I am considering the usual homogeneous 1D wave equation $u_{tt}-u_{xx}=0$ ($c=1$ for the sake of simplicity) on $(x,t)\in\mathbb{R}^2$ together with the condition $u(0,t)=0$. From d'Alembert solution, it is easy to show that $u(x,t)=f(t+x)-f(t-x)$ where $f$ is any function. The question is on the periodicity of $u(x,t)$ in $t$. It is easy to show that if $f$ is a periodic function of period $T>0$ so is $u(x,t)$ in $t$. However, the converse is not clear. Consider that $u(x,t)$ is periodic in time of period $T>0$, then $u(x,t+T)=u(x,t)$ implies $f(t+T+x)-f(t+T-x)=f(t+x)-f(t-x)$ which does not say anything about the periodicity of $f$… Hints?
1D wave equation: d’Alembert solution and periodicity
partial differential equationswave equation
Related Solutions
Have you thought of using spatial Fourier series? Using the spatial periodicity hypothesis, we may expand $$u(x,t) =\sum_n a_n(t)\cos (nkx) + b_n(t)\sin (nkx)$$ as a spatial Fourier series with period $2\pi/k$. The PDE $u_{tt}=u_{xx}$ leads to a set of second-order ODEs for $a_n$, $b_n$, while the initial conditions of the PDE give the initial conditions $a_n(0)$, $b_n(0)$, $a'_n(0)$, $b'_n(0)$ of the ODEs.
Alternatively, using the method of characteristics, one may be able to develop a modified d'Alembert formula (see chap. 12-* of the book (1)).
(1) R. Habermann, Applied Partial Differential Equations; with Fourier Series and Boundary Value Problems, 5th ed., Pearson Education Inc., 2013.
It seems I was confused or made mistakes the first time trying to solve this, so I will post my solution now that it makes sense.
Assume $u(x,t)=F(x+2t)+G(x-2t)$. Then, the initial conditions give us: $$ u(x,0)=F(x)+G(x)=e^{-x}\\ u_t(x,0)=2F'(x)-2G'(x)=2e^{-x} $$
which hold for all $x>0$.
Hence, in the last equation, we may divide both sides by 2 and integrate with respect to x, then solve the resulting system: $$ \begin{cases} F(x)+G(x)=e^{-x}\\ F(x)-G(x)=-e^{-x}+C \end{cases} $$
where $C$ is some constant.
We thus obtain $F(x)=C/2$ and $G(x)=e^{-x}-C/2$. Since the only condition is that $x$ is positive, we may replace it by any positive quantity (I think this is where I was previously confused). Thus, we obtain $u(x,t)=F(x+2t)+G(x-2t)=C/2+e^{-(x-2t)}-C/2=e^{2t-x}$ which holds for all $x>2t$. It is clear, as I noted in my original post, that this corresponds to the same solution that we get from d'Alembert's formula.
Now we handle the case when the argument to $G$ is negative. For this, we need to use the boundary condition.
We have $u_x(0,t)=F'(2t)+G'(-2t)=-\cos t$ for all $t>0$. Make the substitution $z=-2t$ to obtain $G'(z)=-\cos(z/2)-F'(-z)=-\cos(z/2)$ by noting that $F'(-z)=0$ from our previous work. Integrating, we obtain $G(z)=-2\sin(z/2)+\tilde{C}$.
Applying the continuity condition, we must have $G(0)=\tilde{C}=1-C/2$. Thus, we have $u(x,t)=F(x+2t)+G(x-2t)=1-2\sin(x/2-t)$ for $0<x<2t$.
Hence, the complete solution is:
$$u(x,t)=\begin{cases}1-2\sin(x/2-t),&0<x\leq 2t\\e^{2t-x},&x>2t\end{cases}.$$
We can now verify that the initial conditions, boundary condition, and continuity are satisfied.
Best Answer
Taking $x=t$: $$f(T+2t) - f(T) = f(2t) - f(0)$$ So $\forall u,$ $f(T+u) = f(u) + k$ with $k$ some constant. So, $f'$ is $T$-periodic.
Also, note that $f$ is not uniquely defined, it can vary by an additive constant. You can define this constant such that $\int_0^T f'(u)du=0$, then the corresponding $f$ will be periodic, too.