Show that the function equals to the zero function

[evinda]

Hello! (Wave)

Let $u(x) \geq 0$ be harmonic in $\Omega$ and $u|_{\partial{\Omega'}}=0$ where the space $\Omega'$ is a proper subset of $\Omega$. I want to prove that $u(x) \equiv 0$ in $\Omega$.

If we suppose that $u \in C^0(\overline{\Omega'})$ we can deduce that $u(x)=0$ in $\Omega'$.

In order $\Delta u$ to be continuous we have to assume that $u \in C^2(\Omega)$, right?

So, suppose that $u \in C^2(\Omega)$.

Since $\Delta u$ is continuous, if there is a $p \in \Omega$ such that $\Delta u(p)>0$ then there is a ball $B(p,R)$ such that $\Delta u>0$ in $B(\rho,R)$.

So we have to show somehow that $B(\rho,R) \subset \Omega'$, right? (Thinking)

 

[I like Serena]

Hello! (Wave)

Let $u(x) \geq 0$ be harmonic in $\Omega$ and $u|_{\partial{\Omega'}}=0$ where the space $\Omega'$ is a proper subset of $\Omega$. I want to prove that $u(x) \equiv 0$ in $\Omega$.

If we suppose that $u \in C^0(\overline{\Omega'})$ we can deduce that $u(x)=0$ in $\Omega'$.

In order $\Delta u$ to be continuous we have to assume that $u \in C^2(\Omega)$, right?

So, suppose that $u \in C^2(\Omega)$.

Since $\Delta u$ is continuous, if there is a $p \in \Omega$ such that $\Delta u(p)>0$ then there is a ball $B(p,R)$ such that $\Delta u>0$ in $B(\rho,R)$.

So we have to show somehow that $B(\rho,R) \subset \Omega'$, right? (Thinking)

Hey evinda! (Smile)

Isn't it given that $\Delta u=0$? (Wondering)

Instead I think we should look at a $p \in \Omega$ such that $u(p)>0$.
And we will want to prove that $\Delta u \ne 0$ somewhere then. (Thinking)

 

[evinda]

Hey evinda! (Smile)

Isn't it given that $\Delta u=0$? (Wondering)

Instead I think we should look at a $p \in \Omega$ such that $u(p)>0$.
And we will want to prove that $\Delta u \ne 0$ somewhere then. (Thinking)

Oh yes, you are right! (Smirk)

So suppose that there is a $p \in \Omega$ such that $u(p)>0$.
Since $u$ is continuous, there is a ball $B(p,R)$ such that $u(x)>0$ in $B(p,R)$.
Is it right so far? How can we continue? (Thinking)

 

[I like Serena]

Oh yes, you are right! (Smirk)

So suppose that there is a $p \in \Omega$ such that $u(p)>0$.
Since $u$ is continuous, there is a ball $B(p,R)$ such that $u(x)>0$ in $B(p,R)$.
Is it right so far? How can we continue? (Thinking)

It's right, but I wouldn't know how to continue from there.

I was thinking more of applying the mean value theorem for functions of several variables. (Thinking)

 

[evinda]

So do we pick $x, y \in \Omega$ and a $c \in (0,1)$? Then since $u$ is differentiable do we have the following?

$$u(y)-u(x)=\nabla{u}((1-c)x+c y) \cdot (y-x)$$

Or isn't this the mean value theorem for functions of several variables?

 

[I like Serena]

So do we pick $x, y \in \Omega$ and a $c \in (0,1)$? Then since $u$ is differentiable do we have the following?

$$u(y)-u(x)=\nabla{u}((1-c)x+c y) \cdot (y-x)$$

Or isn't this the mean value theorem for functions of several variables?

Yes. That's the one I meant.

If we pick $\mathbf y=\mathbf p$ and $\mathbf x\in \partial \Omega'$ it follows that $(\mathbf y-\mathbf x)\ne \mathbf 0$ and $u(\mathbf y)-u(\mathbf x) > 0$, and therefore $\nabla{u}((1-c)\mathbf x+c \mathbf y) \ne \mathbf 0$. (Thinking)

 

[evinda]

Yes. That's the one I meant.

If we pick $\mathbf y=\mathbf p$ and $\mathbf x\in \partial \Omega'$ it follows that $(\mathbf y-\mathbf x)\ne \mathbf 0$ and $u(\mathbf y)-u(\mathbf x) > 0$, and therefore $\nabla{u}((1-c)\mathbf x+c \mathbf y) \ne \mathbf 0$. (Thinking)

We have that $(\mathbf y-\mathbf x)\ne \mathbf 0$ since $p \in \Omega \setminus{\overline{\Omega'}}$, right?

