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### Messages - Biao Zhang

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1
##### HA4 / Re: HA4 problem 1
« on: March 19, 2015, 11:21:26 PM »
(a)
$$X''+\lambda X=0 \Rightarrow X(x)=C\cos \omega x +D\sin \omega x$$
Let $\lambda = \omega^2$
$$X'(x)=-C\omega \sin \omega x + D \omega \cos \omega x$$
Since $X'(0)=D\omega$ and $X(0)=C$, we have:
$$X'(0)=\alpha X(0)$$
$$\Rightarrow$$

D\omega =\alpha C \rightarrow
D=\frac{\alpha C}{\omega}

$$X'(l)=-c\omega \sin \omega l+D\omega \cos \omega l$$
and
$$X(l)=c\cos \omega l + D \sin \omega l$$
$$\implies$$
$$X'(l)=-\beta X(l)$$
$$-C\omega \sin(\omega l)+D\omega \cos(\omega l)=-\beta C\cos(\omega l)-\beta D\sin(\omega l)$$
$$\Rightarrow$$
$$(D\omega + \beta C)\cos(\omega l)=(C\Omega -\beta D)\sin(\omega l)$$
$$\Rightarrow$$
$$(\alpha C+\beta C)\cos(\omega l)=(C \omega -\frac{\alpha C\beta}{\omega})\sin(\omega l)$$
$$\rightarrow (\alpha + \beta)\omega \cos(\omega l)=(\omega^2-\alpha \beta)\sin(\omega l)$$
$$\Rightarrow \tan(\omega l)=\frac{(\alpha + \beta)\omega}{\omega^2-\alpha \beta}$$
Sub to $X(x)=C\cos \omega x +D\sin \omega x$ and By 8
$$X_n=\omega_n \cos(\omega_n x)+\alpha\sin(\omega_n x)$$
(b)
$\lambda = -\gamma$,$-X''=-\gamma^2\implies X''-\gamma^2X=0$,\\
char. is $k^2-\gamma^2=0\implies$ \\
$k=\pm \gamma \implies X(x)=Ae^{\gamma x}+Be^{-\gamma x}=A\cosh \gamma x + B\sinh \gamma x$
$$X'(x)=A\gamma \sinh \gamma x +B\gamma \cosh \gamma x$$
$$X'(0)=B\gamma \textbf{ and } X(0)=A$$
$$X'(0)=\alpha X(0)\implies$$

B\gamma = \gamma A \implies B=\frac{\alpha A}{\gamma}

$$X'(l)=A\gamma \sinh \gamma l + B\gamma \cosh \gamma l$$
and
$$X(l)=A\cosh \gamma l +B \sinh \gamma l$$
$$\implies$$
$$A\gamma \sinh \gamma l +B\gamma \cosh \gamma l=-\beta A\cosh \gamma l-\beta B \sinh \gamma l$$
$$\implies$$
$$(A\gamma +\beta B)\sinh\gamma l = -(\beta A+B\gamma)\cosh \gamma l$$
$$By (9) \implies$$
$$(A\gamma^2+\beta \alpha A)\sinh \gamma l =-(\beta \gamma A+ \alpha A \gamma )\cosh \gamma l$$
$$\implies$$
$$\frac{\sinh \gamma l}{\cosh \gamma l}=-\frac{(\alpha + \beta)\gamma }{\gamma^2 +\alpha \beta}\implies \tanh(\gamma l)=\frac{(\alpha +\beta)\gamma}{\gamma^2+\alpha \beta}$$
$$X(x)=A\cosh (\gamma x)+B\sinh(\gamma x)$$
By (9)
$$X_n(x)=\cosh(\gamma_n x)+\frac{\alpha}{\gamma_n}\sinh (\gamma_n x)$$

2
##### HA8 / Re: Solution of Question 2
« on: March 19, 2015, 09:36:50 PM »
(a)

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

In $\Delta u \geqq 0$, $u$ is twice differentiable, so the spherical mean is continuous and we have:

\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}

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:

