# Homework Help: Method of Characteristics

1. Jan 13, 2010

### Kreizhn

1. The problem statement, all variables and given/known data
I'm trying to solve the following via the method of characteristics.

$$\frac{\partial P}{\partial t} + k\omega \frac{\partial P}{\partial\omega} = -\frac12 D\omega^2 P$$

2. Relevant equations
By the method of characteristics, we assume that $P(\omega, t)$ can be parameterized as $P(\omega(s), t(s) )$. Then using the chain rule, we get that

$$\frac{dP}{ds} = \frac{\partial P}{\partial t} \frac{ dt}{ds} + \frac{\partial P}{\partial \omega} \frac{ d\omega}{ds}$$

So if we have a PDE of the form

$$a(\omega, t)\frac{\partial P}{\partial t} + b(\omega, t) \frac{\partial P}{\partial\omega} = c(\omega, t)$$

we can rewrite this as a system of first order ODEs

$$\frac{dt}{ds} = a(\omega, t), \quad \frac{d\omega}{ds} = b(\omega, t), \quad \frac{ dP}{ds} = c(\omega, t)$$

3. The attempt at a solution
So I allow $\frac{dt}{ds} = 1, t(0) = 0$ so that $s=t$. Next I take $\frac{d\omega}{ds} = k \omega$ so that $\omega(s) = \omega(t) = C_1 e^{kt}$. Finally, I have $\frac{dP}{ds} = -\frac12 D\omega^2 P$ which implies that

$$P(s) = P(t) = C_2 e^{-\frac12 D \omega^2 t}$$

Now I could substitute my value for $\omega(t)$ into here, but it doesn't give me a correct answer. If I leave $\omega$ as a variable, it still doesn't work. So I figured that $\omega$ is actually a function of s and that should be accounted for, so I got $\frac{dP}{ds} = -\frac12 D\omega^2 P = -\frac12 D C_1^2 e^{2ks} P$ which gives me a solution of

[tex] P(s) = -C_2 e^{-\frac{DC_1^2 e^{2ks} }{4k} } [/itex]

But then not only is this not correct, it seems to me that there should also be an $\omega$ dependency somewhere, and I've removed that.