Consider now the inverse problem with one measured output data f(t) at x=0. In order to formulate the solution of parabolic problem (1) by using the Fourier method of the separation of variables, let us first introduce an auxiliary function v(x,t) as follows:

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

by which we transform problem (1) into a problem with homogeneous boundary conditions. Hence the initial boundary value problem (1) can be rewritten in terms of v(x,t) in the following form:

\{\begin{array}{l}{D}_{t}^{\alpha}v(x,t)-{v}_{xx}(x,t)={((k(x)-1){v}_{x}(x,t))}_{x}-x{D}_{t}^{\alpha}{\psi}_{1}(t)-{D}_{t}^{\alpha}{\psi}_{0}(t)+{k}^{\prime}(x){\psi}_{1}(t),\\ \phantom{\rule{1em}{0ex}}(x,t)\in {\mathrm{\Omega}}_{T},\\ v(x,0)=g(x)-{\psi}_{0}(0)-{\psi}_{1}(0)x,\phantom{\rule{1em}{0ex}}0<x<1,\\ v(0,t)=0,\phantom{\rule{2em}{0ex}}{v}_{x}(1,t)=0,\phantom{\rule{1em}{0ex}}0<t<T.\end{array}

(2)

The unique solution of the initial-boundary value problem can be represented in the following form [12]:

\begin{array}{rcl}v(x,t)& =& \sum _{n=1}^{\mathrm{\infty}}\u3008\zeta (\theta ),{\varphi}_{n}(\theta )\u3009{E}_{\alpha ,1}(-{\lambda}_{n}{t}^{\alpha}){\varphi}_{n}(x)\\ +\sum _{n=1}^{\mathrm{\infty}}\left({\int}_{0}^{t}{s}^{\alpha -1}{E}_{\alpha ,\alpha}(-{\lambda}_{n}{s}^{\alpha})\u3008\xi (\theta ,t-s),{\varphi}_{n}(\theta )\u3009\phantom{\rule{0.2em}{0ex}}ds\right){\varphi}_{n}(x),\end{array}

where

\begin{array}{c}\zeta (x)=g(x)-{\psi}_{0}(0)-{\psi}_{1}(0)x,\hfill \\ \xi (x,t)={((k(x)-1){v}_{x}(x,t))}_{x}-x{D}_{t}^{\alpha}{\psi}_{1}(t)-{D}_{t}^{\alpha}{\psi}_{0}(t)+{k}^{\prime}(x){\psi}_{1}(t).\hfill \end{array}

Moreover, \u3008\zeta (\theta ),{\varphi}_{n}(\theta )\u3009={\int}_{0}^{1}{\varphi}_{n}(\theta )\zeta (\theta )\phantom{\rule{0.2em}{0ex}}d\theta, {E}_{\alpha ,\beta} being the generalized Mittag-Leffler function defined by

{E}_{\alpha ,\beta}(z)=\sum _{n=0}^{\mathrm{\infty}}\frac{{z}^{n}}{\mathrm{\Gamma}(\beta n+\alpha )}.

Assume that {\varphi}_{n}(x) is the solution of the following Sturm-Liouville problem:

\{\begin{array}{l}-{\varphi}_{xx}(x)=\lambda \varphi (x),\phantom{\rule{1em}{0ex}}0<x<1,\\ \varphi (0)=0,\phantom{\rule{2em}{0ex}}{\varphi}_{x}(1)=0,\phantom{\rule{1em}{0ex}}0<t<T.\end{array}

The Neumann-type measured output data at the boundary x=0 in terms of v(x,t) can be written in the following form:

k(0)({v}_{x}(0,t)+{\psi}_{1}(t))=f(t),\phantom{\rule{1em}{0ex}}t\in (0,T].

