Semigroup approach for identification of the unknown diffusion coefficient in a linear parabolic equation with mixed output data

Abstract

This article presents a semigroup approach for the mathematical analysis of the inverse coefficient problems of identifying the unknown coefficient $k(x)$ in the linear parabolic equation $u_t(x,t)={(k(x)u_x(x,t))}_x$ with mixed boundary conditions $k(0)u_x(0,t)={\psi }_0$, $u(1,t)={\psi }_1$. The aim of this paper is to investigate the distinguishability of the input-output mappings $\mathrm{\Phi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$, $\mathrm{\Psi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$ via semigroup theory. In this paper, we show that if the null space of the semigroup $T(t)$ consists of only zero function, then the input-output mappings $\mathrm{\Phi }[\cdot ]$ and $\mathrm{\Psi }[\cdot ]$ have the distinguishability property. It is shown that the types of the boundary conditions and the region on which the problem is defined have a significant impact on the distinguishability property of these mappings. Moreover, in the light of measured output data (boundary observations) $f(t):=u(0,t)$ or/and $h(t):=k(1)u_x(1,t)$, the values $k(0)$ and $k(1)$ of the unknown diffusion coefficient $k(x)$ at $x=0$ and $x=1$, respectively, can be determined explicitly. In addition to these, the values $k^{\prime }(0)$ and $k^{\prime }(1)$ of the unknown coefficient $k(x)$ at $x=0$ and $x=1$, respectively, are also determined via the input data. Furthermore, it is shown that measured output data $f(t)$ and $h(t)$ can be determined analytically by an integral representation. Hence the input-output mappings $\mathrm{\Phi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$, $\mathrm{\Psi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$ are given explicitly in terms of the semigroup.

1 Introduction

Consider the following initial boundary value problem:

$$\{\begin{array}{c}{u}_t(x,t)={(k(x)u_x(x,t))}_x,(x,t)\in {\mathrm{\Omega }}_T,\hfill \\ u(x,0)=g(x),0<x<1,\hfill \\ k(0)u_x(0,t)={\psi }_0,u(1,t)={\psi }_1,0<t<T,\hfill \end{array}$$

(1)

where ${\mathrm{\Omega }}_T=\{(x,t)\in R^2:0<x<1,0<t\le T\}$. The left flux ${\psi }_0$ and the right boundary condition ${\psi }_1$ are assumed to be constants. The functions $c_1>k(x)\ge c_0>0$ and $g(x)$ satisfy the following conditions:

(C1) $k(x)\in H^{1,2}[0,1]$;

(C2) $g(x)\in H^{2,2}[0,1]$, $g^{\prime }(0)={\psi }_0$, $g(1)={\psi }_1$.

Under these conditions, the initial boundary value problem (1) has the unique solution $u(x,t)\in H^{2,2}[0,1]\cap H^{1,2}[0,1]$ [1–4].

Consider the inverse problem of determining the unknown coefficient $k=k(x)$ [5–9] from the following observations at the boundaries $x=0$ and $x=1$:

$$u(0,t)=f(t),k(1)u_x(1,t)=h(t),t\in (0,T].$$

(2)

Here $u=u(x,t)$ is the solution of the parabolic problem (1). The functions $f(t)$, $h(t)$ are assumed to be noisy free measured output data. In this context, the parabolic problem (1) will be referred to as a direct (forward) problem with the inputs $g(x)$ and $k(x)$. It is assumed that the functions $f(t)$ and $h(t)$ belong to $H^{1,2}[0,T]$ and satisfy the consistency conditions $f(0)=g(0)$, $h^{\prime }(0)=k(1)g^{\prime }(1)$.

