Renormalized energies for unit-valued harmonic maps in multiply connected domains

Proposition 3.1.

The infimum ( 1.5 ) is attained and every minimizer u ρ ∈ H 1 ( Ω ρ , S 1 ) satisfies the following Euler–Lagrange equation: (3.1) − Δ u ρ = | ∇ u ρ | 2 u ρ in Ω ρ , u ρ = g on ∂ G , u ρ ∧ ∂ ν u ρ = 0 on ∂ B ρ ( a i ) , i = 1 , … , k . Furthermore, u ρ ∈ C ∞ ( Ω ρ , S 1 ) ∩ C 1 ( Ω ‾ ρ , S 1 ) and u ρ is also smooth up to the boundary of every B ρ ( a i ) for i = 1 , … , k .

When one uses the direct method of calculus of variations to prove Proposition 3.1, the difficulty is that the degree is not continuous with respect to the H 1 2 -weak convergence. However, since we work with S 1 -valued maps, it is possible to show that, in this particular case, we can recover weak continuity of the degree. This follows for example from a result of White [40], but we will give a direct proof relying on Lemma 3.1 below. We first introduce functions V i for i = 1 , … , k defined by (3.2) − Δ V i = 0 in Ω ρ , V i = 0 on ∂ Ω ρ ∖ ∂ B ρ ( a i ) , V i = 1 on ∂ B ρ ( a i ) .

Lemma 3.1.

Let u ∈ H 1 ( Ω ρ , S 1 ) , then (3.3) deg ( u , ∂ B ρ ( a i ) ) = 1 2 π ∫ Ω ρ u ∧ ( ∂ y V i ∂ x u − ∂ x V i ∂ y u ) . Furthermore if u , v ∈ H 1 ( Ω ρ , S 1 ) then (3.4) | deg ( u , ∂ B ρ ( a i ) ) − deg ( v , ∂ B ρ ( a i ) ) | ⩽ 2 π ‖ V i ‖ C 1 ‖ u − v ‖ L 2 ( ‖ ∇ u ‖ L 2 + ‖ ∇ v ‖ L 2 ) .

The proof of this lemma can be found in [6, Section 3] and [16, Proposition 1]. It is also a particular case of [12, Theorem 1]. We give the details for the comfort of the reader.

Proof.

For i = 1 , … , k , an integration by parts gives 1 2 π ∫ Ω ρ u ∧ ( ∂ y V i ∂ x u − ∂ x V i ∂ y u ) = − 1 2 π ∫ Ω ρ V i [ ∂ y ( u ∧ ∂ x u ) − ∂ x ( u ∧ ∂ y u ) ] − 1 2 π ∫ Γ 0 V i ( u ∧ ∂ τ u ) + 1 2 π ∫ ∂ Ω ρ ∖ Γ 0 V i ( u ∧ ∂ τ u ) = deg ( u , ∂ B ρ ( a i ) ) . We have used that, since u ∈ H 1 ( Ω ρ , S 1 ) , we have ∂ x u ∧ ∂ y u = 0 almost everywhere.

For the second point we observe that since V i is locally constant on ∂ Ω ρ , an integration by parts gives ∫ Ω ρ v ∧ ( ∂ y V i ∂ x u − ∂ x V i ∂ y u ) = ∫ Ω ρ u ∧ ( ∂ y V i ∂ x v − ∂ x V i ∂ y v ) − ∫ Γ 0 ∂ τ V ( v ∧ u ) + ∫ ∂ Ω ρ ∖ Γ 0 ∂ τ V ( v ∧ u ) = ∫ Ω ρ u ∧ ( ∂ y V i ∂ x u − ∂ x V i ∂ y u ) . Hence, by using the first point we find 2 π | deg ( u , ∂ B ρ ( a i ) ) − deg ( v , ∂ B ρ ( a i ) ) | = | ∫ Ω ρ ( u − v ) ∧ [ ( ∂ y V i ∂ x u − ∂ x V i ∂ y u ) + ( ∂ y V i ∂ x v − ∂ x V i ∂ y v ) ] | ⩽ 4 ‖ u − v ‖ L 2 ‖ V i ‖ C 1 ( ‖ ∇ u ‖ L 2 + ‖ ∇ v ‖ L 2 ) . □

Proof.

