Non-trivial solution to certain elliptic PDE implies exponential growth. In a paper I am reading there is a line which I don't quite understand. The setup is the following. We have a compact Riemannian manifold $N$, then we consider the cylinder $M=\mathbb{R} \times N$ and we have a nontrivial solution $u:M\rightarrow \mathbb{R}$ of the problem
$(D_t^2+\Delta_N)u-Ku=0$ 
where $\Delta_N$ is the Laplace-Beltrami operator on the cross section $N$ and $K>0$ is a constant. It then says "since the solution is non-trivial, it must exhibit exponential growth in either the $t\rightarrow \infty$ or $t\rightarrow -\infty$ directions. 
Could somebody explain why this is necessarily true? Thanks in advance! 
 A: If your sign convention is such that $\Delta_N$ has positive spectrum (e.g. $\Delta_{\mathbb R} = -\partial_x^2$), giving a wave-like equation, this is not necessarily true: the bounded solution $u = \sin(2 x)\sin(t)$ of $(\partial_t^2-\partial_x^2)u-3u=0$ on $\mathbb R/2 \pi \mathbb Z$ is a counterexample.
Thus I'll assume you use the opposite convention. Some well-known functional analysis tells us that there's an orthonormal basis of $\Delta_N$-eigenfunctions $e_i$ (for some sensible Hilbert space - I think this is true for $\mathcal D(\Delta)\subset L^2$ from memory?) with corresponding negative eigenvalues $\lambda_i;$ so writing $u(x,t)=\sum_i u_i(t) e_i(x)$ the PDE becomes $$u_i''(t)+(\lambda_i-K)u_i(t)=0.$$
Since the coefficient $-k^2 := \lambda_i - K$ of $u_i(t)$ is negative, the general solution is $u_i(t) = c_1 e^{kt} + c_2 e^{-kt},$ which must grow exponentially in a direction unless it vanishes identically. Thus, if $u$ is non-trivial, one of its projections $u_i$ must grow exponentially, and since $\|u(\cdot,t)\|^2=\sum_i|u_i(t)|^2$ we conclude that $u(t)$ must grow exponentially, at least in the $L^2$ norm. I think you should be able to get exponential growth of $\sup|u|$ without too much extra effort.
