Toronto Math Forum

(a)
\begin{equation}
u(y) \leq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S
\end{equation}
In $\Delta u \geqq 0$, $u$ is twice differentiable, so the spherical mean is continuous and we have:
\begin{equation}
\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \rightarrow u(y) \label{2.a.1}
\end{equation}
Thus, to prove $ u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S $, it is sufficient to show that $\frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S$ increases with $r > 0$. By the divergence theorem we have for a $C^1$ function $F$ on a compact set $\Omega$ with boundary $\partial\Omega = \Sigma$:

$$ \int_\Omega (\nabla \cdot F) \,dV = \int_\Sigma (F \cdot n) \, dS \text{, namely: } $$

$$\int_{B(y,r)} (\nabla \cdot \nabla )(x) \,dV = \int_{B(y,r)} \Delta u(x) \,dV = \int_{S(y,r)} (\nabla u \cdot n)(x) \, dS = \int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS $$

As $\Delta u \geqq 0$, we then have:

$$ 0 \leqq \int_{B(y,r)} \Delta u(x) \,dV = \int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS $$

Parametrizing by the normal $\nu = \frac{x-y}{|x-y|}$ on the boundary where $|x-y| = r$, we have: $\frac{\partial u}{\partial n} (x) = \partial_r u(y+ r \nu)$ on $ T : \{ \nu : | \nu | = 1 \} $, and $ dS = r^{n-1} d \nu $

Then:

$$ \int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS = \int_{T} \partial_r( u(y+ r \nu)) r^{n-1} \, d \nu $$

By differentiating under the sign with $\partial_r$ and pulling the $r^{n-1}$ term out of the integral, we have:

$$ = r^{n-1} \partial_r \int_{T} u(y+ r \nu) \, d \nu = r^{n-1} \partial_r (r^{1-n} \int_{S(y,r)} u(x) \, d S) $$

$$ = r^{n-1} \partial_r(\sigma_n \frac{r^{1-n}}{\sigma_n } \int_{S(y,r)} u(x) \, d S ) = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S ) $$

As $r>0$ and $\sigma_n>0$ by (2) we have:

$$ 0 \leqq \int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S ) \implies $$

\begin{equation}
0 \leqq \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )
\end{equation}
Then by (2) as $r \rightarrow 0$ we have equality in (1), and as $r$ increases the spherical mean is non-decreasing by (3), $u(y)$ constant. Thus we have (1) for $r \geqq 0$:

$$ u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \phantom{O}  $$

(b)

$$u(y) = \frac{\sigma_n (n r^n)}{(n \sigma_n) r^n} u(y) = \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u(y) $$

By (1) we have that:

$$ u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S $$

So because $ \{r, \omega_n, \sigma_n\} \geqq 0 $ we have that:

$$\implies \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u(y) \leqq \frac{\sigma_n \int_{0}^{r} t^{n-1}\, dt}{\omega_n r^n} \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S $$

$$ = \frac{1}{\omega_n r^n} \sigma_n\int_{0}^{r}t^{n-1} \,d t \frac{1}{\sigma_n r^{n-1}} \int_T u(y +r \nu) r^{n-1} \, d \nu $$

Where we parameterized as in part a. with $\nu = \frac{x-y}{|x-y|}$ on the boundary $|x-y| = r$, and $ (x) = u(y+ r \nu)$ on $ T : \{ \nu : | \nu | = 1 \} $, $ dS = r^{n-1} d \nu $. Canceling terms gives:

$$ = \frac{1}{\omega_n r^n} \int_{0}^{r}t^{n-1} \,d t \int_T u(y +r \nu) \, d \nu $$

Which allows us to change the order of integration to yield our result:

$$ = \frac{1}{\omega_n r^n} \int_{0}^{r} t^{n-1} \int_T u(y +r \nu) \, d \nu d t = \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u \, d V $$

$$ \implies u(y) \leqq \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u \, d V \phantom{O} $$

