Boundary spike of the singular limit of an energy minimizing problem

In this paper, we consider the singular limit of an energy minimizing problem which is a semi-limit of a singular elliptic equation modeling steady states of thin film equation with both Van der Waals force and Born repulsion force. We show that the singular limit of energy minimizers is a Dirac mass located on the boundary point with the maximum curvature.


1.
Introduction and main results. The equation has been used to model the dynamics of long wave unstable thin films of viscous fluids. Here u is the thickness of the thin film. The nonlinear mobility is given by with λ ≥ 0 and b ∈ (0, 3) where λ = 0 corresponds to the no-slip boundary condition. And we assume the pressure is a sum of contributions from disjoining pressure due to an attractive van der Waals force, a Born repulsion force and a linearized curvature term corresponding to surface tension effects. Here α > 0, β > 0 and > 0. Abundant research on thin film equations including the stability of the solutions has been done by lots of authors [6,7,8,9,10,11,22,23,24,25,26,27,28,33,35,36]. Following [8,15,21], we consider viscous fluids in a cylindrical container whose bottom is represented by Ω, a bounded smooth domain in R n , n ≥ 2 where n = 2 is the physically meaningful dimension. Since there is no flux across the boundary, we have the Neumann boundary condition ∂p ∂ν = 0 on ∂Ω (1. 3) where ν is the unit outer normal to ∂Ω. We also ignore the wetting or nonwetting effect, and assume that the fluid surface is perpendicular to the boundary of the container, i.e., ∂u ∂ν = 0 on ∂Ω. (1.4)

XINFU CHEN, HUIQIANG JIANG AND GUOQING LIU
We can associate (1.1) with energy (1.5) Then, using (1.3) , (1.4), we have formally Hence, for a thin film fluid at rest, p must be a constant, and u satisfies (1.2), i.e., u is a critical point for the energy E (u). If we prescribe the total volume of viscous fluids in the thin film, we obtain an energy minimizing problem with volume constraint. Rewriting the energy in the following form, the authors in [15,21] considered the singular limit of the energy as → 0 with volume constraint and proved that the energy minimizing solutions converge to the limiting profile which is a Dirac measure located on the boundary. Such behavior has also been verified in one dimensional space in [8].
Our goal in this paper is to understand the location of Dirac measure in the limit. If we let approach 0 in the energy E [u], F u− tends to F * χ {u>0} formally which leads to the well-known energy functional where ε = α/2 and we have dropped the constant term − F * α . Adding the mass constrain, we consider the admissible function space H (M ) = u ∈ H 1 (Ω) : u ≥ 0 a.e. in Ω andˆΩ udx = M where M > 0 is a given constant.
Such energy functional without mass constraint has been extensively studied. Caffarelli and Alt [2] showed the Lipschitz continuity for the minima and proved singularities cannot occur for minimizer in two dimensional space. Later, Alt, Caffarelli and Friedman [4] extended the result to the case with two phases using monotonicity formula and developed the full regularity theory of the free boundary ∂{u > 0} in dimension 2 and partial regularity theory in higher dimension. In 1999, Weiss [34] claimed the existence of critical dimension k such that the free boundary is smooth if n < k. Later, Caffarelli, Jerison and Kenig [13] proved the full regularity result in dimension 3. And in 2015, Jerison and Savin [20] extended the result to dimension 4, hence k > 4. Moreover, the work completed by De Silva and Jerison pointed out k < 7 since in 7-dimensional space the singular axisymmetric critical point of the functional is an energy minimizer. Till now, the cases 5 ≤ k ≤ 6 remain open. Also more general energy functionals have been studied in [1,3,14,16,29].
The existence of energy minimizers of E ε in H (M ) follows from the direct method in calculus of variation. Moreover, we have the following asymptotic behavior of the energy minimizers: Theorem 1. Let M > 0 and {ε k } ∞ k=1 be a positive sequence converging to 0. For each k ≥ 1, let u ε k ∈ H (M ) be an energy minimizer of E ε k in H (M ). Then up to a subsequence if necessary, {u ε k } ∞ k=1 approaches a Dirac mass supported on the boundary; that is, there exists p ∈ ∂Ω such that ∀ ϕ ∈ C(Ω).
Next, we want to understand the microscopic structure of the energy minimizer near its concentration point. Let u ε ∈ H (M ) be a minimizer of E ε . Let x ε ∈ Ω be a point where u ε attains its maximum and p ε ∈ ∂Ω be such that (1.7) Let δ = ε 1/(n+1) . We define Then one can verify that v δ is an energy minimizer of Theorem 2. Under the assumption of Theorem 1, passing to a subsequence if necessary, as k → ∞, p ε k → p for some point p ∈ ∂Ω and v δ k → v * , locally uniformly in and ω n denotes the volume of the unit ball in R n .
Note that v * is the global minimizer of

