MHB Uniqueness of Solution for $\Delta u=0$ in a Ball with Boundary Condition $\phi$

[evinda]

Hello! (Wave)


Let $(\star)\left\{\begin{matrix}
\Delta u=0 & \text{ in } B_R \\
u|_{\partial{B_R}}=\phi &
\end{matrix}\right.$.

Theorem: If $\phi \in C^0(\partial{B_R})$ then there is a unique solution of the problem $(\star)$ and $u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{\phi(\xi) dS}{|x- \xi|^n}$.

Proof: if $x_0 \in \partial{B_R}$ then it has to hold that $\lim_{x \to x_0 } u(x)=\phi(x_0)$.

$P(x, \xi)=\frac{R^2-|x|^2}{w_n R |x-\xi|^n}$

It holds that $\int_{\partial{B_R}} P(x, \xi) dS=1$.

$$\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} P(x, \xi) (\phi(\xi)-\phi(x_0)) dS\right| \leq \int_{\partial{B_R}} P(x, \xi) |\phi(\xi)-\phi(x_0)| dS= \int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS+\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS $$

Let $I_1=\int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$ and $I_2=\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$.

$\forall \epsilon >0 \exists \delta>0$ so that if $|\xi-x_0| \leq \delta$ then $|I_1|< \frac{\epsilon}{2}$.

$I_2 \leq 2M \frac{R^2-|x|^2}{w_n R \delta^n} \int_{|\xi-x_0|=R} dS=2M \frac{R^2-|x|^2}{\delta^n} R^{n-2} \leq \frac{\epsilon}{2}$ if $x$ is near to $ x_0 $.

So we have the following: $\forall \epsilon>0 \exists \delta>0$ such that $|u(x)-\phi(x_0)| \leq \epsilon \Leftrightarrow \lim_{x \to x_0} u(x)=\phi(x_0)$.

First of all, why does it hold that $\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} P(x, \xi) (\phi(\xi)-\phi(x_0)) dS\right| $ and not $\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} ( P(x, \xi) \phi(\xi)-\phi(x_0) ) dS\right|$ ?Also why does it hold that $\int_{\partial{B_R}} P(x, \xi) |\phi(\xi)-\phi(x_0)| dS= \int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS+\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$ ?

Could you also explain to me how we get the inequalities for $I_1$ and $I_2$ ?

 

[I like Serena]

Hey evinda! (Smile)

First of all, why does it hold that $\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} P(x, \xi) (\phi(\xi)-\phi(x_0)) dS\right| $ and not $\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} ( P(x, \xi) \phi(\xi)-\phi(x_0) ) dS\right|$ ?

Isn't:
$$\int_{\partial{B_R}} \phi(x_0)dS = w_nR^{n-1}\phi(x_0) \ne \phi(x_0)$$
(Wondering)

While:
$$\phi(x_0) = u(x_0) = \int_{\partial{B_R}} P(x_0,\xi) \phi(x_0)dS
$$

Also why does it hold that $\int_{\partial{B_R}} P(x, \xi) |\phi(\xi)-\phi(x_0)| dS= \int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS+\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$ ?

We can split up the domain of an integral into 2 subsets that combine together to the domain.
Compare for instance:
$$\int_0^1 f(x)dx = \int_0^\delta f(x)dx + \int_\delta^1 f(x)dx$$
Note that we're still constrained to $\partial{B_R}$. (Nerd)

Could you also explain to me how we get the inequalities for $I_1$ and $I_2$ ?

Let's start with $I_1$.

$$|I_1|=\left|\int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS\right| \\
\le \sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \cdot \sup_{|\xi-x_0| \leq \delta} |\phi(\xi)-\phi(x_0)| \cdot \int_{\partial{B_R}}dS \\
\le M \cdot \epsilon' \cdot w_nR^{n-1} = \frac\epsilon 2
$$
where $\epsilon'$ is introduced since $\phi$ is given to be continuous.
So given an $\epsilon$ we can find an $\epsilon'$ and from the continuity of $\phi$, we can find the appropriate $\delta$.

