Exploring the Lemma and Theorem for $Lu=f$ in $\Omega$

[evinda]

Hello! (Wave)

We consider the following problem.

$$Lu=f(x) \text{ in } \Omega \\ u|_{\partial{\Omega}}=0$$

I want to show that if $c(x) \leq -c_0<0$ in $\overline{\Omega}$, then it holds that $\min\{ 0, \frac{\min_{\Omega}f(x)}{-c_0}\}\leq u(x) \leq \max_{\Omega} \{ 0, \frac{\max_{\Omega}f(x)}{-c_0} \}$.

( In general, if $L$ is an elliptic operator, then $Lu=\sum_{i,j=1}^n a_{ij}(x) u_{x_i x_j}+ \sum_{i=1}^n \beta_i(x) u_{x_i}+cu$)
I thought that we could modify somehow the proof of the following lemma:

Lemma: Let $L$ be an elliptic operator in a bounded space $\Omega$ and $u \in C^2(\Omega) \cap C^0(\overline{\Omega})$. If $Lu \geq 0$ ($Lu \leq 0$) , $c \leq 0$ in $\Omega$ then

$$\sup_{\Omega} u \leq \max \left( \sup_{\partial{\Omega}} u, 0\right)$$

$$\left( \inf_{\Omega} u \geq \min \{ \inf_{\partial{\Omega}} u,0\}\right)$$

The proof is the following:$e^{\gamma x_1}$, $\beta_0=\sup \frac{|\beta_1|}{\lambda}, c_0=\sup \frac{|c|}{\lambda}$

$L e^{\gamma x_1}= a_{11} \gamma^2 e^{\gamma x_1}+ \beta_1 \gamma e^{\gamma x_1}+c e^{\gamma x_1} \geq e^{\gamma x_1}( a_{11} \gamma^2- \lambda \beta_0 \gamma - \lambda c_0) \geq e^{\gamma x_1} \lambda (\gamma^2-\beta_0 \gamma-c_0)>0$, we choose $\gamma$ to be large enough.$L(u+ \epsilon e^{\gamma x_1})=Lu+ \epsilon Le^{\gamma x_1}>0$

$\sum_{i,j=1}^n a_{ij} (u+ \epsilon e^{\gamma x_1})_{x_i x_j}+ \sum_{i=1}^n \beta_i (u+ \epsilon e^{\gamma x_1})_{x_i}+ c(u+ \epsilon e^{\gamma x_1})>0 $

$u+ \epsilon e^{\gamma x_1}$ does not achieve its positive maximum in $\overline{\Omega} \setminus{\partial{\Omega}}$

so $u+ \epsilon e^{\gamma x_1} \leq 0$ or it achieves its positive maximum in $\partial{\Omega}$

$u \leq u+ \epsilon e^{\gamma x_1} \leq \max \{ 0, \sup_{\partial{\Omega}}(u+ \epsilon e^{\gamma x_1}) \} \ \forall x \in \Omega$

$u \leq \max \{ 0, \sup_{\partial{\Omega}} (u+ \epsilon e^{\gamma x_1}) \} \ \forall x$

$\epsilon \to 0$In our case, we would have $L(u+ \epsilon e^{\gamma x_1})=f+Le^{\gamma x_1}$.

But can we write an inequality for the above, although nothing is given for $f$ ?

Or could we maybe use somehow the following theorem?

Theorem: Let $Lu=f$ in a bounded space $\Omega$, $L$ an elliptic operator and $u \in C^2(\Omega) \cap C^0(\overline{\Omega})$. Then

$$\sup_{\overline{\Omega}} |u| \leq \sup_{\partial{\Omega}} |u|+ C \sup \frac{|f|}{\lambda}, \text{ C constant}$$

$\lambda$ is such that $0< \lambda |\xi|^2 \leq \sum_{i,j=1}^n a_{ij}(x) \xi_i \xi_j \ \ \ \ \xi \in \mathbb{R}^n, x \in \Omega$

 

[evinda]

If the maximum is achieved on the boundary, then it is zero.

