
MHB Master
#1
October 9th, 2016,
11:37
Hello!!!
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 $u(x)=\frac{R^2x^2}{w_n R} \int_{\partial{B_R}} \frac{\phi(\xi) dS}{x \xi^n}$.
Proof: if $x_0 \in \partial{B_R}$ then it has to hold that $\lim_{x \to x_0 } u(x)=\phi(x_0)$.
$P(x, \xi)=\frac{R^2x^2}{w_n R x\xi^n}$
It holds that $\int_{\partial{B_R}} P(x, \xi) dS=1$.
$$\left u(x) \phi(x_0)\right=\left \int_{\partial{B_R}} P(x, \xi) (\phi(\xi)\phi(x_0)) dS\right \leq \int_{\partial{B_R}} P(x, \xi) \phi(\xi)\phi(x_0) dS= \int_{\xix_0 \leq \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS+\int_{\xix_0 > \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS $$
Let $I_1=\int_{\xix_0 \leq \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS$ and $I_2=\int_{\xix_0 > \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS$.
$\forall \epsilon >0 \exists \delta>0$ so that if $\xix_0 \leq \delta$ then $I_1< \frac{\epsilon}{2}$.
$I_2 \leq 2M \frac{R^2x^2}{w_n R \delta^n} \int_{\xix_0=R} dS=2M \frac{R^2x^2}{\delta^n} R^{n2} \leq \frac{\epsilon}{2}$ 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 $\int_{\partial{B_R}} P(x, \xi) \phi(\xi)\phi(x_0) dS= \int_{\xix_0 \leq \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS+\int_{\xix_0 > \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS$ ?
Could you also explain to me how we get the inequalities for $I_1$ and $I_2$ ?

October 9th, 2016 11:37
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MHB Seeker
#2
October 9th, 2016,
13:26
Hey evinda!!
Originally Posted by
evinda
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:
$$\int_{\partial{B_R}} \phi(x_0)dS = w_nR^{n1}\phi(x_0) \ne \phi(x_0)$$
While:
$$\phi(x_0) = u(x_0) = \int_{\partial{B_R}} P(x_0,\xi) \phi(x_0)dS
$$
Quote:
Also why does it hold that $\int_{\partial{B_R}} P(x, \xi) \phi(\xi)\phi(x_0) dS= \int_{\xix_0 \leq \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS+\int_{\xix_0 > \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS$ ?
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 $\partial{B_R}$.
Quote:
Could you also explain to me how we get the inequalities for $I_1$ and $I_2$ ?
Let's start with $I_1$.
$$I_1=\left\int_{\xix_0 \leq \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS\right \\
\le \sup_{\xix_0 \leq \delta} P(x,\xi) \cdot \sup_{\xix_0 \leq \delta} \phi(\xi)\phi(x_0) \cdot \int_{\partial{B_R}}dS \\
\le M \cdot \epsilon' \cdot w_nR^{n1} = \frac\epsilon 2
$$
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 $x\in B_R$.

MHB Master
#3
October 9th, 2016,
15:06
Thread Author
Originally Posted by
I like Serena
Isn't:
$$\int_{\partial{B_R}} \phi(x_0)dS = w_nR^{n1}\phi(x_0) \ne \phi(x_0)$$
While:
$$\phi(x_0) = u(x_0) = \int_{\partial{B_R}} P(x_0,\xi) \phi(x_0)dS
$$
Oh yes, right...
Originally Posted by
I like Serena
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 $\partial{B_R}$.
Ok...
Originally Posted by
I like Serena
Let's start with $I_1$.
$$I_1=\left\int_{\xix_0 \leq \delta} P(x, \xi)\phi(\xi)\phi(x_0) dS\right
\le \sup_{\xix_0 \leq \delta} P(x,\xi) \cdot \sup_{\xix_0 \leq \delta} \phi(\xi)\phi(x_0) \cdot \int_{\partial{B_R}}dS
\le M \cdot \epsilon' \cdot w_nR^{n1} = \frac\epsilon 2
$$
where $\epsilon'$ is introduced since $\phi$ is given to be continuous.
I see.
Originally Posted by
I like Serena
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 $\xix_0 \leq \delta$ then $\phi(\xi)\phi(x_0) \leq \epsilon'$ and then we set $\epsilon=2M \epsilon' w_n R^{n1}$ ? Or am I wrong?
How can find the appropriate $\delta$?
Originally Posted by
I like Serena
Note that we can only guarantee the bound on $P(x, \xi)$ if $x\notin \partial{B_R}$, which is given since $x\in B_R$.
Could you explain to me further why $P(x, \xi)$ is only bounded if $x\notin \partial{B_R}$?

