Let $(M,g)$ be a closed Einstein manifold of dimension $m>2$ and
$$
\mathrm{Ricc}(g)=\lambda g,
$$
$h$ a symmetric $2$-covariant tensor, $\Delta=\nabla^\*\nabla$ the Laplacian on functions as well as the rough or connection Laplacian on symmetric $2$-covariant tensors, and $\delta\_g$ the divergence on symmetric $2$-covariant tensors as well as on $1$-forms, so $\delta\_g\delta\_gh$ is the double-divergence and is a function. Write $\mathrm{tr}h$ for the metric trace of $h$ wrt $g$. Consider the following PDE:
$$
\Delta(\mathrm{tr}h)=\lambda\mathrm{tr}h+\frac{m-2}{2}\delta\_g\delta\_gh.
$$
*Questions*: Does this PDE have only the trivial solution $h=0$? Are there conditions under which it has only the trivial solution? I'm interested in the case that $\lambda<0$.



Does this PDE only have the trivial solution?
There are lots of nontrivial solutions for any negative $\lambda$. Here's how to construct them all.


First, we can decompose an arbitrary symmetric $2$-tensor $h$ as $h=fg+u$, where $f$ is a scalar function and $u$ is trace-free. It follows that $\operatorname{tr} h = mf$, and $\delta\_g\delta\_g h = -\Delta f + \delta\_g\delta\_g u$. After some simplification, therefore, your equation becomes
$$
\left(\Delta - \frac{2m\lambda }{3m - 2}\right) f = \frac{m-2}{3m-2} \delta\_g\delta\_g u.\tag{$\*$}
$$
Since the eigenvalues of $\Delta$ are all nonnegative, the fact that $\lambda<0$ ensures that the operator in parentheses on the left-hand side is invertible on $C^\infty(M)$. Thus we can choose $u$ to be an arbitrary trace-free symmetric $2$-tensor field, and then let $f$ be the unique solution to ($\*$).


Notice that this has nothing to do with the manifold being Einstein. This argument works on any compact Riemannian manifold for any negative constant $\lambda$. 


The case $\lambda\ge 0$ is a bit more complicated. In that case, if $2m\lambda /(3m-2)$ isn't an eigenvalue of $\Delta$ the same argument works; but if it happens to be an eigenvalue, then there will be solutions only for those $u$ for which $\delta\_g\delta\_g u$ is $L^2$-orthogonal to the kernel of the operator on the left-hand side. 