(c)

In part a. we used the fact that $ \Delta u \geqq 0 $, and $ \{ \sigma_n, r \} \geqq 0$ to show that the spherical mean is non-decreasing in $r$:

$$ 0 \leqq \int_{B(y,r)} \Delta u(x) \,dV = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )$$

$$ \implies 0 \leqq \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S ) $$

Because we have equality at $r \rightarrow 0$:

$$ \lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \rightarrow u(y) $$
and $u(y)$ is constant with respect to $r$, we showed that for all $r > 0$

$$ u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S $$

If instead $ \Delta u \leqq 0$, we would have the inequalities:

$$ 0 \geqq \int_{B(y,r)} \Delta u(x) \,dV = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )$$

$$ \implies 0 \geqq \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S ) $$

And $ \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S$ is non-increasing with $r$, which gives us by the same argument that for all $r>0$

$$ u(y) \geqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \phantom{O} \square$$

In part b. we merely integrated this result radially to extend the inequality for the interior of the sphere. Doing the same in the case of $ \Delta u \leqq 0$ yields us:

$$ u(y) \geqq \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u \, d V \phantom{O} $$

(a)
Let $a > 0$. With BC $h(\theta) = f(\theta)$. It was derived that by changing to polar coordinates $(x,y) \mapsto (r,\theta)$ has a solution of the form:

$$ u(r,\theta) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n (A_n \cos(n \theta) + B_n \sin(n \theta)) $$

Differentiating this equation for $u(r,\theta)$ with respect to $r$ gives us our boundary condition at $r=a$, $f(\theta)$.

$$ u_{r}(r,\theta) = \sum_{n=1}^{\infty} n r^{n-1} (A_n \cos(n \theta) + B_n \sin(n \theta)) $$

$$ u_{r}(a,\theta) = \sum_{n=1}^{\infty} n a^{n-1} (A_n \cos(n \theta) + B_n \sin(n \theta)) = f(\theta) $$

$$ \text{Notice that: }\int_{0}^{2\pi}f(\theta) d \theta = \int_{0}^{\pi} \sin (\theta)d \theta + \int_{\pi}^{2\pi}0 d \theta = 1-1+0 = 0 $$

So $f$ has a free coefficient and we may find a unique solution up to a constant $C \in \mathbb{R}$. We have Fourier coefficients at $r = a$:

$$ A_n = \frac{1}{\pi n a^{n-1}} \int_{0}^{2\pi}h(\phi)\cos(n\phi)d\phi $$

$$ B_n = \frac{1}{\pi n a^{n-1}} \int_{0}^{2\pi}h(\phi)\sin(n\phi)d\phi $$

Substituting in $f(\theta)$ and splitting the integral between $(0,\pi)$ and $(\pi,2\pi)$:

$$ \implies A_n = \frac{1}{\pi n a^{n-1}} \int_{0}^{2\pi}f(\phi)\cos(n\phi)d\phi = \frac{1}{\pi n a^{n-1}}( \int_{0}^{\pi}\sin(\phi)\cos(n\phi)d\phi + \int_{\pi}^{2\pi}0\cos(n\phi)d\phi) $$

$$= \frac{1}{\pi n a^{n-1}}(\frac{-1-(-1)^n}{n^2-1}) \text{, as } n \in \mathbb{N}$$

and $A_0=0$

$$ \implies B_n = \frac{1}{\pi n a^{n-1}} \int_{0}^{2\pi}f(\phi)\sin(n\phi)d\phi = \frac{1}{\pi n a^{n-1}}( \int_{0}^{\pi}\
sin(\phi)\sin(n\phi)d\phi + \int_{\pi}^{2\pi}0\sin(n\phi)d\phi) $$

$$= \frac{1}{\pi n a^{n-1}}(0) = 0 \text{, as } n \in \mathbb{N}$$

$$ \implies u(r,\theta) = \sum_{n=1}^{\infty} (\frac{r^n}{a^{n-1}}) \frac{-1-(-1^2)}{ \pi (n^3-n) } \cos(n \theta) + C, \phantom{\ } C \in \mathbb{R} \phantom{\ } $$