XINFU CHEN, HUIQIANG JIANG AND GUOQING LIU
Now, we are about to investigate the location of the boundary spike. After translation and rotation if necessary, we could assume the concentration point p is the origin. Locally the boundary of ∂Ω can be written as Consequently, the boundary of Ω δ near the origin can be expressed as y n = 1 δ ψ (δy ) , y = (y 1 , · · · , y n−1 ) , |y | ≤ η δ .
Based on the limit profile of v * , we apply the asymptotic analysis and assume the energy minimizer has the asymptotic expansion as follows, Here D = {y : v(y) > 0}, v and R are some constants depending on δ, and λ depending on δ through R is Lagrange multiplier corresponding to the mass constraint. We know, in general, the solution does not necessarily have its mass concentrated near original point so some additional constraints have to be added: where ν is the unit exterior normal vector and κ i is the principal curvature. We are able to show the above linear system has a unique pair of solution. Moreover, we have Theorem 3. The energy of the Quasi-stationary solution (v, D) has the asymptotic expansion where c (n) = (n − 1) (n + 2) (n + 7) ω n−1 √ 2(n + 1) (n + 3) ω n is a positive constant.
The above formula implies the peak should be situated near the most curved part of ∂Ω. This type of behavior has been seen before in [30,31] where Ni and Takagi proved that a type of semilinear elliptic equation with homogeneous Neumann boundary condition admits a least energy solution which attains exactly one peak on the boundary with the maximum of the peak uniformly bounded. We remark here that the maximum of our solutions tends to be unbounded. Later, related Then one can verify that v δ is an energy minimizer of in the space The goal in this section is to understand the dependence of energy e δ (M ) on δ and mass constraint M . Formally, letting δ → 0, we obtain the limit energy in the limit admissible class Chen and Jiang [15] have proved the following result: and ω n denotes the volume of unit ball in R n . Moreover, the minimum energy We start with the dependence of minimum energy e δ (M ) on M .
In particular, e δ (M ) is continuous in M.
Denoting by B r the ball of radius r centered at the origin, using the Taylor expansion we can conclude, for r ∈ (0, R] with fixed R independent of small δ, Remark 1. The above estimate implies that the boundedness of e δ (M ) is uniform in δ. We can pick up δ small such that e δ (M ) ≤ 2e * (M ). Note that The upper bound for e δ (M ) implies that for δ small enough, the minimum for v δ (y) is equal to 0. Meanwhile, the measure of set {x : v δ (x) > 0} is bounded above by 2e * (M ).
Proof. Since we have max .

XINFU CHEN, HUIQIANG JIANG AND GUOQING LIU
Analogously to [15], we use a rearrangement argument to establish a lower bound for e δ (M ) and then obtain the limit of e δ (M ) . For convenience, we provide the detail after little revision here. For δ small enough, from the above remark, we know v = 0. We define for any t ∈ [0, ∞), For any open interval (a, b), we have from the coarea formula [17] , andˆ{ x∈Ω δ ,a<v<b} |∇v (y)| dH n−1 (y) dt. Using we derive from (3.7),ˆΓ Thus,ˆ{ x∈Ω δ ,a<v<b} Let P (·) be the best constant of isoperimetric inequality: . P (α) is decreasing in α since the infimum is taken on a bigger set for larger α. Now for small > 0, since

BOUNDARY SPIKE 3263
Also as Ω δ has almost flat and smooth boundary ∂Ω δ = ∂Ω/δ, we see that Now define the symmetric decreasing rearrangement function w by Then It then follows that And then, Finally, we defineŵ Then we haveˆ∞ Thus, we obtain Letting → 0, we obtain . The assertion is completely proved due to Lemma 2.

