A theoretical study for RTE-based parameter identification problems

We assume that the spatial domain $X\subset \mathbb {R}^{3}$ is bounded with a C¹ boundary ∂X. The outward normal vector is denoted by ${\boldsymbol {\nu }}({\boldsymbol {x}})$ for ${\boldsymbol {x}}\in \partial X$. By $\Omega :=\left\lbrace {\boldsymbol {\omega }}\in {\mathbb {R}^{3}}\mid |{\boldsymbol {\omega }}|=1\right\rbrace$, we denote the unit sphere. We will need the following incoming and outgoing boundaries as subsets of Γ = ∂X × Ω:

$$\begin{equation*} \Gamma _{-}=\lbrace ({\boldsymbol {x}},{\boldsymbol {\omega }})\in \partial X\times \Omega \mid {\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}({\boldsymbol {x}})<0\rbrace ,\qquad \Gamma _{+}=\lbrace ({\boldsymbol {x}},{\boldsymbol {\omega }})\in \partial X\times \Omega \mid {\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}({\boldsymbol {x}})>0\rbrace . \end{equation*}$$

Denote by ${\rm d}\sigma ({\boldsymbol {\omega }})$ the infinitesimal area element on the unit sphere Ω. In the spherical coordinate system, ${\boldsymbol {\omega }}=(\sin \theta \cos \psi ,\,\sin \theta \sin \psi ,\,\cos \theta )^T$, for 0 ⩽ θ ⩽ π and 0 ⩽ ψ < 2π, we have ${\rm d}\sigma ({\boldsymbol {\omega }})=\sin \theta \,{\rm d}\theta \,{\rm d}\psi$. We need an integral operator S defined by

$$\begin{equation} (Su)({\boldsymbol {x}},{\boldsymbol {\omega }})=\int _{\Omega }\eta ({\boldsymbol {x}},{\boldsymbol {\omega }}\cdot \hat{{\boldsymbol {\omega }}})\,u({\boldsymbol {x}},\hat{{\boldsymbol {\omega }}})\,{\rm d}\sigma (\hat{{\boldsymbol {\omega }}}) \end{equation} \tag{ 2.1 }$$

with η a nonnegative normalized phase function:

$$\begin{equation} \int _{\Omega }\eta ({\boldsymbol {x}},{\boldsymbol {\omega }}\cdot \hat{{\boldsymbol {\omega }}})\,{\rm d}\sigma (\hat{{\boldsymbol {\omega }}})=1 \quad \forall \,{\boldsymbol {x}}\in X,\ {\boldsymbol {\omega }}\in \Omega . \end{equation} \tag{ 2.2 }$$

In many applications, the function η is independent of ${\boldsymbol {x}}$. However, in our general study, we allow η to depend on ${\boldsymbol {x}}$. Moreover, we can allow η to be a general function of ${\boldsymbol {x}}$, ${\boldsymbol {\omega }}$ and $\hat{{\boldsymbol {\omega }}}$, i.e., in the form $\eta ({\boldsymbol {x}},{\boldsymbol {\omega }},\hat{{\boldsymbol {\omega }}})$. One well-known example is the Henyey–Greenstein phase function (cf [27]):

$$\begin{equation} \eta (t)=\frac{1-g^2}{4\pi (1+g^{2}-2g t)^{3/2}},\quad t\in [-1,1], \end{equation} \tag{ 2.3 }$$

where the parameter g ∈ ( − 1, 1) is the anisotropy factor of the scattering medium. Note that g = 0 for isotropic scattering, g > 0 for forward scattering and g < 0 for backward scattering.

With the above notation, we can write the boundary value problem (BVP) of the RTE:

$$\begin{eqnarray} {\boldsymbol {\omega }}\cdot \boldsymbol{\nabla }u({\boldsymbol {x}},{\boldsymbol {\omega }})+\mu _{t}({\boldsymbol {x}})u({\boldsymbol {x}},{\boldsymbol {\omega }}) &=\mu _{s}({\boldsymbol {x}}) Su({\boldsymbol {x}},{\boldsymbol {\omega }}) + f({\boldsymbol {x}},{\boldsymbol {\omega }}) \quad {\rm in\ } X\times \Omega , \end{eqnarray} \tag{ 2.4 }$$

$$\begin{eqnarray} u({\boldsymbol {x}},{\boldsymbol {\omega }}) & =u_{\rm in}({\boldsymbol {x}},{\boldsymbol {\omega }})\quad {\rm on\ }\Gamma _{-}. \end{eqnarray} \tag{ 2.5 }$$

Here μ_t = μ_a + μ_s, μ_a is the macroscopic absorption cross section, μ_s is the macroscopic scattering cross section, u_in is the incoming flow on the incoming boundary and $f({\boldsymbol {x}},{\boldsymbol {\omega }})$ is the internal source function. The expression ${\boldsymbol {\omega }}\cdot \boldsymbol{\nabla }u({\boldsymbol {x}},{\boldsymbol {\omega }})$ is a directional derivative,

$$\begin{eqnarray*} &&{\boldsymbol {\omega }}\cdot \boldsymbol{\nabla }u({\boldsymbol {x}},{\boldsymbol {\omega }}) =\frac{\partial }{\partial s}u({\boldsymbol {$z$}}+s{\boldsymbol {\omega }},{\boldsymbol {\omega }})|_{s=0}. \end{eqnarray*}$$

In this paper, we consider the inverse problem of parameter identification through knowledge of the incoming flow and corresponding measurement data u_m on the outgoing boundary. There is no internal source so that the forward problem is

$$\begin{eqnarray} {\boldsymbol {\omega }}\cdot \boldsymbol{\nabla }u({\boldsymbol {x}},{\boldsymbol {\omega }})+\mu _{t}({\boldsymbol {x}})u({\boldsymbol {x}},{\boldsymbol {\omega }}) &=\mu _{s}({\boldsymbol {x}}) Su({\boldsymbol {x}},{\boldsymbol {\omega }}) \quad {\rm in\ } X\times \Omega , \end{eqnarray} \tag{ 2.6 }$$

$$\begin{eqnarray} u({\boldsymbol {x}},{\boldsymbol {\omega }}) & =u_{\rm in}({\boldsymbol {x}},{\boldsymbol {\omega }})\quad {\rm on\ }\Gamma _{-}. \end{eqnarray} \tag{ 2.7 }$$

We impose the following assumptions on the coefficients.

Assumptions

• (A1) The function $\mu _{t}({\boldsymbol {x}})$ is uniformly positive and bounded; i.e., there exist two positive constants $\mu _{t}^{1}$ and $\mu _{t}^{2}$, such that $0<\mu _{t}^{1}\le \mu _{t}\le \mu _{t}^{2}<\infty$ a.e. in X.
• (A2) The function $\mu _{s}({\boldsymbol {x}})$ is uniformly positive and bounded; i.e., there exist two positive constants $\mu _{s}^{1}$ and $\mu _{s}^{2}$, such that $0<\mu _{s}^{1}\le \mu _{s}\le \mu _{s}^{2}<\infty$ a.e. in X.
• (A3) There is a constant c₀ > 0 such that μ_t − μ_s ⩾ c₀ > 0 a.e. in X.

Next, we will introduce a function space for $u({\boldsymbol {x}},{\boldsymbol {\omega }})$, and show the regularity of solution of the above BVP under this space frame.

We introduce some function spaces that will be needed later in the paper:

$$\begin{equation*} H^{1,2}(X\times \Omega ):= \lbrace v\in {L^{2}(X\times \Omega )}\mid {\boldsymbol {\omega }}\cdot {\boldsymbol {$\nabla $}}{v} \in {L^{2}(X\times \Omega )}\rbrace \end{equation*}$$

is a Hilbert space with the inner product

$$\begin{equation*} (u,v)_{H^{1,2}(X\times \Omega )}:=\int _{X\times \Omega }\left({\boldsymbol {\omega }}\cdot {\boldsymbol {$\nabla $}}{u} {\boldsymbol {\omega }}\cdot {\boldsymbol {$\nabla $}}{v}+u v\right) {\rm d}x \,{\rm d}\sigma ({\boldsymbol {\omega }}) \end{equation*}$$

and the corresponding norm $\Vert v\Vert _{H^{1,2}(X\times \Omega )}:=(v,v)_{H^{1,2}(X\times \Omega )}^{1/2}$. We also need the function space L²(Γ_±) on Γ_±, which are Hilbert spaces of functions v on Γ_± with the inner products

$$\begin{equation*} (u,v)_{L^2(\Gamma _{\pm })}:=\int _{\Gamma _{\pm }} |{\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}({\boldsymbol {x}})| u({\boldsymbol {x}},{\boldsymbol {\omega }}) v({\boldsymbol {x}},{\boldsymbol {\omega }}) \,{\rm d}a \,{\rm d}\sigma ({\boldsymbol {\omega }}) \end{equation*}$$

and corresponding norms $\Vert v\Vert _{L^{2}(\Gamma {\pm })}$.

