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Home Assignment 2 / Re: Problem 2
« on: September 27, 2012, 11:11:15 AM »It looks like I should use the result from part 3 for part 4. However if so, how should I use the initial values?
Not really. You use parts (a),(b)
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Not really. You use parts (a),(b)
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Home Assignment 2 / Re: problem 1 typo?
« on: September 27, 2012, 11:10:32 AM »Hi - In the pdf of home assignment 2 in problem 1 the inequalities throughout are different than the other version of the assignment (i.e. pdf version has 'greater than or equal to' and the other version in only '>'). Which one is correct?
Really does not matter, but I changed pdf to coincide
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Home Assignment 1 / Re: Problem 4
« on: September 26, 2012, 09:40:44 PM »
Do you mean:
$$-u_\theta=r^2 \\
\Rightarrow u=-r^2\theta
$$ and since it's not periodic we are done?
Basically yes, except certain things should be clarified: since our trajectories are closed each is parametrized by some parameter ($\theta$) running from $0$ to $T$ (in our case $T=2\pi$ but it may depend on trajectory) and equation looks like $\partial_\theta u= g(\theta,r)$. So we are looking for periodic solution
$u(\theta,r)=\int g(\theta,r)\,d\theta$.
Note that primitive of periodic function $g(\theta)$ is periodic if and only if average of $g$ over period is $0$:
$\int_0^T g(\theta)\,d\theta=0$. Otherwise this primitive is the sum of a periodic function and a linear function.
Finally, in (a) $g(\theta)=r^2 \sin(\theta)\cos(\theta)$ and integral over period is $0$; in (b) $g(\theta)=r^2$ and integral over period is not $0$.
So, the source of trouble is: periodic trajectories.
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Home Assignment 1 / Re: Problem 4
« on: September 26, 2012, 04:34:04 PM »
Here we are talking about first order equation. There is no wave equation. Stick to the problem! How our original equation
\begin{equation}
yu_x-xu_y=f(x,y)
\label{eq-1}
\end{equation}
looks in the polar coordinates?
Hint: the l.h.e. is $-\partial_\theta u$. Prove it using chain rule $u_\theta = u_x x_\theta + u_y y_\theta$ and calculate $x_\theta$, $y_\theta$.
Note that $\theta$ is defined modulo $2\pi \mathbb{Z}$ and all functions must be $2\pi$-periodic with respect to $\theta$ (assuming that we consider domains where one can travel around origin)
\begin{equation}
yu_x-xu_y=f(x,y)
\label{eq-1}
\end{equation}
looks in the polar coordinates?
Hint: the l.h.e. is $-\partial_\theta u$. Prove it using chain rule $u_\theta = u_x x_\theta + u_y y_\theta$ and calculate $x_\theta$, $y_\theta$.
Note that $\theta$ is defined modulo $2\pi \mathbb{Z}$ and all functions must be $2\pi$-periodic with respect to $\theta$ (assuming that we consider domains where one can travel around origin)
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Misc Math / Re: Heat Wave Equation Solution Integral
« on: September 26, 2012, 12:55:13 PM »Sorry I meant the Heat Equation. What does the integral of the solution to that equation represent?
If we talking about heat equation then it is a total heat energy in $(-\infty,x)$. But for us it mainly a trick to get a a formula for a solution.
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Home Assignment 2 / Re: Problem 3
« on: September 26, 2012, 05:21:59 AM »Part (a) and (b), part (c) and (d) the same questions? Thanks
There was misprint in (c) fixed almost immediately. Compare fourth lines in (a) vs (b) , (c) vs (d)
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Home Assignment 1 / Re: Problem 4
« on: September 26, 2012, 05:19:38 AM »
Peishan, the question now boils down to: why solution with r.h.e. $x^2+y^2$ does not exist and one really needs to use polar coordinates (or to use equivalent geometric motivation)
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Misc Math / Re: Heat Wave Equation Solution Integral
« on: September 26, 2012, 02:28:48 AM »After we determined the solution to the heat wave equation, we took the integral from negative infinity to some x. The solution was in the form of a Normal Distribution function where in statistics the integral indicates the probability that we randomly choose a value within that interval. I was wondering what are the physical implications of this same integral in the wave equation? Does the area under the solution indicate the energy in the system up to that point or does it have some other meaning?