Suppose that the maximum is achieved at some $x_0 \in \Omega$.

Then we have $Lu(x_0)=\sum_{i,j=1}^n a_{ij}u_{x_ix_j}(x_0)+cu(x_0)=f(x_0)$ and thus $c(x_0) u(x_0) \geq f(x_0)$.

From this we get that $u(x) \leq u(x_0) \leq \frac{f(x_0)}{c(x_0)} \leq \max_{\Omega} \frac{f(x)}{c(x)}$.

But I think that it does not hold that $\frac{f(x_0)}{c(x_0)} \leq \frac{\max_{\Omega} f(x)}{-c_0}$ since $\frac{1}{c(x_0)} \geq \frac{1}{-c_0}$. What do you think? (Thinking)

 

[I like Serena]

If the maximum is achieved on the boundary, then it is zero.

Suppose that the maximum is achieved at some $x_0 \in \Omega$.

Then we have $Lu(x_0)=\sum_{i,j=1}^n a_{ij}u_{x_ix_j}(x_0)+cu(x_0)=f(x_0)$ and thus $c(x_0) u(x_0) \geq f(x_0)$.

From this we get that $u(x) \leq u(x_0) \leq \frac{f(x_0)}{c(x_0)} \leq \max_{\Omega} \frac{f(x)}{c(x)}$.

But I think that it does not hold that $\frac{f(x_0)}{c(x_0)} \leq \frac{\max_{\Omega} f(x)}{-c_0}$ since $\frac{1}{c(x_0)} \geq \frac{1}{-c_0}$. What do you think? (Thinking)

As part of the elliptic operator, what can we say about $\sum_{i,j=1}^n a_{ij}u_{x_ix_j}(x_0)$?
Is it $> 0$? Or merely $\ne 0$? (Wondering)Suppose we pick an example.

Say $\Omega = (-1,1),\quad Lu=u''-u,\quad f(x)=x,\quad c(x)=-c_0=-1$.
\begin{tikzpicture}[scale=4, ultra thick, font=\Large, >=stealth']
\draw[gray,thin,->] (0,-1) -- (0,1.1);
\draw (-1,0) -- (1,0) node[below] at (-0.35,0) {$\Omega$};
\draw[green] (-1,-1) -- (1,1) node

{$f(x)$};
\draw[red,domain=-1:1] plot (\x,{(exp(2)*\x-\x+exp(1-\x)-exp(\x+1))/(1-exp(2))}) node[below] at (0.5,-0.1) {$u(x)$};
\draw[fill] (-1,0) circle (0.02) node

{$\partial\Omega$} (1,0) circle (0.02) node

{$\partial\Omega$};
\draw[thin,red] (-0.6,0) node[below left] {$x_0$} -- (-0.6,0.056) [fill] circle (0.015) node[above] {$u(x_0)$};
\draw[thin,green] (-0.6,0) -- (-0.6,-0.6) [fill] circle (0.015) node[below right] {$f(x_0)$};
\end{tikzpicture}
Do you think this is a proper example? Or do you know a better example? (Wondering)​

 

[evinda]

As part of the elliptic operator, what can we say about $\sum_{i,j=1}^n a_{ij}u_{x_ix_j}(x_0)$?
Is it $> 0$? Or merely $\ne 0$? (Wondering)

If at $x_0$ , the function $u$ achieves its maximum then $\sum_{i,j=1}^n a_{ij}(x_0) u_{x_i x_j}(x_0) \leq 0$ and if $u$ achieves its minimum at $x_0$ then $\sum_{i,j=1}^n a_{ij}(x_0) u_{x_i x_j}(x_0) \geq 0$.

Suppose we pick an example.

