Messages

1

Final Exam / Re: FE-4

« on: December 19, 2015, 12:06:34 AM »

\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin n \theta - \frac{1}{r} \frac{2}{3} \sin n \theta
\end{equation}

Sorry, there aren't supposed to be n's, those are supposed to be just $\sin \theta$.

So the solution should be:
\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin \theta - \frac{1}{r} \frac{2}{3} \sin\theta
\end{equation}

2

Final Exam / Re: FE-6

« on: December 18, 2015, 11:50:18 PM »

Actually $$\lim_{r\rightarrow 0}\frac{\sinh(r)}{r}=1$$
So that does not blow up at the origin either. However the problem with that function (corresponding to positive eigenvalues) is that it can never be $0$ at $r\neq 0$, so the boundary conditions can never be satisfied.

Thank you Bruce Wu, I can't believe I missed that.

3

Final Exam / Re: FE-3

« on: December 18, 2015, 11:42:17 PM »

Sorry, I meant n, odd! As in all the odd terms, my bad!

4

Final Exam / Re: FE-6

« on: December 18, 2015, 11:28:56 PM »

$\frac{T''}{T}=\frac{R''}{R}+\frac{2}{r}\frac{R'}{R}=-\omega^2$. One can check that there are only negative eigenvalues, so I have defined it this way. Let's look at the $R$ equation first
$$rR''+2R'=(rR)''=-\omega^2 rR$$
This is a second order ODE in $rR$, with solution $rR=A\cos(\omega r)+B\sin(\omega r)$. However, for $u$ to be bounded as $r\rightarrow 0$, we must have $A=0$. This also gives some insight as to why there were only negative eigenvalues to begin with: out of all possible functions, only $\frac{\sin(r)}{r}$ does not blow up at the origin. The boundary condition implies that $\omega=n\pi$. Therefore $R=B\frac{\sin(n\pi r)}{r}$.
Then solving the $T$ equation gives $T=C\cos(n\pi t)+D\sin(n\pi t)$, but the second boundary condition forces $D$ to be $0$. The general solution is, after absorbing and redefining constants:
$$u(r,t)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$
$u(r,0)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}=1\Rightarrow\sum_{n=1}^{\infty}A_n\sin(n\pi r)=r\Rightarrow A_n=2\int_{0}^{1}r\sin(n\pi r)dr=\frac{(-1)^{n+1}}{n\pi}$. Finally:
$$u(r,t)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$

5

Final Exam / Re: FE-3

« on: December 18, 2015, 11:24:32 PM »

We separate variables. So $u(x,t) = T(t)X(x)$. Our equation is then $\frac{T''}{9T} = \frac{X''}{X} =-  \lambda$.

So we solve the $X$ equation. We have $X'' = - \lambda X$. First consider $\lambda = \omega^2 > 0$. Then the solution is $X(x) = A\cos\omega x + B \sin \omega X$. Using the boundary conditions, we see that $A=0$, and $\omega 2 = n \pi \longrightarrow \omega = \frac{n \pi}{2}$.

Now consider negative eigenvalues. The solution will be $X(x) = A\cosh\omega x + B \sinh \omega X$. The boundary conditions clearly show that there is only a trivial solution, so we have no negative eigenvalues.

Likewise with the zero eigenvalue: the solution will be $X(x) = Ax + B$, and we can see that there is only a trivial solution. So no zero eigenvalues.

So then the eigenvalues are $\frac{n^2 \pi^2}{4}$ with eigenfunctions $X(x) = B \sin ( \frac{n \pi x}{2})$.

We can then solve the $T$ equation.

\begin{equation}
\frac{T''}{9T} = - \frac{n^2 \pi^2}{4} \longrightarrow T'' = - \frac{9 n^2 \pi^2 T}{4}
\end{equation}

So the solution is

\begin{equation}
T(t) = C\cos(3n\pi t/2) + D\ sin(3n\pi t/2)
\end{equation}

So overall the solution is:

\begin{equation}
u(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) \left( A_n\cos(3n\pi t/2) + B_n\sin(3n\pi t/2) \right)
\end{equation}

When we use the initial condition at $u(x,0)$, we see right away that $A_n = 0$. So:

\begin{equation}
u(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) B_n\sin(3n\pi t/2)
\end{equation}

\begin{equation}
u_t(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) B_n \frac{3 n \pi}{2} \cos(3n\pi t/2)
\end{equation}

Then using the other condition:

\begin{equation}
u_t(x,0) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) \frac{3 n \pi}{2} B_n = x(2-x)
\end{equation}

We then solve for the $B_n$'s:

\begin{equation}
B_n = \frac{2}{3 n \pi} \int_0^2 (2x-x^2) \sin( \frac{n \pi x}{2}) dx = -( \frac{2}{3 n \pi} ) \frac{8(\pi \sin(n \pi) + 2\cos(n \pi) - 2)}{\pi^3 n^3} =( \frac{2}{3 n \pi} ) \frac{16 ((-1)^n - 1)}{\pi^3 n^3}
\end{equation}

