MHB Find the function at the point (0,0)

[evinda]

Hello! (Wave)

Let $v(x,y)$ be a harmonic function in $\mathbb{R}^2$. I want to compute $v(0,0)$ given that $v|_{x^2+y^2=1}=\sin{\phi}+1$, where $x= \cos{\phi}, y=\sin{\phi}, \phi \in [0,2 \pi)$.

I I thought that we could use the following theorem.

$(\star) \Delta u=0 \text{ in } B_R(0), u|_{\partial{B_R(0)}}=\phi $

($ B_R(0)$ is a sphere with radius $R$) Theorem: Let $ \phi \in C^0(\partial{B_R}(0))$ , then the function $ u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{\phi(\xi)}{|x-\xi|^n} dS $ is the only (classical) solution of $(\star)$.

($ w_n $ is the area of the unit ball in $ \mathbb{R}^n $)

From the theorem we have that the only solution of the problem $ \Delta v=0, v|_{x^2+y^2=1}=\sin{\phi}+1=y+1 $ is the following:

$ v(x,y)=\frac{1-(x^2+y^2)}{2 \pi} \int_{x^2+y^2=1} \frac{(\xi_2+1)}{|x-\xi|^2} dS =\frac{1-(x^2+y^2)}{2 \pi} \int_{x^2+y^2=1} \frac{(\xi_2+1)}{(x- \xi_1)^2+ (y-\xi_2)^2} dS $

Is it right so far? Have I applied correctly the theorem? If so, can we compute the integral?

 

[I like Serena]

Hello! (Wave)

Let $v(x,y)$ be a harmonic function in $\mathbb{R}^2$. I want to compute $v(0,0)$ given that $v|_{x^2+y^2=1}=\sin{\phi}+1$, where $x= \cos{\phi}, y=\sin{\phi}, \phi \in [0,2 \pi)$.

I I thought that we could use the following theorem.

$(\star) \Delta u=0 \text{ in } B_R(0), u|_{\partial{B_R(0)}}=\phi $

($ B_R(0)$ is a sphere with radius $R$) Theorem: Let $ \phi \in C^0(\partial{B_R}(0))$ , then the function $ u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{\phi(\xi)}{|x-\xi|^n} dS $ is the only (classical) solution of $(\star)$.

($ w_n $ is the area of the unit ball in $ \mathbb{R}^n $)

From the theorem we have that the only solution of the problem $ \Delta v=0, v|_{x^2+y^2=1}=\sin{\phi}+1=y+1 $ is the following:

$ v(x,y)=\frac{1-(x^2+y^2)}{2 \pi} \int_{x^2+y^2=1} \frac{(\xi_2+1)}{|x-\xi|^2} dS =\frac{1-(x^2+y^2)}{2 \pi} \int_{x^2+y^2=1} \frac{(\xi_2+1)}{(x- \xi_1)^2+ (y-\xi_2)^2} dS $

Is it right so far? Have I applied correctly the theorem? If so, can we compute the integral?

Hey evinda! (Smile)

I think there's a mix-up in the usage of $\phi$ that has 2 different meanings now.
In your problem statement $\phi$ is the angle with the x-axis, which corresponds to $\xi$ in the theorem.
And in the theorem $\phi(\xi)$ is the value on the boundary, which correspond to $\sin\phi+1$ in the problem statement.

Let's replace $\phi$ in the theorem by $\psi$.
Then the theorem reads:

Theorem
$(\star) \Delta u=0 \text{ in } B_R(0), u|_{\partial{B_R(0)}}=\psi $

Let $ \psi \in C^0(\partial{B_R}(0))$ , then the function $ u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{\psi(\xi)}{|x-\xi|^n} dS $ is the only (classical) solution of $(\star)$. Now we can:
- replace $\psi(\xi)$ by $\sin\phi + 1$,
- replace $\xi$ by $(R\cos\phi,R\sin\phi)$,
- and replace $dS$ by $R\,d\phi$.
(Happy)

 

[evinda]

Hey evinda! (Smile)

I think there's a mix-up in the usage of $\phi$ that has 2 different meanings now.
In your problem statement $\phi$ is the angle with the x-axis, which corresponds to $\xi$ in the theorem.
And in the theorem $\phi(\xi)$ is the value on the boundary, which correspond to $\sin\phi+1$ in the problem statement.