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

\begin{array}{r}{z}_{n}(t)=\u3008\zeta (\theta ),{\varphi}_{n}(\theta )\u3009{E}_{\alpha ,1}(-{\lambda}_{n}{t}^{\alpha}),\\ {w}_{n}(t)={\int}_{0}^{t}{s}^{\alpha -1}{E}_{\alpha ,\alpha}(-{\lambda}_{n}{s}^{\alpha})\u3008\xi (\theta ,t-s),{\varphi}_{n}(\theta )\u3009\phantom{\rule{0.2em}{0ex}}ds.\end{array}

(3)

The solution in terms of {z}_{n}(t) and {w}_{n}(t) can then be rewritten in the following form:

v(x,t)=\sum _{n=1}^{\mathrm{\infty}}{z}_{n}(t){\varphi}_{n}(x)+\sum _{n=1}^{\mathrm{\infty}}{w}_{n}(t){\varphi}_{n}(x).

Differentiating both sides of the above identity with respect to *x* and substituting x=0 yields

{v}_{x}(0,t)=\sum _{n=1}^{\mathrm{\infty}}{z}_{n}(t){\varphi}_{n}^{\prime}(0)+\sum _{n=1}^{\mathrm{\infty}}{w}_{n}(t){\varphi}_{n}^{\prime}(0).

Taking into account the over-measured data k(0)({v}_{x}(0,t)+{\psi}_{1}(t))=f(t),

f(t)=k(0)({\psi}_{1}(t)+\sum _{n=1}^{\mathrm{\infty}}{z}_{n}(t){\varphi}_{n}^{\prime}(0)+\sum _{n=1}^{\mathrm{\infty}}{w}_{n}(t){\varphi}_{n}^{\prime}(0))

(4)

is obtained, which implies that f(t) can be determined analytically. Substituting t=0 into this yields

f(0)=k(0)({\psi}_{1}(0)+\sum _{n=1}^{\mathrm{\infty}}{z}_{n}(0){\varphi}_{n}^{\prime}(0)).

Hence we obtain the following explicit formula for the value k(0) of the unknown coefficient k(x)

k(0)=\frac{f(0)}{{\psi}_{1}(0)+{\sum}_{n=1}^{\mathrm{\infty}}{z}_{n}(0){\varphi}_{n}^{\prime}(0)}.

Under the determined value k(0), the set of admissible coefficients can be defined as follows:

{\mathcal{K}}_{0}:=\{k(x)\in {C}^{1}[0,1]:{c}_{1}>k(x)\ge {c}_{0}>0,x\in [0,1],k(0)=\frac{f(0)}{{\psi}_{1}(0)+{\sum}_{n=1}^{\mathrm{\infty}}{z}_{n}(0){\varphi}_{n}^{\prime}(0)}\}.

The right-hand side of identity (4) defines the input-output mapping \mathrm{\Phi}[k] on the set of admissible source functions \mathcal{K}

\mathrm{\Phi}[k](t):=k(0)({\psi}_{1}(t)+\sum _{n=1}^{\mathrm{\infty}}{z}_{n}(t){\varphi}_{n}^{\prime}(0)+\sum _{n=1}^{\mathrm{\infty}}{w}_{n}(t){\varphi}_{n}^{\prime}(0)),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,T].

The following lemma implies the relation between the parameters {k}_{1}(x),{k}_{2}(x)\in {\mathcal{K}}_{0} at x=0 and the corresponding outputs {f}_{j}(t):=k(0){u}_{x}(0,t;{p}_{j}), j=1,2.

**Lemma 1** *Let* {\upsilon}_{1}(x,t)=\upsilon (x,t;{k}_{1}) *and* {\upsilon}_{2}(x,t)=\upsilon (x,t;{k}_{2}) *be the solutions of direct problem* (2), *corresponding to the admissible parameters* {k}_{1}(x),{k}_{2}(x)\in {\mathcal{K}}_{0}. *If* {f}_{j}(t)={u}_{x}(0,t;{k}_{j}), j=1,2, *are the corresponding outputs*. *If the condition* {k}_{1}(0)={k}_{2}(0)=k(0), *then the outputs* {f}_{j}(t), j=1,2, *satisfy the following integral identity*:

\mathrm{\Delta}f(t)=k(0)\sum _{n=1}^{\mathrm{\infty}}\mathrm{\Delta}{w}_{n}(t){\varphi}_{n}^{\prime}(0)

*for each* t\in (0,T], *where* \mathrm{\Delta}f(t)={f}_{1}(t)-{f}_{2}(t), \mathrm{\Delta}{w}_{n}(t)={w}^{1}(t)-{w}^{2}(t).

