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### Messages - Elliot Jarmain

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1
##### Term Test 2 / Re: TT2--P4
« on: March 23, 2018, 08:43:13 AM »
Making a seperation of variables $u(r, \theta) = R(r)\Theta(\theta)$ yields
\begin{gather*}
R''\Theta + \frac{1}{r}R'\Theta + \frac{1}{r^2}R\Theta'' = 0 \\
r^2\frac{R''}{R} + r\frac{R'}{R} + \frac{\Theta''}{\Theta} = 0 \\
r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda, \quad \frac{\Theta''}{\Theta} = -\lambda
\end{gather*}
Solving the $\Theta$ equation:
\begin{equation*}
\Theta(\theta) =
\begin{cases}
A\theta + B, &\text{if $\lambda = 0$;}\\
Ae^{\sqrt{-\lambda}\theta} + Be^{-\sqrt{-\lambda}\theta} &\text{if $\lambda < 0$;}\\
A \cos{\sqrt{\lambda}\theta} + B \sin{\sqrt{\lambda}\theta} &\text{if $\lambda > 0$.}
\end{cases}
\end{equation*}
Plugging the boundary conditions (2) yields the trivial solution for $\lambda \leq 0$. And for $\lambda > 0$:
\begin{align*}
u|_{\theta=0}=0 &\implies A = 0 \\
u|_{\theta=2\pi}=0 &\implies B \sin{\sqrt{\lambda}2\pi} = 0 \\ &\implies
\sqrt{\lambda} 2\pi = n \pi, \quad n = 1,2,3, \dots
\end{align*}
Therefore $\lambda_n = \frac{n^2}{4}$ and $\Theta_n(\theta) = \sin{\frac{n\theta}{2}}$ for $n = 1,2,3, \dots$
To solve the $R$ equation we first assume that $R$ has the form $R(r) = r^m$ for some $m$. This yields
\begin{equation*}
m(m-1) +m = \lambda \implies m^2 = \lambda
\implies m = \pm \frac{n}{2}
\end{equation*}
So $R(r) = A_n r^{\frac{n}{2}} + B_n r^{-\frac{n}{2}}$ and
\begin{equation*}
u(r, \theta) =
\sum_{n=1}^\infty (A_n r^{\frac{n}{2}} + B_n r^{-\frac{n}{2}}) \sin{\frac{n\theta}{2}}
\end{equation*}
Since $r < 9$, we set $B_n = 0$ for all $n$ to avoid singularities at the origin. The boundary condition (3) yields:
\begin{equation*}
u_r|_{r=9}= \sum_{n=1}^\infty (A_n \left(\frac{n}{2}\right)9^{\frac{n}{2} - 1}) \sin{\frac{n\theta}{2}} =
\sum_{n=1}^\infty \left(\frac{A_n3^{n-2}n}{2}\right) \sin{\frac{n\theta}{2}} = 1
\end{equation*}
So
\begin{equation*}
A_n = \frac{2}{3^{n-2}n\pi}
\int_0^{2\pi}\sin{\frac{n\theta}{2}}
= \frac{2}{3^{n-2}n\pi} \frac{2}{n}
\left. \left(
-\cos{\frac{n\theta}{2}}
\right) \right|_0^{2\pi}
= \frac{4}{3^{n-2}n^2\pi}
(1 - (-1)^n)
\end{equation*}
\begin{equation*}
A_n =
\begin{cases}
\frac{8}{3^{n-2}n^2\pi} &\text{if $n$ odd;}\\
0 &\text{if $n$ even.}
\end{cases}
\end{equation*}
Therefore
\begin{equation*}
u(r, \theta) =
\sum_{n\geq1, \, n \, \text{odd}} \frac{8r^{\frac{n}{2}}}{3^{n-2}n^2\pi}  \sin{\frac{n\theta}{2}}
\end{equation*}

2
##### Quiz-3 / Re: Q3-T5102
« on: February 11, 2018, 09:13:10 AM »
For the problem to be well-posed, $\alpha \neq \frac{1}{c}$.

3
##### Quiz-3 / Re: Q3-T5102
« on: February 11, 2018, 08:46:36 AM »
Based on examples from the online textbook, I am pretty sure we are allowed to integrate at that stage. Also Example 2 b from Section 2.6 seems to discard the constants of integration so I will follow that convention. Here is my solution to part 1:

We know $u(x,t)$ is of the form $u(x,t) = \varphi(x+ct)+\psi(x-ct)$.
For $x>ct$ we use the initial conditions:
\begin{align}
u|_{t=0}= \phi (x)
&\implies \varphi(x)+\psi(x) = \phi(x)  \\
u_t|_{t=0}= c\phi'(x)
&\implies \varphi'(x)-\psi'(x)=\phi'(x)
\end{align}
Therefore $\varphi(x)=\phi(x)$ and $\psi(x)=0$ when $x>0$. So

\text{when } x>ct

For $0<x<ct$ we use the boundary condition:
\begin{align}
(u_x+\alpha u_t)|_{x=0}=0
&\implies \varphi'(ct) + \psi'(-ct)
+ \alpha c\varphi'(ct) -\alpha c\psi'(-ct) = 0 \\
&\implies \frac{1}{c}\varphi(ct) -                              \frac{1}{c}\psi(-ct)
+ \alpha \varphi(ct) +\alpha\psi(-ct) = k
\end{align}
Letting $t=-\frac{x}{c}$ and rearranging yields
\begin{align}
\psi(x) &= - \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\text{for } x<0 \\
&= - \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\phi(-x) + k
\end{align}
So

u(x,t) = \phi(x+ct)- \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\text{when } 0<x<ct

(we set $k=0$ so that $u$ is continuous at $x=ct$)

Therefore,

u(x,t) =
\left\{ \begin{aligned}
&\phi(x+ct), &&x>ct \\
&\phi(x+ct) - \left(
\frac{\alpha c + 1}{\alpha c -1} \right)
\phi(ct-x), &&0<x<ct
\end{aligned}
\right.

4
##### Quiz-1 / Re: Q1-T5102-P2
« on: January 25, 2018, 10:37:02 AM »
Along the characteristic curves:
\begin{gather*}
\frac{dt}{1} = \frac{dx}{t^2+1} \\
\int{t^2+1 \, dt} = \int{dx}\\
\frac{t^3}{3} + t + C = x
\end{gather*}
Also $f(x,t,u) = 0$ implies $u$ is constant along the characteristic curves,
therefore:
\begin{equation*}
u = \phi(x-\frac{t^3}{3}-t)
\end{equation*}

5
##### Quiz-1 / Re: Q1-T5102-P2
« on: January 25, 2018, 09:49:52 AM »
My plot for the characteristics $C = t^3 +t - x$ is attached

6
##### Quiz-1 / Re: Q1-T5102-P1
« on: January 25, 2018, 08:29:42 AM »
\begin{equation*}
u_{xy} = 2u_x
\end{equation*}

Let $v = u_x$:

\begin{equation*}
v_y = 2v
\end{equation*}

Solving this equation:

\begin{gather*}
\frac{d v}{d y} = 2v\\
\implies  \int{\frac{d v}{v}}
= \int{2 \, d y} \\
\implies ln|v| = 2y + \phi(x)\\
\implies v = \pm e^{\phi(x)}e^{2y}\\
\implies v = \varphi'(x) e^{2y}\\
\end{gather*}

Plugging $u_x$ back in for $v$:
\begin{gather*}
\implies  u_x = \varphi'(x) e^{2y} \\
\implies u = \varphi(x) e^{2y} + \psi(y)
\end{gather*}

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