2
$\begingroup$

Let $H$ be the one dimensional heat kernel, i.e the function $H(t,x,y)$ such that the Dirichlet problem: $$ \begin{cases} u_t - u_{xx} = 0 & x\in(0,1) , t \in (0,\infty) \\ u(t,1) = u(t,0) =0 & t>0 \\ u(0,x) = f(x) \end{cases} $$ is solved by: $$ u(t,x) = \int_0^1 f(y) H(t,x,y) \mathrm{d}y $$ Prove that:

  1. $H$ is nonnegative
  2. The integral of $H$ with respect to $y$ is non-increasing with respect to $t$.

I believe that these properties of $H$ can be derived without using the explicit (Fourier sine series) representation of $H$. I am not sure however how to derive the first (I have not started on the second). Would anyone be able to provide a hint (I am not looking for a complete answer, at least not right now). I know that $H$ itself solves the heat equation, but I am not entirely sure how to use this property, as taking derivatives of $H$ without using the explicit representation seems to be an arduous task.

$\endgroup$
1
$\begingroup$

For the second part: we have $H_t = H_{yy}$ so we have that $\frac{d}{dt}\int_{0}^{1}H = H_y(x,1,t)-H_y(x,0,t)$. Now, since $H(x,1,t)=H(x,0,t)=0$ and $H$ is non-negative, you have minimums at $y=1,y=0$. Thus, $H_y(x,1,t) \leq 0$ and $H_y(x,0,t)\geq 0$. (If you don't believe this, look at $f(x) = -x(1-x)$ as an example.) The result follows.

$\endgroup$
0
$\begingroup$

Do you know how to you prove that $f \geq 0$ implies $u \geq 0$? If that's true no matter what continuous function $f$ you choose, then you can prove that $H(t,x,y) \geq 0$ independently of the choice of $(t,x,y)$.

$\endgroup$
  • $\begingroup$ I will try to prove that fact. $\endgroup$ – rubikscube09 Apr 28 '18 at 19:18
  • $\begingroup$ I am having some issues proving it. Would you happen to have any hints? $\endgroup$ – rubikscube09 Apr 28 '18 at 19:39
  • $\begingroup$ I have figured this out. It is an application of the minimum (in the guise of the maximum ) principle to $u$. $\endgroup$ – rubikscube09 Apr 29 '18 at 3:18

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.