Euler-Lagrange equation.
In this section we are going to derive the Euler-Lagrange equation for the minimizer v δ . Firstly, we prove that v δ is continuous inside Ω δ .
Theorem 7. For any compact set K ⊂ Ω δ , there exists a constant C such that Proof. Let B r (y) ⊂ Ω δ be any ball of radius r with center y and u ∈ H 1 (Ω δ ) be the unique function satisfying ∆u = 0 in B r (y) and u = v δ in Ω δ \B r (y).
where C = C(M, n) is some constant depending on the total mass M and dimension n. Here (4.2) follows from the Poincaré Inequality and the last step (4.4) holds according to Lemma 2. We choose r small such that where C 1 , C 2 = C 2 (M, n) are constants following Lemma 2. Therefore, Plug (4.2) into (4.5), we obtain For convenience, we will suppress the subscript δ here and let v = v δ . D := {y ∈ Ω δ : v > 0} is an open set as a result of the continuity. By the standard calculus of variation, for ∀ζ ∈ C ∞ 0 (D) with´D ζdy = 0 and ε sufficiently small so that v + εζ > 0 in D,

XINFU CHEN, HUIQIANG JIANG AND GUOQING LIU
We can derive that ∆v = −λ δ in D where λ δ is the Lagrange multiplier. The next theorem shows that, in a generalized sense, on the free boundary ∂D ∩ Ω δ , The linear term in ε must vanish, giving Using standard variation argument, we also have where ν is the unit outer normal and the constant λ δ is the Lagrange multiplier such that for δ small,

Uniform Hölder Continuity.
In this section, we prove the Uniform Hölder Continuity for the minimizer v δ . we need the following uniform Poincaré inequality.

Lemma 4.
For any open connected domain Ω in class Θ defined by where there exists a uniform C such that for ∀u ∈ H 1 (Ω) with u = 0 on ∂B r (0) ∩ Ω and r < 1.
Proof. Apply the Corollary 3 in [12]. One can check that B 1 (0) ∩ Ω satisfies the ε-cone property. Then there exists a uniform C such that (ˆB for all Ω in class Θ and ∀u ∈ H 1 (Ω) with u = 0 on ∂B 1 (0) ∩ Ω. Rescaling argument shows that (ˆB where Ω is the transformation of Ω after scaling which still belongs to class Θ.
It is ready to make a comparison with a harmonic function in any small ball and obtain the growth of local integrals.
Lemma 5. For any y ∈ Ω δ and r > 0, Proof. Let B r (y) be any ball of radius r centered at a point y in Ω δ and define 2) follows from the uniform Poincaré Inequality (Lemma 4) due to the fact that the smooth boundary of Ω δ is almost flat if we take r so small. The last step (5.3) holds according to Lemma 2. We choose δ to be small so that where C 1 , C 2 = C 2 (M ) are constants following Lemma 2. Therefore, Consequently, solving the above quadratic equation yieldŝ Checking the harmonic function v in the right hand side of (5.1) which satisfies the Neumann boundary condition on the boundary, we quote Lemma 9 in [15] which gives the estimate. Lemma 6. Let 0 < ε ≤ 1. For anyα ∈ (0, 1), there exist r 0 > 0 and Kα > 1 such that for any y ∈ Ω δ and r ∈ (0, r 0 ] and for any v satisfying we have for any σ ∈ (0, 1), (5.5) Combining (5.1) and (5.5) gives the core lemma regarding the growth of the Dirichlet integral for v δ . This is the key step to show C α continuity.
We have from Lemma 5,B For any σ ∈ (0, 1), we have ˆB Here in the second inequality, we have applied Lemma 6 to the second term on the right-hand side. Divide both sides by (σr) n/2−1+α and define