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

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

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:

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

3
##### Test 2 / Re: TT1 problem 2
« on: March 13, 2015, 07:41:59 AM »
(a)

If $\lambda = 0$ is an eigenvalue than: $X' = 0$ $\implies$ $X'(x) = Ax + B$, and
$X'(x) = A$,$X(0) = 0 = B$. Therefore $X(x) = Ax$.
$$X'(l)+\alpha X(l)=A+\alpha Al=A(1+\alpha l) =0$$
In order for $X(x)$ not to be trivial we must have $A \neq 0$. Therefore $(1 + \alpha l) = 0$
$\implies$ $Î± = âˆ’1/l$.

(b)

If $\lambda <$0. Let $\lambda = -\beta ^2$ where $\beta > 0$. Than $X'' - \beta ^2X = 0$, where
$X(x) = A\cosh(\beta x) + B\sinh(\beta x)$. If $X(0) = A = 0$, than $X(x) = B\sinh(\beta x)$
and $X' = B\beta cosh(\beta x$).
$$X'(l)+\alpha X(l)=B(\beta \cosh(\beta l)+\alpha\sinh(\beta l))$$
For $X(x)$ not to be trivial we must have $B\neq 0$ then $\beta \cosh(\beta l)+\alpha \sinh(\beta l)=0$. Then since $\alpha \neq 0$ we have that $\beta \cos (\beta l)>0$. Therefore
$$-\frac{\beta}{\alpha}=\tanh$$
Then if $\alpha < 0\implies y=\frac{1}{\alpha}$ intersects $\tanh(\beta l)$
$$(\tanh(\beta l))'|_{\beta=0}=l$$
When $\beta =0$. Note $(-\frac{\beta}{\alpha})'=\frac{1}{\alpha}$. Than $l>\frac{1}{\alpha}$ . Therefore the only eigenvalue if found when $\alpha < \frac{1}{l}$.

(c)

If $\lambda =\beta^2>0$ we have:
$$X'(l)+\alpha X(l) = B\beta \cos(\beta l) + \alpha B \sin(\beta l)=0$$
For $X$ to be non-trivial, we have $-\frac{\beta}{\alpha}=\tan (\beta l)$. Note $\alpha \neq 0$. We get:
$$\lambda _m=\frac{\beta^2_n \pi}{2l}+\frac{n\pi}{l}\leq\beta_n\leq\frac{3\pi}{2l}+\frac{(n-1)\pi}{n}$$

(d)

$$\tan(\beta l)'=l\sec^2(\beta l)=\frac{1}{\cos^2(\beta l)}$$
If $\frac{-1}{\alpha}>l$ we have an intersection with the first branch. If $\frac{-1}{l}<\alpha<0$ we have that
$$\frac{\pi}{2l}+\frac{(n-1)\pi}{l}<\beta_n<\frac{3\pi}{2l}+\frac{(n-1)\pi}{n}$$
also $0<\beta_0<\frac{\pi}{2}$.\\
If $\alpha<0$, There is no intersection.

4
##### HA7 / Re: HA7 problem 4
« on: March 12, 2015, 09:23:24 PM »
(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 $(\pi,2\pi)$:

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

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

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

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

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

5
##### HA7 / Re: HA7 problem 3
« on: March 12, 2015, 09:22:17 PM »
(a)
Take  $r \ge 0$. Let $a >0$.  $(x,y) \mapsto (r,\theta)$ with Dirichlet BC: $\{u = h(\theta) : r = a\}$, $u(r,\theta)$ has a solution of the form:

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

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

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

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

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

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

$$A_0=\frac{-(-1)^0-1}{\pi a^0(0-1)}=\frac{2}{\pi}$$

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

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

$$\implies u(r,\theta) =\frac{1}{\pi} + \sum_{n=1}^{\infty} (\frac{r}{a})^n \frac{-1-(-1)^n}{(n^2-1) \pi } \cos(n \theta) \phantom{\ }$$
For $n=2m-1$,$m=1,2,3,...$ as 'odd' coefficients are 0.