We denote by $\mathcal{K}:=\{k(x)\in H^{1,2}[0,1]:c_1>k(x)\ge c_0>0,x\in [0,1]\}\subset H^{1,2}[0,1]$, the set of admissible coefficients $k=k(x)$ and introduce the input-output mappings $\mathrm{\Phi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$, $\mathrm{\Psi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$, where

$$\mathrm{\Phi }[k]=u(x,t;k){|}_{x=0},\mathrm{\Psi }[k]=k(x)u_x(x,t;k){|}_{x=1},k\in \mathcal{K},f(t),h(t)\in H^{1,2}[0,T].$$

(3)

Then the inverse problem [10] with the measured data $f(t)$ and $h(t)$ can be formulated as the following operator equations:

$$\mathrm{\Phi }[k]=f,\mathrm{\Psi }[k]=h,k\in \mathcal{K},f,h\in H^{1,2}[0,T].$$

(4)

We denote by $\mathcal{K}:=\{k(x)\in H^{1,2}[0,1]:c_1>k(x)\ge c_0>0,x\in [0,1]\}\subset H^{1,2}[0,1]$, the set of admissible coefficients $k=k(x)$. The monotonicity, continuity and hence invertibility of the input-output mappings $\mathrm{\Phi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$ and $\mathrm{\Psi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$ are given in [3, 4].

The aim of this paper is to study a distinguishability of the unknown coefficient via the above input-output mappings. We say that the mapping $\mathrm{\Phi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$ (or $\mathrm{\Psi }[\cdot ]:\mathcal{K}\to H^{1,2}[0,T]$) has the distinguishability property if $\mathrm{\Phi }[k_1]\ne \mathrm{\Phi }[k_2]$ ($\mathrm{\Psi }[k_1]\ne \mathrm{\Psi }[k_2]$) implies $k_1(x)\ne k_2(x)$. This, in particular, means injectivity of the inverse mappings ${\mathrm{\Phi }}^{-1}$ and ${\mathrm{\Psi }}^{-1}$.

The purpose of this paper is to study the distinguishability of the unknown coefficient via the above input-output mappings. The results presented here are the first ones, to the knowledge of authors, from the point of view of semigroup approach [11] to inverse problems. This approach sheds more light on the identifiability of the unknown coefficient [12] and shows how much information can be extracted from the measured output data, in particular in the case of constant flux and boundary data [12–15].

The paper is organized as follows. In Section 2, the analysis of the semigroup approach is given for the inverse problem with the measured data $f(t)$. A similar analysis is applied to the inverse problem with the single measured output data $h(t)$ given at the point $x=1$ in Section 3. The inverse problem with two Neumann measured data $f(t)$ and $h(t)$ is discussed in Section 4. Finally, some concluding remarks are given in Section 5.

2 Analysis of the inverse problem with measured output data $f(t)$

Consider now the inverse problem with one measured output data $f(t)$ at $x=0$. In order to formulate the solution of the parabolic problem (1) in terms of a semigroup, let us first arrange the parabolic equation as follows:

$$u_t(x,t)-{(k(0)u_x(x,t))}_x={((k(x)-k(0))u_x(x,t))}_x,(x,t)\in {\mathrm{\Omega }}_T.$$

Then the initial boundary value problem (1) can be rewritten in the following form:

(5)

Here we assume that $k(0)$ was known. Later we will determine the value $k(0)$. In order to formulate the solution of the parabolic problem (5) in terms of a semigroup, we need to define the following function:

$$v(x,t)=u(x,t)-\frac{{\psi }_0}{k(0)}x+{\psi }_0-{\psi }_1,x\in [0,1]$$

(6)

which satisfies the following parabolic problem:

(7)

Here $A[v(x,t)]:=-k(0)d^{2}v(x,t)/d{x}^2$ is a second-order differential operator, its domain is $D_A=\{u\in H^{2,2}(0,1)\cap H^{1,2}[0,1]:u_x(0)=u(1)=0\}$. Since the initial value function $g(x)$ belongs to $C^2[0,1]$, it is obvious that $g(x)\in D_A$.

