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Messages - Shaghayegh A

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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

Chapter 2 / Solving the Burgers equation
« on: September 23, 2016, 07:29:07 PM »
In example 7 of chapter 2.1, we wish to solve $$u_{t}+uu_{x}=0.$$
The textbook says
So far correct  The rest here are just your fantasies V.I.

We know $$\frac{\partial x}{\partial t}=u\;and\;\frac{du}{dt}=0\;so\;du=0\implies\frac{du}{0}=1$$
Why is $$\frac{dx}{u}=1?$$

Chapter 2 / question from 2.1 of textbook
« on: September 18, 2016, 07:45:39 PM »
Section 2.1 of the textbook states $$u_t a+u_x b$$ is the directional derivative of u in the direction l=(a,b). But there's an extra factor of $$\frac{1}{\sqrt{a^2+b^2}}$$ right? (which disappears if we set $$u_t a+u_x b$$ to 0). As in:

$$\nabla_{l}u \;. \frac{\bar{l}}{|\bar{l}|}=(\partial u/ \partial t \;\hat{t}\;+\;\partial u/ \partial x \; \hat{x}).\frac{ (a\hat{t}+b\hat{x})}{\sqrt{a^2+b^2}}=(u_t a+u_x b)\frac{1}{\sqrt{a^2+b^2}}$$?

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