(proof of Proposition 3.1) We take a minimizing sequence ( u n ) n for the Dirichlet energy E in the class E g , ρ . Since it is bounded in H 1 , we can extract a subsequence weakly converging to some u ρ ∈ H 1 ( Ω ρ , R 2 ) . Up to other subsequences, we can assume that u n converges strongly to u ρ in L 2 ( Ω ρ ) and u n converges almost everywhere to u ρ . Hence u ∈ H 1 ( Ω ρ , S 1 ) and by using Lemma 3.1 we find that deg ( u ρ , ∂ B ρ ( a i ) ) = deg ( u n , ∂ B ρ ( a i ) ) for all n ∈ N ∗ and i = 1 , … , k . With the weak continuity of the trace operator and the lower semi-continuity of the Dirichlet energy we are able to conclude to the existence. To derive the Euler–Lagrange equations we can make variations of the form u ρ + t φ | u ρ + t φ | for t small and φ ∈ C c ∞ ( Ω ρ , R 2 ) and u ρ e i t ψ for t small and ψ ∈ C ∞ ( Ω ‾ ρ , R ) with ψ vanishing on ∂ G . These variations do preserve the class E g , ρ . The regularity of u ρ follows from the regularity for minimizing harmonic maps due to [30] (see also [22]). The regularity up to the boundaries ∂ B ρ ( a i ) can be proved as in [4, Lemma 4.4]. □

In the rest of the paper we will make an intensive use of the current of a function.

Definition 3.1.

If U ⊂ R 2 is a bounded open set, for u ∈ L p ∩ W 1 , p ′ ( U , C ) , with 1 ⩽ p , p ′ ⩽ + ∞ and 1 / p + 1 / p ′ = 1 , we define the current associated to u by (3.5) j ( u ) : = u ∧ ∇ u = ( u ∧ ∂ x u , u ∧ ∂ y u ) .

Lemma 3.2.

If u ρ is a solution of the minimization problem ( 1.5 ) given by Proposition  3.1 then its current j ( u ρ ) satisfies (3.6) div j ( u ρ ) = 0 and curl j ( u ρ ) = 0 in Ω ρ , j ( u ρ ) · ν = 0 on ∂ B ρ ( a i ) , i = 1 , … , k , and j ( u ρ ) · τ = g ∧ ∂ τ g on ∂ G .

Proof.

We compute div j ( u ρ ) = ∂ x ( u ρ ∧ ∂ x u ρ ) + ∂ y ( u ρ ∧ ∂ y u ρ ) = u ρ ∧ Δ u ρ = 0 . In the same way ∂ x ( u ρ ∧ ∂ y u ρ ) − ∂ y ( u ρ ∧ ∂ x u ρ ) = 2 ∂ x u ρ ∧ ∂ y u ρ = 0 . The information for j ( u ρ ) on the boundary comes from the information on u ρ on the boundary. □

We now use the generalized Poincaré Lemma 2.1 to derive a Hodge decomposition of the current j ( u ρ ) . First we prove

Proposition 3.2.

There exists a unique function Φ ρ ∈ H 1 ( Ω ρ , R ) , satisfying (3.7) Δ Φ ρ = 0 in Ω ρ , Φ ρ = cst. on ∂ B ρ ( a i ) , i = 1 , … , k , ∫ ∂ B ρ ( a i ) ∂ ν Φ ρ = 2 π d i for i = 1 , … , k , ∂ ν Φ ρ = g ∧ ∂ τ g on ∂ G , ∫ ∂ G Φ ρ = 0 . We also have Φ ρ ∈ C ∞ ( Ω ρ , R ) ∩ C 1 ( Ω ‾ ρ , R ) and is smooth up to the boundaries ∂ B ρ ( a i ) , i = 1 , … , k .

Moreover, there exist a unique v ρ ∈ H 1 ( Ω ρ , S 1 ) and a unique θ l , ρ = θ l , ρ ( g , { ( a i , d i ) } ) ∈ [ − π , π [ , l = 1 , … , n , such that j ( v ρ ) = ∇ ⊥ Φ ρ in Ω ρ , v ρ = g on Γ 0 ⊂ ∂ G and v ρ = e − i θ l , ρ g on Γ l , l = 1 , … , n .

Proof.

