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30 September 2026

28 Pages

Blow-Up Phenomena for a Boundary Yamabe Problem with Umbilic Boundary

Dipartimento di Matematica, Universitá degli Studi di Bari Aldo Moro, 70125 Bari, Italy
This article belongs to the Section C: Mathematical Analysis

Abstract

We study a linear boundary perturbation of the problem of prescribing the scalar curvature K and the boundary mean curvature H on a Riemannian manifold with umbilic boundary. Assuming that the Weyl tensor is nowhere vanishing along the boundary, we treat separately the cases in which K and H are constant and non-constant. In the negative scalar-curvature regime K < 0 , and in dimensions n ≥ 8 , we prove the existence of positive solutions that blow up at the boundary as the perturbation parameter tends to zero.

1. Introduction

One of the central problems in differential geometry is the prescribed scalar curvature problem, i.e., given ( M , g ) being a Riemannian closed manifold of dimension n ≥ 3 and a smooth function K : M → R , finding a metric g ˜ conformal to the original metric g whose scalar curvature is K .
The problem was studied in the foundational work of Kazdan and Warner [1]. A broad treatment of scalar curvature and related geometric questions can be found in Schoen and Yau [2], while the relevant analytical framework for nonlinear elliptic equations is discussed by Hebey [3]. Further developments concerning prescribed curvature equations are presented by Chen and Li [4].
As is well known, since g ˜ = u 4 n − 2 g , this is equivalent to finding a positive solution of the semilinear elliptic equation:
− 4 ( n − 1 ) n − 2 Δ g u + k g u = K u n + 2 n − 2 , u > 0 , in M ,
where k g denotes the scalar curvature of M with respect to g and Δ g is the Laplace–Beltrami operator.
If M is a manifold with boundary, given a smooth function H : ∂ M → R , it is natural to ask if there exists a conformal metric whose scalar curvature and boundary mean curvature can be prescribed as K and H respectively. As in (1), the geometric problem turns out to be equivalent to a semilinear elliptic equation with a Neumann boundary condition:
− 4 ( n − 1 ) n − 2 Δ g u + k g u = K u n + 2 n − 2 , u > 0 , in M , 2 n − 2 ∂ ν u + h g u = H u n n − 2 , on ∂ M ,
where, k g and h g denote the scalar and boundary mean curvatures of M with respect to g and ν is the outward normal unit vector with respect to the metric g.
When K and H are constants, the problem is known as the Escobar problem, since it was first proposed and studied by Escobar in 1992 in the case of H = 0 [5,6] and in the case of K = 0 [7]. Afterwards, many subsequent contributions for this problem were given in [8,9,10,11].
The case of non-zero constants K and H (with K > 0 ) was first studied by Han & Li in [12,13] and then it was completed by Chen, Ruan & Sun in [14].
In all these results, the existence of solutions for the problem (2) strongly depends on the dimension of the manifold, on the properties of the boundary (i.e., being umbilic or not) and on vanishing properties of the Weyl tensor.
The case of non-constant functions K and H is less studied and all the available results are for special manifolds (typically the unit ball and the half-sphere). We refer to [15,16,17,18,19] for the case of H = 0 and to [20,21,22,23] for the case of K = 0 .
When both K and H are not constants and not zero, the problem becomes more difficult. Djadli, Malchiodi & Ahmedou consider problem (2) in [24] on the three-dimensional half-sphere proving some existence and compactness results. Chen, Ho & Sun proved the existence of solutions for (2) when K and H are negative functions and the boundary ∂ M has negative Yamabe invariant (see [25]). In [26], Ambrosetti, Li & Malchiodi considered the perturbation problem in the unit ball when both K and H are positive. That is, they consider K = K 0 + ε K > 0 and H = H 0 + ε H > 0 , where K 0 > 0 , H 0 > 0 , ε > 0 is small, and K and H are smooth functions. They proved an existence result when K and H satisfy some conditions.
The first result concerning the case of negative prescribed scalar curvature (namely K < 0 ) is due to Cruz-Blázquez, Malchiodi & Ruiz in [27]. They introduce the scaling-invariant quantity
D ( p ) : = n ( n − 1 ) H ( p ) | K ( p ) | , p ∈ ∂ M .
They established the existence of a solution to (2) whenever D ( p ) < 1 along the whole boundary. On the other hand, if D ( p ) > 1 at some boundary points, they obtained a solution only on a three-dimensional manifold, for a generic choice of K and H . Their proof relies on a careful blow-up analysis: first they show that blow-up occurs at boundary points p with D ( p ) ≥ 1 , with different behaviors depending on whether D ( p ) = 1 or D ( p ) > 1 . To deal with the loss of compactness at points with D ( p ) > 1 , where bubbling of solutions occurs, it is shown that in dimension three all the blow-up points are isolated and simple. As a consequence, the number of blow-up points is finite and the blow-up is excluded via integral estimates. In that regard, n = 3 is the maximal dimension for which one can prove that the blow-up points with D ( p ) > 1 are isolated and simple for generic choices of K and H . In the closed case, such a property is assured up to dimension four (see [19]) but, as observed in [24], the presence of the boundary produces a stronger interaction of the bubbling solutions with the function K .
Afterwards, in [28], the authors considered the perturbation problem in the ball under the condition K < 0 and H > 0 . I.e., K = K 0 + ε K < 0 and H = H 0 + ε H > 0 , where K 0 < 0 , H 0 > 0 , ε > 0 is small, and K and H are smooth functions showing the existence of solutions with some constraint of K and H .
Recently, in [29] problem (2) in the unit ball showed the existence of infinitely many non-radial solutions under some suitable assumptions on the functions K and H (see also [30]) for the closed case, [31,32] for K = 0 and H > 0 ).
The existence in the general case in dimension n ≥ 4 is not known at the moment because the difficulties that arise in proving the compactness condition. In [33,34], a linear perturbation of the geometric problem (2) is studied; namely,
− 4 ( n − 1 ) n − 2 Δ g v + k g v = K v n + 2 n − 2 in M 2 n − 2 ∂ v ∂ ν + h g v + ε γ v = H v n n − 2 on ∂ M .
where ε is a small and positive parameter and γ is a given smooth function. By using the Escobar metric (namely by letting g so that h g = 0 ) then, in [33], the existence of clustering-type solutions for problem (4) is shown in the case in which K and H are constants, 4 ≤ n ≤ 7 and γ = 1 , while, in [34], the existence of a blowing-up solution when K and H are not constants, n ≥ 4 and γ = 1 , is proven.
Here we continue the study of problem (4) when the manifold M has an umbilic boundary, and we want to show the existence of a blowing-up solution also in this case.
For every p ∈ ∂ M such that K ( p ) < 0 , we define
D ( p ) : = n ( n − 1 ) H ( p ) | K ( p ) | .
Throughout the paper we work in the bubbling regime D ( p ) > 1 . Since K ( p ) < 0 , the denominator in the definition of D ( p ) is strictly positive; consequently, D ( p ) > 1 implies H ( p ) > 0 .
Let g ¯ denote the metric induced by g on ∂ M , and set K ∂ : = K | ∂ M . We use ∇ g K and ∇ g 2 K for the ambient gradient and Hessian, while ∇ g ¯ and ∇ g ¯ 2 denote the intrinsic gradient and Hessian on ∂ M .
We use the following hypotheses:
( Hyp 1 )
K ∈ C ∞ ( M ¯ ) , H ∈ C ∞ ( ∂ M ) are such that K < 0 . The function H may have arbitrary sign on ∂ M , but necessarily H ( p ) > 0 at every point at which D ( p ) > 1 .
( Hyp 2 )
There exists p 0 ∈ ∂ M such that
D ( p 0 ) > 1 , ∇ g K ( p 0 ) = 0 , ∇ g ¯ H ( p 0 ) = 0 .
Moreover, ∇ g 2 K ( p 0 ) is positive definite on T p 0 M , and ∇ g ¯ 2 H ( p 0 ) is positive definite on T p 0 ∂ M . In particular, ∂ ν K ( p 0 ) = 0 . Since ∇ g K ( p 0 ) = 0 , we have
∇ g ¯ 2 K ∂ ( p 0 ) = ∇ g 2 K ( p 0 ) T p 0 ∂ M × T p 0 ∂ M .
Thus p 0 is a common nondegenerate local minimum of K ∂ and H on ∂ M .
We remark that, if K and H are constants, ( Hyp 1 ) means only that K < 0 and H > 0 are such that D > 1 .
Before stating the result we also recall the following: we say that ( M , g ) is of positive type if the quadratic form
Q g ( u ) : = ∫ M c n | ∇ g u | g 2 + k g u 2 d ν g + 2 ( n − 1 ) ∫ ∂ M h g u 2 d σ g , c n = 4 ( n − 1 ) n − 2 ,
is coercive on H 1 ( M ) ; namely, there exists c > 0 such that
Q g ( u ) ≥ c ∥ u ∥ H 1 ( M ) 2 for every u ∈ H 1 ( M ) .
The main result of the paper is stated as follows.
Theorem 1.
Let ( M , g ) be a smooth, compact, n-dimensional Riemannian manifold of positive type with regular umbilic boundary ∂ M .
Suppose n ≥ 8 and that the Weyl tensor is nowhere vanishing along the boundary and let us assume ( Hyp 1 ) .
If K and H are constants, let γ : M → R be smooth with γ > 0 on ∂ M .
If K and H are non-constant, we assume in addition ( Hyp 2 ) .
Then, for all sufficiently small ε > 0 , there exists a positive solution u ε that blows up at a point p ∈ ∂ M as ε → 0 .
Let us make some comments.
  • The proof of Theorem 1 relies on a finite dimensional Lyapunov–Schmidt reduction method. Here the main difficulty is due to the fact that the umbilicity of the boundary forces us to consider higher-order expansion in the metric g that, together with a different kind of bubble, makes the computations hard. Moreover, when K and H are constants we also need to correct the main part of the ansatz by adding a function V p (given in Proposition 1) in order to have a good error. We remark that, when K and H are not constants, then the correction is not needed and this is a great difference with respect to the non-umbilic case (the result contained in [34] with K and H not constants needs a correction which is different from V p ).
  • The result provides the exact location of the blow-up point when K and H are non-constant: it is their common non-degenerate critical point. Here γ can be taken to be equal to 1 because it does not affect the leading-order reduced problem. When K and H are constant, the situation is more delicate and the present argument does not determine the precise location of the blow-up point. The Weyl tensor on the boundary is expected to be the geometric quantity responsible for the existence and location of the blowing-up solution. Establishing this directly would require a more explicit characterization of the function V p from Proposition 1.
  • In [35] the problem with K = 0 was considered. We remark that even if one can think that these problems are similar, the form of the bubble, namely the classification result for the limit problem, makes the computations completely different.
The structure of the paper is as follows. In Section 2 we collect notation and preliminary results. In Section 3 we construct an approximate solution, and in Section 4 we perform the finite-dimensional reduction. In Section 5 we analyze the reduced problem and prove Theorem 1. We summarize the results and discuss further questions in Section 6.

