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Messages - Shaghayegh A

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FE / Re: FE6
« on: December 18, 2016, 02:18:34 PM »
Okay, I just figured this out. I believe this is what you do: we know solution is spherically symmetric by the hint and because the boundary conditions show that the solution has no $\theta$ or $\phi$ dependence. So converting to spherical coordinates, the problem looks like:
$$u_{tt}-u_{rr}-\frac{2}{r} u_r  =0$$
We separate variables: $u(x,t)= T(t) X(x)$. So

$$\frac{T_{tt}}{T}-\frac{R_{rr}}{R}-\frac{2}{r}\frac{R_r}{R}=0 \implies \\
\frac{T_{tt}}{T}=-\lambda \\
\frac{R_{rr}}{R}+\frac{2}{r}\frac{R_r}{R}+\lambda = 0$$where $\lambda \in R$

We know $$T(t)=A\sin(\sqrt\lambda t) +B\cos(\sqrt\lambda t)$$ The initial condition $$u_t(r,0) = 0 \implies T_t(t) = 0 \implies T(t)=B\cos(\sqrt\lambda t)$$

We also have $$\frac{R_{rr}}{R}+\frac{2}{r}\frac{R_r}{R}+\lambda = 0$$ By the hint, we write this as $$(rR)'' = -\lambda (rR)$$ Making a change of variables with $$L(r) =rR(r)$$ we get
$$L_{rr} = -\lambda L \implies L(r)=E\sin(\sqrt\lambda r)+F\cos(\sqrt \lambda r) \implies R(r) = \frac{E\sin(\sqrt\lambda r)}{r}+\frac{F\cos(\sqrt \lambda r)}{r}$$ We rule out the cos term because we get a 1/0 as r approached infinity, which is bad. So
$$R(r) = \frac{E\sin(\sqrt\lambda r)}{r}$$

Now we combine the R(r) and T(t) solutions. We call $n = \lambda$ for the sake of notation.

$$u(r,t)=\sum_{n= 1} ^{\infty} \frac{C_n \cos(nt) \sin(nr)}{r}$$

By the initial condition (equation 2) in the problem, we have $$\begin{align} &u(r,0)=f(r)=\left\{\begin{aligned} &1\quad &&r<1,\\ &0 &&r\ge 1,\end{aligned}\right.\qquad  \end{align}$$

So we have
$$u(r,0)=\sum_{n= 1} ^{\infty} C_n \sin(nr) = r f(r)$$ This is a sine fourier series with $L=\pi$, so
$$C_n = \frac{2}{\pi} \int_0^{\pi} sin(nr) r f(r) dr = \frac{2}{\pi} \int_0^1 sin(nr) r dr $$
We use integration by parts here and get
$$C_n = \frac{2}{\pi} \frac{\sin(n)-n\cos(n)}{n^2}$$
$$u(r,t)=\sum_{n= 1} ^{\infty} \frac{\frac{2}{\pi} \frac{\sin(n)-n\cos(n)}{n^2} \cos(nt) \sin(nr)}{r}$$

FE / Re: FE5
« on: December 18, 2016, 12:03:03 PM »
By separation of variables, we get $u(x,y) = X(x) Y(y)$. Plugging this in $ u_{xx} +u_{yy}=0$ we get
 $$\frac{X_{xx}}{X} +\frac{Y_{yy}}{Y}=0  \implies \\ \begin{align}&\frac{X_{xx}}{X}=-\lambda,\label{5-4}\\ &\frac{Y_{yy}}{Y}=\lambda \label{5-5} \end{align}$$ where $\lambda \in R$.
From the boundary conditions given, we must have
$$\begin{align}&X(0) = 0\\
&X(\pi) = 0\end{align}$$ in order to have a nontrivial solution for u(x,y).
Equations (2)-(4) are the associated eigenvalue problem.

We know $$X(x) = A \sin(\sqrt\lambda x)+B\cos(\sqrt\lambda x)$$ but we rule out the cosine term because of the first boundary condition on x ( equation 3). Equation 4 implies $$X(\pi) = A\sin(\sqrt\lambda \pi)= 0 \implies \sqrt\lambda \pi = n \pi, n \in N \implies \lambda = n^2 \implies X(x) = A sin(nx)$$ where $\sin(nx)$ are the X eigenfunctions and $-n^2$ is the eigenvalue.