Denote by $T(t)$ the semigroup of linear operators generated by the operator −A [5, 6]. Note that we can easily find the eigenvalues and eigenfunctions of the differential operator A. Furthermore, the semigroup $T(t)$ can be easily constructed by using the eigenvalues and eigenfunctions of a differential operator A. For this reason, we first consider the following eigenvalue problem:

This problem is called a Sturm-Liouville problem. We can easily determine that the eigenvalues are ${\lambda }_n=k(0){(2n-1)}^2{\pi }^2/4$ for all $n=1,\dots$ and the corresponding eigenfunctions are ${\varphi }_n(x)=\sqrt{2}cos((2n-1)x\pi /2)$. In this case, the semigroup $T(t)$ can be represented in the following way:

$$T(t)U(x,s)=\sum _{n=0}^{\mathrm{\infty }}〈{\varphi }_n(x),U(x,s)〉e^{-{\lambda }_{n}t}{\varphi }_n(x),$$

where $〈{\varphi }_n(x),U(x,s)〉={\int }_0^1{\varphi }_n(x)U(x,s)dx$. The null space of the semigroup $T(t)$ of the linear operators can be defined as follows:

$$N(T)=\{U(x,s):〈{\varphi }_n(x),U(x,s)〉=0,\text{ for all }n=1,2,3,\dots \}.$$

From the definition of the semigroup $T(t)$, we can say that the null space of it is an empty set, i.e., $N(T)=\{0\}$. This result is very important for the uniqueness of the unknown coefficient $k(x)$.

The unique solution of the initial value problem (7) in terms of a semigroup $T(t)$ can be represented in the following form:

$$v(x,t)=T(t)v(x,0)+{\int }_0^{t}T(t-s){\left((k(x)-k(0))(v_x(x,t)+\frac{{\psi }_0}{k(0)})\right)}_{x}ds.$$

Hence, by using identity (6), the solution $u(x,t)$ of the parabolic problem (5) in terms of a semigroup can be written in the following form:

$$\begin{array}{rcl}u(x,t)& =& \frac{{\psi }_0}{k(0)}x+{\psi }_1-{\psi }_0+T(t)(g(x)-\frac{{\psi }_0}{k(0)}x+{\psi }_0-{\psi }_1)\\ +{\int }_0^{t}T(t-s){((k(x)-k(0))u_x(x,s))}_{x}ds.\end{array}$$

(8)

In order to arrange the above solution representation, let us define the following:

(9)

(10)

Then we can rewrite the solution representation in terms of $\zeta (x)$ and $\xi (x,s)$ in the following form:

$$u(x,t)=\frac{{\psi }_0}{k(0)}x+{\psi }_1-{\psi }_0+T(t)\zeta (x)+{\int }_0^{t}T(t-s)\xi (x,s)ds.$$

Substituting $x=0$ into this solution representation yields

$$u(0,t)={\psi }_1-{\psi }_0+T(t)\zeta (0)+{\int }_0^{t}T(t-s)\xi (0,s)ds.$$

Taking into account the overmeasured data $u(0,t)=f(t)$, we get

$$f(t)=({\psi }_1-{\psi }_0+T(t)\zeta (0)+{\int }_0^{t}T(t-s)\xi (0,s)ds),$$

(11)

which implies that $f(t)$ can be determined analytically.

Differentiating both sides of the above identity with respect to x and using semigroup properties at $x=0$ yield

$$u_x(0,t)=\frac{{\psi }_0}{k(0)}+z(0,t)+{\int }_0^{t}w(0,t-s,s)ds.$$

Using the boundary condition $k(0)u_x(0,t)={\psi }_0$, we can write $k(0)={\psi }_0/u_x(0,t)$ for all $t\ge 0$ which can be rewritten in terms of a semigroup in the following form:

$$k(0)={\psi }_0/(-{\psi }_0+{\psi }_1+z(0,t)+{\int }_0^{t}w(0,t-s,s)ds).$$

Taking limit as $t\to 0$ in the above identity, we obtain the following explicit formula for the value $k(0)$ of the unknown coefficient $k(x)$:

$$k(0)={\psi }_0/(-{\psi }_0+{\psi }_1+z(0,0)).$$

The right-hand side of identity (11) defines explicitly the semigroup representation of the input-output mapping $\mathrm{\Phi }[k]$ on the set of admissible unknown diffusion coefficients $\mathcal{K}$:

$$\mathrm{\Phi }[k](x):={\psi }_1-{\psi }_0+T(t)\zeta (0)+{\int }_0^{t}T(t-s)\xi (0,s)ds,\mathrm{\forall }t\in [0,T].$$