2. Preliminaries and Variational Framework

Notations: Here we collect our main notations.
We will use the indices 1 ≤ i , j , k , ℓ , m , q , r , s ≤ n − 1 and 1 ≤ a , b , c , d ≤ n .
We denote by g the Riemannian metric, by R a b c d the full Riemannian curvature tensor, by R a b the Ricci tensor and by k g the scalar curvature of ( M , g ) . Moreover the Weyl tensor of ( M , g ) will be denoted by Weyl g .
Let ( h i j ) i j ( p ) be the tensor of the second fundamental form in a point p ∈ ∂ M . We recall that the boundary ∂ M is umbilic (namely composed only of umbilic points) when, for all p ∈ ∂ M , h i j ( p ) = 0 for all i ≠ j and h i i ( p ) = h g ( p ) where h g ( p ) is the mean curvature of ∂ M at the point p.
The bar over an object (e.g., Weyl g ¯ ) will mean the restriction of this object to the metric of ∂ M . We will often use the notation
L g : = − 4 ( n − 1 ) ( n − 2 ) Δ g + k g , B g : = ∂ ∂ ν + n − 2 2 h g
to denote the conformal Laplacian and the conformal boundary operator respectively.
When we derive a tensor, e.g., T i j , with respect to a coordinate y ℓ we use the usual shortened notation T i j , ℓ for ∂ ∂ y ℓ T i j .
Finally, for a tensor T and a number N ∈ N , we use
Sym i 1 … i N T i 1 … i N = 1 N ! ∑ σ ∈ S N T i σ ( 1 ) … i σ ( N )
with S N being the group of all permutations of q elements.
In the case where K and H are constants, then D is also a constant that does not depend on the point p ∈ ∂ M . For simplicity we omit the dependence on the point and we understand that the two situations differ.
Remark 1.
Since ∂ M is umbilic for any p ∈ ∂ M , there exists a metric g ˜ p = g ˜ conformal to g, namely g ˜ p = Λ p 4 n − 2 g , such that
| det g ˜ p ( y ) | = 1 + O ( | y | n )
| h ˜ i j ( y ) | = o ( | y | 3 )
g ˜ i j ( y ) = δ i j + 1 3 R ¯ i k j ℓ y k y ℓ + R n i n j y n 2 + 1 6 R ¯ i k j ℓ , m y k y ℓ y m + R n i n j , k y n 2 y k + 1 3 R n i n j , n y n 3 + 1 20 R ¯ i k j ℓ , m q + 1 15 R ¯ i k s ℓ R ¯ j m s q y k y ℓ y m y q + 1 2 R n i n j , k ℓ + 1 3 Sym i j ( R ¯ i k s ℓ R n s n j ) y n 2 y k y ℓ + 1 3 R n i n j , n k y n 3 y k + 1 12 ( R n i n j , n n + 8 R n i n s R n s n j ) y n 4 + O ( | y | 5 )
| k ¯ g ˜ p ( y ) | = O ( | y | 2 ) and ∂ i i 2 k ¯ g ˜ p ( p ) = − 1 6 | Weyl g ¯ ( p ) | 2
k g ˜ p ( p ) = 0 ; k g ˜ p , a ( p ) = 0 , and ∂ i i 2 k g ˜ p ( p ) = − 1 6 | Weyl g ¯ ( p ) | 2
R ¯ k ℓ ( p ) = R n n ( p ) = R n k ( p ) = 0
uniformly with respect to p ∈ ∂ M and y ∈ T p M . Also Λ p ( p ) = 1 and ∇ Λ p ( p ) = 0 . These results are contained in [10].
The conformal Laplacian and the conformal boundary operator transform under the change of metric g ˜ p = Λ p 4 n − 2 g in the following way:
L g ˜ p φ = Λ p − n + 2 n − 2 L g ( Λ p φ ) B g ˜ p φ = Λ p − n n − 2 B g ( Λ p φ ) .
Then we can rewrite our initial problem (4) in the following way: let v : = Λ p u . Then v is a positive solution of (4) if and only if u is a positive solution of
L g ˜ p u = K u n + 2 n − 2 in M B g ˜ p u + n − 2 2 ε Λ p − 2 n − 2 γ u = n − 2 2 H u n n − 2 on ∂ M .
From now on we set γ ˜ : = Λ p − 2 n − 2 γ .
By the positive-type assumption, the bilinear form associated with Q g defines a scalar product on H 1 ( M ) , whose induced norm is equivalent to the standard Sobolev norm, so we endow the Sobolev space H 1 ( M ) with the equivalent scalar product
⟨ u , v ⟩ g : = ∫ M ( c n ∇ g u ∇ g v + k g u v ) d ν g + 2 ( n − 1 ) ∫ ∂ M h g u v d σ g ,
where d ν g is the volume element of the manifold, d σ g is the volume element of the boundary and c n : = 4 ( n − 1 ) ( n − 2 ) . This scalar product induces a norm in H 1 ( M ) which is equivalent to the standard one, and which we denote by ∥ · ∥ g . We remark also that Λ p is an isometry in the sense that for any u , v ∈ H 1 ( M )
⟨ Λ p u , Λ p v ⟩ g = ⟨ u , v ⟩ g ˜ p
and, consequently,
∥ Λ p u ∥ g = ∥ u ∥ g ˜ p .
Moreover, for any u ∈ L q ( M ) and v ∈ L q ( ∂ M ) , we put
∥ u ∥ L q ( M ) : = ∫ M | u | q d ν g 1 q and ∥ v ∥ L q ( ∂ M ) : = ∫ ∂ M | v | q d σ g 1 q .
For notational convenience, we will often omit the volume or surface elements in integrals.
We have the well-known embedding continuous maps
i ∂ M : H 1 ( M ) → L 2 ♯ ( ∂ M ) , i M : H 1 ( M ) → L 2 * ( M ) , i ∂ M * : L 2 ( n − 1 ) n ( ∂ M ) → H 1 ( M ) , i M * : L 2 n n + 2 ( M ) → H 1 ( M ) ,
where 2 * = 2 n n − 2 and 2 ♯ = 2 ( n − 1 ) n − 2 denote the critical Sobolev exponents for M and ∂ M , respectively. Now, given q ∈ L 2 ( n − 1 ) n ( ∂ M ) , the function w 1 = i ∂ M * ( q ) in H 1 ( M ) is defined as the unique solution of the equation
L g w 1 = 0 in M , B g w 1 = q on ∂ M .
Similarly, if we let r ∈ L 2 n n + 2 ( M ) , w 2 = i M * ( r ) denotes the unique solution of the equation
L g w 2 = r in M , B g w 2 = 0 on ∂ M .
By continuity of i M and i ∂ M , we get
∥ i ∂ M * ( q ) ∥ g ≤ C 1 ∥ q ∥ L 2 ( n − 1 ) n ( ∂ M ) and ∥ i M * ( r ) ∥ g ≤ C 2 ∥ r ∥ L 2 n n + 2 ( M ) ,
for some C 1 > 0 and independent of q and some C 2 > 0 and independent of r. Then, we are able to rewrite the problem (4) as
v = i M * ( K N M ( v ) ) + i ∂ M * n − 2 2 H N ∂ M ( v ) − ε γ v ,
where we set N M ( v ) : = ( v + ) n + 2 n − 2 and N ∂ M ( v ) = ( v + ) n n − 2 .
We also define the energy J ε , g : H 1 ( M ) → R associated with
J ε , g ( v ) : = ∫ M c n 2 | ∇ g v | 2 + 1 2 k g v 2 − K P M ( v ) d ν g + ( n − 1 ) ∫ ∂ M h g v 2 d σ g − c n n − 2 2 ∫ ∂ M H P ∂ M ( v ) d σ g + ( n − 1 ) ε ∫ ∂ M γ v 2 d σ g ,
as being
P M ( s ) = ∫ 0 s N M ( t ) d t , P ∂ M ( s ) = ∫ 0 s N ∂ M ( t ) d t .
Notice that, if we define
J ˜ ε , g ˜ p ( u ) : = ∫ M c n 2 | ∇ g ˜ p u | 2 + 1 2 k g ˜ p u 2 − K P M ( u ) d ν g ˜ p + ( n − 1 ) ∫ ∂ M h g ˜ p u 2 d σ g ˜ p − c n n − 2 2 ∫ ∂ M H P ∂ M ( u ) d σ g ˜ p + ( n − 1 ) ε ∫ ∂ M γ ˜ u 2 d σ g ˜ p ,
then we have
J ε , g ( Λ p u ) = J ˜ ε , g ˜ p ( u ) .
Now we introduce some integral quantities that will appear in our computations: let
I m α : = ∫ 0 + ∞ ρ α ( 1 + ρ 2 ) m d ρ , with α + 1 < 2 m
It is useful to recall the following relations:
I n n = n − 3 n + 1 I n n + 2 , I n − 2 n − 2 = 4 ( n − 2 ) n + 1 I n n + 2 .
Moreover, for p ∈ ∂ M with D ( p ) > 1 , we denote
φ m ( p ) : = ∫ D ( p ) + ∞ 1 ( t 2 − 1 ) m d t ; φ ^ m ( p ) : = ∫ D ( p ) + ∞ ( t − D ( p ) ) 2 ( t 2 − 1 ) m d t ; φ ˜ m ( p ) : = ∫ D ( p ) + ∞ ( t − D ( p ) ) 4 ( t 2 − 1 ) m d t .
Here and in the sequel we agree that f ≲ g means | f | ≤ C | g | for some positive constant C which is independent of f and g, and f ∼ g means f = g ( 1 + o ( 1 ) ) . We use the letter C to denote a positive constant that may change from line to line.