The existence follows from the fact that a solution to (3.7) is a minimizer of 1 2 ∫ Ω ρ | ∇ ϕ | 2 + 2 π ∑ i = 1 k d i ϕ | ∂ B ρ ( a i ) − ∫ ∂ G ϕ ( g ∧ ∂ τ g ) in the space V ρ : = { ϕ ∈ H 1 ( Ω ρ , R ) ; φ = cst. = ϕ | ∂ B ρ ( a i ) on each ∂ B ρ ( a i ) , i = 1 , … , k , ∫ ∂ G φ = 0 } . The uniqueness follows because the functional to be minimized is strictly convex in V ρ . The smoothness of Φ ρ follows from the regularity for harmonic functions. Since the vector field ( − ∂ y Φ ρ , ∂ x Φ ρ ) satisfies the assumption of Lemma 2.3, we can find a function v ρ ∈ H 1 ( Ω ρ , S 1 ) satisfying j ( v ρ ) = ∇ ⊥ Φ ρ in Ω ρ , furthermore v ρ is unique up to a phase. Since v ρ satisfies v ρ ∧ ∂ τ v ρ = ∂ ν Φ ρ = g ∧ ∂ τ g on ∂ G we can choose an appropriate phase to prescribe v ρ = g on Γ 0 , and then we have v ρ = e − i θ l , ρ g on each Γ l , l = 1 , … , n for some θ l , ρ = θ l , ρ ( g , { ( a i , d i ) } ) ∈ [ − π , π [ . Note that we can write v ρ ( x ) = e i ∫ γ x ∇ ⊥ Φ ρ · τ g ( x 0 ) where x 0 ∈ Γ 0 , γ x is any smooth path joining x 0 to x, and τ denotes the tangent vector to this path (oriented from x 0 to x). Then, for any point x l ∈ Γ l , v ρ ( x l ) = e i ∫ γ x l ∇ ⊥ Φ ρ · τ g ( x 0 ) = e − i θ l , ρ g ( x l ) . □

Proposition 3.3.

Let u ρ be a solution of ( 1.5 ), let Φ ρ be a solution of ( 3.7 ) and v ρ be the map given by Proposition  3.2 , then there exists H ρ ∈ C ∞ ( Ω ‾ ρ , R ) such that (3.8) j ( u ρ ) = ∇ ⊥ Φ ρ + ∇ H ρ , u ρ = v ρ e i H ρ , (3.9) Δ H ρ = 0 in Ω ρ , ∂ ν H ρ = 0 on ∂ B ρ ( a i ) , i = 1 , … , k , H ρ = α l , ρ ( g , { ( a i , d i ) } ) on Γ l , l = 1 … , n , H ρ = 0 on Γ 0 , where α l , ρ ( g , { ( a i , d i ) } ) = θ l , ρ + 2 π Z , l = 1 , … , n , with θ l , ρ defined in Proposition  3.2 .

Proof.

This follows from Lemma 2.1 since j ( u ρ ) − ∇ ⊥ Φ ρ verifies that curl ( j ( u ρ ) − ∇ ⊥ Φ ρ ) = 0 , ( j ( u ρ ) · τ − ∂ ν Φ ρ ) = g ∧ ∂ τ g − g ∧ ∂ τ g = 0 on ∂ G and ∫ ∂ B ρ ( a i ) ( j ( u ρ ) · τ − ∂ ν Φ ρ ) = 2 π ( d i − d i ) = 0 for i = 1 , … , k . Note that H ρ is defined up to a constant and that is why we can impose H ρ = 0 on Γ 0 . Now we have that j ( v ρ e i H ρ ) = v ρ e i H ρ ∧ ∇ ( v ρ e i H ρ ) = v ρ e i H ρ ∧ ( ∇ v ρ e i H ρ + i v ρ ∇ H ρ e i H ρ ) = v ρ ∧ ∇ v ρ + ∇ H ρ = ∇ ⊥ Φ ρ + ∇ H ρ . From Lemma 2.3, this means that u ρ = v ρ e i H ρ up to a phase, but since H ρ = 0 on Γ 0 and v ρ = g on Γ 0 we have u ρ = v ρ e i H ρ . Hence, since from Proposition 3.2 v ρ = e − i θ l , ρ g on Γ l , l = 1 , … , n we find that α l , ρ = θ l , ρ + 2 π Z . □

We recall that Φ 0 ∈ C ∞ ( G ∖ { a 1 , … , a k } ) is the solution of (1.8).

Lemma 3.3.

There exists a unique v 0 ∈ C ∞ ( G ∖ { a 1 , … , a k } , S 1 ) such that ( 1.13 ) and ( 1.14 ) hold.

Proof.

The proof of Lemma 2.3 can be adapted to this context to find this v 0 . Note that ∂ ν Φ 0 = g ∧ ∂ τ g = v 0 ∧ ∂ τ v 0 on ∂ G implies that on every connected component Γ 0 , Γ 1 , … , Γ n of ∂ G we can write v 0 = e − i θ l g , furthermore we can choose θ 0 = 0 . □

Proposition 3.4.

