# Ngô Quốc Anh

## August 29, 2014

### Prescribed Q-curvature and scalar curvature problems in the null case

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 20:24

On a 2-dimensional compact Riemannian manifold $(M, g)$ without boundary, the prescribed scalar curvature problem in the flat case is equivalent to solving the following PDE

$\displaystyle -\Delta_g u = fe^{2u}$

with $f$ is a given non-constant smooth function on $M$ and $\Delta_g$ is the Laplace-Beltrami operator associated with the metric $g$.

Simply by integrating both sides of the PDE, it is immediate to see that if $u$ solves the PDE, it would satisfy $\int_M f e^{2u} dv =0$; hence the candidate function $f$ must change sign in $M$. In their elegant paper published in 1974, Kazdan and Warner showed that in addition to the sign-changing property of $f$, it must also satisfy the following inequality

$\displaystyle \int_M f dv <0.$

This is just a simple observation from integration by parts if we multiply both sides of the PDE by $e^{-2u}$. Interestingly, Kazdan and Warner were able to show that the above two properties are also sufficient in the sense that it is enough to prove that the PDE is solvable.

In higher dimensions, perhaps, the most natural generalization of the operator $\Delta_g$ is the well-known Paneitz operator and its corresponding Q-curvature, see this link.

Assume that $(M,g)$ is a compact Riemannian 4-manifold without boundary. We denote by $P_g^4$ the so-called Paneitz operator acting on any smooth function $u$ via the following rule

$\displaystyle P_g^4(u) = \Delta _g^2u + {\rm div}\left( {\frac{2}{3}{R_g} - 2{\rm Ric}_g} \right)du ,$

where by ${\rm Ric}$ and $R$ we mean the Ricci tensor and the scalar curvature of $g$, respectively.

Then the prescribed Q-curvature problem in the null case can be formulated as solving the following PDE

$\displaystyle P_g^4(u)=fe^{4u}$

where again $f$ is the candidate, or the prescribed function. Notice that by flatness, we mean that the Q-curvature of the background metric $g$ is identically zero, i.e. $Q_g =0$ in $M$, see this link for details.

As far as we know, the prescribed Q-curvature problem in the null case was first studied by Ye and Xu in 2008. In their paper, they proved the same result as that obtained by Kazdan and Warner in 1974. Precisely, they showed that if

• The total integral $\int_M f dv<0$ and
• $\sup_M f>0$

then the PDE $P_g^4(u)=fe^{4u}$ always admits a smooth solution.

Ye and Xu’s proof is variational. First, we try to minimize the associated energy functional to the problem given by

$\displaystyle E(v)=\int_M vP_g^4 (v) dv$

over the set

$\displaystyle F=\Big\{v \in H^{2,2}(M): \int_M v dv=0, \int_M fe^{4v} dv=0 \Big\}.$

Then the condition $\int_M f dv<0$ helps us to conclude that the set $F$ is non-empty. Standard arguments plus the Lagrange multiplier then implies that there exist a constant $\beta$ and a function $u$ solving the following equation

$\displaystyle P_g^4(u)=\beta fe^{4u}.$

Since $P_g^4$ is invariant up to an additive constant, we would expect that the new function $U=u+(\log \beta )/4$ will solve our PDE. However, to this end, it must be shown that $\beta>0$. To this purpose, Ye and Xu showed that the minimum value of the above minimizing problem is in fact belong to a wider set

$\displaystyle \widetilde F=\Big\{v \in H^{2,2}(M): \int_M v dv=0, \int_M fe^{4v} dv \geqslant 0 \Big\}.$

Then, if $\beta<0$, they can construct a new test function belonging to the set $\widetilde F$ which admits smaller energy. Hence, the proof follows.

However, as they already mentioned in the paper, the two conditions $\int_M f dv<0$ and $\sup_M f>0$ are not necessary. Indeed, let $M = \mathbb T^4 = (\mathbb S^1)^4$ be the standard 4-dimensional torus and $g$ be the standard flat metric. Then there exists a conformal change $\widetilde g = e^{2u} g$ in such a way that the new Q-curvature $Q_{\widetilde g}$ is sing-changing with non-negative total integral $\int_M Q_{\widetilde g} dv_g \geqslant 0$.

As a consequence of a result due to Gursky, see this link, we expect that $\int_M f dv_g \leqslant 8\pi^2$ must hold. Notice that, only $\int_M Q_{\widetilde g} dv_{\widetilde g}$ is invariant in the sense that $\int_M Q_{\widetilde g} dv_{\widetilde g}= \int_M Q_g dv_g$ for any ${\widetilde g}=e^{2u}g$. However, simply by multiplying by a large constant, we know that if the PDE is solvable for some $f$ with $\int_M f dv>0$, then it is also solvable for any $f$ with arbitrarily large $\int_M f dv$. Hence, any upper bound for $\int_M f dv$ should depend on $f$ as well.