$L_\infty$-stability for forward heat equation $u_t=\frac12u_{xx}$

I am considering the forward heat equation

$$\frac{\partial u}{\partial t}=\frac12\frac{\partial^2u}{\partial x^2}$$

with $$0, $$t>0$$ under the explicit Euler scheme such that

$$U_j^{(n+1)}=U_j^{(n)}+\frac{\Delta t}{2\Delta x^2}\left(U_{j+1}^{(n)}-2U_j^{(n)}+U_{j-1}^{(n)}\right).$$

I have already shown conditional $$L_2$$-stability using von Neumann analysis, and now would like to prove $$L_\infty$$-stability conditioned on $$\Delta t\leq \Delta x^2$$. My attempt at a proof is as follows: denoting $$\lambda=\frac{\Delta t}{2\Delta x^2}$$, we have

\begin{align*} |U^{(n+1)}_j|&\leq|U^{(n}_j|+\lambda\left(|U^{(n)}_{j+1}|-2|U^{(n)}_j|+|U^{(n)}_{j-1}|\right)\\ &\leq\max_j|U^{(n)}_j|+\lambda\left(\max_j|U^{(n)}_{j+1}|-2\max_j|U^{(n)}_j|+\max_j|U^{(n)}_{j-1}|\right)\\ &=\max_j|U^{(n)}_j|, \end{align*}

where I have used $$|x+y|\leq|x|+|y|$$. However, my attempt therefore leads to unconditional $$L_\infty$$-stability. Where have I gone wrong?

In general, how should I go about deriving the conditions for $$L_\infty$$-stability? I have just taken a closer look at the hints provided and it mentions that since

$$U_j^{(n+1)}=pU_{j+1}^{(n)}+(1-2p)U_j^{(n)}+pU_{j-1}^{(n)},$$

where $$p=\lambda/2$$, I must justify that for $$2p\leq1$$, $$\min_j|U^{(n+1)}_j|\leq U^{(n+1)}_j\leq\max_j|U^{(n+1)}_j|$$ $$\forall n$$, and therefore $$\max_j|U^{(n+1)}_j|\leq\max_j|U^{(n)}_j|$$. However, where this justification comes from (I suspect some form of discrete maximum principle) and it contradicts what was provided in the comment by @John_Krampf. What has happened here?

• For the heat equation the maximum and minimum values $u$ will ever take exist at the initial time. By induction this means $$||U^{(n+1)}||_{L_{\infty}} \leq ||U^{(n)}||_{L_{\infty}}$$ Thus $\{||U^{(n)}||_{L_{\infty}}\}_n$ is a positive and monotonically decreasing sequence of positive numbers so by the Monotone Convergence Theorem it converges to a limit. Why do you need your proof to be conditioned? Dec 28, 2020 at 4:25
• @John_Krampf Hi, I managed to find a hint for my question, but it contradicts your comment that it is unconditionally stable in the maximum norm; could you compare it with my update? Thanks! Dec 28, 2020 at 18:11
• You've incorrectly estimated the middle term in the parentheses. It has negative sign and you're taking max. Try collecting terms first Dec 30, 2020 at 14:39

Since $$2p\le 1$$ implies that $$1-2p\ge0$$, we have that $$|1-2p| = 1-2p$$, and $$p\ge 0$$ implies that $$|p|=p$$, and so
\begin{align*} |U^{n+1}_j| & = |pU^{n}_{j+1}+(1-2p)U^{n}_{j}+pU^{n}_{j-1}|\\ &\le|pU^{n}_{j+1}|+|(1-2p)U^{n}_{j}|+|pU^{n}_{j-1}|\\ &=|p||U^{n}_{j+1}|+|(1-2p)||U^{n}_{j}|+|p||U^{n}_{j-1}|\\ &=p|U^{n}_{j+1}|+(1-2p)|U^{n}_{j}|+p|U^{n}_{j-1}|\\ &\le p\max_k|U^{n}_{k}|+(1-2p)\max_k|U^{n}_{k}|+p\max_k|U^{n}_{k}|\\ &=(p+1-2p+p)\max_k|U^{n}_{k}|\\ &=\max_k|U^{n}_{k}|. \end{align*} Thus, $$|U^{n+1}_j|\le \max_k|U^{n}_{k}|$$ for all $$j$$, and therefore $$\max_k|U^{n+1}_k|\le \max_k|U^{n}_{k}|$$.