Good afternoon,
Here's the question I'm currently working on:
Determine the solution of the following transport equation: $$u_t = (f(t)-1)u_x \space \space \space ,\space x \in \mathbb{R}, \space 0 < t < T $$
$$ u(x,0) = x $$
I've already solved for some arbitrary functions of $f(t)$, and I've concluded that the solution should look something like this,
$$ u(x,t) = x + \int_{0}^{t} f(s)-1\space ds $$
I've also calculated the characteristic curves for this problem and I can't develop the equations:
$$\frac{dx}{ds} = -(f(t)-1) \implies x(s) = - \int_{0}^{s} f(t) - 1 \space dt$$
$$\frac{dt}{ds} = 1 \implies t(s) = s + \alpha \space , \space \alpha \in \mathbb{R} $$
Could I have some hint on how to follow up and come up with the solution for the problem? I also don't have any particular information on the function $f$.
Thanks for reading this far!
The characteristics are paths of the form $x(t;x_0)=x_0 - \int_0^t f(s) - 1 ds$. You then have $\frac{d}{dt} u(x(t;x_0),t)=0$. Therefore $u(x(t;x_0),t)=u(x(0;x_0),0)=x_0$ so it suffices to find $x_0(x,t)$. But you know $x=x_0 - \int_0^t f(s) - 1 ds$ so $x_0=x+\int_0^t f(s) - 1 ds$ as desired.