Let Φ ρ be the solution to ( 3.7 ) and Φ 0 the solution to ( 1.8 ), then for every m ∈ N and every compact set K ⊂ G ‾ ∖ { a 1 , … , a k } there exists C m , K such that (3.10) ‖ Φ ρ − Φ 0 ‖ C m ( K ) ⩽ C m , K ρ .

Proof.

We apply Lemma A.1 to v ρ = Φ 0 − Φ ρ which satisfies, Δ v ρ = 0 in Ω ρ , ∂ ν v ρ = 0 on ∂ G , ∫ ∂ B ρ ( a i ) ∂ ν v ρ = 0 for i = 1 , … , k . Hence, since Φ ρ is constant on ∂ B ρ ( a i ) we find sup Ω ρ v ρ − inf Ω ρ v ρ ⩽ ∑ i = 1 k ( sup ∂ B ρ ( a i ) v ρ − inf ∂ B ρ ( a i ) v ρ ) ⩽ ∑ i = 1 k ( sup ∂ B ρ ( a i ) Φ 0 − inf ∂ B ρ ( a i ) Φ 0 ) ⩽ C ρ . Since ∫ ∂ G ( Φ ρ − Φ 0 ) = 0 , there exists a point x ∈ ∂ G such that ( Φ ρ − Φ 0 ) ( x ) = 0 , thus we find that ‖ Φ ρ − Φ 0 ‖ L ∞ ( Ω ρ ) ⩽ C ρ . By elliptic estimates, see e.g. [19, Theorem 2.10], we obtain (3.10). □

We introduce (3.11) E ˜ g , ρ : = { v ∈ H 1 ( Ω ρ , S 1 ) ; tr | ∂ G v = g ; tr | ∂ B ρ ( a i ) v = ( x − a i | x − a i | ) d i , i = 1 , … , k } (3.12) W ˜ g , d 1 , … , d k ρ ( a 1 , … , a k ) : = inf v ∈ E ˜ g , ρ 1 2 ∫ Ω ρ | ∇ v | 2 .

Lemma 3.4.

The map ρ ↦ W g , d 1 , … , d k ρ ( a 1 , … , a k ) − π ( ∑ i = 1 k d i 2 ) | log ρ | is non-increasing and the map ρ ↦ W ˜ g , d 1 , … , d k ρ ( a 1 , … , a k ) − π ( ∑ i = 1 k d i 2 ) | log ρ | is non-decreasing. Furthermore W g , d 1 , … , d k ρ ( a 1 , … , a k ) ⩽ W ˜ g , d 1 , … , d k ρ ( a 1 , … , a k ) and W g , d 1 , … , d k ( a 1 , … , a k ) : = lim ρ → 0 ( W g , d 1 , … , d k ρ ( a 1 , … , a k ) − π ( ∑ i = 1 k d i 2 ) | log ρ | ) exists and is finite.

This lemma follows from Proposition 2.10 and Lemma 2.11 in [28]. We reproduce the proof for the comfort of the reader.

Proof.

Let 0 < ρ < σ with σ small enough so that the balls B ‾ σ ( a i ) are disjoints and included in G. We can write that ∫ G ∖ ⋃ i = 1 k B ‾ ρ ( a i ) | ∇ u ρ | 2 = ∫ G ∖ ⋃ i = 1 k B ‾ σ ( a i ) | ∇ u ρ | 2 + ∫ ⋃ i = 1 k ( B σ ( a i ) ∖ B ‾ ρ ( a i ) ) | ∇ u ρ | 2 . Now, by using polar coordinates centred at a i and the Cauchy-Schwarz inequality, we have ∫ B σ ( a i ) ∖ B ‾ ρ ( a i ) | ∇ u ρ | 2 ⩾ ∫ ρ σ ∫ 0 2 π | ∂ θ i u ρ | 2 r i d θ i d r i ⩾ 1 2 π ∫ ρ σ 1 r i | ∫ 0 2 π ( u ρ ∧ ∂ θ i u ρ ) d θ i | 2 d r i ⩾ 2 π d i 2 log σ ρ . Thus we find that (3.13) W g ρ − π ( ∑ i = 1 k d i 2 ) log 1 ρ ⩾ W g σ − π ( ∑ i = 1 k d i 2 ) log 1 σ which proves the first assertion (here and in the rest of the proof, for simplicity, we do not write the dependence of the singularities { ( a i , d i ) } , i = 1 , … , k ). For the second assertion, if 0 < ρ < σ and if u ˜ σ is a minimizer for the problem W ˜ g σ , then the map v ( x ) = u ˜ σ ( x ) if x ∈ Ω σ ( x − a i | x − a i | ) d i if x ∈ B σ ( a i ) ∖ B ‾ ρ ( a i ) is a comparison map from the minimization problem W ˜ g ρ . Thus W ˜ g ρ ⩽ 1 2 ∫ Ω ρ | ∇ v | 2 = W ˜ g σ + π ( ∑ i = 1 k d i 2 ) log σ ρ . This proves that ρ ↦ W ˜ g ρ is non-decreasing. We can easily see that W g ρ ⩽ W ˜ g ρ for every ρ. Hence both quantities admit a limit when ρ goes to zero and their limits are finite. □

