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Today I came across a question on PDE which makes me really frustrating.

The question is to solve this initial boundary value problem using method of separation variables:

$$u_{tt}=9u_{xx}\text{ for } x>0, t>0$$

$$u(x,0)=x^2 ,\ u_t(x,0)=e^{-x} \text{ for } x>0$$

$$u(0,t)=0 \text{ for }t>0$$

Given answer: $u(x,t)=x^2+9t^2+\dfrac{1}{3}e^{-x}\sinh3t$ for $x-3t>0$

$$u(x,t)=\begin{cases}x^2+9t^2+\dfrac{1}{3}e^{-x}\sinh3t&x-3t>0\\x^2+9t^2+\dfrac{1}{3}e^{-3t}\sinh x&x-3t<0\end{cases}$$

I really dont know how to solve this question because they only provide 1 boundary condition only.

Could someone please show some workings on this problem so that I can undesrtand clearly. Furthermore, I dont have example of this question in my textbook. And my teacher also not able to do it because he said he is not good in PDE chapter. Im really dead now because no one able to help me and my friends also not sure about this. So please please help me to solve this only question so that I can use this solution as reference to solve my other PDEs problems by my ownself.

Lastly, if you can help me, I will really be very grateful to you.

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    $\begingroup$ Given the solution, do you think separation of variables is going to do much good here? $\endgroup$
    – Ron Gordon
    Jan 5, 2013 at 17:52
  • $\begingroup$ I got the answer using the D'Alembert Solution for the first part of the soluiton x-3t>0 but i cant get the second line of the solution which is x-3t<0.And i cant understand why there are 2 solution?? $\endgroup$
    – Garett
    Jan 5, 2013 at 18:16
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    $\begingroup$ I was going to outline a solution using D'Alembert. SoV is not going to help here, as the boundary conditions are not amenable to them. $\endgroup$
    – Ron Gordon
    Jan 5, 2013 at 18:18
  • $\begingroup$ Thanks rlgordonma.That helps me alot for this fuzzy problem. But how am I suppose to get to the 2nd line of the solution which is for x-3t<0 ?? $\endgroup$
    – Garett
    Jan 5, 2013 at 18:30
  • $\begingroup$ This is a reflection principle problem. I solved one such problem recently; you should try to follow the same here. The idea is to extend the initial condition ($x^2$, $e^{-x}$) to the entire line as odd functions. This will enforce $u(x,t)=-u(-x,t)$ for all time, in particular $u(0,t)=0$ will hold for all times. The solution comes directly from d'Alembert's formula. Since it involves the piecewise defined initial conditions $x^2\operatorname{sign}x$ and $e^{-x}\operatorname{sign}x$, the solution also comes out to be piecewise defined. $\endgroup$
    – user53153
    Jan 5, 2013 at 19:36

1 Answer 1

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Note that when without the condition $u(0,t)=0$ this is in fact a just-determining problem and the solution can be expressed by using D’Alembert’s formula $u(x,t)=\dfrac{(x+3t)^2+(x-3t)^2}{2}+\dfrac{1}{6}\int_{x-3t}^{x+3t}e^{-s}~ds=\dfrac{x^2+6xt+9t^2+x^2-6xt+9t^2}{2}-\dfrac{1}{6}[e^{-s}]_{x-3t}^{x+3t}=\dfrac{2x^2+18t^2}{2}-\dfrac{e^{-x-3t}-e^{-x+3t}}{6}=x^2+9t^2+\dfrac{e^{-x}\sinh3t}{3}$

Check for $u(0,t)$ :

$u(0,t)=9t^2+\dfrac{\sinh3t}{3}\neq0$

$\therefore$ We should need to use the result in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf:

The solution is $u(x,t)=\begin{cases}\dfrac{(x+3t)^2+(x-3t)^2}{2}+\dfrac{1}{6}\int_{x-3t}^{x+3t}e^{-s}~ds&\text{when}~x-3t>0\\\dfrac{(x+3t)^2-(3t-x)^2}{2}+\dfrac{1}{6}\int_{3t-x}^{x+3t}e^{-s}~ds&\text{when}~x-3t<0\end{cases}=\begin{cases}\dfrac{x^2+6xt+9t^2+x^2-6xt+9t^2}{2}-\dfrac{1}{6}[e^{-s}]_{x-3t}^{x+3t}&\text{when}~x-3t>0\\\dfrac{x^2+6xt+9t^2-(9t^2-6xt+x^2)}{2}-\dfrac{1}{6}[e^{-s}]_{3t-x}^{x+3t}&\text{when}~x-3t<0\end{cases}=\begin{cases}\dfrac{2x^2+18t^2}{2}-\dfrac{e^{-x-3t}-e^{-x+3t}}{6}&\text{when}~x-3t>0\\\dfrac{12xt}{2}-\dfrac{e^{-x-3t}-e^{-3t+x}}{6}&\text{when}~x-3t<0\end{cases}=\begin{cases}x^2+9t^2+\dfrac{e^{-x}\sinh3t}{3}&\text{when}~x-3t>0\\6xt+\dfrac{e^{-3t}\sinh x}{3}&\text{when}~x-3t<0\end{cases}$

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  • $\begingroup$ Thank you very much doraemonpaul. I got the same answer as you also for the 2nd line of the solution which is $ 6xt+ (e^{-3t}sinhx/3) $. I think the answer proposed by the question is wrong. Once again you showed me great help on this question and previous question on PDE. Thank you man!!! $\endgroup$
    – Garett
    Jan 6, 2013 at 3:59
  • $\begingroup$ +1 nice one. May I ask you to see this one math.stackexchange.com/a/275105/8581, and tell me if it is true? Thanks for the time. ;) $\endgroup$
    – Mikasa
    Jan 10, 2013 at 13:54

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