Note that we can only guarantee the bound on $|P(x, \xi)|$ if $x\notin \partial{B_R}$, which is given since $x\in B_R$. (Thinking)

 

[evinda]

Isn't:
$$\int_{\partial{B_R}} \phi(x_0)dS = w_nR^{n-1}\phi(x_0) \ne \phi(x_0)$$
(Wondering)

While:
$$\phi(x_0) = u(x_0) = \int_{\partial{B_R}} P(x_0,\xi) \phi(x_0)dS
$$

Oh yes, right... (Nod)

We can split up the domain of an integral into 2 subsets that combine together to the domain.
Compare for instance:
$$\int_0^1 f(x)dx = \int_0^\delta f(x)dx + \int_\delta^1 f(x)dx$$
Note that we're still constrained to $\partial{B_R}$. (Nerd)

Ok...

Let's start with $I_1$.

$$|I_1|=\left|\int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS\right|
\le \sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \cdot \sup_{|\xi-x_0| \leq \delta} |\phi(\xi)-\phi(x_0)| \cdot \int_{\partial{B_R}}dS
\le M \cdot \epsilon' \cdot w_nR^{n-1} = \frac\epsilon 2
$$
where $\epsilon'$ is introduced since $\phi$ is given to be continuous.

I see.

So given an $\epsilon$ we can find an $\epsilon'$ and from the continuity of $\phi$, we can find the appropriate $\delta$.

Don't we have from the continuity of $\phi$ that $\forall \epsilon'>0 \exists \delta>0$ such that if $|\xi-x_0| \leq \delta$ then $|\phi(\xi)-\phi(x_0)| \leq \epsilon'$ and then we set $\epsilon=2M \epsilon' w_n R^{n-1}$ ? Or am I wrong? (Thinking)

How can find the appropriate $\delta$?

Note that we can only guarantee the bound on $|P(x, \xi)|$ if $x\notin \partial{B_R}$, which is given since $x\in B_R$. (Thinking)

Could you explain to me further why $|P(x, \xi)|$ is only bounded if $x\notin \partial{B_R}$?

 

[I like Serena]

Don't we have from the continuity of $\phi$ that $\forall \epsilon'>0 \exists \delta>0$ such that if $|\xi-x_0| \leq \delta$ then $|\phi(\xi)-\phi(x_0)| \leq \epsilon'$ and then we set $\epsilon=2M \epsilon' w_n R^{n-1}$ ? Or am I wrong? (Thinking)

How can find the appropriate $\delta$?

We just did. (Thinking)

For any $\epsilon>0$ we define $\epsilon'=\frac{\epsilon}{2M w_n R^{n-1}}$.
From the continuity of $\phi$ we know that for this $\epsilon'>0$ we can find a $\delta>0$, such that if $|\xi-x_0| <\delta$ then $|\phi(\xi)-\phi(x_0)| < \epsilon'$.
It follows that for this same $\delta$ we now also have $|I_1|<\frac\epsilon 2$.

Could you explain to me further why $|P(x, \xi)|$ is only bounded if $x\notin \partial{B_R}$?

Because $P(x,\xi)=\frac{R^2-|x|^2}{w_nR |x-\xi|^n}$, which is not defined if $x=\xi$.
But that can't be because $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball. (Nerd)

 

[evinda]

We just did. (Thinking)

For any $\epsilon>0$ we define $\epsilon'=\frac{\epsilon}{2M w_n R^{n-1}}$.
From the continuity of $\phi$ we know that for this $\epsilon'>0$ we can find a $\delta>0$, such that if $|\xi-x_0| <\delta$ then $|\phi(\xi)-\phi(x_0)| \leq \epsilon'$.
It follows that for this same $\delta$ we now also have $|I_1|<\frac\epsilon 2$.

I understand.

Because $P(x,\xi)=\frac{R^2-|x|^2}{w_nR |x-\xi|^n}$, which is not defined if $x=\xi$.
But that can't be because $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball. (Nerd)

Do we assume that $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball? Or how do we know it? (Worried)

I thought that we would have at the boundary $|\xi-x|=R$. So do we have $|\xi-x|=0$ ?

 

[I like Serena]

Do we assume that $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball? Or how do we know it? (Worried)

The given formula for $u(x)$ has an integral in it that is not defined if $x$ is on the boundary.
That's because it has $|x-\xi|$ in the denominator, which would become zero when $\xi=x$ ($\xi$ is always a point on the boundary).
Therefore we have to conclude that $x$ is intended to be a point in the interior of the ball. (Thinking)

I thought that we would have at the boundary $|\xi-x|=R$. So do we have $|\xi-x|=0$ ?

