Thread: Partial Differential Equation Problem

Solve the PDE

$\displaystyle \frac {\partial u}{\partial t} $ + $\displaystyle 3t \frac {\partial u}{\partial x} $ = $\displaystyle u$ and $\displaystyle u(x,0)$ = $\displaystyle \exp^(-x^2)$

Plot characteristc and solution at t = 0 and 1

Originally Posted by r2dee6

Solve the PDE

$\displaystyle \frac {\partial u}{\partial t} $ + $\displaystyle 3t \frac {\partial u}{\partial x} $ = $\displaystyle u$ and $\displaystyle u(x,0)$ = $\displaystyle \exp^(-x^2)$

Plot characteristc and solution at t = 0 and 1

If

$\displaystyle \frac{du}{dt} = \frac{\partial u}{\partial t} + \frac{dx}{dt} \frac{\partial u}{\partial x}$ then your characteristic curve is given by

$\displaystyle \frac{dx}{dt} = 3t$.

Using the method of characteristics

$\displaystyle \frac{dt}{1} = \frac{dx}{3t} = \frac{du}{u}$

which we solve in pairs giving

$\displaystyle dx = 3 t \, dt\;\;\; c_1 = x - \frac{3}{2} t^2$ (you can plot this)

$\displaystyle du = dt\;\;\; c_2 = u e^{-t}$

The solution is

$\displaystyle u = e^t f(x - \frac{3}{2} t^2)$

Imposing the initial condition gives

$\displaystyle f(x) = e^{-x^2} $

and the final solution

$\displaystyle u = e^{t -\left( x - \frac{3}{2} t^2 \right)^2}$

hey thanks but how do i get teh characteristic equations?i mean how do i plot them ?