Proposition 3.5.

Let u ρ be a solution to the minimization problem ( 1.5 ). Then there exist u 0 ∈ H loc 1 ( G ‾ ∖ { a 1 , … , a k } , S 1 ) , H 0 ∈ H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) and a sequence ρ p → 0 such that u ρ p ⇀ u 0 in H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) , H ρ p ⇀ H 0 in H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) .

Proof.

The proof follows the idea of [28, Proposition 8.1]. By Lemma 3.4, for 0 < ρ < σ we have ∫ G ∖ ⋃ i = 1 k B ‾ σ ( a i ) | ∇ u ρ | 2 2 ⩽ W g , d 1 , … , d k ρ ( a 1 , … , a k ) − π ( ∑ i = 1 k d i 2 ) log σ ρ . By using Lemma 3.4 again we arrive at (3.14) ∫ G ∖ ⋃ i = 1 k B ‾ σ ( a i ) | ∇ u ρ | 2 2 ⩽ W g , d 1 , … , d k ( a 1 , … , a k ) − π ( ∑ i = 1 k d i 2 ) log σ . Thanks to the boundedness condition (3.14) we can use a diagonal argument to find a subsequence ρ p → 0 and a map u 0 ∈ H loc 1 ( G ‾ ∖ { a 1 , … , a k } , S 1 ) such that u ρ p ⇀ u 0 in H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) . Now since we know from Proposition 3.4 that Φ ρ → Φ 0 in C loc m ( G ‾ ∖ { a 1 , … , a k } ) we find that ∇ H ρ p = j ( u ρ p ) − ∇ ⊥ Φ ρ p converges weakly in L loc 2 ( G ‾ ∖ { a 1 , … , a k } ) . From the Poincaré inequality, which is valid here since H ρ = 0 on Γ 0 , we infer that there exists H 0 ∈ H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) such that H ρ p ⇀ H 0 in H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) . □

In particular, from the previous proposition and the weak continuity of the trace operator, there exist a subsequence ρ p → 0 and α l = α l ( g , { ( a i , d i ) } ) such that (3.15) α l , ρ → α l , for l = 1 , … , n .

Lemma 3.5.

Let Φ ρ be the solution to ( 3.7 ). Then, we have lim ρ → 0 ( 1 2 ∫ Ω ρ | ∇ Φ ρ | 2 − π ( ∑ i = 1 k d i 2 ) | log ρ | ) = − π ∑ i ≠ j d i d j log | a i − a j | + 1 2 ∫ ∂ G Φ 0 ( g ∧ ∂ τ g ) − π ∑ i = 1 k d i R 0 ( a i ) < + ∞ .

Proof.

An integration by parts gives ∫ Ω ρ | ∇ Φ ρ | 2 = ∫ Γ 0 ∂ ν Φ ρ Φ ρ − ∑ l = 1 n ∫ Γ l ∂ ν Φ ρ Φ ρ − ∑ i = 1 k ∫ ∂ B ρ ( a i ) ∂ ν Φ ρ Φ ρ . Now we use (3.7) and more particularly we use ∂ ν Φ ρ = g ∧ ∂ τ g on Γ l , l = 0 , 1 , … , n , Φ ρ = c s t . on ∂ B ρ ( a i ) , i = 1 , … , k and ∫ ∂ B ρ ( a i ) ∂ ν Φ ρ = 2 π d i to obtain ∫ Ω ρ | ∇ Φ ρ | 2 = ∫ Γ 0 ( g ∧ ∂ τ g ) Φ ρ − ∑ l = 1 n ∫ Γ l ( g ∧ ∂ τ g ) Φ ρ − ∑ i = 1 k 2 π d i Φ ρ ( ∂ B ρ ( a i ) ) = ∫ Γ 0 ( g ∧ ∂ τ g ) Φ 0 − ∑ l = 1 n ∫ Γ l ( g ∧ ∂ τ g ) Φ 0 − ∑ i = 1 k 2 π d i Φ 0 ( x i ) + O ( ρ ) where x i is a point in ∂ B ρ ( a i ) . Since R 0 ( x ) = Φ 0 ( x ) − ∑ i = 1 k d i log | x − a i | we can write ∫ Ω ρ | ∇ Φ ρ | 2 = ∫ Γ 0 ( g ∧ ∂ τ g ) Φ 0 − ∑ l = 1 n ∫ Γ l ( g ∧ ∂ τ g ) Φ 0 + ∑ i = 1 k 2 π d i 2 | log ρ | − ∑ i ≠ j 2 π d i d j log | a i − a j | − ∑ i = 1 k 2 π d i R 0 ( a i ) + O ( ρ ) . This yields the result. □

