Regularity of harmonic maps with values into the sphere

Transcription

1 Université Paris-Sud 11 Regularity of harmonic maps with values into the sphere Supervisor: Student: Radu Ignat Daniel Drimbe Orsay 2013

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3 Contents Introduction 2 1 Regularity of critical points of a nonlocal energy The description of the energy The equation satisfied by any critical point of the energy The regularity of the critical points which are with values into S The continuity of the critical points in higher dimension Basic results and some notations The regularity of critical points in higher dimension Regularity of critical states in thin ferromagnetic films The equation satisfied by the critical points Some remarks and notations The regularity of the critical points Bibliography 24 1

4 Introduction In this dissertation we study the interior regularity of critical points valued to sphere for different energies. In the first chapter the energy is composed by the Dirichlet energy and a nonlocal energy, which comes from a pseudo-differential operator of order 0. The first step in proving C regularity is to prove the continuity. Using the uniform continuity on a precompact set, we obtain by classical technics the transition from H 1 regularity to Hloc 2. The C regularity is obtained once we gain Hloc k regularity for every k, which is obtained inductively by elliptic regularity. The proof becomes easier if we are dealing with the particular case S 1, using a lifting theorem. In the second chapter we are dealing just with maps valued to S 1, but the perturbation is a pseudo-differential operator of order 1. Again, we benefit from the lifting theorem, but we need to use more technics (e.g. isolating the boundary with cut-off functions). 2

5 Chapter 1 Regularity of critical points of a nonlocal energy 1.1 The description of the energy We start by defining the subspaces of divergence-free and curl-free vector fields as follows: V div = {F L 2 (R 2, R 2 ); divf = 0} V curl = {F L 2 (R 2, R 2 ); curlf = 0} The well-known Hodge descomposition theorem states that: V div V curl = L 2 (R 2, R 2 ) Actually this is true in every dimension, but we will use it just for dimension two. Let n 1 be an integer number. For u H 1 (B 2, S n ), we define ũ = (u 1, u 2 ) and ū = 1 B 2ũ. Define the energy E(u) = 1 u 2 1 H u (1.1) 2 B 2 2 B 2 where H = ( H, 0,..., 0) : R 2 R n+1 and H L 2 (R 2, R 2 ) is again the solution of the following problem: { curl H = 0 in D (R 2 ) div( H + ū) = 0 in D (R 2 (1.2) ). By Poincare Lemma, exists φ D (R 2 ) such that H = φ and then φ = div ū in D (B 2 ). 3

6 More general we may define a linear bounded operator H by the same formula (1.2), replacing ū by an arbitrary map v L 2 (R 2, R 2 ). Let s observe that H(v) is none other than the projection of v on the closed subspace V curl. In particular, we have H(v) L 2 (R 2 ) v L 2 (R 2 ), for every v L 2 (R 2, R 2 ) and H(v) w = H(w) v = H(v) H(w), for every v, w L 2 (R 2, R 2 ). Thus, H defines a linear bounded operator in L 2 (R 2, R n+1 ) by H(v) = ( H(v 1, v 2 ), 0,..., 0) = ( ( ) 1 (v 1, v 2 ), 0,..., 0), for every v = (v 1,..., v n+1 ) L 2 (R 2, R n+1 ). This operator H is satisfying similiar properties: H(v) L 2 (R 2,R n+1 ) v L 2 (R 2,R n+1 ) (1.3) for every v L 2 (R 2, R n+1 ) and H(v) w = H(w) v = H(v) H(w), (1.4) for every v, w L 2 (R 2, R n+1 ). Remark 1. Using Fourier Transform properties, the expression of H may be given by F( H(ū))(ξ) = ξ ξ ξ 2 F(ū)(ξ) which emphasizes that H is a pseudo differentiable operator of order 0. All these observations will be useful in the next sections. Let s notice now that we have at least one critical point for this energy. Proposition Exists a minimizer for our energy which is a constant. Proof. By (1.4), we have H 2 = R 2 H u. B 2 (1.5) We deduce now, that the energy E 0 and observe that the constant map u = (1, 0...0) satisfy E(u) = 0. This shows the existence of a minimizer of E, so we have at least a critical point. 4

