A conformal Yamabe problem with potential on the euclidean space

We consider, in the Euclidean setting, a conformal Yamabe-type equation related to a potential generalization of the classical constant scalar curvature problem and which naturally arises in the study of Ricci solitons structures. We prove existence and nonexistence results, focusing on the radial case, under some general hypothesis on the potential.


introduction
In [4] the first and the third author considered "potential" generalizations of some canonical metrics on smooth complete Riemannian manifolds. In this paper we focus our attention on one of those classes, namely f -Yamabe metrics. We recall that, given a n-dimensional Riemannian manifold (M, g), where g is the metric, and a smooth function f ∈ C ∞ (M ), we say that the triple (M, g, f ) ∈ Y f if and only if it satisfies the condition (1.1) ∇R = 2 Ric(∇f, ·) , where Ric and R are, respectively, the Ricci and the scalar curvature of g and ∇ denotes the Levi-Civita connection associated to g. In a local orthonormal frame {e i }, i = 1, . . . , n, (1.1) becomes where R ij = Ric(e i , e j ). Note that we are using the Einstein summation convention over repeated indices. This equation is a meaningful generalization of the one for constant scalar curvature metrics and naturally arises in the study of Ricci solitons structures (for a general overview see [3]). Moreover, it is clear that any Ricci flat metric satisfy (1.1), for any function f and, more in general, any product of a Ricci flat metric with a metric with constant scalar curvature solves (1.1), for any function f which depends only on the first factor. In the same spirit of the classical Yamabe problem it is natural to address the following questions: (A) having fixed f ∈ C ∞ (M ), does there exist a metric g such that (M, g, f ) ∈ Y f ? (B) having fixed f ∈ C ∞ (M ) and a metric g, does there exist a conformal metric g in the conformal class [g] such that (M, g, f ) ∈ Y f ? More generally, one could ask the question (C) does there exist a metric g and a smooth function Clearly the answer to (C) is positive, since it is always possible to construct a (complete) metric with constant (negative) scalar curvature ( [1] and [2]). Furthermore, when f is constant, (B) boils down to the well known Yamabe problem, which is completely solved when M is compact (see e.g. [9]). We will refer to (B) as the conformal f -Yamabe problem. In this paper we consider problem (B) (when f is not constant) on the Euclidean space R n endowed with the standard flat metric g R n . In particular, in dimension four, we prove the following:

ODE formulation of the conformal f -Yamabe problem
Let (M, g) be a smooth n-dimensional Riemannian manifold, n ≥ 2, and let f ∈ C ∞ (M ). It is well known (see for instance [5]) that, if g = e 2w g ∈ [g] for some w ∈ C ∞ (M ), then the following formulas hold: where ∇ 2 is the Hessian and ∆ = g ij ∇ 2 ij is the Laplace-Beltrami operator of g. A computation shows that (M, g, f ) = (M, e 2w g, f ) ∈ Y f if and only if the function w solves the system of PDEs In particular, since (R n , g R n ) is Ricci flat, then (R n , g, f ) = (R n , e 2w g R n , f ) ∈ Y f if and only if w solves the system of PDEs To fully exploit the symmetries of the Euclidean space, it is reasonable to start our analysis by considering radial solutions w = w(r) of (2.2) for a given radial function f = f (r), where r denotes the distance function from the origin. In this case, in standard polar coordinates, one has g R n = dr 2 + r 2 g S n−1 , dw = w (r)dr and df = f (r)dr , and a computation shows that system (2.2) boils down to the following second order nonlinear ODE for the function u(r) := w (r) where h(r) := f (r). Note that if n = 2, then the cubic term disappears in (2.3). We then impose the initial conditions , then u(r) = αr + O(r 2 ) as r → 0 and, in turn, as r → 0 which shows that, by combining suitably the terms in (2.3), we obtain finite limits as r → 0. The existence and uniqueness of a solution of (2.3)-(2.4) can then be proved rigorously by adapting the arguments of Proposition 1 in [11]: one needs to combine the Ascoli-Arzelà Theorem with the Schauder fixed point Theorem in order to obtain existence of a solution.
Then the solution is unique as long as it can be continued [6,Proposition 4.2]. Before stating our existence and nonexistence results, let us discuss heuristically the structure of (2.3).

Heuristic preliminaries
We first notice that, if n ≥ 3, then there exist exactly two singular (negative) solutions of (2.3) of the type cr −1 given by regardless of the explicit form of h. This fact suggests that the "interesting dynamics" for (2.3) occurs when u(r) < 0 and global solutions of (2.3) are more likely to be prevalently negative. If n = 2 then the functions u(r) = cr −1 are singular solutions of (2.3) for any c = 0; in particular, there exist infinitely many positive singular solutions and the dynamics appears much more chaotic. It is quite useful to consider the two functions defined for all (r, y) ∈ R + × R: Then (2.3) may be written in normal form as Depending on h ∈ C 0 [0, ∞), we define the two regions Clearly, I h contains a right neighborhood of r = 0 and is therefore nonempty for all h, while It is also straightforward that: • if r ∈ I h , then P (r, y) = 0 if and only if y = 0, moreover P has the same sign as y; • if r ∈ I h , then we may write and hence P (r, y) vanishes if and only if one of the following facts occurs: Note that ψ(r) > ϕ(r) for all r ∈ I h but, while ϕ(r) < 0 for all r ∈ I h , the sign of ψ(r) may vary and it is the opposite of the sign of rh(r) + n − 1; in particular, ψ(r) < 0 in a right neighborhood of r = 0. One expects a crucial role for the existence results to be played by the signs of Q and P . However, the overall picture is not completely clear. To see this, consider the trivial case h ≡ 0 for which the function First notice that if a < 0 (so that u a (0) > 0), then u a blows up as r → 1/ |a|. Therefore, u a is a global solution of (2.3) if and only if a > 0. Simple computations then show that if n = 3, 4, 5 then ∃ρ > 0 , ϕ(r) < u 1 (r) < ψ(r) ∀r > ρ , if n ≥ 6 then ∃ρ > 0 , u 1 (r) < ϕ(r) ∀r > ρ .
In the left picture we see that the graph of u 1 (thick line) eventually lies inside Γ while in the right picture it eventually lies outside. Therefore, the function P (r, u 1 (r)) does not always have the same sign as r → ∞.

