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