(12)

Let us differentiate now both sides of identity (8) with respect to t:

$$\begin{array}{rcl}{u}_t(x,t)& =& T(t)A(g(x)-\frac{{\psi }_0}{k(0)}x+{\psi }_0-{\psi }_1)+{((k(x)-k(0))u_x(x,t))}_x\\ +{\int }_0^{t}AT(t-s){((k(x)-k(0))u_x(x,s))}_{x}ds.\end{array}$$

Using the semigroup property $-{\int }_0^{t}AT(s)u(x,s)ds=T(t)u(x,t)-T(0)u(x,t)$, we obtain

$$\begin{array}{rcl}{u}_t(x,t)& =& -k(0)T(t)g^{″}(x)-2T(0){((k(x)-k(0))u_x(x,t))}_x\\ +T(t){((k(x)-k(0))u_x(x,0))}_x.\end{array}$$

Taking $x=0$ in the above identity, we get

$$\begin{array}{rcl}{u}_t(0,t)& =& -k(0)T(t)g^{″}(0)-T(0)k^{\prime }(0)u_x(0,0)\\ +T(t)(k^{\prime }(0)u_x(0,0))-T(0)(k^{\prime }(0)u_x(0,t)).\end{array}$$

Since $u(0,t)=f(t)$, we have $u_t(0,t)=f^{\prime }(t)$. Taking into account this and substituting $t=0$ yield

$$f^{\prime }(0)=-k(0)g^{″}(0)-k^{\prime }(0)\frac{g^{\prime }(0)}{k(0)}.$$

Solving this equation for $k^{\prime }(0)$ and substituting $u_x(0,0)=g^{\prime }(0)/k(0)$, we obtain the following explicit formula for the value $k^{\prime }(0)$ of the first derivative $k^{\prime }(x)$ of the unknown coefficient at $x=0$:

$$k^{\prime }(0)=\frac{-k^2(0)g^{″}(0)-k(0)f^{\prime }(0)}{g^{\prime }(0)}.$$

(13)

Under the determined values $k(0)$ and $k^{\prime }(0)$, the set of admissible coefficients can be defined as follows:

$${\mathcal{K}}_0:=\{k\in \mathcal{K}:k(0)=\frac{{\psi }_0}{-{\psi }_0+{\psi }_1+z(0,0)},k^{\prime }(0)=\frac{-k^2(0)g^{″}(0)-k(0)f^{\prime }(0)}{g^{\prime }(0)}\}.$$

The following lemma implies the relationship between the diffusion coefficients $k_1(x),k_2(x)\in \mathcal{K}$ at $x=0$ and the corresponding outputs $f_j(t):=u(0,t;k_j)$, $j=1,2$.

Lemma 2.1 Let $u_1(x,t)=u(x,t;k_1)$ and $u_2(x,t)=u(x,t;k_2)$ be solutions of the direct problem (5) corresponding to the admissible coefficients $k_1(x),k_2(x)\in \mathcal{K}$. Suppose that $f_j(t)=u(0,t;k_j)$, $j=1,2$, are the corresponding outputs and denote by $\mathrm{\Delta }f(t)=f_1(t)-f_2(t)$, $\mathrm{\Delta }\xi (x,t)={\xi }^1(x,t)-{\xi }^2(x,t)$. If the condition

$$k_1(0)=k_2(0):=k(0)$$

holds, then the outputs $f_j(t)$, $j=1,2$, satisfy the following integral identity:

$$\mathrm{\Delta }f(\tau )={\int }_0^{\tau }T(\tau -s)\mathrm{\Delta }\xi (0,s)dsds$$

(14)

for each $\tau \in (0,T]$.

