Question: I have the following problem:
given a rod of length $L$, the heat equation describing the behaviour of the temperature $u(x,t)$ on this rod, is:
$$\partial\_t u(x,t)=k(x)\partial\_{xx}u(x,t)$$
in which the function $k(x)$ is unknown. We have also the following boundary and initial conditions:
$$u(x,0)=T\_0$$
$$u(0,t)=u\_0(t)$$
$$u(L,t)=u\_L(t)$$
and we have also the following functions in two points $x\_0$ and $x\_1$ on the rod:
$$u(x\_0,t)=u\_{x\_0}(t)$$
and:
$$u(x\_1,t)=u\_{x\_1}(t)$$

The question now is: is it possible to find the function $k(x)$ with only this informations available?

Answer: 
Not always. Suppose the boundary conditions are 

$$ u(0,t) = u(L,t) = T\_0 $$

Then $u \equiv T\_0$ is a solution to the heat equation for **any** diffusion coefficient $k(x)$.
Note that the boundary conditions are inhomogeneous. So we should first apply the method in <http://maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=38> , i.e. let $u(x,t)=v(x,t)+u\_0(t)+\dfrac{x}{L}(u\_L(t)−u\_0(t))$ ,

Then $\partial\_tu(x,t)=\partial\_tv(x,t)+\partial\_tu\_0(t)+\dfrac{x}{L}(\partial\_tu\_L(t)−\partial\_tu\_0(t))$

$\partial\_xu(x,t)=\partial\_xv(x,t)+u\_L(t)−u\_0(t)$

$\partial\_{xx}u(x,t)=\partial\_{xx}v(x,t)$

$\therefore\partial\_tv(x,t)+\partial\_tu\_0(t)+\dfrac{x}{L}(\partial\_tu\_L(t)−\partial\_tu\_0(t))=k(x)\partial\_{xx}v(x,t)$

$\partial\_tv(x,t)-k(x)\partial\_{xx}v(x,t)=-\partial\_tu\_0(t)-\dfrac{x}{L}(\partial\_tu\_L(t)−\partial\_tu\_0(t))$

with $v(0,t)=0$ , $v(L,t)=0$ and $v(x,0)=T\_0-u\_0(0)-\dfrac{x}{L}(u\_L(0)−u\_0(0))$