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.
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 $L^2(Q_T)$.