7 1.2 The equation satisfied by any critical point of the energy Let us see which is the Euler-Lagrance equation satisfied by a critical point of this energy. Take ψ D(B 2, R n+1 ) for n N, n 1 and denote by u t = u+tψ u+tψ. We normalize u + tψ in order to have a map valued to S n. We remark that because ψ is bounded, u + tψ does not vanish, for small enough t. It is easy to see that we have u t = u + t(ψ < u, ψ > u) + O(t 2 ), by the Taylor expansion at t = 0. So any critical point must satisfy the equation d dt E(u t) t=0 = E(u)(ψ < u, ψ > u) = 0. (1.6) Let s devolep this. Denote by F (v) = 1 v 2 2 B 2 the Dirichlet energy and by G(v) = 1 2 B 2 H v the part which involves the non-local operator H. We need to compute F (v) and G(v). For the Dirichlet energy it is classical that F (v) = v. Observe that because u has modulus one, we get by derivation that u u = 0 where we have applied the operation between a matrix and a vector. follows that F (u)(ψ < u, ψ > u) = u (ψ < u, ψ > u) = u ψ u < u, ψ > u It =< u u 2 u, ψ > D,D, 5

8 because we have observed above that u is orthogonal on u. For the second term, we observe that G is a bilinear functional, so G(w)(v) = 1 H(w)v + 1 H(v)w = H(w)v 2 2 for every v, w H 1 (B 2, R n+1 ) using formula (1.4); this implies G(u)(ψ < u, ψ > u) = H ψ < u, H > u ψ. Finally, we obtain the following equation for any critical point: u u u 2 H+ < u, H > u = 0, D (B 2 ). (1.7) Once we have obtained this equation, we should expect that the critical point is smooth, because we make a perturbation with a pseudo differentiable operator of order 0 as we explained in Remark The regularity of the critical points which are with values into S 1 In this section we prove that any critical point valued to S 1 of the energy E is smooth in B 2. The proof is based on the existence of a lifting, granted to Bethuel and Zheng (see [1]); more precisely the theorem states that: Theorem For any u H 1 (B 2, S 1 ), it exists ϕ H 1 (B 2, R) such that u = e iϕ and ϕ is unique up to a constant which belongs to 2πZ. The energy can be written E(u) = Ê(ϕ) = 1 ϕ B 2 2 with u and ϕ like in the above theorem. Ĥ(ϕ) := H(u). B 2 Ĥ(ϕ) (cos ϕ, sin ϕ), We ve also made the notation Suppose u is a critical point for E and u = e iϕ. Then ϕ is a critical point for Ê. Indeed, take ψ D(B2 ) a scalar function and observe that Ê(ϕ + tψ) Ê(ϕ)ψ = lim Ê(ϕ) t 0 t = E(u)ψu = 0 because of relation (1.6) (u u = 0) and because of d dt t=0(e i(ϕ+tψ) ) = iψu = ψu 6 E(e i(ϕ+tψ) ) E(e iϕ ) = lim = t 0 t

9 Let s identify now the Euler-Lagrange equation satisfied by any critical point ϕ of the energy Ê. Take again a scalar function ψ D(B2 ) and let s compute first the gradient of the application ϕ 1 H (cos ϕ, sin ϕ) =: Ĝ(ϕ) = G(u). 2 B 2 For obtaining the equation (1.7), we have computed G(u), so Ĝ(ϕ)(ψ) = G(u)(ψu ) = H(u)ψu So the critical point must satisfy the equation ϕ = H ( sin ϕ, cos ϕ) = H u, D (B 2 ). (1.8) Theorem The map u = e iϕ chosen above as critical point is smooth. Proof. We have remarked in the Section 1.1 that φ = div u, D (B 2 ). This implies using the elliptic regularity that if u W k,p loc (B2, S 1 ), then φ W k+1,p loc (B 2 ) and H W k,p loc (B2, R 2 ) for any k 1, integer number and p > 1. Fix q > 2. We will use in the following that W k,q loc (B2 ) is an algebra, if k 1, since we are working in dimension two. Now, we observe that the right hand side of equation (1.8) is in L q loc (B2 ), so we get that ϕ W 2,q loc (B2 ). This implies H W 2,q loc (B2, R 2 ). Inductively we obtain that ϕ W k,q loc (B2 ), for every positive integer k: Suppose ϕ W k,q loc (B2 ) for a fixed integer k 2. This implies that u, H W k,q loc (B2, R 2 ) and therefore ϕ W k,q loc (B2, R 2 ). We conclude the induction using elliptic regularity. This finishes the proof, showing that u = e iϕ is smooth in B 2. Remark 2. In this dimension the proof was relatively easy, based just on the lifting theorem and the standard elliptic interior regularity. The reason is that in the equation solved by critical point in higher dimension (1.7) contains a term of the form u u 2, which with the initial regularity u H 1 (B 2, S n ), we have u u 2 L 1 (B 2 ). This is one of the critical cases when elliptic regularity can t be applied. On the other hand, the equation solved by the critical point in dimension 2 (1.8) does not contains a term of that form, allowing us to apply the elliptic regularity. 7

