Messages

46

HA8 / Re: HA8-P1

« on: November 06, 2015, 03:10:20 PM »

(b) This part is very similar to the first, so I will omit several steps for the sake of convenience. Recall that for a solution that depends only on $r$, we can reduce the problem to:

\begin{equation}
\Delta u(r) = u_{rr} + \frac{2}{r}u_r = -k^2u
\end{equation}

By the same method of making a substitution $u(r) = \frac{v(r)}{r}$ as in part (a), we will arrive at the same derivatives as before. This will give us an equation in terms of $v(r)$:

\begin{equation}
v''(r) = -k^2v
\end{equation}

Again, this is a simple ODE. By the usual methods, this will lead to a solution:

\begin{equation}
v(r) = Ce^{ikr} + De^{-ikr}
\end{equation}

Plugging this back in terms of $u(r)$, we arrive at a solution:

\begin{equation}
u(r) = Ce^{ikr}r^{-1} + De^{-ikr}r^{-1}
\end{equation}

47

HA8 / HA8-P1

« on: November 06, 2015, 02:55:24 PM »

Problem 1. http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html

(a) We are asked to find the solutions of $\Delta u = u_{xx} + u_{yy} + u_{zz} = k^2u$ that depend only on $r$. So, we must turn this into spherical coordinates. By the methods of the textbook section 6.3.2, we arrive at:

\begin{equation}
\Delta u(r, \phi{}, \theta{}) = u_{rr} + \frac{2u_r}{r} + \frac{u_{\phi{} \phi{}}}{r^2} + \frac{\cot(\phi{})u_{\phi}}{r^2} + \frac{u_{\theta{} \theta{}}}{\sin^2(\phi{})r^2} = k^2u
\end{equation}

Note that these are defined as in the textbook: $r$ is the radius, $\phi$ is the latitude, and $\theta$ is the longitude.

Now, the problem asks for a solution that depends only on r. In this case, any derivatives with respect to the other variables should be zero. So canceling these terms, we can write the equation as:

\begin{equation}
\Delta u(r) = u_{rr} + \frac{2}{r}u_r = k^2u
\end{equation}

Now we can make use of the hint provided in the problem, namely that we should make the substitution $u(r) = \frac{v(r)}{r}$. Then $u''(r)$ and $u'(r)$ are:

\begin{equation}
\begin{cases}
u_r = \frac{v'(r)}{r} - \frac{v(r)}{r^2} \\
u_{rr} = \frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3}
\end{cases}
\end{equation}

Now plugging these all into the problem, we have:

\begin{equation}
\frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3} + \frac{2}{r} \Big( \frac{v'(r)}{r} - \frac{v(r)}{r^2} \Big) = k^2\frac{v(r)}{r} \longrightarrow
\end{equation}
\begin{equation}
\frac{v''(r)}{r} - 2\frac{v'(r)}{r^2} + 2\frac{v}{r^3} + 2\frac{v'(r)}{r^2} - 2\frac{v(r)}{r^3} = k^2\frac{v(r)}{r} \longrightarrow
\end{equation}
\begin{equation}
\frac{v''(r)}{r} = k^2\frac{v(r)}{r} \longrightarrow
\end{equation}
\begin{equation}
v''(r) = k^2v
\end{equation}

We now have a simple ODE which can be solved by the usual methods. Recall that $k>0$ as defined in the problem. The solution to the ODE is:

\begin{equation}
v(r) = Ae^{kr} + Be^{-kr}
\end{equation}

Now we put this back in terms of $u(r)$, where we recall that $u(r) = \frac{v(r)}{r}$. So:

\begin{equation}
u(r) = Ae^{kr}r^{-1} + Be^{-kr}r^{-1}
\end{equation}

48

HA7 / Re: HA7-P1

« on: November 05, 2015, 10:37:35 AM »

Professor, would you be able to provide a hint for part (a)? I'm not sure how to get started on it.

