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Messages - Victor Ivrii

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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!!!

Misc Math / Re: characteristic vs. integral lines
« on: September 21, 2012, 06:36:18 PM »
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
then characteristics of the equation (\ref{eq-1}) are integral lines of the vector field $(a_0,a_1,a_2)$ i.e. curves
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)
and here $u$ is no more constant along characteristics but solves ODE
Further, notion of characteristics generalizes to higher order equations. Definition curves along which solution is constant goes to the garbage bin almost immediately.

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.

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

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

Technical Questions / Testing Math
« on: September 18, 2012, 11:59:02 PM »
Testing how MathJax was hooked up
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]
\iint_{\Delta (x,t)} f(x',t' )\,dx\,d t' }_{=u_1}.

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