Existence and orbital stability of standing waves to a nonlinear Schr\"odinger equation with inverse square potential on the half-line

We investigate the properties of standing waves to a nonlinear Schr\"odinger equation with inverse-square potential on the half-line. We first establish the existence of standing waves. Then, by a variational characterization of the ground states, we establish the orbital stability of standing waves for mass sub-critical nonlinearity. Owing to the non-compactness and the absence of translational invariance of the problem, we apply a profile decomposition argument. We obtain convergent minimizing sequences by comparing the problem to the problem at"infinity"(i.e., the equation without inverse square potential). Finally, we establish orbital instability by a blow-up argument for mass super-critical nonlinearity.

There has been considerable interest recently in the study of the Schrödinger equation with inverse-square potential in three and higher dimensions. Classification of the so-called minimal mass blow-up solutions, global well-posedness, and stability of standing wave solutions were studied in [1,6,8,22]. In the papers by A. Bensouilah, V. D. Dinh, and S. Zhu [1], and by G. P. Trachanas and N. B. Zographopoulos [22] the authors establish orbital stability of ground state solutions in the Hardy subcritical (c < (N − 2) 2 /4) and Hardy critical (c = (N − 2) 2 /4) case respectively for dimensions higher that three. In both cases, orbital stability is proved by showing the precompactness of minimizing sequences of the energy functional on an L 2 constraint. Local well-posedness was established for the two-dimensional space by T. Suzuki in [21], and in three and higher dimensions by N. Okazawa, T. Suzuki, and T. Yokota in [18]. The presence of the inverse square potential in one-dimensional space has also attracted attention. In [13] the H. Kovarik and F. Truc established dispersive estimates for ∂ 2 x + c/x 2 . The dynamics of the equation is closely related to Hardy's inequality (see [7]) where c 1/4. We introduce the Hardy functional which is closely related to our problem. We will mainly focus on the case 0 < c < 1/4, when the natural energy space associated to (1.1) is H 1 0 (R + ), and the semi-norm u ′ 2 L 2 is equivalent to H(u).

Let us consider the operator
Owing to the Hardy inequality, if c < 1/4 the quadratic form H c ϕ, ϕ is positive definite on C ∞ 0 (R + ). It is natural to take the Friedrichs extension of H c , thereby defining a self-adjoint operator in L 2 (R + ), which generates an isometry group in H 1 0 (R + ). Local well-posedness for parameters 1 < p < ∞ and 0 < c < 1 4 follows by standard arguments (see e.g. in [3] Chapter 4). In particular, the following holds. Theorem 1.1. Let 1 < p < ∞ and c < 1/4. For any initial value u 0 ∈ H 1 0 (R + ), there exist T min , T max ∈ (0, ∞] and a unique maximal solution u ∈ C((−T min , T max ), H 1 0 (R + )) of (1.1), which satisfies for all t ∈ (−T min , T max ) the conservation laws where the energy is defined as Moreover, the so-called blow-up alternative holds: if T max < ∞ then lim t→Tmax u ′ (t) L 2 = ∞, (or T min < ∞ then lim t→−T min u ′ (t) L 2 = ∞).
In this work we address the existence of standing wave solutions and their orbital stability/instability. By introducing the ansatz u(t, x) = e iωt ϕ(x), the standing wave profile equation to (1.1) reads as (1.5) ϕ ′′ − c x 2 ϕ + ωϕ − |ϕ| p−1 ϕ = 0. First we will prove regularity of standing waves and the Pohozaev identities. To establish the existence of standing waves we carry out a minimization procedure on the Nehari manifold for the so-called action functional Owing to the non-compactness of the problem, we have to use a profile decomposition lemma, in the spirit of the article by L. Jeanjean and K. Tanaka [11]. To establish strong convergence of the minimizing sequence on the Nehari manifold we compare the minimization problem with the problem "at infinity", i.e. when c = 0. Hence, we obtain that the set of bound states is not empty: A = {u ∈ H 1 0 (R + ) \ {0} : u ′′ + cu/x 2 − ωu + |u| p−1 u = 0} = ∅. We are in particular interested in the orbital stability/instability of ground states, i.e., solutions which minimize the action We use Lions' concentration-compactness principle to obtain a variational characterization of ground states on an L 2 -constraint, thereby establishing the orbital stability of the set of ground states for nonlinearities with power 1 < p < 5. Finally, for p 5 we establish strong instability by a convexity argument. In conclusion we prove the following theorem: Theorem 1.2. Let 0 < c < 1/4, 0 < ω and 1 < p. Then (1.5) admits a nontrivial solution in H 1 0 (R + ), which decays exponentially at infinity. Moreover, if 1 < p < 5 the set of ground states is orbitally stable, and if p 5, the set of ground states is strongly unstable by blow-up.