From a result on the trace of an H^{1, 2}(X × Ω) function ([1]), we know that if v ∈ H^{1, 2}(X × Ω) and $v|_{\Gamma _{-}}\in {L^{2}(\Gamma _{-})}$, then $v|_{\Gamma _{+}}\in {L^{2}(\Gamma _{+})}$, and for some constant c depending only on X, the following inequality holds:

$$\begin{equation} \Vert v\Vert _{L^2(\Gamma _{+})}\le c[\Vert v\Vert _{H^{1,2}(X\times \Omega )}+\Vert v\Vert _{L^2(\Gamma _{-})}]. \end{equation} \tag{ 2.8 }$$

It is known that for each u_in ∈ L²(Γ₋), the problem (2.6)–(2.7) has a unique solution u ∈ H^{1, 2}(X × Ω). The solution of the BVP (2.6)–(2.7) will be sought from the function space

$$\begin{equation*} H_{A}(X\times \Omega ):=\lbrace v\in {H^{1,2}(X\times \Omega )\mid v|_{\Gamma _{-}}}\in {L^2(\Gamma _{-})}\rbrace \end{equation*}$$

with the norm

$$\begin{equation*} \Vert v\Vert _{H_A(X\times \Omega )}:=\big[\Vert v\Vert _{H^{1,2}(X\times \Omega )}^2 +\Vert v|_{\Gamma _{-}}\Vert _{L^2(\Gamma _{-})}^2\big]^{1/2}. \end{equation*}$$

Proposition 2.1. If v ∈ H_A(X × Ω), then $v|_{\Gamma _{\pm }}\in {L^{2}(\Gamma _{\pm })}$, and for some constant c depending only on X,

$$\begin{equation*} \Vert v|_{\Gamma _{\pm }}\Vert _{L^{2}(\Gamma _{\pm })}\le c \Vert v\Vert _{H_{A}(X\times \Omega )}. \end{equation*}$$

Obviously, we can take c = 1 in the above inequality in bounding $\Vert v|_{\Gamma _{-}}\Vert _{L^{2}(\Gamma _{-})}$.

Corresponding to equation (2.6), we introduce the linear integro-differential operator L by the formula

$$\begin{equation*} Lv({\boldsymbol {x}},{\boldsymbol {\omega }})={\boldsymbol {\omega }}\cdot {\boldsymbol {$\nabla $}}v({\boldsymbol {x}},{\boldsymbol {\omega }})+\mu _{t}({\boldsymbol {x}})v({\boldsymbol {x}},{\boldsymbol {\omega }})-\mu _{s}({\boldsymbol {x}})(Sv)({\boldsymbol {x}},{\boldsymbol {\omega }}). \end{equation*}$$

Then we have the following result (cf [1, theorem 3.1]).

Proposition 2.2. Under the assumption on coefficients, in the space H_A(X × Ω), the norm $\Vert \cdot \Vert _{H_{A}(X\times \Omega )}$ is equivalent to the following norm:

$$\begin{equation*} [v] _{A}= \big[\Vert v\Vert _{L^{2}(\Gamma _{-})}^{2}+\Vert Lv\Vert _{L^{2}(X\times \Omega )}^{2}\big]^{1/2}. \end{equation*}$$

Consider the BVP (2.6)–(2.7), assuming that u_in ∈ L²(Γ₋). Introduce the following variational problem.

Variational problem. Find a function u ∈ H_A(X × Ω), such that

$$\begin{equation} J(u)=\inf \left\lbrace J(v)\mid v\in H_{A}(X\times \Omega )\right\rbrace , \end{equation} \tag{ 2.9 }$$

where

$$\begin{equation} J(v)=\Vert v-u_{\rm in}\Vert _{L^{2}(\Gamma _{-})}^{2}+\Vert Lv\Vert _{L^{2}(X\times \Omega )}^{2}. \end{equation} \tag{ 2.10 }$$

The following result is standard (cf [1]).

Theorem 2.1. Assume (A1)–(A3). Let u_in ∈ L²(Γ₋). Then

• (1)  
  there exists a unique solution u ∈ H_A(X × Ω) to the variational problem (2.9);
• (2)  
  the solution u satisfies the equation (2.6) a.e. in X × Ω and the boundary condition (2.7) a.e. on Γ₋;
• (3)  
  for some constants c and $\tilde{c}>0,$ independent of u and u_in, the following bounds hold:

  $$\begin{equation} c\Vert u_{\rm in}\Vert _{L^{2}(\Gamma _{-})}\le \Vert u\Vert _{H_{A}(X\times \Omega )} \le \tilde{c}\Vert u_{\rm in}\Vert _{L^{2}(\Gamma _{-})} . \end{equation} \tag{ 2.11 }$$

Based on (2.9), we introduce the weak formulation of the BVP (2.6)–(2.7):

$$\begin{equation} {find}\ u\in {H_{A}(X\times \Omega )}:\quad a(u,v)=(u_{\rm in},v)_{L^{2}(\Gamma _{-})} \quad \forall \,v\in {H_{A}(X\times \Omega )}, \end{equation} \tag{ 2.12 }$$

where

$$\begin{equation} a(u,v)=(u,v)_{L^{2}(\Gamma _{-})}+(Lu,Lv)_{L^{2}(X\times \Omega )}. \end{equation} \tag{ 2.13 }$$

Recall the assumptions (A1)–(A3). With the norm equivalence result, proposition 2.2, we can show the boundedness and $\mathit {V}$-ellipticity of the bilinear form a( ·, ·):

$$\begin{eqnarray*} &&|a(u,v)| \le M\Vert u\Vert _{H_{A}}\Vert v\Vert _{H_{A}} \quad \forall u,v\in {H_{A}(X\times \Omega )},\\ &&a(v,v) \ge \alpha \Vert v\Vert _{H_{A}}^{2} \quad \forall v\in {H_{A}(X\times \Omega )}. \end{eqnarray*}$$

Applying the Lax–Milgram lemma, we conclude that the weak formulation (2.12) has a unique solution u ∈ H_A(X × Ω). We will use the weak formulation to analyze the properties of the forward operator.

The measured light intensity in the optical tomography experiments is generally taken on the boundaries of the regions of interest. There are mainly two types of imaging systems for measurements frequently used in optical tomography, namely, optical fiber-based imaging systems and CCD camera-based imaging systems ([2, 33]). We can treat the measurement data from the imaging system as angularly averaged data or angularly resolved data. Let B be the measurement operator acting in the space H_A(X × Ω). For angularly averaged data, the measurement operator can be defined as

$$\begin{equation} B: H_{A}(X\times \Omega )\rightarrow L^{2}(\partial X), \quad (Bu)({\boldsymbol {x}})=\int _{\Omega _{\boldsymbol{x},+}}{\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}({\boldsymbol {x}})\,u({\boldsymbol {x}},{\boldsymbol {\omega }})\,{\rm d}\sigma ({\boldsymbol {\omega }}), \end{equation} \tag{ 2.14 }$$

where

$$\begin{equation*} \Omega _{\boldsymbol{x},+}:=\left\lbrace {\boldsymbol {\omega }}\in \Omega \mid {\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}({\boldsymbol {x}})>0\right\rbrace . \end{equation*}$$

This is, the outgoing current of photons on the boundary. For the angularly resolved data, the measurement operator can be defined as ([33])

$$\begin{equation*} B : H_{A}(X\times \Omega )\rightarrow L^{2}(\Gamma _{+}), \quad (Bu)({\boldsymbol {x}},{\boldsymbol {\omega }})=u({\boldsymbol {x}},{\boldsymbol {\omega }})|_ {\Gamma _{+}}. \end{equation*}$$

This is simply the outgoing boundary value of the solution of the transport equation. Obviously, the angularly resolved data are richer than the angularly averaged data because angular dependence is allowed. However, in practice, collecting angularly resolved data is too expensive and angularly averaged data are more popularly in use. In this paper, we will focus on angularly averaged measurement data (2.14).

