Can we just use a part of the boundary?

[evinda]

Hello! (Wave)

We consider an elliptic operator $L$ in the space $\Omega$ with $c(x) \leq 0$. We suppose that $\partial{\Omega}=S_1 \cup S_2$. What can we say about the solution of the following problem?

$$Lu=0 \text{ in } \Omega \\ u|_{S_1}=0, \frac{\partial{u}}{\partial{\mathcal{v}}}|_{S_2}=0$$

The space $\Omega$ satisfies the interior sphere condition and $\mathcal{v}$ is unit normal to $S_2$.

( 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 to use the following theorem:

Theorem: Suppose that $u \in C^2(\Omega)$ satisfies in $\Omega$ the relation $Lu \geq 0$ ( $Lu \leq 0$). We suppose that $\Omega$ satisfies the interior sphere condition.
If $c \leq 0$ then $u$ does not achieve its positive maximum in $\Omega$, i.e. in $\overline{\Omega} \setminus{\partial{\Omega}}$ (negative minimum) if it is not constant.

Applying the above theorem, we have that $u$ does not achieve its positive maximum in $\Omega$ if it is not constant.
That means that either $u \leq 0$ or $u=0$ because of the fact that $u|_{S_1}=0$.

We also have that $u$ does not achive its negative maximum in $\Omega$ if it is not constant.
That means that either $u \geq 0$ or $u=0$ because of the fact that $u|_{S_1}=0$.

So we deduce that $u=0$ in $\overline{\Omega}$.

But is this right, given that $S_1$ is just a part of the boundary? (Thinking)

 

[I like Serena]

Hello! (Wave)

We consider an elliptic operator $L$ in the space $\Omega$ with $c(x) \leq 0$. We suppose that $\partial{\Omega}=S_1 \cup S_2$. What can we say about the solution of the following problem?

$$Lu=0 \text{ in } \Omega \\ u|_{S_1}=0, \frac{\partial{u}}{\partial{\mathcal{v}}}|_{S_2}=0$$

The space $\Omega$ satisfies the interior sphere condition and $\mathcal{v}$ is unit normal to $S_2$.

( 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 to use the following theorem:

Theorem: Suppose that $u \in C^2(\Omega)$ satisfies in $\Omega$ the relation $Lu \geq 0$ ( $Lu \leq 0$). We suppose that $\Omega$ satisfies the interior sphere condition.
If $c \leq 0$ then $u$ does not achieve its positive maximum in $\Omega$, i.e. in $\overline{\Omega} \setminus{\partial{\Omega}}$ (negative minimum) if it is not constant.

Applying the above theorem, we have that $u$ does not achieve its positive maximum in $\Omega$ if it is not constant.
That means that either $u \leq 0$ or $u=0$ because of the fact that $u|_{S_1}=0$.

We also have that $u$ does not achive its negative maximum in $\Omega$ if it is not constant.
That means that either $u \geq 0$ or $u=0$ because of the fact that $u|_{S_1}=0$.

So we deduce that $u=0$ in $\overline{\Omega}$.

But is this right, given that $S_1$ is just a part of the boundary? (Thinking)

Hey evinda! (Smile)

Let's pick an example.
Suppose we pick $\Omega=(0,1)$, $S_1=\{0\}$, $S_2=\{1\}$, and $u$ as in:
[TIKZ][scale=5,font=\large]
\draw (0,0) -- node[below] {$\Omega$} (1,0);
\draw (0,0.05) -- (0,-0.05) node[below] {$S_1$} (1,0.05) -- (1,-0.05) node[below] {$S_2$};
\draw[ultra thick,blue] (0,0) parabola bend (1,0.5) (1,0.5) node

{$u$};
[/TIKZ]
Could $u$ be a solution? (Wondering)​

 

[evinda]

Hey evinda! (Smile)

