 Solution

Use the Dirichlet Laplacian eigenvalues and eigenfunctions on $(-1,1)$:An expansion $u={\sum }_j{q}_j(t)e_j$ reduces the wave equation to independent ordinary differential equationsWhen $\alpha <{\lambda }_1$, put ${\omega }_j=\sqrt{{\lambda }_j-\alpha }$; each coefficient isFor finite-energy data $\varphi \in H_0^1(-1,1)$ and $\psi \in L^2(-1,1)$, a uniform spatial bound follows from the energy method, rather than from unjustified absolute summation of the Fourier series. The conserved quantity isThe Poincare inequality gives $\Vert u{\Vert }_2^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_2^2$. Hence $2E\ge c_{\alpha }\Vert u_x{\Vert }_2^2$ with $c_{\alpha }=1-\alpha /{\lambda }_1>0$ for $0\le \alpha <{\lambda }_1$, and with $c_{\alpha }=1$ for $\alpha <0$. Since the Dirichlet boundary condition gives $u(x,t)={\int }_{-1}^x{u}_x(s,t)ds$, the Cauchy-Schwarz inequality yieldsAt $\alpha ={\lambda }_1$, the first coefficient is $q_1(t)={\varphi }_1+t{\psi }_1$, which is unbounded for some admissible data. For $\alpha >{\lambda }_1$, that coefficient has an exponentially growing component for some admissible data. Thus all-time boundedness of a wave equation with a reaction term requiresAt equality, boundedness for a particular data set requires ${\psi }_1=0$; the remaining modes are bounded by their spectral gap. Above the threshold, every unstable mode must have its growing component canceled, and any zero-frequency mode must have zero initial velocity. There is no condition on $\alpha$ alone for arbitrary specially chosen data.

The printed reference to a limit needs qualification. Bounded oscillations generally have no limit as $t\to \infty$. If actual existence of that limit for every admissible initial datum is required, no real $\alpha$ works: for every $\alpha$, choose $j$ with ${\lambda }_j>\alpha$ and nonzero pure oscillatory data in that eigenfunction. The boxed inequality answers the intended long-time boundedness question.