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I'd like to solve the following PDE: for $x\in[0,L]$ and $t\in\mathbb{R^+}$,

$$\left\lbrace \begin{array}{l} \partial^2_{tt} u - c^2\partial^2_{xx}u = 0 \\ u(x,0)=0 \\ \partial_t u(x,0)=\varphi(x) \\ \partial_x u(0,0)=0\\ \partial_x u(L,0)=0 \end{array}\right. $$

Originally, I tried the following. The solutions are of the form $u(x,t)=f(x+ct)+g(x-ct)$, where $f,g$ are piecewise $\mathcal{C}^2$ over $\mathbb{R}$.

From the initial conditions, $\forall x\in [0,L]$, $f(x)+g(x)=0$ and $cf'(x)-cg'(x)=\varphi(x)$, or: $$\left\lbrace\begin{array}{l} f(x)=\frac{1}{2c}\int_0^x \varphi(\xi)\,\text{d}\xi\\ g(x)=-\frac{1}{2c}\int_0^x \varphi(\xi)\,\text{d}\xi \end{array}\right.$$

From the boundary conditions, $\forall t\geq 0$, $f'(ct)+g'(-ct)=0$ and $f'(L+ct)+g'(L-ct)=0$ so $f'(L+ct)=-g'(L-ct)=-g'(-L-ct)$ so $f$ and $g$ are $2L$-antiperiodic.

Putting everything together, $$u(x,t)=\dfrac{1}{2c}\int_0^{x+ct} \varphi(\xi)\,\text{d}\xi - \dfrac{1}{2c}\int_0^{x-ct} \varphi(\xi)\,\text{d}\xi = \dfrac{1}{2c}\int_{x-ct}^{x+ct} \varphi(\xi)\,\text{d}\xi$$

I have some doubts that this is the solution over $[0,L]\times \mathbb{R}^+$ because it does not seem to much the initial $t$-derivative and the Neumann conditions:

$$2c\partial_t u(x,t)=c\varphi(x+ct)+c\varphi(x-ct)+\int_{x-ct}^{x+ct} \partial_t\varphi(\xi)\,\text{d}\xi$$ so $\partial_t u(x,0)=\varphi(x)\neq 0$. Similarly, $$2c\partial_x u(x,t)=\varphi(x+ct)-\varphi(x-ct)+\int_{x-ct}^{x+ct}\varphi'(\xi)\,\text{d}\xi$$ so $\partial_x u(0,t)=2(\varphi(ct)-\varphi(-ct))\neq 0$ and $\partial_x u(L,t)=2(\varphi(L+ct)-\varphi(L-ct))\neq 0$.

Additionally, $\varphi$ is defined over $[0,L]$ so that's another reason to believe there's something wrong, but where? Lastly, as commented by Ian, the form $f(x+ct)+g(x-ct)$ does not account for the wave reflections. Maybe making the problem infinite with $L$-periodic initial conditions would do the trick?

Some questions indicate separating variables but this would lead to stationnary solutions. I am interested in any way of finding the general solution.

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    $\begingroup$ The D'Alambert solution does not exactly work on a finite domain... $\endgroup$
    – Ian
    Aug 31, 2016 at 1:57
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    $\begingroup$ Yes; in some sense the signal hits the boundary and must get "reflected" instead of propagating outside the domain. $\endgroup$
    – Ian
    Aug 31, 2016 at 2:05
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    $\begingroup$ That's no loss; any $L^2$ function has a Fourier series that converges in the sense of $L^2$. $\endgroup$
    – Ian
    Aug 31, 2016 at 2:45
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    $\begingroup$ @Ian Actually, d'Alembert solution works on a finite domain, it first the domain is extended to $\mathbb{R}$ by (anti-)periodicity and symmetry. $\endgroup$
    – anderstood
    Sep 4, 2016 at 17:28
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    $\begingroup$ It can be modified, but there is nontrivial extension to be handled, and the success of the technique depends on the relevant boundary conditions. $\endgroup$
    – Ian
    Sep 4, 2016 at 17:31

1 Answer 1

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I figured out the answer. The solution was correct, but details were missing. The trick was to extend all the functions $f,g,\varphi$ to $\mathbb{R}$. The boundary conditions imply that they can be extended by $2L$-periodic even functions. Then, $\partial_x u(0,t)=\frac{1}{2}\varphi(ct)-\varphi(-ct)=0$ and similarly for $\partial_x u(L,t)$.

So, after extending the domains of definition, $u(x,t)=f(x-ct)+g(x+ct)$ holds.

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