Proposition 2.3. The angularly averaged measurement data-based measurement operator B: H_A(X × Ω) → L²(∂X) is well defined, linear and bounded, and for some constant c > 0 depending only on X, we have

$$\begin{equation} \Vert Bu\Vert _{L^{2}(\partial X)}\le c\,\Vert u\Vert _{H_{A}(X\times \Omega )}\quad \forall \,u\in H_A(X\times \Omega ). \end{equation} \tag{ 2.15 }$$

Proof. From proposition 2.1,

$$\begin{equation*} \Vert u\Vert _{L^{2}(\Gamma _{+})}\le \Vert u\Vert _{H_{A}(X\times \Omega )}\quad \forall u\in H_A(X\times \Omega ). \end{equation*}$$

Let us bound $\Vert Bu\Vert _{L^{2}(\partial X)}$ in terms of $\Vert u\Vert _{L^{2}(\Gamma _{+})}$:

$$\begin{eqnarray*} \Vert Bu\Vert _{L^{2}(\partial X)}^{2}&=\int _{\partial X}\left|\int _{\Omega _{\boldsymbol{x},+}} {\boldsymbol {\nu }}({\boldsymbol {x}})\cdot {\boldsymbol {\omega }}u({\boldsymbol {x}},{\boldsymbol {\omega }}) \,{\rm d}\sigma ({\boldsymbol {\omega }})\right|^{2}\,{\rm d}a \\ &\le 4 \pi \int _{\partial X}\int _{\Omega _{\boldsymbol{x},+}}\left|{\boldsymbol {\nu }}({\boldsymbol {x}})\cdot {\boldsymbol {\omega }}\right| u({\boldsymbol {x}},{\boldsymbol {\omega }})^{2}\,{\rm d}\sigma ({\boldsymbol {\omega }}) \,{\rm d}a \\ &=4 \pi \Vert u\Vert _{L^{2}(\Gamma _{+})}^{2}. \end{eqnarray*}$$

Thus, (2.15) holds. □

To prove the continuity and differentiability of the forward operator F, we first recall the solvability of the governing BVP

$$\begin{equation*} Lu=f \ {\rm in}\ X\times \Omega , \quad u=u_{\rm in} \ {\rm on} \ \Gamma _{-}, \end{equation*}$$

where f ∈ L²(X × Ω). Similar to theorem 2.3, the following result holds.

Theorem 3.1. Assume (A1)–(A3) and f ∈ L²(X × Ω), u_in ∈ L²(Γ₋). Then there is a unique solution u ∈ H_A(X × Ω) to the variational problem:

$$\begin{equation*} (u,v)_{L^{2}(\Gamma _{-})}+(Lu,Lv)_{L^{2}(X\times \Omega )}=(f,Lv)_{L^{2}(X\times \Omega )} +(u_{\rm in},v)_{L^{2}(\Gamma _{-})}\quad \forall v\in {H_{A}(X\times \Omega )}. \end{equation*}$$

Moreover,

$$\begin{equation*} \Vert u\Vert _{H_{A}(X\times \Omega )}\le C(\Vert f \Vert _{L^{2}(X\times \Omega )}+\Vert u_{\rm in}\Vert _{L^{2}(\Gamma _{-})}) \end{equation*}$$

with a constant C depending only on the domain and the bounds of the coefficients.

Theorem 3.2. Assume (A1)–(A3) on (μ_t, μ_s) and $(\tilde{\mu }_{t},\tilde{\mu }_{s})$, and let u and $\tilde{u}$ denote the solutions of (2.6)–(2.7) with u_in ∈ L²(Γ₋) corresponding to coefficients (μ_t, μ_s) and $(\tilde{\mu }_{t},\tilde{\mu }_{s})$, respectively. Then

$$\begin{equation} \Vert u-\tilde{u}\Vert _{H_{A}(X\times \Omega )}\le C(\Vert \mu _{t}-\tilde{\mu }_{t}\Vert _{L^{\infty }(X)} +\Vert \mu _{s}-\tilde{\mu }_{s}\Vert _{L^{\infty }(X)})\Vert u_{\rm in}\Vert _{L^{2}(\Gamma _{-})} , \end{equation} \tag{ 3.2 }$$

with a constant C depending only on the domain and the bounds of the coefficients.

Proof. We use the weak formulation (2.12) for the solution $\tilde{u}$ with the parameters $(\tilde{\mu }_{t},\tilde{\mu }_{s})$ and the solution u with the parameters (μ_t, μ_s). Then the difference $w:=u-\tilde{u}$ satisfies

$$\begin{equation*} (w,v)_{L^{2}(\Gamma _{-})}+(Lw,Lv)_{L^{2}(X\times \Omega )} =-((L-\tilde{L})\tilde{u},Lv)_{L^{2}(X\times \Omega )}\quad \forall v\in {H_{A}(X\times \Omega )}, \end{equation*}$$

where $\tilde{L}$ denotes the linear operator in (2.6) corresponding to $(\tilde{\mu }_{t},\tilde{\mu }_{s})$ and we used the fact that $\tilde{L}\tilde{u}=0$ a.e. in X × Ω.

Applying theorem 3.1,

$$\begin{equation*} \Vert w\Vert _{H_{A}(X\times \Omega )}\le c \Vert (L-\tilde{L})\tilde{u}\Vert _{L^{2}(X\times \Omega )}. \end{equation*}$$

Observe that

$$\begin{equation*} (L-\tilde{L})(\cdot )=(\mu _{t}-\tilde{\mu }_{t})(\cdot )-(\mu _{s}-\tilde{\mu }_{s})S(\cdot ). \end{equation*}$$

We can then obtain the inequality (3.2). □

As a corollary of theorem 3.2 and proposition 2.4, we can obtain the Lipsichitz continuity property of the forward operator.

Theorem 3.3 (Lipsichitz continuity). Let the assumptions (A1)–(A3) hold. Then the forward operator F: D(F) → L²(∂X) is Lipschitz continuous with respect to the topologies of L^∞(X) × L^∞(X) and L²(∂X).

The Lipschitz continuity of the forward operator F indicates a certain differentiability. Differentiability is very important in analyzing convergence properties of the regularization method.

Theorem 3.4 (Differentiability). Let u_in ∈ L²(Γ₋) be given, and let (μ_t, μ_s) ∈ D(F), (δμ_t, δμ_s) ∈ D(F) such that (μ_t + αδμ_t, μ_s + αδμ_s) ∈ D(F) for all $\alpha \in {\mathbb {R}}$ with |α| sufficiently small. Then the Fr$\acute{e}$chet derivative of the forward operator F at (μ_t, μ_s) in the direction (δμ_t, δμ_s) is given by

$$\begin{equation*} F^{\prime }(\mu _{t},\mu _{s})[\delta \mu _{t},\delta \mu _{s}] =\int _{\Omega _{\boldsymbol{x},+}} |{\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}({\boldsymbol {x}})| w({\boldsymbol {x}},{\boldsymbol {\omega }}) \,{\rm d}\sigma ({\boldsymbol {\omega }}), \end{equation*}$$

where w is the solution of the following problem:

$$\begin{equation} (w,v)_{L^{2}(\Gamma _{-})}+(Lw,Lv)_{L^{2}(X\times \Omega )} =-(\delta \mu _{t}u,Lv)_{L^{2}(X\times \Omega )} +(\delta \mu _{s}Su,Lv)_{L^{2}(X\times \Omega )} \end{equation} \tag{ 3.3 }$$

for all v ∈ H_A(X × Ω), and u is the solution of BVP (2.6)–(2.7). Moreover,

$$\begin{equation} \Vert F^{\prime }(\mu _{t},\mu _{s})[\delta \mu _{t},\delta \mu _{s}]\Vert _{L^{2}(\partial X)} \le C(\Vert \delta \mu _{t}\Vert _{L^{\infty }(X)}+\Vert \delta \mu _{s}\Vert _{L^{\infty }(X)}) \Vert u_{\rm in}\Vert _{L^{2}(\Gamma _{-})} \end{equation} \tag{ 3.4 }$$

with a constant C depending only on the domain and the bounds of the coefficients.

The proof of theorem 3.4 is similar to that of theorem 3.2 and that of theorem 3.5 below, and is hence omitted. Thus, F'(μ_t, μ_s) defines a bounded linear operator mapping from L^∞(X) × L^∞(X) to L²(∂X) and the estimate (3.4) holds.

Theorem 3.5. The linear Taylor expansion of the forward operator F around (μ_t, μ_s) ∈ D(F) is second-order accurate, i.e., for $(\tilde{\mu }_{t},\tilde{\mu }_{s})\in {D(F)}$ the following bound holds:

$$\begin{equation} \Vert \tilde{u}-u-w\Vert _{L^{2}(\Gamma _{+})}\le C_{L}(\Vert \tilde{\mu }_{t}-\mu _{t}\Vert _{L^{\infty }(X)}^{2} +\Vert \tilde{\mu }_{s}-\mu _{s}\Vert _{L^{\infty }(X)}^{2})\Vert u_{\rm in}\Vert _{L^{2}(\Gamma _{-})} \end{equation} \tag{ 3.5 }$$

with a constant C_L > 0 depending only on the domain and the bounds of the coefficients, where $\tilde{u}$ and u are the solutions of the BVP (2.6)–(2.7) corresponding to $(\tilde{\mu }_{t},\tilde{\mu }_{s})$ and (μ_t, μ_s), respectively, and w is the solution to the problem (3.3).

