2
$\begingroup$

I need to solve the following by using the method of characteristics $u\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1~,~u|_{x=y}=\frac{x}{2}$

I have the following characteric equations:

$\frac{dx}{ds}=u~;\frac{dy}{ds}=1~;\frac{du}{ds}=1$

from the above I get

$ x=us+x_{0} $ $ y=s +y_{0} $ $ u=s+u_{0} $

I am now thinking I should go with the standard conditions $y_0=0$ and $u(x,0)=f(x_0)$

this now gives me: $ x=uy+x_o $ $ y=s $ $ u=y+f(x_0) $

Im confused because of the $u$ term in my equation for $x$

Can anyone please help.

Thanks a mil

  • 1
    http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node8.html2011-10-05

1 Answers 1

1

Your solution to characteristic equations is incorrect, which you can easily check by plugging your current solution back in. The source of the problem is that $u(s)$ is not constant, thus $x^\prime(s) = u(s)$ is not solved by $x(s) = u(s) s + x_0$.

It may help to note that $x^{\prime\prime}(s) = u^\prime(s) = 1$. Now that you are back into ODE with constant coefficients, finding solutions should be easy.

  • 0
    its easier to put the math stuff here. But It says I should move it to chat. :| ... ok, so I put in the chat that I get $s=y_0-2u_0$ so where do I sub this into? I feel like Im going round in circles.2011-10-05