Show stability with respect to the initial data for the heat equation, in other words, if $u^1$ and $u^2$ solve $$\begin{cases} u_t^i - \Delta u^i = 0 & \mbox{ in} \space\ U_T \\ u^i = \rho^i(x) & \mbox{ on} \space\ \partial U_T \end{cases}$$ for $i=1,2$, then: $$\sup_{0\leq t \leq T} \int_U |u^1(x,t) - u^2(x,t)|^2 dx \leq \int_U |\rho^1(x) - \rho^2(x)|^2 dx.$$

My (incomplete) attempt:

Define $w:= u^1 - u^2$ and $g:= \rho^1 - \rho^2$. We'll show if $$\begin{cases} w_t - \Delta w = 0 & \mbox{ in} \space\ U_T \\ w=g & \mbox{ on} \space\ \partial U_T \end{cases}$$ then $$\sup_{0\leq t \leq T} \int_U |w(x,t)|^2 dx \leq \int_U |g(x,t)|^2 dx $$ Consider energy methods of the heat equation. Define: $$e(t) = \dfrac{1}{2} \int_U w^2 dx$$ Then, \begin{align*} e'(t) &= \dfrac{1}{2} \int_U 2ww_t dx \\ &= \int_U w \Delta w dx \hspace{2cm} \text{(since $w_t = \Delta w$ in $U_T$)}\\ &= - \int_U |\nabla w|^2 dx\hspace{1.5cm} \text{(integration by parts)} \\ &\leq 0 \end{align*} This shows that the $L^2$ norm of $w$ is negative.

I am unsure if I may conclude my proof at this point. Is showing that the $L^2$ norm negative sufficient for establishing the inequality: $$\sup_{0\leq t \leq T} \int_U |w(x,t)|^2 dx \leq \int_U |g(x,t)|^2 dx $$


1 Answer 1


You did not show, that the $L^2$-Norm of $w$ is negativ, but you showed $e$ is monotonically decreasing in time $t$. Hence $$\int_U |w(x,t)|^2\,dx\leq \int_U |w(x,0)|^2\,dx$$ Since your initial conditions are $w(x,0)=g(x,0)$ you have $$\int_U |w(x,t)|^2\,dx\leq\int_U |g(x,0)|^2\,dx,$$ which is what you want.

Now I have to point out a possible mistake: Your data is described as $u^i=p^i(x)$ on $\partial U_T$. Hence if $x\in \partial U$ you might not have $p^i(x)=0$, which would be needed in your integration by parts formula. It seems to me you neglected some boundary conditions.

  • $\begingroup$ What boundary condition would I need to prescribe? Just $u^i = g(x,t)$? $\endgroup$
    – Dragonite
    Nov 7, 2017 at 15:41
  • $\begingroup$ Usually for these kind of equations $U_T=U\times [0,T)$ and $\partial U_T=\partial U \times [0,T)\cup U\times\{0\}$. Now what you need if $g:\partial U_T\rightarrow \mathbb{R}$ is your initial/boundary datum, is that $g$ has to satisfy $g(x,t)=0$ for $x\in\partial U$ and $t\in[0,T)$ arbitrary. This would imply that the boundary terms in the integration by parts formula do not appear. $\endgroup$ Nov 7, 2017 at 15:48

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