# Proof of Mean Value Property

1. Aug 30, 2013

### yungman

Attached is a copy of p181 of Strauss Partial Differential Equation. This is part of proof of Mean Value Property using Green's 1st identity:
$$\int_Dv\nabla u+\nabla v\cdot\nabla u \;dV=\oint_A v\frac {\partial u}{\partial r}dA$$

Let $v=1$ and let $u(r,\theta,\phi)$ be a harmonic function ie $\nabla^2 u=0$ and $u$ has continuous 1st and 2nd partial derivatives.
$$\Rightarrow\;\oint_A \frac {\partial u}{\partial r}dA=0$$
This is (5) on the top of the copy of the book.

If you follow the steps, the next step shows integration of
$$\int_0^{2\pi}\int_0^{\pi}\frac {\partial u(a,\theta,\phi)}{\partial r}\;a^2\;\sin\theta\; d\theta \;d\phi\;=0,\hbox{ Where the integration is on surface at }\;r=a$$

My question is in the very next step when the book claimed since it is equal to zero, then $u(a,\theta,\phi)$ is not a function of $r$, therefore you can move the $\frac{\partial}{\partial r}$ outside the integral. That's all reasonable. But then the next line you see the equation using $u(r,\theta,\phi)$, which is function of $r$!! How can you do that?

Please explain to me why, Thanks

Alan

#### Attached Files:

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2. Aug 30, 2013

### Mandelbroth

(Bolding by me.)

Riddle me this: is that why you can pull out the $\frac{\partial}{\partial r}$?

3. Aug 30, 2013

### yungman

No, the book pull it out as shown in the copy. You see the second equation using $u_r(a,\theta,\phi)$? That's the $\frac{\partial}{\partial r}$ inside the integral.

Thanks

4. Aug 30, 2013

### Mandelbroth

I'm not familiar with the book, but I didn't see any indication of "therefore" for pulling it out. I'm betting the referenced section ("Section A.3") is something about differentiation under the integral sign.

At least, that's what I'm seeing. I haven't had my chocolate-covered espresso beans tonight, so trusting my judgement might be a little precarious. :tongue:

5. Aug 30, 2013

### yungman

Did you see in the copy, right above the 3rd equation, it said "Then we pull $\partial /\partial r$ outside the integral ( see Section A.3),........

Tell me about it, the doctor tell me to cut out diet Coke.....Caffeine!!! this is my second day!!!! Who said caffeine is not addictive?!!

6. Aug 30, 2013

### Mandelbroth

Yes. Did you refer to Section A.3 in the book? It probably gives the justification for the pulling out $\frac{\partial}{\partial r}$ (differentiation under the integral sign).

Paul Erdős claimed a mathematician was a machine for turning coffee into theorems.

7. Aug 30, 2013

### yungman

I think I found it reason. There might be a typo:

$$\int_0^{2\pi}\int_0^{\pi}\frac {\partial u(a,\theta,\phi)}{\partial r}\;a^2\;\sin\theta\; d\theta \;d\phi\;=0$$
should be
$$\int_0^{2\pi}\int_0^{\pi}\frac {\partial u(r,\theta,\phi)}{\partial r}\;a^2\;\sin\theta\; d\theta \;d\phi\;=0$$

The reason is this is a surface integral at $r=a$ where $dA=a^2\;\sin\theta\; d\theta \;d\phi\;$. The integral does not involve $r$, so even if $u(r,\theta,\phi)$ is function of $r$, you can pull $\frac{\partial}{\partial r}$ out of the integral. Does that make sense.

Maybe it's the caffeine......or the lack of it, I should have seen this.

Thanks

8. Aug 30, 2013

### yungman

But still there is something that does not make sense. because even if you pull $\frac {\partial}{\partial r}$ out,
$$\int_0^{2\pi}\int_0^{\pi} u(r,\theta,\phi)\;a^2\;\sin\theta\; d\theta \;d\phi$$
Still a function of $r$. It just mean:
$$\frac{\partial}{\partial r}\left[\int_0^{2\pi}\int_0^{\pi} u(r,\theta,\phi)\;a^2\;\sin\theta\; d\theta \;d\phi\right]=0$$

9. Aug 31, 2013

### hellofolks

Your confusion is caused just by a matter of notation.

Strauss wrote $u(r,\theta,\phi)$ because function $u$ depends on $r$.
What is meant is that the expression

$$\frac{1}{4\pi}\int_0^{2\pi}\!\!\!\int_0^\pi u(r,\theta,\phi)\,d\theta\,d\phi$$

does not depend on $r$. Notice it doesn't depend on $\theta$ and $\phi$, either, since we are integrating with respect to those variables. So what he actually proves is that the above integral is a constant. That is, whatever value of $r$
we take in the above integration, we will obtain the same value. Now let $r=0$ (or more rigorously $r \to 0$), and the rest of the proof follows easily.

10. Aug 31, 2013

### hellofolks

Just another detail: when $\frac{\partial f}{\partial r}=0$, this means that a function doesn't depend on variable $r$. So, if it is a multivariate function, it will depend only on the other variables.

