1d-heat equation $L_{\infty}$

Question

Consider the heat equation

$\begin{align} u_t -u_{xx}=0 && (x,t) \in(0,1) \times\mathbb R_+ \\u(x,0)=u_o(x) && x \in(0,1) \\u(x,t)=0 && (x,t)\in\{0,1\} \times \mathbb R_+\end{align}$

and $u_o$(1)=$u_0(0)$=0

Prove the estimate: $\lVert u \rVert_{L^\infty} \le \lVert u_o \rVert_{L^\infty}$

By the maximum principle,$u$ obtains it's maximum at the parabolic boundary. $u(x,t) \le 0$ or $u(x,t) \le max(u_o(x))$

I don't know how I conclude the result from that

would appreciate any help

Answer 1

From the maximum principle we obtain $$ u(x,t)\le\max\{0,\max_x(u_0(x))\}\le\max_x|u_0(x)|=\|u_0\|_{L_\infty}. $$ Now apply the maximum principle to $-u(x,t)$, which is the solution to the same homogeneous heat equation with the boundary condition $-u_0$ and use $\pm u_0(x)\le|u_0(x)|$ $$ -u(x,t)\le\max\{0,\max_x(-u_0(x))\}\le\max_x|u_0(x)|=\|u_0\|_{L_\infty}. $$ Combining the two inequalities and passing to the maximum gives the result.