3. The Ansatz

We want to find a solution u of problem (12) by a finite-dimensional reduction.
The main ingredient to cook up our solution is the so-called bubble, whose expression is given by
U δ , x 0 ( δ ) ( x ) : = α n | K ( p ) | n − 2 4 δ n − 2 2 | x − x 0 ( δ ) | 2 − δ 2 n − 2 2
where α n : = 4 n ( n − 1 ) n − 2 4 , x 0 ( δ ) : = ( x ˜ 0 , − D δ ) ∈ R n , x ˜ 0 ∈ R n − 1 and δ > 0 . When D > 1 , the n-dimensional family of functions defined above describe all the solutions to the following problem in R + n (see [36]):
− c n Δ U = − | K ( p ) | U n + 2 n − 2 in R + n 2 n − 2 ∂ U ∂ ν = H ( p ) U n n − 2 on ∂ R + n .
We set
U ( x ) = U 1 , x 0 ( 1 ) ( x ˜ , x n ) = α n | K ( p ) | n − 2 4 1 | x ˜ | 2 + ( x n + D ) 2 − 1 n − 2 2 ,
where x ˜ = ( x 1 , … , x n − 1 ) ∈ R n − 1 and x n > 0 . Moreover α n : = 4 n ( n − 1 ) n − 2 4 .
We also need to introduce the linear problem
− c n Δ v + | K ( p ) | n + 2 n − 2 U 4 n − 2 v = 0 in R + n 2 n − 2 ∂ v ∂ ν − n n − 2 H ( p ) U 2 n − 2 v = 0 on ∂ R + n .
In Section 2 of [33] we have shown that the n-dimensional space of solutions of (19) is generated by the functions
j i ( x ) : = ∂ U ∂ x i ( x ) = α n | K ( p ) | n − 2 4 ( 2 − n ) x i | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 n 2 , i = 1 , … , n − 1
and
j n ( x ) : = 2 − n 2 U ( x ) − ∇ U ( x ) · ( x + D ( p ) e n ) + D ( p ) ∂ U ∂ x n = α n | K ( p ) | n − 2 4 n − 2 2 | x | 2 + 1 − D 2 ( p ) | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 n 2 .
As it happens in the Yamabe problem, bubbles are not a good enough approximating solution due to some error terms arising from the geometry of M. Therefore, they need to be corrected by a higher-order term V p : R + n → R , whose main properties are collected in the next proposition.
Proposition 1.
Let U be as in (18). We set
E p ( x ) = c n 1 3 R ¯ i k j ℓ ( p ) x k x ℓ + R n i n j ( p ) x n 2 ∂ i j 2 U ( x ) .
Then the problem
− c n Δ V + n + 2 n − 2 | K ( p ) | U 4 n − 2 V = E p i n R + n , 2 n − 2 ∂ V ∂ ν − n n − 2 H ( p ) U 2 n − 2 V = 0 o n ∂ R + n ,
admits a solution V p satisfying the following properties:
(i) 
∫ R + n V p ( x ) j i ( x ) d x = 0 for any i = 1 , … , n (see (20) and (21)),
(ii) 
| ∇ α V p | ( x ) ≲ 1 + | x | 4 − n − α for any x ∈ R + n and α = 0 , 1 , 2 ,
(iii) 
| K ( p ) | ∫ R + n U n + 2 n − 2 V p d x = ( n − 1 ) H ( p ) ∫ ∂ R + n U n n − 2 V p d x ˜ ,
(iv) 
We have that
∫ R + n − c n Δ V p + n + 2 n − 2 | K ( p ) | U 4 n − 2 V p V p ≥ 0
(v) 
The map p ↦ V p is C 2 ( ∂ M ) .
Proof. 
Fix p ∈ ∂ M such that D ( p ) > 1 . Throughout the proof, all curvature tensors are evaluated at p, and repeated tangential indices are summed from 1 to n − 1 . We write
r = | x ˜ | , t = x n , Q ( x ) = r 2 + ( t + D ( p ) ) 2 − 1 , C p = α n | K ( p ) | ( n − 2 ) / 4 .
Thus
U ( x ) = C p Q ( x ) − ( n − 2 ) / 2 .
We first simplify the forcing term. For tangential indices i , j , direct differentiation gives
∂ i j 2 U = − ( n − 2 ) C p δ i j Q − n / 2 + n ( n − 2 ) C p x i x j Q − ( n + 2 ) / 2 .
By the curvature symmetries and (10),
∑ i R ¯ i k i ℓ = R ¯ k ℓ = 0 , R ¯ i k j ℓ x i x k x j x ℓ = 0 .
Consequently,
R ¯ i k j ℓ x k x ℓ ∂ i j 2 U = 0 .
Moreover,
∑ i R n i n i = R n n = 0 .
Notice that this is a contraction identity; no vanishing of the individual components R n i n i is required. It follows that
E p ( x ) = c n n ( n − 2 ) C p t 2 Q ( x ) − ( n + 2 ) / 2 R n i n j x i x j .
The polynomial
P p ( x ˜ ) : = R n i n j x i x j
is homogeneous of degree two and harmonic in R n − 1 , since
Δ x ˜ P p = 2 ∑ i R n i n i = 0 .
Thus E p belongs to the degree-two tangential spherical harmonic sector. In particular,
| E p ( x ) | ≤ C ( 1 + | x | ) 2 − n ,
and, because n ≥ 8 ,
E p ∈ L 2 n n + 2 ( R + n ) .
We construct the solution in the same angular sector. Let
D 1 , 2 ( R + n ) = C c ∞ ( R + n ¯ ) ¯ ∥ ∇ · ∥ L 2 ,
and let X 2 be its closed subspace consisting of functions whose dependence on θ = x ˜ / | x ˜ | ∈ S n − 2 contains only spherical harmonics of degree two. Set
a p ( x ) = n + 2 n − 2 | K ( p ) | U ( x ) 4 / ( n − 2 ) , b p ( x ˜ ) = n 2 H ( p ) U ( x ˜ , 0 ) 2 / ( n − 2 ) .
The weak form of (22) is associated with
B p ( v , w ) : = c n ∫ R + n ∇ v · ∇ w d x + ∫ R + n a p v w d x − c n ∫ ∂ R + n b p v w d x ˜ .
We claim that B p is coercive on X 2 .
To prove this, recall that the eigenvalues of − Δ S n − 2 in degrees one and two are
λ 1 = n − 2 , λ 2 = 2 ( n − 1 ) .
Differentiating the equation and boundary condition for U in a tangential direction shows that ∂ i U solves the homogeneous problem (19). Since
∂ i U = − ( n − 2 ) C p r Q − n / 2 θ i ,
the positive function
h ( r , t ) = r Q ( r , t ) − n / 2 , r > 0 , t ≥ 0 ,
satisfies
− c n h r r + n − 2 r h r + h t t − λ 1 r 2 h + a p h = 0
and
− h t ( r , 0 ) = b p ( r ) h ( r , 0 ) .
Let Y be an L 2 ( S n − 2 ) -normalized spherical harmonic of degree two. For v ( r , t , θ ) = ψ ( r , t ) Y ( θ ) , the ground-state identity associated with h gives
B p ( v , v ) = c n ∫ 0 ∞ ∫ 0 ∞ h 2 ∇ r , t ψ h 2 r n − 2 d r d t + c n ( λ 2 − λ 1 ) ∫ 0 ∞ ∫ 0 ∞ ψ 2 r n − 4 d r d t .
The identity follows first for smooth profiles supported away from r = 0 and then by density. Since λ 2 − λ 1 = n > 0 , it implies
B p ( v , v ) > 0 for every nonzero v ∈ X 2 .
Here and below, the same argument applies after summing over an orthonormal basis of degree-two spherical harmonics.
Furthermore,
a p ( x ) = O ( ( 1 + | x | ) − 4 ) , b p ( x ˜ ) = O ( ( 1 + | x ˜ | ) − 2 ) .
The two lower-order terms in B p are therefore compact quadratic forms relative to the Dirichlet norm, by local compactness and the Sobolev and trace inequalities. Strict positivity consequently implies coercivity: there exists κ p > 0 such that
B p ( v , v ) ≥ κ p ∫ R + n | ∇ v | 2 d x for every v ∈ X 2 .
Since E p ∈ L 2 n / ( n + 2 ) , the Lax–Milgram theorem provides a unique V p ∈ X 2 satisfying
B p ( V p , w ) = ∫ R + n E p w d x for every w ∈ X 2 .
All coefficients depend only on r and t. Hence different tangential spherical harmonic degrees are orthogonal for B p , and the right-hand side has degree two by (23). The preceding weak identity therefore holds for every test function in D 1 , 2 ( R + n ) . Thus V p solves (22). Local elliptic regularity gives smoothness up to the boundary at every finite point.
We next establish the decay in (ii). Consider the Kelvin transform
V ^ p ( y ) = | y | 2 − n V p y | y | 2 .
In a sufficiently small half-ball around the origin, V ^ p satisfies
− c n Δ V ^ p + a ^ p V ^ p = E ^ p , ∂ ν V ^ p = b ^ p V ^ p on the flat boundary ,
where
a ^ p ( y ) = | y | − 4 a p y | y | 2 , b ^ p ( y ) = | y | − 2 b p y | y | 2 ,
and
E ^ p ( y ) = | y | − n − 2 E p y | y | 2 .
Indeed,
Q y | y | 2 = | y | − 2 1 + 2 D ( p ) y n + ( D ( p ) 2 − 1 ) | y | 2 .
It follows that a ^ p and b ^ p extend smoothly across the origin, whereas
| ∇ α E ^ p ( y ) | ≤ C α | y | − 4 − α .
The Kelvin transform has finite Dirichlet energy. Moreover, E ^ p belongs locally to L 2 n / ( n + 2 ) because n > 6 . Thus the transformed equation holds weakly across the origin. On a small half-ball, the Green function for the operator with Robin condition on the flat boundary and Dirichlet condition on the hemispherical boundary has the bound
| G ( y , z ) | ≤ C | y − z | 2 − n .
The smooth data on the hemispherical boundary contribute a bounded term. Consequently,
| V ^ p ( y ) | ≤ C + C ∫ B ρ + | y − z | 2 − n | z | − 4 d z ≤ C ( 1 + | y | − 2 ) .
Scaled interior and boundary elliptic estimates on annuli then yield
| ∇ α V ^ p ( y ) | ≤ C | y | − 2 − α , α = 0 , 1 , 2 .
Transforming back and using local regularity gives
| ∇ α V p ( x ) | ≤ C ( 1 + | x | ) 4 − n − α , α = 0 , 1 , 2 ,
which proves (ii).
We now verify (i). For each fixed r , t , the angular dependence of V p has degree two. By (20), j s , s = 1 , … , n − 1 , has angular degree one, whereas j n in (21) has degree zero. Orthogonality of spherical harmonics gives
∫ R + n V p j s d x = 0 , s = 1 , … , n .
These integrals are absolutely convergent by (ii) and the explicit formulas for j s . This proves (i). In particular, the solution is normalized directly in the original variables.
Since U depends only on r and t, the same angular orthogonality gives
∫ R + n U n + 2 n − 2 V p d x = 0 , ∫ ∂ R + n U n n − 2 V p d x ˜ = 0 .
Both integrals are absolutely convergent. Hence
| K ( p ) | ∫ R + n U n + 2 n − 2 V p d x = ( n − 1 ) H ( p ) ∫ ∂ R + n U n n − 2 V p d x ˜ ,
which is (iii).
Taking w = V p in the weak formulation and using coercivity on X 2 , we obtain
∫ R + n − c n Δ V p + n + 2 n − 2 | K ( p ) | U 4 n − 2 V p V p d x = ∫ R + n E p V p d x = B p ( V p , V p ) ≥ 0 .
The integrability of the displayed expressions follows from (ii) and the estimate for E p . This proves (iv).
Finally, choose local orthonormal frames and conformal Fermi coordinates depending smoothly on p. In these coordinates the space X 2 is fixed. On every compact subset of
{ p ∈ ∂ M : D ( p ) > 1 } ,
the coefficients of B p and the forcing term E p depend C 2 on p, respectively in the operator norm on X 2 and in its dual. The coercivity constants can be chosen uniformly on such compact subsets. Differentiating
B p ( V p , w ) = ∫ R + n E p w d x
with respect to p, or equivalently differentiating the inverse of the associated coercive operator, shows that p ↦ V p is C 2 with values in D 1 , 2 ( R + n ) . Local elliptic regularity gives the corresponding C 2 dependence in C loc 2 ( R + n ¯ ) . This proves (v). □
Remark 2.
It is hard to show (but it is reasonable that it is like this) that there exists a positive constant F n such that
∫ R + n − c n Δ V p + n + 2 n − 2 | K | U 4 n − 2 V p V p = F n − 2 R n i n s 2 S
where S is defined in (38). This means that the reduced functional given in Lemma 3 when K and H are constants is given by
J ε , g ( Θ ˜ δ , p ) = E + A γ ( p ) ε δ − δ 4 F ˜ n | Weyl g ¯ ( p ) | 2 + O ( δ 5 )
and hence the location of the blow-up depends on the non-degenerate critical points of the Weyl tensor on the boundary of the manifold.
We are now able to define the good ansatz of the solution we are looking for. With this aim, take p ∈ ∂ M with D > 1 and consider ψ p ∂ : R + n → M the Fermi coordinates in a neighborhood of p. Then, let us define
W δ , p ( ξ ) : = χ ( ψ p ∂ ) − 1 ( ξ ) 1 δ n − 2 2 U ( ψ p ∂ ) − 1 ( ξ ) δ , V δ , p ( ξ ) : = χ ( ψ p ∂ ) − 1 ( ξ ) 1 δ n − 2 2 V p ( ψ p ∂ ) − 1 ( ξ ) δ
where χ is a radial cut-off function with support in a ball of radius r 0 .
Moreover, for any i = 1 , … , n , we also set
Z δ , p , i ( ξ ) : = 1 δ n − 2 2 j i ( ψ p ∂ ) − 1 ( ξ ) δ χ ( ψ p ∂ ) − 1 ( ξ ) ,
with j i being the functions defined in (20) and (21).
We look for a solution of the form
u ε = W δ , p + δ 2 V δ , p + Φ
where Φ is a remainder term. So we will find a solution of the original problem (4) of the form
v ε : = Λ p W δ , p + δ 2 V δ , p + Φ .
In the following we simply use W ˜ δ , p , V ˜ δ , p , and Φ ˜ , Z ˜ δ , p , i to denote Λ p W δ , p , Λ p V δ , p , Λ p Φ , and Λ p Z δ , p , i , respectively.
In order to simplify the notation we let
Θ δ , p : = W δ , p + δ 2 V δ , p a n d Θ ˜ δ , p : = W ˜ δ , p + δ 2 V ˜ δ , p .
Let us decompose H 1 ( M ) into the direct sum of the following two subspaces
K ˜ : = span Z ˜ δ , p , i : i = 1 , … , n
and
K ˜ ⊥ : = φ ∈ H 1 ( M ) : ⟨ φ , Z ˜ δ , p , i ⟩ g = 0 , i = 1 , … , n .
In order to solve (12) we will use the following finite-dimensional reduction: define the projections
Π : H 1 ( M ) → K ˜ , Π ⊥ : H 1 ( M ) → K ˜ ⊥ .
Therefore, solving (12) is equivalent to solving the system
Π ⊥ v ε − i M * ( K N M ( v ε ) ) − i ∂ M * n − 2 2 ( H N ∂ M ( v ε ) − ε γ v ε ) = 0 ,
Π v ε − i M * ( K N M ( v ε ) ) − i ∂ M * n − 2 2 ( H N ∂ M ( v ε ) − ε γ v ε ) = 0 ,
with v ε defined in (24).
Equations (25) and (26) are called the auxiliary and the bifurcation equations, respectively. Solving (25) will give us the error term Φ ˜ . With the size of the error in mind, we check that our choice of the parameters leads to a solution of (26).

