Suppose that $\Omega=(a,b)\times\mathbb{R}^+$ and that $\phi:\Omega\to\mathbb{R}$ and $\phi=\phi(x,t)$ satisfies \begin{equation} L(\phi)=0,\;\;\;\;\;(1) \end{equation} where $L$ is a 2nd order differential operator in the first variable $x$ only. Furthermore, suppose that $$\phi(a,t)=\phi(b,t)=f(t),\;\;\text{ for all }t\in\mathbb{R}^+.$$ If $f$ is bounded, can there be a differential operator $L$ and a function $\phi\in\ker L$ such that $|\phi|$ is unbounded? What if $f$ is unbounded?
Note 1: The function $\phi$ is in $H^2\left((a,b\right))$ for all $t$ and $C(\mathbb{R}^+)$ for all $x\in(a,b)$ but these conditions can be changed for the sake of an example/counter-example.
Note 2: Of course, a suitable choice $\phi_0(x)=\phi(x,0)$ is to be given also as part of the problem.
My try: Suppose that we take $f\equiv 0$. If $\phi(x,t)$ is a solution for (1), I claim that $t\cdot\phi(x,t)$ is also a solution for (1).
I can propose a following counterexample. Let $a=0$, $b=\pi$, and $f\equiv 0$. For any function $g:\Bbb R^+\to\Bbb R$, a function $\phi:\Omega\to\Bbb R$ such that $\phi(x,t)=g(t)\sin x$ for each $(x,t)\in\Omega$, satisfies the boundary condition. If the function $g$ is unbounded then the function $\phi$ is unbounded too. If $L=\tfrac{\partial^2}{\partial x^2}+1$ then $L(\phi)=0$.