Partial Differential Equation Problem

**r2dee6** · January 29th 2009, 01:21 PM

Solve the PDE

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

Plot characteristc and solution at t = 0 and 1

**Jester** · January 29th 2009, 01:46 PM

Originally Posted by r2dee6

Solve the PDE

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

Plot characteristc and solution at t = 0 and 1

If

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

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

Using the method of characteristics

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

which we solve in pairs giving

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

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

The solution is

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

Imposing the initial condition gives

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

and the final solution

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

**r2dee6** · January 29th 2009, 02:16 PM

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

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