How Do We Get the Last Two Relations in the Mean Value Proposition Proof?

[evinda]

Hello! (Wave)

We set $L_k \equiv u_{tt}+\frac{k}{t}-\Delta u=0, k \in \mathbb{N}_0$.

I am looking at the proof of the following proposition:

We suppose that $g(x)$ is twice differentiable in $\mathbb{R}^n$. Then the mean value of $g(x)$ at the sphere with radius $t$ and center at $x$, which we symbolize with $M(t,x,g)$, is a solution of the problem
$L_{n-1} M=0, M(0,x,g)=g(x), M_t(0,x,g)=0$.

I am given the proof for $n=2$, but I have some questions.

For $n=2$ we have $M(t,x,g)=\frac{1}{2 \pi t} \int_{|x-\xi|=t} g(\xi) ds=\frac{1}{2 \pi} \int_0^{2 \pi} g(x_1+t \cos{\phi}, x_2+ t \sin{\phi}) d{\phi}$.

Let $\xi_1=x_1+ t \cos{\phi}$ and $\xi_2=x_2+ t \sin{\phi}$.

We have:

$$\frac{\partial{M}}{\partial{t}}=\frac{1}{2 \pi} \int_0^{2 \pi} (g_{\xi_1} \cos{\phi}+g_{\xi_2} \sin{\phi}) d{\phi}=\frac{1}{2 \pi} \int_0^{2 \pi} \nabla_{\xi}g \cdot (\cos{\phi}, \sin{\phi}) d{\phi}=\frac{1}{2 \pi t} \int_{|x-\xi|=t} \nabla_{\xi}g \cdot \mathcal{v} ds=\frac{1}{2 \pi} \int_{|x-\xi| \leq t} \Delta_{\xi}gd{\xi}$$

Could you explain to me how we get the last two relations? (Thinking)

 

[Euge]

The normal $\nu$ to the circle is the unit radial vector from the origin, so in coordinates it's $(\cos \phi, \sin \phi)$. Also, on the circle, the arc length parameter $s$ satisfies $s = t\phi$, and so $ds = t\, d\phi$, or $d\phi = \frac{1}{t}\, ds$. This is why, altogether, you get

$$\frac{1}{2\pi}\int_0^{2\pi} \nabla_\xi g\cdot(\cos \phi, \sin \phi)\, d\phi = \frac{1}{2\pi t}\int_{\lvert x - \xi\rvert = t} \nabla_\xi g \cdot \nu \, ds.$$

The last equation follows from the divergence theorem, since $\operatorname{div}(\nabla_\xi g) = \Delta_\xi g$.

 

[evinda]

The normal $\nu$ to the circle is the unit radial vector from the origin, so in coordinates it's $(\cos \phi, \sin \phi)$.

Could you explain further to me why the coordinates of $\nu$ are $(\cos \phi, \sin \phi)$ ? (Thinking)

The last equation follows from the divergence theorem, since $\operatorname{div}(\nabla_\xi g) = \Delta_\xi g$.

I see... (Nod)

 

[evinda]

Also, then it says the following:$$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$

First of all, why don't we have :

$$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|\leq t+ \Delta t} \Delta_{\xi}g d{\xi}$$

?

Secondly, how do we get the following equality?

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$

And... does it hold that $\Delta M= \frac{1}{2 \pi} \int_0^{2 \pi} (g_{\xi_1 \xi_1} \cos^2{\phi}+g_{\xi_2 \xi_2} \sin^2{\phi}) d{\phi}$ ? (Thinking)

 

[Euge]

Could you explain further to me why the coordinates of $\nu$ are $(\cos \phi, \sin \phi)$ ? (Thinking)

I made a slight error in my description of $\varphi$. What I meant to say is that for every point $\xi$ on your circle, $\nu(\xi)$ is the unit radial vector from the center of the circle (which is $x$) to $\xi$. Thus $\nu(\xi) = \frac{\xi - x}{\|\xi - x\|}$. Now $\xi = x + t(\cos \phi, \sin \phi)$ for some $\phi$, and so $\nu(\xi) = \frac{t(\cos \phi, \sin \phi)}{t} = (\cos \phi, \sin \phi)$.

Also, then it says the following:$$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$

First of all, why don't we have :

$$\frac{\partial}{\partial{t}} \int_{|x-\xi|\leq t} \Delta_{\xi}g d{\xi}=\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|\leq t+ \Delta t} \Delta_{\xi}g d{\xi}$$

?

Your equation is correct, but so is the author's. Note the domain of integration in your integral differs from the domain of integration of the authors by a null set.

Secondly, how do we get the following equality?

