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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.

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Is the coefficient on $u$ the indicator function on the interval $[-1,1]$? There's some formatting oddity there, I think. –  Christopher A. Wong Nov 25 '11 at 4:00
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is that $1|_{[-1,1]}$ as a function of $t$ or as a function of $x$? –  Pietro Majer Nov 25 '11 at 10:27
    
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 $t\to \infty$? –  Ori Gurel-Gurevich Nov 27 '11 at 6:00
    
@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]. –  George Lowther Nov 27 '11 at 22:47
    
There are explicit formulas for one-sided intervals such as $(-\infty,1]$, I'm not sure about bounded intervals though. –  George Lowther Nov 27 '11 at 22:55
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1 Answer

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.

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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. –  Matt Cooper Nov 27 '11 at 4:04
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