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The fractional Hopf differential and a weak formulation of stationarity for the half Dirichlet energy

  • Filippo Gaia ORCID logo EMAIL logo
Published/Copyright: November 20, 2024

Abstract

We obtain a weak formulation of the stationarity condition for the half Dirichlet energy, which can be expressed in terms of a fractional quantity, related to the trace of the Hopf differential of the harmonic extension of the original map. As an application we show that conformal harmonic maps from the disc are precisely the harmonic extensions of stationary points of the half Dirichlet energy on the circle. We also derive a Noether theorem and a Pohozaev identity for stationary points of the half Dirichlet energy.

MSC 2020: 58E20; 35R11; 35J20

Communicated by Yannick Sire


A Commutator estimates and fractional divergences

In this appendix we will derive some estimate for commutators, which will allow us to extend the definition of the operator D s ( a , b ) in a distributional sense to a wide family of maps a , b . We will then see how the operator D s ( a , b ) is related to the fractional divergence operator introduced in [14].

We will make use of the following function space.

Definition A.1.

Let 𝔸 ( 𝕊 1 , k ) be the space of all the elements u in 𝒟 ( 𝕊 1 , k ) such that

u 𝔸 := k | u ^ ( k ) | < .

𝔸 ( 𝕊 1 ) is called Wiener algebra.

Remark 3.

Recall that for any s > 1 2 there is a continuous embedding H s ( 𝕊 1 , k ) 𝔸 ( 𝕊 1 , k ) . Indeed, for any u H s ( 𝕊 1 , k ) ,

(A.1) k k 0 | u ^ ( k ) | ( k k 0 | u ^ ( k ) | 2 | k | 2 s ) 1 2 ( k k 0 | k | - 2 s ) 1 2

by the Cauchy–Schwarz inequality. As s > 1 2 , the second factor is finite. Moreover, | u ^ ( 0 ) | | u H s . In particular, all the results of this section have a slightly weaker formulation in terms of Sobolev spaces only.

Lemma A.1.

Let s ( 0 , 1 2 ) . For any a , b , φ C ( S 1 , R k ) there holds

| D s ( a , b ) φ | C φ 𝔸 a H 2 s - 1 b L 2 .

Proof.

We define the action of the commutator [ ( - Δ ) s , a ] on φ as follows:

[ ( - Δ ) s , a ] φ = ( - Δ ) s ( a φ ) - a ( - Δ ) s φ .

To prove the lemma, it will be enough to show the following estimate:

[ ( - Δ ) s , a ] φ L 2 C φ 𝔸 a H 2 s - 1 .

For any n we have

( [ ( - Δ ) s , a ] φ ) ( n ) = | n | 2 s a φ ^ ( n ) - k | n - k | 2 s a ^ ( n - k ) φ ^ ( k )
= | n | 2 s k a ^ ( n - k ) φ ^ ( k ) - k | n - k | 2 s a ^ ( n - k ) φ ^ ( k )
= k ( | n | 2 s - | n - k | 2 s ) φ ^ ( k ) a ^ ( n - k ) .

By Plancherel’s identity,

[ ( - Δ ) s , a ] φ L 2 2 = 4 π 2 n | k ( | n | 2 s - | n - k | 2 s ) φ ^ ( k ) a ^ ( n - k ) | 2 .

Observe that if | k | 2 | n | > 0 , there holds | k | 2 | k + n | ; thus

| | k + n | 2 s - | k | 2 s | s | k | | n + k | | k | | n + k | r 2 s - 1 𝑑 r s | n | ( | k | 2 ) 2 s - 1 3 1 - 2 s s | n | | k - n | 2 s - 1 .

Therefore, Young’s inequality yields

(A.2)

k ( 0 < | n | | k | 2 | φ ^ ( n ) a ^ ( k - n ) | | | k + n | 2 s - | k | 2 s | ) 2 3 2 ( 1 - 2 s ) s 2 k ( 0 < | n | | k | 2 | φ ^ ( n ) n | | a ^ ( k - n ) | | k - n | 2 s - 1 ) 2
3 2 ( 1 - 2 s ) s 2 ( n | φ ^ ( n ) n | ) 2 ( k | a ^ ( k ) | 2 | k | 4 s - 2 )
= 3 2 ( 1 - 2 s ) s 2 φ 𝔸 2 [ a ] H ˙ 2 s - 1 2 .

On the other hand, we always have

| | n + k | 2 s - | k | 2 s | | n | 2 s ,

and if | k | < 2 | n | , then | k - n | 3 | n | . Therefore,

(A.3)

k ( | k | 2 < | n | | φ ^ ( n ) a ^ ( k - n ) | | | k + n | 2 s - | k | 2 s | ) 2 k ( | k | 2 < | n | | φ ^ ( n ) | | n | 2 s | k - n | 1 - 2 s | a ^ ( k - n ) | | k - n | 2 s - 1 ) 2
3 2 ( 1 - 2 s ) k ( n | φ ^ ( n ) | | n | | a ^ ( k - n ) | | | k - n | 2 s - 1 ) 2
3 2 ( 1 - 2 s ) ( n | φ ^ ( n ) n | ) 2 ( k | a ^ ( k ) | 2 | k | 4 s - 2 )
= 3 2 ( 1 - 2 s ) φ 𝔸 2 [ a ] H ˙ 2 s - 1 2 ,

where the second-last steps follow again from Young’s inequality. Therefore, combining (A.2) and (A.3), we obtain

(A.4) [ ( - Δ ) s , a ] φ L 2 2 4 π 2 3 2 ( 1 - 2 s ) ( 1 + s 2 ) φ 𝔸 2 [ a ] H ˙ 2 s - 1 2

This concludes the proof of the lemma. ∎

Lemma A.2.