Let's pick an example.
Suppose we pick $\Omega=(0,1)$, $S_1=\{0\}$, $S_2=\{1\}$, and $u$ as in:
[TIKZ][scale=5,font=\large]
\draw (0,0) -- node[below] {$\Omega$} (1,0);
\draw (0,0.05) -- (0,-0.05) node[below] {$S_1$} (1,0.05) -- (1,-0.05) node[below] {$S_2$};
\draw[ultra thick,blue] (0,0) parabola bend (1,0.5) (1,0.5) node

{$u$};
[/TIKZ]
Could $u$ be a solution? (Wondering)​

How can we find the value of $\frac{\partial{u}}{\partial{v}}|_{S_2}$ ? (Thinking)​

 

[I like Serena]

How can we find the value of $\frac{\partial{u}}{\partial{v}}|_{S_2}$ ? (Thinking)

Don't we have $v=(1)$ and $\frac{\partial{u}}{\partial{v}}=u'$? (Wondering)

 

[evinda]

Don't we have $v=(1)$ and $\frac{\partial{u}}{\partial{v}}=u'$? (Wondering)

So we have that $u'=0$. So if it holds $Lu=0$ in $\Omega$ then in the way that I used the proposition, we would get that $u=0$ or $u \leq 0$, which is wrong. (Worried)

 

[I like Serena]

So we have that $u'=0$. So if it holds $Lu=0$ in $\Omega$ then in the way that I used the proposition, we would get that $u=0$ or $u \leq 0$, which is wrong. (Worried)

What went wrong? (Crying)

 

[evinda]

What went wrong? (Crying)

Having $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$ doesn't imply that $u=0$ in $S_2$.
And we cannot just use the fact that $u|_{S_1}=0$, right? But how can we use the fact that $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$ at the theorem? (Thinking) Or can't we use it?

 

[I like Serena]

Having $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$ doesn't imply that $u=0$ in $S_2$.
And we cannot just use the fact that $u|_{S_1}=0$, right? But how can we use the fact that $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$ at the theorem? (Thinking) Or can't we use it?

Yes. (Worried)

Which theorems do you have that include a boundary condition with the derivative? (Wondering)

 

[evinda]

There are the following lemmas:

Lemma 1: Let $L$ be an elliptic operator , $c \equiv 0, Lu \geq 0$ in $\Omega \subset \mathbb{R}^n$. We also suppose that $\Omega$ satisfies the interior sphere condition in $x_0 \in \partial{\Omega}$ and

1. at $x_0$ u is continuous
2. $u(x_0)> u(x) \forall x \in \Omega$

then if at $x_0$ $\exists \frac{\partial{u}}{\partial{v}}$ then $\frac{\partial{u}}{\partial{v}}(x_0)>0$.

If $c \leq 0$ the same holds if $u(x_0) \geq 0$.Lemma 2: Let $L$ be an elliptic operator , $c \equiv 0, Lu \leq 0$ in $\Omega \subset \mathbb{R}^n$. We also suppose that $\Omega$ satisfies the interior sphere condition in $x_0 \in \partial{\Omega}$ and

1. at $x_0$ u is continuous
2. $u(x_0)< u(x) \forall x \in \Omega$

then if at $x_0$ $\exists \frac{\partial{u}}{\partial{v}}$ then $\frac{\partial{u}}{\partial{v}}(x_0)<0$.

If $c \leq 0$ the same holds if $u(x_0) \leq 0$.But... in our case we have $\partial{\Omega}=S_1 \cup S_2$. What can we do? (Worried)

 

[I like Serena]

Suppose we try to apply Lemma 1 to the example I gave... (Thinking)

[TIKZ][scale=5,font=\Large]
\draw[very thick] (0,0) -- node[below] {$\Omega$} (1,0);
\draw[ultra thick,green!70!black] (0,0) parabola bend (1,0.5) (1,0.5) node

{$u$};
\foreach \x/\label in {0/S_1,1/S_2} \filldraw (\x,0) circle (0.5pt) [yshift=-0.3pt] node[below] {$\x$} [yshift=0.6pt] node[above] {$\label$};[/TIKZ]

That is, $x_0=1$ and we suppose that $u(x_0)>0$.
Since $\frac{\partial{u}}{\partial{v}}(x_0)=0$ it follows from the lemma that $u(x_0)\le u(x)$.
But $u(0)=0$, so we get a contradiction!​