Existence of bound states
We start by investigating the standing wave equation, . First, we prove the regularity of solutions to (2.1) by a bootstrap argument.
Let ε > 0 and θ ε (x) = e x 1+εx , for x 0. It is easy to see that θ ε is bounded, Lipschitz continuous, and |θ ′ ε (x)| θ ε (x) for all x ∈ R + . Additionally, θ ε (x) → e x uniformly on bounded sets of R + . Taking the scalar product of the equation Let R > 0 such that if x > R, then c x 2 1 8 and |ϕ(x)| p−1 1 8 . Then we get easily that From the last two inequalities it follows that By taking ε ↓ 0 we get Since both ϕ and ϕ ′ are Lipschitz continuous we deduce that |ϕ(x)|e x and |ϕ ′ (x)|e x are bounded.
We now prove that there exists a solution to (2.1). We define the action functional associated to (2.1) as follows Therefore, to prove the existence of a solution to (2.1) amounts to show that S has a nontrivial critical point. A simple calculation yields the following identities.
To prove the second equality, let us put ϕ λ (x) = λ 1/2 ϕ(λx) for λ > 0. We have that We also have that , since ϕ and ϕ ′ are exponentially decaying at infinity by Proposition 2.1. We obtain that the right hand-side is well-defined. Since ϕ is a critical point of S, we obtain S ′ (ϕ) = 0, which concludes the proof.
Let us define for all u ∈ H 1 0 (R + ) the following functional: : J(u) = 0} contains all nontrivial critical points of S. We aim to show that the infimum of the following minimization problem is attained First we prove the following lemma.
Let m ∈ R. We say that {u n } n∈N is a Palais-Smale sequence for S at level m, if as n → ∞.
Lemma 2.6. Let c < 1/4, and p > 1. There exists a bounded Palais-Smale sequence {u n } n∈N ⊂ N for S at the level m. Namely, there is a bounded sequence {u n } n∈N ⊂ N such that, as n → ∞, Proof. Since N is a closed manifold in H 1 0 (R + ), it is a complete metric space. Hence, Ekeland's variational principle (see pp. 51-53 in [20]) directly yields the existence of a Palais-Smale sequence at level m in N .
We now show that if {u n } n∈N ⊂ N and u n 2 H 1 → ∞, then S(u n ) → ∞. Indeed, since u n ∈ N from Hardy's inequality we get that Therefore, any Palais-Smale sequence {u n } n∈N is bounded in H 1 0 (R + ). Before proceeding to our next lemma, let us recall some classical results, see e.g. [3], concerning the case c = 0. It is well-known that the set of solutions of is given by {e iθ q(·+y) : y ∈ R, θ ∈ R}, where q is a symmetric, positive solution of (2.5), explicitly given by .

E. CSOBO
Moreover, up to translation and phase invariance, it is the unique solution of the minimization problem where the functionals S ∞ and J ∞ are defined by Proof. It is not hard to see that m m ∞ , we only need to prove that m = m ∞ . Let us first note Indeed, if J(u) < 0, then let us define .
For large enough A we obtain the following estimates (see Lemma 5.1 in the Appendix): Since 0 < c < 1/4, we obtain for A > 0 large enough which concludes the proof.
We need the following lemma, which describes the behavior of bounded Palais-Smale sequences. We note that H 1 0 (R + ) functions can be extended to functions in H 1 (R) by setting u ≡ 0 on R − . The proof of the following statement is presented in the appendix.
where in case k = 0, the above holds without q i and x i n . We only need to show that the critical point of S provided by Lemma 2.8 is non-trivial.
Proof. We only have to prove that the {u n } n∈N bounded Palais-Smale sequence obtained in Lemma 2.6 admits a strongly convergent subsequence. Assume that it is not the case. Using Lemma 2.8 we see that k 1 and u n is weakly convergent to u 0 in H 1 0 (R + ) up to a subsequence. Then Now, S(u 0 ) 0 since J(u 0 ) = 0. Thus m m ∞ , which contradicts Lemma 2.7. Hence k = 0 and u n → u 0 in H 1 0 (R + ). Lemma 2.10. Let p > 1 and ω > 0. There exists a µ > 0 such that ∞ 0 |u| 2 dx = µ, for every u ∈ G.

