Solving a first-order transport PDE with initial/boundary conditions on a halfline

Question

I have another question that I need help with.

My main issue concerns the following problem (to which I came up with a general solution for - see below...further, the question is given exactly how it is presented):

Problem:

Solve the initial-boundary value problem:

$\begin{cases}u_{t}+3u_{x}+2u=0,~~~t,x\in\mathbb{R}^{+}\\ ~\\ u(0,x)=\sin(x),~~u(t,0)=t.\end{cases}$

The general solutions I came up with is $u(t,x)=e^{-\frac{1}{5}(t+3x)}F(3t-x)$, where $F\in C^{1}(\mathbb{R})$. This can be derived in a number of ways, but the way I did it is via the following change of variables, $\xi=t+3x,~\eta=3t-x$. Details are omitted, however, using the change-of-variables previously mentioned will provide the transformed PDE of $u_{\xi}+\frac{1}{5}u=0$ which can be easily solved with the integrating factor $I=e^{\frac{1}{5}\int d\xi}=e^{\frac{1}{5}\xi}$, and then partially integrating the equation after multiplying both sides by $I$...this gives the general solution I provided above. Furthermore, using the initial condition $u(0,x)=\sin(x)$ will give a unique solution of $u(t,x)=-e^{-2t}\sin(3t-x)$ - note that the terms invovling the variable $x$ in the exponent of $e$ will cancel after applying this initial condition.

This is where I need help, as applying the second, boundary condition of $u(t,0)=t$ will yield $t=-e^{-2t}\sin(3t)$. I've also tried using this condition, $u(t,0)=t$, with the general solution first to come up with its, respective, unique solution (because we get a different $F$), but then I run into the same problem when trying to use the remaining boundary condition.

I mainly just need a hint on how the remaining mechanics work.

Answer 1

One can try to deal with this using a special reflection from half-line to the entire line, but I prefer a direct approach. The characteristics are lines $t=t_0+s$, $x=x_0+3s$, with parameter $s$. Along any such line we have $du/ds = -2u$, and solving this ODE yields $$u(t_0+s,x_0+3s)=e^{-2s}u(t_0,x_0)$$ It remains to figure out what $(t_0,x_0)$ should be here. Think of going backward along a characteristic line: you hit either $t=0$ or $x=0$ Accordingly, there are two cases:

1. If $x>3t$, then the initial point is $(0,x-3t)$, at which the initial value is $\sin(x-3t)$. Also, $s=t$ here, leading to $u(x,t)=e^{-2t}\sin(x-3t)$.
2. If $x<3t$, then the initial point is $(t-x/3,0)$, at which the initial value is $t-x/3$. Also, $s=x/3$ here, leading to $u(x,t)=e^{-2x/3}(t-x/3)$.

These give the same result on the boundary $3x=t$, which is nice, and is a reflection of the fact that $(\sin 3x)_{x=0} = (t)_{t=0}$. The solution is not differentiable on $x=3t$, which is not unusual for the transport equation: it carries forward whatever singularities and corners were present in the initial data. Here the data had a "corner" due to the combination of initial and boundary conditions.