Partial Differential Equation -MOC

[El_Nino]

Homework Statement

The PDE: [itex]{\partial \rho \over \partial t} + \rho {\partial \rho \over \partial x} =-x\rho[/itex]
and [itex]\rho(x,0) = f(x)[/itex]
Determine a parametric representation of the solution.

Homework Equations

[itex]{dx\over dt} = \rho[/itex]

[itex]{d\rho \over dt} = -x\rho[/itex]

The Attempt at a Solution

Using method of characteristics I get

[itex]\rho = \rho(x_0,0)\exp(-xt) = f(x_0)\exp(-xt)[/itex]

[itex]x=-f(x_0)\exp(-xt)/x + c[/itex]

And since at t=0 x=x_0
[itex]x=-{f(x_0)\over x}(1-\exp(-xt))+x_0[/itex]

This is my solution, but if i plug in f.x f(x) = x and substitute ρ into the PDE, i get rubish.

If f(x)=x, then

[itex]x_0 = {x\over (1-\exp(-xt))/x +1}[/itex]

[itex]\rho = x_0 \exp(-xt)[/itex]

And this substituting this into
[itex]{\partial \rho \over \partial t} + \rho {\partial \rho \over \partial x} +x\rho[/itex]

Gives a non-zero term.

Any idea what I'm doing wrong?

 

[mighty2000]

It looks like you may have made a mistake in your calculation for x_0. When f(x)=x, x_0 should just equal x. So your parametric solution should be ρ = x exp(-xt). Plugging this into the PDE, we get:

{\partial \rho \over \partial t} + \rho {\partial \rho \over \partial x} = -x exp(-xt) - x exp(-xt) = -2x exp(-xt) = -x ρ

which satisfies the PDE. So your solution is correct, you just made a small error in calculating x_0.