MHB Seeker
#4
October 9th, 2016,
15:19
Originally Posted by
evinda
Don't we have from the continuity of $\phi$ that $\forall \epsilon'>0 \exists \delta>0$ such that if $\xix_0 \leq \delta$ then $\phi(\xi)\phi(x_0) \leq \epsilon'$ and then we set $\epsilon=2M \epsilon' w_n R^{n1}$ ? Or am I wrong?
How can find the appropriate $\delta$?
We just did.
For any $\epsilon>0$ we define $\epsilon'=\frac{\epsilon}{2M w_n R^{n1}}$.
From the continuity of $\phi$ we know that for this $\epsilon'>0$ we can find a $\delta>0$, such that if $\xix_0 <\delta$ then $\phi(\xi)\phi(x_0) < \epsilon'$.
It follows that for this same $\delta$ we now also have $I_1<\frac\epsilon 2$.
Quote:
Could you explain to me further why $P(x, \xi)$ is only bounded if $x\notin \partial{B_R}$?
Because $P(x,\xi)=\frac{R^2x^2}{w_nR x\xi^n}$, 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.

MHB Master
#5
October 10th, 2016,
11:46
Thread Author
Originally Posted by
I like Serena
We just did.
For any $\epsilon>0$ we define $\epsilon'=\frac{\epsilon}{2M w_n R^{n1}}$.
From the continuity of $\phi$ we know that for this $\epsilon'>0$ we can find a $\delta>0$, such that if $\xix_0 <\delta$ 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.
Originally Posted by
I like Serena
Because $P(x,\xi)=\frac{R^2x^2}{w_nR x\xi^n}$, 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.
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?
I thought that we would have at the boundary $\xix=R$. So do we have $\xix=0$ ?

MHB Seeker
#6
November 13th, 2016,
16:08
Originally Posted by
evinda
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?
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.
Quote:
I thought that we would have at the boundary $\xix=R$. So do we have $\xix=0$ ?
$x$ 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 $0<\xix<2R$.

MHB Master
#7
January 10th, 2017,
09:56
Thread Author
Originally Posted by
I like Serena
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.
I see.
Originally Posted by
I like Serena
$x$ 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 $0<\xix<2R$.
Ok. So we have
$$\sup_{\xix_0 \leq \delta} P(x,\xi)=\sup_{\xix_0 \leq \delta} \left \frac{R^2x^2}{w_n R x\xi^n} \right=\frac{R}{w_n} \sup_{\xix_0 \leq \delta} \frac{1}{x\xi^n}$$
Since $x\xi>0$ we have that $\frac{1}{x\xi^n} \to 0$ while $n \to +\infty$ and so $\sup_{\xix_0 \leq \delta} P(x,\xi) \to 0$ and so it is bounded. Right?

MHB Seeker
#8
January 10th, 2017,
17:37
Originally Posted by
evinda
Ok. So we have
$$\sup_{\xix_0 \leq \delta} P(x,\xi)=\sup_{\xix_0 \leq \delta} \left \frac{R^2x^2}{w_n R x\xi^n} \right=\frac{R}{w_n} \sup_{\xix_0 \leq \delta} \frac{1}{x\xi^n}$$
Since $x\xi>0$ we have that $\frac{1}{x\xi^n} \to 0$ while $n \to +\infty$ and so $\sup_{\xix_0 \leq \delta} P(x,\xi) \to 0$ and so it is bounded. Right?
Erm... shouldn't we pick $n$ to be some unknown constant? I thought it wasn't supposed to go to infinity. Or is it?
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:
$$\sup_{\xix_0 \leq \delta} \frac{1}{x\xi^n} < \frac{1}{\epsilon^n}$$

MHB Master
#9
January 10th, 2017,
18:04
Thread Author
Originally Posted by
I like Serena
Erm... shouldn't we pick $n$ to be some unknown constant? I thought it wasn't supposed to go to infinity. Or is it?
Oh yes, you are right.
Originally Posted by
I like Serena
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:
$$\sup_{\xix_0 \leq \delta} \frac{1}{x\xi^n} < \frac{1}{\epsilon^n}$$
I see... But $\frac{1}{\epsilon^n}$ isn't bounded for any $\epsilon>0$. So how can we show that $\sup_{\xix_0 \leq \delta} P(x,\xi)$ is bounded?

MHB Seeker
#10
January 10th, 2017,
18:11