XINFU CHEN, HUIQIANG JIANG AND GUOQING LIU
We have Choose σ so that √ Kασα −α < 1. A simple induction then gives for any r ∈ (σr 0 , r 0 ] and for any k ∈ N.
and furthermore, A uniform Hölder bound for v δ follows from the decay estimate of Lemma 7: Theorem 9. Let M > 0, δ > 0 and v δ be a minimizer of E δ in H δ (M ). There exists a constant C such that for given small δ 0 and any 0 < δ ≤ δ 0 , v δ C α (Ω δ ) ≤ C.
Therefore, v δ is uniformly bounded in δ.
Proof. Applying Poincaré's inequality, we have for any y ∈ Ω δ and r ∈ (0, r 0 ], Hence, v δ is in Campanato space L 2,n+2α (Ω δ ). Following from Theorem 1.2 in page 70 of [18], v δ is Hölder continuous with αth Hölder seminorm bounded by constant C 4 which is independent of δ. Since Therefore, sup v δ is uniformly bounded which ends the proof of the theorem. 6. Uniform Lipschitz continuity. The main goal of this section is to prove the uniform Lipschitz continuity of v δ . We remark here that uniform Hölder continuity obtained in the previous section is sufficient for the results of this paper, but we think such an estimate may be useful in our future research in this topic. The idea is based on the work by Caffarelli and H.W.Alt [2] with Dirichlet boundary setting. The mass constraint is the new technical difficulty here and we also require uniform global estimate involving boundary under Neumann boundary setting which is not covered in [2]. In this section, we assume δ is small and all the constants are independent of such uniformly small δ.
Firstly, we prove the following lemma.
Then for any small ball Proof. Take harmonic function u such that ∆u = 0 in B r and u = v in Ω δ \B r .
It follows thatˆB Note that the map x −→ (1 − |x|)z + x is an isomorphism from B 1 (0) to itself. Also for ∀ξ ∈ ∂B 1 , define if the set is nonempty and r ξ = 1 if the set is empty. For almost all ξ ∈ ∂B 1 , Also using Green function G(x, y) with ∆G(x, y) = δ x for x ∈ B 1 (0) , Integrating over ξ and then integrating over z, Together with (6.1), we obtain, where λ > 0 is chosen such that´B rũ =´B r v. It is easy to check that Therefore for both cases, we have (6.5), that is, Recall that D = {x ∈ Ω δ : v(x) > 0} and the measure of D is bounded above by 2e * (M ). Define Σ = {x ∈ Ω δ : v(x) = 0}. We have the following interior estimate, Proof. For any x ∈ Ω δ , take the maximal ball B r (x) ⊂ D = Ω δ \Σ. Since the measure of D is bounded above, then r ≤ C(M, n). Since ∆v = −λ δ in B r (x), then we can rewrite v as For dist(x, Σ) < dist(x, ∂Ω δ ), ∂B r (x) does not touch ∂Ω δ . Then for arbitrary small , B r+ (x) ∩ D is nonempty which follows, Take → 0, then 1 Consequently, For dist(x, Σ) = dist(x, ∂Ω δ ), we see x as the limit of a sequence of points {x n } ∞ n=1 with dist(x n , Σ) < dist(x n , ∂Ω δ ). By applying the continuity of |∇v| in D, we will be able to finish our proof for this lemma.
In order to prove the uniform Lipschitz continuity, we have to make some boundary estimates and prove the boundedness of |∇v(x)| for dist(x, Σ) > dist(x, ∂Ω δ ). So we divide into two cases dist(x, Σ) ≥ r 0 and r 0 > dist(x, Σ) > dist(x, ∂Ω δ ) for fixed small r 0 .
Note that the Harnack inequality we used here is for function ∆u = −λ δ R ≤ 0 in a ball B R (0). The proof is to take the classical the Harnack inequality on harmonic function u * = u − λ δ R 2n (R 2 − |x| 2 ). Then, Now take ball B 1 2 (x) and then x 0 ∈ B 1 2 (x). Apply the Harnack inequality again in Therefore, apply the same estimate as above for w( whereC only depends on M, n and r 0 .
Remark 2. One referee suggested that with the estimates we obtained, it is possible to show that the free boundary has uniform C 1,α regularity following Alt-Caffarelli theory. We decide to focus on the singular limit of the variational problem and will not pursue the free boundary regularity here.