Proof The solutions of the direct problem (5) corresponding to the admissible coefficients $k_1(x),k_2(x)\in \mathcal{K}$ can be written at $x=0$ as follows:

respectively, by using representation (11). From identity (9) it is obvious that ${\zeta }^1(0,\tau )={\zeta }^2(0,\tau )$ for each $\tau \in (0,T]$. Hence the difference of these formulas implies the desired result. □

This lemma with identity (14) implies the following.

Corollary 2.1 Let conditions of Lemma 2.1 hold. Then $f_1(t)=f_2(t)$, $\mathrm{\forall }t\in [0,T]$, if and only if

$$〈{\varphi }_n(x),\mathrm{\Delta }\xi (0,s)〉=〈{\varphi }_n(x),{\xi }^1(x,t)-{\xi }^2(x,t)〉=0,\mathrm{\forall }t\in (0,T],n=0,1,\dots$$

Since the Strum-Liouville problem generates a complete orthogonal family of eigenfunctions, the null space of a semigroup contains only zero function, i.e., $N(T)=\{0\}$. Thus Corollary 2.1 states that $f_1\equiv f_2$ if and only if ${\xi }^1(x,t)-{\xi }^2(x,t)=0$ for all $(x,t)\in {\mathrm{\Omega }}_T$. The definition of $\xi (x,t)$ implies that $k_1(x)=k_2(x)$ for all $x\in [0,1]$.

The combination of the conclusions of Lemma 2.1 and Corollary 2.1 can be given by the following theorem which states the distinguishability of the input-output mapping $\mathrm{\Phi }[\cdot ]:{\mathcal{K}}_0\to H^{1,2}[0,T]$.

Theorem 2.1 Let conditions (C1) and (C2) hold. Assume that $\mathrm{\Phi }[\cdot ]:{\mathcal{K}}_0\to H^{1,2}[0,T]$ is the input-output mapping defined by (3) and corresponding to the measured output $f(t):=u(0,t)$. Then the mapping $\mathrm{\Phi }[k]$ has the distinguishability property in the class of admissible coefficients ${\mathcal{K}}_0$, i.e.,

$$\mathrm{\Phi }[k_1]\ne \mathrm{\Phi }[k_2]\mathrm{\forall }{k}_1,k_2\in {\mathcal{K}}_0,k_1(x)\ne k_2(x).$$

3 Analysis of the inverse problem with measured output data $h(t)$

Consider now the inverse problem with one measured output data $h(t)$ at $x=1$. As in the previous section, let us arrange the parabolic equation as follows:

$$u_t(x,t)-{(k(1)u_x(x,t))}_x={((k(x)-k(1))u_x(x,t))}_x,(x,t)\in {\mathrm{\Omega }}_T.$$

Then the initial boundary value problem (1) can be rewritten in the following form:

(15)

In order to formulate the solution of the above parabolic problem in terms of a semigroup, let us use the same variable $v(x,t)$ in identity (6), which satisfies the following parabolic problem:

(16)

Here $B[v(x,t)]:=-k(1)d^{2}v(x,t)/d{x}^2$ is a second-order differential operator, its domain is $D_B=\{u\in H^{2,2}(0,1)\cap H^{1,2}[0,1]:u_x(0)=u(1)=0\}$. Since the initial value function $g(x)$ belongs to $H^{2,2}[0,1]$, it is obvious that $g(x)\in D_B$.

