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### Messages - Victor Ivrii

Pages: 1 ... 152 153 [154]
2296
##### 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?

2297
##### 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.

2298
##### 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

2299
##### 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!!!

2300
##### 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.

2301
##### Misc Math / MOVED: Do we get extra credit for participating in online discussion?
« on: September 21, 2012, 01:59:12 PM »

2302
##### 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.

2303
##### 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

2304
##### 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

2305
##### Technical Questions / Testing Math
« on: September 18, 2012, 11:59:02 PM »
Testing how MathJax was hooked up
\begin{align*}
u(x,t)= &\underbracket{\frac{1}{2}\bigl[ g(x+ct)+g(x-ct)\bigr]+\frac{1}{2c}\int_{x-ct}^{x+ct} h(y)\,dy}_{=u_2}+\\[3pt]
&\underbracket{\frac{1}{2c}
\iint_{\Delta (x,t)} f(x',t' )\,dx\,d t' }_{=u_1}.
\label{eq-4}
\end{align*}

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