(b)

Let $a >0$. BC $h(\theta) = f(\theta)$.  $r\mapsto \frac{1}{r}$ : $\{ z : z \in \mathbb{R}_\infty , z \ge a \} \mapsto \{ \frac{1}{z} : \frac{1}{z} \in \mathbb{R}_{\infty} , \frac{1}{z} \le \frac{1}{a}\} $. Then $\frac{1}{r} \rightarrow 0$ as $r \rightarrow \infty$.

Now we have: $ \int_{0}^{2\pi}f(\theta) d \theta $ so our BC has a free coefficient, and the Neumann problem on the interior of the disk with BC: $\{u = h(\theta) : r = a\}$, $u(r,\theta)$ has a unique solution up to a constant $C \in \mathbb{R}$ of the form:

$$ u(r,\theta) = \sum_{n=1}^{\infty} r^n (A_n \cos(n \theta) + B_n \sin(n \theta)) + C $$

$$ A_n = \frac{1}{\pi n a^{n-1}} \int_{0}^{2\pi}h(\phi)\cos(n\phi)d\phi $$

$$ B_n = \frac{1}{\pi n a^{n-1}} \int_{0}^{2\pi}h(\phi)\sin(n\phi)d\phi $$

Mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$ gives us the equivalent solution on the exterior:

$$ \implies u(r,\theta) = \sum_{n=1}^{\infty} r^{-n} (A_n \cos(n \theta) + B_n \sin(n \theta)) + C $$

BC at $\frac{1}{r} = \frac{1}{a}$, $u_r = f(\theta)$ so we have:

$$u_{r} = \sum_{n=1}^{\infty} (-n) r^{-n-1} (A_n \cos(n \theta) + B_n \sin(n \theta)) $$

$$u_{r} \bigr|_{\frac{1}{a}} = \sum_{n=1}^{\infty} (-n) a^{-n-1} (A_n \cos(n \theta) + B_n \sin(n \theta))= f(\theta) $$

we have Fourier coefficients:

$$ A_n = - \frac{1}{\pi n a^{-n-1}} \int_{0}^{2\pi}h(\phi)\cos(n\phi)d\phi $$

$$ B_n = - \frac{1}{\pi n a^{-n-1}} \int_{0}^{2\pi}h(\phi)\sin(n\phi)d\phi $$

Substituting in $f(\theta)$ and splitting the integral between $(0,\pi)$ and $(\pi,2\pi)$:


$$ \implies A_n = - \frac{1}{\pi n a^{-n-1}} \int_{0}^{2\pi}f(\phi)\cos(n\phi)d\phi = - \frac{1}{\pi n a^{-n-1}}( \int_{0}^{\pi}\sin(\phi)\cos(n\phi)d\phi + \int_{\pi}^{2\pi}0\cos(n\phi)d\phi) $$

$$= - \frac{1}{\pi n a^{-n-1}}( \frac{-1-(-1)^2}{n^2-1}) = 0 \text{, as } n \in \mathbb{N}$$

$$ \implies B_n = - \frac{1}{\pi n a^{-n-1}} \int_{0}^{2\pi}f(\phi)\sin(n\phi)d\phi = - \frac{1}{\pi n a^{-n-1}}( \int_{0}^{\pi}\sin(\phi)\sin(n\phi)d\phi + \int_{\pi}^{2\pi}0\sin(n\phi)d\phi) $$

$$= - \frac{1}{\pi n a^{-n-1}}(0) = 0 \text{, as } n \in \mathbb{N}$$

$$ \implies u(r,\theta) = \sum_{n=1}^{\infty} (\frac{a^{n+1}}{r^n}) \frac{-1-(-1)^2}{ \pi (n^3-n) } \cos(n \theta) + C, \phantom{\ } C \in \mathbb{R} \phantom{\ } $$

