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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.

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  • $\begingroup$ 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... $\endgroup$ – Rafa Budría Jun 27 '17 at 20:37
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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 $1=e^{2\pi}$.

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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.

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

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