Solving $\Delta \text{tr}(h) - \mathrm{div}(\mathrm{div}(h)) + \text{tr}(h) = f$ on $S^2$ $\DeclareMathOperator\ddiv{div}\DeclareMathOperator\tr{tr}\newcommand{\conf}{\mathrm{conf}}$Consider this PDE on a symmetric tensor $h$ on $S^2$:
$$\Delta \text{tr}(h) - \ddiv(\ddiv(h)) + \tr(h) = f$$
where $f \in L^2(S^2)$ and $\Delta$, $\ddiv$ and $\tr$ are with respect to the round metric on $S^2$.
I wish to show that there exists at least one solution to this.
If we assume for simplicity that $h = \frac{1}{2} \tr(h) g_{S^2}$, then the PDE becomes
$$\Delta \tr(h) + 2\tr(h) = 2f$$
which doesn't have a solution for every $f$ since $-2$ is an eigenvalue of $\Delta$ (I am assuming that $\Delta + \lambda$ is not surjective if $-\lambda$ is an eigenvalue; is this correct?). I am not sure how to approach this.
One approach is decomposing $h$ into its trace part and a conformal Lie derivative of a vector field $X$: $h = \frac{1}{2} \tr(h) g_{S^2} + \mathcal{L}_{\conf}X$. Then the PDE becomes:
$$\frac{1}{2}\Delta \tr(h) - \ddiv(\Delta_{\conf}X) + \text{tr}(h) = f$$
where $\Delta_{\conf}$ is the conformal laplacian on vector fields.
I am not able to continue.
Any help is appreciated.
 A: Your second-order differential operator appears when one takes the variation of the scalar curvature of the sphere by a symmetric tensor $h\mapsto \frac{d}{dt}|_{t=0}\mathrm{Sc}_{g+th}$. In constant sectional curvature, such operators were studied for instance by Calabi (60'), and then in constant scalar curvature by Ebin (69').
Edit: I have taken another look at Ebin's paper and figured out a similar, yet simpler argument than my original one (you can still find it below). I think this argument provides an answer to your question in any geometry: let $(M,g)$ be an arbitrary closed Riemmanian manfiold. Define a linear second order operator $\gamma:S^{2}(T^{*}M)\rightarrow C^{\infty}(M)$ by $\gamma h=-\Delta_{g}tr_{g}h+\mathrm{div}\mathrm{div}h-(h,Ric_{g})_{g}$. Consider its $L^{2}$-dual $\gamma^{*}:C^{\infty}(M)\rightarrow S^{2}(T^{*}M)$. It can be checked to assume the form $\gamma^{*}f=\mathrm{Hess}_{g}f-fRic_{g}-\Delta_{g}fg$. The symbol of $\gamma$ is $|\xi|^{2}tr+i_{\xi}i_{\xi}$ while the symbol of $\gamma^{*}$ is $\xi\otimes\xi+|\xi|^{2}I$. Thus the symbol of $\gamma\gamma^{*}$ is a constant multiplied by $|\xi|^{4}$, so $\gamma\gamma^{*}$ is an elliptic fourth order operator (a "bilaplacian"). Therefore, if you want to solve $\gamma h=f$ then replace $h=\gamma^{*}g$ and solve $\gamma\gamma^{*} g=f$. This has a solution if and only if $f$ satisfies a compatibility condition: it must be $L^{2}$-orthogonal to the finite dimensional kernel of $\gamma\gamma^{*}$, which in this case is equal to the kernel of $\gamma^{*}$.
In your original case of the sphere, $Ric_{g}=g$ and so $(h,Ric_{g})_{g}=tr_{g}h$.
My original answer is below:
Here is what I suggest to solve your question in the special case of $S^{2}$.
If you have a Riemmanian manifold with constant sectional curvature, $(M,g)$, consider the second order differential operator $H_{g}=\frac{1}{2}{(d^{\nabla^{g}}d^{\nabla^{g}}_{V}+d^{\nabla^{g}}_{V}d^{\nabla^{g}}})-\frac{1}{2}\kappa g\wedge$.
$\kappa$ here is the sectional curvature of the space, so in your case it can be taken to be $\kappa=1$. The wedge product here is known as the Kulkarni-Numizu product of symmetric tensor fields (there is a more generlized version of this wedge product, operating on "double forms", also adressed by Kulkarni, but this notion will do). $d^{\nabla^{g}}$ and $d^{\nabla^{g}}_{V}$ here are the exterior covariant derviatvie of the first and second index of a symmetric tensor, respectively, where we think of symmetric tensors as examples of vector-valued differential forms $\Omega^{1}(M ; T^{*}M)$. The image of $H_{g}$ lies in $\Omega^{2}(M;\Lambda^{2}T^{*}M)$, and in the case where $M$ is two-dimensional every element in this space is fully determined by its "scalar curvature", namely if $\sigma\in\Omega^{2}(M;\Lambda^{2}T^{*}M)$ then $\sigma=\frac{1}{4}(tr_{g}tr_{g}\sigma) g\wedge g$.
In the case where $\kappa=1$, a direct calcultion shows that taking the trace twice from $H_{g}$ yields the operator $tr_{g}tr_{g}H_{g}=-\Delta_{g}tr_{g}+\mathrm{div}\mathrm{div}-tr_{g}$, so solving your equation is equivelent of solving $H_{g}h=\frac{1}{4} f g\wedge g$, where $g$ is the metric of the sphere. Replacing $h=H^{*}_{g}\psi$ for $\psi\in\Omega^2(M;\Lambda^{2}T^{*}M)$ where $H_{g}^{*}$ is the formal $L^{2}$ dual of $H_{g}$ yields the equation $H_{g}H^{*}_{g}\psi=\frac{1}{4} f g\wedge g$. Note how since $H_{g}$ operates on symemtric tensors, the image of $H^{*}_{g}$ is a symmetric tensor in $\Omega^{1}(M;T^{*}M)$.
Another calculation then shows that the principle symbol of $H_{g}H^{*}_{g}$ in the case where $M$ is two dimensional is $|\xi|^{4}$. Thus this is an elliptic fourth order differential opertor (a "bilaplcian"), and so the equation $H_{g}H^{*}_{g}\psi=\frac{1}{4} f g\wedge g$ is solvable for $\psi$ if and only if $ \frac{1}{4} f g\wedge g$ is orthogonal to the kernel of $H_{g}H^{*}_{g}$. By duality, this kernel is equal to $\mathrm{ker} H^{*}_{g}$. I am not sure if this kernel is trivial when $M$ is simply connected. If not, then this orthogonality yields a compatability condition which $f$ must satisfy in order for the equation to have a solution.