(a)
 The form of Bessel's function,
 
 $$\frac {\partial^2 y}{\partial x^2}+\frac{1}{x} \frac{\partial y}{\partial x}+(1+\frac{n^2}{x^2})y=0$$
 
we can rewrite the equation:

$$ u_{rr} + \frac{1}{r} u_{r} +(1-\frac{s^2}{r^2})u = 0 $$

$$ \text{Let: } u(r) = v(i k r) \implies u_{r} = i k v_{r}, \phantom{\ } u_{rr} = - k^2 v_{rr} $$

$$ \implies \Delta_{(r,\theta)} u(r) := u_{rr} + \frac{1}{r} u_{r} - k^2 u = 0$$

$$- k^2 v_{rr} + \frac{1}{r} i k v_{r} - k^2 v = 0 $$

Since $ k >0$, $ k^2 \ne 0 $, so:

$$ v_{rr} - \frac{i}{k r} v_{r} + v = 0 $$

$ - \frac{i}{k r} = \frac{(-i)}{(1) k r} = \frac{(-i)}{(-i * i) k r} = \frac{\overline{i}}{\overline{i}} \frac{1}{i k r} = \frac{1}{i k r} $ so our ODE in $v(i k r)$, say $v(z)$ where $z = i k r$ is equivalent to:

$$ v_{rr} + \frac{1}{i k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$

Since $v_{rr} + \frac{1}{i k r} v_{r} = -v $ is a linear combination of Bessel funtion of $J_0 \& N_0$

$$ \implies v(z) = v(i k r) = u(r) = A_0 J_0(i k r) + B_0 N_0(i k r) \phantom{\ } $$


(b)
As (a):

$$ \Delta_{(r,\theta)} u(r,\theta) := u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^2} u_{\theta\theta} = u_{rr} + \frac{1}{r} u_{r} = - k^2 u $$

rewrite in Bessel equation:

$$ u_{zz} + \frac{1}{z} u_{z} +(1-\frac{s^2}{z^2})u = 0 $$

$$ \text{Let: } u(r) = v(k r) \implies u_{r} = k v_{r}, \phantom{\ } u_{rr} = k^2 v_{rr} $$

$$ \implies \Delta_{(r,\theta)} u(r) := u_{rr} + \frac{1}{r} u_{r} + k^2 u = 0 arrow k^2 v_{rr} + \frac{1}{r} k v_{r} + k^2 v = 0 $$

Let $z = k r$. Again, $k >0$ so $k^2 \ne 0$ :

$$ v_{rr} + \frac{1}{k r} v_{r} + v = v_{zz} + \frac{1}{z} v_{z} + v = 0 $$

Which is clearly a Bessel's differential equation with $s = n = 0$:

$$ v_{zz} + \frac{1}{z} v_{z} + v = v_{zz} + \frac{1}{z} v_{z} +(1-\frac{s^2}{z^2})v = 0 $$

So our solution for $v(z)$ is again $ v(z) = A J_0(z) + B N_0(z) $ for some $ \{A,B\} \in \mathbb{R} $.

