_{p}\leq C(b^{5}\epsilon^{2}+\frac{M^{2}}{b^{\frac{4}{3}}|\ln\epsilon|^{\frac{1}{2}}}),$$

where $C>0$ depends on $n$, $\delta$, $T_{0}$ and $R$. Combining the above estimates and (3.18), we complete the proof.