The mass of ground state solutions is
Proof. Since u ∈ G is a solution of (2.1), it satisfies (2.2) and (2.3). By subtracting the two identities we get Additionally, since u is a ground state solution, it also solves the minimization problem (2.4). From (2.4) and (2.3) we get From (2.7) and (2.8) it follows which concludes the proof.

Stability
In this section we consider nonlinearities with 1 < p < 5. Our aim is to prove orbital stability of the standing waves. To do so, we investigate the minimization problem: . and the energy E is defined by (1.4). We will rely on a of Lions' concentration-compactness principle [15] and the arguments by Cazenave and Lions [4], see also in [3]. The main problem is to obtain compactness of minimizing sequences owing to the absence of translation invariance. We define the problem at infinity by We recall some well-known facts about the minimization problem (3.2) (see [3,Chapter 8.]). For every µ > 0, there exists a unique, positive, symmetric function q = q(µ) ∈ H 1 (R), such that and q solves the nonlinear equation We proceed by proving the following lemma: , then the following inequality holds: Proof. For A > 0, let C(A) be a normalizing factor specified later. Let us define Since q is even, we obtain Ψ A ∈ H 1 0 (R + ) and We estimate the second integral by (see Lemma 5.1) We define

C(A) is a continuous function of A, C(A)
1, and C(A) → 1 exponentially fast as A → ∞. Thus, Ψ A L 2 = µ for all A > 0. By Lemma 5.1 in the Appendix, we obtain for A > 0 large enough that Hence for A large enough we get Owing to the exponential decay of the last term, for large A we get We need the following version of the concentration-compactness principle. The proof follows the same way as in the classical case (see [15]). Then there exists a subsequence {u n } n∈N such that it satisfies one of the following alternatives.
(Dichotomy) There are sequences {v n } n∈N , {w n } n∈N in H 1 0 (R + ) and a constant α ∈ (0, 1) such that: There exists a sequence y n ∈ R + , such that for any ε > 0 there is an R > 0 with the property that for all n ∈ N.
We are now in a position to prove the following lemma. Lemma 3.3. Let 1 < p < 5, 0 < c < 1/4, and ω > 0. Then the infimum in (3.1) is attained. Additionally, all minimizing sequences are relatively compact, that is if {u n } n∈N satisfies u n 2 L 2 → µ and E(u n ) → I then there exists a subsequence {u n } n∈N which converges to a minimizer u ∈ We get from the Gagliardo-Nirenberg inequality that there exists C > 0 such that for all u ∈ H 1 Since 1 < p < 5, this yields that there exists δ > 0 and K > 0 such that Every minimizing sequence is bounded in H 1 0 (R + ) and bounded from below in L p+1 (R + ). Indeed, let {u n } n∈N ⊂ Γ be a minimizing sequence, then by (3.3) it is bounded in H 1 0 (R + ). Furthermore, for n large enough we have E(u n ) < I/2, thus Now I < 0, hence the result follows.
Step 2. We now verify that all minimizing sequences have a subsequence which converges to a limit u in H 1 0 (R + ). Let {u n } n∈N satisfying u n 2 L 2 → µ and E(u n ) → I. Since every minimizing sequence is bounded in H 1 0 (R + ), {u n } n∈N has a weak-limit u ∈ L p (R + ) . We can apply the concentration-compactness principle (see Lemma 3.2) to the sequence {u n } n∈N . We note that since the sequence is bounded from below in L p+1 (R + ) vanishing cannot occur. Now let us assume that dichotomy occurs. Let α ∈ (0, 1), {v n } n∈N and {w n } n∈N sequences as in Lemma 3.2. It follows from (5) and (6) Observe that for u ∈ H 1 0 (R + ), and a > 0, we have Let a n = √ µ/ v n L 2 and b 2 k = √ µ/ w n L 2 . Hence, a n v n ∈ Γ and b n w n ∈ Γ, which implies Now we observe a −2 n → α and b −2 n → (1 − α) by (4) of Lemma 3.2. Since α ∈ (0, 1), we get that θ = min{α −(p−1)/2 ; (1 − α) −(p−1)/2 )} > 1. Property (5) of Lemma 3.2 and (3.4) implies which contradicts with (3.5). Hence the following holds: there exists a sequence y n ∈ R + , such that for any ε > 0 there exists R > 0 with the property that for all k ∈ N.
