- #1
Kreizhn
- 743
- 1
Homework Statement
I'm trying to solve the following via the method of characteristics.
[tex] \frac{\partial P}{\partial t} + k\omega \frac{\partial P}{\partial\omega} = -\frac12 D\omega^2 P [/tex]
Homework Equations
By the method of characteristics, we assume that [itex] P(\omega, t) [/itex] can be parameterized as [itex] P(\omega(s), t(s) ) [/itex]. Then using the chain rule, we get that
[tex] \frac{dP}{ds} = \frac{\partial P}{\partial t} \frac{ dt}{ds} + \frac{\partial P}{\partial \omega} \frac{ d\omega}{ds} [/tex]
So if we have a PDE of the form
[tex] a(\omega, t)\frac{\partial P}{\partial t} + b(\omega, t) \frac{\partial P}{\partial\omega} = c(\omega, t) [/tex]
we can rewrite this as a system of first order ODEs
[tex] \frac{dt}{ds} = a(\omega, t), \quad \frac{d\omega}{ds} = b(\omega, t), \quad \frac{ dP}{ds} = c(\omega, t) [/tex]
The Attempt at a Solution
So I allow [itex] \frac{dt}{ds} = 1, t(0) = 0 [/itex] so that [itex] s=t [/itex]. Next I take [itex] \frac{d\omega}{ds} = k \omega [/itex] so that [itex] \omega(s) = \omega(t) = C_1 e^{kt} [/itex]. Finally, I have [itex] \frac{dP}{ds} = -\frac12 D\omega^2 P [/itex] which implies that
[tex] P(s) = P(t) = C_2 e^{-\frac12 D \omega^2 t} [/tex]
Now I could substitute my value for [itex] \omega(t) [/itex] into here, but it doesn't give me a correct answer. If I leave [itex] \omega [/itex] as a variable, it still doesn't work. So I figured that [itex] \omega [/itex] is actually a function of s and that should be accounted for, so I got [itex] \frac{dP}{ds} = -\frac12 D\omega^2 P = -\frac12 D C_1^2 e^{2ks} P[/itex] 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 [itex] \omega [/itex] dependency somewhere, and I've removed that.