If n is even we see that this is zero, and if n is odd we see that this is $\frac{-64}{3 \pi^4 n^4}$

So we see that

\begin{equation}
u(x,t) = -\frac{64}{3\pi^4}\sum_{n odd}^{\infty}\frac{1}{n^4}\sin(\frac{n \pi x}{2})\sin(3n\pi t/2)
\end{equation}

6

Final Exam / Re: FE-2

« on: December 18, 2015, 11:07:55 PM »

Use $G_D(x-y, t) - G_D(x+y,t)$:

\begin{equation}
G_D(x-y, t) - G_D(x+y,t) = \frac{1}{\sqrt{4\pi t}}e^{\frac{-(x-y)^2}{4t}} - \frac{1}{\sqrt{4\pi t}}e^{\frac{-(x+y)^2}{4t}}
\end{equation}

Now integrate them from 0 to infinity over y:

\begin{equation}
\frac{1}{\sqrt{4\pi t}}\int^{\infty}_0 e^{\frac{-(x-y)^2}{4t}} dy - \frac{1}{\sqrt{4\pi t}}\int^{\infty}_0 e^{\frac{-(x+y)^2}{4t}} dy
\end{equation}

Change variables to $z_1$ and $z_2$:

$z_1 = \frac{x-y}{2\sqrt{t}}$ , $dz_1 = \frac{-1}{2\sqrt{t}}dy$,
$z_2 = \frac{x+y}{2\sqrt{t}}$ , $dz_2 = \frac{1}{2\sqrt{t}}dy$,

\begin{equation}
-\frac{1}{\sqrt{\pi}}\int^{\infty}_{\frac{x}{2\sqrt{t}}} e^{-(z_1)^2} dz_1 - \frac{1}{\sqrt{\pi}}\int^{\infty}_{\frac{x}{2\sqrt{t}}} e^{-(z_2)^2} dz_2
\end{equation}

\begin{equation}
u(x,t) = xerf(\frac{x}{2\sqrt{t}}) + 2\sqrt{\frac{t}{\pi}}e^{-\frac{x^2}{4t}}
\end{equation}

7

Final Exam / Re: FE-5

« on: December 18, 2015, 10:56:24 PM »

$T''=4n^2 T\Rightarrow T=Ce^{2ny}+De^{-2ny}$. However, $y>0$, so for the solution to be bounded we must have $C=0$.
Then the general solution is, after absorbing and redefining some constants:
$$u(x,y)=\sum_{n=1}^{\infty}A_n \sin(2nx)e^{-2ny}$$
$u(x,0)=\cos(x)=\sum_{n=1}^{\infty} A_n\sin(2nx)\Rightarrow A_n=\frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\cos(x)\sin(2nx)dx=\frac{8n}{\pi(4n^2 -1)}$
Therefore the final solution is:
$$u(x,y)=\sum_{n=1}^{\infty}\frac{8n}{\pi(4n^2 -1)}\sin(2nx)e^{-2ny}$$

8

Final Exam / Re: FE-4

« on: December 18, 2015, 10:55:16 PM »

(d) We then have a general solution $u(r, \theta) = \sum_1^{\infty}$, and we have to make use of the boundary conditions. So we have:

\begin{equation}
u(r, \theta) =  \sum_1^{\infty} \left[ r^n \left(A_n\cos n \theta + B_n \sin n \theta \right) + r^{-n} \left( C_n \cos n \theta + D_n \sin n \theta \right) \right]
\end{equation}

As $r=1$:

\begin{equation}
u(1, \theta) =  \sum_1^{\infty} \left[ \left(A_n\cos n \theta + B_n \sin n \theta \right) + \left( C_n \cos n \theta + D_n \sin n \theta \right) \right] = \sin\theta
\end{equation}

We see that this is only matched when all terms that aren't $n=1$ are zero. Also, $A_1$ and $C_1$ are zero. So we're left with:

\begin{equation}
B_1 + D_1 = 1
\end{equation}

Likewise for $r=2$:

\begin{equation}
u(2, \theta) =  \sum_1^{\infty} \left[ 2^n \left(A_n\cos n \theta + B_n \sin n \theta \right) + 2^{-n} \left( C_n \cos n \theta + D_n \sin n \theta \right) \right] = 3\sin\theta
\end{equation}

By the same logic as before, notice that we must have all coefficients zero, except for $B_1$ and $D_1$, where we have that:

\begin{equation}
2B_1 + \frac{1}{2}D_1 = 3
\end{equation}

We can solve these two equations for $B_1$ and $D_1$ to get that $D_1 = -\frac{2}{3}$ and $B_1 = \frac{5}{3}$. Our answer is then:

\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin n \theta - \frac{1}{r} \frac{2}{3} \sin n \theta
\end{equation}

9

Final Exam / Re: FE-5

« on: December 18, 2015, 10:43:25 PM »

b) There are only strictly positive eigenvalues, so let $\lambda=\omega^2$. Then $X''=-\omega^2 X\Rightarrow X=A\cos(\omega x)+B\sin(\omega x)$. The boundary conditions imply that $A=0$, and $\omega\frac{\pi}{2}=n\pi\Rightarrow\omega=2n$. The eigenfunctions and eigenvalues are:
$$\lambda_n=-4n^2,~X_n (x)=B_n\sin(2nx)$$

