Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given the heat equation:

$$\partial_{t}{\varPhi(x,t)}=k^2\partial_{xx}{\varPhi(x,t)}$$

with the boundary conditions:

$$\Phi(x,0)=\Phi_0$$

and a Neumann boundary condition of the kind:

$${\partial_{x}}{\Phi(0,t)=\nu(t)+C}$$

where $\nu(t)$ is a stochastic variable with gaussian distribution ${\sigma=\sigma_0,\mu=0}$ and $C$ a constant, what is the distribution of the $\Phi(L,t)$?

Thanks in advance

share|improve this question
    
It's not clear to me what you mean by "$\nu(t)$ is a stochastic variable". You have to specify what the covariance is between $\nu(t)$ and $\nu(s)$. In his answer, Jon has interpreted $\nu(t)$ to be Gaussian white noise. Is this what you meant? –  Paul Tupper Feb 13 '12 at 18:55
    
@Paul Tupper: Yes, the Jon's interpretation is correct. –  Riccardo.Alestra Feb 14 '12 at 10:43
add comment

1 Answer 1

up vote 5 down vote accepted

In this case $\Phi(x,t)$ is itself a stochastic process and this equation should be rewritten in a proper way. There is some literature as this and a more general theory of stochastic pde due to John Walsh (a tutorial can be found here). In this case, a general solution can be written down using the fundamental solution of the heat equation given by

$$\Delta(x,t)=\frac{1}{\sqrt{4\pi k^2 t}}e^{-\frac{x^2}{4k^2 t}}$$

and then one has

$$\Phi(x,t)=\int dx'\Phi_0(x')\Delta(x-x',t)+k^2\int dt'[\nu(t')+C]\Delta(x,t-t')$$

and we can easily compute

$$\langle\Phi(x,t)\rangle = \int dx'\Phi_0(x')\Delta(x-x',t)+k^2C\int dt'\Delta(x,t-t')$$

$$\langle\Phi(x,t)\Phi(y,s)\rangle=\int dx'\Phi_0(x')\Delta(x-x',t)\int dy'\Phi_0(y')\Delta(y-y',s)+$$ $$k^2C\int dx'\Phi_0(x')\Delta(x-x',t)\int ds'\Delta(y,s-s')+$$ $$k^2C\int dt'\Delta(x,t-t')\int dy'\Phi_0(y')\Delta(y-y',s)+$$ $$k^4C^2\int dt'\Delta(x,t-t')\int ds'\Delta(x,s-s')+k^4\sigma^2_0\int dt'\Delta(x,t-t')\Delta(y,s-t')$$

where I used the fact that $\langle\nu(s)\nu(t)\rangle=\sigma^2_0\delta(t-s)$. It is interesting to note the simplest case $\Phi_0=0$ and $C=0$ producing immediately

$$\langle\Phi(x,t)\rangle = 0$$

and

$$\langle\Phi(x,t)\Phi(y,s)\rangle=k^4\sigma^2_0\int dt'\Delta(x,t-t')\Delta(y,s-t').$$

So, all higher order even correlation functions are given by products of the fundamental solution of the heat equation properly integrate in intermediate times.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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