Let's replace $\phi$ in the theorem by $\psi$.
Then the theorem reads:

Theorem:
$(\star) \Delta u=0 \text{ in } B_R(0), u|_{\partial{B_R(0)}}=\psi $

Let $ \psi \in C^0(\partial{B_R}(0))$ , then the function $ u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{\psi(\xi)}{|x-\xi|^n} dS $ is the only (classical) solution of $(\star)$.

I see...

Now we can:
- replace $\xi$ by $\phi$,

But isn't $\xi$ a vector while $\phi$ is a real number? Or am I wrong?

- and replace $dS$ by $R\,d\phi$.

Why do we replace $dS$ by $R\,d\phi$ ?

 

[I like Serena]

But isn't $\xi$ a vector while $\phi$ is a real number? Or am I wrong?

You're right. (Tmi)

Instead we should replace $\boldsymbol\xi$ by $(R\cos\phi, R\sin\phi)$.

Why do we replace $dS$ by $R\,d\phi$ ?

Because $dS$ is the area of an infinitesimal surface element on an n-dimensional sphere at location $\boldsymbol \xi$.
In the case of a circle that corresponds to an infinitesimal arc with length $R\,d\phi$. (Nerd)

 

[evinda]

So in other words, do we use polar coordinates in order to compute the integral?

But like that, doesn't $\boldsymbol\xi$ get the value of the given $(x,y)$? (Worried)

 

[I like Serena]

So in other words, do we use polar coordinates in order to compute the integral?

But like that, doesn't $\boldsymbol\xi$ get the value of the given $(x,y)$? (Worried)

Indeed, I believe it's easiest to use polar coordinates.

And yes, $\boldsymbol\xi = \boldsymbol\xi(\phi) = (x(\phi),y(\phi)) = (R\cos\phi, R\sin\phi)$. (Thinking)

 

[evinda]

At the integral, the limit of integration is $\partial{B_R(0)}$. Is this the condition $|\xi|=1$ or $|x|=1$?

- - - Updated - - -

Also having $\boldsymbol\xi = \boldsymbol\xi(\phi) = (x(\phi),y(\phi)) = (R\cos\phi, R\sin\phi)$, we divide by $0$ at the integral, don't we? (Sweating) Or am I wrong?

 

[I like Serena]

At the integral, the limit of integration is $\partial{B_R(0)}$. Is this the condition $|\xi|=1$ or $|x|=1$?

- - - Updated - - -

Also having $\boldsymbol\xi = \boldsymbol\xi(\phi) = (x(\phi),y(\phi)) = (R\cos\phi, R\sin\phi)$, we divide by $0$ at the integral, don't we? (Sweating) Or am I wrong?

It's the condition $\|\boldsymbol\xi\|=1$.
And $\|\boldsymbol x\|$ can't be $1$, otherwise we'd indeed be dividing by 0 when we'd get to $\boldsymbol \xi = \boldsymbol x$.
At best we can only take a limit where $\|\boldsymbol x\|\to 1$.

 

[evinda]

It's the condition $\|\boldsymbol\xi\|=1$.
And $\|\boldsymbol x\|$ can't be $1$, otherwise we'd indeed be dividing by 0 when we'd get to $\boldsymbol \xi = \boldsymbol x$.
At best we can only take a limit where $\|\boldsymbol x\|\to 1$.

So the $\boldsymbol x$ of the solution that we find from the theorem isn't $(x,y)$ where $x= \cos{\phi}$ and $y=\sin{\phi}$, right? We have $v|_{\|\boldsymbol\xi\|=1}= \sin{\phi}+1$, right?

 

[I like Serena]

So the $\boldsymbol x$ of the solution that we find from the theorem isn't $(x,y)$ where $x= \cos{\phi}$ and $y=\sin{\phi}$, right? We have $v|_{\|\boldsymbol\xi\|=1}= \sin{\phi}+1$, right?

Correct. (Nod)
$\boldsymbol x$ is some point inside the ball for which we want to know $u(\boldsymbol x)$.

In this particular case we want to know $v(0,0)$, so we have $\boldsymbol x = (0,0)$.

 

[evinda]

Correct. (Nod)
$\boldsymbol x$ is some point inside the ball for which we want to know $u(\boldsymbol x)$.

