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 ;)

Re: method of characteristic

hi,

I advanced in my problem but I'm again in trouble

I transformed my expression with polar coordinates $\displaystyle v(\frac{dv}{dr})=r$

so $\displaystyle v=r+c_{3}$ (c is my constant) $\displaystyle u=sqrt(x^2+y^2)+c_{3}$

Now I have to find the value of my constant by caracteristics method

so I continue my reasoning:

$\displaystyle \frac{dr}{dt}=v=r+c_{3}$ such as $\displaystyle r=e^(t)c_{1}-c_{3}$

$\displaystyle \frac{d(\theta)}{dt}=0 $ such as $\displaystyle \theta=c_{2}$

$\displaystyle \frac{dv}{dt}=r $ we find before that $\displaystyle v=r+c_{3}$

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

so i guess $\displaystyle r=(\frac{s}{cos(\theta)})$

$\displaystyle \theta=1$

$\displaystyle v=sqrt{(s^2+1)}$

for t=0, $\displaystyle c_{1}-c_{3}=\frac{s}{cos(\theta)}$

$\displaystyle c_{2}=1$

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

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

in short I feel that's wrong

someone can help me please??

thanks:)