49

HA7 / Re: HA7-P1

« on: November 03, 2015, 01:54:02 PM »

(c) We want to show that applying the Fourier transform operator four times will recover the original function (i.e. the fourth power of the Fourier transform operator is equivalent to the Identity). This is easily built off of part (b). For consider the result of part (b): \begin{equation}

F^2(f(x)) = f(-x) \end{equation}

Now let's apply this same result again: \begin{equation}

F^2[F^2(f(x))] = F^2(f(-x)) = f(-(-x)) = f(x) \longrightarrow F^4(f(x)) = f(x) \end{equation}

Since this operation recovers the original function for all functions, it must be that the fourth power of the Fourier transform operator is the Identity: \begin{equation}

F^4 = I \end{equation}

50

HA7 / Re: HA7-P1

« on: November 03, 2015, 01:47:02 PM »

(b) We start by considering $(F^2f)(x) = F(Ff)(x)$.

We have that: \begin{equation}
F(f(x)) = \hat{f}(k) = \frac{1}{\sqrt{2\pi{}}}\int_{-\infty}^{\infty} f(x)e^{-ikx}dx \end{equation}

Now consider that: \begin{equation}
f(x) = \frac{1}{\sqrt{2\pi{}}}\int_{-\infty}^{\infty} \hat{f}(k)e^{ikx}dk \end{equation}

Then \begin{equation}
F^2(f(x)) = F(F(f(x)) = F(\hat{f}(k)) = \frac{1}{\sqrt{2\pi{}}}\int_{-\infty}^{\infty} \hat{f}(k)e^{-ikx}dk = f(-x) \end{equation}

Since an even function is defined such that $f(-x) = f(x)$, the above operation will recover the original function. For an odd function, which has the property that $f(-x) = -f(x)$, the operation will recover the negative of the function.

51

Web Bonus = Oct / Re: Web Bonus Problem to Week 7 (#1)

« on: October 30, 2015, 05:32:29 PM »

We start by assuming a solution: \begin{equation}

U(x,t) = X(x)T(x) \end{equation}

Then plugging this into the wave equation: \begin{equation}

X(x)T''(t) - c^2T(t)X''(x) = 0 \end{equation}

Dividing through by $U(x,t)$ gives: \begin{equation}

\frac{T''(t)}{T(t)} = c^2\frac{X''(x)}{X(x)} = \lambda \end{equation}

In this case we will first consider $\lambda = \omega{}^2 > 0$. Then this gives us two ODEs. Solving for $T(t)$: \begin{equation}

\frac{T''(t)}{c^2T(t)} = \omega{}^2 \rightarrow T''(t) - \omega{}^2c^2T(t) = 0 \end{equation}

The characteristic equation is: \begin{equation}

r^2 = \omega{}^2 = 0 \rightarrow r^2 = \omega{}^2 \rightarrow r = \pm \omega \end{equation}

We solve this to get the solution for $T(t)$: \begin{equation}

T(t) = Ae^{c\omega{}t} + Be^{-c\omega{}t} \end{equation}

Likewise, we will have the following equation for $X(x)$: \begin{equation}

\frac{X''(x)}{X(x)} = \omega{}^2 \end{equation}

We can solve this to show that: \begin{equation}

X(x) = Ce^{\omega{}x} + De^{-\omega{}x} \end{equation}

I will continue the rest of the problem soon, or anyone else can contribute if they want to.

52

HA6 / Re: HW6-P2

« on: October 29, 2015, 11:55:21 AM »

(b) We let $\lambda = - \gamma^2$. We can again proceed as in the first problem to get that: \begin{equation}
X(x) = A\cosh(\gamma{}x) + B\sinh(\gamma{}x) \end{equation}
With the derivative being: \begin{equation}
X'(x) = A\gamma{}\sinh(\gamma{}x) + B\gamma{}\cosh(\gamma{}x)
\end{equation}

Plugging in the Dirichlet condition as in part (a): \begin{equation}
X(0) = A = 0 \end{equation}

Therefore $A=0$. Thus the solution is: \begin{equation}
X_n = \sinh(\gamma{}_nx) \end{equation}

We have chosen $B=1$ for convenience.

