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:

${u}_{t}(x,t)-{(k(0){u}_{x}(x,t))}_{x}={((k(x)-k(0)){u}_{x}(x,t))}_{x},\phantom{\rule{1em}{0ex}}(x,t)\in {\mathrm{\Omega}}_{T}.$

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

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:

$v(x,t)=u(x,t)-\frac{{\psi}_{0}}{k(0)}x+{\psi}_{0}-{\psi}_{1},\phantom{\rule{1em}{0ex}}x\in [0,1]$

(6)

which satisfies the following parabolic problem:

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:

$T(t)U(x,s)=\sum _{n=0}^{\mathrm{\infty}}\u3008{\varphi}_{n}(x),U(x,s)\u3009{e}^{-{\lambda}_{n}t}{\varphi}_{n}(x),$

where

$\u3008{\varphi}_{n}(x),U(x,s)\u3009={\int}_{0}^{1}{\varphi}_{n}(x)U(x,s)\phantom{\rule{0.2em}{0ex}}dx$. The null space of the semigroup

$T(t)$ of the linear operators can be defined as follows:

$N(T)=\{U(x,s):\u3008{\varphi}_{n}(x),U(x,s)\u3009=0,\text{for all}n=1,2,3,\dots \}.$

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:

$v(x,t)=T(t)v(x,0)+{\int}_{0}^{t}T(t-s){\left((k(x)-k(0))({v}_{x}(x,t)+\frac{{\psi}_{0}}{k(0)})\right)}_{x}\phantom{\rule{0.2em}{0ex}}ds.$

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:

$\begin{array}{rcl}u(x,t)& =& \frac{{\psi}_{0}}{k(0)}x+{\psi}_{1}-{\psi}_{0}+T(t)(g(x)-\frac{{\psi}_{0}}{k(0)}x+{\psi}_{0}-{\psi}_{1})\\ +{\int}_{0}^{t}T(t-s){((k(x)-k(0)){u}_{x}(x,s))}_{x}\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

(8)

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

Then we can rewrite the solution representation in terms of

$\zeta (x)$ and

$\xi (x,s)$ in the following form:

$u(x,t)=\frac{{\psi}_{0}}{k(0)}x+{\psi}_{1}-{\psi}_{0}+T(t)\zeta (x)+{\int}_{0}^{t}T(t-s)\xi (x,s)\phantom{\rule{0.2em}{0ex}}ds.$

Substituting

$x=0$ into this solution representation yields

$u(0,t)={\psi}_{1}-{\psi}_{0}+T(t)\zeta (0)+{\int}_{0}^{t}T(t-s)\xi (0,s)\phantom{\rule{0.2em}{0ex}}ds.$

Taking into account the overmeasured data

$u(0,t)=f(t)$, we get

$f(t)=({\psi}_{1}-{\psi}_{0}+T(t)\zeta (0)+{\int}_{0}^{t}T(t-s)\xi (0,s)\phantom{\rule{0.2em}{0ex}}ds),$

(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

${u}_{x}(0,t)=\frac{{\psi}_{0}}{k(0)}+z(0,t)+{\int}_{0}^{t}w(0,t-s,s)\phantom{\rule{0.2em}{0ex}}ds.$

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:

$k(0)={\psi}_{0}/(-{\psi}_{0}+{\psi}_{1}+z(0,t)+{\int}_{0}^{t}w(0,t-s,s)\phantom{\rule{0.2em}{0ex}}ds).$

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

$\mathcal{K}$:

$\mathrm{\Phi}[k](x):={\psi}_{1}-{\psi}_{0}+T(t)\zeta (0)+{\int}_{0}^{t}T(t-s)\xi (0,s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,T].$

(12)

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

*t*:

$\begin{array}{rcl}{u}_{t}(x,t)& =& T(t)A(g(x)-\frac{{\psi}_{0}}{k(0)}x+{\psi}_{0}-{\psi}_{1})+{((k(x)-k(0)){u}_{x}(x,t))}_{x}\\ +{\int}_{0}^{t}AT(t-s){((k(x)-k(0)){u}_{x}(x,s))}_{x}\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

Using the semigroup property

$-{\int}_{0}^{t}AT(s)u(x,s)\phantom{\rule{0.2em}{0ex}}ds=T(t)u(x,t)-T(0)u(x,t)$, we obtain

$\begin{array}{rcl}{u}_{t}(x,t)& =& -k(0)T(t){g}^{\u2033}(x)-2T(0){((k(x)-k(0)){u}_{x}(x,t))}_{x}\\ +T(t){((k(x)-k(0)){u}_{x}(x,0))}_{x}.\end{array}$

Taking

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

$\begin{array}{rcl}{u}_{t}(0,t)& =& -k(0)T(t){g}^{\u2033}(0)-T(0){k}^{\prime}(0){u}_{x}(0,0)\\ +T(t)({k}^{\prime}(0){u}_{x}(0,0))-T(0)({k}^{\prime}(0){u}_{x}(0,t)).\end{array}$

Since

$u(0,t)=f(t)$, we have

${u}_{t}(0,t)={f}^{\prime}(t)$. Taking into account this and substituting

$t=0$ yield

${f}^{\prime}(0)=-k(0){g}^{\u2033}(0)-{k}^{\prime}(0)\frac{{g}^{\prime}(0)}{k(0)}.$

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

$x=0$:

${k}^{\prime}(0)=\frac{-{k}^{2}(0){g}^{\u2033}(0)-k(0){f}^{\prime}(0)}{{g}^{\prime}(0)}.$

(13)

Under the determined values

$k(0)$ and

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

${\mathcal{K}}_{0}:=\{k\in \mathcal{K}:k(0)=\frac{{\psi}_{0}}{-{\psi}_{0}+{\psi}_{1}+z(0,0)},{k}^{\prime}(0)=\frac{-{k}^{2}(0){g}^{\u2033}(0)-k(0){f}^{\prime}(0)}{{g}^{\prime}(0)}\}.$

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*:

$\mathrm{\Delta}f(\tau )={\int}_{0}^{\tau}T(\tau -s)\mathrm{\Delta}\xi (0,s)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}ds$

(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* $\u3008{\varphi}_{n}(x),\mathrm{\Delta}\xi (0,s)\u3009=\u3008{\varphi}_{n}(x),{\xi}^{1}(x,t)-{\xi}^{2}(x,t)\u3009=0,\phantom{\rule{1em}{0ex}}\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*.,

$\mathrm{\Phi}[{k}_{1}]\ne \mathrm{\Phi}[{k}_{2}]\phantom{\rule{1em}{0ex}}\mathrm{\forall}{k}_{1},{k}_{2}\in {\mathcal{K}}_{0},\phantom{\rule{2em}{0ex}}{k}_{1}(x)\ne {k}_{2}(x).$