From equation (2), we get $$Y(y) = C e^{\sqrt\lambda y} +De^{-\sqrt\lambda y} $$ but we rule out $C e^{\sqrt\lambda y}$ because this reaches infinity as y approaches infinity. We are left with $$Y(y)=De^{-n y}$$ where $e^{-ny}$ are the y eigenfunctions and $n^2$ is the eigenvalue.

So $$u(x,y) = X(x) Y(y) = \sum_{n = 1}^{\infty} C_n sin(nx) e^{-n y}$$

The boundary condition $$u_y(x,0)=\cos(x) \implies cos(x) =\sum_{n = 1}^{\infty} -n C_n sin(nx) $$ This is a sine fourier series with L = $\pi$, we call $A_n = -n C_n$, and solve for $A_n$:

$$A_n=\frac{2}{\pi} \int_0^{\pi} \cos(x) \sin(nx) dx=\frac{2}{2 \pi} \int_0^{\pi} \Bigl(\sin(n-1)x + \sin(n+1)x \Bigr)\;dx \tag{$\checkmark$}\\ \\ = \frac{-1}{\pi} \left.\frac{\cos(n-1)x}{n-1} \right|_{0}^{\pi} + \left.\frac{\cos(n+1)x)}{n+1} \right|_{0}^{\pi} \implies$$

$A_n = 0$           n is odd
$A_n = \frac{4n}{\pi (n^2-1)}$    n is even

$C_n = 0$                  n is odd
$C_n=-\frac{4}{\pi(n^2-1)}$        n is even

So $$ u(x,y) = \sum_{n = 1, \;n \;even}^{\infty} -\frac{4}{\pi (n^2-1)}  \sin(nx) e^{-n y} \;for \;y>0, 0<x<\pi$$ which can be written as
$$u(x,y) = \sum_{m = 1}^{\infty} -\frac{4}{\pi \Bigl((2m)^2-1 \Bigr)}  \sin(2mx) e^{-2m y}, \;\; for \;y>0, 0<x<\pi$$

Edit: my final answer has remained the same, except it's written in a different form (I'm not sure what error I've made after Prof. Ivrii's checkmark, perhaps someone can point it out). I've also formatted the latex a bit differently

Chapter 4 / problem involving fourier series
« on: December 11, 2016, 10:31:05 PM »
Problem: Solve by Fourier method $$\begin{align} & u_{tt}-u_{xx}=0\qquad -\frac{\pi}{2}<x<\frac{\pi}{2},\label{1-1}\\ & u_x|_{x=-\pi/2}=u_x|_{x=\pi/2}=0,\label{1-2}\\ &u| _{t=0}=x^2,\qquad u_t|_{t=0}=0.\label{1-3} \end{align}$$

I have $X(x) = A\sin(\sqrt\lambda x) +B\cos(\sqrt\lambda x)$ Using (2), I can either take

$X(x) = A\sin(nx), \lambda = n^2$ OR $X(x) = B\cos(2nx), \lambda = 4n^2$ Is it correct to take either one? Or is there a right one?

Chapter 2 / energy integrals
« on: December 10, 2016, 06:49:42 PM »
Problem 4 of the 2012 final exam:
It asks to prove that the total energy (kinetic + potential) is constant with time. I get up to $$\frac{d}{dt} k(t) + p(t) = u_x(\infty) u_t(\infty) - u_x(-\infty) u_t(-\infty) $$ How do I prove $$u_x(\infty) u_t(\infty) - u_x(-\infty) u_t(-\infty) = 0 $$ using the boundary conditions? right now I know $u_t =0$ only when t = 0 (and x is large)

Chapter 5 / Properties of fourier transforms
« on: December 10, 2016, 02:55:06 PM »
I'm stuck on Problem 7 of the 2015 S final exam:    (link also includes prof's solution)

He gets $$\begin{equation*} \hat{u}(k,t)=-ik (2\pi)^{-1} e^{-k^2a^2 /2} \end{equation*}$$ and he's trying to solve for u(x,t). I don't understand how he gets u(x,t); I know he's using the properties of fourier transforms, but I don't know how to go backward from the fourier transform to the inverse fourier transforms! Thanks

FE / Final exam coverage
« on: November 25, 2016, 12:05:57 PM »
The website says HA 1-10 is a good preparation. Does that mean the final exam doesn't cover chapter 10? (i.e  variational methods)

Chapter 8 / Re: HA 10, problem 3a
« on: November 22, 2016, 08:53:19 PM »
Oops! I corrected my solution

Chapter 8 / HA 10, problem 3c
« on: November 22, 2016, 01:15:06 PM »
Does 3c want us to go through steps a and b in spherical coordinates, or just write u as a function of r, theta, and phi?