*Proof* By using identity (4), the measured output data {f}_{j}(t):=k(0)({v}_{x}(0,t)+{\psi}_{1}(t)-{\psi}_{0}(t)), j=1,2, can be written as follows:

\begin{array}{r}{f}_{1}(\tau )=k(0)({\psi}_{1}(\tau )+\sum _{n=1}^{\mathrm{\infty}}{z}_{n}^{1}(t){\varphi}_{n}^{\prime}(0)+\sum _{n=1}^{\mathrm{\infty}}{w}_{n}^{1}(t){\varphi}_{n}^{\prime}(0)),\\ {f}_{2}(\tau )=k(0)({\psi}_{1}(\tau )+\sum _{n=1}^{\mathrm{\infty}}{z}_{n}^{2}(t){\varphi}_{n}^{\prime}(0)+\sum _{n=1}^{\mathrm{\infty}}{w}_{n}^{2}(t){\varphi}_{n}^{\prime}(0)),\end{array}

respectively. Note that the definition of {z}_{n}(t) implies that {z}_{n}^{1}(t)={z}_{n}^{2}(t). Hence, the difference of these formulas implies the desired result. □

The lemma and the definitions of {w}_{n}(t) and {z}_{n}(t) given above enable us to reach the following conclusion.

**Corollary 1** *Let the conditions of Lemma * 1 *hold*. *If in addition*

\u3008{\xi}_{1}(x,t)-{\xi}_{2}(x,t),{\varphi}_{n}(x)\u3009=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in (0,T],\mathrm{\forall}n=0,1,\dots

*holds*, *where*

\begin{array}{c}{\xi}_{1}(x,t)={(({k}_{1}(x)-1){v}_{1x}(x,t))}_{x}-x{D}_{t}^{\alpha}{\psi}_{1}(t)-{D}_{t}^{\alpha}{\psi}_{0}(t)+{k}_{1}^{\prime}(x){\psi}_{1}(t),\hfill \\ {\xi}_{2}(x,t)={(({k}_{2}(x)-1)={v}_{2x}(x,t))}_{x}-x{D}_{t}^{\alpha}{\psi}_{1}(t)-{D}_{t}^{\alpha}{\psi}_{0}(t)+{k}_{2}^{\prime}(x){\psi}_{1}(t),\hfill \end{array}

*then* {f}_{1}(t)={f}_{2}(t), \mathrm{\forall}t\in [0,T].

*Proof* If \u3008{\xi}_{1}(x,t)-{\xi}_{2}(x,t),{\varphi}_{n}(x)\u3009=0, \mathrm{\forall}n=0,1,\dots , then {k}_{1}(x)={k}_{2}(x). If {k}_{1}(x)={k}_{2}(x), then {u}_{1}(x,t)={u}_{2}(x,t). Since f(t) depends on u(x,t), then from the uniqueness of solution {f}_{1}(t)={f}_{2}(t).

Since {\varphi}_{n}(x), \mathrm{\forall}n=0,1,2,\dots form a basis for the space and {\varphi}_{n}^{\prime}(0)\ne 0, \mathrm{\forall}n=0,1,2,\dots , then {k}_{1}(x)\ne {k}_{2}(x) implies that \u3008{\xi}_{1}(x,t)-{\xi}_{2}(x,t),{\varphi}_{n}(x)\u3009\ne 0 at least for some n\in \mathcal{N}. Hence by Lemma 1 we conclude that {f}_{1}(t)\ne {f}_{2}(t), which leads us to the following consequence: {k}_{1}(x)\ne {k}_{2}(x) implies that \mathrm{\Phi}[{k}_{1}]\ne \mathrm{\Phi}[{k}_{2}]. □

**Theorem 1** *Let conditions* (C1), (C2) *hold*. *Assume that* \mathrm{\Phi}[\cdot ]:{\mathcal{K}}_{0}\to {C}^{1}[0,T] *is the input*-*output mapping defined by* (4) *and corresponding to the measured output* f(t):=k(0){u}_{x}(0,t). *In this case the mapping* \mathrm{\Phi}[k] *has the distinguishability property in the class of admissible parameters* {\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{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{k}_{1}(x)\ne {k}_{2}(x).