In this particular case we want to know $v(0,0)$, so we have $\boldsymbol x = (0,0)$.

So we have $v(\boldsymbol x)=\frac{1-(x^2+y^2)}{2 \pi} \int_{|| \boldsymbol \xi||=1} \frac{\sin{\phi}+1}{\sqrt{(x- \cos{\phi})^2+(y- \sin{\phi})^2}} d{\phi}$.

Right?

By taking $\boldsymbol \xi=(R \cos{\phi}, R \sin{\phi})=(\cos{\phi}, \sin{\phi})$ the equality $|| \boldsymbol \xi||=1$ always holds, right?

 

[I like Serena]

So we have $v(\boldsymbol x)=\frac{1-(x^2+y^2)}{2 \pi} \int_{|| \boldsymbol \xi||=1} \frac{\sin{\phi}+1}{\sqrt{(x- \cos{\phi})^2+(y- \sin{\phi})^2}} d{\phi}$.

Right?

By taking $\boldsymbol \xi=(R \cos{\phi}, R \sin{\phi})=(\cos{\phi}, \sin{\phi})$ the equality $|| \boldsymbol \xi||=1$ always holds, right?

Yes, $|| \boldsymbol \xi||=1$ will always hold, but don't we have $n=2$? (Wondering)

I'd make it:
$$v(x,y)=\frac{1-(x^2+y^2)}{2 \pi} \int_0^{2\pi} \frac{\sin{\phi}+1}{(x- \cos{\phi})^2+(y- \sin{\phi})^2} d{\phi}$$
(Thinking)

 

[evinda]

Yes, $|| \boldsymbol \xi||=1$ will always hold, but don't we have $n=2$? (Wondering)

I'd make it:
$$v(x,y)=\frac{1-(x^2+y^2)}{2 \pi} \int_0^{2\pi} \frac{\sin{\phi}+1}{(x- \cos{\phi})^2+(y- \sin{\phi})^2} d{\phi}$$
(Thinking)

Yes, in our case we have that $n=2$.
The $\phi$ of $d{\phi}$ is a vector and an other than the one given by the problem statement, right?

 

[evinda]

No, it's just a variable... But $\phi \in [0,2 \pi)$ so was the previous limit of integration wrong? $\int_{||\boldsymbol \xi||=1} \dots d{\phi}$ ?

 

[I like Serena]

No, it's just a variable... But $\phi \in [0,2 \pi)$ so was the previous limit of integration wrong? $\int_{||\boldsymbol \xi||=1} \dots d{\phi}$ ?

Not wrong... just not consistent.
We've been replacing $\boldsymbol\xi$ by $(\cos\phi,\sin\phi)$ everywhere.
That also means replacing $\int_{\|\boldsymbol \xi\|=1}...dS$ by $\int_0^{2\pi}...d\phi$. (Thinking)

 

[evinda]

A ok... So we have

$$v(0,0)=\frac{1}{2 \pi} \int_0^{2\pi} (\sin{\phi}+1) d{\phi}=1$$

Right?

 

[I like Serena]

A ok... So we have

$$v(0,0)=\frac{1}{2 \pi} \int_0^{2\pi} (\sin{\phi}+1) d{\phi}=1$$

Right?

Yep! (Nod)

 

[evinda]

Nice! In order to apply the theorem we say that $\Delta v=0 \text{ in } \mathbb{R}^2$ and this implies that $\Delta v=0 \text{ in } \{ (x,y) \in \mathbb{R}^2: x^2+y^2<1 \}$, right?

 

[evinda]

Then the solution we have found for $v$ will hold for elements in $\{ (x,y) \in \mathbb{R}^2: x^2+y^2<1\}$ and we can use it since $(0,0)$ belongs to this set. Right? (Smile)

 

[I like Serena]

Nice! In order to apply the theorem we say that $\Delta v=0 \text{ in } \mathbb{R}^2$ and this implies that $\Delta v=0 \text{ in } \{ (x,y) \in \mathbb{R}^2: x^2+y^2<1 \}$, right?

Then the solution we have found for $v$ will hold for elements in $\{ (x,y) \in \mathbb{R}^2: x^2+y^2<1\}$ and we can use it since $(0,0)$ belongs to this set. Right?

Correct. (Smile)

 

[evinda]

Correct. (Smile)

Great... Thank you very much! (Smirk)