Nonexistence results
We can the prove the following (partial) nonexistence results. The proof of Theorem 4.1 is given in Section 7. As a by-product, the very same proof enables us to obtain Concerning nonexistence of negative solutions, a weaker result holds. First of all, we put together the three static terms This shows that the static term changes sign whenever the graph of u crosses one of the two hyperbola: Then rewrite (2.3) as If we assume that and that then the above equation tells us that r → r n−1 u (r) is decreasing for r > R. In particular, we have that u (r) < 0 and u(r) < h 1 (r) for all r > R. Finally, this yields the existence of γ > 0 such that 1 r n−1 r n−1 u (r) ≤ γu(r) 3 ∀r > R .

Existence results
We start with a simple but interesting example. If 3) is more likely to have negative solutions whenever h itself is negative. We prove that this is the case, at least in dimension n = 4.

Remarks and open problems
We discuss some open problem related to conformal f -Yamabe metrics and to solutions of equation (2.1).
(1) In Theorem 4.1 we stated a partial nonexistence results for radial solutions in the Euclidean space, while Theorem 5.1 provides a general existence result. It would be interesting to prove a sharp condition on the potential function f (or on its derivative) ensuring existence of global solutions to (2.3).
(2) It is well known [8] that global positive solution the Yamabe equation have to be radial (and thus classified). We could ask the same question for (general) solutions to (2.2). For a given f ∈ C ∞ (R n ), are there any nonradial solutions w? If f is radial, are all solutions to (2.2) radial?
(3) In this paper we studied conformal f -Yamabe metrics for (R n , g R n ). What about other rotationally symmetric spaces? In particular, what we can say for the hyperbolic space (H n , g H n ) or the round sphere (S n , g S n )?
(4) In the existence result (Theorem 5.1) the dimension n = 4 seems to be peculiar, at least from the analytic point of view. Is there a geometric interpretation of this fact?

Proof of Theorem 4.1
Throughout this proof we will need the following particular class of test functions.
Definition 7.1. Let ρ > 0. We say that a nonnegative function φ ∈ C 2 c [0, ∞) satisfies the ρ-property if and if It is clear that such functions exist; to see this, it suffices to replace any φ satisfying (7.1) with a power φ k for k sufficiently large so that (7.2) will be satisfied.
For the proof of Theorem 4.1, we first observe that, since α > 0 the solution of (2.3)-(2.4) is positive and strictly increasing in a right neighborhood of r = 0, say in some maximal interval (0, R). Clearly, among u and u the first one which can vanish is u . But if u (R) = 0 then, using the lower bound for h, we see that (2.3) yields giving a contradiction. Therefore, u cannot vanish and two cases may occur: The proof will be complete if we show that (ii) occurs. At this point, we distinguish two cases.
In order to prove (ii) in (7.3), we argue for contradiction by assuming that R = ∞ so that u, u > 0 for all r > 0. From the assumptions and (2.3), we then infer that (recall n ≤ 4) To reach a contradiction we need the following estimate, inspired to the method developed by Mitidieri-Pohožaev [10] (see also the proof of [7,Proposition 5]).
Take a function φ 1 satisfying the 1-property and observe that the function φ ρ (r) := φ 1 r ρ ∀ρ > 1 satisfies the ρ-property. Therefore, for all ε > 0, from (7.4) and Lemma 7.2 we infer that Take 0 < ε < 3(n − 2), then the latter inequality yields With the change of variable r = ρt this becomes Since u is increasing on R + , we have u(ρt) ≥ u(t) for all ρ > 1 so that the left hand side of this inequality is positive and increasing for ρ ≥ 1. By letting ρ → ∞, the right hand side tends to 0 and this leads to a contradiction which rules out case (i). Hence, case (ii) occurs and the solution u of (2.3)-(2.4) with α > 0 cannot be continued to all the interval [0, ∞). This completes the proof of Theorem 4.1 in the case n = 3, 4.
Proof. In this proof we will use the Young inequality in the following form:
Let v(r) = ru(r), then v satisfies the equation If u satisfies (2.3)-(2.4) with α < 0, then u(r) and u (r) are strictly negative in a right neighborhood of r = 0. By definition of v, also v(r) is strictly negative in a right neighborhood of r = 0. We claim that −2 < v(r) < 0 for all r > 0. If not, let R > 0 be the first time where (8.2) either v(R) = 0 or v(R) = −2 .
Multiply (8.1) by r 2 v (r) and integrate over [0, R] to obtain If n = 4 and h(r) ≤ 0 we get a contradiction which shows that R does not exist and therefore −2 < v(r) < 0 for all r > 0. This proves the claim. Hence, by (8.1), also v and v remain bounded and the solutions exists. This concludes the proof of Theorem 5.1.