We now show that {y n } n∈N is bounded in R + . First we show that if y n → ∞, then Let us take ξ ∈ C ∞ (R + ), such that forR > 0 and a > 0 we have that ξ(r) = 1 for 0 r R , ξ(r) = 0 for r R + a, and ξ ′ L ∞ 2/a. We introduce u n,1 = u n · ξ and u n,2 = u n · (1 − ξ). Clearly, u n,1 ∈ H 1 0 (R + ), u n,2 ∈ H 1 0 (R + ) and u n = u n,1 + u n,2 . Moreover, the following inequalities hold |u ′ n,1 | 2 2(4a −2 |u n | 2 + |u ′ n | 2 ), |u ′ n,2 | 2 2(4a −2 |u n | 2 + |u ′ n | 2 ). We obtain by direct calculation that We show that there existsR > 0 and a > 1, such that for n large enough |ρ n | (1/4 − c) δ 4 . First we observe by the properties of the cut-off that We claim that there existR > 0 and a > 1 such that for a subsequence {u n k } we have Suppose that this claim does not hold, that is for all R > 0, a > 1 there exists k ∈ N such that for all n k the following holds Let (R 1 , R 1 + a 1 ). There exists k 1 ∈ N, such that for all n k 1 we have Now let R 2 > R 1 + a 1 and a 2 > 1. Then by our assumption there exists k 2 ∈ N, such that for all n k 2 it holds that Hence, there exists a subsequence {v n k } k∈N such that for all j ∈ {1, 2} it holds that for all k ∈ N. Therefore, we can construct for all l ∈ N a subsequence {u n k } k∈N , such that for all 1 j l there are disjoint intervals A j = (R j , R j + a j ), such that Hence for all l ∈ N there exists a subsequence {u n k } k∈N , such that for all k ∈ N we have This implies that ∞ 0 |u ′ n k | 2 dx → ∞, which is a contradiction since {u n } n∈N is bounded in H 1 0 (R + ). Hence the assertion (3.10) is true. Now we note that Since {u n } n∈N is bounded in L ∞ (R + ), in view of (3.6) we obtain for R > 0 given in (3.6) that For large n we haveR + a < y n − R, since y n → ∞ by our assumption. Now (3.11) implies Let us observe that u n,1 L p+1 → 0 by (3.11). Hence E(u n,1 ) = 1 2 H(u n,1 ) + o(1).
Now let us notice that supp(u n,2 ) ⊂ (R, ∞). Moreover, in view of (3.6), Thus, From the properties of the cut-off and (3.6), we get Since 1 2 H(u n,1 ) + ρ n > 0 by (3.9) and (3.13), we obtain which is a contradiction, hence (3.7) follows. Now, from (3.7) we obtain which is again a contradiction. Thus {y n } n∈N is bounded and has an accumulation point y * ∈ R + . Therefore, it follows that for any ε > 0 there is R > 0 such that for all k ∈ N. Hence u n → u strongly in L 2 (R + ). Moreover, since {u n } is bounded in H 1 0 (R + ) it is also strongly convergent in L p+1 (R + ). By the weak-lower semicontinuity of H (see [17]), it follows that E(u) lim n→∞ E(u n ) = I. Hence E(u) = I, and E(u n ) → E(u) implies that H(u n ) → H(u), which concludes that proof.
Remark 3.4. If c < 0, the infimum is not attained on the L 2 constraint. Indeed, let us assume that there exists v ∈ H 1 0 (R + ), such that v 2 L 2 = µ and E(v) = I. Then taking translates of v, i.e. v(· − y) for y > 0, we get E(v(· − y)) < I, which is a contradiction. and m Γ = inf{S(u) : u ∈ Γ}. If u ∈ G, then S(u) = m Γ . By Lemma 2.10 we know that u ∈ Γ, hence m A m Γ .