$x$ is an arbitrary point inside $B_R$, which is the ball with center $0$ and radius $R$, while $\xi$ is a point on the boundary.
So we have $|\xi - 0|=R$ and $0<|\xi-x|<2R$. (Thinking)

 

[evinda]

The given formula for $u(x)$ has an integral in it that is not defined if $x$ is on the boundary.
That's because it has $|x-\xi|$ in the denominator, which would become zero when $\xi=x$ ($\xi$ is always a point on the boundary).
Therefore we have to conclude that $x$ is intended to be a point in the interior of the ball. (Thinking)

I see.

$x$ is an arbitrary point inside $B_R$, which is the ball with center $0$ and radius $R$, while $\xi$ is a point on the boundary.So we have $|\xi - 0|=R$ and $0<|\xi-x|<2R$. (Thinking)

Ok. So we have

$$\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|=\sup_{|\xi-x_0| \leq \delta} \left| \frac{R^2-|x|^2}{w_n R |x-\xi|^n} \right|=\frac{R}{w_n} \sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n}$$

Since $|x-\xi|>0$ we have that $\frac{1}{|x-\xi|^n} \to 0$ while $n \to +\infty$ and so $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \to 0$ and so it is bounded. Right? (Thinking)

 

[I like Serena]

Ok. So we have

$$\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|=\sup_{|\xi-x_0| \leq \delta} \left| \frac{R^2-|x|^2}{w_n R |x-\xi|^n} \right|=\frac{R}{w_n} \sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n}$$

Since $|x-\xi|>0$ we have that $\frac{1}{|x-\xi|^n} \to 0$ while $n \to +\infty$ and so $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \to 0$ and so it is bounded. Right? (Thinking)

Erm... shouldn't we pick $n$ to be some unknown constant? I thought it wasn't supposed to go to infinity. Or is it? (Wondering)

We do have that there is some $\epsilon > 0$ such that $|x-\xi|>\epsilon >0$.
That's because $x$ is supposed to be inside the ball, while $\xi$ is supposed to be on the boundary of the ball.
Isn't therefore:
$$\sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n} < \frac{1}{\epsilon^n}$$
(Wondering)

 

[evinda]

Erm... shouldn't we pick $n$ to be some unknown constant? I thought it wasn't supposed to go to infinity. Or is it? (Wondering)

Oh yes, you are right.

We do have that there is some $\epsilon > 0$ such that $|x-\xi|>\epsilon >0$.
That's because $x$ is supposed to be inside the ball, while $\xi$ is supposed to be on the boundary of the ball.
Isn't therefore:
$$\sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n} < \frac{1}{\epsilon^n}$$
(Wondering)

I see... But $\frac{1}{\epsilon^n}$ isn't bounded for any $\epsilon>0$. So how can we show that $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded? (Thinking)

 

[I like Serena]

I see... But $\frac{1}{\epsilon^n}$ isn't bounded for any $\epsilon>0$. So how can we show that $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded? (Thinking)

Not for any $\epsilon>0$... (Thinking)
For any given $x$ in the inside of the ball there exists an $\epsilon>0$ such that for each $\xi$ on the boundary with $|\xi - x_0|\le\delta$ we have $|\xi - x|>\epsilon$.

So for every given $x$ in the inside of the ball $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded. (Thinking)

 

[evinda]

Not for any $\epsilon>0$... (Thinking)
For any given $x$ in the inside of the ball there exists an $\epsilon>0$ such that for each $\xi$ on the boundary with $|\xi - x_0|\le\delta$ we have $|\xi - x|>\epsilon$.

So for every given $x$ in the inside of the ball $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded. (Thinking)

Ah I see... (Nod)

Can you also explain to me how we get the inequality for $I_2$ ? (Thinking)

 

[I like Serena]

$P(x, \xi)=\frac{R^2-|x|^2}{w_n R |x-\xi|^n}$

$I_2=\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$.

$\forall \epsilon >0 \exists \delta>0$ so that if $|\xi-x_0| \leq \delta$ then $|I_1|< \frac{\epsilon}{2}$.

$I_2 \leq 2M \frac{R^2-|x|^2}{w_n R \delta^n} \int_{|\xi-x_0|=R} dS=2M \frac{R^2-|x|^2}{\delta^n} R^{n-2} \leq \frac{\epsilon}{2}$ if $x$ is near to $ x_0 $.