4. The Finite Dimensional Reduction

Equation (25) is equivalent to
L ( Φ ˜ ) = N ( Φ ˜ ) + R
where L : K ˜ ⊥ → K ˜ ⊥ is the linear operator defined as
L ( Φ ˜ ) : = Π ⊥ Φ ˜ − i M * ( K N M ′ ( Θ ˜ δ , p ) Φ ˜ ) − i ∂ M * n − 2 2 H N ∂ M ′ ( Θ ˜ δ , p ) Φ ˜ − ε γ Φ ˜ ,
while the nonlinear term N ( Φ ˜ ) and the error R are given, respectively, as
N ( Φ ˜ ) : = Π ⊥ i M * K N M ( Θ ˜ δ , p + Φ ˜ ) − N M ( Θ ˜ δ , p ) − N M ′ ( Θ ˜ δ , p ) Φ ˜ + + Π ⊥ { i ∂ M * n − 2 2 H N ∂ M ( Θ ˜ δ , p + Φ ˜ ) − N ∂ M ( Θ ˜ δ , p ) − N ∂ M ′ ( Θ ˜ δ , p ) Φ ˜
R : = Π ⊥ i M * ( K N M ( Θ ˜ δ , p ) ) + i ∂ M * n − 2 2 H N ∂ M ( Θ ˜ δ , p ) − ε γ Θ ˜ δ , p − Θ ˜ δ , p

