Computing the Integral of a Harmonic Function with Boundary Conditions.

[evinda]

Hello! (Wave)

Let $L=\{ (x,y): x^2+y^2<5\}$.
Suppose that we have a function $v(x,y)$ that is harmonic in $L$ and $\Omega \subset L$ an arbitrary space such that $(0,0)$ belongs to that space.
Suppose that $v$ is equal to $1$ at the boundary of $\Omega$.
How can we compute the integral $\int_{x^2+y^2 \leq 4} v(x,y) dxdy$ ?

First I thought that we could use the second version of the mean value theorem. But then we couldn't use the fact that $v$ is equal to $1$ at the boundary of $\Omega$. Could we? If not, could you give me a hint how else we could compute the integral?

 

[HallsofIvy]

Do you know of any relation between the value of a harmonic function on the boundary of a set and its value inside the set?

 

[evinda]

Do you know of any relation between the value of a harmonic function on the boundary of a set and its value inside the set?

We have the following theorem.

Let $u$ be harmonic in $\Omega$ and $x \in \Omega$ an arbitrary point. Then $\forall \rho \in (0,d)$ where $d=dist(x, \partial{\Omega})$ it holds that $u(x)=\frac{1}{w_n \rho^{n-1}} \int_{|x-\xi|=\rho} u ds$.In order to apply this theorem, we have to assume that $\Omega$ is the unit ball of radius $2$, right?

If so, then we have $u(x,y)=\frac{1}{ w_2 2^{2-1}} \int_{|(x,y)-(\xi_1, \xi_2)|=2} u ds \Rightarrow u(x,y)=\frac{1}{2w_2} \int_{\sqrt{(x- \xi_1)^2+(x- \xi_2)^2}=2} u ds $.

For $(x,y)=(0,0)$ we have that $u(0,0)=\frac{1}{2 w_2} \int_{\xi_1^2+ \xi_2^2=4} u ds$.

Is it right so far? How could we continue?

 

[evinda]

Using polar coordinates we have the following:

$$\int_{x^2+y^2 \leq 4} u(x,y) dx dy=\int_{0}^2 \int_{0}^{2 \pi} u(r \cos{\phi}, r \sin{\phi}) r dr d{\phi}$$

Right?

Can we integrate now as for $\phi$? How can we do so?
Or how else can we continue?

 

[I like Serena]

Hello! (Wave)

Let $L=\{ (x,y): x^2+y^2<5\}$.
Suppose that we have a function $v(x,y)$ that is harmonic in $L$ and $\Omega \subset L$ an arbitrary space such that $(0,0)$ belongs to that space.
Suppose that $v$ is equal to $1$ at the boundary of $\Omega$.
How can we compute the integral $\int_{x^2+y^2 \leq 4} v(x,y) dxdy$ ?

First I thought that we could use the second version of the mean value theorem. But then we couldn't use the fact that $v$ is equal to $1$ at the boundary of $\Omega$. Could we? If not, could you give me a hint how else we could compute the integral?

Hey evinda! (Smile)

What is $\Omega$?
And can you state the second version of the mean value theorem?

If $\Omega$ is what I suspect it is, the mean value property should help us out. (Happy)

 

[evinda]

What is $\Omega$?

It is now given any further. But I think that we have to pick $\Omega=\{ (x,y): x^2+y^2< 2^2\}$, do you agree?

And can you state the second version of the mean value theorem?

If $u$ is harmonic in $\Omega$ and $x \in \Omega$ then $u(x)=\frac{n}{w_n r^n} \int_{|x-\xi|\leq r} u d{\xi}, r \in (0,d)$ where $d=dist(x, \partial{\Omega})$.

I have thought the following:

From the mean value theorem we have that $u(x,y)=\frac{1}{4 \pi} \int_{\sqrt{(x- \xi_1)^2+(y- \xi_2)^2}=2} u ds$ and so we get that $u(0,0)=\frac{1}{4 \pi} \int_{\xi_1^2+\xi_2^2=4} 1 ds=1$.

Then from the second version of the mean value theorem we get that $\int_{x^2+y^2 \leq 4} u(x,y) dx dy=4 \pi$.

Is my idea right?

 

[I like Serena]

It is now given any further. But I think that we have to pick $\Omega=\{ (x,y): x^2+y^2< 2^2\}$, do you agree?

I guess we'll have to assume that yes.

If $u$ is harmonic in $\Omega$ and $x \in \Omega$ then $u(x)=\frac{n}{w_n r^n} \int_{|x-\xi|\leq r} u d{\xi}, r \in (0,d)$ where $d=dist(x, \partial{\Omega})$.

I have thought the following:

From the mean value theorem we have that $u(x,y)=\frac{1}{4 \pi} \int_{\sqrt{(x- \xi_1)^2+(y- \xi_2)^2}=2} u ds$ and so we get that $u(0,0)=\frac{1}{4 \pi} \int_{\xi_1^2+\xi_2^2=4} 1 ds=1$.

Then from the second version of the mean value theorem we get that $\int_{x^2+y^2 \leq 4} u(x,y) dx dy=4 \pi$.

Is my idea right?

The idea is right, but I think we have:
$$v(0,0)=\frac{1}{4 \pi} \oint_{\xi_1^2+\xi_2^2=4} 1 \,ds=1$$
and:
$$v(0,0)=\frac{3}{4 \pi} \iint_{\xi_1^2+\xi_2^2\le4} v(\xi_1,\xi_2)\,d\xi_1\,d\xi_2$$
Thus:
$$\iint_{x^2+y^2 \leq 4} v(x,y)\, dx\, dy=\frac{4 \pi}{3}$$
Don't we? (Wondering)

 

[evinda]

I guess we'll have to assume that yes.The idea is right, but I think we have:
$$v(0,0)=\frac{1}{4 \pi} \oint_{\xi_1^2+\xi_2^2=4} 1 \,ds=1$$
and:
$$v(0,0)=\frac{3}{4 \pi} \iint_{\xi_1^2+\xi_2^2\le4} v(\xi_1,\xi_2)\,d\xi_1\,d\xi_2$$
Thus:
$$\iint_{x^2+y^2 \leq 4} v(x,y)\, dx\, dy=\frac{4 \pi}{3}$$
Don't we? (Wondering)

I found the following from the second version of the mean value theorem:

$$u(x,y)=\frac{2}{2 \pi \cdot 4} \int_{\sqrt{(x- \xi_1)^2+(y-\xi_2)^2} \leq 2} u d{\xi}$$

and so $\int_{\xi_1^2+ \xi_2^2 \leq 4} u d{\xi}=4 \pi u(0,0)=4 \pi$

Am I wrong? (Thinking)

 

[I like Serena]

I found the following from the second version of the mean value theorem:

$$u(x,y)=\frac{2}{2 \pi \cdot 4} \int_{\sqrt{(x- \xi_1)^2+(y-\xi_2)^2} \leq 2} u d{\xi}$$

and so $\int_{\xi_1^2+ \xi_2^2 \leq 4} u d{\xi}=4 \pi u(0,0)=4 \pi$

Am I wrong? (Thinking)

My mistake. You're quite right. (Nod)

 

[evinda]

My mistake. You're quite right. (Nod)

No problem... Thank you! (Smile)