10 1.4 The continuity of the critical points in higher dimension In the paper Regularity for critical points of a non local energy, by Carbou (see [3]), it is proved the continuity for the critical points of this energy. It is based on elliptic regularity and also the following lemma granted to Wente (see [2]). Lemma Wente Let u, v H 1 (B 2 ) be two scalar functions. There exists a unique weak solution ϕ W 1,1 0 (B 2 ) of { ϕ = v u on B 2 ϕ = 0 on B 2 (1.9). This solution also satisfies ϕ C( B 2 ) H 1 0 (B2 ) and ϕ L (B 2 ) + ϕ L 2 (B 2 ) C u L 2 (B 2 ) v L 2 (B 2 ). Moreover, if u k, v k H 1 (B 2 ), for 1 k l, there exists a unique weak solution ϕ W 1,1 0 (B 2 ) of { ϕ = l k=1 vk u k on B 2 ϕ = 0 on B 2 (1.10). This solution also satisfies ϕ C( B 2 ) H 1 0 (B2 ) and ϕ L (B 2 ) + ϕ L 2 (B 2 ) C l u k L 2 (B 2 ) v k L 2 (B 2 ). Proof. The existence and the uniqueness follows from the classical elliptic regularity theorem. This can be seen easy, writting for every 0 < ɛ < 2. k=1 v u = (v u) W 1,2 ɛ (B 2 ), Step I The smooth case First, we will prove the lemma assuming that u, v D(R 2 ). Let denote by ψ = E ( v u) where E(x, y) = 1 2π log 1, is the fundamental solution of. Using x 2 +y2 polar coordinates (r, θ), we have v u = u x v y u y v x = 1 r (u rv θ u θ v r ). 8

11 Denote by ū(r) the mean of u w.r.t θ. Thus ψ(0) = 1 (log 1 2π (0,2π) R + r )(u rv θ u θ v r )drdθ = 1 (log 1 2π (0,2π) R + r )((uv θ) r (uv r ) θ )drdθ = 1 (log 1 2π (0,2π) R + r )(uv θ) r drdθ = 1 1 2π (0,2π) R + r (uv θ)drdθ = 1 1 2π (0,2π) R + r (u ū)v θdrdθ, using (log 1 r )uv θ = r(log r) u(v x sin θ v y cos θ) which converges to 0 for r 0, since u and v are smooth. Furthemore ψ(0) 1 2π 1 2π 1 2π 0 0 ( 0 1 u ū L 2 (0,2π) v θ L 2 (0,2π) r dr 1 u θ L 2 (0,2π) v θ L 2 (0,2π) u θ 2 1 L 2 (0,2π) r dr 1 2π u L 2 (R 2 ) v L 2 (R 2 ). r dr ) 1/2 ( 0 ) v θ 2 1 1/2 L 2 (0,2π) r dr If we try to estimate ψ(x), we have to replace u, v with their translatations u( x), v( x),. That s why we obtain, in same manner ψ L (R 2 ) 1 2π u L 2 (R 2 ) v L 2 (R 2 ). Because (ϕ ψ) = 0 on B 2, we can use the maximum principle to obtain that So we have that ϕ ψ L (B 2 ) ϕ ψ L ( B 2 ) = ψ L ( B 2 ). ϕ L (B 2 ) 2 ψ L (B 2 ) 1 π u L 2 (R 2 ) v L 2 (R 2 ). Using now the hypothesis (1.9), multiplying it with ϕ, we obtain that B 2 ϕ 2 ϕ L (B 2 ) u L 2 (B 2 ) v L 2 (B 2 ) 1 π u 2 L 2 (R 2 ) v 2 L 2 (R 2 ). 9

12 Exactly the same, we obtain the relations (1.10) for the smooth case. Step II The general case Let be u n D(R 2 ), v n D(R 2 ), which aproximate u, respectively v in the H 1 (B 2 ) norm and ϕ n the associate solution for the problem (1.9). Let now observe that ϕ n ϕ m solve a problem similar to that of (1.9). Indeed: { (ϕ n ϕ m ) = (v n v m ) u n + v m (u n u m ), on B 2 ϕ n ϕ m = 0, on B 2. Using Step I, we obtain that ϕ n ϕ m L (B 2 ) + ϕ n ϕ m L 2 (B 2 ) (1.11) C u n L 2 (B 2 ) v n v m L 2 (B 2 ) + C v m L 2 (B 2 ) u n u m L 2 (B 2 ), which implies that exists η C( B 2 ) H0 1(B2 ) such that ϕ n η and in H 1 norm. uniform Passing to limit, we obtain that { η = v u, in D (B 2 ) η = 0, on B 2 because convergence in L 1 implies convergence in D and all ϕ n are zero on B 2. Using the uniqueness, we obtain that ϕ = η, finishing the proof. With this result, we are able to prove the continuity of any critical point of this energy. Theorem Let u H 1 (B 2, S n ) be a critical point of E. Then u is continous on B 2. Proof. Because u has modulus one, we obtain that n+1 u j u j = 0 j=1 so writting on components, we get u i = j u j (u i u j u j u i ) + ψ i 10

