# Method of characteristics

by gtfitzpatrick
Tags: characteristics, method
 P: 376 1. The problem statement, all variables and given/known data $x \frac{ \partial u}{ \partial x} + y \frac{ \partial u}{ \partial y}= -x^2u^2$ 2. Relevant equations 3. The attempt at a solution characteristics are given by $\frac{ dy}{ dx} = \frac{ y}{ x}$ (a) and $\frac{ du}{ dx} = -\frac{x^2u^2 }{ x}$ (b) So i integrate both equations but for (a) do i bring the y across which ends up giving ln(y) = ln(x) + k or leave it where it is and i get y = -yln(x) + k ??
 HW Helper P: 1,583 Hello, still doing characteristics? Why not write the characteristic equations as: $$\dot{x}=x,\quad\dot{y}=y,\quad\dot{u}=-x^{2}u^{2}$$ You won't be able to write down a complete solution as you have no initial condition to work from. For your equation for characteristics, your first answer was correct, the characteristics are given by $y=kx$ for k constant.
 HW Helper P: 1,583 There are essentially two ways for the method of characteristics (I recognise your name from a number of MOC posts, did I post my notes on the subject?) From one of your calculations you have: $$\frac{du}{dx}=-xu^{2}$$ Integrating this equation shows that: $$\frac{1}{u}=\frac{x^{2}}{2}+F(\xi )$$ We now paramatrise the initial data, so take $(\xi ,1)$ as the point which the characteristic passes through, this will give the initial values as $u(\xi ,1)=\xi$, evaluating the characteristic at this point yields $1=k\xi$, giving a value for k which can then be inserted back into the equation for the characteristic. We now evaluate $u$ at the point $(\xi ,1)$ to obtain $$\frac{1}{\xi}=\frac{\xi^{2}}{2}+F(\xi )$$ From this you can compute $F(\xi )$ and then from there you can substitute for $\xi$ by using the equation of the characteristic. Now to find u you have the solution