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Messages - Bruce Wu

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Website errors / Lecture notes
« on: November 14, 2015, 11:54:30 AM »

At the bottom of the page I think it was meant to say HA1-11


HA9 / Re: HA9-P5
« on: November 12, 2015, 09:43:09 PM »
Let $M$ and $R$ be the mass and radius of the Earth, respectively. If we assume that the Earth is a homogeneous sphere, then it has a constant density $\rho=\frac{3M}{4\pi R^3}$. For some radius $r<R$ from the centre of the Earth, all of the mass located at a radius greater than $r$ would not contribute anything to the gravitational field according to Newton's shell theorem. Therefore by Newton's law of gravity, the gravitational field at such an $r$ is $$\vec{g}=-\frac{GM_{enc}}{r^2}\hat{r}$$
Where $M_{enc}$ is the mass enclosed within the sphere of radius $r$.
$$M_{enc}=\rho\frac{4}{3}\pi r^3=M\frac{r^3}{R^3}$$ Finally
Which is proportional to $r$.

HA8 / Re: HA8-P3
« on: November 08, 2015, 09:46:49 AM »
To me the Fourier method looked easier than using the formula. Explicit forms are important but aren't always so nice to work with.

Textbook errors / Equation 6.5.3
« on: November 08, 2015, 01:54:56 AM »

HA8 / Re: HA8-P3
« on: November 08, 2015, 01:46:54 AM »
b) General solution is given by Fourier coefficients are the same as before, except we replace $A_n$ with $B_n$, $C_n$ with $D_n$, and $a^{-n}$ with $a^{n}$. The final solution is
$$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{-n}a^{n}}{n}\sin(n\theta)$$

HA8 / Re: HA8-P3
« on: November 08, 2015, 01:37:18 AM »
a) The general solution is given by, and Fourier coefficients by
We have $$A_{n}=\frac{a^{-n}}{\pi}\left[\int^{\pi}_{0}\cos(n\theta')d\theta'-\int^{2\pi}_{\pi}\cos(n\theta')d\theta'\right]=0$$
$$C_{n}=\frac{a^{-n}}{\pi}\left[\int^{\pi}_{0}\sin(n\theta')d\theta'-\int^{2\pi}_{\pi}\sin(n\theta')d\theta'\right]=\frac{a^{-n}}{\pi}\left[\frac{1-\cos(n\pi)}{n}-\frac{\cos(n\pi)-\cos(2n\pi)}{n}\right]=\frac{2a^{-n}[1-\cos(n\pi)]}{\pi n}$$
Only odd $n$ terms survive. In that case $$C_n=\frac{4a^{-n}}{n\pi}$$
The final solution is $$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{n}a^{-n}}{n}\sin(n\theta)$$

Textbook errors / Equation 6.4.12
« on: November 08, 2015, 12:52:15 AM »

We are already evaluating $u$ at $r=a$. So after the summation sign it should be $a^n$ instead of $r^n$


HA8 / Re: HA8-P2
« on: November 08, 2015, 12:31:44 AM »
a) In 2-dimensional polar coordinates $$\Delta u = u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta}$$
If $u=u(r)$ only, then $u_{\theta}=0$ and $u_{r}=u'$. Our problem transforms to $$u''+\frac{1}{r} u'=k^{2}u$$
Note that it is $\frac{1}{r}u'$, not $\frac{2}{r}u'$ as was in problem 1. This prevents us from applying the same trick of letting $u=\frac{v}{r}$. If we attempt to, the result is
Which does not simplify our original problem. We stop here and say that $u(r)$ must satisfy the ODE
b) Same as above except $u(r)$ must satisfy the ODE

Website errors / Section 6.5 wrong link
« on: November 07, 2015, 04:30:02 PM »

HA6 / Re: HA6-P3
« on: October 29, 2015, 01:14:04 AM »
After this I am not quite sure how to continue. Row reducing the matrix to solve for three of $C,D,E,F$ in terms of the other one seems like a daunting task, but without them I cannot find the eigenfunctions. The hint says to consider separately even and odd eigenfunctions, but I'm not certain what forms they should take. The exponentials cannot have any symmetry unless $D=\pm C$ so that they turn into $2C\cosh(\omega r)$ and $2C\sinh(\omega r)$, respectively.
If my logic is correct, then the even eigenfunctions should be $$2C\cosh(\omega r) + E\cos(\omega r)$$
and the odd eigenfunctions $$2C\sinh(\omega r) + F\sin(\omega r)$$
However the relationship between $C,E,F$ is unknown.

HA6 / Re: HA6-P3
« on: October 29, 2015, 12:53:53 AM »
b) One can see a plot of the general form of the function $\cos(x)\cosh(x)$ here
It is a very weird function. The different $\omega_{n}$'s are the values of $\frac{x}{l}$ when the function intersects the horizontal line $y=1$

HA6 / Re: HA6-P3
« on: October 29, 2015, 12:49:01 AM »
a) Following the hint, I made a change of variables $r=x-\frac{l}{2}$. The beam equation and its boundary conditions then become:
$$u_{tt}+Ku_{rrrr}=0,~-\frac{l}{2}<r<\frac{l}{2}\\u(-\frac{l}{2},t)=u_{r}(-\frac{l}{2},t)= u(\frac{l}{2},t)=u_{r}(\frac{l}{2},t)=0$$
Letting $u=T(t)R(r)$, we see that $$T''+K\lambda T=0\\ R''''-\lambda R=0$$
Here we are assuming that $\lambda>0$, so let $\lambda=\omega^{4}$. Solving the corresponding characteristic equations and ODEs we get
$$T=A\cos(\sqrt{K}\omega^{2}t)+B\sin(\sqrt{K}\omega^{2}t)\\ R=Ce^{\omega r}+De^{-\omega r}+E\cos(\omega r)+F\sin(\omega r)$$
Using the 4 boundary conditions, the fact that $T$ is a constant with respect to $r$ and is not identically $0$ we arrive at the system of 4 equations below
We can convert this into a matrix, and it only has non-trivial solutions when the determinant of the matrix is $0$. This computation is tedious, but the final condition that $\omega$ must satisfy is
$$\cos(\omega l)\cosh(\omega l)=1$$

HA6 / Re: hm6 Q1
« on: October 28, 2015, 04:06:50 PM »
I don't see why $A_{n}=\omega_{n}$

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