Proof. Let $m:=\tilde{u}-u-w$. Using (2.12) and (3.3), we can obtain

$$\begin{eqnarray*} &&(m,v)_{L^{2}(\Gamma _{-})}+(Lm,Lv)_{L^{2}(X\times \Omega )} ={-}(\delta \mu _{t}(\tilde{u}-u),Lv)_{L^{2}(X\times \Omega )} +(\delta \mu _{s}S(\tilde{u}-u),Lv)_{L^{2}(X\times \Omega )} \end{eqnarray*}$$

for all v ∈ H_A(X × Ω). Applying theorems 3.1 and 3.2, we obtain from the previous equality that

$$\begin{eqnarray*} \Vert m\Vert _{H_{A}(X\times \Omega )}& \le C(\Vert \delta \mu _{t}\Vert _{L^{\infty }(X)}+\Vert \delta \mu _{s}\Vert _{L^{\infty }(X)})\Vert \tilde{u}-u\Vert _{L^{2}(X\times \Omega )}\\ &\le C(\Vert \delta \mu _{t}\Vert _{L^{\infty }(X)}+\Vert \delta \mu _{s}\Vert _{L^{\infty }(X)})\Vert \tilde{u}-u\Vert _{H_{A}(X\times \Omega )}\\ &\le C(\Vert \delta \mu _{t}\Vert _{L^{\infty }(X)}^{2}+\Vert \delta \mu _{s}\Vert _{L^{\infty }(X)}^{2})\Vert u_{\rm in}\Vert _{L^{2}(\Gamma _{-})}. \end{eqnarray*}$$

The result (3.5) then follows. □

A formula for the derivative operator F'(μ_t, μ_s) was derived in theorem 3.4. Here, we derive an adjoint equation for the calculation of the adjoint of the derivative operator F'(μ_t, μ_s). The adjoint equation is important in numerical calculations as well as in analyzing the back transport effect in application. Recall that the $Fr\acute{e}chet$ derivative of the forward operator F can be computed by solving the following BVP (cf (3.3)):

$$\begin{eqnarray} {\boldsymbol {\omega }}\cdot \boldsymbol{\nabla }w+\mu _{t}w-\mu _{s} Sw &= -\delta \mu _{t}u+\delta \mu _{s} Su \quad {\rm in\ } X\times \Omega , \end{eqnarray} \tag{ 3.6 }$$

$$\begin{eqnarray} w & =0\quad {\rm on\ }\Gamma _{-}. \end{eqnarray} \tag{ 3.7 }$$

Theorem 3.6 (Adjoint problem). Let (μ_t, μ_s) ∈ D(F). Assume u_in ∈ L^∞(Γ₋). Then the adjoint of the derivative operator F'(μ_t, μ_s) is given by

$$\begin{eqnarray*} &&F^{\prime }(\mu _{t},\mu _{s})^{*}: L^{2}(\partial X) \rightarrow L^{\infty }(X)^{*}\times L^{\infty }(X)^{*},\\ &&\zeta \mapsto ({\rm d}\mu _{t},{\rm d}\mu _{s}), \end{eqnarray*}$$

where

$$\begin{eqnarray} &&{\rm d}\mu _{t}:= {-}\int _{\Omega }u({\boldsymbol {x}},{\boldsymbol {\omega }})\varphi ({\boldsymbol {x}},{\boldsymbol {\omega }})\,{\rm d}\sigma ({\boldsymbol {\omega }}), \end{eqnarray} \tag{ 3.8 }$$

$$\begin{eqnarray} {\rm d}\mu _{s}&:= \int _{\Omega }(Su)({\boldsymbol {x}},{\boldsymbol {\omega }})\varphi ({\boldsymbol {x}},{\boldsymbol {\omega }})\,{\rm d}\sigma ({\boldsymbol {\omega }}). \end{eqnarray} \tag{ 3.9 }$$

Here, u denotes the solution to the forward problem (2.6)–(2.7) and $\varphi ({\boldsymbol {x}},{\boldsymbol {\omega }})$ is the solution of the following adjoint problem:

$$\begin{eqnarray} {-}{\boldsymbol {\omega }}\cdot {\boldsymbol {$\nabla $}}\varphi ({\boldsymbol {x}},{\boldsymbol {\omega }})+\mu _{t}\varphi ({\boldsymbol {x}},{\boldsymbol {\omega }})-\mu _{s}(S\varphi )({\boldsymbol {x}},{\boldsymbol {\omega }}) &=0\quad {\rm in} \ X\times \Omega \end{eqnarray} \tag{ 3.10 }$$

$$\begin{eqnarray} &&\varphi =\zeta \quad {\rm on} \ \Gamma _{+}. \end{eqnarray} \tag{ 3.11 }$$

Proof. We multiply both sides of the equation (3.6) by φ and integrate

$$\begin{eqnarray} &&\int _{X}{\rm d}x \int _{\Omega }{\rm d}\sigma ({\boldsymbol {\omega }}) \left({\boldsymbol {\omega }}\cdot \boldsymbol{\nabla }w+\mu _{t}w-\mu _{s} Sw\right)\varphi = \int _{X}{\rm d}x \int _{\Omega }\,{\rm d}\sigma ({\boldsymbol {\omega }}) \left(-\delta \mu _{t}u+\delta \mu _{s} Su \right)\varphi. \end{eqnarray} \tag{ 3.12 }$$

The left side of (3.12) is equal to

$$\begin{eqnarray*} &&\int _\Gamma {\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}w \varphi \,{\rm d}\sigma ({\boldsymbol {\omega }}) \,{\rm d}a +\int _{X\times \Omega }\left(-{\boldsymbol {\omega }}\cdot {\boldsymbol {$\nabla $}}\varphi +\mu _t \varphi -\mu _sS\varphi \right) w \,{\rm d}x \,{\rm d}\sigma ({\boldsymbol {\omega }}) \\ &&= \int _{\Gamma _{+}} {\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}w \varphi \,{\rm d}\sigma ({\boldsymbol {\omega }}) \,{\rm d}a\\ &&= \int _{\partial X} \Biggl (\int _{\Omega _{\boldsymbol{x},+}}|{\boldsymbol {\omega }}\cdot {\boldsymbol {\nu }}| w \,{\rm d}\sigma ({\boldsymbol {\omega }})\Biggr ) \varphi \,{\rm d}a, \end{eqnarray*}$$

whereas the right side of (3.12) is

$$\begin{equation*} \int _{X}\left\lbrace{-}\delta \mu _{t}\int _{\Omega } u \varphi \,{\rm d}\sigma ({\boldsymbol {\omega }}) +\delta \mu _{s}\int _{\Omega } Su \varphi \,{\rm d}\sigma ({\boldsymbol {\omega }})\right\rbrace \,{\rm d}x. \end{equation*}$$

So

$$\begin{equation*} F^{\prime }(\mu _{t},\mu _{s})^{*}\zeta = ({\rm d}\mu _{t},{\rm d}\mu _{s}) \end{equation*}$$

with dμ_t defined by (3.8) and dμ_s defined by (3.9).

Since u_in ∈ L^∞(Γ₋), by [1, lemma 3.2], u ∈ L^∞(X × Ω). By a variant of theorem 2.3, φ ∈ L²(X × Ω). Then it is easy to see dμ_t ∈ L²(X). Similarly, dμ_s ∈ L²(X). □

The main result here is on the solution existence to both problems (P_β) and (P). First, we recall some properties of the BV(X) space ([5, 19]).

Proposition 4.1. 

• (i)  
  Let {μⁿ} be a bounded sequence in BV(X). Then there are a subsequence $\lbrace \mu ^{n_j}\rbrace$ and an element μ ∈ BV(X) such that $\lbrace \mu ^{n_j}\rbrace$ converges to μ in L¹(X).
• (ii)  
  Let {μⁿ} be a sequence in BV(X) which converges to μ in L¹(X). Then μ ∈ BV(X) and

  $$\begin{equation*} \int _{X}|\nabla \mu _{t}|+\int _{X}|\nabla \mu _{s}|\le \liminf _{n\rightarrow \infty }\int _{X}|\nabla \mu _{t}^{n}|+\int _{X}|\nabla \mu _{s}^{n}|. \end{equation*}$$

Note that if {μⁿ}⊂BV(X) converges to μ ∈ BV(X), then

$$\begin{equation} \lim _{n\rightarrow \infty }\int _{X}|\nabla \mu _{t}^{n}|+\int _{X}|\nabla \mu _{s}^{n}| =\int _{X}|\nabla \mu _{t}|+\int _{X}|\nabla \mu _{s}|. \end{equation} \tag{ 4.3 }$$

Theorem 4.1. 

• (i)  
  A solution μ_β of the problem (P_β) exists.
• (ii)  
  A solution μ† of the problem (P) exists.

