7
$\begingroup$

I am interested in the following damped heat equation on $\mathbf{R}$, $u_t = u_{xx} - 1_{x \in [-1,1]} u$ with initial data $u(0,x) = \delta(x-x_0)$ for some $x_0 \in \mathbf{R}$.

In particular I am interested in obtaining non-trivial bounds on $u(t,0)$. Of course the heat kernel gives a trivial bound on $u(t,0)$ but I am struggling to obtain anything stronger.

Perhaps the equation has a closed form solution from which it is easy to read such information off?

Added later: Of course appropriate growth conditions at infinity are assumed to ensure a unique solution.

Correction: The indicator function is a function of the $x$ variable only.

$\endgroup$
8
  • $\begingroup$ Is the coefficient on $u$ the indicator function on the interval $[-1,1]$? There's some formatting oddity there, I think. $\endgroup$ Nov 25, 2011 at 4:00
  • 1
    $\begingroup$ is that $1|_{[-1,1]}$ as a function of $t$ or as a function of $x$? $\endgroup$ Nov 25, 2011 at 10:27
  • $\begingroup$ Am I right to rephrase this probabilistically as Brownian motion killed at rate 1 when inside $[-1,1]$? Also, are you mostly interested in bounding when $x_0$ is fixed and $t\to \infty$? $\endgroup$ Nov 27, 2011 at 6:00
  • $\begingroup$ @Ori: Yes, that seems right. As the question concerns the bound at $x=0$ then it reduces to computing the Laplace transform of the time that a Brownian bridge spends in [-1,1]. $\endgroup$ Nov 27, 2011 at 22:47
  • $\begingroup$ There are explicit formulas for one-sided intervals such as $(-\infty,1]$, I'm not sure about bounded intervals though. $\endgroup$ Nov 27, 2011 at 22:55

1 Answer 1

1
$\begingroup$

Let us assume that $1_{[-1,1]}$ is the identity function for $t$ but here I consider a generic function $f(t)$. Let us consider the Dirichlet problem on a bounded domain $D$

$$\Delta\phi_n+\lambda_n\phi_n=0 \qquad \phi=0\ on\ \partial D$$

You can write down the exact solution to your equation as

$$u(t,x,y)=\sum_n a_n(t)\phi_n(x)\phi_n(y)$$

so that $u(0,x,y)=\delta(x-y)$ implies $a_n(0)=1$. By a direct substitution you get the equations to be solved

$$\dot a_n+\lambda_na_n(t)+f(t)a_n(t)=0$$

that admits the solution

$$a_n(t)=e^{-\lambda_n t-\int_0^t dt'f(t')}.$$

In this way you should be able to get a better bound on the solution.

Now, let us assume that $1_{[-1,1]}$ is the identity function for $x$. The problem is reduced to the one of a Schroedinger equation for a rectangular potential barrier. Let us search for eigenfunctions to the problem

$$\partial^2\phi_E(x)-1_{[-1,1]}\phi_E(x)=-E\phi_E(x)$$

We expect a continuous spectrum in this case and will get

$$\phi^L_E(x)=A_1e^{ik_0x}+A_2e^{-ik_0x}\qquad x<-1$$ $$\phi^C_E(x)=B_1e^{ik_1x}+B_2e^{-ik_1x}\qquad x\in [-1,1]$$ $$\phi^R_E(x)=C_1e^{ik_0x}+C_2e^{-ik_0x}\qquad x>1$$

being $k_1=\sqrt{E-1}$ for $x\in [-1,1]$ and $k_0=\sqrt{E}$ otherwise. You now impose a continuity condition on the derivative and the eigenfunctions to get the coefficients. The final solution will take an integral form as

$$u(t,x,x_0)=\int_C dEe^{-Et}\phi_E(x)\phi_E(x_0)$$

with a properly chosen contour $C$. Please, note that is also $1_{[-1,1]}=\theta(x+1)-\theta(x-1)$ being $\theta(x)$ the Heaviside function.

$\endgroup$
1
  • $\begingroup$ I apologize for the confusion but I intended the indicator function to be a function of $x$. The original post has been updated to reflect this. Of course, as you pointed out, if it is a function of $t$ alone then one can solve via an integrating factor. $\endgroup$ Nov 27, 2011 at 4:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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