Singular limit profile. Given the total mass
be a sequence such that lim k→∞ ε k = 0. Let u ε k ∈ H M be an energy minimizer of E ε k in H M . For simplicity, we will suppress the k subscript whenever there is no confusion. Let x ε ∈Ω be a point where u ε attains its maximum and p ε ∈ ∂Ω be such that Passing to a subsequence if necessary, we can assume lim k→∞ p ε k = p * ∈ ∂Ω and we denote ν * = ν (p * ), the unit outer normal of ∂Ω at p * . Let Ω δ and v δ be defined in (1.8) and (1.9). Then v δ is a minimizer of E δ in H (M, Ω δ ) and as k → ∞, Ω δ → R n ν * := {y ∈ R n | y · ν * < 0}. For simplicity, after a rotation if necessary, we assume ν * = (0, · · · , 0, −1) and hence R n ν * = R n + . Proposition 2. There exist constants Proof. When k is sufficiently large, we have δ k ≤ m 2|Ω| and from Theorem 6, e δ (M ) ≤ 2e * (M ), hence Lemma 3 implies .
On the other hand, the uniform Hölder norm of v δ k follows from Theorem 9 and uniform Lipschitz continuity follows from Theorem 10. Note that the constant is independent of δ.
Due to the uniform bound of Hölder continuity, passing to a subsequence if necessary, we can assume v δ converges locally uniformly to a limit v * in R n + . The main goal in this section is to show that v * is the unique energy minimizer of E * with´R n Lemma 10. There exists a constant C > 0, such that for any k ∈ N, Proof. If such constant doesn't exist, passing to a subsequence if necessary, we can assume For simplicity, we again suppress the k subscript. We define a blow up sequence along x ε byΩ Thenṽ δ is a minimizer of E δ in the space where in the definition of energy E δ , Ω δ is replaced byΩ δ . Since |pε−xε| δ → ∞, we haveΩ δ →R n as k → ∞. Noticing that for each k,ṽ δ is a translation of v δ , the uniform bound of Hölder norms of v δ implies that, passing to a subsequence if necessary,ṽ δ → v * locally uniformly in R n as k → ∞, which implies the uniform Hölder continuity ofṽ δ implies For any σ > 0 sufficiently small, we can choose R 0 > 0, such that for some K > 0 which is independent of small ε, we can choose 1 ≤ l ≤ N so that Now let η ∈ C ∞ (R n ) be a cutoff function, such that η ∈ [0, 1] and |∇η| ≤ 2 for any x ∈ R n .
We haveṽ Hence, we conclude when k sufficiently large, we havêΩ Letting k → ∞, we have Next, we show there is no loss of mass in the limiting process.
Proof. Let Since |pε−xε| δ is uniformly bounded, the uniform Hölder bound for v δ and uniformly positive lower bounds for Proof of Theorem 2. Up to a subsequence, we also assume v δ k converges to v * weakly in H 1 loc R n + as k → ∞ and hence, the lower semi-continuity of norms implies 1 2ˆRn On the other hand, let |{v * > 0}| = µ * > 0.
For each σ > 0, there exists N > 0 such that Since σ is arbitrary, The convergence of the blow up sequence v δ implies the convergence of u ε .
Proof of Theorem 1. Since {u ε k } ∞ k=1 is a sequence of positive function with total mass m, there exists a measure µ onΩ such that passing to a subsequence if necessary u ε k * µ in the weak star topology as k → ∞. Passing to a subsequence if necessary , we also have the blow up sequence v δ → v * locally uniformly as k → ∞ and´Ω v * = M . Hence Since´Ω u ε (x) dx = M and p ε → p * , we conclude u ε * µ = M δ p * as k → ∞.
The above theorem implies when ε approaches zero, the energy minimizer converges to a Dirac measure concentrated on the boundary. We are going to show next that the Dirac measure should be located near the point with maximal mean curvature.

