I found the following PDE in a book, describing a kind of a heating system:
$\partial_tu=f(t)-\partial_xu-u$
where $u=u(t,x)$ and $(t,x)\in(0,\infty)\times(0,1)$.
After own research i found out that is a special type of so called 'transport equation'. I tried to solve this by method of characteristics, but without success (i'm not familiar with transport equations). Then i tried to treat this equation as parabolic equation
$\partial_tu-Au=f(t), \text{ where } A:=-(\partial_x+\mathrm{id})$
but here also without success(i considered the associated semi group...).
So has someone any idea? Is this equation solvable at all?
I intentionally make no assumtion on the function $f$, i tried to solve the equation in classical sense. But if you have any proof in weak/sobolev sense, please share it with me. I also omit boundary conditions.
Thank you
The PDE $u_t + u_x = f-u$ is a non-homogeneous linear advection equation. Let us solve the initial value problem $u(x,0)=u_0(x)$ by using the method of characteristics. We introduce the parametrization $x(t)$ of $x$ such that the total time-derivative of $u(x(t),t)$ satisfies $$ \frac{\text d u}{\text d t} = u_x \frac{\text d x}{\text d t} + u_t = f-u \, . $$ According to the PDE, we therefore choose $\frac{\text d x}{\text d t} = 1$. Hence, the characteristic curves $(x(t),t)$ are straight parallel lines $x(t) = t + x_0$ in the $x$-$t$ plane, along which $u$ is not constant. More precisely, $u(x(t),t)$ is obtained by integration of the above differential equation, with the initial condition $u(x_0,0) = u_0(x_0)$. Finally, after elimination of $x_0 = x(t) - t$, we obtain $$ u(x,t) = u_0(x-t)\, e^{-t} + \int_0^t e^{-(t-\tau)} f(\tau)\,\text d \tau\, . $$