Say $\Omega = (-1,1),\quad Lu=u''-u,\quad f(x)=x,\quad c(x)=-c_0=-1$.
\begin{tikzpicture}[scale=4, ultra thick, font=\Large, >=stealth']
\draw[gray,thin,->] (0,-1) -- (0,1.1);
\draw (-1,0) -- (1,0) node[below] at (-0.35,0) {$\Omega$};
\draw[green] (-1,-1) -- (1,1) node

{$f(x)$};
\draw[red,domain=-1:1] plot (\x,{(exp(2)*\x-\x+exp(1-\x)-exp(\x+1))/(1-exp(2))}) node[below] at (0.5,-0.1) {$u(x)$};
\draw[fill] (-1,0) circle (0.02) node

{$\partial\Omega$} (1,0) circle (0.02) node

{$\partial\Omega$};
\draw[thin,red] (-0.6,0) node[below left] {$x_0$} -- (-0.6,0.056) [fill] circle (0.015) node[above] {$u(x_0)$};
\draw[thin,green] (-0.6,0) -- (-0.6,-0.6) [fill] circle (0.015) node[below right] {$f(x_0)$};
\end{tikzpicture}
Do you think this is a proper example? Or do you know a better example? (Wondering)​

You mean in order to check the inequality?​

 

[I like Serena]

If at $x_0$ , the function $u$ achieves its maximum then $\sum_{i,j=1}^n a_{ij}(x_0) u_{x_i x_j}(x_0) \leq 0$ and if $u$ achieves its minimum at $x_0$ then $\sum_{i,j=1}^n a_{ij}(x_0) u_{x_i x_j}(x_0) \geq 0$.

Ah yes.

You mean in order to check the inequality?

Yes.
It seems to me that we have 2 cases:
Either $f(x_0)\ge 0$, in which case $\frac{f(x_0)}{c(x_0)} \le 0$.
Or $f(x_0) < 0$, as in the example, in which case $
\frac{f(x_0)}{c(x_0)} \le \frac{\inf_\Omega f(x)}{\sup_\Omega c(x)} \le \frac{\inf_\Omega f(x)}{-c_0}$. (Thinking)

 

[evinda]

Yes.
It seems to me that we have 2 cases:
Either $f(x_0)\ge 0$, in which case $\frac{f(x_0)}{c(x_0)} \le 0$.
Or $f(x_0) < 0$, as in the example, in which case $
\frac{f(x_0)}{c(x_0)} \le \frac{\inf_\Omega f(x)}{\sup_\Omega c(x)} \le \frac{\inf_\Omega f(x)}{-c_0}$. (Thinking)

So in order to show the inequality, we have to distinguish cases for $f$, right?

The solution of your example is $u(x)=c_1 e^x-c_1 e^{-x}-x$, right?

 

[I like Serena]

So in order to show the inequality, we have to distinguish cases for $f$, right?

The solution of your example is $u(x)=c_1 e^x-c_1 e^{-x}-x$, right?

I think so yes. (Thinking)

Oh, and the solution for my example is:
$$u(x)=\frac{e^2x-x+e^{1-x}-e^{x+1}}{1-e^2}$$
Courtesy of Wolfram. (Blush)

 

[evinda]

I think that this:

If the maximum is achieved on the boundary, then it is zero.

Suppose that the maximum is achieved at some $x_0 \in \Omega$.

Then we have $Lu(x_0)=\sum_{i,j=1}^n a_{ij}u_{x_ix_j}(x_0)+cu(x_0)=f(x_0)$ and thus $c(x_0) u(x_0) \geq f(x_0)$.

From this we get that $u(x) \leq u(x_0) \leq \frac{f(x_0)}{c(x_0)}$

holds in any case.

Now if $f(x) \geq 0$ then $\frac{f(x_0)}{c(x_0)} \leq \frac{\max_{\Omega} f(x)}{c(x_0)}$ and if $f(x)<0$ then $\frac{f(x_0)}{c(x_0)} \leq \frac{\min_{\Omega} f(x)}{c(x_0)}$ .

