# Explicit solution to the advection-reaction equation?

Consider the advection-equation equation $$\frac{\partial u}{\partial t}(x,t) + \frac{\partial u}{\partial x}(x,t) + u(x,t) = 0, \quad (x,t) \in (0,2\pi)\times (0,T),$$ with boundary conditions $u(x,0) = \sin(x)$, $u(0,t) = u(2\pi,t) = -\sin(t)$.

I'm trying to get wolfram alpha to give me a solution but i can't, how can I know if there is in fact a solution given those boundary conditions or not?

More precisely, I'm trying to find an explicit solution, but don't know how to proceed. I already found a numerical solution, so ideally I should be able to compute its accuracy.

Thank you.

• The boundary conditions, $u$ at $x=0,2\pi$, are incompatible with the initial condition, $u$ for $t=0$. Can you give us some broader context, e.g. what's the problem origin... – Rafa Budría Jun 27 '17 at 20:37

Setting $u(x,t)=v(t-x,t+x)$ transforms the equation to $$2\partial_2v+v=0$$ so that $$u(x,t)=e^{-(t+x)/2}w(t-x)$$ This implies that $$e^{t/2+s}u(s, t+s)=w(t)=e^{t/2}u(0,t)=e^{t/2+2\pi}u(2\pi,t+2\pi)$$ which is not compatible with the boundary conditions, as they would require $1=e^{2\pi}$.

Actually, it turns out there is a solution:

$$u(x,t) = \begin{cases} e^{-x}\sin(x-t), &x<t\\ e^{-t}\sin(x-t), &x \geq t \end{cases}.$$

This is in fact a $C^1$ function, which solves the original PDE, and satisfies all the boundary conditions.

• Well, except the $u(2\pi,t)=sin(-t)$ is not satisfied... Oh well. – user45453 Jun 28 '17 at 2:02

The advection-reaction equation needs only one boundary condition, because the highest $$x$$ derivative has order of 1. Based on the velocity, which is the coefficient of the $$\frac{\partial u}{\partial x}$$ term which is equal to 1, the direction of advection is from left to right. As a result, you will need a boundary condition on the left side at $$x=0$$ and the boundary condition at the right side $$x=2\pi$$ is superfluous.