11. Aug 31, 2013

### yungman

I understand, but $u(r,\theta,\phi)$ is specified to be a function of $r$. There lies the contradiction. The integration cannot be zero if $u(r,\theta,\phi)$ is a function of $r$.

Put it in another way, if $u$ is not a function of $r$, the usefulness of this Mean Value is quite useless as not too many function is not a function of $r$ in EM.

12. Aug 31, 2013

### hellofolks

It is not the integral that is zero, but the derivative of the integral divided by $4\pi$. Thus, the integral is a constant (not necessarily zero). Also, $u$ IS a function of $r$. The mean value IS NOT a function of $r$

13. Aug 31, 2013

### yungman

Thanks for the reply, I still don't get if the $u$ inside the integral is a function of $r$, how can the integral of $u$ not a function of $r$ if the integral does the touch the $r$ part of $u$.

14. Aug 31, 2013

### hellofolks

That's the beauty of the mean value property! It doesn't matter if you're calculating the average of a harmonic function in huge or in a tiny ball -- the average is always the same and equals the harmonic function evaluated at the center of the ball. This has a lot of consequences, besides having interesting physical interpretations, too. The mean value property is perhaps the reason why everybody loves the qualitative theory of solutions of Laplace's equations.

15. Aug 31, 2013

### Mandelbroth

Consider a function $$c:\mathbb{R}^5\to\mathbb{R},~(x,y,z,u,v)\mapsto c.$$ Call it a constant function of 5 real variables.

What is $\frac{\partial c}{\partial x}$?

16. Aug 31, 2013

### yungman

Thanks
Can you give an example of $u$ is a function of $r$, but after the integration, it becomes independent of $r$ so the $\partial /\partial r$ =0?

17. Aug 31, 2013

### yungman

Thanks
But you now limits the function $u$ to be a constant function. That will not be useful for electromagnetics. All the potentials and fields depend on distance $r$. None are constant.

Green's function is part of the advanced EM material, I just don't think it meant to be a constant function.

18. Sep 1, 2013

### hellofolks

Take $u(r,\theta,\phi)=r^2\sin^2 \theta\cos 2\phi$. The function $u$ is not constant, depends on $r$ and is unbounded. Now calculate the surface integral of $u$ on a sphere of radius $r$. You'll see that it equals zero, independently of $r$. The same is true of the average of $u$ on that sphere.(Remark: in accordance to the notation of Strauss, take $\theta \in [0,\pi]$ (polar angle or colatitude) and $\phi \in [0,2\pi]$ (azimuthal angle or longitude)).

Try to understand why this happens in terms of the mean value property. (Hint: show that $u$ satisfies Laplace's equation.)

If you have trouble with the details, do not hesitate to let me know.

19. Sep 1, 2013

### yungman

I worked through $u=r^2\sin^2\theta\cos 2\phi$, $\nabla^2 u \;\neq \;0$.

I am using
$$\nabla ^2 u=\frac {1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial u}{\partial \theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2u}{\partial \phi^2}$$
I have
$$\nabla^2 u=-4\cos 2\phi\;\neq\;0$$
So this is not a Harmonic function. This Mean Value property require $u$ being a Harmonic function.

Last edited: Sep 1, 2013
20. Sep 1, 2013

### yungman

I have even more confusion: $\oint_A \frac {\partial u}{\partial r}dA=0$ is based on $\nabla^2 u=0$.

Let $u=\frac{1}{r}$
$$\nabla ^2 u=\frac {1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial u}{\partial \theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2u}{\partial \phi^2}$$
$$\Rightarrow\;\nabla^2 u=\frac {1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \left(\frac{1}{r}\right)}{\partial r}\right)+0+0=\frac {1}{r^2}\frac{\partial}{\partial r}(-1)=0$$

to check 1st and 2nd partial derivatives are continuous:
$$\frac {\partial u}{\partial r}+\frac {\partial u}{\partial \theta}+\frac {\partial u}{\partial \phi}=\frac {\partial u}{\partial r}=-\frac{1}{r^2}$$
$$\frac {\partial^2 u}{\partial r^2}=2\frac{1}{r^3}$$
Both are continuous for $r\neq 0$. Therefore $u=1/r$ is a Harmonic Function.

Green's first identity:
$$\int_Dv\nabla u+\nabla v\cdot\nabla u \;dV=\oint_A v\frac {\partial u}{\partial r}dA$$
Let $v=1$
$$v=1\Rightarrow\;\int_Dv\nabla^2 u+\nabla v\cdot\nabla u \;dV=0$$
But
$$\int_A\frac{\partial u}{\partial r}dA=-\int_A\frac{1}{r^2} a^2\sin\theta\;d\theta\;d\phi\;\neq\;0$$

I can't even get the Green's first identity.
$$\int_Dv\nabla u+\nabla v\cdot\nabla u \;dV=\oint_A v\frac {\partial u}{\partial r}dA$$