13 where ψ = H < u, H > u L 2 (B 2, R n+1 ). By direct computations we obtain that (u i u j u j u i ) = u i ψ j + u j ψ i. Because the right-hand side is in L 2 (B 2 ), it exists a map b ij H 1 (B 2, R 2 ), such that b ij = u i ψ j + u j ψ i. Using the last two equations, we duduce that exists c ij H 1 (B 2 ) such that: u i u j u j u i b ij = c ij. So, we can write the Euler-Lagrange equation like u i = j u j c ij + j u j b ij + ψ i. The solution α i of { α i = j uj c ij α i = 0 on B 2. is continous on B 2, using Wente s lemma. The solution β i of the problem { β i = j uj b ij + ψ i β i = 0 on B 2 is in W 2, 3 2 (B 2 ), by elliptic regularity, which is embeded in C 0,γ (B 2 ). Now, because u α β is smooth, we have obtained the continuity of u. 1.5 Basic results and some notations The next theorem provides the full regularity for the critical point. Before this, we need to recall some results and to make some remarks. Let us introduce some notations and make some observations which will be useful further. Fix e 1, e 2 the canonical base of R 2, h R and take the functions d, f, g : R 2 R. We define h j g(x) = g(x + he j) g(x). h 11

14 By linearity we have h j g(x) = h j g(x) and by change of variable we have a formura similar to integratian by parts f h j g = h j fg. Let s also point out the following formula h j (dfg)(x) = (df)(x+he j ) h j g(x)+d(x+he j ) h j f(x)g(x)+ h j d(x)(fg)(x). (1.12) The following two propositions are classical. See for example [4], chapter five. Proposition Assume 1 p <, Ω R N ω Ω, f W 1,p (Ω) a scalar function. Then a bounded open set, h j f L p (ω) f L p (Ω), for all 0 < h < d(ω, Ω)/2 and 1 j N. Proposition Assume 1 < p <, Ω R N a bounded open set, ω Ω, f L p (Ω) a scalar function and there exists C > 0 such that h j f L p (ω) C, for all 0 < h < d(ω, Ω)/2 and 1 j N. Then f W 1,p (ω), with f x j L p (ω) C, for all 1 j N. 1.6 The regularity of critical points in higher dimension The technics for obtaing smoothness are used from [6]. The ideea is to obtain the H 2 regularity with the Proposition and then we use a simple induction based on the elliptic regularity in the ecuation (1.7). For emphasizing the operations used in the next lemma, let s denote by A : B, the scalar product between the matrix A, B. We denote by A o v the action of matrix A over the vector v. We denote by v w the matrix obtained by the tensorial product of the vectors v, w. And classicaly, we use for the scalar product between the vectors v, w the notation v w. 12

15 Lemma Let be B B 2 a ball. Let ɛ > 0. Then exist r > 0 such that for every ψ D(B(x 1, r)) with B(x 1, r) B and every 0 < h < d(b, B 2 )/2 =: d, we have u 2 h j u 2 ψ 2 ɛ h j u 2 ψ 2 + K 0 u 2, B(x 1,r) B(x 1,r) where K 0 depends just on ψ L, ψ L. B(x 1,r+d) Proof. First, we recall that in the first section we have defined H = ( H, 0,..., 0) : R 2 R n+1 and we have proved that H = φ, with φ H 2 (B), by laplacian regularity. So, let s denote by M := φ L (B), which is finite because Sobolev embeddings. Let s also denote by ũ = (u 1, u 2 ) the first two components of the map u. In the equation (1.7) we take ϕ = (u u(x 1 )) h j u 2 ψ 2, which belongs to H 1, because u H 1 L. We obtain u : [(u u(x 1 )) h j u 2 ψ 2 ] = (u u(x 1 )) u u 2 h j u 2 ψ φ (ũ ũ(x 1 )) h j u 2 ψ 2 φ ũ u (u u(x 1 )) h j u 2 ψ 2, with the above notation. Let s denote the left hand side with LHS and the right hand side with RHS. Before getting any further with computations and estimations, let s compute ( h j u 2 ψ 2 ). For any p {1, 2}, we have obtaining ( h j u 2 ψ 2 ) xp = 2 h j u xp h j uψ h j u 2 ψψ xp ( h j u 2 ψ 2 ) = 2 h j u 0 h j uψ h j u 2 ψ ψ. (1.13) We compute first LHS, second RHS and then we make estimations. LHS = u 2 h j u 2 ψ 2 + u : ((u u(x 1 )) ( h j u 2 ψ 2 )). RHS = (u u(x 1 )) u u 2 h j u 2 ψ 2 φ[ (ũ ũ(x 1 )) h j u 2 ψ 2 + (ũ ũ(x 1 )) ( h j u 2 ψ 2 )]+ + φ [u (u u(x 1 )) h j u 2 ψ 2 ũ], using integration by parts. Let s observe that the divegence appearing at the third integral splits in three parts: [u (u u(x 1 )) h j u 2 ψ 2 ũ] = u (u u(x 1 )) h j u 2 ψ 2 ũ+ + (u (u u(x 1 ))) ũ h j u 2 ψ 2 + u (u u(x 1 )) ( h j u 2 ψ 2 ) ũ. 13