Heat wave equation? WTH are you talking about? There is a wave equation (and its ilk) and the heat equation (and its ilk) describing either very different processes or the same process under very different assumption. The properties of these equations are really different.
PS Sure more complex phenomena can be described by systems containing both equations but for the purpose of this class ...
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Home Assignment 1 / Re: Problem 6
« on: September 26, 2012, 02:20:39 AM »
Intuitively we know that driver's speed has negative correlation with traffic density. A more realistic choice for $c$ is to let it be a monotone decreasing function of $\rho$. In this case however, conservation of cars equation is not linear anymore. This is discussed in detail by Prof. Ivrii in last year's forum.
The really interesting thing is a distinction between the speed of the individual cars $c(\rho)$ and the group speed $v(\rho)= (c(\rho)\rho)'= c'(\rho)\rho + c(\rho)$ of the group of cars. In fact the group does not have a constant "crew": if there is a place with higher density of cars than an average, it moves but the leading cars in the group have a lesser density in front, accelerate and leave the group while cars behind the group catch with it, and join it.
Such distinction is common in wave motion
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Home Assignment 1 / Re: Problem 4
« on: September 26, 2012, 02:11:08 AM »
c) In the latter case, if $(x,y)=(0,0)$ is in domain of $u$, general solution does not exist as $\arcsin{\frac{x}{\sqrt{x^2+y^2}}}$ is not well-defined.
Not persuasive: may be we just were not smart enough? You need to demonstrate that solution really does not exist and explain what is an obstacle.
Hint Use polar coordinates
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Home Assignment 1 / Re: Problem 2
« on: September 25, 2012, 06:53:47 PM »
The difference between two cases is that in one of them all trajectories have $(0,0)$ as the limit points and in another only those with $x=0$ or $y=0$ (node vs saddle).
Actually since in the saddle case $x^4y=C$ for $C\ne 0$ consists of two disjoint parts (with $x>0$ and with $x<0$), the values of $u$ on these parts are not necessarily equal and $u=f(x^4y)+x|x|^{-1}g(x^4y)$ with $g(0)=0$.
Actually since in the saddle case $x^4y=C$ for $C\ne 0$ consists of two disjoint parts (with $x>0$ and with $x<0$), the values of $u$ on these parts are not necessarily equal and $u=f(x^4y)+x|x|^{-1}g(x^4y)$ with $g(0)=0$.
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Home Assignment 1 / Re: Problem 5
« on: September 25, 2012, 06:42:11 PM »
The previous post is definitely an improvement over the preceding one
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Home Assignment 1 / Re: Problem 5
« on: September 25, 2012, 03:12:39 PM »
Scanning is barely passable
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Home Assignment 1 / Re: Problem 1
« on: September 25, 2012, 03:08:21 PM »
So, what are the answers for (c), (d)? Especially (d). Formulate them explicitly
Scanning is passable but not spectacular
Scanning is passable but not spectacular
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Home Assignment 1 / Re: WTH?
« on: September 25, 2012, 10:14:20 AM »Dear professor Ivrii,
Thank you for remembering meBut I spent too much time to resize them and post them; Some of them is readable, can you please consider them?
Next time I will use scanner.
Thank you,
Aida
Barely readable. You don't need to resize (may be just rotate). However correct initial settings of the scanner/camera are crucial.
I definitely do not want to encourage this type of submission. If it was from some clueless newbie I could consider, but from veteran-poster -- no way
BTW, has anyone any idea how much time it takes to prepare 1 hour of lecture notes or 1 home assignment? 2-3 hours :-)