Linearization.
To understand the location of the boundary spike, we consider the free boundary problem (4.6) associated to the scaled energy minimizing problem: Since true solution should have spikes near specific boundary points, here for a fixed point p ∈ ∂Ω, we seek a pair (v, D) such that the "center" of D is the origin and that (v, D) only approximately solves the free boundary problem (8.1), e.g., having error O δ 2 . Then we compare the energy when p is moving around the boundary. By shifting and rotation, we assume that p = 0 and the unit normal of ∂Ω at p is (0 , −1). The boundary near p is represented in local coordinates as We call κ i the principal curvature of ∂Ω at p and denote by κ = n−1 i=1 κ i /(n−1) the mean curvature of ∂Ω at p. Locally the boundary of ∂Ω δ near q := p/δ is expressed as In general, (8.1) does not have a solution that has mass concentrated near q. To overcome this difficulty, we add an extra constraint in the class of minimization to ensure that the mass is near q. Hence we consider the minimization of E * δ in the space where N (q) is the normal vector of Ω δ at q and the symbol represents parallel relation between two vectors. In the current notation, the second set of constraints meanˆΩ δ y i vdy = 0 ∀ i = 1, · · · , n − 1. (8. 2) The corresponding free boundary problem can be written as where λ, λ 1 , · · · , λ n−1 are Lagrange multipliers.
We search a solution of (8.3) that can be expanded in the δ-power series as follows where R and λ are constants depending on δ, R 1 and v 1 are unknown functions that depend on δ only through the constants λ and R. We derive the equations for (R, λ, v 1 , R 1 ) as follows.
(1) The free boundary condition v = 0 on the free boundary implies (2) The normal of the free boundary is The free boundary condition which can be achieved by setting and we have N = (δκ 1 y 1 , · · · , δκ n−1 y n−1 , −1) + O δ 2 , and the boundary condition ∂ n v = 0 on ∂Ω δ can be written as This can be achieved only by setting Thus we see that (v 1 , R 1 ) needs to be a solution of the linearized problem given by It is sufficient to consider only the equation for v 1 . Note that We derive from (8.1) that Theorem 11. The mixed boundary condition problem (8.4) with the constraint (8.5) admits a unique solution.
First we establish the lemma for Robin boundary condition problem on a ball.
Lemma 12. Assume f ∈ L 2 (S R ), the Robin boundary condition problem on a ball with radius R given by admits a solution u ∈ H 1 (B R ) if and only if f satisfies the compatibility condition The solution is unique if we add the constraintŝ B R y i udy = 0, ∀ i = 1, · · · , n.
Secondly, suppose f ∈ L 2 (S R ) satisfying the compatibility condition. Let H m (R n ) denote the subspace of all the homogeneous harmonic polynomials on R n of degree m. and H m (S R ) represent the subspace of all the homogeneous harmonic polynomials in H m (R n ) with restriction to S R of degree m. Since L 2 (S R ) = ⊕ ∞ m=0 H m (S R ) (Theorem 5.12 and Theorem 5.29 in [5]), Using the homogeneity, we see, where d m is to be determined. Formal calculation gives According to the Robin boundary condition, we define Then u M is harmonic in B R and for N > M > 1, Moreover, . Therefore u M is a Cauchy sequence in L 2 (B R ) and ∇u M is a Cauchy sequence in (L 2 (B R )) n . Then let we can obtain that as the limit of Regarding the uniqueness, we consider solution u ∈ H 1 (B R ) to the homogeneous system, where p m (y) ∈ H m (R n ). Applying the robin boundary condition, we obtain, Due to the orthogonality of H m (R n ) on S R in sense of L 2 inner product, then, p m (y) = 0, ∀ m = 1.

XINFU CHEN, HUIQIANG JIANG AND GUOQING LIU
We have, Therefore, the constraints´B R y i udy = 0 implies c i = 0 which is the uniqueness of the solution. Now we are ready to prove the existence of the solution to the non-homogeneous problem on half ball.
Theorem 12. Given f ∈ L 2 (B + R ), g ∈ L 2 (Γ R ) and h ∈ L 2 (B R × {0}), the mixed boundary condition problem given by admits a solution w ∈ H 1 (B + R ), if and only if (f, g, h) satisfies the compatibility conditionŝ If there is a solution w sp , then the general solution is given by where c 1 , · · · , c n−1 are arbitrary constants. The solution is unique if we requirê B + R y i w(y)dy = 0 ∀ i = 1, · · · , n − 1.
We first consider the homogeneous system   Due to the Neumann boundary condition on B R × {0}, even reflection gives Applying Lemma 12 and the fact that w is even in y n , we have the general solutions for the homogeneous system are given by Next, for the non-homogeneous problem, we choose functions F ∈ H 2 (B + R ) and Here G ∈ L 2 (Γ R ). Similarly, applying even reflection for u and making use of Lemma 12, solution u ∈ H 1 (B R ) exists if and only if for any i = 1, , · · · , n − 1, Hence, for any i = 1, , · · · , n − 1, In order to obtain the explicit solution, we can compute the basis for H m (R n ) using zonal harmonics (See Chapter 5 in [5]). Then given compatibility condition, the special solution for u can be calculated using inner product. Therefore it gives the special solution w sp = u + F + H. The general solution is given by It is easy to see problem (8.4) together with the constraint (8.5) is a special case of (8.7). Then the proof of Theorem 11 naturally follows from theorem 12. Hence, v 1 is uniquely solvable. 9. Location of spike. Though it is hard to find explicit solution of (8.4) and (8.5), we can still proceed to find quantities of our interest. In this section, we will focus on the energy expansion which helps to locate the position of the spike. Applying the asymptotic analysis, we consider the effect of the mass constraint. Later as the theorem 3 stated, the energy of the Quasi-stationary solution (v, D) has the asymptotic expansion where c (n) = (n − 1) (n + 2) (n + 7) ω n−1 √ 2(n + 1) (n + 3) ω n .