(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 $r \rightarrow \infty$.

BC: $\{u = h(\theta) : r = a\}$, $u(r,\theta)$ has a solution of the form:

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

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

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

Substituting in $f(\theta)$, mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $(0,\pi)$ and $(\pi,2\pi)$, gives us the solution on the exterior:

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

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

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

so we have::

$$A_0=\frac{2}{\pi}$$

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

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

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

6
##### HA7 / Re: HA7 problem 2
« on: March 12, 2015, 09:20:56 PM »
(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 $\{A,B\} \in \mathbb{R}$.

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

7
##### HA7 / Re: HA7 Problem 1
« on: March 12, 2015, 09:20:08 PM »
(a)
in the polar variables:\\

$$\Delta u(r,\theta,\phi) := u_{rr} + \frac{2}{r} u_{r} + \frac{1}{r^2}(u_{\theta\theta} +\cot(\theta)u_\theta + \frac{1}{\sin(\theta)^2}u_{\phi\phi}) = k^2 u$$

Since we are looking for $u(r,\theta,\phi)=u(r)$ . Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, so we have:

$$\Delta u(r) := u_{rr} + \frac{2}{r} u_{r} = k^2 u$$

Substituting in $u(r) = \frac{v(r)}{r}$ then:

$$u_{r} = \frac{v_r}{r} - \frac{v}{r^2}$$,
$$u_{rr} = \frac{v_{rr}}{r} - 2\frac{v_{r}}{r^2}+ 2\frac{v}{r^3}$$

$$\implies$$

$$\Delta u := (\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}) + \frac{2}{r} ( \frac{v_r}{r} - \frac{v}{r^2} ) = k^2 \frac{v}{r}$$

$$\implies$$

$$\frac{v_{rr}}{r}=k^2\frac{v}{r}\implies v_{rr} = k^2 v$$

Since we have that $k >0$, $k^2 >0$,  :

$$v(r) = c_1 e^{k r} + c_2 e^{-k r} = c_3 \cosh(k r) + c_4 \sinh(k r)$$

$$\implies u(r) = \frac{1}{r} v(r) = c_1\frac{e^{k r}}{r}+c_2\frac{e^{-k r}}{r} =c_3\frac{\cosh(k r)}{r} + c_4 \frac{\sinh(k r)}{r}\phantom{\ }$$

where $\{c_1,c_2,c_3,c_4\} \in \mathbb{R}$

(b)

Similar to a:\\

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

Since we are looking for $u(r,\theta,\phi)=u(r)$ . Then $u_{\theta\theta} = u_{\theta} = u_{\phi\phi} = 0$, so we have:

$$\Delta u(r) := u_{rr} + \frac{2}{r} u_{r} = -k^2 u$$

Substituting in $u(r) = \frac{v(r)}{r}$ then:

$$u_{r} = \frac{v_r}{r} - \frac{v}{r^2}$$,
$$u_{rr} = \frac{v_{rr}}{r} - 2\frac{v_{r}}{r^2}+ 2\frac{v}{r^3}$$

$$\implies$$

$$\Delta u := (\frac{v_{rr}}{r} - \frac{2 v_{r}}{r^2} + \frac{2 v}{r^3}) + \frac{2}{r} ( \frac{v_r}{r} - \frac{v}{r^2} ) = -k^2 \frac{v}{r}$$

$$\implies$$

$$v_{rr} = - k^2 v$$

Since we have that $k >0$, $k^2 >0$,  :

$$v(r) = c_1 \sin (k r) + c_2 \cos (k r)$$

$$\implies u(r) = \frac{v}{r} = c_1\frac{\sin (k r)}{r}+c_2\frac{\cos (k r)}{r}$$

where $\{c_1,c_2\} \in \mathbb{R}$

8
##### HA5 / Re: HA5 Problem 3
« on: February 27, 2015, 07:16:47 PM »
[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.

