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?
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Are you assuming $u_t(x,0) = 0$? Otherwise, the solution is different, see wikipedia. – anderstood Apr 05 '20 at 19:51
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@anderstood Is there a confusion with $f$ and $u(x,0)$? The solution I provide is general, with no specific assumption on $u_t(x,0)$ I believe. – pluton Apr 05 '20 at 20:08
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1@Harry49 $u(x,t)=F(x+t)+G(x-t)$ together with $u(0,t)=0$ implies $F(t)=-G(-t)$ that is $G(x-t)=-F(t-x)$, ie $u(x,t)=F(x+t)-F(t-x)$. Do you agree? – pluton Apr 05 '20 at 20:19
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@pluton Ah ok, my bad – anderstood Apr 05 '20 at 20:37
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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.
PinkyWay
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anderstood
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The general solution to $f(T+u)=f(u)+k$ is $f(u)=h(t)+p_T(u)$ where $p_T(u)$ is any $T$-periodic function and $h(t)$ is a particular solution to $h(T+u)=h(u)+k$, for instance $h(u)=(k/T)u$ (https://math.stackexchange.com/questions/3608687/functional-equation-with-delay). – pluton Apr 05 '20 at 22:44
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@pluton What is $t$ in $h(t)$? Do you mean $h(T)$? Then, $h(T)=(k/T)T = k$. In my answer, there is nothing such as a general solution. I just set $k=f(T)-f(0)$. – anderstood Apr 06 '20 at 07:00
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I meant $h(u)$ sorry, with $h(u)=(k/T)u$. Also, your solution assumes that $f$ is differentiable in the usual sense. I think your statement generalizes to the distributional sense as well, to be checked. However, your solution misses the particular solution $h(u)$. – pluton Apr 06 '20 at 15:10
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Your PDE, written as such, only makes sense if $u$ in $\mathcal{C}^2$, hence $f'$ is defined (in the classical sense). Also, I think you should read the answer more carefully; what part exactly do you claim is wrong? @pluton – anderstood Apr 06 '20 at 15:17
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Since $f(u)$ is not periodic, are you sure that $\int_0^Tf=0$ enforces periodicity? For instance, if I take $f(u)=k/Tu+1+\cos u$ where $p_T(u)=1+\cos u$, $\int_0^Tf=0$ does not imply $k=0$ if I am not mistaken... and so $f$ is not periodic – pluton Apr 07 '20 at 00:27
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1@pluton Indeed, I meant $\int_0^T f'$, edited. And that gives $k=0$. – anderstood Apr 07 '20 at 12:02