Let a , b , φ C ( S 1 , R k ) . Then

(A.5) | D 1 2 ( a , b ) φ | C ( - Δ ) 3 4 φ 𝔸 a H - 1 2 b H 1 2 .

Proof.

Proceeding as in the proof of Lemma A.1, we see that it is enough to show the following estimate:

(A.6) [ ( - Δ ) 1 2 , a ] φ H - 1 2 C ( - Δ ) 3 4 φ 𝔸 a H - 1 2 .

In term of Fourier coefficients, we would like to obtain a bound for

(A.7) [ ( - Δ ) 1 2 , a ] φ H - 1 2 2 = n | ( [ ( - Δ ) 1 2 , a ] φ ) ( n ) | 2 ( 1 + | n | 2 ) - 1 2
= n | k ( | n | - | n - k | ) ( 1 + | n | 2 ) - 1 4 φ ^ ( k ) a ^ ( n - k ) | 2 .

Observe that if | n | 2 | k | , then | n - k | 3 2 | n | and therefore

| n | - | n - k | ( 1 + | n | 2 ) 1 4 ( 3 2 ) 1 2 | k | ( 1 + | n - k | 2 ) 1 4 .

Thus, by Young’s inequality,

(A.8)

n | | k | | n | 2 φ ^ ( k ) a ^ ( n - k ) ( | n | - | n - k | ) ( 1 + | n | 2 ) - 1 4 | 2 3 2 n ( k | φ ^ ( k ) | | k | | a ^ ( n - k ) | | ( 1 + | n - k | 2 ) - 1 4 | ) 2
3 2 ( k | φ ^ ( k ) k | ) 2 k | a ^ ( k ) | 2 ( 1 + | k | 2 ) - 1 2
= 3 2 φ 𝔸 2 a H - 1 2 2 .

On the other hand, if | n | < 2 | k | , then | n - k | 3 | k | and therefore

| n | - | n - k | ( 1 + | n | 2 ) 1 4 | k | ( 1 + | n - k | 2 ) 1 4 ( 1 + | n - k | 2 ) 1 4 2 | k | 3 2 ( 1 + | n - k | 2 ) 1 4 .

Thus, by Young’s inequality,

(A.9)

n | | n | < 2 | k | φ ^ ( k ) a ^ ( n - k ) ( | n | - | n - k | ) ( 1 + | n | 2 ) - 1 4 | 2 4 n ( | n | < 2 | k | | φ ^ ( k ) | | k | 3 2 | a ^ ( n - k ) | | ( 1 + | n - k | 2 ) 1 4 ) 2
4 ( k | φ ^ ( k ) | | k | 3 2 ) 2 k | a ^ ( k ) | 2 ( 1 + | k | 2 ) - 1 2
= 4 ( - Δ ) 3 4 φ 𝔸 2 a H - 1 2 2 = 4 ( - Δ ) 3 4 φ 𝔸 2 w H - 1 2 .

Combining (A.8) and (A.9), we obtain estimate (A.6). ∎

Next we discuss the link of the operator D s with the fractional divergence operator introduced in [14]. For s ( 0 , 1 2 ) denote K s the kernel of the s-fractional Laplacian, so that for any φ C ( 𝕊 1 ) ,

( - Δ ) s φ ( x ) = 𝕊 1 ( φ ( x ) - φ ( y ) ) K s ( x - y ) 𝑑 y .

For a description of K s , see [18]. Following [14], for any F : 𝕊 1 × 𝕊 1 we define the s-fractional divergence of F to be the distribution

div s F [ φ ] = 𝕊 1 𝕊 1 F ( x , y ) ( φ ( x ) - φ ( y ) ) K s 2 ( x - y ) 𝑑 x 𝑑 y

whenever the integral is well defined.

Lemma A.3.

Let s ( 0 , 1 2 ) . Let a , b , φ C ( S 1 ) . Then

𝕊 1 D s ( a , b ) φ = 2 div 2 s ( a ( x ) b ( y ) ) [ φ ] .

Proof.

We compute

𝕊 1 D s ( a , b ) φ = 𝕊 1 𝕊 1 ( ( a ( x ) - a ( y ) ) b ( x ) - a ( x ) ( b ( x ) - b ( y ) ) ) φ ( x ) K s ( x - y ) 𝑑 x 𝑑 y
= 1 2 𝕊 1 𝕊 1 ( ( a ( x ) - a ( y ) ) b ( x ) - a ( x ) ( b ( x ) - b ( y ) ) ) φ ( x ) K ( x - y ) d x d y
+ 1 2 𝕊 1 𝕊 1 ( ( a ( y ) - a ( x ) ) b ( y ) - a ( y ) ( b ( y ) - b ( x ) ) ) φ ( y ) K s ( x - y ) 𝑑 x 𝑑 y
= 𝕊 1 𝕊 1 ( a ( x ) b ( y ) - a ( y ) b ( x ) ) ( φ ( x ) - φ ( y ) ) K s ( x - y ) 𝑑 x 𝑑 y
= 2 𝕊 1 𝕊 1 a ( x ) b ( y ) ( φ ( x ) - φ ( y ) ) K s ( x - y ) 𝑑 x 𝑑 y
= 2 div 2 s ( a ( x ) b ( y ) ) [ φ ] ,

where in the second step we interchanged the variables x and y. ∎

Acknowledgements

The present work is based on some chapters of the author’s Master thesis [9]. The author would like to thank Professor Tristan Rivière, Professor Xavier Ros-Oton and Alessandro Pigati for their guidance and their support.

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Received: 2024-03-15
Accepted: 2024-09-24
Published Online: 2024-11-20
Published in Print: 2025-04-01

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