16 We continue the computation now: RHS = (u u(x 1 )) u u 2 h j u 2 ψ 2 φ[ ũ h j u 2 ψ 2 + (ũ ũ(x 1 )) (2 h j u o h j uψ h j u 2 ψ ψ)]+ + φ[u (u u(x 1 )) h j u 2 ψ 2 ũ + u o (u u(x 1 )) ũ h j u 2 ψ u o u ũ h j u 2 ψ 2 + u (u u(x 1 ))(2 h j u o h j uψ h j u 2 ψ ψ) ũ]. To simplify notations, let s denote by U = U(x 1, r) = sup x B(x1,r) u u(x 1 ). Using ũ u, ũ ũ(x 1 ) u u(x 1 ), we obtain RHS U u 2 h j u 2 ψ M u h j u 2 ψ 2 + U(2 h j Du h j u ψ h j u 2 ψ ψ )+ + MU h j u 2 ψ 2 u + u h j u 2 ψ M u h j u 2 ψ 2 + MU 2 h j u h j u ψ h j u 2 ψ ψ. Using Young s inequality, we obtain RHS U u 2 h j u 2 ψ Mɛ 1 u 2 h j u 2 ψ 2 + M h j u 2 ψ 2 + 4ɛ 1 + MU h j u 2 ψ 2 + h j u 2 ψ h j u 2 ψ ψ + + MU h j u 2 ψ 2 u 2 + h j u 2 ψ M ɛ 1 u 2 h j u 2 ψ h j u 2 ψ ɛ 1 + MU ( h j u 2 ψ 2 + h j u 2 ψ h j u 2 ψ ψ ). From the expresion of LHS, we obtain u 2 h j u 2 ψ 2 LHS + U u (2 h j u h j u ψ h j u 2 ψ ψ ) LHS + U( u 2 h j u 2 ψ 2 + h j u 2 ψ u 2 h j u 2 ψ 2 + h j u 2 ψ 2 ). 14

17 Now we majorate LHS with the estimation of RHS proved above. Let s use the uniform continuity of u on the compact B B 2. We can take r sufficient small such that U(x 1, r) < ɛ 1 for every x 1 B with the property B(x 1, r) B. To conclude it remains to pick a sufficient small ɛ 1. Theorem Let u H 1 (B 2, S n ) be a critical point of E. Then u is smooth on B 2. Proof. We will prove the theorem in 2 main steps: Step I : Let us prove that u H 2 loc (B2, S n ). Let ω B 2. Take B, a ball, such that ω B B 2 and denote by d = d(b, B 2 )/2. Let ɛ > 0. We aim to prove u H 2 (ω, S n ). Take r small enough as is Lemma 1.6.1, such that r < 1 2 d(ω, B). Let be x 1 ω such that B(x 1, r) B. Using formula (1.12) we obtain the estimate h j (u u 2 ) (x) u(x + he j ) h j u + u h j u + u 2 h j u (1.14) Let s observe that we may extend by liniarity (from functions to maps) the formula (1.12) to h j (< u, H > u)(x) =< u, H > (x + he j ) h j u(x)+ + < u(x + he j ), h j H(x) > u(x)+ < h j u(x), H(x) > u(x). (1.15) Let us take in the equation (1.7) ϕ = h j (ψ 2 h j u), where ψ D(B(x 1, r)) satisfying the following properties: 0 ψ 1, ψ = 1 on B(x 1, r/2) (1.16) ψ 4/r So we get that h j u 2 ψ 2 + h j u h j u (ψ 2 ) = h j u (ψ 2 h j u) = = u ( h j (ψ 2 h j u)) = = h j Gψ 2 h j u, G h j (ψ 2 h j u) = where G = u u 2 + H < u, H > u. 15