4.1. The Size of the Error

Lemma 1.
Assume n ≥ 8 ; then it holds that
∥ R ∥ g ≲ δ 2 + ε δ i f K a n d H a r e n o t c o n s t a n t s δ 3 + ε δ i f K a n d H a r e c o n s t a n t s
Proof. 
Let
Γ M : = i M * ( K N M ( Θ ˜ δ , p ) ) and Γ ∂ M : = i ∂ M * n − 2 2 H N ∂ M ( Θ ˜ δ , p ) − ε γ Θ ˜ δ , p .
Hence it follows that
L g Γ M = K N M ( Θ ˜ δ , p ) in M B g Γ M = 0 on ∂ M L g Γ ∂ M = 0 in M B g Γ ∂ M = n − 2 2 H N ∂ M ( Θ ˜ δ , p ) − ε γ Θ ˜ δ , p on ∂ M
Since
d ν g ˜ p = det g ˜ p d x = det ( Λ p 4 n − 2 g ) d x = Λ p 2 n n − 2 det g d x = Λ p 2 * d ν g
and similarly
d σ g ˜ p = Λ p 2 ♯ d σ g
we get that
∥ R ∥ g 2 = ∥ Γ M + Γ ∂ M − Θ ˜ δ , p ∥ g 2 = ∫ M L g ( Γ M + Γ ∂ M − Θ ˜ δ , p ) ( Γ M + Γ ∂ M − Θ ˜ δ , p ) d ν g + ∫ ∂ M c n B g ( Γ M + Γ ∂ M − Θ ˜ δ , p ) ( Γ M + Γ ∂ M − Θ ˜ δ , p ) d σ g = ∫ M − L g ( Θ ˜ δ , p ) + K N M ( Θ ˜ δ , p ) R d ν g
+ c n ∫ ∂ M − B g ( Θ ˜ δ , p ) + B g ( Γ ∂ M ) R d σ g = ∫ M − L g ˜ p ( Θ δ , p ) + K N M ( Θ δ , p ) Λ p − 1 R d ν g ˜ p + c n ∫ ∂ M − B g ˜ p ( Θ δ , p ) + B g ˜ p ( Λ p − 1 Γ ∂ M ) Λ p − 1 R d σ g ˜ p = ∫ M c n Δ g ˜ p Θ δ , p − k g ˜ p Θ δ , p + K N M ( Θ δ , p ) Λ p − 1 R d ν g ˜ p + c n ∫ ∂ M − ∂ Θ δ , p ∂ ν − n − 2 2 h g ˜ p Θ δ , p + n − 2 2 H N ∂ M ( Θ δ , p ) − n − 2 2 ε γ ˜ Θ δ , p Λ p − 1 R d σ g ˜ p ≲ ∥ c n Δ g ˜ p Θ δ , p + K N M ( Θ δ , p ) ∥ L 2 n n + 2 ( M , g ˜ p ) ∥ Λ p − 1 R ∥ g ˜ p + ∥ k g ˜ p Θ δ , p ∥ L 2 n n + 2 ( M , g ˜ p ) ∥ Λ p − 1 R ∥ g ˜ p + − ∂ Θ δ , p ∂ ν + n − 2 2 H N ∂ M ( Θ δ , p ) L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ∥ Λ p − 1 R ∥ g ˜ p + ∥ h g ˜ p Θ δ , p ∥ L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ∥ Λ p − 1 R ∥ g ˜ p + ∥ ε γ ˜ Θ δ , p ∥ L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ∥ Λ p − 1 R ∥ g ˜ p
Now, we recall the expression for the Laplace–Beltrami operator in local charts:
Δ g ˜ p : = Δ + g ˜ p i j ( y ) − δ i j ∂ i j 2 + ∂ i g ˜ p i j ( y ) + g ˜ i j ( y ) ∂ i | g ˜ p | 1 2 ( y ) | g ˜ p | 1 2 ( y ) ∂ j + ∂ n | g ˜ p | 1 2 ( y ) | g ˜ p | 1 2 ( y ) ∂ n .
Then, in variables y = δ x
c n Δ g ˜ p W δ , p = δ − n − 2 2 − 2 c n Δ ( U ( x ) χ ( δ x ) ) + δ − n − 2 2 c n 1 3 R ¯ i k j ℓ x k x ℓ + R n i n j x n 2 + δ O ( | x | 3 ) ∂ i j 2 ( U ( x ) χ ( δ x ) ) + δ − n − 2 2 c n 3 R ¯ i i j ℓ x ℓ + R ¯ i k j i x k + δ O ( | x | 2 ) ∂ j ( U ( x ) χ ( δ x ) ) + δ − n − 2 2 δ 2 O ( | x | 3 ) ∂ n ( U ( x ) χ ( δ x ) ) . = δ − n + 2 2 | K ( p ) | U 2 * − 1 ︸ b y ( ) χ ( δ x ) + δ − n − 2 2 c n 1 3 R ¯ i k j ℓ x k x ℓ + R n i n j x n 2 ∂ i j 2 U ( x ) ︸ : = δ − n − 2 2 E p ( x ) χ ( δ x ) + O ( δ − n − 2 2 + 1 | x | 3 ∂ i j 2 U ( x ) ) + O ( δ − n − 2 2 + 1 | x | 2 ∂ j U ( x ) )
We remark that, by symmetry, R ¯ i i j ℓ = 0 and also R ¯ i k j i = − R ¯ j k = 0 (see [10]). Now
δ 2 c n Δ g ˜ p V δ , p = δ − n − 2 2 Δ ( V p ( x ) χ ( δ x ) ) + δ − n − 2 2 + 2 O ( | x | 2 ) ∂ i j 2 ( V p ( x ) χ ( δ x ) ) + O ( | x | ) ∂ j ( V p ( x ) χ ( δ x ) ) = δ − n − 2 2 | K ( p ) | N M ′ ( U ) V p − E p ( x ) ︸ b y ( ) χ ( δ x ) + δ − n − 2 2 + 2 O ( | x | 2 ) ∂ i j 2 ( V p ( x ) χ ( δ x ) ) + O ( | x | ) ∂ j ( V p ( x ) χ ( δ x ) ) .
Hence, in variables y = δ x
c n Δ g ˜ p Θ δ , p + K N M ( Θ δ , p ) = c n Δ g ˜ p W δ , p + δ 2 c n Δ g ˜ p V δ , p + K N M ( W δ , p + δ 2 V δ , p ) = δ − n + 2 2 K ( δ x ) N M ( U + δ 2 V p ) − N M ( U ) − δ 2 N M ′ ( U ) V p ( χ ( δ x ) ) 2 * − 1 + δ − n + 2 2 K ( δ x ) − K ( p ) N M ( U ) ( χ ( δ x ) ) 2 * − 1 + δ − n + 2 2 K ( p ) N M ( U ) ( χ ( δ x ) 2 * − 1 − χ ( δ x ) + δ − n − 2 2 K ( δ x ) − K ( p ) N M ′ ( U ) V p ( χ ( δ x ) ) 2 * − 1 + δ − n − 2 2 K ( p ) N M ′ ( U ) V p ( χ ( δ x ) 2 * − 1 − χ ( δ x ) + O ( δ − n − 2 2 + 1 | x | 3 ∂ i j 2 U ( x ) ) + O ( δ − n − 2 2 + 1 | x | 2 ∂ j U ( x ) ) + δ − n − 2 2 + 2 O ( | x | 2 ) ∂ i j 2 ( V p ( x ) χ ( δ x ) ) + O ( | x | ) ∂ j ( V p ( x ) χ ( δ x ) ) .
Now
N M ( U + δ 2 V p ) − N M ( U ) − δ 2 N M ′ ( U ) V p ≲ U 6 − n n − 2 ( δ 2 V p ) 2 sin ce n ≥ 7
Hence we get
∥ δ − n + 2 2 K ( δ x ) N M ( U + δ 2 V p ) − N M ( U ) − δ 2 N M ′ ( U ) V p ( χ ( δ x ) ) 2 * − 1 ∥ L 2 n n + 2 ( M , g ˜ p ) ≲ δ 4 .
Now if K is constant then δ − n + 2 2 K ( δ x ) − K ( p ) N M ( U ) ( χ ( δ x ) ) 2 * − 1 = 0 , while if K is not a constant, since p is a non-degenerate critical point of K , then we get
∥ δ − n + 2 2 K ( δ x ) − K ( p ) N M ( U ) ( χ ( δ x ) ) 2 * − 1 ∥ L 2 n n + 2 ( M , g ˜ p ) ≲ δ 2 .
Moreover
∥ δ − n + 2 2 K ( p ) N M ( U ) ( χ ( δ x ) ) 2 * − 1 − χ ( δ x ) ∥ L 2 n n + 2 ( M , g ˜ p ) ≲ δ 3 .
Again, if K is constant then δ − n − 2 2 K ( δ x ) − K ( p ) N M ′ ( U ) V p ( χ ( δ x ) ) 2 * − 1 = 0 while, if K is not constant we get
∥ δ − n − 2 2 K ( δ x ) − K ( p ) N M ′ ( U ) V p ( χ ( δ x ) ) 2 * − 1 ∥ L 2 n n + 2 ( M , g ˜ p ) ≲ δ 4
and
∥ δ − n − 2 2 K ( p ) N M ′ ( U ) V p ( χ ( δ x ) 2 * − 1 − χ ( δ x ) ∥ L 2 n n + 2 ( M , g ˜ p ) ≲ δ 4 .
Hence
∥ c n Δ g ˜ p Θ δ , p + K N M ( Θ δ , p ) ∥ L 2 n n + 2 ( M , g ˜ p ) ≲ δ 3 if K is constant δ 2 if K is not a constant
Now since k g ˜ p ( p ) = k g ˜ p , i ( p ) = k g ˜ p , n ( p ) = 0 (see [10]) then
∥ k g ˜ p Θ δ , p ∥ L 2 n n + 2 ( M , g ˜ p ) ≲ δ 2 ∫ R + n k g ˜ p ( δ x ) U ( x ) χ ( δ x ) 2 n n + 2 n + 2 2 n + δ 4 ∫ R + n k g ˜ p ( δ x ) V p ( x ) χ ( δ x ) 2 n n + 2 n + 2 2 n ≲ δ 4 ∫ R + n | x | 2 U ( x ) 2 n n + 2 n + 2 2 n + δ 6 ∫ R + n | x | 2 V p ( x ) χ ( δ x ) 2 n n + 2 n + 2 2 n ≲ δ 4 if n ≥ 10 δ 3 if n = 8 , 9 .
Now
∥ ε γ ˜ Θ δ , p ∥ L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ ε δ ∫ R n − 1 | U ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x n 2 ( n − 1 ) + ε δ 3 ∫ R n − 1 | V p ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x n 2 ( n − 1 ) ≲ ε δ
since for n ≥ 5
∫ R n − 1 | U ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x < + ∞
while
∫ R n − 1 | V p ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x n 2 ( n − 1 ) ≲ c if n ≥ 9 | log δ | 4 7 if n = 8 .
Since h g ˜ p ( p ) = h g ˜ p , i ( p ) = h g ˜ p , i k ( p ) = 0
∥ h g ˜ p Θ δ , p ∥ L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ δ 4 ∫ R n − 1 | | x ˜ | 3 U ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x n 2 ( n − 1 ) + δ 6 ∫ R n − 1 | | x ˜ | 3 V p ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x n 2 ( n − 1 ) ≲ δ 3
since for n ≥ 8
∫ R n − 1 | | x ˜ | 3 U ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x n 2 ( n − 1 ) ≲ c
and
∫ R n − 1 | | x ˜ | 3 V p ( x ˜ , 0 ) χ ( δ x ˜ , 0 ) | 2 ( n − 1 ) n d x n 2 ( n − 1 ) ≲ c if n ≥ 12 δ − 3 20 if n = 11 δ − 2 3 if n = 10 δ − 19 16 if n = 9 δ − 12 7 if n = 8 .
At the end
− ∂ Θ δ , p ∂ ν + n − 2 2 H N ∂ M ( Θ δ , p ) = − ∂ W δ , p ∂ ν + n − 2 2 H ( p ) W δ , p n n − 2 + n − 2 2 H − H ( p ) W δ , p n n − 2 + n − 2 2 H N ∂ M ( W δ , p + δ 2 V δ , p ) − N ∂ M ( W δ , p ) − δ 2 ∂ V δ , p ∂ ν
Since U satisfies (17),
− ∂ W δ , p ∂ ν + n − 2 2 H ( p ) W δ , p n n − 2 L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ δ 3
Now if H is constant then H − H ( p ) = 0 , while letting p a non-degenerate critical point of H , we get that
∥ H − H ( p ) W δ , p n n − 2 ∥ L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ δ 2 .
At the end we get
n − 2 2 H N ∂ M ( W δ , p + δ 2 V δ , p ) − N ∂ M ( W δ , p ) − δ 2 ∂ V δ , p ∂ ν L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ n − 2 2 H N ∂ M ( U + δ 2 V p ) − N ∂ M ( U ) χ n n − 2 ( δ x ˜ , 0 ) − δ 2 ∂ V p ∂ ν χ ( δ x ˜ , 0 ) L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ δ 2 H N ∂ M ′ ( U + δ 2 θ V p ) V p χ ( δ x ˜ , 0 ) − n n − 2 H U 2 n − 2 ∂ V p χ ( δ x ˜ , 0 ) L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ δ 2 ∥ H ( χ n n − 2 − χ ) N ∂ M ′ ( U + δ 2 θ V p ) V p ∥ L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) + δ 2 ∥ H ( N ∂ M ′ ( U + δ 2 θ V p ) − N ∂ M ′ ( U ) ) V p ∥ L 2 ( n − 1 ) n ( ∂ M , g ˜ p ) ≲ δ 3
□

4.2. Solving the Auxiliary Equation

At this point we can use the same strategy of Proposition 4.1 of [34] to prove the following result
Proposition 2.
There exists a positive constant C such that for ε , δ small, for any p ∈ ∂ M , there exists a unique Φ ˜ : = Φ ˜ ε , δ , p ∈ K ˜ ⊥ which solves the projected auxiliary Equation (27) such that
∥ Φ ˜ ∥ g = ∥ Λ p Φ ∥ g ≲ δ 2 + ε δ i f K a n d H are not constants δ 3 + ε δ i f K a n d H are constants
At this stage, the function
v ε , δ , p = Θ ˜ δ , p + Φ ˜ ε , δ , p
satisfies only the auxiliary equation obtained by projection onto K ˜ ⊥ . It solves the full Equation (12) if and only if the remaining equation obtained by projection onto K ˜ is also satisfied. This finite-dimensional equation will be addressed through the reduced functional.

