# Random diffusion coefficient in the Fourier equation

I'm stuck on the following simple problem: It's given the Fourier equation: $$\partial_t{u(x,t)}=\partial_x[k(t)\partial_xu(x,t)]$$where the diffusion coefficient $k(t)$ is a random variable with a gaussian probability density function: $N(0,\sigma)$. Because $k(t)$ is not depending on $x$ we can write the previous equation as: $$\partial_t{u(x,t)}=k(t)\partial_{xx}u(x,t)$$ Given the boundary conditions: $$u(0,t)=u_0,u(L,0)=u_L=u(x,0)$$ How can I solve (if it's possible) this stochastic PDE? Thanks in advance.

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As far as I understand your question, you can just solve the equation as if $k$ was deterministic and then integrate with respect to the law of $k$.

By the way, you say that

$k(t)$ is a random variable with a gaussian probability density function: $N(0,\sigma)$.

Does this mean that $k(t)$ is a (random) constant ?

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I don't understand what do you mean for 'random constant'. If $k$ was a constant it was not a function of the time. In any case, I'm interested on the pdf of the solution $u(x,t)$ – Riccardo.Alestra Feb 24 '12 at 13:34
In this case, can you make the dependance on $t$ precise? – Pascal Feb 24 '12 at 13:46
$k(t)=W_i(t)$ where $W_i(t)$ is a $Wiener$ $process$ – Riccardo.Alestra Feb 24 '12 at 14:12
OK, now it is clearer. Still, you can first pretend that $k$ is deterministic, solve the equation for $u$ (I don't know the solution of the equation for time-dependent diffusion constant, you might look it up somewhere), then integrate over the law of $k$ (in this case the Wiener measure). – Pascal Feb 24 '12 at 15:03

Of course we use separation of variables:

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=k(t)X''(x)T(t)$

$\dfrac{T'(t)}{k(t)T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2k(t)\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-s^2\int_0^tk(t)~dt}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+\int_0^\infty C_4(s)e^{-s^2\int_0^tk(t)~dt}\cos xs~ds$

$u(0,t)=u_0$ :

$C_2+\int_0^\infty C_4(s)e^{-s^2\int_0^tk(t)~dt}~ds=u_0$

$\int_0^\infty C_4(s)e^{-s^2\int_0^tk(t)~dt}~ds=u_0-C_2$

$C_4(s)=(u_0-C_2)\delta(s)$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+\int_0^\infty(u_0-C_2)\delta(s)e^{-s^2\int_0^tk(t)~dt}\cos xs~ds=C_1x+C_2+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+u_0-C_2=C_1x+u_0+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds$

$u(x,0)=u(L,0)=u_L$ :

$C_1x+u_0+\int_0^\infty C_3(s)\sin xs~ds=u_L$

$\int_0^\infty C_3(s)\sin xs~ds=-C_1x+u_L-u_0$

$C_3(s)=C_1\delta'(s)+\dfrac{2(u_L-u_0)}{\pi s}$

$\therefore u(x,t)=C_1x+u_0+\int_0^\infty\biggl(C_1\delta'(s)+\dfrac{2(u_L-u_0)}{\pi s}\biggr)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds=C_1x+u_0+\int_0^\infty C_1\delta'(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds=C_1x+u_0-\biggl.\dfrac{d}{ds}(C_1e^{-s^2\int_0^tk(t)~dt}\sin xs)\biggr|_{s=0}+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds=C_1x+u_0-C_1x+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds=u_0+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds$

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