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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?

Thanks in advance.

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What do you mean by the "information available"? That is, what are we assuming we know? The solution u(x,t)? Some or all of the functions $T_0,u_0,u_L,u_{x_0},u_{x_1}$? Are we allowed to run the "experiment" multiple times, changing some parameters, measuring others, etc.? –  BaronVT Sep 6 '12 at 8:16
    
The only thing I know is that you should look into "inverse problems for (degenerate) parabolic equations / the heat equation". Hope this helps at least a bit. –  vanguard2k Sep 6 '12 at 8:48
    
@BaronVT: the only informations available are that shown in the question. Eventually, we can assume we can measure the temperature on the rod and so we could know the $u(x,t)$ –  Riccardo.Alestra Sep 6 '12 at 9:15
    
Can you do $k(x)=\partial_t u(x,t)/\partial_{xx}u(x,t)$ ? –  timur Sep 6 '12 at 22:06

2 Answers 2

up vote 3 down vote accepted

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 $k(x)$.

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I see. It would be interesting to know if is it possible to find $k(x)$ in the case $u(0,t)=T_0(t)$ and $u(L,t)=T_1(t)$ –  Riccardo.Alestra Sep 6 '12 at 12:37

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 $v(x,0)=T_0-u_0(0)-\dfrac{x}{L}(u_L(0)−u_0(0))$

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