Solving PDEs Using Method of Characteristics

[Kreizhn]

Homework Statement

I'm trying to solve the following via the method of characteristics.

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

Homework 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

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

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.

 

[mighty2000]

Thank you for your post. I understand that you are trying to solve the given PDE using the method of characteristics. Your approach is correct, but there are a few things that need to be corrected.

Firstly, the parameterization of P(\omega,t) should be P(\omega(s),t(s)) instead of P(\omega(s),t). This is because both \omega and t are functions of s.

Secondly, for the equation \frac{d\omega}{ds} = k\omega, the solution should be \omega(s) = C_1e^{ks} instead of \omega(s) = C_1e^{kt}.

Thirdly, when substituting the values of \omega(s) and t(s) into the equation for P(s), you should not use the same variable s, as it represents the parameter along the characteristic curve. Instead, you should use a new variable, say \tau, to represent the parameter along the characteristic curve. This will give you a solution of P(\tau) = C_2e^{-\frac{DC_1^2e^{2k\tau}}{4k}}.

Lastly, you are correct in assuming that \omega is a function of s, but this is already accounted for in the equation \frac{dP}{ds} = -\frac{1}{2}D\omega^2P. So you do not need to add an extra \omega dependency in your solution.

I hope this helps clarify your doubts. Keep up the good work in solving PDEs using the method of characteristics!