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.


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 ?

| cite | improve this answer | |
  • $\begingroup$ 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 $u(x,t)$ $\endgroup$ – Riccardo.Alestra Feb 24 '12 at 13:34
  • $\begingroup$ In this case, can you make the dependance on $t$ precise? $\endgroup$ – Pascal Feb 24 '12 at 13:46
  • $\begingroup$ $k(t)=W_i(t)$ where $W_i(t)$ is a $Wiener$ $process$ $\endgroup$ – Riccardo.Alestra Feb 24 '12 at 14:12
  • 1
    $\begingroup$ 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). $\endgroup$ – Pascal Feb 24 '12 at 15:03

Of course we use separation of variables:

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

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



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

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

$u(0,t)=u_0$ :

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

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


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

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

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

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

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

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.