 

[evinda]

Suppose we try to apply Lemma 1 to the example I gave... (Thinking)

[TIKZ][scale=5,font=\Large]
\draw[very thick] (0,0) -- node[below] {$\Omega$} (1,0);
\draw[ultra thick,green!70!black] (0,0) parabola bend (1,0.5) (1,0.5) node

{$u$};
\foreach \x/\label in {0/S_1,1/S_2} \filldraw (\x,0) circle (0.5pt) [yshift=-0.3pt] node[below] {$\x$} [yshift=0.6pt] node[above] {$\label$};[/TIKZ]

That is, $x_0=1$ and we suppose that $u(x_0)>0$.
Since $\frac{\partial{u}}{\partial{v}}(x_0)=0$ it follows from the lemma that $u(x_0)\le u(x)$.
But $u(0)=0$, so we get a contradiction!​

Isn't $u(x_0)>u(x)$ a condition so that we can apply the lemma?

- - - Updated - - -

From the lemma, we get that it has to hold that $\frac{\partial{u}}{\partial{v}}|_{S_2}>0$, that is a contradiction. Right? (Thinking)​

 

[I like Serena]

Isn't $u(x_0)>u(x)$ a condition so that we can apply the lemma?

From the lemma, we get that it has to hold that $\frac{\partial{u}}{\partial{v}}|_{\partial{\Omega}}=0$, that is a contradiction. Right? (Thinking)

Let's suppose that $u(x_0)>u(x)$, as is the case in the example, then we can apply the lemma and get that contradiction.
So one of the conditions of the lemma must be false.
The only one that qualifies is: $u(x_0)> u(x) \forall x \in \Omega$.
So we get that $\exists x \in\Omega: u(x_0)\le u(x)$.

To be fair, that's only at least one $x$ and not all of them, so my argument that $u(x_0) > u(0)$ is not sufficient. (Blush)

Still, we now have a supposed maximum $u(x_0) > 0$ on the boundary, and a point $x\in\Omega$ that is at least as high.
Can't we apply the lemma that says that if we have an internal maximum, that then the only solution is a constant function? (Wondering)

 

[evinda]

So for the general case:

We suppose that $u(x_0)>u(x)$ for some $x_0 \in S_2$ and for all $x \in \Omega$.

Then applying the lemma we get that if $\exists \frac{\partial{u}}{\partial{v}}$ at $x_0$ then $ \frac{\partial{u}}{\partial{v}}(x_0)>0$.

Since $x_0 \in S_2$ is arbitrary, we get that it has to hold $\frac{\partial{u}}{\partial{v}}|_{S_2}>0$.

But we are given that $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$.

So one of the conditions of the lemma must be false.

The only one that qualifies is: $u(x_0)>u(x), \forall x \in \Omega $.

So we get that $\exists x \in \Omega$: $u(x_0) \leq u(x)$.

Since we have found an internal maximum, we deduce that $u$ is constant.

Since $u|_{S_1}=0$ and $u$ is continuous, we deduce that $u=0$ in $\overline{\Omega}$.

Is it right? Could I improve something? (Thinking)

 

[I like Serena]

So for the general case:

We suppose that $u(x_0)>u(x)$ for some $x_0 \in S_2$ and for all $x \in \Omega$.

Then applying the lemma we get that if $\exists \frac{\partial{u}}{\partial{v}}$ at $x_0$ then $ \frac{\partial{u}}{\partial{v}}(x_0)>0$.

Since $x_0 \in S_2$ is arbitrary, we get that it has to hold $\frac{\partial{u}}{\partial{v}}|_{S_2}>0$.

But we are given that $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$.

So one of the conditions of the lemma must be false.

The only one that qualifies is: $u(x_0)>u(x), \forall x \in \Omega $.

So we get that $\exists x \in \Omega$: $u(x_0) \leq u(x)$.

Since we have found an internal maximum, we deduce that $u$ is constant.