Proposition 3.6.

Let H ρ be the solution to ( 3.9 ), then up to a subsequence ρ p → 0 , we can find α l = α l ( g , { ( a i , d i ) } ) for l = 1 , … , n and H 0 ∈ C ∞ ( G ‾ ∖ { a 1 , … , a k } ) such that (3.16) H ρ p → H 0 in C loc m ( G ‾ ∖ { a 1 , … , a k } ) for every m ∈ N with H 0 satisfying (3.17) Δ H 0 = 0 in G , H 0 = 0 on Γ 0 , H 0 = α l on Γ l , l = 1 , … , n .

Proof.

We already know from Proposition 3.5 that there exist α l , l = 1 , … , n and H 0 ∈ H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) such that, up to a subsequence not labelled, H ρ ⇀ H 0 in H loc 1 ( G ‾ ∖ { a 1 , … , a k } ) . It remains to show that H 0 satisfies (3.17). First by elliptic estimates, cf. e.g. [18, Theorem 5.21], we have find that (3.16) holds and Δ H 0 = 0 in G ∖ { a 1 , … , a k } , H 0 = 0 on Γ 0 and H 0 = α l on Γ l , l = 1 , … , n . But we can use (3.8), | ∇ u ρ | 2 = | j ( u ρ ) | 2 and an integration by parts to write ∫ Ω ρ | ∇ u ρ | 2 = ∫ Ω ρ | ∇ Φ ρ | 2 + ∫ Ω ρ | ∇ H ρ | 2 + 2 ∫ Ω ρ ∇ ⊥ Φ ρ · ∇ H ρ = ∫ Ω ρ | ∇ Φ ρ | 2 + ∫ Ω ρ | ∇ H ρ | 2 + 2 ∑ l = 1 n ∫ Γ l α l , ρ ∂ τ Φ ρ . By Proposition 3.4 and since α l , ρ → α l as ρ → 0 we find that ∫ Γ l α l , ρ ∂ τ Φ ρ → ∫ Γ l α l ∂ τ Φ 0 as ρ → 0 . From Lemma 3.4 and Lemma 3.5 we obtain that sup ρ > 0 ∫ Ω ρ | ∇ H ρ | 2 < + ∞ . By lower semi-continuity of the Dirichlet energy, for every σ > 0 we have ∫ Ω σ | ∇ H 0 | 2 ⩽ lim inf ρ → 0 ∫ Ω σ | ∇ H ρ | 2 ⩽ lim sup ρ → 0 ∫ Ω σ | ∇ H ρ | 2 . But if ρ < σ then ∫ Ω σ | ∇ H ρ | 2 ⩽ ∫ Ω ρ | ∇ H ρ | 2 and hence we arrive at sup σ > 0 ∫ Ω σ | ∇ H 0 | 2 < + ∞ . By monotone convergence, it implies that ∇ H 0 ∈ L 2 ( G ) , and by the Poincaré inequality we find that H 0 ∈ L 2 ( G ) . Then it can be show that the singularities a 1 , … , a k are removable1

¹

To prove this we can take a cut-off function η such that η ≡ 1 in B ε ( a i ) and η ≡ 0 in B 2 ε ( a i ) c , i = 1 , … , k , then we write that ∫ G ∇ H 0 ∇ ψ = ∫ G ∇ H 0 ∇ [ ψ ( 1 − η ) + ψ η ] . By using that ‖ ∇ η ‖ L ∞ ⩽ C / ε , and that ‖ ∇ H 0 ‖ L 2 ( B ε ( a i ) ) → 0 as ε → 0 , we arrive at the result.

for H 0 and thus Δ H 0 = 0 in G. □

Proposition 3.7.