18 In the previous relation we keep in mind the equality between the first term and the last term and using the relations (1.14) and (1.15) we obtain h j u 2 ψ 2 2 h j u h j u ψ ψ + u(x + he j ) h j u h j u ψ 2 + u(x) h j u h j u ψ 2 + u(x) 2 h j u 2 ψ 2 + h j H h j uψ 2 + u(x + he j ) h j H u h j uψ (u H)(x + he j ) h j u h j uψ 2 + h j u H u h j uψ 2 ɛ 1 h j u 2 ψ 2 + 1/ɛ 1 h j u 2 ψ 2 + ɛ 1 h j u 2 ψ u(x + he j ) 2 h j u 2 ψ 2 4ɛ 1 + ɛ 1 h j u 2 ψ 2 + ( 1 + 1) u(x) 2 h j u 2 ψ 2 4ɛ h j H h j u h j H h j u φ(x + he j ) ( h j u h j uψ 2 (x) u(x + he j )) + φ (u h j uψ 2 h j u) (1.17) using Young s inequality and integration by parts. Let s denote by I 1 and I 2 the last two terms in the last expression (1.17). I 1 = φ(x + he j )[ h j u o h j uψ 2 u(x + he j )+ + ( h j uψ 2 ) o h j u u(x + he j ) + h j u h j uψ 2 u(x + he j )] M h j u h j u ψ 2 + h j u ψ 2 h j u + h j u 2ψ ψ h j u + + h j u h j u ψ 2 (x) u (x + he j ) M 2(ɛ 1 h j u 2 ψ h j u 2 ψ 2 ) + 2 h j u 2 ψ L + 4ɛ h j u 2 (x) u 2 (x + he j )ψ h j u 2 ψ 2, where we denote by M = φ L. (1.18) 16

19 I 2 = φ[ u o h j uψ 2 h j u + ( h j uψ 2 ) o u h j u+ + u h j uψ 2 h j u] M u h j u ψ 2 h j u + h j u ψ 2 h j u + h j u 2ψ ψ h j u + + h j u ψ 2 h j u 1 M 2 u 2 h j u 2 ψ h j u 2 ψ 2 + 2(ɛ 1 h j u 2 ψ h j u 2 ψ 2 ) 4ɛ h j u 2 ψ L. (1.19) We use the estimations (1.17), (1.18), (1.19) and take ɛ 1 sufficient small. We obtain that exists constants K 1 and K 2 which does not depend on h such that h j u 2 ψ 2 K 1 u 2 h j u 2 ψ 2 + u 2 (x + he j ) h j u 2 ψ 2 + (1.20) + K 2 H 2 + u 2, B(x 1,r+d) using also Proposition We apply Lemma for the function ψ selected in (1.16) and obtain that u 2 h j u 2 ψ 2 ɛ h j u 2 ψ 2 + K 3 u 2, (1.21) B(x 1,r+d) with K 3 a constant which does not depend on h. We apply again Lemma for the translatation of ψ selected in (1.16) with he j and replacing h with h we obtain that u 2 h j u 2 (x)ψ 2 (x he j ) ɛ + K 3 h j u 2 (x)ψ 2 (x he j ) B(x 1,r+d) u 2. (1.22) (this is possible since supp ψ( he j ) B(x 1 + he j, r) B for h < d(ω, B)/2.) By change of variable the relation (1.22) is transformed to the one we need: u 2 (x + he j ) h j u 2 (x)ψ 2 (x) ɛ h j u 2 (x)ψ 2 (x) (1.23) + K 3 u 2. B(x 1,r+d) 17

20 Now we are done. Using the relations (1.20), (1.21) and (1.23) and taking ɛ sufficient small, we obtain that exists K 4 which does not depend on h such that h j u 2 ψ 2 K 4 H 2 + u 2. (1.24) B(x 1,r+d) Using that ψ = 1 on B(x 1, r/2), we get by Proposition that Du Hloc 1 (B(x 1, r/2), R 2 (n+1) ). By a covering argument, we get that Du H 1 (ω, R 2 (n+1) ). Step 1 is proved. Step II : : Let us prove now that u C (B 2, R n+1 ). We have that { H = Φ on R 2 Φ = div ũ on B 2. with Φ H 1 (R 2 ). From here, we may conclude that if u W k,p loc (B2, R n+1 ), we obtain that Φ W k,p loc (B2, R 2 ). Fix q > 2. Because u Hloc 2 (B2, R n+1 ), we obtain that u L p loc (B2, R 2 (n+1) ) for every p < and u, H W 1,q loc (B2, R n+1 ). So u u 2 L q loc (B2, R n+1 ) (choosing a large enough p) which implies u L q loc (B2, R n+1 ). This gives u W 2,q loc (B2, R n+1 ). By induction we obtain that u W k,q loc (B2, R n+1 ) for every k N. Indeed, suppose u W k,q loc (B2, R n+1 ) for k 2. The expression u u 2 W k 1,q loc (B 2, R n+1 ). The other term in the right hand side of the ecuation (1.7) satifsfied by u belongs to W k,q loc (B2, R n+1 ), so u W k 1,q loc (B 2, R n+1 ). We apply again laplacian regularity. This finishes the induction and the theorem is proved. 18