The relation is: \begin{equation}
\tanh(\gamma{}l) = -\frac{\gamma{}}{\beta{}} \end{equation}

53

HA6 / Re: HW6-P2

« on: October 29, 2015, 11:40:47 AM »

This problem follows similar logic to the first problem.

(a) We let $\lambda = \omega^2$. We can then proceed as in the first problem (check that post if you would like to see the steps) to get that: \begin{equation}
X(x) = A\cos(\omega{}x) + B\sin(\omega{}x) \end{equation}
With the derivative being: \begin{equation}
X'(x) = -A\omega{}\sin(\omega{}x) + B\omega{}\cos(\omega{}x)
\end{equation}

This time we have a Dirichlet condition that $X(0) = 0$. So plugging this in: \begin{equation}
X(0) = A = 0 \end{equation}

Therefore $A=0$. Thus the solution is: \begin{equation}
X_n = \sin(\omega{}_nx) \end{equation}

We have chosen $B=1$ for convenience.

Edit: the similar tangent relation is, as you can easily show using the boundary condition on the right end, is: \begin{equation}
\tan(\omega{}l) = -\frac{\omega{}}{\beta{}} \end{equation}

54

HA6 / Re: hm6 Q1

« on: October 29, 2015, 11:27:54 AM »

Could someone clarify what we are supposed to do in part c? It seems like there is no question being asked.

55

HA6 / Re: hm6 Q1

« on: October 28, 2015, 06:07:27 PM »

After speaking to Professor Ivrii about it we found out that you simplify choose the value of $A_n$ for convenience. You do not need to plug in any tangent term.

56

Web Bonus = Oct / Re: Web bonus problem : Week 4 (#3)

« on: October 22, 2015, 12:18:56 PM »

Sorry, forgot to mention that we probably need to make the assumption that u is fast-decaying.

57

Web Bonus = Oct / Re: Web bonus problem : Week 4 (#3)

« on: October 22, 2015, 12:13:29 PM »

I will start this one.

We take the time derivative of the energy as given in the problem. So: \begin{equation}

\frac{1}{2}\int_0^{\infty}[(2u_tu_{tt} + 2ku_{xx}u_{xxt}]dx = \int_0^{\infty}[u_t(-ku_{xxxx}) + ku_{xx}u_{xxt}]dx \end{equation}

Now we will integrate by parts twice on the second term in this integral. I will omit the steps and state the result: \begin{equation}

-k\int_0^{\infty}u_tu_{xxxx}dx +ku_{xx}u_{xt}|_0^{\infty} - ku_tu_{xxx}|_0^{\infty} + k\int_0^{\infty}u_tu_{xxxx}dx \end{equation}

The first and last terms cancel. Let's make use of the first boundary condition. \begin{equation}

u|_{x=0} = u_x|{x=0} = 0 \longrightarrow u_t|_{x=0} = u_{xt}|_{x=0} = 0 \end{equation}

We plug this in to get the final result: \begin{equation}
 \frac{\partial{}E(t)}{\partial{}t} = 0 \end{equation}

I will add the other BCs later; or if someone wants to collaborate and add those ones please do!

58

Web Bonus = Oct / Re: Web bonus problem : Week 4 (#2)

« on: October 22, 2015, 11:44:33 AM »

I believe it should perhaps be $u_t|_{x=0}=0$ for the last part.

59

Quiz 4 / Will Quiz 4 be next week?

« on: October 22, 2015, 09:11:39 AM »

Just wondering if we will have quiz 4 on the 29th of October, or if we will have no quiz next week. Thank you!

60

HA5 / Re: HA5-P2

« on: October 20, 2015, 08:05:11 PM »

Professor, could you verify the answer to this problem? I couldn't figure it out myself but I am not sure about this solution.