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t}\int_{t<|x-\xi|<t+ \Delta t} \Delta_{\xi}g d{\xi}=\int_{|x-\xi|=t} \Delta_{\xi} g ds$$

Switch to polar coordinates centered at $x$. At some point you'll compute the limits

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} \nabla^2 g(x_1 + r\cos \phi, x_2 +r\sin \phi)\, r\, dr,$$

which evaluate to $\nabla^2 g(x_1 + r\cos \phi, x_2 + r\sin \phi)\,t$ by the fundamental theorem of calculus.

And... does it hold that $\Delta M= \frac{1}{2 \pi} \int_0^{2 \pi} (g_{\xi_1 \xi_1} \cos^2{\phi}+g_{\xi_2 \xi_2} \sin^2{\phi}) d{\phi}$ ? (Thinking)

How did you derive this formula?

 

[evinda]

I made a slight error in my description of $\varphi$. What I meant to say is that for every point $\xi$ on your circle, $\nu(\xi)$ is the unit radial vector from the center of the circle (which is $x$) to $\xi$. Thus $\nu(\xi) = \frac{\xi - x}{\|\xi - x\|}$. Now $\xi = x + t(\cos \phi, \sin \phi)$ for some $\phi$, and so $\nu(\xi) = \frac{t(\cos \phi, \sin \phi)}{t} = (\cos \phi, \sin \phi)$.

A ok... (Nod)

Your equation is correct, but so is the author's. Note the domain of integration in your integral differs from the domain of integration of the authors by a null set.

I see...

Switch to polar coordinates centered at $x$. At some point you'll compute the limits

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} \nabla^2 g(x_1 + r\cos \phi, x_2 +r\sin \phi)\, r\, dr,$$

which evaluate to $\nabla^2 g(x_1 + r\cos \phi, x_2 + r\sin \phi)\,t$ by the fundamental theorem of calculus.

Switching to polar coordinates we get the limit $\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{t<|r|<t+\Delta t} \Delta_{\xi} g r dr$, which is equal to that what you wrote , right?
Then from the fundamental theorem of calculus, we have that:

$$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} \Delta_{\xi} g r dr=\lim_{\Delta t \to 0} \frac{1}{\Delta t} \left[\Delta_{\xi}g(x_1+(t+\Delta t) \cos{\phi}, x_2+ (t+\Delta t)\sin{\phi})(t+\Delta t)-\Delta_{\xi}g(x_1+t \cos{\phi}, x_2+t \sin{\phi})t\right]$$

Right? How do we continue?

How did you derive this formula?

I don't remember how I got this.
Do we have:

$$\Delta M=\frac{1}{2 \pi t} \int_{|x-\xi|=t} \Delta_{\xi}g d{\xi}$$

although $\xi$ appears at the limit of integration? (Thinking)

 

[Euge]

Switching to polar coordinates we get the limit $\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{t<|r|<t+\Delta t} \Delta_{\xi} g r dr$, which is equal to that what you wrote , right?

No. It should be

$$\lim_{\Delta t\to 0} \int_0^{2\pi} \int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi)\, r\, dr\, d\phi$$

I don't remember how I got this.
Do we have:

$$\Delta M=\frac{1}{2 \pi t} \int_{|x-\xi|=t} \Delta_{\xi}g d{\xi}$$

although $\xi$ appears at the limit of integration? (Thinking)

The $d\xi$ should be $ds$. Otherwise, the formula is correct.

 

[evinda]

No. It should be

$$\lim_{\Delta t\to 0} \int_0^{2\pi} \int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi)\, r\, dr\, d\phi$$

Could you explain further to me how you got this?

The $d\xi$ should be $ds$. Otherwise, the formula is correct.

Having $ds$ instead of $d\xi$ , in respect to which variable do we integrate?

 

[Euge]

Could you explain further to me how you got this?

The region $t < \lvert x - \xi \lvert < t + \Delta t$ is an annulus centered at $x =(x_1,x_2)$ with inner radius $t$ and outer radius $t + \Delta t$. So the radial variable $r$ ranges from $t$ to $t + \Delta t$ and the polar variable $\phi$ ranges from $0$ to $2\pi$. This leads to the double integral representation

$$\int_{t < \lvert x - \xi \rvert < t + \Delta t} \Delta_\xi g\, d\xi = \int_0^{2\pi}\int_t^{t + \Delta t} \nabla^2g(x_1 + r\cos \phi, x_2 + r\sin \phi) \,r\, dr\, d\phi$$

Having $ds$ instead of $d\xi$ , in respect to which variable do we integrate?

The integration is done with respect to arclength; the integral over the circle $\lvert x - \xi\rvert = t$ is a line integral. If you use $d\xi$ instead, the integral would be zero because the circle is a null set in the plane.