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.


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.


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 $(t,x,y)$.

  • $\begingroup$ 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

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