Denote by $S(t)$ the semigroup of linear operators generated by the operator −A [5, 6]. As in the previous section, we can easily find the eigenvalues and eigenfunctions of the differential operator B. Furthermore, the semigroup $S(t)$ can be easily constructed by using the eigenvalues and eigenfunctions of the differential operator B. For this reason, we first consider the following eigenvalue problem:

$$B\varphi (x)=\lambda \varphi (x),{\varphi }_x(0)=0;\varphi (1)=0.$$

This problem is called a Sturm-Liouville problem. We can easily determine that the eigenvalues are ${\lambda }_n=k(1){(2n-1)}^2{\pi }^2/4$ for all $n=0,1,\dots$ and the corresponding eigenfunctions become ${\varphi }_n(x)=\sqrt{2}cos((2n-1)\pi /2x)$. Hence the semigroup $S(t)$ can be represented in the following form:

$$S(t)U(x,s)=\sum _{n=0}^{\mathrm{\infty }}〈{\varphi }_n(x),U(x,s)〉e^{-{\lambda }_{n}t}{\varphi }_n(x).$$

The null space of the semigroup $S(t)$ of the linear operators can be defined as follows:

$$N(S)=\{U(x,s):〈{\varphi }_n(x),U(x,s)〉=0,\text{ for all }n=1,2,3,\dots \}.$$

Since the Sturm-Liouville problem generates a complete orthogonal family of eigenfunctions, we can say that the null space of the semigroup $S(t)$ is an empty set, i.e., $N(S)=\mathrm{\varnothing }$. This result is very important for the uniqueness of the unknown coefficient $k(x)$.

The unique solution of the initial value problem (16) in terms of a semigroup $S(t)$ can be represented in the following form:

$$v(x,t)=S(t)v(x,0)+{\int }_0^{t}S(t-s){\left((k(x)-k(1))(v_x(x,s)+\frac{{\psi }_0}{k(0)})\right)}_{x}ds.$$

Hence, by using identity (6), the solution $u(x,t)$ of the parabolic problem (15) in terms of a semigroup can be written in the following form:

$$\begin{array}{rcl}u(x,t)& =& \frac{{\psi }_0}{k(0)}x+{\psi }_1-\frac{{\psi }_0}{k(0)}+S(t)(g(x)-\frac{{\psi }_0}{k(0)}x+\frac{{\psi }_0}{k(0)}-{\psi }_1)\\ +{\int }_0^{t}S(t-s){((k(x)-k(1))u_x(x,s))}_{x}ds.\end{array}$$

(17)

Defining the following:

(18)

(19)

(20)

The solution representation of the parabolic problem (17) can be rewritten in the following form:

$$u(x,t)=\frac{{\psi }_0}{k(0)}x+{\psi }_1-\frac{{\psi }_0}{k(0)}+S(t)\zeta (x)+{\int }_0^{t}S(t-s)\chi (x,s)ds.$$

Differentiating both sides of the above identity with respect to x and substituting $x=1$ yield

$$u_x(1,t)=\frac{{\psi }_0}{k(0)}+z_1(1,t)+{\int }_0^t{w}_1(1,t-s,s)ds.$$

Taking into account the overmeasured data $k(1)u_x(1,t)=h(t)$, we get

$$h(t)=k(1)(\frac{{\psi }_0}{k(0)}+z_1(1,t)+{\int }_0^t{w}_1(1,t-s,s)ds).$$

(21)

Now we can determine the value $k(1)$. From the overmeasured data $k(1)u_x(1,t)=h(t)$, the identity $k(1)=h(t)/u_x(1,t)$ for all $t>0$ can be rewritten in terms of a semigroup in the following form:

$$k(1)=\frac{h(t)}{(\frac{{\psi }_0}{k(0)}+z_1(1,t)+{\int }_0^t{w}_1(1,t-s,s)ds)}.$$

Taking limit as $t\to 0$ in the above identity yields

$$k(1)=h(0)/(\frac{{\psi }_0}{k(0)}+z_1(1,0)).$$

The right-hand side of the above identity defines the semigroup representation of the input-output mapping $\mathrm{\Psi }[k]$ on the set of admissible unknown diffusion coefficient $\mathcal{K}$:

$$\mathrm{\Psi }[k](x):=k(1)(\frac{{\psi }_0}{k(0)}+z_1(1,t)+{\int }_0^t{w}_1(1,t-s,s)ds),\mathrm{\forall }t\in [0,T].$$