We have u λ ∈ Γ. Since u 1 is a solution of (3.14), we get from (3.15) and Lemma 2.2 that We can deduce directly from (3.15) and (3.16) that which implies that λ > 0. Let us define v by By (3.16), v ∈ A, hence S(v) m A . We obtain simple calculation that Hence, Since u is a solution of (3.15), we obtain from Lemma 2.2 that m A 0. By Lemma 2.2 and Lemma 2.10 we have that The right hand side is always strictly positive, except if λ = 1. Thus, λ = 1, which implies together with (3.16) that u ∈ A.
Set v n = u n (t n ). Applying Lemma 3.5, we obtain By the conservation of charge and energy, we obtain v n 2 L 2 → µ, and E(v n ) → I. Hence {v n } n∈N is a minimizing sequence of (3.1). It follows from Lemma 3.3, that there exists a solution u of the problem (3.1), such that v n − u H 1 → 0. By Lemma 3.5 we obtain that u ∈ G, which contradicts (3.17).

Instability
In this section we assume that p 5. Let us define for v ∈ H 1 0 (R + ) the functional In Lemma 2.2 we have shown that if v is a solution of (2.1), then Q(v) = 0. First, we prove the virial identities.
Proposition 4.1. Let u 0 ∈ H 1 0 (R + ) be such that xu 0 ∈ L 2 (R + ) and u be the corresponding maximal solution to (1.1). Then xu(t) ∈ L 2 (R + ) for any t ∈ (−T min , T max ). Moreover, the following identities hold for all v ∈ H 1 0 (R + ): Proof. The proof follows the same line as in [6].
Then the maximal solution u to (1.1) with initial condition u 0 blows up in finite time.
Since p 5, we get by the conservation of the energy that Q(u(t)) 2E(u 0 ) < 0 for all t ∈ (−T min , T max ).
Hence, Proposition 4.1 implies that Integrating twice, we get The main coefficient of the second order polynomial on the right hand side is negative. Thus, it is negative for |t| large, what contradicts with xu(t) 2 L 2 0 for all t. Therefore, −T min > −∞ and T max < +∞. Proof. Since p = 5, we have for all v ∈ H 1 0 (R + ), that 2E(v) = Q(v). Hence from Lemma 2.2 we get that Let us define ϕ n,0 = 1 + 1 n ϕ. It is easy to see that E(ϕ n,0 ) < 0. By Lemma 2.1 we know that xϕ n,0 ∈ L 2 (R + ). The conclusion follows from Proposition 4.2.
Proof. Let u 0 ∈ J and u ∈ C((−T min , T max ), H 1 0 (R + )) the corresponding maximal solution. Since S is conserved under the flow of (1.1) we have for all t ∈ (−T min , T max ) that S(u(t)) = S(u 0 ) < d.
We prove the assertion by contradiction. Suppose that there exists t ∈ (−T min , T max ) such that J(u(t)) 0.
Then, since J and u are continuous, there exists t 0 ∈ (−T min , T max ) such that J(u(t 0 )) = 0, thus u(t 0 ) ∈ N . Then by Proposition 2.9 we have that which is a contradiction, thus J(u(t)) < 0 for all t ∈ (−T min , T max ). Let us suppose now that for some t ∈ (−T min , T max ) we have Q(u(t)) 0. Again, by continuity, there exists t 1 ∈ (−T min , T max ) such that Q(u(t 1 )) = 0.

Appendix
We prove the following Lemma: (2.6). Then ψ A ∈ H 1 0 (R + ) and for large A > 0, we have the following approximations: Proof. We will use the fact that q(x) Me − √ ω|x| and q ′ (x) Me − √ ω|x| for some M > 0.
We get (5.1) by using the symmetry of q and q ′ : We estimate the second term by hence (5.1) follows. We get (5.2) the same way. We now show (5.3). From Hardy's inequality we get which is the estimate in (5.3).
To show (5.4), we use the fact that We get This concludes the proof.
We now state the proof of Lemma 2.8. The proof follows the arguments of the paper [11], with some important modifications. We introduce the norm which is equivalent to the standard norm on H 1 0 (R + ) if 0 < c < 1/4. Proof. Proof of Lemma 2. 8 Step 1. There exists u 0 ∈ H 1 0 (R + ), such that, up to a subsequence, u n is weakly convergent to u 0 in H 1 0 (R + ), and S ′ (u 0 ) = 0. Since {u n } n∈N is bounded in H 1 0 (R + ), it admits a weakly convergent subsequence in H 1 0 (R + ) with a weak limit u 0 ∈ H 1 0 (R + ). We only need to show that S ′ (u 0 ) = 0. Since by our assumption S ′ (u n ) → 0, it suffices to show that for all ϕ ∈ C ∞ 0 (R + ) we have Since u n ⇀ u 0 in H 1 0 (R + ) and strongly in L q loc (R + ) for all q 1, our statement follows.
