Consider now the inverse problem with one measured output data u({x}_{0},t)={\psi}_{1} at x={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(u(0,0)){u}_{x}(x,t))}_{x}={([k(u)-k(u(0,0))]{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:

\begin{array}{r}{u}_{t}(x,t)-k(u(0,0)){u}_{xx}(x,t)={((k(u)-k(u(0,0))){u}_{x}(x,t))}_{x},\phantom{\rule{1em}{0ex}}(x,t)\in {\mathrm{\Omega}}_{T},\\ u(x,0)=g(x),\phantom{\rule{1em}{0ex}}0<x<1,\\ u(0,t)={\psi}_{0},\phantom{\rule{2em}{0ex}}u(1,t)=f(t),\phantom{\rule{1em}{0ex}}0<t<T.\end{array}

(4)

In order to determine the unknown boundary condition u(1,t)=f(t), we need to determine the solution of the following parabolic problem:

\begin{array}{r}{u}_{t}(x,t)-k(u(0,0)){u}_{xx}(x,t)={((k(u)-k(u(0,0))){u}_{x}(x,t))}_{x},\phantom{\rule{1em}{0ex}}(x,t)\in {\mathrm{\Omega}}_{{T}_{0}},\\ u(x,0)=g(x),\phantom{\rule{1em}{0ex}}0<x<{x}_{0},\\ u(0,t)={\psi}_{0},\phantom{\rule{2em}{0ex}}u({x}_{0},t)={\psi}_{1},\phantom{\rule{1em}{0ex}}0<t<T,\end{array}

(5)

where {\mathrm{\Omega}}_{{T}_{0}}=\{(x,t)\in {R}^{2}:0<x<{x}_{0},0<t\le T\}. To formulate the solution of the above problem in terms of a semigroup, we need to define a new function

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

(6)

which satisfies the following parabolic problem:

\begin{array}{r}{v}_{t}(x,t)+A[v(x,t)]=\left(\right(k(v(x,t)+{\psi}_{0}-\frac{({\psi}_{0}-{\psi}_{1})x}{{x}_{0}})\\ \phantom{{v}_{t}(x,t)+A[v(x,t)]=}-k(u(0,0))){({v}_{x}(x,t)-\frac{({\psi}_{0}-{\psi}_{1})}{{x}_{0}}))}_{x},\phantom{\rule{1em}{0ex}}(x,t)\in {\mathrm{\Omega}}_{T},\\ v(x,0)=g(x)+\frac{({\psi}_{0}-{\psi}_{1})x}{{x}_{0}}-{\psi}_{0},\phantom{\rule{1em}{0ex}}0<x<{x}_{0},\\ v(0,t)=0,\phantom{\rule{2em}{0ex}}v({x}_{0},t)=0,\phantom{\rule{1em}{0ex}}0<t<T.\end{array}

(7)

Here, A[\cdot ]:=-k(u(0,0))\frac{{d}^{2}[\cdot ]}{d{x}^{2}} is a second-order differential operator and its domain is {D}_{A}=\{v(x)\in {H}_{0}^{2,2}(0,1)\cap {H}_{0}^{3,2}[0,1]:v(0)=0=v(1)\}, where {H}_{0}^{2,2}(0,1)=\overline{{C}_{0}^{2}(0,1)} and {H}_{0}^{1,2}[0,1]=\overline{{C}_{0}^{1}[0,1]} are Sobolev spaces. Obviously, by completion g(x)\in {D}_{A}, since the initial value function g(x) belongs to {C}^{3}[0,1]. Hence, {D}_{A} is dense in {H}_{0}^{2,2}[0,1], which is a necessary condition for being an infinitesimal generator.

In the following, despite doing the calculations in the smooth function space, by completion they are valid in the Sobolev space.

Let us denote the semigroup of linear operators by T(t) generated by the operator *A* [8, 9]. We can easily find the eigenvalues and eigenfunctions of the differential operator *A*. Moreover, the semigroup T(t) can be easily constructed by using the eigenvalues and eigenfunctions of the infinitesimal generator *A*. Hence, we first consider the following eigenvalue problem:

\begin{array}{r}A\varphi (x)=\lambda \varphi (x),\\ \varphi (0)=0;\phantom{\rule{2em}{0ex}}\varphi ({x}_{0})=0.\end{array}

(8)

We can easily determine that the eigenvalues are {\lambda}_{n}=k(u(0,0))\frac{{n}^{2}{\pi}^{2}}{{x}_{0}^{2}} for all n=1,\dots and the corresponding eigenfunctions are {\varphi}_{n}(x)=sin(\frac{n\pi x}{{x}_{0}}). 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),

(9)

where \u3008{\varphi}_{n}(x),U(x,s)\u3009={\int}_{0}^{1}{\varphi}_{n}(x)U(x,s)\phantom{\rule{0.2em}{0ex}}dx. Under this representation, 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=0,1,2,3,\dots \}.

From the definition of the semigroup T(t), we can say that the null space of it consists of only zero function, *i.e.*, N(T)=\{0\}. This result is very important for the uniqueness of the unknown boundary condition u(1,t).

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

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

Now, by using the identity (6) and taking the initial value u(x,0)=g(x) into account, the integral equation for 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)& =& {\psi}_{0}-\frac{({\psi}_{0}-{\psi}_{1})x}{{x}_{0}}+T(t)(g(x)+\frac{({\psi}_{0}-{\psi}_{1})x}{{x}_{0}}-{\psi}_{0})\\ +{\int}_{0}^{t}T(t-s){((k(u(x,s))-k(u(0,0))){u}_{x}(x,s))}_{x}\phantom{\rule{0.2em}{0ex}}ds.\end{array}

(10)

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

\begin{array}{c}\zeta (x)=(g(x)+\frac{({\psi}_{0}-{\psi}_{1})x}{{x}_{0}}-{\psi}_{0}),\hfill \\ \xi (x,t)={((k(u(x,t))-k(u(0,0))){u}_{x}(x,t))}_{x}.\hfill \end{array}

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

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

(11)

This is the integral representation of a solution of the initial-boundary value problem (5) on {\mathrm{\Omega}}_{{T}_{0}}=\{(x,t)\in {R}^{2}:0<x<{x}_{0},0<t\le T\}. It is obvious from the eigenfunctions {\varphi}_{n}(x), the domain of eigenfunctions can be extended to the closed interval [0,1]. Moreover they are continuous on [0,1]. Under this extension, the uniqueness of the solutions of the initial-boundary value problems (4) and (5) imply that the integral representation (11) becomes the integral representation of a solution of the initial-boundary value problem (4) on {\mathrm{\Omega}}_{T}=\{(x,t)\in {R}^{2}:0<x<1,0<t\le T\}.

At this stage, it is obvious that the solution of the inverse problem can easily be obtained by substituting x=1 into the integral representation (11) of the solution u(x,t),

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

(12)

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

The right-hand side of the identity (12) defines *the semigroup representation of the unknown boundary condition* u(1,t) *at* x=1.