MHB Can the Maximum Principle be Used to Estimate Solutions of Elliptic Equations?

[mathmari]

Hey!

Let $S=\{x \in \mathbb{R}^2 \mid |x| <1\}$. Using the maximum principle I have to show that the solution of the problem $$-\Delta u(x)=f(x), x \in S \\ u(x)=0, x \in \partial{S}$$ satisfies the estimation $$|u(x)| \leq \frac{1}{4}\max_{x \in \overline{S}} |f(x)|, x \in S$$
To use the maximum principle shouldn't it stand that $$\Delta u \geq 0$$ ?? (Wondering)

Do we have to take cases for $f$, if it is positive or negative?? (Wondering)

 

[I like Serena]

Hey!

Let $S=\{x \in \mathbb{R}^2 \mid |x| <1\}$. Using the maximum principle I have to show that the solution of the problem $$-\Delta u(x)=f(x), x \in S \\ u(x)=0, x \in \partial{S}$$ satisfies the estimation $$|u(x)| \leq \frac{1}{4}\max_{x \in \overline{S}} |f(x)|, x \in S$$
To use the maximum principle shouldn't it stand that $$\Delta u \geq 0$$ ?? (Wondering)

Do we have to take cases for $f$, if it is positive or negative?? (Wondering)

Hi! (Smile)

From wiki, to use the maximum principle we need that $Lu \ge 0$ where $L$ is an elliptic differential operator.
We could pick $Lu=\Delta x$, which satisfies the criteria.
However, since we don't know whether $\Delta u \geq 0$, we need to add or subtract something to ensure it becomes positive. (Thinking)Perhaps we should look at a 1-dimensional problem first.
Suppose we pick $S' = \{x \in \mathbb{R} \mid |x| <1\}$.

Let's define $M = \max_{x \in \overline{S'}} |f(x)|$.
What would be the solution for:
$$-\Delta u = M$$
? (Wondering)

 

[I like Serena]

Let's rewrite your problem to:
$$\Delta u = -f(x,y) \tag 1$$
Then $\Delta u$ is an elliptic operator. (Thinking)

Let $M = \frac 14 \max |f(x,y)| \tag 2$

Let $v = u - M(1-x^2-y^2) \tag 3$

Then: $\Delta v = \Delta u + 4M = -f(x,y) + \max |f(x,y)| \ge 0\tag 4$

Now we can apply the maximum principle to $v$. (Emo)