Proof. (i) It is clear that J_β(μ) ⩾ 0 on the admissible set Q_ad. There exists a minimizing sequence {μⁿ} ∈ Q_ad such that

$$\begin{eqnarray} &&\lim _{n\rightarrow \infty }J_{\beta }(\mu ^{n})=\inf _{\mu \in {Q_{\rm ad}}}J_{\beta }(\mu ). \end{eqnarray} \tag{ 4.4 }$$

Hence {J_β(μⁿ)} is bounded, J_β(μⁿ) ⩽ J₀ for some constant J₀. By the definition of J_β(μ) with μ replaced by μⁿ,

$$\begin{eqnarray*} &&\frac{1}{2}\sum _{k=1}^{k_{0}}\Vert m_{k}^{n}-m_{k}\Vert _{L^{2} (\partial X)}^{2} +\beta \left(\int _{X}|\nabla \mu _{t}^{n}|+\int _{X}|\nabla \mu _{s}^{n}|\right)\le J_{0}. \end{eqnarray*}$$

Here, we denote by $u_{k}^{n}=U_{k}(\mu ^{n})$ the solution of the BVP (2.6)–(2.7) with μ replaced by μⁿ, and

$$\begin{equation*} m_{k}^{n}({\boldsymbol {x}})=\int _{\Omega _{\boldsymbol{x},+}}{\boldsymbol {\nu }}\cdot {\boldsymbol {\omega }}u_{k}^{n} \,{\rm d}\sigma ({\boldsymbol {\omega }}) \end{equation*}$$

is the corresponding measurement on the boundary. From the last inequality, $\lbrace m_{k}^{n}\rbrace$ is bounded in L²(∂X), $\lbrace \mu _{t}^{n}\rbrace$ and $\lbrace \mu _{s}^{n}\rbrace$ are bounded in BV(X). On the other hand, from theorem 2.3,

$$\begin{equation*} \Vert u_{k}^{n}\Vert _{H_{A}(X\times \Omega )}\le C\Vert u_{\rm in,\mathit {k}}\Vert _{L^{2}(\Gamma _{-})}. \end{equation*}$$

So $\lbrace u_{k}^{n}\rbrace$ is uniformly bounded in H_A(X × Ω). By possibly extracting a subsequence, there exist $\mu _{t}^{*}$, $\mu _{s}^{*}$ and $u^*_k$ satisfying

$$\begin{equation} \mu _t^{n_1}\rightarrow \mu _t^{*} \ {\rm and\ } \mu _s^{n_1}\rightarrow \mu _s^{*} \ {\rm in}\ L^{1}(X),\quad u_{k}^{n_1}\rightharpoonup u_k^{*}\ {\rm in} \ H_A(X\times \Omega ). \end{equation} \tag{ 4.5 }$$

Then, there exist further subsequences $\lbrace \mu _t^{n_2}\rbrace \subset \lbrace \mu _t^{n_1}\rbrace$ and $\lbrace \mu _s^{n_2}\rbrace \subset \lbrace \mu _s^{n_1}\rbrace$ such that $\mu _t^{n_2}\rightarrow \mu _t^{*}$ and $\mu _s^{n_2}\rightarrow \mu _s^{*}$ a.e. on X. Note that

$$\begin{eqnarray} &&u_{k}^{n_{2}}\rightharpoonup u_{k}^{*}\quad {\rm in} \ H_{A}(X\times \Omega ). \end{eqnarray} \tag{ 4.6 }$$

Since $\mu _{t}^{n_{2}}\in {Q_{1}}$ for all n₂, we have $\mu _{t}^{*}\in {Q_{1}}$, and hence $\mu _{t}^{*}\in {Q_{1}\cap BV(X)}$. Similarly, $\mu _{s}^{*}\in {Q_{2}\cap BV(X)}$. From proposition 4.1,

$$\begin{eqnarray} &&\int _{X}|\nabla \mu _{t}^{*}|\le \liminf _{n_2\rightarrow \infty }\int _{X}|\nabla \mu _{t}^{n_{2}}|, \quad \int _{X}|\nabla \mu _{s}^{*}|\le \liminf _{n_2\rightarrow \infty }\int _{X}|\nabla \mu _{s}^{n_{2}}|. \end{eqnarray} \tag{ 4.7 }$$

Now let us show that for k = 1, ..., k₀, $F_k(\mu ^{*})=m_k^{*}$ on ∂X. We first note that

$$\begin{eqnarray} &&(u_{k}^{n_{2}},v)_{L^{2}(\Gamma _{-})}+(L^{n_{2}}u_{k}^{n_{2}},L^{n_{2}}v)_{L^{2}(X\times \Omega )} =(u_{\rm in,\mathit {k}},v)\quad \forall v\in {H_{A}(X\times \Omega )}. \end{eqnarray} \tag{ 4.8 }$$

Here, $L^{n_{2}}$ denotes the linear integro-differential operator corresponding to the coefficients $\mu ^{n_{2}}$. Next, let us denote by L* the linear integro-differential operator with the coefficients μ*. Let v ∈ H_A(X × Ω) be arbitrary but fixed. We will show that

$$\begin{equation} (u_{k}^{n_{2}},v)_{L^{2}(\Gamma _{-})}+(L^{n_{2}}u_{k}^{n_{2}},L^{n_{2}}v)_{L^{2}(X\times \Omega )}\rightarrow (u_{k}^{*},v)_{L^{2}(\Gamma _{-})}+(L^{*}u_{k}^{*},L^{*}v)_{L^{2}(X\times \Omega )}. \end{equation} \tag{ 4.9 }$$

Since $u_{k}^{n_{2}}\rightharpoonup u_{k}^{*}$ in H_A(X × Ω), by the definition of space H_A(X × Ω), $u_{k}^{n_{2}}\rightharpoonup u_{k}^{*}$ also in L²(Γ₋), which implies that

$$\begin{equation*} (u_{k}^{n_{2}},v)_{L^{2}(\Gamma _{-})}\rightarrow (u_{k}^{*},v)_{L^{2}(\Gamma _{-})}. \end{equation*}$$

Moreover,

$$\begin{eqnarray} &&\left| (L^{n_{2}}u_{k}^{n_{2}},L^{n_{2}}v)-(L^{*}u_{k}^{*},L^{*}v)\right| \le \left| \left(L^{n_{2}}u_{k}^{n_{2}},(L^{n_{2}}-L^{*})v\right)\right|\nonumber\\ &&+\,\left|\left((L^{n_{2}}-L^{*})u_{k}^{n_{2}},L^{*}v\right)\right| +\left|\left(L^{*}(u_{k}^{n_{2}}-u_{k}^{*}),L^{*}v\right)\right|. \end{eqnarray} \tag{ 4.10 }$$

For the first item on the right,

$$\begin{eqnarray*} \left| \left(L^{n_{2}}u_{k}^{n_{2}},(L^{n_{2}}-L^{*})v\right)\right| &=\left| \left(L^{n_{2}}u_{k}^{n_{2}},(\mu _{t}^{n_{2}}-\mu _{t}^{*})v-(\mu _{t}^{n_{2}}-\mu _{t}^{*})Sv\right)\right| \\ &\le \left| \left(L^{n_{2}}u_{k}^{n_{2}},(\mu _{t}^{n_{2}}-\mu _{t}^{*})v\right)\right| + \left| \left(L^{n_{2}}u_{k}^{n_{2}}, (\mu _{s}^{n_{2}}-\mu _{s}^{*})Sv\right)\right|. \end{eqnarray*}$$

Now,

$$\begin{equation*} \left| \left(L^{n_{2}}u_{k}^{n_{2}},(\mu _{t}^{n_{2}}-\mu _{t}^{*})v\right)\right| \le \left\Vert L^{n_{2}}u_{k}^{n_{2}}\right\Vert _{L^2(X\times \Omega )} \left\Vert (\mu _{t}^{n_{2}}-\mu _{t}^{*})v\right\Vert _{L^2(X\times \Omega )}. \end{equation*}$$

The values $\lbrace \Vert L^{n_{2}}u_{k}^{n_{2}}\Vert _{L^2(X\times \Omega )}\rbrace$ are uniformly bounded. Since $|\mu _{t}^{n_{2}}-\mu _{t}^{*}|\le \mu _{t}^{2}-\mu _t^1$ and $\mu _{t}^{n_{2}}$ converge to $\mu _{t}^{*}$ almost everywhere, an application of the Lebesgue dominant convergence theorem on the term $\left\Vert (\mu _{t}^{n_{2}}-\mu _{t}^{*})v\right\Vert _{L^2(X\times \Omega )}$ shows that

$$\begin{equation*} \left| \left(L^{n_{2}}u_{k}^{n_{2}},(\mu _{t}^{n_{2}}-\mu _{t}^{*})v\right)\right|\rightarrow 0. \end{equation*}$$

Similarly,

$$\begin{equation*} \left| \left(L^{n_{2}}u_{k}^{n_{2}},(\mu _{s}^{n_{2}}-\mu _{s}^{*})Sv\right)\right| \rightarrow 0. \end{equation*}$$

Thus, we obtain

$$\begin{eqnarray*} &&\left| \left(L^{n_{2}}u_{k}^{n_{2}},(L^{n_{2}}-L^{*})v\right)\right| \rightarrow 0. \end{eqnarray*}$$

For the second term on the right of (4.10), write

$$\begin{equation*} \left((L^{n_{2}}-L^{*})u_{k}^{n_{2}},L^{*}v\right) =\left(u_{k}^{n_{2}},(\mu _t^n-\mu _t) L^{*}v\right) +\left(Su_{k}^{n_{2}},(\mu _s^n-\mu _s) L^{*}v\right). \end{equation*}$$

By the same argument, we then conclude that

$$\begin{eqnarray*} &&\left| \left((L^{n_{2}}-L^{*})u_{k}^{n_{2}},L^{*}v\right) \right| \rightarrow 0. \end{eqnarray*}$$

For the last term on the right of (4.10), we have

$$\begin{eqnarray*} &&\left| \left(L^{*}(u_{k}^{n_{2}}-u_{k}^{*}),L^{*}v\right)\right| \rightarrow 0 \end{eqnarray*}$$

due to the condition that $u_{k}^{n_{2}}\rightharpoonup u_{k}^{*}$ in H_A(X × Ω).

