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Messages - Victor Ivrii

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2536
Home Assignment 1 / HA1-pdf
« on: September 22, 2012, 01:40:44 PM »
Here is HA1.pdf - Home assignment 1 printed to pdf (updated Mon 24 Sep 2012 05:03:22 EDT)

2537
APM346 Misc / Re: about the MathJax
« on: September 22, 2012, 12:47:24 PM »
i had try to follow the instruction below but don't know wehre to find <head> or <body> code so i can insert the code to.

...snip...

To see how to enter mathematics in your web pages, see Putting mathematics in a web page below.

You do not need to follow the instructions you copied, they are for hooking up MathJax and we did it.

Since you do not see a properly rendered math, I presume that something is wrong with your set-up

Is Javascript enabled?
If you use laptop, or handheld can you come with it to my office during office hours?

2538
Home Assignment 1 / Re: Problem 2
« on: September 22, 2012, 12:40:30 PM »
Hi, for number 2, say we get $u = f(g(x,y))$ for our general solution when $(x,y) \ne (0,0)$. Then are we just finding a $u(0,0)$ that is equal to the limit of $u = f(g(x,y))$ as $(x,y)$ approaching $(0,0)$ in order to make u continuous at $(0,0)$?

Well, you need to be sure that this limit exists, right?

2539
APM346 Announcements / Reminder: Home Assignment 1 is due Mon September 24
« on: September 22, 2012, 07:59:26 AM »

2540
Misc Math / Re: characteristic vs. integral lines
« on: September 22, 2012, 02:20:38 AM »
I feel like I understand what is being said here, but I am confused by the second set of lecture notes,  specifically equation (7) which seems to be doing the same thing as equation (4) in the reply to this post except with a + sign instead of a equal. Am I misunderstanding and these are actually different scenarios, or is one a misprint?

Miranda, thanks!  There was a misprint (mistype) in Lecture 2 (equation (7)), should be =.
Now it has been corrected.

2541
Home Assignment 1 / Re: Problem 4
« on: September 22, 2012, 02:15:54 AM »
In this settings you need to select a solution having certain properties

It may happen that

• Is what one expects
• Unusually narrow

2542
APM346 Misc / Re: about the MathJax
« on: September 22, 2012, 02:12:11 AM »
where can I find the code from my document to start from OR to insert the copied code from the MathJax website to?

What do you want and what is your problem?
• Do you see mathematical expressions I typed on our forum properly?
• Yes
• No
• Have you troubles typing math by yourself?
• Yes
• No
• Are you trying to set MathJax somewhere else?
• Yes
• No

Only in the last case you need to worry about setting MathJax.

2543
Home Assignment 1 / Re: Problem 2
« on: September 21, 2012, 07:30:41 PM »
Thank you for your hint but I still didn't get the point..

For example if the general solution has the form $f(x/y)$, how can I make it continuous at (0,0)? Thanks!!

You are almost done (but check the general solutions!) Think - and don't post solutions!!!

2544
Misc Math / Re: characteristic vs. integral lines
« on: September 21, 2012, 06:36:18 PM »
Heythere,
I am confused by the terms characteristic lines and integral lines. The book introduces characteristic lines as the curves along which a function is constant. Now in the notes integral lines are curves to which the vector field is tangential, i.e. in the case of the gradient vector field the lines along which the function changes most (in abs. value).

So I thought these two should be orthogonal in the case of $au_t+bu_x=0$.
So are integral lines the same as characteristic lines?

If we consider 1-st order PDEs in the form

a_0\partial_t u + a_1\partial_x u + a_2\partial_y u=0
\label{eq-1}

then characteristics of the equation (\ref{eq-1}) are integral lines of the vector field $(a_0,a_1,a_2)$ i.e. curves

\frac{dt}{a_0}=\frac{dx}{a_1}=\frac{dy}{a_2}.
\label{eq-2}

There could be just two variables $(t,x)$ or more ... and coefficients are not necessary constant. Yes, for (\ref{eq-1}) characteristics are lines along which $u$ is constant.

But we preserve the same definition of characteristics as integral lines for more general equation f.e.

a_0\partial_t u + a_1\partial_x u + a_2\partial_y u =f (t,x,y,u)
\label{eq-3}

and here $u$ is no more constant along characteristics but solves ODE

\frac{dt}{a_0}=\frac{dx}{a_1}=\frac{dy}{a_2}=\frac{du}{f}.
\label{eq-4}

Further, notion of characteristics generalizes to higher order equations. Definition curves along which solution is constant goes to the garbage bin almost immediately.

2545
APM346 Misc / Re: about the MathJex
« on: September 21, 2012, 03:09:09 PM »
Hi,
I have problem of using the mathjex in my computer, could anyone help me out for this?

First, it is MathJax as it is related to Ajax http://en.wikipedia.org/wiki/Ajax_(programming)

Do you have troubles seeing math? Or there is a raw code? In the latter case see

Or do you have problems writing math?

2546
APM346 Misc / Re: Do we get extra credit for participating in online discussion?
« on: September 21, 2012, 02:14:08 PM »
Like helping other students to answer questions? Thanks

Yes, the credits will be given for helping.

However JT cannot get any credits as according to BlackBoard there is no such student in the class. Sure, I can try to restore your real name through emailâ€”but why should I? (Please, change your screen name to reflect your BlackBoard name).

Finally, do not post in the wrong boards (note the board description; I moved your post)

HTH

2547
Misc Math / MOVED: Do we get extra credit for participating in online discussion?
« on: September 21, 2012, 01:59:12 PM »

2548
Home Assignment 1 / Re: Problem 2
« on: September 21, 2012, 01:57:34 PM »
First, I need to apologize: this problem contained a misprint which I just corrected. The source of errors is simple: everything was done in the extreme rush and nobody checked it.

Now hint: it may happen in some problems that the solution (having certain properties) does not exist or must have a very special form. Then the your presented solution should state this.

I am not claiming that this is the case with this problem but your statement is wrong anyway: there is at least one continuous solution $u=0$ identically.

2549
Technical Questions / Re: Testing Math
« on: September 21, 2012, 02:55:35 AM »
This is just a test to see whether I can copy and paste from LyX:

u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy

Yes, you can but need to switch on math (inline or display respectively)

Code: [Select]
$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$

Code: [Select]
$$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$$$$u(x, t) = \varphi(x-ct)+\phi(x+ct)-\intop_{x-ct}^{x+ct}g(y)dy$$

Actually I never saw \intop (just \int) and double dollars are deprecated (see my code), but your example works

2550
Home Assignment 1 / Re: Problem 4 [corrected]
« on: September 20, 2012, 07:42:53 PM »
To both: for one of them solution does not exist.

There was an error in the left-hand expression, now it has been corrected

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