16
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 $v=\frac{x}{t}$?
Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
Pages: 1 [2]
17
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 :
$$\begin{align*}
& u_{tt}-c^2u_{xx}=0, &&t>0, x>0, \\\
&u|_{t=0}= \phi (x), &&x>0, \\
&u_t|_{t=0}= c\phi'(x), &&x>0, \\
&u|_{x=0}=\chi(t), &&t>0.
\end{align*}$$
My solution: $u=f(x+ct)+g(x-ct)$ where f and g are some functions. By the boundary conditions,
$$\begin{align*}
& f(x)+g(x)=\phi(x) \\\
& f'(x)-g'(x)=\phi ' (x) \implies f(x)-g(x)=\phi(x)\\\
\end{align*}$$ 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
$$\begin{align*}
& u_{tt}-c^2u_{xx}=0, &&t>0, x>0, \\\
&u|_{t=0}= \phi (x), &&x>0, \\
&u_t|_{t=0}= c\phi'(x), &&x>0, \\
&u|_{x=0}=\chi(t), &&t>0.
\end{align*}$$
My solution: $u=f(x+ct)+g(x-ct)$ where f and g are some functions. By the boundary conditions,
$$\begin{align*}
& f(x)+g(x)=\phi(x) \\\
& f'(x)-g'(x)=\phi ' (x) \implies f(x)-g(x)=\phi(x)\\\
\end{align*}$$ 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
18
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 $Ï_t+vÏ_x=0\;(6)$ 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
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
19
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
$$
au_t+bu_x=f\tag{6}
$$
Then we have
$$
\frac{\partial u}{\partial t}dt+ \color{orange}{\frac{\partial x}{\partial t}}dt \frac{\partial u}{\partial x}=u
\tag{*}$$
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
Thanks
$$
au_t+bu_x=f\tag{6}
$$
Then we have
$$
\frac{\partial u}{\partial t}dt+ \color{orange}{\frac{\partial x}{\partial t}}dt \frac{\partial u}{\partial x}=u
\tag{*}$$
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
Thanks
20
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
$$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}.$$
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 $$\frac{dx}{u}=1?$$
The textbook says
$$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}.$$
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 $$\frac{dx}{u}=1?$$
21
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:
$$\nabla_{l}u \;. \frac{\bar{l}}{|\bar{l}|}=(\partial u/ \partial t \;\hat{t}\;+\;\partial u/ \partial x \; \hat{x}).\frac{ (a\hat{t}+b\hat{x})}{\sqrt{a^2+b^2}}=(u_t a+u_x b)\frac{1}{\sqrt{a^2+b^2}}$$?
$$\nabla_{l}u \;. \frac{\bar{l}}{|\bar{l}|}=(\partial u/ \partial t \;\hat{t}\;+\;\partial u/ \partial x \; \hat{x}).\frac{ (a\hat{t}+b\hat{x})}{\sqrt{a^2+b^2}}=(u_t a+u_x b)\frac{1}{\sqrt{a^2+b^2}}$$?
Pages: 1 [2]