Since $u|_{S_1}=0$ and $u$ is continuous, we deduce that $u=0$ in $\overline{\Omega}$.

Is it right? Could I improve something? (Thinking)

We assume that $S_1 \ne \varnothing$, which I believe is not given is it? (Wondering)
And we also assume that $S_2 \ne \varnothing$.

Oh, and don't we also assume that $u(x_0)>0$?
Because otherwise it $u(x_0)$ might still actually be a minimum.

Btw, which lemma are we applying exactly to deduce that $u$ is constant?
I've gotten lost between the harmonic, subharmonic, and elliptical operators. (Sweating)

Can we assume that $\Omega$ is open and connected? Because I think that matters as well.
Suppose $S_1$ and $S_2$ are boundaries of unconnected parts of $\Omega$! (Worried)

I think we need to distinguish a couple more cases. (Thinking)

 

[evinda]

We assume that $S_1 \ne \varnothing$, which I believe is not given is it? (Wondering)
And we also assume that $S_2 \ne \varnothing$.

No, it is not given.

Oh, and don't we also assume that $u(x_0)>0$?
Because otherwise it $u(x_0)$ might still actually be a minimum.

And do we apply afterwards the lemma 2 and suppose that $u(x_0)<0$ ? Or can't we do so? (Worried)

Btw, which lemma are we applying exactly to deduce that $u$ is constant?
I've gotten lost between the harmonic, subharmonic, and elliptical operators. (Sweating)

We have the following theorem:

Theorem: We suppose that $u \in C^2(\Omega)$ satisfies in the space $\Omega$ the relation $Lu \geq 0$ ($Lu \leq 0$). We suppose that $\Omega$ satisfies the interior sphere condition.
If $c \leq 0$ then $u(x)$ does not achieve its positive maximum in $\Omega$, i.e. in $\overline{\Omega} \setminus{\partial{\Omega}}$ (negative minimum) if it is not constant.
If $c \equiv 0$ then $u$ does not achieve its maximum in $\Omega$ (minimum) if it is not constant.

Can we assume that $\Omega$ is open and connected? Because I think that matters as well.
Suppose $S_1$ and $S_2$ are boundaries of unconnected parts of $\Omega$! (Worried)

I think that we assume that $\Omega$ is connected, as also at the proof of the maximum principle value.

I think we need to distinguish a couple more cases. (Thinking)

What cases? (Thinking)

 

[I like Serena]

I think that we assume that $\Omega$ is connected, as also at the proof of the maximum principle value.

And bounded? So that we'll have a maximum and a minimum? (Wondering)

What cases? (Thinking)

I think:

1. $S_1=\varnothing$
2. $S_2=\varnothing$
3. $S_1 \cap S_2 \ne \varnothing \wedge \exists x_0\in S_2:u(x_0)=\displaystyle\max_{\xi\in S_2}(u(\xi)) \ge 0$
4. $S_1 \cap S_2 \ne \varnothing \wedge \exists x_0\in S_2:u(x_0)=\displaystyle\min_{\xi\in S_2}(u(\xi)) < 0$

What do you think? (Thinking)

 

[evinda]

And bounded? So that we'll have a maximum and a minimum? (Wondering)

At some propositions, it is stated that $\Omega$ is bounded , at some others only that it satisfies the interior sphere condition.

I think:

1. $S_1=\varnothing$
2. $S_2=\varnothing$
3. $S_1 \cap S_2 \ne \varnothing \wedge \exists x_0\in S_2:u(x_0)=\displaystyle\max_{\xi\in S_2}(u(\xi)) \ge 0$
4. $S_1 \cap S_2 \ne \varnothing \wedge \exists x_0\in S_2:u(x_0)=\displaystyle\min_{\xi\in S_2}(u(\xi)) < 0$

What do you think? (Thinking)

Couldn't it also hold that $S_1 \cap S_2= \varnothing$ ? (Thinking)

Also couldn't we have at the boundary both a maximum and a minimum? Or do we always need just one such condition?

Do this four cases cover all the ones that we could have? (Thinking)

 

[I like Serena]

At some propositions, it is stated that $\Omega$ is bounded , at some others only that it satisfies the interior sphere condition.

