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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.

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  • $\begingroup$ Sorry that I forgot the first few words of the problem-statement upfront. I just edited it, and it is exact as of now. $\endgroup$ – Procore Oct 31 '15 at 23:48
  • $\begingroup$ @Winther I used Mathematica to confirm what you said...it provided an imaginary answer. I asked my Professor about it, and he recommended to used the characteristic method, and then look at the geometry of things to develop the solution. I have yet to return the this problem, but I at least know my solution is correct. $\endgroup$ – Procore Nov 1 '15 at 23:34
  • $\begingroup$ @Winter (Sorry for not mentioning this) The problem is from a take-home midterm that we turned in last week, and I know I couldn't provide the full answer. I unfortunately don't have a link for the problem, as I have a hard copy of it, but the problem is exact. $\endgroup$ – Procore Nov 2 '15 at 0:23
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    $\begingroup$ I think I was wrong. You need to define a different function $F(z)$ for $z<0$ and $z>0$ and then you get the correct solution I think. However the drawback is that $F$ will not be $C^1$ at $z=0$. $\endgroup$ – Winther Nov 2 '15 at 0:27
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    $\begingroup$ For reference I find $u(t,x) = e^{-2t} F(x-3t)$ where $F(z) = \left\{\matrix{\sin(z) & z\geq 0\\ -\frac{z}{3}e^{-\frac{2z}{3}} & z \leq 0}\right.$. $\endgroup$ – Winther Nov 2 '15 at 0:32
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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.

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  • $\begingroup$ Norman, tank you for your answer! I was having trouble setting up the ODE in regards to the characteristic method (I, at least, had the right-hand side of the ODE correct...woo hoo). We've covered in class what you mean about "re-tracing" the characteristic lines backwards. I'm applying the finishing touches now, and thank you (again) for your solution...this helps A LOT! $\endgroup$ – Procore Nov 2 '15 at 18:08

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