What do we get from the fact that $\nabla{u}((1-c)\mathbf x+c \mathbf y) \ne \mathbf 0$? (Thinking)

 

[I like Serena]

We have that $(\mathbf y-\mathbf x)\ne \mathbf 0$ since $p \in \Omega \setminus{\overline{\Omega'}}$, right?

We have $u(\mathbf y)>0$ and $u(\mathbf x) =0$.
Therefore $\mathbf y \ne \mathbf x$.

I thought we only knew that $\mathbf p \in \Omega \setminus{\partial\Omega'}$? (Wondering)

What do we get from the fact that $\nabla{u}((1-c)\mathbf x+c \mathbf y) \ne \mathbf 0$? (Thinking)

We found a point where $\nabla$ is non-zero.
Can we take it one step further, and show that we can find a point where $\Delta=\nabla\cdot \nabla$ is non-zero?
That is, applying the mean value theorem again? (Wondering)

 

[evinda]

We have $u(\mathbf y)>0$ and $u(\mathbf x) =0$.
Therefore $\mathbf y \ne \mathbf x$.

I thought we only knew that $\mathbf p \in \Omega \setminus{\partial\Omega'}$? (Wondering)

Oh yes, right. (Nod)

We found a point where $\nabla$ is non-zero.
Can we take it one step further, and show that we can find a point where $\Delta=\nabla\cdot \nabla$ is non-zero?
That is, applying the mean value theorem again? (Wondering)

Previously we had to suppose that $u$ is differentiable and now that $\nabla{u}$ is. So in general we have to suppose that $u \in C^2(\Omega)$, right?

So we pick the same $\boldsymbol x, \boldsymbol y, c$ as before and since $\nabla{u}$ is differentiable we have:

$$\nabla u(\boldsymbol x)- \nabla u(\boldsymbol y)=\nabla \cdot \nabla u((1-t) \boldsymbol x+ t \boldsymbol y) \cdot (\boldsymbol y- \boldsymbol x)= \Delta u((1-t) \boldsymbol x+t \boldsymbol y) \cdot (\boldsymbol x-\boldsymbol y)$$Is it right so far?

We have that $\boldsymbol x-\boldsymbol y \neq \boldsymbol 0$ .
But we can not deduce from $u(\boldsymbol x) \neq u(\boldsymbol y)$ that $\nabla u(\boldsymbol x)-\nabla u(\boldsymbol y) \neq \boldsymbol 0 $, can we? (Thinking)

 

[I like Serena]

Previously we had to suppose that $u$ is differentiable and now that $\nabla{u}$ is. So in general we have to suppose that $u \in C^2(\Omega)$, right?

It's given that $u$ is harmonic in $\Omega$.
Doesn't that imply that $u \in C^2(\Omega)$? (Wondering)

So we pick the same $\boldsymbol x, \boldsymbol y, c$ as before and since $\nabla{u}$ is differentiable we have:

$$\nabla u(\boldsymbol x)- \nabla u(\boldsymbol y)=\nabla \cdot \nabla u((1-t) \boldsymbol x+ t \boldsymbol y) \cdot (\boldsymbol y- \boldsymbol x)= \Delta u((1-t) \boldsymbol x+t \boldsymbol y) \cdot (\boldsymbol x-\boldsymbol y)$$Is it right so far?

We have that $\boldsymbol x-\boldsymbol y \leq \boldsymbol 0$ .
But we can not deduce from $u(\boldsymbol x) \leq u(\boldsymbol y)$ that $\nabla u(\boldsymbol x)-\nabla u(\boldsymbol y) \neq \boldsymbol 0 $, can we? (Thinking)

Let's reduce the problem to a one-dimensional problem first.
Suppose we have a function $f:\mathbb R \to \mathbb R$ with $f(x) \ge 0$ and $f\in C^2(\mathbb R)$.
And suppose we know that $f(0)=f(1)=0$.