Couldn't it also hold that $S_1 \cap S_2= \varnothing$ ? (Thinking)

Also couldn't we have at the boundary both a maximum and a minimum? Or do we always need just one such condition?

Do this four cases cover all the ones that we could have? (Thinking)

Oh yes. It should be:

1. $S_1=\varnothing$
2. $S_2=\varnothing$
3. $S_1\ne \varnothing \wedge S_2 \ne \varnothing \wedge \exists x_0\in S_2:u(x_0)=\displaystyle\max_{\xi\in S_2}(u(\xi)) \ge 0$
4. $S_1\ne \varnothing \wedge S_2 \ne \varnothing \wedge \exists x_0\in S_2:u(x_0)=\displaystyle\min_{\xi\in S_2}(u(\xi)) < 0$
5. $S_2 \ne \varnothing \wedge (S_2$ does not contain its supremum or $S_2$ does not contain its infimum$)$

Did we miss anything?
Or would this cover all possible situations?
Well... if $S_2$ is not bounded it could be that there is no maximum or minimum...
Oh! And if $S_2$ is open, it can be that its supremum or infimum is not part in the set... (Thinking)

And yes, we can and will always have a maximum and a minimum, so 3 and 4 are not mutually exclusive.
It's just a matter if we have at least one point above 0 or not.

Anway, we'll still have to see if we can solve each of these cases. (Sweating)

They look like this (no interior minimum or maximum):
[TIKZ][scale=5,font=\Large]
%preamble \usetikzlibrary{arrows}
\draw[gray,thin,-triangle 60] (0,-0.3) -- (0,0.5) node

{$u$};
\draw[very thick] (0,0) -- node[below] {$\Omega$} (1,0) node

{2};
\draw[ultra thick,red!70!black] (0,0.2) cos (0.5,0.3) sin (1,0.4) node

{1};
\draw[ultra thick,green!70!black] (0,0) parabola bend (1,0.25) (1,0.25) node

{3};
\draw[ultra thick,blue] (0,0) parabola bend (1,-0.25) (1,-0.25) node

{4};
\foreach \x/\label in {0/S_1,1/S_2} \filldraw (\x,0) circle (0.5pt);
[/TIKZ]​

 

[evinda]

For the case 3, we have the following:

We suppose that $u(x_0)>u(x)$ , $\forall x \in \Omega$.

Then applying the lemma $1$ we get that if $\exists \frac{\partial{u}}{\partial{v}}$ at $x_0$ then $ \frac{\partial{u}}{\partial{v}}(x_0)>0$.

But we are given that $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$, so $ \frac{\partial{u}}{\partial{v}}(x_0)=0$.

So one of the conditions of the lemma must be false.

The only one that qualifies is: $u(x_0)>u(x), \forall x \in \Omega $.

So we get that $\exists x \in \Omega$: $u(x_0) \leq u(x)$.

Since we have found an internal maximum, we deduce that $u$ is constant.

Since $u|_{S_1}=0$ and $u$ is continuous, we deduce that $u=0$ in $\overline{\Omega}$.And for the case 4, we have the following:

We suppose that $u(x_0)<u(x)$ , $\forall x \in \Omega$.

Then applying the lemma $2$ we get that if $\exists \frac{\partial{u}}{\partial{v}}$ at $x_0$ then $ \frac{\partial{u}}{\partial{v}}(x_0)<0$.

But we are given that $\frac{\partial{u}}{\partial{v}}|_{S_2}=0$, so $ \frac{\partial{u}}{\partial{v}}(x_0)=0$.

So one of the conditions of the lemma must be false.

The only one that qualifies is: $u(x_0)<u(x), \forall x \in \Omega $.

So we get that $\exists x \in \Omega$: $u(x_0) \geq u(x)$.

Since we have found an internal minimum, we deduce that $u$ is constant.

Since $u|_{S_1}=0$ and $u$ is continuous, we deduce that $u=0$ in $\overline{\Omega}$.Right so far? (Thinking)

 

[I like Serena]

I think we also need to assume, without loss of generality, that $\Omega$ is an open set.
Otherwise our $x$ is not necessarily an internal point.