5. The Reduced Functional

In this section we perform the expansion of the functional with respect to the parameter ε and δ . First, reasoning as in [34] we get that
Lemma 2.
The energy of the bubble is
E ( p ) : = a n | K ( p ) | n − 2 2 − ( n − 1 ) φ n + 1 2 ( p ) + D ( D 2 − 1 ) n − 1 2 ,
where
a n : = α n 2 ♯ ω n − 1 I n − 1 n n − 3 ( n − 1 ) n ( n − 1 ) .
Moreover, again as in [34] we can show that
J ε , g ( Θ ˜ δ , p + Φ ˜ ) − J ε , g ( Θ ˜ δ , p ) = O δ 3 if K , H are   not   constants O δ 5 if K , H are   constants
C 0 − uniformly for p ∈ ∂ M . Now we need to expand the energy on the ansatz.
Lemma 3.
If H , K are not constants, then, for ε sufficiently small, it holds that
J ε , g ( Θ ˜ δ , p ) = E ( p ) + A ( p ) ε δ − B ( p ) δ 2 + O ( δ 3 )
where E ( p ) is the energy of the bubble evaluated in Lemma 2 and A ( p ) and B ( p ) are defined in (36) and (37) respectively.
If, instead, H , K are constants, then, for ε sufficiently small, it holds
J ε , g ( Θ ˜ δ , p ) = E + A γ ( p ) ε δ − δ 4 B ( p ) + O ( δ 5 )
where E is the energy of the bubble evaluated in Lemma 2 and A and B ( p ) are defined in (36) and (39) respectively.
Proof. 
We have
J ε , g ( Θ ˜ δ , p ) = J ˜ ε , g ˜ p ( Θ δ , p ) = c n 2 ∫ M | ∇ g ˜ p Θ δ , p | 2 d ν g ˜ p + 1 2 ∫ M k g ˜ p Θ δ , p 2 d ν g ˜ p + ( n − 1 ) ε ∫ ∂ M γ ˜ Θ δ , p 2 d σ g ˜ p − c n ( n − 2 ) 2 ∫ ∂ M H P ∂ M ( Θ δ , p ) − P ∂ M ( W δ , p ) d σ g ˜ p − c n ( n − 2 ) 2 ∫ ∂ M H P ∂ M ( W δ , p ) d σ g ˜ p + ( n − 1 ) ∫ ∂ M h g ˜ p Θ δ , p 2 d σ g ˜ p − ∫ M K P M ( Θ δ , p ) − P M ( W δ , p ) d ν g ˜ p − ∫ M K P M ( W δ , p ) d ν g ˜ p = A 1 + A 2 + A 3 + A 4 + A 5 + A 6 + A 7 + A 8
Now, by (6) we get
A 6 = ( n − 1 ) δ 4 ∫ R n − 1 ∂ i j k 3 h g ˜ p ( x ˜ , 0 ) x ˜ i x ˜ j x ˜ k U 2 ( x ˜ , 0 ) d x ˜ ︸ : = 0 by symmetry + O ( δ 5 ) = O ( δ 5 ) .
By (9) we get
A 2 = 1 2 δ 4 ∫ R + n ∂ a b 2 k g ˜ p ( p ) x a x b U 2 ( x ) d x + O ( δ 5 ) = 1 4 δ 4 ∫ R + n ∂ i i 2 k g ˜ p | x ˜ | 2 U 2 ( x ˜ , x n ) n − 1 d x ˜ d x n + ∂ n n 2 k g ˜ p ∫ R + n x n 2 U 2 ( x ˜ , x n ) d x ˜ d x n + O ( δ 5 ) = − 1 24 ( n − 1 ) δ 4 | Weyl g ¯ ( p ) | 2 ∫ R + n | x ˜ | 2 U 2 ( x ˜ , x n ) d x ˜ d x n + 1 4 δ 4 ∂ n n 2 k g ˜ p ∫ R + n x n 2 U 2 ( x ˜ , x n ) d x ˜ d x n + O ( δ 5 )
Analogously we have, since Λ p ( p ) = 1 and ∇ Λ p ( p ) = 0 , that
A 3 = ( n − 1 ) ε γ ( p ) δ ∫ R n − 1 U 2 ( x ˜ , 0 ) d x ˜ + O ( ε δ 3 ) .
Now, using the fact that p is a non-degenerate critical point of H when H is not constant
A 5 = − c n ( n − 2 ) 2 ∫ R n − 1 H ( δ x ˜ , 0 ) U 2 ♯ ( x ˜ , 0 ) d x ˜ + O ( δ 5 ) = − c n ( n − 2 ) 2 H ( p ) ∫ R n − 1 U 2 ♯ ( x ˜ , 0 ) d x ˜ + δ 2 2 ∫ R n − 1 ∇ g ¯ 2 H ( p ) [ x ˜ , x ˜ ] U 2 ♯ ( x ˜ , 0 ) d x ˜ + O ( δ 3 ) if H is not constant − c n ( n − 2 ) 2 H ∫ R n − 1 U 2 ♯ ( x ˜ , 0 ) d x ˜ + O ( δ 5 ) if H is constant
Analogously, using the fact that p is a non-degenerate critical point of K when K is not constant
A 8 = − 1 2 * ∫ R + n K ( δ x ) U 2 * ( x ) d x + O ( δ 5 ) = − 1 2 * K ( p ) ∫ R + n U 2 * ( x ) d x + δ 2 2 ∫ R + n ∇ g 2 K ( p ) [ x , x ] U 2 * ( x ) d x + O ( δ 3 ) if K is not constant − 1 2 * K ∫ R + n U 2 * ( x ) d x + O ( δ 5 ) if K is constant
For the term A 4 , expanding twice by Taylor formula we get
A 4 = − c n ( n − 2 ) 2 H ( p ) ∫ R n − 1 P ∂ M ( U + δ 2 V p ) − P ∂ M ( U ) d x ˜ + O ( δ 4 ) if H is not constant − c n ( n − 2 ) 2 H ∫ R n − 1 P ∂ M ( U + δ 2 V p ) − P ∂ M ( U ) d x ˜ + O ( δ 5 ) if H is constant
Then, if H is constant then
A 4 = − c n ( n − 2 ) 2 H δ 2 ∫ R n − 1 U 2 ♯ − 1 ( x ˜ , 0 ) V p ( x ˜ , 0 ) d x ˜ − c n ( n − 2 ) ( 2 ♯ − 1 ) 4 H δ 4 ∫ R n − 1 U 2 ♯ − 2 ( x ˜ , 0 ) V p 2 ( x ˜ , 0 ) d x ˜ + O ( δ 5 )
while if H is not constant then
A 4 = − c n ( n − 2 ) 2 H ( p ) δ 2 ∫ R n − 1 U 2 ♯ − 1 ( x ˜ , 0 ) V p ( x ˜ , 0 ) d x ˜ + O ( δ 4 ) .
Analogously, if K is constant then
A 7 = − K δ 2 ∫ R + n U 2 * − 1 ( x ) V p ( x ) d x − 2 * − 1 2 K δ 4 ∫ R + n U 2 * − 2 ( x ) V p 2 ( x ) d x + O ( δ 5 )
while if K is not constant then
A 7 = − K ( p ) δ 2 ∫ R + n U 2 * − 1 ( x ) V p ( x ) d x + O ( δ 4 ) .
At the end we evaluate A 1 . First we have that
A 1 = c n 2 ∫ M | ∇ g ˜ p W δ , p | 2 d ν g ˜ p + c n δ 2 ∫ M ∇ g ˜ p W δ , p ∇ g ˜ p V δ , p d ν g ˜ p + c n 2 δ 4 ∫ M | ∇ g ˜ p V δ , p | 2 d ν g ˜ p = L 1 + L 2 + L 3 .
By using (5) and (7) and integrating by parts we get
L 3 = c n 2 δ 4 ∫ R + n | ∇ V p | 2 d x + O ( δ 5 ) = − c n 2 δ 4 ∫ R + n V p Δ V p d x + c n 2 δ 4 ∫ ∂ R + n V p ∂ ∂ ν V p d x ˜ + O ( δ 5 ) = − c n 2 δ 4 ∫ R + n V p Δ V p d x + c n n 4 δ 4 ∫ R n − 1 H ( p ) U 2 ♯ − 2 V p 2 d x ˜ + O ( δ 5 )
while
L 2 = c n δ 2 ∫ R + n ∇ U ∇ V p d x + δ 4 ∫ R + n 1 3 R ¯ i k j ℓ x ˜ k x ˜ ℓ ∂ i U ∂ j V p + R n i n j x n 2 ∂ i U ∂ j V p d x + O ( δ 5 ) = − c n δ 2 ∫ R + n Δ U V p d x + c n δ 2 ∫ R n − 1 ∂ ∂ ν U V p d x ˜ + δ 4 ∫ R + n 1 3 R ¯ i k j ℓ x ˜ k x ˜ ℓ ∂ i U ∂ j V p + R n i n j x n 2 ∂ i U ∂ j V p d x + O ( δ 5 ) = δ 2 ∫ R + n K ( p ) U 2 * − 1 V p d x + c n ( n − 2 ) 2 δ 2 ∫ R n − 1 H ( p ) U 2 ♯ − 1 V p d x ˜ + δ 4 ∫ R + n 1 3 R ¯ i k j ℓ x ˜ k x ˜ ℓ ∂ i U ∂ j V p + R n i n j x n 2 ∂ i U ∂ j V p d x ︸ : = L 2 1 + O ( δ 5 )
Integrating by parts we get
L 2 1 = ∫ ∂ R + n 1 3 R ¯ i k j ℓ x ˜ k x ˜ ℓ + R n i n j x n 2 V p ∂ i U ν j ︸ : = 0 since ν j = 0 j = 1 , … n − 1 − ∫ R + n 1 3 R ¯ i k j ℓ x ˜ k x ˜ ℓ + R n i n j x n 2 ∂ i j 2 U ︸ : = E p V p − ∫ R + n ∂ j 1 3 R ¯ i k j ℓ x ˜ k x ˜ ℓ + R n i n j x n 2 ∂ i U V p = − ∫ R + n E p V p − 1 3 R ¯ i ℓ ∫ R + n x ˜ ℓ ∂ i U V p − 1 3 R ¯ i k j j ∫ R + n x ˜ ℓ ∂ i U V p = − ∫ R + n E p V p
by using the symmetries of the curvature tensor and (10). Hence
L 2 = δ 2 ∫ R + n K ( p ) U 2 * − 1 V p d x + c n ( n − 2 ) 2 δ 2 ∫ R n − 1 H ( p ) U 2 ♯ − 1 V p d x ˜ − δ 4 ∫ R + n E p V p + O ( δ 5 )
Finally, by (5) and (7) and since the terms of odd degree disappear by symmetry, we get
L 1 = c n 2 ∫ R + n | ∇ U | 2 + c n 2 δ 2 ∫ R + n 1 3 R ¯ i k j ℓ x ˜ k x ˜ ℓ + R n i n j x n 2 ∂ i U ∂ j U + c n 2 δ 4 ∫ R + n 1 20 R ¯ i k j ℓ , m q + 1 15 R ¯ i k s ℓ R ¯ j m s q x ˜ k x ˜ ℓ x ˜ m x ˜ q ∂ i U ∂ j U + c n 2 δ 4 ∫ R + n 1 2 R n i n j , k ℓ + 1 3 Sym i j ( R ¯ i k s ℓ R n s n j ) x n 2 x ˜ k x ˜ ℓ ∂ i U ∂ j U ︸ G 1 + c n 2 δ 4 ∫ R + n 1 3 R n i n j , n k x n 3 x ˜ k + 1 12 ( R n i n j , n n + 8 R n i n s R n s n j ) x n 4 ∂ i U ∂ j U ︸ : = G 2 + O ( δ 5 ) .
Reasoning as in the proof of Lemma 6 in [35], one can show that all the terms of order δ 2 vanish.
Moreover, by the symmetries of the curvature tensor (see [10], page 1614 formula C) we get
∫ R + n 1 20 R ¯ i k j ℓ , m q + 1 15 R ¯ i k s ℓ R ¯ j m s q x ˜ k x ˜ ℓ x ˜ m x ˜ q ∂ i U ∂ j U = 0 .
Moreover
G 2 = α n 2 ( n − 2 ) 2 12 ( n − 1 ) | K ( p ) | n − 2 2 ∫ R + n R n i n i , n n ︸ : = R n n , n n = − 2 R n i n s 2 + 8 R n i n s R n s n i ︸ : = R n i n s 2 x n 4 | x ˜ | 2 ( | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 ) n d x = α n 2 ( n − 2 ) 2 2 ( n − 1 ) | K ( p ) | n − 2 2 R n i n s 2 ∫ R + n x n 4 | x ˜ | 2 ( | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 ) n d x
It remains to estimate G 1 . For symmetry reasons we have only to consider the cases i = j = k = ℓ , i = j ≠ k = ℓ , i = k ≠ j = ℓ and i = ℓ ≠ j = k . Then the Sym term gives no contribution. Hence
G 1 = α n 2 ( n − 2 ) 2 | K ( p ) | n − 2 2 ∫ R + n R n i n j , k ℓ x n 2 x ˜ k x ˜ ℓ x ˜ i x ˜ j ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n d x ˜ d x n = α n 2 ( n − 2 ) 2 | K ( p ) | n − 2 2 ∑ i R n i n i , i i ∫ R + n x n 2 x ˜ i 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n d x ˜ d x n + ∑ i ≠ k R n i n i , k k + ∑ i ≠ j R n i n j , i j + ∑ i ≠ j R n i n j , j i ∫ R + n x n 2 x ˜ i 2 x ˜ j 2 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n = α n 2 ( n − 2 ) 2 | K ( p ) | n − 2 2 ∑ i R n i n i , i i + 1 3 ∑ i ≠ k R n i n i , k k + ∑ i ≠ j R n i n j , i j + ∑ i ≠ j R n i n j , j i × × ∫ R + n x n 2 x ˜ i 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n = α n 2 ( n − 2 ) 2 3 | K ( p ) | n − 2 2 3 ∑ i R n i n i , i i + ∑ i ≠ k R n i n i , k k + ∑ i ≠ j R n i n j , i j + ∑ i ≠ j R n i n j , j i × × ∫ R + n x n 2 x ˜ i 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n = α n 2 ( n − 2 ) 2 ( n 2 − 1 ) | K ( p ) | n − 2 2 3 ∑ i R n i n i , i i + ∑ i ≠ k R n i n i , k k + ∑ i ≠ j R n i n j , i j + ∑ i ≠ j R n i n j , j i × × ∫ R + n x n 2 | x ˜ | 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n
