# method of characteristics

by gtfitzpatrick
Tags: characteristics, method
 P: 378 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: 1,584 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.
 P: 72 To expand on hunt_mat's answer: Don't we need to separate the variables before integrating? 1/y dy = 1/x dx Hmmm, then again you're using partial deriv. notation, which I've only studied Calc. for, not yet Differential Equations.
P: 378

## method of characteristics

thanks for the replies lads. I guess i never total got to grips with methods of chars.!!!
I also have initial conditions of u(x,1) = x for -infinity<x<infinity.

y=kx becomes k=y/x

then from (b) we have $\frac{ du}{ dx} = -\frac{x^2u^2 }{ x}$ which when differentiated gives $u = \frac{ 2}{ x^2} + F(k)$

and transfer in our earlier value of k gives $u = \frac{ 2}{ x^2} + F(\frac{ y}{ x})$
im getting confused now i think in different methods?
 HW Helper P: 1,584 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

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