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Topics - Shaghayegh A

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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 / 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 / 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?

Chapter 2 / HA #4, problem 3
« on: October 11, 2016, 04:59:07 PM »
In problem 3 of HA #4, are the functions u(x,t) and v(x,t) separate functions? Or is $v=\frac{x}{t}$?

Chapter 2 / HA #4, problem 1
« on: October 10, 2016, 01:25:42 PM »
I am having trouble with problem 1 of home assignment 1, it asks to find u(x,t) for :

& u_{tt}-c^2u_{xx}=0, &&t>0, x>0,  \\\
&u|_{t=0}= \phi (x),   &&x>0, \\
&u_t|_{t=0}= c\phi'(x),  &&x>0, \\
&u|_{x=0}=\chi(t), &&t>0.

My solution:  $u=f(x+ct)+g(x-ct)$ where f and g are some functions. By the boundary conditions,
& f(x)+g(x)=\phi(x) \\\
& f'(x)-g'(x)=\phi ' (x) \implies f(x)-g(x)=\phi(x)\\\
\end{align*}$$ So $f(x)=\phi(x)$ and $g(x)=0$ so $f(x+ct)=\phi(x+ct)$ , but is this true for all x>0? Because it seems that t can be negative here and we must say $$f(x+ct)=\phi(x+ct),  x>ct $$
Thank you

Chapter 2 / derivation of a PDE describing traffic flow
« on: September 25, 2016, 03:41:49 PM »
In example 8 of chapter 2.1 where we derive a PDE describing traffic flow, how do we derive $ρ_t+vρ_x=0\;(6)$ from $p_t+q_x=0\;(3)\;?$

It seems that $q_x$ some how equals $vp_x=[c(\rho)+ c' (\rho)\rho] \;p_x=c(p) \frac{\partial p}{\partial x}+\frac{d c(p)}{p} p \frac{\partial p}{\partial x}$? Can someone please explain how we get equation (6)? Thanks

Chapter 2 / Deriving equation 7 of section 2.1
« on: September 24, 2016, 03:20:01 PM »
In the section variable coefficients of section 2.1, we have
Then we have
\frac{\partial u}{\partial t}dt+ \color{orange}{\frac{\partial x}{\partial t}}dt \frac{\partial u}{\partial x}=u
No, $\frac{d x}{d t}$
I  assume  the  $dt$  cancels  with  the  $\partial t$   in  the  $\frac{\partial x}{\partial t}dt \frac{\partial u}{\partial x} $  part  because the  textbook says we get
$$u_t dt+dx u_x =du$$
Wrong conclusion due to your error in (*)
Why doesn't the $dt$ cancel the  $\partial t $  in  $\frac{\partial u}{\partial t}dt$   to give us   $du+dxu_x =du$?
Calculus II
Also,  to  derive  $$\frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}\tag{7}$$ from   (6)  why don't we just  compare   (6) to   
$\frac{du}{dt}=\frac{\partial u}{\partial t}+\color{orange}{\frac{\partial x}{\partial t}}\frac{\partial u}{\partial x}$    (chain   rule)   and conclude that  $\frac{\partial t}{a}=\frac{\partial x}{b}$   and $\frac{dt}{a}=\frac{du}{f}  \implies    \frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}$  (7)   instead of doing all that work?
The same mistake; also there should be  $\frac{d t}{a}=\frac{d x}{b}$and if corrected it would be exactly what we do

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