21 Chapter 2 Regularity of critical states in thin ferromagnetic films 2.1 The equation satisfied by the critical points Let introduce the energy E(u) = u 2 + u1 B 2 2 B 2 H 1/2 (R 2 ), for u H 1 (B 2, S 1 ). We aim to prove the regularity of any critical point of this energy. In this case we will use a different method in gaining regularity. The proof for regularity uses ideas found in a paper by Ignat and Knüpfer (see [5]). Before getting any further, let s see why does the term u1 B 2 2 H 1/2 is finite. Proposition It exists a constant k such that for every f L 2 (R 2 ) with supp f B 2, we have that f H 1/2 (R 2 ) k f L 2 (R 2 ). Proof. We decompose the H 1/2 (R 2 ) seminorm into f 2 H 1/2 (R 2 ) F(f)(x) 2 F(f)(x) 2 + x 1 x x <1 x Ff 2 L 2 (R 2 ) + 1 Ff 2 L (R 2 ) x k 0 ( f 2 L 2 (R 2 ) + f 2 L 1 (B 2 ) ) k f 2 L 2 (R 2 ), x 1 19

22 since supp f B 2 and f L 1 (B 2 ) f L 2 (B 2 ) B In our situation we have that u L 2 (B 2 ). We just take f = u 1 B 2 L 2 (R 2 ) and obtain u 1 B 2 H 1/2 (R 2 ) k u1 B 2 L 2 (R 2 ) k u L 2 (B 2 ). Using again Theorem 1.3.1, we have that exists ϕ : B 2 R, ϕ H 1 (B 2 ) such that m = e iϕ. Further, we may express the energy as E(ϕ) = ϕ B 2 R 2 x F([ ϕ x 1 sin ϕ + ϕ x2 cos ϕ]1 B 2) 2 Let s compute the equation satisfied by any critical point. Let ψ Cc (B 2 ) be a scalar function. We get that E(ϕ + ɛψ) = ϕ + ɛ ψ B 2 R 2 x F([ (ϕ x 1 + ɛψ x1 )(sin ϕ + ɛψ cos ϕ) + (ϕ x2 + ɛψ x2 )(cos ϕ ɛψ sin ϕ)]1 B 2) 2 + O(ɛ 2 ) If ϕ is a critical point for this energy, it has to satisfy 1 ϕ ψ + B 2 R 2 x F([ ϕ x 1 sin ϕ + ϕ x2 cos ϕ]1 B 2) F([ ψ x1 sin ϕ + ψ x2 cos ϕ ϕ x1 ψ cos ϕ ϕ x2 ψ sin ϕ]1 B 2) = 0 which implies that ϕ satisfies the equation ϕ ( b sin ϕ, b cos ϕ) b(ϕ x1 cos ϕ + ϕ x2 sin ϕ) = 0, D (B 2 ), where b = 1 f and f = ( ϕ x1 sin ϕ + ϕ x2 cos ϕ)1 B 2 = u 1 B 2, where m = e iϕ. After reducing some terms we obtain that ϕ satisfies the equation ϕ b (e iϕ ) = 0, D (B 2 ) (2.1) Remark 3. In Fourier we obtain that F( b)(ξ) = iξ ξ F( u 1 B 2)(ξ), which shows that we are dealing this time with a perturbation of a pseudooperator of order 1. 20