Other than that it looks good to me! (Mmm)

 

[evinda]

Ok. At the first case, when $S_1=\varnothing$ we have the following problem

$$Lu=0 \text{ in } \Omega \\ \frac{\partial{u}}{\partial{v}}|_{S_2}=0$$

We suppose that for some $x_0 \in S_2$ we have that $u(x_0)>u(x) , \forall x \in \Omega$.

Then by lemma 1, if at $x_0$ $\exists \frac{\partial{u}}{\partial{v}}$ then $\frac{\partial{u}}{\partial{v}}(x_0)>0$.

But we are given that $ \frac{\partial{u}}{\partial{v}}|_{S_2}=0$ and so $\frac{\partial{u}}{\partial{v}}(x_0)=0$.

So we deduce that $\exists x \in \Omega$ such that $u(x_0) \leq u(x)$.

So we have found an internal maximum and so we deduce that $u$ is constant.

Applying the lemma 2, we would get the same, right?

In this case, we cannot find the value of the constant function $u$, right?

 

[I like Serena]

Agreed. The solution is the set of constant functions. (Mmm)

 

[evinda]

Agreed. The solution is the set of constant functions. (Mmm)

Nice (Smirk)

At the second case when $S_2=\varnothing$, we have the following problem

$$Lu=0 \text{ in } \Omega \\ u|_{S_1}=0$$

the only solution of which is $u(x)=0$. Right?

- - - Updated - - -

What can we say about the case 5? (Thinking)

 

[I like Serena]

Nice (Smirk)

At the second case when $S_2=\varnothing$, we have the following problem

$$Lu=0 \text{ in } \Omega \\ u|_{S_1}=0$$

the only solution of which is $u(x)=0$. Right?

- - - Updated - - -

What can we say about the case 5? (Thinking)

Right! (Nod)

- - - Updated - - -

If $\Omega$ is not bounded, we could actually also have yet another case: $S_1=S_2=\varnothing$.
As an example, pick $\Omega=\mathbb R$ and $Lu=u''-u$.
The solution is $u(x)=c_1 e^x + c_2 e^{-x}$, which is unbounded as well.
[TIKZ][scale=3,font=\Large]
%preamble \usetikzlibrary{arrows}
\draw[gray,thin,-triangle 60] (0,-0.3) -- (0,1) node

{$u$};
\draw[very thick,triangle 60-triangle 60] (-1,0) -- node[below] {$\Omega$} (1,0);
\draw[ultra thick,red!70!black,domain=-1:1,variable=\x] plot (\x,{e^(\x)/3});
[/TIKZ]
Does that match the problem statement and is it a proper solution?
Or am I wrong? (Wondering)

Or as another example with unbounded $\Omega$, we could have $\Omega=(0,\infty), S_1=\{0\},Lu=u''-u$ with solution $u(x)=c_1 \sinh x$.
That is:
[TIKZ][scale=2,font=\Large]
%preamble \usetikzlibrary{arrows}
\draw[gray,thin,-triangle 60] (0,-0.3) -- (0,1.2) node

{$u$};
\draw[very thick,-triangle 60] (0,0) -- node[below] {$\Omega$} (3,0);
\draw[ultra thick,red!70!black,domain=0:3,variable=\x] plot (\x,{(e^\x-e^(-\x))/20});
\foreach \x/\label in {0/S_1} \filldraw (\x,0) circle (1pt) node

{$S_1$};
[/TIKZ]

Maybe we should suppose that $\Omega$ is bounded? (Wondering)
Do your notes perchance say anything about $\Omega$ in a prologue, in chapter 1, or at the beginning of the chapter? (Wondering)

- - - Updated - - -

Alternatively, $S_2$ could be an open set with its supremum on its boundary.
Assuming that $\Omega$ is connected, that would mean that the supremum is in $S_1$.
What could we say then? (Wondering)​

 

[evinda]