Or am I wrong? (Thinking)

 

[I like Serena]

Now if $f(x) \geq 0$ then $\frac{f(x_0)}{c(x_0)} \leq \frac{\max_{\Omega} f(x)}{c(x_0)}$ and if $f(x)<0$ then $\frac{f(x_0)}{c(x_0)} \leq \frac{\min_{\Omega} f(x)}{c(x_0)}$ .

Or am I wrong? (Thinking)

Shouldn't that be $f(x_0) \geq 0$ respectively $f(x_0)<0$? (Wondering)Anyway, let's consider $f(x) \geq 0$.
Suppose $f(x_0)=1$, $\sup_\Omega f(x)=2$, and $c(x_0)=-1$.
Then we'd get:
$$-1=\frac{1}{-1} = \frac{f(x_0)}{c(x_0)} \leq \frac{\sup_{\Omega} f(x)}{c(x_0)} = \frac 2{-1} = -2$$
That can't be right, can it? (Wondering)

 

[evinda]

Shouldn't that be $f(x_0) \geq 0$ respectively $f(x_0)<0$? (Wondering)

Oh yes, right... (Nod)

Anyway, let's consider $f(x) \geq 0$.
Suppose $f(x_0)=1$, $\sup_\Omega f(x)=2$, and $c(x_0)=-1$.
Then we'd get:
$$-1=\frac{1}{-1} = \frac{f(x_0)}{c(x_0)} \leq \frac{\sup_{\Omega} f(x)}{c(x_0)} = \frac 2{-1} = -2$$
That can't be right, can it? (Wondering)

No, it can't... (Shake)

Can we maybe only consider the maximum value of the whole expression $\frac{f}{c}$ ?

Or what else have I done wrong? (Thinking)

 

[I like Serena]

Oh yes, right... (Nod)
No, it can't... (Shake)

Can we maybe only consider the maximum value of the whole expression $\frac{f}{c}$ ?

Or what else have I done wrong? (Thinking)

Let's see... (Thinking)

We suppose that $u$ has a maximum in $\overline\Omega$ at $x_0$.
Then we have:
$$u(x) \le u(x_0) \le \frac{f(x_0)}{c(x_0)} = \frac{-f(x_0)}{|c(x_0)|}$$

Case 1. Suppose $f(x_0) \ge 0$, then at least $\frac{-f(x_0)}{|c(x_0)|} \le 0$.
But maybe we can make it sharper.

Case 1a. Suppose additionally that for all $x \in \overline\Omega: -f(x)<0$, so $\sup_\Omega(-f(x))<0$, then:
$$\frac{-f(x_0)}{|c(x_0)|} \le \frac{\sup_\Omega(-f(x))}{|c(x_0)|}=\frac{-\inf_\Omega(f(x))}{|c(x_0)|}
\le \frac{-\inf_\Omega(f(x))}{\sup_\Omega|c(x)|} = \frac{-\inf_\Omega(f(x))}{-\inf_\Omega c(x)}
<0$$

Case 1b. If we don't have case 1a, then $0$ is the sharpest upper boundary.

How does it look so far?
And how could we continue? (Wondering)

 

[evinda]

Let's see... (Thinking)

We suppose that $u$ has a maximum in $\overline\Omega$ at $x_0$.
Then we have:
$$u(x) \le u(x_0) \le \frac{f(x_0)}{c(x_0)} = \frac{-f(x_0)}{|c(x_0)|}$$

Case 1. Suppose $f(x_0) \ge 0$, then at least $\frac{-f(x_0)}{|c(x_0)|} \le 0$.
But maybe we can make it sharper.

Case 1a. Suppose additionally that for all $x \in \overline\Omega: -f(x)<0$, so $\sup_\Omega(-f(x))<0$, then:
$$\frac{-f(x_0)}{|c(x_0)|} \le \frac{\sup_\Omega(-f(x))}{|c(x_0)|}=\frac{-\inf_\Omega(f(x))}{|c(x_0)|}
\le \frac{-\inf_\Omega(f(x))}{\sup_\Omega|c(x)|} = \frac{-\inf_\Omega(f(x))}{-\inf_\Omega c(x)}
<0$$

Case 1b. If we don't have case 1a, then $0$ is the sharpest upper boundary.