Now suppose there is a $p \in (0,1)$ such that $f(p)>0$.
Then it follows from the mean value theorem that there are points $c \in (0,p)$ and $d \in (p,1)$, such that:
$$f'(c) = \frac{f(p) - 0}{p-0} > 0 \\ f'(d)= \frac{0-f(p)}{1-p} < 0$$
Right? (Wondering)

Now let's define $g: \mathbb R \to \mathbb R$ with $g(x) = f'(x)$.
So we have $g(c) > 0$ and $g(d) < 0$.
Then it follows from the mean value theorem that there is an $e \in (c,d)$, such that:
$$g'(e) = \frac{g(d)-g(c)}{d-c} < 0$$
Right? (Wondering)

In other words, there is a point $e \in (0,1)$, such that $f''(e) < 0$.
More generally, if a smooth function first goes up and then goes down, there must be a point where the second derivative is negative. (Thinking)

 

[evinda]

It's given that $u$ is harmonic in $\Omega$.
Doesn't that imply that $u \in C^2(\Omega)$? (Wondering)

Oh yes, right... (Nod)

Let's reduce the problem to a one-dimensional problem first.
Suppose we have a function $f:\mathbb R \to \mathbb R$ with $f(x) \ge 0$ and $f\in C^2(\mathbb R)$.
And suppose we know that $f(0)=f(1)=0$.

Now suppose there is a $p \in (0,1)$ such that $f(p)>0$.
Then it follows from the mean value theorem that there are points $c \in (0,p)$ and $d \in (p,1)$, such that:
$$f'(c) = \frac{f(p) - 0}{p-0} > 0 \\ f'(d)= \frac{0-f(p)}{1-p} < 0$$
Right? (Wondering)

Now let's define $g: \mathbb R \to \mathbb R$ with $g(x) = f'(x)$.
So we have $g(c) > 0$ and $g(d) < 0$.
Then it follows from the mean value theorem that there is an $e \in (c,d)$, such that:
$$g'(e) = \frac{g(d)-g(c)}{d-c} < 0$$
Right? (Wondering)

In other words, there is a point $e \in (0,1)$, such that $f''(e) < 0$.
More generally, if a smooth function first goes up and then goes down, there must be a point where the second derivative is negative. (Thinking)

I understand. But applying the mean value theorem for functions of several variables we just get one equation.
How can we get something about $\nabla{u}$? (Thinking)

 

[I like Serena]

Oh yes, right... (Nod)

I understand. But applying the mean value theorem for functions of several variables we just get one equation.
How can we get something about $\nabla{u}$? (Thinking)

Erm... I'm not sure any more.
I can tell that with $u(p)>0$ there will be a point where $\nabla u \ne 0$, but now I'm unsure how to turn that into a point where $\Delta u \ne 0$. (Worried)

 

[I like Serena]

I think we can apply the maximum principle:
[box=green]
Let $u = u(x), x = (x_1, …, x_n)$ be a $C^2$ function which satisfies the differential inequality
$$Lu = \sum_{ij} a_{ij}(x)\frac{\partial^2 u}{\partial x_i\partial x_j} +
\sum_i b_i\frac{\partial u}{\partial x_i} \geq 0$$
in an open domain $Ω$, where the symmetric matrix $a_{ij} = a_{ij}(x)$ is locally uniformly positive definite in $Ω$ and the coefficients $a_{ij}, b_i = b_i(x)$ are locally bounded. If $u$ takes a maximum value $M$ in $Ω$ then $u ≡ M$.
[/box]

If we pick $Lu=\Delta (-u) = 0 \ge 0$, it follows from $(-u) \le 0$ that $-u\equiv 0$. (Thinking)

 

[evinda]

I haven't seen directly this theorem.
Could we use for example also the following?

Let $u \in C^2(\Omega)$ satisfies in the space $\Omega$ the relation $Lu \geq 0$. We suppose that $\Omega$ satifies the interior sphere condition.
If $c=0$ then $u$ does not achieve its maximum in $\Omega$ if it is not constant.

 

[evinda]

And we also have the following:

Suppose that $L$ is an elliptic operator in $\Omega$, $c \equiv 0$.
If $Lu \geq 0$ then $u$ achieves its maximum in $\partial{\Omega}$, i.e. $\sup_{\overline{\Omega}} u= \sup_{\partial{\Omega}} u$.

From that we can deduce that $u \equiv =0$ in $\Omega'$. Right?
But how can we deduce that $u=0$ in $\Omega$?

Or isn't it as I say? (Thinking)

 

[I like Serena]

I haven't seen directly this theorem.
Could we use for example also the following?

Let $u \in C^2(\Omega)$ satisfies in the space $\Omega$ the relation $Lu \geq 0$. We suppose that $\Omega$ satifies the interior sphere condition.
If $c=0$ then $u$ does not achieve its maximum in $\Omega$ if it is not constant.

And we also have the following:

Suppose that $L$ is an elliptic operator in $\Omega$, $c \equiv 0$.
If $Lu \geq 0$ then $u$ achieves its maximum in $\partial{\Omega}$, i.e. $\sup_{\overline{\Omega}} u= \sup_{\partial{\Omega}} u$.