Combining the above results, we see that (4.9) is valid, implying that for 1 ⩽ k ⩽ k₀, $u_{k}^{*}$ is the solution of the BVP (2.6)–(2.7) with the coefficients $\mu _t^*$ and $\mu _s^*$ corresponding to the inflow value $u_{\rm in,\mathit {k}}$. Furthermore, $F_{k}(\mu ^{*})=m_{k}^{*}$ on ∂X.

Now, let us check that $u_{k}^{*}$ is a minimizer of problem (P_β). Note that $u_{k}^{n_{2}}\rightharpoonup u_{k}^{*}$ in H_A(X × Ω), so by the definition of space H_A(X × Ω) and [1, lemma 2.2], we can conclude that $m_{k}^{n_{2}}\rightharpoonup m_{k}^{*}$ in L²(∂X). Then, by [4, exercise 2.7.2], for 1 ⩽ k ⩽ k₀,

$$\begin{equation*} \Vert m_k^{*}-m_{k}\Vert _{L^2(\partial X)}\le \liminf _{n_2\rightarrow \infty }\Vert m_k^{n_{2}}-m_{k}\Vert _{L^2(\partial X)}, \end{equation*}$$

so

$$\begin{equation*} \frac{1}{2}\sum _{k=1}^{k_{0}}\Vert m_{k}^{*}-m_{k}\Vert _{L^{2}(\partial X)}^{2}\le \liminf _{n_2\rightarrow \infty } \frac{1}{2}\sum _{k=1}^{k_{0}}\Vert m_{k}^{n_{2}}-m_{k}\Vert _{L^{2}(\partial X)}^{2}. \end{equation*}$$

Together with (4.7), we have

$$\begin{equation*} J_{\beta }(\mu _{t}^{*},\mu _{s}^{*})\le \liminf _{n_2\rightarrow \infty }J_{\beta } (\mu _{t}^{n_2},\mu _{s}^{n_2})=\inf _{\mu \in {Q_{\rm ad}}}J_{\beta }(\mu ). \end{equation*}$$

This completes the proof of (i).

(ii) Let {μⁿ} be a sequence in Q_ad({m_{t, k}}) such that

$$\begin{eqnarray} &&\lim _{n} J_{2}(\mu ^{n})=\inf _{\mu \in {Q_{\rm ad}(\lbrace m_{t,k}\rbrace )}}J_{2}(\mu ). \end{eqnarray} \tag{ 4.11 }$$

From the assumption on the absorption and scattering coefficients, it is obvious that {μⁿ} is bounded in L¹(X). Following from (4.11), we know that {μⁿ} is bounded in BV(X). Hence, there exist a subsequence $\lbrace \mu ^{n_{1}}\rbrace$ of {μⁿ} and an element μ† such that $\lbrace \mu ^{n_{1}}\rbrace$ converges to μ† in L¹(X) and

$$\begin{eqnarray} &&J_{2}(\mu ^{\dagger })\le \liminf _{n}J_{2}(\mu ^{n_{1}}). \end{eqnarray} \tag{ 4.12 }$$

A subsequence $\lbrace \mu ^{n_{2}}\rbrace$ of $\lbrace \mu ^{n_{1}}\rbrace$ exists, converging to μ† a.e. on X. Since $\lbrace \mu ^{n_{2}}\rbrace \subset Q_{1}\times Q_{2}$, μ† ∈ Q_ad. On the other hand, for k = 1, ..., k₀, we have

$$\begin{equation*} (u_{\rm in,\mathit {k}},v)=(u_{k}^{n_{2}},v)_{L^{2}(\Gamma _{-})} +(L^{n_{2}}u_{k}^{n_{2}},L^{n_{2}}v)_{L^{2}(X\times \Omega )}\quad \forall v\in {H_{A}(X\times \Omega )}. \end{equation*}$$

As in the proof of (i), we have $u_{k}^{n_{2}}\rightharpoonup u_{k}^{\dagger }$ in H_A(X × Ω), $u_{k}^{\dagger }$ is the solution of the BVP:

$$\begin{equation*} (u_{\rm in,\mathit {k}},v)=(u_{k}^{\dagger },v)_{L^{2}(\Gamma _{-})} +(L^{\dagger }u_{k}^{\dagger },L^{\dagger }v)_{L^{2}(X\times \Omega )} \end{equation*}$$

and satisfies the measurement boundary condition. Thus, μ† ∈ Q_ad({m_{t, k}}). Furthermore, from inequalities (4.11) and (4.12), we have

$$\begin{eqnarray*} &&J_{2}(\mu ^{\dagger })\le \inf _{\mu \in Q_{\rm ad}(\lbrace m_{t,k}\rbrace )}J_2(\mu ). \end{eqnarray*}$$

This means μ† is a solution of the problem (P). Thus, the proof is completed. □

Theorem 4.2 (Stability). Let the regularization parameter β > 0 be fixed. For 1 ⩽ k ⩽ k₀, let $\lbrace {m_{k}^{n}}\rbrace$ be a sequence which converges to m_k in L²(∂X), and let ${\mu _{\beta }^{n}}$ be the minimizers of the problems:

$$\begin{eqnarray*} &&\inf _{\mu \in {Q_{\rm ad}}}J_{1}(\mu ;m_{k}^{n})+\beta J_{2}(\mu ). \end{eqnarray*}$$

Here, to show explicitly the dependence of J₁ on the measurement data, we use $J_{1}(\mu ;m_{k}^{n})$ instead of J₁(μ). Then, a subsequence $\lbrace {\mu _{\beta }^{n_2}}\rbrace$ of $\lbrace \mu _{\beta }^{n}\rbrace$ and $\tilde{\mu }\in {Q_{\rm ad}}$ exist, such that $\lbrace {\mu _{\beta }^{n_2}}\rbrace$ converges to $\tilde{\mu }$ in L¹(X) and

$$\begin{eqnarray} &&\lim _{n_2\rightarrow \infty }\left(\int _{X}|\nabla (\mu _{t})_{\beta }^{n_2}|+\int _{X}|\nabla (\mu _{s})_{\beta }^{n_2}|\right) =\int _{X}|\nabla \tilde{\mu }_{t}|+\int _{X}|\nabla \tilde{\mu }_{s}|. \end{eqnarray} \tag{ 4.13 }$$

Furthermore, $\tilde{\mu }$ is a solution to (P_β).

Proof. For all n and μ ∈ Q_ad, we have

$$\begin{eqnarray} &&J_{1}(\mu _{\beta }^{n};m_{k}^{n})+\beta J_{2}(\mu _{\beta }^{n})\le J_{1}(\mu ;m_{k}^{n})+\beta J_{2}(\mu ). \end{eqnarray} \tag{ 4.14 }$$

Since $\lbrace m_{k}^{n}\rbrace$ is a bounded sequence in L²(∂X), from the above inequality, it follows that the sequence $\lbrace \mu _{\beta }^{n}\rbrace$ is bounded in BV(X). By proposition 4.1, a subsequence $\lbrace \mu _{\beta }^{n_{1}}\rbrace$ of $\lbrace \mu _{\beta }^{n}\rbrace$ and $\tilde{\mu }\in {BV(X)}$ exist such that $\lbrace \mu _{\beta }^{n_{1}}\rbrace$ converges to $\tilde{\mu }$ in L¹(X) and

$$\begin{eqnarray} &&J_{2}(\tilde{\mu })\le \liminf _{n_1\rightarrow \infty }J_{2}(\mu _{\beta }^{n_{1}}). \end{eqnarray} \tag{ 4.15 }$$