Here we have used the fact that (see [10])
∫ R + n x n 2 x ˜ i 2 x ˜ j 2 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n d x ˜ d x n = 1 3 ∫ R + n x n 2 x ˜ i 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n d x ˜ d x n
and
∫ R + n x n 2 x ˜ i 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n d x ˜ d x n = 3 n 2 − 1 ∫ R + n x n 2 | x ˜ | 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n d x ˜ d x n .
At this point we also have that R n n , k k = 0 for all k = 1 , … , n − 1 (see Proposition 3.2 of [10]). Then, at the end, we get
G 1 = α n 2 ( n − 2 ) 2 ( n 2 − 1 ) | K ( p ) | n − 2 2 R n i n j , i j ∫ R + n x n 2 | x ˜ | 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n
Collecting all the estimates, we get that
L 1 = c n 2 ∫ R + n | ∇ U | 2 + + c n α n 2 ( n − 2 ) 2 2 ( n − 1 ) | K ( p ) | n − 2 2 δ 4 × × 1 2 R n i n s 2 ∫ R + n x n 4 | x ˜ | 2 ( | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 ) n d x + 1 n + 1 ∫ R + n x n 2 | x ˜ | 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n + O ( δ 5 ) .
So, if K and H are not constants, then we remark again that the correction V p is not necessary and the reduced functional is (putting together the previous estimates and letting γ = 1 )
J ε , g ( Θ ˜ δ , p ) = E ( p ) + A ( p ) ε δ − B ( p ) δ 2 + O ( δ 3 ) + O ( ε δ 3 )
where E ( p ) is the energy of the bubble evaluated in Lemma 2 while
A ( p ) : = ( n − 1 ) ∫ R n − 1 U 2 ( x ˜ , 0 ) d x ˜
and
B ( p ) : = c n ( n − 2 ) 4 ∫ R n − 1 ∇ g ¯ 2 H ( p ) [ x ˜ , x ˜ ] U 2 ♯ ( x ˜ , 0 ) d x ˜ + 1 2 · 2 * ∫ R + n ∇ g 2 K ( p ) [ x , x ] U 2 * ( x ) d x .
If, instead, H and K are constants and using the identity
∂ n n 2 k g ˜ p = − 2 R n i n j , i j − 2 R n i n j 2
then we have that
J ε , g ( Θ ˜ δ , p ) = E + 1 2 δ 4 ∫ R + n c n Δ V p + ( 2 * − 1 ) K ( p ) U 2 * − 2 V p V p d x + ( n − 1 ) γ ( p ) ε δ ∫ R n − 1 U 2 ( x ˜ , 0 ) d x ˜ − δ 4 1 24 ( n − 1 ) | Weyl g ¯ ( p ) | 2 ∫ R + n | x ˜ | 2 U 2 ( x ˜ , x n ) d x ˜ + 1 4 δ 4 ∂ n n 2 k g ˜ p ∫ R + n x n 2 U 2 d x + δ 4 c n α n 2 ( n − 2 ) 2 2 ( n − 1 ) | K ( p ) | n − 2 2 × × 1 2 R n i n s 2 ∫ R + n x n 4 | x ˜ | 2 ( | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 ) n d x + 1 n + 1 R n i n j , i j ∫ R + n x n 2 | x ˜ | 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n + O ( δ 5 ) = E + 1 2 δ 4 ∫ R + n c n Δ V p + ( 2 * − 1 ) K ( p ) U 2 * − 2 V p V p d x + ( n − 1 ) γ ( p ) ε δ ∫ R n − 1 U 2 ( x ˜ , 0 ) d x ˜ − δ 4 1 24 ( n − 1 ) | Weyl g ¯ ( p ) | 2 ∫ R + n | x ˜ | 2 U 2 ( x ˜ , x n ) d x ˜ + δ 4 R n i n s 2 c n α n 2 ( n − 2 ) 2 4 ( n − 1 ) | K ( p ) | n − 2 2 ∫ R + n x n 4 | x ˜ | 2 ( | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 ) n d x − 1 2 ∫ R + n x n 2 U 2 d x ︸ ( I 1 ) + δ 4 R n i n j , i j c n α n 2 ( n − 2 ) 2 2 ( n 2 − 1 ) | K ( p ) | n − 2 2 ∫ R + n x n 2 | x ˜ | 4 ( | x ˜ | 2 + ( x n + D ) 2 − 1 ) n − 1 2 ∫ R + n x n 2 U 2 d x ︸ ( I 2 )
First we remark that by simply evaluating
∫ R + n x n 2 U 2 d x = α n 2 | K | n − 2 2 ∫ 0 + ∞ ∫ R n − 1 x n 2 | x ˜ | 2 + ( x n + D ) 2 − 1 n − 2 d x ˜ d x n = ω n − 1 α n 2 | K | n − 2 2 ∫ D + ∞ ( t − D ) 2 ( t 2 − 1 ) n − 3 2 d t I n − 2 n − 2 = ω n − 1 α n 2 | K | n − 2 2 φ ^ n − 3 2 4 ( n − 2 ) n + 1 I n n + 2 .
Instead
∫ R + n x n 2 | x ˜ | 4 ( | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 ) n d x = ω n − 1 I n n + 2 ∫ D + ∞ ( t − D ) 2 ( t 2 − 1 ) n − 3 2 d t = ω n − 1 I n n + 2 φ ^ n − 3 2 .
At the end
∫ R + n x n 4 | x ˜ | 2 ( | x ˜ | 2 + ( x n + D ( p ) ) 2 − 1 ) n d x = ω n − 1 n − 3 n + 1 I n n + 2 φ ˜ n − 1 2 .
An integration by parts shows that
φ ˜ n − 1 2 = 3 n − 3 φ ^ n − 3 2 − D ∫ D + ∞ ( t − D ) 3 ( t 2 − 1 ) n − 1 2 .
Then
( I 1 ) : = α n 2 | K | n − 2 2 ω n − 1 n − 2 n + 1 I n n + 2 ( n − 3 ) φ ˜ n − 1 2 − 4 φ ^ n − 3 2 = α n 2 | K | n − 2 2 ω n − 1 n − 2 n + 1 I n n + 2 − φ ^ n − 3 2 − ( n − 3 ) D ∫ D + ∞ ( t − D ) 3 ( t 2 − 1 ) n − 1 2 = − S < 0
while
( I 2 ) : = 0
Then
J ε , g ( Θ ˜ δ , p ) = E + A γ ( p ) ε δ − δ 4 B ( p ) + O ( δ 5 )
where E is the energy of the bubble that does not depend on the point p, and A ≡ A ( p ) is defined as in (36), while
B ( p ) : = − 1 2 ∫ R + n c n Δ V p + ( 2 * − 1 ) K ( p ) U 2 * − 2 V p V p d x + 1 24 ( n − 1 ) | Weyl g ¯ ( p ) | 2 ∫ R + n | x ˜ | 2 U 2 ( x ˜ , x n ) d x ˜ + R n i n s 2 S .
We remark that B ( p ) > 0 also in the constant case since, by using Proposition 1-(iv) we have that
− 1 2 ∫ R + n c n Δ V p + ( 2 * − 1 ) K ( p ) U 2 * − 2 V p V p d x = 1 2 ∫ R + n − c n Δ V p + ( 2 * − 1 ) | K ( p ) | U 2 * − 2 V p V p d x ≥ 0 .
□
Now we are ready to prove Theorem 1.
Proof of Theorem 1.
If K and H are constants, let δ = d ε 1 3 , with d ∈ [ d 1 , d 2 ] ⊂ ( 0 , + ∞ ) . By the preceding results,
J ε , g Θ ˜ d ε 1 3 , p + Φ ˜ = E + ε 4 3 A γ ( p ) d − d 4 B ( p ) + O ( ε 5 3 )
uniformly in C 0 for p ∈ ∂ M and d ∈ [ d 1 , d 2 ] , where A , B ( p ) , and E are defined in Lemma 3. Define the reduced functional
F ε ( p , d ) = J ε , g Θ ˜ d ε 1 3 , p + Φ ˜ .
As in the standard Lyapunov–Schmidt reduction, a critical point of F ε yields a solution of (12). The leading reduced function is
Ψ ( p , d ) = A γ ( p ) d − d 4 B ( p ) .
By Theorem 1, A > 0 , γ ( p ) > 0 , and B ( p ) > 0 . Moreover,
∂ d Ψ ( p , d ) = A γ ( p ) − 4 d 3 B ( p ) .
Thus its unique positive critical scale is
d ( p ) = A γ ( p ) 4 B ( p ) 1 / 3 .
Moreover,
∂ d d Ψ ( p , d ( p ) ) = − 12 B ( p ) d ( p ) 2 < 0 ,
so this critical point is a strict maximum with respect to d.
Compactness of ∂ M and positivity of the coefficients allow 0 < d 1 < d 2 to be chosen uniformly. Since Ψ ( p , d ) → − ∞ uniformly as d → + ∞ , the reduced function has a C 0 -stable maximum in ∂ M × ( d 1 , d 2 ) . If, instead, H and K are not constants, then we let δ = d ε , d ∈ [ d 1 , d 2 ] ⊂ ( 0 , + ∞ ) .
By summarizing the previous results we have that
J ε , g Θ ˜ d ε , p + Φ ˜ = E ( p ) + ε 2 A ( p ) d − d 2 B ( p ) + O ( ε 3 )
C 0 − uniformly for p ∈ ∂ M , d ∈ [ d 1 , d 2 ] , where A ( p ) , B ( p ) , E ( p ) are defined in Lemma 3.
We again define the reduced functional
F ε ( p , d ) = J ε , g Θ ˜ d ε , p + Φ ˜ .
Now we set G p ( d ) = d A ( p ) − d 2 B ( p ) . Let p 0 ∈ ∂ M be a non-degenerate minimum point of H and K in the sense of the assumption ( Hyp 2 ) . By Lemma 2, differentiating the explicit expression for E ( p ) at p 0 shows that p 0 is a non-degenerate maximum point of E ( p ) .
Hence, there is a σ 1 − neighborhood of p 0 , say U σ 1 ⊂ ∂ M , such that for any sufficiently small η > 0
E ( p ) ≤ E ( p 0 ) − η ∀ p ∈ ∂ U σ 1 .
Furthermore,
d 0 : = A ( p 0 ) 2 B ( p 0 )
is the unique strict maximum point of G p 0 ( d ) because
G p 0 ′ ( d ) = A ( p 0 ) − 2 d B ( p 0 ) , G p 0 ′ ′ ( d ) = − 2 B ( p 0 ) < 0 .
Thus, there is an open interval I σ 2 such that I σ 2 ¯ ⊂ R + and
G p 0 ( d ) ≤ G p 0 ( d 0 ) − η ∀ d ∈ ∂ I σ 2 .
By continuity, after shrinking U σ 1 if necessary, there exists c 0 > 0 such that
G p ( d ) − G p ( d 0 ) ≤ − c 0 for every p ∈ U σ 1 ¯ , d ∈ ∂ I σ 2 .
We maximize F ε on the compact set
C : = U σ 1 ¯ × I σ 2 ¯ .
By continuity and the Weierstrass theorem, F ε attains a maximum on C . Let ( p ε , d ε ) such point. We want to show that it is in the interior of C .
By contradiction suppose that the point ( p ε , d ε ) ∈ ∂ C . There are two possibilities:
(a)
p ε ∈ ∂ U σ 1 , d ε ∈ I σ 2 ¯
(b)
p ε ∈ U σ 1 , d ε ∈ ∂ I σ 2 .
If (a) holds, by using the fact that ( p ε , d ε ) is a maximum point for F ε , Lemma 3 and (40), we have
0 ≤ F ε ( p ε , d ε ) − F ε ( p 0 , d ε ) = E ( p ε ) − E ( p 0 ) + O ( ε 2 ) ≤ − η + O ( ε 2 ) < 0
for ε sufficiently small, which is a contradiction.
If now (b) holds, then by using Lemma 3, again the fact that ( p ε , d ε ) is a maximum point for F ε and (42), we have
0 ≤ F ε ( p ε , d ε ) − F ε ( p ε , d 0 ) = ε 2 G p ε ( d ε ) − G p ε ( d 0 ) + o ( 1 ) ≤ − η ε 2 + o ( ε 2 ) < 0
for any ε sufficiently small, which is again a contradiction.
It remains to show that ( p ε , d ε ) → ( p 0 , d 0 ) as ε → 0 . Indeed, by using the fact that ( p ε , d ε ) is a maximum point for F ε and Lemma 3 we get
F ε ( p 0 , d ε ) ≤ F ε ( p ε , d ε ) ⇔ E ( p 0 ) ≤ E ( p ε ) .
Moreover by (40)
E ( p ε ) ≤ E ( p 0 )
and hence, passing to the limit it follows
lim ε → 0 E ( p ε ) = E ( p 0 ) .
Up to a subsequence, since p ε is a local maximum for E , it follows that p ε → p 0 .
In the same way one can show that d ε → d 0 as ε → 0 . □