Chapter 8 / Re: HA 10, problem 3a
« on: November 22, 2016, 11:03:22 AM »
My solution is $U=x^2+y^2-z^2-\frac{1}{3} (x^2+y^2+z^2-1)$, the laplacian of this equation is zero and it equal g(x,y,z) at the boundary. How can I write this as a sum of harmonic homogenous polynomials, since there is a factor of 1/3 : $U=2/3 x^2+ 2/3 y^2- 4/3 z^2+ \frac{1}{3}$

By the way, g is $x^2+y^2-z^2$

Chapter 8 / HA 10, problem 3a
« on: November 21, 2016, 10:58:36 PM »
link:     (3a)

Since P(x,y,z) is a polynomial of degree 0, it is a constant. So $U=x^2+y^2+z^2-c_0 (x^2+y^2+z^2)$, but you can't write this as a sum of homogenous harmonic polynomials since there is a term c_0 remaining?

Chapter 8 / HA 10, trouble with question 2
« on: November 21, 2016, 09:54:12 PM »

I have a partial differential equation for my function $\Phi(\phi)$ which depends on l and m (l is a natural number, |m| < l ). For the case when l=0, I have  $\Phi(\phi)=c_0$, which solves my PDE for any c0. Do we have
$\Phi(\phi)=c_1 cos(\phi) +c_2$ for l=1
$\Phi(\phi)=c_3 cos^2(\phi) +c_4 cos(\phi) +c_5 $ for l=2
$\Phi(\phi)=c_6 cos^3(\phi) +c_7 cos^2(\phi) +c_8 cos(\phi) +c_9 $ for l=3? What do I do now? Do I plug in each  $\Phi(\phi)$ into its corresponding PDE and find the constants? Thanks

Chapter 4 / HA 6, sections 4.3-4.5, problem 6
« on: November 14, 2016, 02:18:44 PM »

For problem 6f, I get $b_n=0$, which I know is wrong.
I have $$b_n=\frac{2}{\pi} \int_0 ^{\pi} \sin((m-1/2)x) \sin((n+1/2)x) dx \\
=1/pi \int_0 ^{\pi} \cos((m-n-1)x) -\cos((m+n)x) dx =0$$
Separately for $n \neq m-1$ and $n=m-1$. Why am I getting b_n=0?

Chapter 4 / HA 6, problem 3c (sections 4.1 and 4.2)
« on: November 14, 2016, 02:09:32 PM »

For 3c: I assume that M(y) and N(y) are two arbitrary eigenfunctions with the same eigenvalues $\omega$. Then, M and N satisfy
$$Y^{(4)} (y)=\omega^4 Y(y) \\
Y(-L)=Y_y (-L)=0 \\
$$ where I've switched coordinate systems so that $y=x-l/2=x-L$. I want to prove
$$\int_{-L}^{L} M(y) N(y) dy=0$$ but I'm not sure how to do that. Any advise?
Thank you

Chapter 4 / HA 6, problem 1c of sections 4.1 and 4.2
« on: November 14, 2016, 02:03:24 PM »
Problem 1c asks to investigate how many negative eigenvalues there are:

I understand that we have the hyperbola $$\alpha + \beta+ \alpha \beta l=0$$ which divides the $(\alpha,\beta)$ plane into three zones, as he problem states. But how does that actually help us find the number of negative eigenvalues?

Chapter 4 / HA6, problem 5 (problems to 4.1 and 4.2)
« on: November 14, 2016, 01:55:31 PM »
Problem 5 of the problems in 4.1 and 4.2 asks to "Consiser oscillations of the beam with the clamped left and the free right end. The boundary conditions are then $$u_{to}+Ku_{xxxx}=0, 0<x<l (9)$$and $$u_{xx}(l,t)=u_{xxx}(l,t)=0 (13)$$. After separating variables and pluggin in boundary counditikns, I get
$$-A \sin(\omega L)-B \cos(\omega L)+ C \sinh(\omega L)+D \cosh(\omega L)=0$$ and
$$A\cos(\omega L)+B \sin(\omega L)+C \cosh(\omega L)+ D \sinh(\omega L)=0$$
This is two equations and four unknowns. How can I solve for A, B, C, D?

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