9
##### HA5 / Re: HA5 Problem 1
« on: February 27, 2015, 07:15:34 PM »
[a:]
\begin{equation*}
a_0 = \frac{1}{l} \int_{-l}^{l} e^{zx}dx\\
= \frac{1}{zl} e^{zx}\big|_{-l}^{l}\\
= \frac{1}{zl}(e^{zl} - e^{-zl} )
\end{equation*}
Then we have
\begin{equation*}
a_n = \frac{1}{2l} \int_{-l}^{l} e^{zx}\big( \exp{(\frac{in\pi x}{l})} + \exp{(-\frac{in\pi x}{l})} \big) dx \\
= \frac{1}{2l} \int_{-l}^{l} \exp{((z + \frac{in\pi }{l})x)} dx +\frac{1}{2l} \int_{-l}^{l} \exp{((z -\frac{in\pi }{l})x)} dx \\
= \frac{1}{2l} \frac{\exp{((z +\frac{in\pi}{l})x)}}{z +\frac{in\pi }{l}} \big|_{-l}^{l} + \frac{1}{2l} \frac{\exp{((z -\frac{in\pi}{l})x)}}{z -\frac{in\pi }{l}} \big|_{-l}^{l}\\
= \frac{1}{2l} \frac{e^{zl + in\pi }}{z +\frac{in\pi }{l}} -\frac{1}{2l} \frac{e^{-zl - in\pi }}{z +\frac{in\pi }{l}} + \frac{1}{2l} \frac{e^{zl - in\pi }}{z -\frac{in\pi }{l}} - \frac{1}{2l} \frac{e^{-zl + in\pi }}{z -\frac{in\pi }{l}}\\
=\frac{1}{2l}(-1)^n \big[e^{zl} - e^{-zl} \big]\big(\frac{1}{z +\frac{in\pi }{l}} + \frac{1}{z -\frac{in\pi }{l}}\big)\\
=\frac{(-1)^n}{l} \big(e^{zl} - e^{-zl} \big) \frac{z}{z^2 + (\frac{n \pi}{l})^2}\\
= \frac{(-1)^nzl \big(e^{zl} - e^{-zl} \big)}{(zl)^2 + (n \pi)^2}
\end{equation*}
Similarly, for $b_n$:\\
\begin{equation*}
= {- }\frac{(-1)^n {\pi }{n } \big(e^{zl} - e^{-zl} \big)}{(zl)^2 + (n \pi)^2 }
\end{equation*}
So, exception values are $lz=in\pi$ or $lz=-in\pi$

e^{zx} = (e^{zl} - e^{-zl}) [\frac{1}{2l} + \sum_{n=1}^{\infty} \frac{(-1)^nzl }{(zl)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{(zl)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}}  ].

[b:]