6. Conclusions

We have established the existence of positive blowing-up solutions for a linearly perturbed boundary Yamabe problem on Riemannian manifolds with umbilic boundary in dimensions n ≥ 8 . We considered negative prescribed scalar curvature and the bubbling regime D ( p ) > 1 , treating separately the cases of constant and non-constant prescribed curvatures.
In the non-constant case, the solution concentrates at a common non-degenerate local minimum of K and H . In the constant case, the umbilicity of the boundary eliminates lower-order geometric contributions, so the higher-order correction V p is essential and the existence argument is governed by the fourth-order coefficient B ( p ) in the reduced energy.
The main difference from [34] is the treatment of the umbilic-boundary setting. In the nonumbilic case studied there, the leading geometric error appears at a lower order and a different correction term is required. Here the vanishing of the trace-free second fundamental form forces higher-order expansions of the metric and curvature terms. In particular, no correction is needed at leading order in the present non-constant case, whereas V p is indispensable in the constant case. These differences also yield distinct concentration scales: δ ∼ ε in the non-constant case and δ ∼ ε 1 / 3 in the constant case.
The present result therefore extends the blow-up analysis for linearly perturbed boundary Yamabe equations to the geometrically degenerate umbilic case. A natural direction for future work is to determine whether positivity of the fourth-order coefficient in the constant case can be characterized solely in terms of boundary curvature invariants, in particular the boundary Weyl tensor.

Funding

This work was partially supported by the MUR-PRIN-P2022YFAJH “Linear and Nonlinear PDE’s: New directions and Applications” and by the INdAM-GNAMPA project “Fenomeni non lineari: problemi locali e non locali e loro applicazioni”, CUP E5324001950001.

Data Availability Statement

All data generated or analyzed during this study are included in this article.

Conflicts of Interest

The author declares no conflicts of interest.

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