(22)

Differentiating both sides of identity (17) with respect to t, we get

$$\begin{array}{rcl}{u}_t(x,t)& =& S(t)B(g(x)-\frac{{\psi }_0}{k(0)}x+\frac{{\psi }_0}{k(0)}-{\psi }_1)\\ -S(0){((k(x)-k(1))u_x(x,t))}_x+{\int }_0^{t}BS(t-s){((k(x)-k(1))u_x(x,s))}_{x}ds.\end{array}$$

Using semigroup properties, we obtain

$$\begin{array}{rcl}{u}_t(x,t)& =& -S(t)k(1)g^{″}(x)-S(0)(k^{\prime }(x)u_x(x,t)+(k(x)-k(1))u_{xx}(x,t))\\ +S(t){((k(x)-k(1))u_x(x,0))}_x-S(0){((k(x)-k(1))u_x(x,t))}_x.\end{array}$$

Taking $x=1$ in the above identity, we get

$$u_t(1,t)=-S(t)k(1)g^{″}(1)-2S(0)k^{\prime }(1)u_x(1,t)+S(t)k^{\prime }(1)u_x(1,0).$$

Since $u(1,t)={\psi }_1$, we have $u_t(1,t)=0$. Taking into account this and substituting $t=0$, we get

$$0=-k(1)g^{″}(1)-k^{\prime }(1)u_x(1,0).$$

Solving this equation for $k^{\prime }(1)$ and substituting $u_x(1,0)=h(0)/k(1)$, we reach the following result:

$$k^{\prime }(1)=-\frac{k^2(1)g^{″}(1)}{h(0)}.$$

(23)

Then we can define the admissible set of diffusion coefficients as follows:

$${\mathcal{K}}_1:=\{k\in \mathcal{K}:k(1)=\frac{h(0)}{(\frac{{\psi }_0}{k(0)}+z_1(1,0))},k^{\prime }(1)=-\frac{k^2(1)g^{″}(1)}{h(0)}\}.$$

The following lemma implies the relation between the coefficients $k_1(x),k_2(x)\in \mathcal{K}$ at $x=1$ and the corresponding outputs $h_j(t):=k_j(1)u_x(1,t;k_j)$, $j=1,2$.

Lemma 3.1 Let $u_1(x,t)=u(x,t;k_1)$ and $u_2(x,t)=u(x,t;k_2)$ be solutions of the direct problem (16) corresponding to the admissible coefficients $k_1(x),k_2(x)\in \mathcal{K}$. Suppose that $h_j(t)=u(1,t;k_j)$, $j=1,2$, are the corresponding outputs and denote by $\mathrm{\Delta }h(t)=h_1(t)-h_2(t)$, $\mathrm{\Delta }{w}_1(x,t,s)=w_1^1(x,t,s)-w_1^2(x,t,s)$. If the condition

$$k_1(1)=k_2(1):=k(1)$$

holds, then the outputs $h_j(t)$, $j=1,2$, satisfy the following integral identity:

$$\mathrm{\Delta }h(\tau )=k(1){\int }_0^{\tau }\mathrm{\Delta }{w}_1(1,\tau -s,s)dsds$$

(24)

for each $\tau \in (0,T]$.

Proof The solutions of the direct problem (15) corresponding to the admissible coefficients $k_1(x),k_2(x)\in \mathcal{K}$ can be written at $x=1$ as follows:

respectively, by using formula (20). From definition (18), it is obvious that $z_1^1(1,\tau )=z_1^2(1,\tau )$ for each $\tau \in (0,T]$. Hence the difference of these formulas implies the desired result. □

This lemma with identity (23) implies the following conclusion.

Corollary 3.1 Let the conditions of Lemma 3.1 hold. Then $h_1(t)=h_2(t)$, $\mathrm{\forall }t\in [0,T]$, if and only if

$$〈{\varphi }_n(x),{\chi }^1(x,t)-{\chi }^2(x,t)〉=0,\mathrm{\forall }t\in (0,T],n=0,1,\dots$$

hold.