Since $\mu _{\beta }^{n_{1}}\in {Q_{1}\times Q_{2}}$ for all n₁, it follows that $\tilde{\mu }\in {Q_{1}\times Q_{2}}$. Hence, $\tilde{\mu }\in Q_{\rm ad}$. By the argument used in the proof of theorem 4.2, a subsequence $\lbrace \mu _{\beta }^{n_{2}}\rbrace$ of $\lbrace \mu _{\beta }^{n_{1}}\rbrace$ exists such that $U_{k}(\mu _{\beta }^{n_{2}})$ weakly converges to $U_{k}(\tilde{\mu })$ in H_A(X × Ω), and $F_{k}(\mu _{\beta }^{n_{2}})$ weakly converges to $F_{k}(\tilde{\mu })$ in L²(∂X). Thus, we have

$$\begin{eqnarray*} \Vert F_{k}(\tilde{\mu })-m_{k}\Vert _{L^{2}(\partial X)}&\le \liminf _{n_{2}} \Vert F_{k}(\mu _{\beta }^{n_{2}})-m_{k}\Vert _{L^{2}(\partial X)}\\ &\le \liminf _{n_{2}} \left(\Vert F_{k}(\mu _{\beta }^{n_{2}})-m_{k}^{n_{2}}\Vert _{L^{2}(\partial X)} +\Vert m_{k}^{n_{2}}-m_{k}\Vert _{L^{2}(\partial X)}\right)\\ &=\liminf _{n_{2}} \Vert F_{k}(\mu _{\beta }^{n_{2}})-m_{k}^{n_{2}}\Vert _{L^{2}(\partial X)}. \end{eqnarray*}$$

So

$$\begin{eqnarray*} &&\frac{1}{2}\sum _{k=1}^{k_{0}}\Vert F_{k}(\tilde{\mu })-m_{k}\Vert _{L^{2}(\partial X)}^{2}\le \liminf _{n_{2}} \frac{1}{2}\sum _{k=1}^{k_{0}} \Vert F_{k}(\mu _{\beta }^{n_{2}})-m_{k}^{n_{2}}\Vert _{L^{2}(\partial X)}^{2}, \end{eqnarray*}$$

which means that

$$\begin{eqnarray} &&J_1(\tilde{\mu };m_{k})\le \liminf _{n_2\rightarrow \infty } J_1(\mu _{\beta }^{n_2};m_{k}^{n_2}). \end{eqnarray} \tag{ 4.16 }$$

Together with (4.14) and (4.15), we obtain that for any μ ∈ Q_ad,

$$\begin{eqnarray*} J_{1}(\tilde{\mu };m_{k})+\beta J_{2}(\tilde{\mu }) &\le \liminf _{n_{2}} J_{1}(\mu _{\beta }^{n_{2}};m_{k}^{n_{2}})+\beta J_{2}(\mu _{\beta }^{n_{2}})\\ &\le \limsup _{n_{2}} J_{1}(\mu _{\beta }^{n_{2}};m_{k}^{n_{2}})+\beta J_{2}(\mu _{\beta }^{n_{2}})\\ &\le \limsup _{n_{2}} J_{1}(\mu ;m_{k}^{n_{2}})+\beta J_{2}(\mu )\\ &=\lim _{n_{2}} J_{1}(\mu ;m_{k}^{n_{2}})+\beta J_{2}(\mu )\\ &=J_{1}(\mu ;m_{k})+\beta J_{2}(\mu ). \end{eqnarray*}$$

This implies that $\tilde{\mu }$ is a solution of the problem (P_β) and

$$\begin{eqnarray} &&\lim _{n_{2}} J_{1}(\mu _{\beta }^{n_{2}};m_{k}^{n_{2}})+\beta J_{2}(\mu _{\beta }^{n_{2}})=J_{1}(\tilde{\mu };m_{k})+\beta J_{2}(\tilde{\mu }). \end{eqnarray} \tag{ 4.17 }$$

Finally, let us show (4.13) holds, that is, $J_{2}(\mu _{\beta }^{n_{2}})\rightarrow J_{2}(\tilde{\mu })$ as n₂ → ∞. Assume $J_{2}(\mu _{\beta }^{n_{2}})\nrightarrow J_{2}(\tilde{\mu })$. Then from (4.15), we have

$$\begin{eqnarray} &&J_{2}(\tilde{\mu })< \limsup _{n_{2}} J_{2}(\mu _{\beta }^{n_{2}}):=J_{0}. \end{eqnarray} \tag{ 4.18 }$$

From the definition of limit superior, there exists a subsequence $\lbrace \mu _{\beta }^{n_{3}}\rbrace$ of $\lbrace \mu _{\beta }^{n_{2}}\rbrace$ such that $\mu _{\beta }^{n_{3}}$ converges to $\tilde{\mu }$ almost everywhere, $F_{k}(\mu _{\beta }^{n_{3}})\rightharpoonup F_{k}(\tilde{\mu })$ and $J_{2}(\mu _{\beta }^{n_{3}})\rightarrow J_{0}$. Since (4.17) holds, we obtain that

$$\begin{eqnarray*} \lim _{n_{3}} J_{1}(\mu _{\beta }^{n_{3}};m_{k}^{n_{3}})&=J_{1}(\tilde{\mu };m_{k})+\beta (J_{2}(\tilde{\mu })-\lim _{n_{3}} J_2(\mu _{\beta }^{n_{3}}))\\ &=J_{1}(\tilde{\mu };m_{k})+\beta (J_{2}(\tilde{\mu })-J_{0})\\ &< J_{1}(\tilde{\mu };m_{k}), \end{eqnarray*}$$

which contradicts (4.16). Thus $J_2(\mu _{\beta }^{n_2})\rightarrow J_2(\tilde{\mu })$, i.e., (4.13) holds. □

We will show that the solutions of the problem (P_β) with the noisy measurement data $\lbrace m^n_{t,k}\rbrace$ converge toward a solution of the problem (P) as β → 0 and the noise level $\delta _{k}^n\rightarrow 0$, 1 ⩽ k ⩽ k₀. For the noise level vector $\delta ^n=(\delta _1^n,\ldots ,\delta _{k_0}^n)$, the arithmetic mean of the square of k₀ noise levels is $m(\delta ^{n})=\big[\big(\delta _1^n\big)^2+\cdots +\big(\delta _{k_0}^n\big)^2\big]/k_{0}$.

We measure the convergence with respect to the Bregman distance, which is briefly recalled here. Let B be a Banach space with B* being the dual space of it; R: B → ( − ∞, ∞] is a proper convex functional and ∂R(q) stands for the subdifferential of R at q ∈ Dom R := {q ∈ B∣R(q) < ∞} ≠ ∅ defined by

$$\begin{eqnarray*} &&\partial R(q):=\lbrace q^{*}\in {B^{*}}\mid R(p)\ge R(q)+\langle q^{*},p-q\rangle _{(B^{*},B)} \quad \hbox{for all}\ p\in {B}\rbrace . \end{eqnarray*}$$

The set ∂R(q) may be empty; however, if R is continuous at q, then it is nonempty. Furthermore, ∂R(q) is convex and weak* compact. In the case ∂R(q) ≠ ∅, for any fixed p ∈ B, we denote

$$\begin{eqnarray*} &&D_{R}(p,q):=\lbrace R(p)-R(q)+\langle q^{*},p-q\rangle _{(B^{*},B)}\mid q^{*}\in {\partial R(q)}\rbrace . \end{eqnarray*}$$

Then for a fixed element q* ∈ ∂R(q),

$$\begin{eqnarray*} &&D_{R}^{q^{*}}(p,q):=R(p)-R(q)-\langle q^{*},p-q\rangle _{(B^{*},B)} \end{eqnarray*}$$

is called the Bregman distance with respect to R and q* of two elements p, q ∈ B.

The Bregman distance is not a metric on B. However, for each q* ∈ ∂R(q), the $D_{R}^{q^{*}}(p,q)\ge 0$ for any p ∈ B and $D_{R}^{q^{*}}(q,q)= 0$. Furthermore, in the case R is the strictly convex function, $D_{R}^{q^{*}}(p,q)=0$ if and only if p = q.

