Poisson’s equation estimate: $\Vert u\Vert_{L^{\infty}(\Omega)}\leq\frac{1}{8}A+B$

Question

Define $\Omega=(0,1)^3$ and $A,B>0$.
Let $u\in C^2(\Omega)\cap C^0(\bar{\Omega})$ be a solution of the equation

\begin{equation}
\begin{cases}
-\Delta u=A&\text{ in }\Omega\\ \quad u=B& \text{ in } \partial\Omega
\end{cases}
\end{equation}

I want to show that the sup norm of $u$ is bounded like: $\Vert u\Vert_{L^{\infty}(\Omega)}\leq\frac{1}{8}A+B$.

What I have tried so far:

I defined a function $f(x):=u(x)+\frac{1}{2}x_1^2A$.

Then $-\Delta f=-\Delta u-A=0$ in $\Omega$.

I can now use the weak maximum principle to show that

\begin{equation}
\max\limits_{x\in\bar{\Omega}}f(x)=\max\limits_{x\in\partial\Omega}f(x)=B+\frac{1}{2}x_1^2A\leq B+\frac{1}{2}A
\end{equation}
And also since $\frac{1}{2}x_1^2A>0$ I get $\Vert u\Vert_{L^{\infty}(\Omega)}\leq B+\frac{1}{2}A$

This estimate is obviously not strong enough but if I reduce the $\frac{1}{2}$ to a $\frac{1}{8}$ I can not apply the maximum principle anymore.
I think if I choose a better $f$ this could be fixed but I do not really know how so any help would be appreciated.

Answer 1

You can just use $\frac12(x_1-0.5)^2$ instead of $\frac12x_1^2$. It doesn't affect $-\Delta f$ but $$ B+\frac12(x_1-0.5)^2A\leq B+\frac18A $$