# method of characteristic

• Nov 21st 2012, 02:31 PM
metouka
method of characteristic !!!
(Rofl) Hi everyone,
i have a problem for solving a partial differential equation with the method of characteristic. I already spent lot of time and I have not yet found the solution
http://latex.codecogs.com/gif.latex?...+y^2, x>0, y>0
http://latex.codecogs.com/gif.latex?u(x,1)=\sqrt{x^2+1}
In fact, when I try to solve du/dt after I already obtained before dx/dt and dy/dt, I have an expression an expression which depends on t and u. Impossible to solve...
If someone can help me pleasseeeee!!!!

Thankss ;)
• Nov 23rd 2012, 02:14 AM
metouka
Re: method of characteristic
hi,
I advanced in my problem but I'm again in trouble
I transformed my expression with polar coordinates $v(\frac{dv}{dr})=r$
so $v=r+c_{3}$ (c is my constant) $u=sqrt(x^2+y^2)+c_{3}$

Now I have to find the value of my constant by caracteristics method
so I continue my reasoning:

$\frac{dr}{dt}=v=r+c_{3}$ such as $r=e^(t)c_{1}-c_{3}$
$\frac{d(\theta)}{dt}=0$ such as $\theta=c_{2}$
$\frac{dv}{dt}=r$ we find before that $v=r+c_{3}$

moreover with my initial condition $v(rcos(\theta),1)=\sqrt{(r^2.cos^2(\theta)+1)}$
so i guess $r=(\frac{s}{cos(\theta)})$
$\theta=1$
$v=sqrt{(s^2+1)}$

for t=0, $c_{1}-c_{3}=\frac{s}{cos(\theta)}$
$c_{2}=1$
$c1=sqrt{(s^2+1)}$ donc que $c_{3}=sqrt{(s^2+1)}-\frac{s}{cos(\theta)}$

so $u=e^(t)sqrt{(s^2+1)}-(sqrt{(s^2+1)}-\frac{s}{cos(\theta)})=sqrt{(s^2+1)}(e^(t)-1) + (\frac{s}{cos(\theta)}$

in short I feel that's wrong