Since the null space of it consists of only zero function, i.e., $N(S)=\{0\}$, Corollary 3.1 states that $h_1\equiv h_2$ if and only if ${\chi }^1(x,t)-{\chi }^2(x,t)=0$ for all $(x,t)\in {\mathrm{\Omega }}_T$. The definition of $\chi (x,t)$ implies that $k_1(x)=k_2(x)$ for all $x\in (0,1]$.

Theorem 3.1 Let conditions (C1) and (C2) hold. Assume that $\mathrm{\Psi }[\cdot ]:{\mathcal{K}}_1\to C^1[0,T]$ is the input-output mapping defined by (3) and corresponding to the measured output $h(t):=k(1)u_x(1,t)$. Then the mapping $\mathrm{\Psi }[k]$ has the distinguishability property in the class of admissible coefficients ${\mathcal{K}}_1$, i.e.,

$$\mathrm{\Psi }[k_1]\ne \mathrm{\Psi }[k_2]\mathrm{\forall }{k}_1,k_2\in {\mathcal{K}}_1,k_1(x)\ne k_2(x).$$

4 The inverse problem with mixed output data

Consider now the inverse problem (1)-(2) with two measured output data $f(t)$ and $h(t)$. As shown before, having these two data, the values $k(0)$ as well as $k(1)$ can be defined by the above explicit formulas. Based on this result, let us define now the set of admissible coefficients ${\mathcal{K}}_2$ as an intersection:

$$\begin{array}{rcl}{\mathcal{K}}_2:={\mathcal{K}}_0\cap {\mathcal{K}}_1& =& \{k\in \mathcal{K}:k(0)=\frac{{\psi }_0}{-{\psi }_0+{\psi }_1+z(0,0)},k(1)=\frac{h(0)}{{\psi }_0/k(0)+z_1(1,0)},\\ k^{\prime }(0)=\frac{-k^2(0)g^{″}(0)-k(0)f^{\prime }(0)}{g^{\prime }(0)},k^{\prime }(1)=\frac{-k^2(1)g^{″}(1)}{h(0)}\}.\end{array}$$

On this set, both input-output mappings $\mathrm{\Phi }[k]$ and $\mathrm{\Psi }[k]$ have distinguishability property.

Corollary 4.1 The input-output mappings $\mathrm{\Phi }[\cdot ]:{\mathcal{K}}_2\to H^{1,2}[0,T]$ and $\mathrm{\Psi }[\cdot ]:{\mathcal{K}}_2\to H^{1,2}[0,T]$ distinguish any two functions $k_1(x)\ne k_2(x)$ from the set ${\mathcal{K}}_2$, i.e.,

$$\mathrm{\forall }{k}_1(x),k_2(x)\in {\mathcal{K}}_2,k_1(x)\ne k_2(x),\mathrm{\Phi }[k_1]\ne \mathrm{\Phi }[k_2],\mathrm{\Psi }[k_1]\ne \mathrm{\Psi }[k_2].$$

5 Conclusion

The aim of this study was to analyze distinguishability properties of the input-output mappings $\mathrm{\Phi }[\cdot ]:{\mathcal{K}}_2\to H^{1,2}[0,T]$ and $\mathrm{\Psi }[\cdot ]:{\mathcal{K}}_2\to H^{1,2}[0,T]$ which are naturally determined by the measured output data. In this paper we show that if the null spaces of the semigroups $T(t)$ and $S(t)$ include only zero function then the corresponding input-output mappings $\mathrm{\Phi }[\cdot ]$ and $\mathrm{\Psi }[\cdot ]$ have distinguishability property.

This study shows that boundary conditions and the region on which the problem is defined have a significant impact on the distinguishability of the input-output mappings $\mathrm{\Phi }[\cdot ]$ and $\mathrm{\Psi }[\cdot ]$ since these key elements determine the structure of the semigroups $T(t)$ and $S(t)$ of linear operators and their null spaces.