Theorem 4.3. (Convergence) For any positive vector sequence δⁿ → 0 such that for β_n := β(δⁿ) → 0, m(δⁿ)/β_n → 0 as n → ∞, let $\lbrace m_{k}^{n}\rbrace$ be a sequence in L²(∂X) such that for every k, $\Vert m_{k}^n-m_{t,k}\Vert _{L^2(\partial X)} \le \delta _k^n$, and let $\lbrace \mu _{\beta _{n}}^{n}\rbrace$ be the minimizers of the problems

$$\begin{eqnarray*} &&\min _{\mu \in {Q_{\rm ad}}}J_{1}(\mu ; m_{k}^{n})+\beta _{n}J_{2}(\mu ). \end{eqnarray*}$$

Then, a subsequence $\lbrace {\mu _{\beta _{n_2}}^{n_2}}\rbrace$ of $\lbrace {\mu _{\beta _{n}}^{n}}\rbrace$ and an element $\hat{\mu }\in Q_{\rm ad}(\lbrace m_{t,k}\rbrace )$ exist such that

$$\begin{equation} \lim _{n_2\rightarrow \infty }\Vert \mu _{\beta _{n_2}}^{n_2}-\hat{\mu }\Vert _{L^{1}(X)}=0 \end{equation} \tag{ 4.19 }$$

and

$$\begin{equation} \lim _{n_2\rightarrow \infty }J_{2}(\mu _{\beta _{n_2}}^{n_2})=J_{2}(\hat{\mu }). \end{equation} \tag{ 4.20 }$$

Furthermore, $\hat{\mu }$ is a solution to the problem (P) and for all $l\in {\partial (\int _{X}|\nabla (\cdot )|)(\hat{\mu })}$,

$$\begin{equation} \lim _{n_2\rightarrow \infty }D_{J_2}^{l}(\mu _{\beta _{n_2}}^{n_2},\hat{\mu })=0. \end{equation} \tag{ 4.21 }$$

Proof. For all n, by the definition of $\mu _{\beta _{n}}^{n}$, for any μ ∈ Q_ad, we have

$$\begin{eqnarray*} J_{1}(\mu _{\beta _{n}}^{n};m_{k}^{n})+\beta _{n}J_{2}(\mu _{\beta _{n}}^{n}) &=\frac{1}{2}\sum _{k=1}^{k_{0}}\Vert F_{k}(\mu _{\beta _{n}}^{n})-m_{k}^{n}\Vert _{L^{2}(\partial X)}^{2}+\beta _{n}J_{2}(\mu _{\beta _{n}}^{n})\\ &\le \frac{1}{2}\sum _{k=1}^{k_{0}}\Vert F_{k}(\mu )-m_{k}^{n}\Vert _{L^{2}(\partial X)}^{2}+\beta _{n}J_{2}(\mu ). \end{eqnarray*}$$

In particular, for any μ ∈ Q_ad({m_{t, k}}),

$$\begin{eqnarray*} &&\frac{1}{2}\sum _{k=1}^{k_{0}}\Vert F_{k}(\mu _{\beta _{n}}^{n})-m_{k}^{n}\Vert _{L^{2}(\partial X)}^{2}+\beta _{n}J_{2}(\mu _{\beta _{n}}^{n})\le \frac{1}{2}\sum _{k=1}^{k_{0}}\Vert m_{t,k}-m_{k}^{n}\Vert _{L^{2}(\partial X)}^{2}+\beta _{n}J_{2}(\mu ). \end{eqnarray*}$$

From the above inequality, we have

$$\begin{eqnarray} &&J_{2}(\mu _{\beta _{n}}^{n})\le \frac{1}{2}\frac{k_{0}m(\delta ^{n})}{\beta _{n}}+J_{2}(\mu ) \quad \forall \mu \in {Q_{\rm ad}(\lbrace m_{t,k}\rbrace )}. \end{eqnarray} \tag{ 4.22 }$$

Since m(δⁿ)/β_n → 0, $\lbrace \mu _{\beta _n}^n\rbrace$ is bounded in BV(X). By proposition 4.1, we conclude that there exist a subsequence $\lbrace \mu _{\beta _{n_{1}}}^{n_{1}}\rbrace$ of $\lbrace \mu _{\beta _{n}}^{n}\rbrace$ and $\hat{\mu }\in {Q_{\rm ad}}$, such that

$$\begin{eqnarray} \mu _{\beta _{n_{1}}}^{n_{1}} &\rightarrow \hat{\mu } \quad {\rm in} \:L^{1}(X), \end{eqnarray} \tag{ 4.23 }$$

$$\begin{eqnarray} J_{2}(\hat{\mu }) & \le \liminf _{n_1\rightarrow \infty }J_{2}(\mu _{\beta _{n_{1}}}^{n_{1}}). \end{eqnarray} \tag{ 4.24 }$$

By the same reasoning as in the proof of theorem 4.3, a subsequence $\lbrace \mu _{\beta _{n_{2}}}^{n_{2}}\rbrace$ of $\lbrace \mu _{\beta _{n_{1}}}^{n_{1}}\rbrace$ exists, such that, for 1 ⩽ k ⩽ k₀,

$$\begin{eqnarray*} U_{k}(\mu _{\beta _{n_{2}}}^{n_{2}})& \rightharpoonup U_{k}(\hat{\mu }) \ {\rm in}\: H_{A}(X\times \Omega ); \qquad F_{k}(\mu _{\beta _{n_{2}}}^{n_{2}})\rightharpoonup M_{k}(\hat{\mu }) \ {\rm in}\: L^{2}(\partial X). \end{eqnarray*}$$

On the other hand,

$$\begin{eqnarray*} &&\frac{1}{2}\sum _{k=1}^{k_{0}}\Vert F_{k}(\mu _{\beta _{n_{2}}}^{n_{2}})-m_{k}^{n_{2}}\Vert _{L^{2}(\partial X)}^{2}\le \frac{1}{2}\sum _{k=1}^{k_{0}}\Vert m_{t,k}-m_{k}^{n_{2}}\Vert _{L^{2}(\partial X)}^{2}+\beta _{n_{2}}J_{2}(\mu ). \end{eqnarray*}$$

When $\beta _{n_{2}}\rightarrow 0$ and $m(\delta _{n_{2}})\rightarrow 0$, the first item on the right of the above inequality approaches zero, and so

$$\begin{eqnarray*} &&\frac{1}{2}\sum _{k=1}^{k_{0}}\Vert F(\mu _{\beta _{n_{2}}}^{n_{2}})-m_{k}^{n_{2}}\Vert _{L^{2}(\partial X)}^{2}\rightarrow 0. \end{eqnarray*}$$

Moreover, from

$$\begin{equation*} \Vert F_{k}(\mu _{\beta _{n_{2}}}^{n_{2}})-m_{t,k}\Vert _{L^{2}(\partial X)} \le \Vert F_{k}(\mu _{\beta _{n_{2}}}^{n_{2}})-m_{k}^{n_{2}}\Vert _{L^{2}(\partial X)} +\Vert m_{k}^{n_2}-m_{t,k}\Vert _{L^{2}(\partial X)} \end{equation*}$$

and $\Vert m_{k}^{n_{2}}-m_{t,k}\Vert _{L^{2}(\partial X)}\le \delta _k^{n_2}\rightarrow 0$, we conclude that

$$\begin{eqnarray*} &&\frac{1}{2}\sum _{k=1}^{k_{0}}\Vert F_{k}(\mu _{\beta _{n_{2}}}^{n_{2}})-m_{t,k}\Vert _{L^{2}(\partial X)}^{2}\rightarrow 0, \end{eqnarray*}$$

i.e., $F_{k}(\mu _{\beta _{n_{2}}}^{n_{2}})\rightarrow m_{t,k}$ as n → ∞ in L²(∂X). Recall that $F_{k}(\mu _{\beta _{n_{2}}}^{\delta _{n_{2}}})\rightharpoonup M_{k}(\hat{\mu })$ in L²(∂X). By the uniqueness of the weak limit, $M_{k}(\hat{\mu })=m_{t,k}$, i.e., $\hat{\mu }\in {Q_{\rm ad}(\lbrace m_{t,k}\rbrace )}$.

It remains to prove (4.21). For any μ ∈ Q_ad({m_{t, k}}), by (4.22) and (4.24), we have

$$\begin{eqnarray*} &&J_{2}(\hat{\mu })\le \liminf _{n_2\rightarrow \infty }\left( \frac{m(\delta ^{n_{2}})}{\beta _{n_{2}}}+J_{2}(\mu ) \right)= J_{2}(\mu ). \end{eqnarray*}$$

This means that $\hat{\mu }$ is a solution to the problem (P). Again using (4.22) and (4.24), we have

$$\begin{eqnarray*} &&\limsup _{n_2\rightarrow \infty }J_{2}(\mu _{\beta _{n_{2}}}^{n_{2}})\le J_{2}(\hat{\mu }), \quad J_{2}(\hat{\mu })\le \liminf _{n_2\rightarrow \infty }J_{2}(\mu _{\beta _{n_{2}}}^{n_{2}}). \end{eqnarray*}$$

So

$$\begin{eqnarray} &&\lim _{n_2\rightarrow \infty }J_{2}(\mu _{\beta _{n_{2}}}^{n_{2}})= J_{2}(\hat{\mu }). \end{eqnarray} \tag{ 4.25 }$$

From (4.23) and (4.25), by [5, proposition 10.1.2], the sequence $\lbrace \mu _{\beta _{n_{2}}}^{n_{2}}\rbrace$ weakly converges to $\hat{\mu }$ in BV(X) . Thus, for all $l\in {\partial (\int _{X}|\nabla (\cdot )|)(\hat{\mu })}$ ,

$$\begin{eqnarray*} &&\lim _{n_2\rightarrow \infty }D_{J_2}^{l}(\mu _{\beta _{n_{2}}}^{n_{2}},\hat{\mu })=0 \end{eqnarray*}$$

which is (4.21). □