Regarding incompatibilities between boundary conditions and initial values in PDEs, I mostly find discussions about the "corner problem", i.e. the initial data at the boundary violating the boundary conditions (in form of jumps or weaker discontinuities).
For coupled PDEs an IMHO different type of incompatibility puzzles me at the moment. A particularly simple examples reads \begin{align*} u_t&=v_{xx}\\ v_t&=-u_{xx} \quad. \end{align*} where $u$ and $v$ are $[0,2\pi]\times\mathbb R\mapsto\mathbb R$.
With, case (i), the initial values \begin{equation*} u(x,0)=v(x,0)=\sin x \end{equation*} and the boundary conditions \begin{equation*} u(0,t)=u(2\pi,t)=v(0,t)=v(2\pi,t)=0 \quad, \end{equation*} the solution is readily found: \begin{align*} u(x,t)&=(\cos t-\sin t)\,\sin x\\ v(x,t)&=(\cos t+\sin t)\,\sin x \end{align*}
But hoping for a similar success for case (ii), namely \begin{equation*} u(x,0)=\sin x \quad\text{and}\quad v(x,0)=\cos x \end{equation*} with \begin{equation*} u(0,t)=u(2\pi,t)=0 \quad\text{and}\quad v(0,t)=v(2\pi,t)=1 \end{equation*} is in vain: While the (putative) solution \begin{align*} u(x,t)&=\cos t\,\sin x-\sin t\cos x\\ v(x,t)&=\cos t\,\cos x+\sin t\sin x \end{align*} obeys the initial values, it violates the boundary conditions for $\frac{t}{2\pi}\not\in\mathbb Z$. Actually, rewriting it as \begin{align*} u(x,t)&=\sin(x-t)\\ v(x,t)&=\cos(x-t) \end{align*} makes it particularly clear that in this case no (separate, spatial, stationary) boundary conditions can be imposed on $u$ and $v$.
Are there any key words for this type of incompatibility?
What also puzzles me: "Solving" the problem (ii) numerically, there is convergence (regarding ever finer discretization) for finite $t$. But the obtained $u$ and $v$ are solutions to which problem?