If $\Omega$ is not bounded, we could actually also have yet another case: $S_1=S_2=\varnothing$.
As an example, pick $\Omega=\mathbb R$ and $Lu=u''-u$.
The solution is $u(x)=c_1 e^x + c_2 e^{-x}$, which is unbounded as well.
[TIKZ][scale=3,font=\Large]
\draw[gray,thin,-triangle 60] (0,-0.3) -- (0,1) node

{$u$};
\draw[very thick,triangle 60-triangle 60] (-1,0) -- node[below] {$\Omega$} (1,0);
\draw[ultra thick,red!70!black,domain=-1:1,variable=\x] plot (\x,{e^(\x)/3});
[/TIKZ]
Does that match the problem statement and is it a proper solution?
Or am I wrong? (Wondering)

Or as another example with unbounded $\Omega$, we could have $\Omega=(0,\infty), S_1=\{0\},Lu=u''-u$ with solution $u(x)=c_1 \sinh x$.
That is:
[TIKZ][scale=2,font=\Large]
\draw[gray,thin,-triangle 60] (0,-0.3) -- (0,1.2) node

{$u$};
\draw[very thick,-triangle 60] (0,0) -- node[below] {$\Omega$} (3,0);
\draw[ultra thick,red!70!black,domain=0:3,variable=\x] plot (\x,{(e^\x-e^(-\x))/20});
\foreach \x/\label in {0/S_1} \filldraw (\x,0) circle (1pt) node

{$S_1$};
[/TIKZ]

Maybe we should suppose that $\Omega$ is bounded? (Wondering)
Do your notes perchance say anything about $\Omega$ in a prologue, in chapter 1, or at the beginning of the chapter? (Wondering)​

At some theorem, it is required that $\Omega$ is bounded at some others that it satisfies the interior sphere condition.
So we need that $\Omega$ is bounded in order the solution to be bounded? (Thinking)

Alternatively, $S_2$ could be an open set with its supremum on its boundary.
Assuming that $\Omega$ is connected, that would mean that the supremum is in $S_1$.
What could we say then? (Wondering)

Why is in this case the supremum in $S_1$?​

 

[evinda]

By supposing that $\Omega$ is bounded, we have to check just the first four cases, right? (Thinking)

 

[evinda]

I think that if we assume that the maximum and minimum value are achieved, then we get that the solution is constant. So the only solution of the problem is the zero function.

 

[I like Serena]

At some theorem, it is required that $\Omega$ is bounded at some others that it satisfies the interior sphere condition.
So we need that $\Omega$ is bounded in order the solution to be bounded? (Thinking)

I believe so yes.

Why is in this case the supremum in $S_1$?

If we have a supremum, it means we can create a sequence that approaches this supremum in the limit.
Since such a sequence would be entirely inside $S_2$, the limit must be in $\overline {S_2}$, which is a subset of $\partial \Omega$.
So this supremum could be in the boundary of $S_2$ and not in $S_2$ itself, but if so, it must still be in $\partial \Omega$ and therefore in $S_1$.
Note that with continuity of $\frac{\partial u}{\partial v}$, the derivative will still be zero.

So I think we should rephrase cases 3 and 4 a little bit to cover this as well:

3. $S_1\ne \varnothing \wedge S_2 \ne \varnothing \wedge \exists x_0\in \partial\Omega: u(x_0)=\displaystyle\sup_{\xi\in S_2}(u(\xi)) \ge 0$
4. $S_1\ne \varnothing \wedge S_2 \ne \varnothing \wedge \exists x_0\in \partial\Omega:u(x_0)=\displaystyle\inf_{\xi\in S_2}(u(\xi)) < 0$
(Thinking)

By supposing that $\Omega$ is bounded, we have to check just the first four cases, right? (Thinking)

Yes, but with slightly modified cases 3 and 4 as I just mentioned.

I think that if we assume that the maximum and minimum value are achieved, then we get that the solution is constant. So the only solution of the problem is the zero function.

Yes, if minimum and maximum are achieved, then the solution is constant.
For it to be zero we need $S_1\ne\varnothing$ though. (Mmm)