PDE with periodic boundary conditions I am trying to solve
$$u_{t} = \alpha u_{x}$$
for some $\alpha>0$ subject to the boundary conditions $u(0,x) = h(x)$ and $u(t,0) = u(t,L)$ for $t>0$ but I am having far more trouble than I expected.
I've tried using separation of variables, but came to the conclusion that these boundary conditions only admit trivial solutions if we specify that $u(t,x) = \xi(x)\eta(t)$. Other than that, I tried to begin with an ansatz involving complex exponentials, but this only led me think this is a harder problem than I thought it was.
I am sure a question like this has a standard technique to use, but I cannot figure out what it is or find it online.
 A: Standard techniques is to look-around-and-find some self-adjoint differential operator. The case when no self-adjoint differential operator can be found requires much more advanced approach not needed here. Solution is to be periodic  in variable $x$, hence the only possible option around is the self-adjoint differential operator
$$
\mathcal{L}=i\frac{d\,}{dx}\colon D_{\mathcal{L}}\overset{\rm def}{=}\{v\in H^1(0,L)\,\colon\; v(0)=v(L)\}
\subset L^2(0,L)\to L^2(0,L)\tag{$\ast$}
$$
Solution of the Sturm-Liouville problem 
$$
\begin{cases}
\mathcal{L}X=\lambda X,\\
\mathcal{L}\in D_{\mathcal{L}}
\end{cases}
\quad
\begin{cases}
X_n=e^{2\pi inx/L},\\
\lambda_n=-2\pi n/L,\;n\in\mathbb{Z}.
\end{cases}
$$
is an orthogonal basis in $\,L^2(0,L)$, and in accordance with the standard techniques,
the desired solution is to be represented in the form
$$
u(t,x)=\sum_{n=-\infty}^{\infty}T_n(t)e^{2\pi inx/L},\quad 
T_n(t)=\frac{(u,X_n)}{\|X_n\|^2}=
\frac{1}{L}\int\limits_0^L u(t,x)e^{-2\pi inx/L}dx
$$
with the sesquilinear form $(\cdot,\cdot)$ standing for the inner product in complex 
$\,L^2(0,L)$, i.e.,
$$
(v,w)=\int\limits_0^L v\,\overline{w}\,dx.
$$
Multiplying the equation $u_t=\alpha u_x$ by the eigenfunction $\,X_n\,$ w.r.t. the inner product $(\cdot,\cdot)$ we get $i(u_t\,,X_n)= \alpha(iu_x\,,X_n)$, where $(iu_x\,,X_n)=(u,iX'_n)=(u,\lambda_nX_n)=\overline{\lambda}_n(u,X_n)=\lambda_n(u,X_n)$ since the differential operator $(\ast)$ is self-adjoint, while $\,(u_t\,,X_n)=\frac{d\,}{dt}(u,X_n)\,$. Hence, the Fourier coefficients $T_n$ are to be solutions of Cauchy problems
$$
\begin{cases}
iT'_n=\alpha \lambda_nT_n\,,\quad t>0,\; n\in\mathbb{Z},\\
T_n(0)=h_n\overset{\rm def}{=}\frac{1}{L}\int\limits_0^L h(x)e^{-2\pi inx/L}dx.
\end{cases}
$$
It is clear that $T_n(t)= h_ne^{2\pi i\alpha nt/L}$, $n\in\mathbb{Z}$. Therefore, the desired solution is of the form
$$
u(t,x)=\sum_{n=-\infty}^{\infty}h_ne^{2\pi in(\alpha t+x)/L} \tag{$\ast\ast$}
$$
If function $h\in H^1(0,L)$ satisfies a compatibility condition $h(0)=h(L)$, it can be
represented by its Fourier series
$$
h(x)=\sum_{n=-\infty}^{\infty}h_ne^{2\pi inx/L},\quad 
h_n=\frac{1}{L}\int\limits_0^L h(x)e^{-2\pi inx/L}dx,
$$
convergent in $H^1(0,L)$. In this case, the Fourier series representng solution $(\ast\ast)$ sums to $u(t,x)=h(\alpha t+x)$ in $H^1(Q_T)$, where $Q_T=(0,T)\times (0,L)$ with arbitrary given $T>0$. When function $h\in H^1(0,L)$ does not satisfy the compatibility condition $h(0)=h(L)$,  solution $(\ast\ast)$ no longer belongs to $H^1(Q_T)$ while staying a weak solution of the class $L^2$. More precisely, given arbitrary $T>0$, a function $u\in L^2(Q_T)$ is called a weak solution if 
$$
\begin{align*}
\int\limits_{Q_T}u(t,x)\varphi_t(t,x)\,dx\,dt-
\alpha\int\limits_{Q_T}u(t,x)\varphi_x(t,x)\,dx\,dt=
\int\limits_0^L h(x)\varphi(0,x)\,dx\\
\forall\,\varphi\in H^1(Q_T)\,\colon\;\varphi(t,0)=\varphi(t,L)\;
\forall\,t\in (0,T),\;\varphi(T,x)=0\; 
\forall\,x\in (0,L).
\end{align*}
$$
Note that a weak solution of the class $L^2$ does exist even for initial data 
$h\in L^2(0,L)$. Such solution is represented by the same Fourier series $(\ast\ast)$ convergent in $L^2(Q_T)$. 