$$ \implies v(z) = v(k r) = u(r) = A_0 J_0(k r) + B_0 N_0(k r) \phantom{\ } $$

[a:]
The function is even, so $b_n = 0$.  \begin{equation*} a_0 = \frac{2}{\pi} \int_{0}^{\pi} \sin x dx\\
= \frac{4}{\pi}. \end{equation*}  \begin{equation*} a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin x \cos nx dx \end{equation*}  Using $2\sin a \cos b = \sin(a+b) + \sin(a-b)$,  \begin{equation*} =\frac{1}{\pi} \int_{0}^{\pi} \sin ((n+1)x) + \sin ((1-n)x) dx\\
\end{equation*}
Notice that for $n = 1$, the solution is $0$.
\begin{equation*}  =-\frac{1}{\pi}\left[\frac{\cos((n+1)x)}{n+1} - \frac{\cos((n-1)x)}{n-1}  \right]_{0}^{\pi}\\ =-\frac{1}{\pi}\left[\frac{(-1)^{n+1}}{n+1} - \frac{1}{n+1} - \frac{(-1)^{n+1}}{n-1} + \frac{1}{n-1}  \right]_{0}^{\pi}\\ =\frac{1}{\pi}\big((-1)^n + 1 \big) \big(\frac{1}{n+1} + \frac{1}{n-1}  \big)\\ =\frac{-2}{\pi} \big((-1)^n + 1  \big) \big( \frac{1}{n^2 - 1} \big).\\
\end{equation*} 
Let $n=2k$. Thus, for $n>1$, \begin{equation*} |\sin x| = \frac{2}{\pi} -\frac{4}{\pi} \sum_{k=1}^{\infty} \frac{1}{(2k)^2 - 1}\cos (2kx). \end{equation*} 
[b:] Again, the function $|\cos x|$ is even, so $b_n = 0$. We proceed as above:  \begin{equation*} a_0 = \frac{2}{\pi} \int_{0}^{\pi} |\cos x| dx \\ =  \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos x dx -  \frac{2}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos x dx\\ = \frac{4}{\pi}\\ \end{equation*}  \begin{equation*} a_n =  \frac{2}{\pi} \int_{0}^{\pi} |\cos x|\cos{nx} dx \\ = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \cos x  \cos{nx} dx -  \frac{2}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos x \cos{nx} dx\\ = \frac{2}{\pi}\left(-2\frac{\cos{\frac{\pi n}{2}}}{n^2 - 1}\right)\\ \end{equation*}  Therefore,  \begin{equation*} |\cos x| = \frac{2}{\pi} - \frac{4}{\pi}\sum_{n=1}^{\infty} \frac{\cos{\frac{\pi n}{2}}}{n^2 - 1} \cos{nx}.   \end{equation*}  Both functions are continuous, so the Fourier series will converge to each function at every point.

\begin{align*}
\frac{dE(t)}{dt}=\frac{1}{2}\int _0^L \frac{d}{dt}(u_t^2+Ku_{xx}^2+\omega^2u^2)\,dx\\
=\frac{1}{2}\int _0^L (2u_tu_{tt}+2Ku_{xx}u_{xxt}+2\omega^2uu_t)\,dx\\
=\int _0^L (u_tu_{tt}+Ku_{xx}u_{xxt}+\omega^2uu_t)\,dx\\
\end{align*}
Integrate $K\int _0^L u_{xx}u_{xxt}\,dx$ by part.\\
let $a=u_{xx}$, $db=u_{xxt}dx$. we have $da=u_{xxx}dx$,$b=u_{xt}$, then\\
\begin{equation*}
K((u_{xx}u_{xt})\Big|_{0}^L - \int _0^L u_{xxx}u_{xt}\,dx) \\
\end{equation*}
Integrate by part again, let $a'=u_{xxx}$,$db'=u_{xx}dx$. We have $da'=u_{xxxx}dx$ and $b'=u_t$. Then:
\begin{align*}
\rightarrow K(u_{xx}u_{xt})\Big|_{0}^L-K(u_tu_{xxx})\Big|_{0}^L+\int _0^L Ku_{xxxx}u_t\, dx\\
\rightarrow \int _0^L u_t\underbrace{(u_{tt}+Ku_{xxxx}+\omega^2u)}_0 dx+K(u_{xx}u_{xt}-u_tu_{xxx})\Big|_0^L\\
=K(u_{xx}u_{xt}-u_tu_{xxx})\Big|_0^L
\end{align*}
Since $u_{xx}(0,t)=u_{xxx}(0,t)=u(L,t)=u_x(L,t)=0$\\
We find out that $\frac{dE(t)}{dt}=0$, therefore $E(t)$ is not depend on $t$.

HA1-3

HA1-6