From that we can deduce that $u \equiv =0$ in $\Omega'$. Right?
But how can we deduce that $u=0$ in $\Omega$?

Or isn't it as I say?

That looks correct to me. (Nod)

Since the interior sphere condition is not given, I guess we'll need the second theorem.
To deduce that $u=0$ in $\Omega$, perhaps we can use the proof for the theorem and tweak it a bit.
Do you have the proof for the theorem available? (Wondering)

 

[evinda]

You mean the proof of this lemma:

Suppose that $L$ is an elliptic operator in $\Omega$, $c \equiv 0$.
If $Lu \geq 0$ then $u$ achieves its maximum in $\partial{\Omega}$, i.e. $\sup_{\overline{\Omega}} u= \sup_{\partial{\Omega}} u$.

?

The proof is the following:

$a_{11} \geq \lambda >0$, $e^{\gamma x_1}$ $\gamma$ constant

$L e^{\gamma x_1}=(\gamma^2 a_{11}+ \gamma \beta_1) e^{\gamma x_1} \geq \lambda (\gamma^2- \gamma \beta_0) e^{\gamma x_1}>0$ where $\beta_0 \geq \frac{|\beta_1|}{\lambda}$, $\gamma> \beta_0$.

$L(u+ \epsilon e^{\gamma x_1})=Lu+ \epsilon L e^{\gamma x_1} >0$ $(\epsilon>0)$

so $u+ \epsilon e^{\gamma x_1}$ is such that

$\sup_{\overline{\Omega}} (u+ \epsilon e^{\gamma x_1}) \leq \sup_{\partial{\Omega}} (u+ \epsilon e^{\gamma x_1}) \forall \epsilon>0$

And then we take the limit $\epsilon \to 0$.

 

[evinda]

Does this proof somehow help? (Thinking)

 

[I like Serena]

so $u+ \epsilon e^{\gamma x_1}$ is such that

$\sup_{\overline{\Omega}} (u+ \epsilon e^{\gamma x_1}) \leq \sup_{\partial{\Omega}} (u+ \epsilon e^{\gamma x_1}) \forall \epsilon>0$

I don't understand this last step.
Can you explain it? (Wondering)

 

[evinda]

I don't understand this last step.
Can you explain it? (Wondering)

I think that we use this lemma:

If $c<0, Lu \geq 0$ or $c \leq 0, Lu>0$ then u $(C^2(\Omega) \cap C^0(\overline{\Omega}))$ does not achieve its positive maximum at the interior points of $\Omega$ $(\overline{\Omega} \setminus{\partial{\Omega}})$.In our case, $c \leq 0, Lu>0$.

 

[I like Serena]

I think that we use this lemma:

If $c<0, Lu \geq 0$ or $c \leq 0, Lu>0$ then u $(C^2(\Omega) \cap C^0(\overline{\Omega}))$ does not achieve its positive maximum at the interior points of $\Omega$ $(\overline{\Omega} \setminus{\partial{\Omega}})$.

Do you have a proof for that? (Wondering)

In our case, $c \leq 0, Lu>0$.

I presume $c$ is defined as $c = u|_{\partial\Omega}$?

If we have $c=0$ for $u$, don't we have $c>0$ for $u+\epsilon e^{\gamma x_1}$?
Or am I missing something? (Wondering)

 

[evinda]

I presume $c$ is defined as $c = u|_{\partial\Omega}$?

No $c$ is a function that appears at the elliptic operator.

We have that $Lu=\sum_{i,j=1}^n a_{ij}(x) u_{x_i x_j}+ \sum_{i=1}^n \beta_i(x) u_{x_i}+c(x)u$

If $c=0$ we talk about the maximum, but if $c \leq 0$ we talk about the positive maximum.

Do you have a proof for that? (Wondering)

Yes. The proof is the following:

$Lu \equiv \sum_{i,j=1}^n a_{ij}(x) u_{x_i x_j}+ \sum_{i,j=1}^n \beta_i u_{x_i}+ cu$

We suppose that in $x_0 \in \overline{\Omega} \setminus{\partial{\Omega}}$ u achieves its positive maximum.

In $x_0 $ we have $cu \leq 0, \sum_{i=1}^n \beta_i u_{x_i}=0, \sum_{i,j=1}^n a_{ij} u_{x_i x_j} \leq 0$.In $x_0 \in \overline{\Omega} \setminus{\partial{\Omega}}$ where we have the positive maximum , it holds

1) $Lu|_{x_0}<0$ contradiction since $Lu \geq 0$

2) $Lu|_{x_0 } \leq 0$ contradiction since $Lu>0$.