Messages

1756

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

1757

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

1758

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 $g(0)=0$. 

1759

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

1760

Home Assignment 1 / Re: Problem 5

« on: September 25, 2012, 03:12:39 PM »

Scanning is barely passable

1761

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

1762

Home Assignment 1 / Re: WTH?

« on: September 25, 2012, 10:14:20 AM »

Dear professor Ivrii,

Thank you for remembering me But 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 :-)

1763

Home Assignment 1 / Re: Problem 3

« on: September 25, 2012, 10:09:56 AM »

Now it is correct.

1764

Home Assignment 1 / WTH?

« on: September 25, 2012, 02:08:17 AM »

Aida, WTH? In MAT244 your scan was a golden standard of scanning
http://weyl.math.toronto.edu:8888/MAT244-2012S-Forum/index.php?topic=38.msg111#msg111 (see, I remember, so good it was), and now this? Such posts defy the whole purpose of this forum, it is not to submit your papers for grading (you submit it to TA) but even this is difficult as quality of scan is poor (actually you used cellphone without taking care of settings), but to share the solution with your classmates who can comment, find errors or correct them.

From this point the typed solution using MathJax is far the best as I can edit it, just marking the place where I see an error, and anyone can copy-paste code from it. But apart of this your former clean black-white scan with perfect position of the paper was the very best thing.

Here you use colour (not even grayscale) scan and some papers are horizontal and some diagonal ... everyone has a really hard time to read them. I admit, some of your submissions a better than othersbut  from the point of view of this forum I must consider all of them non-existent . It looks like you just decided to capture the space preventing anyone else from posting the solutions.

So, everyone should feel to post solutions.

------

Also:
some people used paper clips despite our request to staple and some even tried to use "poor man paper clips" just folding several times the corner of the paper and adding a bit of saliva. Sorry, no of these constructions is robust enough to prevent separation of pages. I hope I caught every such attempt and stapled, but I am not sure. I am also not sure if Prof Colliander managed with this. So, if you have not stapled and your pages are separated and lost, you know whom to blame.

Not everyone indicated the section where to bring your papers back (so get such papers from TA who graded it).

Please use the standard paper (letter size). Someone used A4 (a bit more narrow and a longer overseas size). It looks like a little thing but also causes problems when dealing with tens of papers.

And finally, again: please, do not come to submit papers during the class -- only before lecture, during the break or after. Everyone in the night class feels tired (instructor in the least degree because of adrenalin rush) and such distractions are rather disruptive.

1765

Home Assignment 1 / Re: HA1-pdf

« on: September 24, 2012, 10:35:46 PM »

Is it an ancient manuscript? Can anyone read this?

1766

Home Assignment 1 / Re: Problem 6

« on: September 24, 2012, 11:46:56 AM »

One-way (no way to reverse the direction)

1767

Home Assignment 1 / Re: Problem 5

« on: September 23, 2012, 11:28:50 PM »

The easiest way is to write the general solution and then trying to satisfy boundary conditions. This works for correct settings (and for those with the typo as well).

1768

Home Assignment 1 / Re: Problem 5

« on: September 23, 2012, 06:15:02 PM »

I guess that there is a typo in the assignment for 5.(c) about the Goursat problem:

the formula (9) should be as below to be a Goursat problem
\begin{equation}  u|_{x=3t}=t, \quad u|_{x=-3t}=2t. \end{equation}

Nice spotting! You are right (copy, paste and correct works faster but is more error prone).


PS. Actually problem as stated originally (with $u_t$ instead of $u$) is not IVP problem (as lines don't coincide) and is well-posed as well, but it is not a Goursat problem and it is what was intended. 

1769

Misc Math / Re: integration constant in wave equation

« on: September 23, 2012, 02:33:46 PM »

Good morning,
in the notes "Homogeneous 1D Wave equation" we get to
$$
u(x,t)=\phi(x+ct)+\psi(x-ct)
$$
as the general solution, but in the very last paragraph it is mentioned that we could add a constant to $\phi$ if we subtract that same constant from $\psi$, and that this constant would be the only arbitrariness of this solution. But why does this have to be the same constant?
I can add for instance 1434 to $psi$ and subtract -12i from $\phi$ and the sub of the two would still satisfy the PDE, as any derivative of 1434-12i is zero...
Thanks for your help!

It is not an arbitrariness of the solution, but arbitrariness of the representation of the given solution in the given form. Really, consider the same solution
$$
u(x,t)=\phi_1(x+ct)+\psi_1(x-ct)=\phi_2(x+ct)+\psi_2(x-ct) \qquad \forall x,t,
$$
Then $\phi=\phi_1-\phi_2$ and $\psi=\psi_2-\psi_1$ satisfy
\begin{equation}\phi (x+ct)=\psi (x-ct) \qquad \forall x,t
\label{V}
\end{equation}
plugging $t=x/c$ we get$ \phi(2x)=\psi(0)$ and therefore $\phi(x)=C$ for all $x$. Thus $\phi=C$. Then (\ref{V}) implies that $\psi=-c_1$.

1770

Home Assignment 1 / Re: Problem 5

« on: September 23, 2012, 02:22:16 PM »

My question was with the Goursat problem, should it be of the form:

$$u_{tt}-c^{2}u_{xx}=0$$

$$u(t=\tau)=g(x)$$

$$u_{t}(t=\tau)=h(x)$$

like the example from class?

Then it would be not a Goursat, but Cauchy problem