23 2.2 Some remarks and notations Let be R (0, 1), δ = 1 R 4 and make the notations f 0 = fψ, f 1 = f(1 ψ), b 0 = f 0, b 1 = f 1, b = f. where Because of the relation ψ C0 (B(R + 2δ)), ψ = 1 on B(R + δ), 0 ψ 1. 1 = ( ) 1, we have that b = b. (2.2) This relation will be useful, using the elliptic regularity. Before starting the regularity theorem, let s make two remarks which will be usefull in the regularity theorem. Remark 4. Using f 1 L 2 (R 2 ) and suppf 1 B 2 \B(R + δ), we have that b 1 H k (B(R)), k N. Proof. To prove this, let s take η C c (B(R)) and evaluate (b 1, where α + β k. k η (b 1, ) x α L 2 1 xβ (R 2 ) x F(f 1 )F( k η k η ) =< f 1, 2 R 2 x α 1 xβ 2 x α 1 xβ 2 = R 2 = 1 π = R 2 suppf 1 (f 1 (x) f 1 (y))( suppf 1 suppη suppη c f 1 L 2 (R 2 ) η L 2 (R 2 ), k η x α 1 xβ 2 2π x y 3 k η (x) f 1 (y) (x) x α 1 xβ 2 x y 3 dxdy k η x α 1 xβ 2 f 1 (y)η(x)r( x y, x j )dxdy k η x α 1 xβ 2 ) L 2 (R 2 ), > H 1/2 (R 2 ) (y)) dxdy where R is a rational function in two variables and we have used that suppf 1 and supp η are disjoint bounded sets. Remark 5. For any p > 2 and k N, we have that W k,p (B 2 ) H k (B 2 ) H k (B 2 ). 21

24 Proof. It easily seen that if the remark is proved for k = 1, by induction is true for every k, using the relation k (fg) = k 1 ( fg) + k 1 (f g) for f W k,p (B 2 ) and g H k (B 2 ) ( denotes here one of the partial derivatives). Now, for proving the statement for k = 1, observe that the distribution derivative of the product fg is f xj g + fg xj L p L q + L L 2, for every q > 1 real number. Because p > 2 is a strict inequality, we may find a big enough number q such that This finishes the proof. L p L q L The regularity of the critical points Let s prove now the regularity theorem: Theorem ϕ satisfying the critical point equation (2.1) is smooth on the unit ball. Proof. Step I : We prove here that ϕ, b H 2 loc (B2 ). Having ϕ H 1 (B 2 ) we obtain that f L 2 (B 2 ), and then b 0 H 1 (R 2 ), because if η Cc (R 2 ), we have that (b 0, η) H 1,H 1 0 = R 2 x ˆf 0ˆη ˆf 0 L 2 η L 2. From Remark (4), we have that b 1 H 1 (B(R)), so we get that b H 1 (B(R)). This implies that b Hloc 1 (B(R)) by the elliptic regularity; and further that b H 1 loc (B 2 ). Using b H 1 loc (B2 ) and u L (B 2 ), we obtain that b L 2 loc (B2, R 2 ) and u L, which implies using equation (2.1) that ϕ L 2 loc (B2 ) and further by laplacian regularity in (2.1) ϕ H 2 loc (B2 ). 22

25 By Remark 5 we obtain f 0 Hloc 1 (B2 ). Because the support of ψ is contained in B 2, we may conclude that f 0 H 1 (R 2 ). From b 0 = f 0, we get that b 0 L 2 (R 2 ); using Remark 4 and the elliptic regularity in (2.2), we obtain b Hloc 2 (B2 ). Step II: We prove here the inductive step. Take k 2, a natural number. Suppose by induction that and prove that ϕ, b Hloc k (B(R + 2δ)) ϕ, b H k+1 loc (B(R)). We have that b H k 1 (B(R+2δ), R 2 ) and sin ϕ, cos ϕ W k 1,q loc 2δ)). Remark 5 shows that ϕ H k 1 loc ϕ H k+1 loc (B(R + 2δ)). (B(R+ (B(R + 2δ)). By elliptic regularity Now, by Remark 5, f Hloc k (B(R+2δ)). By the same remark, we obtain that f 0 Hloc k (B(R + 2δ). Since ψ has the support in B(R + 2δ), we may say that f 0 H k (R 2 ). This implies b 0 H k 1 (R 2 ), which gives us using Remark 4 that b H k 1 loc (B(R)). The laplacian regularity in (2.2) gives now that b H k+1 loc (B(R)). This finishes the induction proof, obtaining at the end that ϕ C (B 2 ). 23

26 Bibliography [1] Fabrice Bethuel, Xiaomin Zheng: Density of smooth functions between two manifolds in Sobolev Spaces, Journal of Functional Analysis 80, 60-75, [2] Haïm Brezis, Jean-Michel Coron: Multiple Solutions of H-Systems and Rellich s Conjecture, Comm. Pure Appli. Math., , [3] Gilles Carbou: Regularity for critical points of a non local energy, Calculus of Variations 5, , [4] Lawrence C.Evans: Partial Differential Equations, second edition, American Mathematical Society, [5] Radu Ignat, Hans Knüpfer: Vortex energy and 360-Nel walls in thin films micromagnetics, Comm. Pure Appl. Math. 63, , [6] Jürgen Jost: Riemannian Geometry and Geometric Analysis, fifth edition, Universitext, Springer-Verlag, Berlin,