Let us set v n = u n − u 0 .
Step 2. Assume that where B 1 (z) is the unit ball centered at z. Then u n → u 0 strongly in H 1 0 (R + ), and Lemma 2.8 holds with k = 0. Using the fact that S ′ (u 0 ) = 0, we get We recall that S ′ (u n ) → 0. Hölder's inequality implies that Assumption (5.5) and Lemma 1.1 in [16] implies that v n L p+1 → 0. Hence We obtain similarly that Re ∞ 0 |u 0 | p−1 u 0vn dx → 0, hence v n 2 → 0, which completes the proof of Step 2.
To show (i), let us assume by contradiction that {z n } n∈N has an accumulation point z * ∈ R + . Then for a subsequence of {v n } n∈N we have which is a contradiction, hence (i) holds. Since u n (· + z n ) is bounded in H 1 (R) the re exists q ∈ H 1 (R) such that u n (· + z n ) converges weakly to q in H 1 (R). We only need to show that q = 0. Since u 0 (· + z n ) ⇀ 0 in H 1 (R), we have that v n (· + z n ) ⇀ q in H 1 (R), and in L 2 loc (R) in particular. Hence |v n (y)| 2 dy d > 0.
This implies that q = 0.
It remains to show that S ∞′ (ũ n )ϕ → 0. For any fixed ϕ ∈ C ∞ 0 (R), ϕ(· − z n ) is in H 1 0 (R + ) for sufficiently big n ∈ N. Hence, we obtain Moreover, since u n is bounded in L ∞ , and ϕ is compactly supported, we get which concludes the proof of Step 3.
Suppose assumption 1) holds. We introduce ξ n = u n − u 0 − k i=1 q a i (· − x i n ), where q a i is a suitable cut-off of q i , such that supp(q a i ) ⊂ (0, ∞). This is possible owing to the exponential decay of q i at infinity, and x i n → ∞ as n → ∞ for all i. We get Since S ′ (u 0 )ξ n = 0, we get S ′ (u n )ξ n = ξ n 2 + Re ∞ 0 (|u 0 | p−1 u 0 − |u n | p−1 u n )ξ n dx + Re Using the fact that ξ n L p+1 → 0 by Lemma 1.1 in [16], we get that the second term of the right hand side converges to zero. Now, from the weak convergence of ξ n to zero and that S ′ (u n ) → 0,we obtain that ξ n → 0. Suppose now that assumption 2) holds. Then (i) and (ii) follows as in Step 3. To show (ii), let us setũ n = u n (· + z n ). We note that S ∞′ (ũ n )ϕ − S ∞′ (q)ϕ → 0, for all ϕ ∈ C ∞ 0 (R). Now S ∞′ (ũ n ) → 0 follows similarly as in Step 3, which concludes the proof.
Step 5. Conclusion By Step 1 we know that u n ⇀ u 0 and S ′ (u 0 ) = 0. Hence (i) of Lemma 2.8 is verified. If the assumption of Step 2 holds, then Lemma 2.8 is true with k = 0. Otherwise, the assumption of Step 3 holds. We have to iterate Step 4. We only need to show that assumption 1 of Step 4 occurs after a finite number of iterations. Let us notice that Moreover, since u n ⇀ u 0 and u n (· + x n i ) ⇀ q i , we get for the last term that Now since u n converges weakly to u 0 , we obtain for k 1 that Since q i is a nontrivial critical point of S ∞ , it is true that q i H 1 ǫ > 0. Hence, after a finite number of iterations assumption 1 of Step 4 must occur.
Finally, we have to verify that We first show that (5.7) S(u n ) → S(u 0 ) + S ∞ (v n ).
A straightforward calculation gives S(u n ) = S(u 0 ) + S ∞ (v n ) + Re From a lemma by Brezis and Lieb (see e.g. Lemme 4.6 [12]) we have Hence (5.7) follows. It only remains to show that We calculate .
We have shown that v n − k i=1 q i (· − x n i ) → 0 strongly in H 1 . Hence the first and third term above converges to zero as n → ∞. By using Sobolev's inequality and A − B which concludes the proof.