Max Principle: Showing $|u_1-u_2|\leq \max_{\partial{\Omega}}|g_1-g_2|$

[mathmari]

Hey!

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 $$|u_1(x)-u_2(x)| \leq \max_{x \in \partial{\Omega}}|g_1(x)-g_2(x)|, x \in \Omega$$ ?? (Wondering)

 

[I like Serena]

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

Hi! (Blush)

Looks good to me. (Nod)

Suppose we distinguish cases... (Thinking)

Suppose that at some point $y \in \Omega$ we have that $u_1(y) - u_2(y) \ge 0$.
Then:
$$|u_1(y) - u_2(y)| = u_1(y) - u_2(y) \le \max_{x \in \partial \Omega} (g_1(x)-g_2(x)) \le \max_{x \in \partial \Omega} |g_1(x)-g_2(x)|$$

Now suppose that $u_1(y) - u_2(y) < 0$... (Thinking)

 

[mathmari]

Looks good to me. (Nod)

(Happy)

Suppose we distinguish cases... (Thinking)

Suppose that at some point $y \in \Omega$ we have that $u_1(y) - u_2(y) \ge 0$.
Then:
$$|u_1(y) - u_2(y)| = u_1(y) - u_2(y) \le \max_{x \in \partial \Omega} (g_1(x)-g_2(x)) \le \max_{x \in \partial \Omega} |g_1(x)-g_2(x)|$$

Now suppose that $u_1(y) - u_2(y) < 0$... (Thinking)

When we have that $u_1(y)-u_2(y)<0$, then:

$$|u_1(y)-u_2(y)|=-(u_1(y)-u_2(y)) \leq \max_{x \in \Omega} (-(u_1(x)-u_2(x)))=\max_{x \in \partial{\Omega}} (-(g_1(x)-g_2(x)))$$

Is this correct so far?? (Wondering)

 

[I like Serena]

When we have that $u_1(y)-u_2(y)<0$, then:

$$|u_1(y)-u_2(y)|=-(u_1(y)-u_2(y)) \leq \max_{x \in \Omega} (-(u_1(x)-u_2(x)))=\max_{x \in \partial{\Omega}} (-(g_1(x)-g_2(x)))$$

Is this correct so far?? (Wondering)

Yup.
Now we only need that $\max\limits_{x \in \partial{\Omega}} (-(g_1(x)-g_2(x))) \le \max\limits_{x \in \partial{\Omega}} |g_1(x)-g_2(x)|$... (Thinking)