How does it look so far?
And how could we continue? (Wondering)

Why does it hold that $\frac{-\inf_\Omega(f(x))}{|c(x_0)|} \le \frac{-\inf_\Omega(f(x))}{\sup_\Omega|c(x)|}$ although $\sup_\Omega|c(x)| \geq |c(x_0)|$ and thus $\frac{1}{\sup_\Omega|c(x)| } \leq \frac{1}{|c(x_0)|}$?

Don't we have the opposite inequality? (Thinking)

 

[I like Serena]

Why does it hold that $\frac{-\inf_\Omega(f(x))}{|c(x_0)|} \le \frac{-\inf_\Omega(f(x))}{\sup_\Omega|c(x)|}$ although $\sup_\Omega|c(x)| \geq |c(x_0)|$ and thus $\frac{1}{\sup_\Omega|c(x)| } \leq \frac{1}{|c(x_0)|}$?

Don't we have the opposite inequality? (Thinking)

Because the numerator is negative.
Don't we have
$$\frac{-1}{1}<\frac{-1}{2}$$
(Wondering)

 

[evinda]

Because the numerator is negative.
Don't we have
$$\frac{-1}{1}<\frac{-1}{2}$$
(Wondering)

Yes. (Nod)

And why does it hold that $ \frac{-\inf_\Omega(f(x))}{-\inf_\Omega c(x)}<0$ ? (Thinking)

 

[I like Serena]

Yes. (Nod)

And why does it hold that $ \frac{-\inf_\Omega(f(x))}{-\inf_\Omega c(x)}<0$ ? (Thinking)

Because $-\inf_\Omega(f(x))$ is negative by assumption, and $-\inf_\Omega c(x)$ is positive by definition. (Thinking)

 

[evinda]

Because $-\inf_\Omega(f(x))$ is negative by assumption, and $-\inf_\Omega c(x)$ is positive by definition. (Thinking)

Ok. And we know that if the maximum is achieved at the boundary, then it is equal to $0$.

So we have that $u(x) \leq \min \{ 0, \frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}\}=0 \Rightarrow u(x) \leq \frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}$.

Right? (Thinking)

 

[I like Serena]

Ok. And we know that if the maximum is achieved at the boundary, then it is equal to $0$.

So we have that $u(x) \leq \min \{ 0, \frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}\}=0 \Rightarrow u(x) \leq \frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}$.

Right? (Thinking)

Actually, with our assumption that $-f(x)>0$, we have that $u(x) \leq \min \{ 0, \frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}\}=\frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}$.
However, we have a boundary value of $0$, so the maximum can't be $\frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}$.
So if we have an internal maximum and if $-f(x)>0$ for all $x$ in $\Omega$, we have $u(x) \le 0$. (Thinking)

 

[evinda]

Actually, with our assumption that $-f(x)>0$, we have that $u(x) \leq \min \{ 0, \frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}\}=\frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}$.
However, we have a boundary value of $0$, so the maximum can't be $\frac{-\inf_{\Omega} f(x)}{-\inf_{\Omega} c(x)}$.
So if we have an internal maximum and if $-f(x)>0$ for all $x$ in $\Omega$, we have $u(x) \le 0$. (Thinking)

You mean with the assumption that $-f(x)<0$ ? (Thinking)

Does it indeed hold that when we have $Lu=f(x)$ in $\Omega$ and $u|_{\partial{\Omega}}=0$ with negative $c(x)$ that we cannot have a positive solution? (Thinking)

 

[evinda]

Is the following inequality right?

$$u(x) \leq \max \{0, \max_{\Omega} \frac{f(x)}{c(x)} \}$$

If so, then $\max_{\Omega} \frac{f(x)}{c(x)}$ could also be positive. Couldn't it? (Thinking)