\cos{\omega x} = \frac{(e^{i\omega} - e^{-i\omega})}{2}

Let $z = i\omega$ or $z=-i\omega$, from Part (a) we have:
\begin{equation*}
\frac{1}{2} [ (e^{i\omega l} - e^{-i\omega l}) (\frac{1}{2i \omega l} + \sum_{n=1}^{\infty} \frac{(-1)^n i\omega l }{-(\omega l)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{-(\omega l)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}}  )  - \\
(e^{i\omega l} - e^{-i\omega l}) (\frac{1}{-2i \omega l} + -\sum_{n=1}^{\infty} \frac{(-1)^n i\omega l }{-(\omega l)^2 + (n \pi)^2 } \cos{\frac{n\pi x}{l}} - \sum_{n=1}^{\infty} \frac{(-1)^n n\pi }{-(\omega l)^2 + (n \pi)^2 } \sin{\frac{n\pi x}{l}}  )  ]\\
= \sin{\omega l} (\frac{1}{\omega l} -  \sum_{n=1}^{\infty} 2(-1)^n  \frac{\omega l \cos{\frac{n\pi x}{l}}}{(n\pi)^2 -( \omega l)^2}   ).
\end{equation*}
Exception values occur when $(\omega l)^2=(n\pi)^2$. When $\omega =0$, we have an indeterminate form which is defined in the limit as $\omega$ approaches 0.
Similarly, for sin $\omega x$:
\begin{equation*}
\sin{\omega x} = -2\sin{\omega l} \sum_{n=1}^{\infty} 2(-1)^n\frac{n \pi \sin{\frac{n\pi x}{l}}}{(n\pi)^2 -( \omega l)^2}
\end{equation*}
[c:]
Just as in (b), we have $\cosh{\eta x} = \frac{e^{\eta x} + e^{-\eta x}}{2}$ and can make the substitution $z = \eta$ or $z = -\eta$. Then  \begin{equation*}  \cosh{\eta x} = \sin{\eta l}\left(\frac{1}{\eta l} -  \sum_{n=1}^{\infty} 2(-1)^n  \frac{\eta l \cos{\frac{n\pi x}{l}}}{(n\pi)^2 +( \eta l)^2}  \right) \end{equation*} And for $\sinh{\eta x}$, we have \begin{equation*} \sinh(\eta x) = \sinh{\eta l}\left(- 2 \sum_{n=1}^{\infty} (-1)^n  \frac{n\pi  \sin{\frac{n\pi x}{l}}}{(n\pi)^2 +( \eta l)^2}  \right). \end{equation*}

10
##### Test 1 / Re: TT1 Problem 3
« on: February 13, 2015, 06:29:14 AM »
\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 $t$.

11
##### Test 1 / Re: TT1 Problem 2
« on: February 13, 2015, 05:28:50 AM »
To solve the question by using D'Alemeber formula:\\
\begin{multline*}
u(x,t)= \underbrace{\frac{1}{2}\bigl( G(x+ct)+G(x-ct) \bigr)}_1
+ \underbrace{\frac{1}{2c}\int_{x-ct} ^{x+ct} H(x')\,dx'}_2 +
\underbrace{\frac{1}{2c} \int_0^t \int _{x-c(t-t')} ^{x+c(t-t')} F(x',t')\, dx' dt'}_3
\end{multline*}
we have
\begin{align*}
c=2\\
G(x)=e^{-x^2}\\
H(x)=xe^{-x^2}\\
F(x,t)=xte^{-x^2}
\end{align*}
\begin{align*}
(1): \frac{1}{2}(e^{-(x+2t)^2}+e^{-(x-2t)^2})\\
(2): \frac{1}{4}\int_{x-2t} ^{x+2t} H(x')\,dx'\\
\hspace{1cm} =\frac{1}{8}(e^{-(x-2t)^2}-e^{-(x+2t)^2})
\end{align*}
combine (1) and (2) we have:
\begin{equation*}
\frac{3}{8}e^{-(x+2t)^2} +\frac{5}{8}e^{-(x-2t)^2}
\end{equation*}
now we need find (3):
\begin{align*}
\frac{1}{4} \int_0^t \int _{x-2(t-t')} ^{x+2(t-t')} x't'e^{-x'^2}\, dx' dt'\\
\frac{1}{8}  \int_0^t t'(e^{-(x+2t-2t')^2}-e^{-(x-2t+2t')^2})\, dt'
\end{align*}
This part i don't know how to integrate this part . but answer should be:
\begin{equation*}
\frac{3}{8}e^{-(x+2t)^2} +\frac{5}{8}e^{-(x-2t)^2}+\frac{1}{8}  \int_0^t t'(e^{-(x+2t-2t')^2}-e^{-(x-2t+2t')^2})\, dt'
\end{equation*}

12
##### HA1 / Re: HA1 problem 3
« on: January 22, 2015, 10:27:42 PM »
HA1-3

13
##### HA1 / Re: HA1 problem 1
« on: January 22, 2015, 10:25:39 PM »
i think equation has some typo. should be y instead of 7

14
##### HA1 / Re: HA1 problem 6
« on: January 22, 2015, 10:23:35 PM »
HA1-6

15
##### HA1 / Re: HA1 problem 4
« on: January 22, 2015, 10:19:41 PM »
HA1-4

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