Let u ρ be the solution to the minimization problem ( 1.5 ), then there exists a sequence ρ p → 0 such that (3.18) u ρ p → u 0 in C loc m ( G ∖ { a 1 , … , a k } ) for every m ∈ N with u 0 ∈ C ∞ ( G ∖ { a 1 , … , a k } , S 1 ) satisfying (3.19) − Δ u 0 = | ∇ u 0 | 2 u 0 in G ∖ { a 1 , … , a k } , u 0 = g on ∂ G . Furthermore we have that (3.20) j ( u 0 ) = ∇ ⊥ Φ 0 + ∇ H 0 and u 0 = v 0 e i H 0 . In particular u 0 satisfies div j ( u 0 ) = 0 , curl j ( u 0 ) = 2 π ∑ i = 1 k d i δ a i in G and j ( u 0 ) · τ = g ∧ ∂ τ g on ∂ G .

Proof.

This result follows from the convergence of Φ ρ in Proposition 3.4 and H ρ in Proposition 3.6. □

We are now in position to obtain Theorem 1.1

Proof of Theorem 1.1.

From Lemma 3.4 we know that the limit of W g , d 1 , … , d k ρ ( a 1 , … , a k ) − π ( ∑ i = 1 k d i 2 ) | log ρ | as ρ → 0 exists and is finite. To compute this limit we can use a special subsequence ρ p → 0 such that Proposition 3.7 and Proposition 3.6 hold. For simplicity of notation, in the rest of the proof we let ρ = ρ p . Let u ρ be the solution to the minimization problem (1.5). We use that | ∇ u ρ | 2 = | j ( u ρ ) | 2 along with (3.8) and an integration by parts to obtain ∫ Ω ρ | ∇ u ρ | 2 = ∫ Ω ρ | ∇ Φ ρ | 2 + ∫ Ω ρ | ∇ H ρ | 2 + 2 ∫ Ω ρ ∇ ⊥ Φ ρ · ∇ H ρ = ∫ Ω ρ | ∇ Φ ρ | 2 − ∑ l = 1 n ∫ Γ l α l , ρ ∂ ν H ρ + 2 ∑ l = 1 n ∫ Γ l α l , ρ ∂ τ Φ ρ . We have used the boundary condition for H ρ in (3.9). Now we use Lemma 3.5 and Proposition 3.6 to obtain ∫ Ω ρ | ∇ u ρ | 2 = ∫ Γ 0 Φ 0 ( g ∧ ∂ τ g ) − ∑ l = 1 n ∫ Γ l Φ 0 ( g ∧ ∂ τ g ) + 2 π ∑ i = 1 k d i 2 | log ρ | − 2 π ∑ i ≠ j d i d j log | a i − a j | − 2 π ∑ i = 1 k d i R 0 ( a i ) − ∑ l = 1 n ∫ Γ l α l ∂ ν H 0 + 2 ∑ l = 1 n ∫ Γ l α l ∂ τ Φ 0 + o ρ ( 1 ) . We can integrate by parts once more and find that (3.21) ∑ l = 1 n ∫ Γ l α l ∂ ν H 0 = ∫ G | ∇ H 0 | 2 . We decompose H 0 = ∑ l = 1 n α l φ l ( x ) where the functions φ l are defined by (1.10) to find (1.16). We also observe that (3.22) ∫ G | ∇ H 0 | 2 = ∑ l = 1 n ∫ Γ l α l ∂ ν H 0 = ∑ l = 1 n ∑ m = 1 n α l α m ∫ Γ l ∂ ν φ m . Now we describe the coefficients α l = α l ( g , { ( a i , d i ) } ) . We recall from Lemma 3.3 that v 0 = e − i θ l g on Γ l , l = 1 , … , n . Then, we observe that j ( u 0 v ‾ 0 ) = ∇ H 0 and thus we can conclude that u 0 = v 0 e i H 0 e i η for some constant η ∈ R . Since u 0 = v 0 on Γ 0 , and H 0 = 0 on Γ 0 we obtain that η ∈ 2 π Z . Besides, on each Γ l , l = 1 , … , n we obtain that g = g e − i θ l + i α l + i η which implies (1.17) and α l = θ l + 2 π Z . The minimality of the coefficients α l comes from the minimality of H ρ the solution to (3.9) for every ρ > 0 and the convergence H ρ | Γ l = α l , ρ → α l for l = 1 , … , n . Next we take the inner product of (3.20) with ∇ φ l to find ∫ G j ( u 0 ) · ∇ φ l = ∫ G ∇ ⊥ Φ 0 · ∇ φ l + ∫ G ∇ H 0 · ∇ φ l = ∫ ∂ G φ l ∂ τ Φ 0 + ∫ ∂ G ∂ ν φ l H 0 = ∑ m = 1 n α m ∫ Γ m ∂ ν φ l for l = 1 , … , n Hence we obtain (1.18). It remains to show that u 0 is given by (1.15). First, by using the same arguments as in Proposition 3.1 we can see that there exists a minimizer U g of the Dirichlet energy in I g where this class is defined in (1.11). To prove the uniqueness up to a multiplication by a constant. We write the Euler–Lagrange equations for U g and we use Lemma 2.1 to prove that j ( U g ) = ∇ ⊥ Φ U for some function Φ U . We use again the Euler–Lagrange equations on U g to obtain that Φ U satisfies (3.23) Δ Φ U = 0 in G , Φ U = cst. on Γ l , l = 0 , 1 , … , n , ∫ Γ l ∂ ν Φ U = deg ( g , Γ l ) l = 1 , … , n , ∫ Γ 0 ∂ ν Φ U = deg ( g , Γ 0 ) − ∑ i = 1 k d i . As in Proposition 3.3 we obtain that Φ U is uniquely determined up to a constant, since it is a minimizer of F ( φ ) = 1 2 ∫ G | ∇ φ | 2 + 2 π ∑ l = 1 n φ | Γ l deg ( g , Γ l ) + 2 π φ | Γ 0 ( deg ( g , Γ 0 ) − ∑ i = 1 k d i ) in the space { φ ∈ H 1 ( G , R ) ; φ = c s t . on Γ l , l = 0 , 1 , … , n } . This minimizer is unique up to a constant by a convexity argument. We then use Lemma 2.3. Now we call V : = u 0 U ‾ g ∏ i = 1 k ( x − a i ‾ | x − a i | ) d i and we compute that (3.24) j ( V ) = j ( u 0 ) − ∑ i = 1 k d i ( ( x − a i ) ⊥ | x − a i | 2 ) − ∇ ⊥ Φ U . Thus we can check that curl j ( V ) = 0 in G, ∫ Γ l V · τ = 0 for l = 0 , 1 , … , n . By applying Lemma 2.1 we find ψ g ∈ C 1 ( G ) such that j ( V ) = ∇ ψ g . Therefore, by using Lemma 2.3 we can write V = e i ψ g which yields (1.15). We can check that div j ( V ) = 0 in G and hence we find that Δ ψ g = 0 in G. □

