How could we continue to show the inequality?

Question

Let $\Omega$ a bounded space. Let $u_1$ the solution of the problem $$-\Delta u_1(x)=f(x), x \in \Omega \\ u_1(x)=g_1(x), x \in \partial{\Omega}$$ and $u_2$ is the solution of the problem $$-\Delta u_2(x)=f(x), x \in \Omega \\ u_2(x)=g_2(x), x \in \partial{\Omega}$$ Using the maximum principle I have to show that $$|u_1(x)-u_2(x)| \leq \max_{x \in \partial{\Omega}}|g_1(x)-g_2(x)|, x \in \Omega$$

I have done the following:

Subtracting the two problems we get the following:

$$\Delta (u_1(x)-u_2(x))=0, x \in \Omega \\ u_1(x)-u_2(x)=g_1(x)-g_2(x), x \in \partial{\Omega} $$ and $$\Delta(-(u_1(x)-u_2(x)))=0, x\in \Omega \\ -(u_1(x)-u_2(x))=-(g_1(x)-g_2(x)), x \in \partial{\Omega}$$ Since $\Delta (u_1-u_2) \geq 0$, from the maximum for $u_1-u_2$ we have that $$\max_{\Omega}(u_1-u_2)=\max_{\partial{\Omega}}(u_1-u_2)=\max_{\partial{\Omega}}(g_1-g_2)$$ Since $\Delta (-(u_1-u_2)) \geq 0$, from the maximum for $-(u_1-u_2)$ we have that $$\max_{\Omega}(-(u_1-u_2))=\max_{\partial{\Omega}}(-(u_1-u_2))=\max_{\partial{\Omega}}(-(g_1-g_2))$$ Since $$\max (-(u_1-u_2))=\min (u_1-u_2)$$ we have that $$\min_{\Omega}(u_1-u_2) \leq u_1-u_2 \leq \max_{\Omega}(u_1-u_2) \\ \Rightarrow \max_{\partial{\Omega}}(-(g_1-g_2)) \leq u_1-u_2 \leq \max_{\partial{\Omega}}(g_1-g_2)$$

Is this correct so far?? How could we continue to show that $$|u_1(x)-u_2(x)| \leq \max_{x \in \partial{\Omega}}|g_1(x)-g_2(x)|, x \in \Omega$$ ??

Answer 1

It looks like you have a slight mistake, in that \begin{equation} \max_{\Omega}(-(u_1 - u_2)) = - \min_{\Omega}(u_1 - u_2) \end{equation} not $+\min_{\Omega}(u_1 - u_2)$ which you have written. This means that \begin{equation} \min_{\Omega}(u_1 - u_2) = -\max_{\partial\Omega}(-(g_1 - g_2)) = \min_{\partial \Omega}(g_1 - g_2). \end{equation}

You already showed \begin{equation} \max_{\Omega}(u_1 - u_2) = \max_{\partial\Omega}(g_1 - g_2). \end{equation} So combining the two, \begin{equation} \min_{\partial\Omega}(g_1 - g_2) \leq u_1 - u_2 \leq \max_{\partial\Omega}(g_1 - g_2). \end{equation}

The last thing is to observe is \begin{equation} \max_{\partial \Omega}(g_1 - g_2) \leq \max_{\partial \Omega}|g_1 - g_2| \end{equation} and \begin{equation} (g_1 - g_2) \geq -|g_1 - g_2|\quad \Longrightarrow \quad\min_{\partial\Omega}(g_1 - g_2) \geq \min_{\partial\Omega}(-|g_1 - g_2|) = -\max_{\partial \Omega}|g_1 - g_2| \end{equation}

So, \begin{equation} -\max_{\partial\Omega}|g_1 - g_2| \leq u_1 - u_2 \leq \max_{\partial \Omega}|g_1 - g_2| \end{equation} which is what we want.