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So I have to demonstrate how to solve $u_{tt} = c^2 u_{xx}$, where $u(0, t) = u(L, t) = u_t(x, 0) = 0$, $$u(x, 0) = f(x) = \begin{cases} 2x/L, &x \in [0, L/2]; \\ 2(L-x)/L, &x \in (L/2, L]. \end{cases}$$ Somehow it's similar to this problem but we don't know the wave speed $c$ and the initial condition $f(x)$ is different.

By using d'Alembert formula, I should get $u(x, t) = 0.5(f(x - ct) + f(x + ct))$ but when I plug in $f(x)$ the solution reduces to $f(x)$ itself. The Fourier series expression has $\cos(n \pi c t/L)$, so I doubt I missed something when I use the formula to find $u(x, t)$. Any insights?

Yuki.F
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  • If $x-ct<0$, then it's not obvious what $f(x-ct)$ is supposed to mean in the first place (and similarly for $f(x+ct)$, and for $x-ct>L$.) My guess would be one must choose $f=0$ in this scenario. – Semiclassical Apr 26 '21 at 15:36
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    @Semiclassical I think alternatively one could also choose to periodically extend $f(x)$ on those spacial domains – Andrew McMillan Apr 26 '21 at 15:38
  • Is it both end $x = 0, x = L$ are fixed at $u = 0$ so we can't just use this formula to get $u(x, t)$? – Yuki.F Apr 26 '21 at 15:40
  • @AndrewMcMillan Good point, especially since this aligns with the Fourier series solution. – Semiclassical Apr 26 '21 at 15:43
  • @AndrewMcMillan e.g. what if I extend $u(x, 0)$ so it becomes odd from $-L$ to $L$ and keeping the nodes $nL$ fixed at $0$? – Yuki.F Apr 26 '21 at 15:45
  • @Yuki.F Why do you want to make $u(x,0)$ odd? – Andrew McMillan Apr 26 '21 at 16:56
  • What do you mean by “the solution reduces to $f(x)$ itself”? You should get the right solution if you extend $f$ to a $2L$-periodic odd function and use d'Alembert. – Hans Lundmark Apr 26 '21 at 17:23
  • @AndrewMcMillan because you suggests periodically extend $f(x)$, so I thought how about making $u(x, 0)$ odd and then extend it to the whole real line. – Yuki.F Apr 27 '21 at 07:45
  • @HansLundmark I mean $0.5(f(x - ct) + f(x + ct)) = f(x)$. Did I do it wrong? – Yuki.F Apr 27 '21 at 07:46
  • That's not right. It's true for some values of $x$ and $t$, but not all. You have two peaks moving apart, and when you superimpose them (after a short time, say), you get a (short) interval around where the initial peak was where the solution $u(x,t)$ is constant, and this constant segment will “sink downwards” and grow longer as the two moving peaks become further separated. When you get down to zero, the whole process will repeat itself in reverse with a negative amplitude, and in the end you get periodic oscillations. – Hans Lundmark Apr 27 '21 at 09:10
  • @HansLundmark Would you mind to show me and the others what the results should be if we extend $f$ to be $2L$-periodic odd and then use d'Alembert? – Yuki.F Apr 27 '21 at 15:19
  • That would be a hassle, since I would need to make pictures in order to explain it properly. But at least I managed to find a video which illustrates the solution $u(x,t)$ that I was trying to describe in my previous comment, even though it doesn't show the individual left- and right-going components, only their sum: https://www.youtube.com/watch?v=7GCwjjyw5dM. – Hans Lundmark Apr 27 '21 at 17:37
  • See also this one, on the same channel: https://www.youtube.com/watch?v=ksscQQvatEc – Hans Lundmark Apr 27 '21 at 17:39

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