We can also prove that every non-trivial lattice can appear in the expression of the renormalized energy with Dirichlet boundary conditions in multiply connected domains. Proposition 3.8.

Let a 1 , … , a k ∈ G and d 1 , … , d k ∈ Z satisfying ( 1.2 ). For all θ 1 , … , θ n in [ − π , π [ there exists a boundary data g ∈ C 1 ( ∂ G , S 1 ) such that W g , d 1 , … , d k ( a 1 , … , a k ) = − π ∑ i ≠ j d i d j log | a i − a j | + 1 2 ∫ ∂ G Φ 0 ( g ∧ ∂ τ g ) − π ∑ i = 1 k d i R 0 ( a i ) + ∑ l = 1 n ∫ Γ l α l ∂ τ Φ 0 + 1 2 ∫ G | ∑ l = 1 n α l ∇ φ l | 2 , with α l = θ l + 2 π Z .

Proof.

Let g ˜ ∈ C 1 ( ∂ G , S 1 ) be any boundary data. We consider Φ 0 the unique solution to Δ Φ 0 = 2 π ∑ i = 1 k d i δ a i in G ∂ ν Φ 0 = g ˜ ∧ ∂ τ g ˜ on ∂ G ∫ ∂ G Φ 0 = 0 . To this Φ 0 we can associate a harmonic map v 0 ∈ H loc 1 ( G ∖ { a 1 , … , a k } ) (unique modulo to a phase) such that j ( v 0 ) = ∇ ⊥ Φ 0 . We set g 0 = v 0 | ∂ G and we define g : = g 0 on Γ 0 , g : = e i θ l g 0 on Γ l , l = 1 , … , n . Then we can check, from (1.17) and the proof giving the expression of the renormalized energy that W g , d 1 , … , d k ( a 1 , … , a k ) satisfies the desired expression. □

4. Renormalized energies with Neumann boundary conditions