Ah I see... (Nod)

Can you also explain to me how we get the inequality for $I_2$ ? (Thinking)

We can write:
$$|I_2|=\left|\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS\right| \\
\le \sup_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} |P(x, \xi)| \cdot \sup_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} |\phi(\xi)-\phi(x_0)| \cdot \left|\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS\right|
$$
How might we continue? (Wondering)

 

[evinda]

How can we get a relation for the integral $\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS$ ?

Also we have that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$. Does this imply that $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds=1$ ? (Thinking)

 

[I like Serena]

How can we get a relation for the integral $\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS$ ?

Also we have that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$. Does this imply that $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds=1$ ? (Thinking)

Don't we have that:
$$\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS \le \int_{\xi \in \partial B_R}dS = w_n R^{n-1}$$
(Wondering)Btw, I don't understand yet why it's supposed to hold that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
Do you?
Is it somewhere in your notes?
Anyway, if it holds, then, since $P(x,\xi) > 0$, we have:
$$\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds \le \int_{\partial{B_R}} P(x,\xi) ds_{\xi} = 1$$
It seems we don't need it though. (Thinking)

 

[evinda]

Don't we have that:
$$\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS \le \int_{\xi \in \partial B_R}dS = w_n R^{n-1}$$
(Wondering)

Oh yes, right.

Btw, I don't understand yet why it's supposed to hold that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
Do you?
Is it somewhere in your notes?

We have that if $u \in C^2(B_R(0)) \cap C^1(\overline{B_R}(0))$ is harmonic in $B_R(0)$, then the following formula holds:

$u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{u}{|x-\xi|^n} ds_{\xi}$.

For $u=1$ we get that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.

Anyway, if it holds, then, since $P(x,\xi) > 0$, we have:
$$\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds \le \int_{\partial{B_R}} P(x,\xi) ds_{\xi} = 1$$
It seems we don't need it though. (Thinking)

So then we have $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| dS \leq \sup_{\xi \in \partial{B_R}, |\xi-x_0|> \delta} |\phi(\xi)-\phi(x_0)| \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} P(x, \xi) dS \leq 2M \int_{\partial{B_R}} P(x,\xi) dS=2M $So we have to pick $4M$ as $\epsilon$ ?

 

[I like Serena]

We have that if $u \in C^2(B_R(0)) \cap C^1(\overline{B_R}(0))$ is harmonic in $B_R(0)$, then the following formula holds:

$u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{u}{|x-\xi|^n} ds_{\xi}$.

For $u=1$ we get that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.

Ok! (Happy)

So then we have $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| dS \leq \sup_{\xi \in \partial{B_R}, |\xi-x_0|> \delta} |\phi(\xi)-\phi(x_0)| \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} P(x, \xi) dS \leq 2M \int_{\partial{B_R}} P(x,\xi) dS=2M $

So we have to pick $4M$ as $\epsilon$ ?

I don't think so. We cannot make $4M$ as small as we want to.

Shouldn't we pick $\xi$ sufficiently close to $x_0$ for the proof?
That is, $|\xi-x_0|<\delta$? (Wondering)

 

[evinda]

Ok! (Happy)
I don't think so. We cannot make $4M$ as small as we want to.

Shouldn't we pick $\xi$ sufficiently close to $x_0$ for the proof?
That is, $|\xi-x_0|<\delta$? (Wondering)

We pick $|x-x_0|$ to be sufficiently small.

I think that I understood the inequality of my notes.

Since $\phi$ is bounded, there is a $M>0$ such that $|\phi| \leq M$.

So $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| ds \leq 2M \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} \frac{R^2-|x|^2}{w_n R |x-\xi|^n} ds < 2M \frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1}, \text{ since } |\xi-x_0|> \delta$.

Since $|x-x_0|$ is sufficiently small , $|x| \to |x_0|=|R|$ and so $\frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1} \to 0$ and so it is $\leq \frac{\epsilon}{2}$.

 

[I like Serena]

We pick $|x-x_0|$ to be sufficiently small.

Oh yes.

I think that I understood the inequality of my notes.

Since $\phi$ is bounded, there is a $M>0$ such that $|\phi| \leq M$.

So $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| ds \leq 2M \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} \frac{R^2-|x|^2}{w_n R |x-\xi|^n} ds < 2M \frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1}, \text{ since } |\xi-x_0|> \delta$.

Since $|x-x_0|$ is sufficiently small , $|x| \to |x_0|=|R|$ and so $\frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1} \to 0$ and so it is $\leq \frac{\epsilon}{2}$.

Good! (Happy)