10

Final Exam / Re: FE-1

« on: December 18, 2015, 10:43:19 PM »

For this I used D'Alembert formulas:

for $ x > \frac{t}{2}$:
\begin{equation}
u(x, t) = \frac{1}{2}[2(x + \frac{t}{2})e^{-2(x+\frac{t}{2})^2} + 2(x - \frac{t}{2})e^{-2(x-\frac{t}{2})^2}] + \int^{x+\frac{t}{2}}_{x-\frac{t}{2}}0 dx'
\end{equation}

\begin{equation}
u(x, t) =  \frac{1}{2}[2(x + \frac{t}{2})e^{-2(x+\frac{t}{2})^2} + 2(x - \frac{t}{2})e^{-2(x-\frac{t}{2})^2}]
\end{equation}

for $ 0 < x < \frac{t}{2}$:
\begin{equation}
u(x, t) = \frac{1}{2}[2(x + \frac{t}{2})e^{-2(x+\frac{t}{2})^2} + 2(\frac{t}{2} - x)e^{-2(\frac{t}{2} - x)^2}]  + (t - 2x)e^{\frac{(t - 2x)^{2}}{2}}
\end{equation}

11

Final Exam / Re: FE-4

« on: December 18, 2015, 10:39:13 PM »

(c) The differential equation that we must solve for $R$ is $r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda$, or $r^2R'' + rR' = \lambda R$, so we assume a solution $r^m$, and then we get:

\begin{equation}
r^2m(m-1)r^{m-2} + rmr^{m-1} - \lambda r^m = 0 \longrightarrow m^2 - m + m - \lambda = 0 \longrightarrow m^2 = n^2
\end{equation}

So we have $m = \pm n$, which means that the solution is $R(r) = r^n + r^{-n}$.

12

Final Exam / Re: FE-5

« on: December 18, 2015, 10:37:20 PM »

a) $\frac{X''}{X}+\frac{Y''}{Y}=0\Rightarrow \frac{X''}{X}=-\lambda,~\frac{Y''}{Y}=\lambda$
The associated eigenvalue problem is in the $x$ component:
$$\frac{X''}{X}=-\lambda,~X(0)=X(\frac{\pi}{2})=0$$

13

Final Exam / Re: FE-4

« on: December 18, 2015, 10:32:56 PM »

(b) The eigenvalue problem for $P$ (as seen in part a) is:

\begin{equation}
P'' = - \lambda P
\end{equation}

First let $\lambda = \omega^2 > 0$. Then the equation is $P'' = - \omega^2 P$, and the solution is $P(\theta) = A\cos(\omega \theta) + B\sin(\omega \theta)$.

Let $\lambda = - \omega^2 < 0$. Then the equation is $P'' = \omega^2 P$, and the solution is $P(\theta) = C\cosh(\omega \theta) + D\sinh(\omega \theta)$.

Then let $\lambda = 0$. Then the equation is $P'' = 0$, and the solution is $P(\theta) = E \theta + F$.

So the eigenvalues are $n^2$, and the eigenfunctions are $P(\theta) = A\cos(n \theta) + B\sin(n \theta)$.

14

Final Exam / Re: FE-7

« on: December 18, 2015, 10:29:00 PM »

$g$ is a degree 3 polynomial, so $P$ will be a degree 1 polynomial. $g$ also has rotational symmetry about the z axis ($x^2+y^2=s^2$ in cylindrical coordinates), so $P$ must does as well. No degree 1 polynomial in $x$ and $y$ can have such symmetry, so $P=az$ only, where $a$ is to be determined. Plugging in this expression for $P$ we obtain:
$$\Delta u=2z+2z-2az-2az-6az=0\Rightarrow a=\frac{2}{5}$$
Therefore the answer is, using $R=1$:
$$u=z(x^2 + y^2)-\frac{2}{5}z(x^2+y^2+z^2-1)$$

15

Final Exam / Re: FE-4

« on: December 18, 2015, 10:23:03 PM »

(a) We solve this question using the two dimensional Laplacian in polar coordinates. So the equation becomes:

\begin{equation}
u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta} = 0
\end{equation}

So the equation is

\begin{equation}
\frac{R''}{R} + \frac{1}{r}\frac{R'}{R} + \frac{1}{r^2}\frac{P''}{P} = 0
\end{equation}

Multiplying through by $r^2$, we see that the equation is:

\begin{equation}
r^2\frac{R''}{R} + r\frac{R'}{R} + \frac{P''}{P} = 0
\end{equation}

The equation is now separable. We have a set of ordinary differential equations:

\begin{gather}
r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda \\
\frac{P''}{P} = - \lambda
\end{gather}

The boundary condition for P is that it is $2\pi$-periodic: $P(\theta) = P(2\pi + \theta)$.