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).