Symmetric Solutions to Minimization of a P-energy Functional with Ellipsoid Value

The author proves the W 1,p convergence of the symmetric minimizers u ε = (u ε1 , u ε2 , u ε3) of a p-energy functional as ε → 0, and the zeros of u 2 ε1 + u 2 ε2 are located roughly. In addition, the estimates of the convergent rate of u 2 ε3 (to 0) are presented. At last, based on researching the Euler-Lagrange equation of symmetric solutions and establishing its C 1,α estimate, the author obtains the C 1,α convergence of some symmetric minimizer.

x 2 3 b 2 = 1} be a surface of an ellipsoid.Assume g(x) = (e idθ , 0) where x = (cos θ, sin θ) on ∂B, d ∈ N .We concern with the minimizer of the energy functional in the function class W = {u(x) = (sin f (r)e idθ , b cos f (r)) ∈ W 1,p (B, E(b)); u| ∂B = g}, which is named the symmetric minimizer of E ε (u, B).When p = 2, the functional E ε (u, B) was introduced in the study of some simplified model of high-energy physics, which controls the statics of planar ferromagnets and antiferromagnets (see [5] [8]).The asymptotic behavior of minimizers of E ε (u, B) has been considered in [3].In particular, they discussed the asymptotic behavior of the symmetric minimizer with E(1)-value of E ε (u, B) in §5.When the term connected tightly with the study of harmonic map with E(1)-value.Due to this we may also research the asymptotic behavior of minimizers of E ε (u, B) by referring to the p-harmonic map with ellipsoid value (which was discussed in [2]).
In this paper, we always assume p > 2. As in [1] and [3], we are interested in the behavior of minimizers of E ε (u, B) as ε → 0. We will prove the W 1,p loc convergence of the symmetric minimizers.In addition, some estimates of the convergent rate of the symmetric minimizer will be presented and we will discuss the location of the points where u 2 3 = b 2 .In polar coordinates, for u(x) = (sin f (r)e idθ , b cos f (r)), we have If we denote V = {f ∈ W 1,p loc (0, 1]; r 1/p f r , r (1−p)/p sin f ∈ L p (0, 1), f (r) ≥ 0, where This shows that u = (sin f (r)e idθ , b cos f (r)) ∈ W is the minimizer of E ε (u, B) if and only if f (r) ∈ V is the minimizer of E ε (f, (0, 1)).Applying the direct method in the calculus of variations we can see that the functional E ε (u, B) achieves its minimum on W by a function u ε (x) = (sin f ε (r)e idθ , b cos f ε (r)), hence f ε (r) is the minimizer of E ε (f, (0, 1)) in V .Observing the expression of the functional E ε (f, (0, 1)), we may assume that, without loss of generality, the function f satisfies 0 ≤ f ≤ π 2 .We will prove the following Theorem 1.1 Let u ε be a symmetric minimizer of E ε (u, B) on W . Then for any small positive constant γ ≤ b, there exists a constant This theorem shows that all the points where u 2 ε3 = b 2 are contained in B(0, hε).Hence as ε → 0, these points converge to 0.
However, there may be several symmetric minimizers of the functional in W .We will prove that one of the symmetric minimizer ũε can be obtained as the limit of a subsequence u τ k ε of the symmetric minimizer u τ ε of the regularized functionals on W as τ k → 0. In fact, there exist a subsequence u τ k ε of u τ ε and ũε ∈ W such that lim Here ũε is a symmetric minimizer of E ε (u, B) in W .The symmetric minimizer ũε is called the regularized minimizer.Recall that the paper [3] studied the asymptotic behavior of minimizers for some α ∈ (0, 1), where u * is a harmonic map, A is the set of singularities of u * .Theorem 1.2 has shown the W 1,p loc (B \ {0}) convergence (weaker than (1.5)) of the symmetric minimizer.We will prove that the convergence of (1.5) is still EJQTDE, 2003 No. 22, p. 3 true for the regularized minimizer.The result holds only for the regularized minimizer, since the Euler-Lagrange equation for the symmetric minimizer u ε is degenerate.To derive the C 1,α convergence of the regularized minimizer ũε , we try to set up the uniform estimate of u τ ε by researching the classical Euler-Lagrange equation which u τ ε satisfies.By this and applying (1.4), one can see the C 1,α convergence of ũε .So, the following theorem holds only for the regularized minimizer.
Theorem 1.4 Let ũε be a regularized minimizer of E ε (u, B).Then for any compact subset K ⊂ B \ {0}, we have At the same time, the estimates of the convergent rate of the regularized minimizer, which is better than (1.3), will be presented as following Theorem 1.5 Let ũε (x) be the regularized minimizer of E ε (u, B).Then for any compact subset K of (0, 1] there exist positive constants ε 0 and C (independent of ε), such that as ε ∈ (0, ε 0 ), sup where λ = 1 2 .Furthermore, if K is any compact subset of (0, 1), then (1.6) holds with λ = 1.
The proof of Theorem 1.1 will be given in §2.In §3, we will set up the uniform estimate of E ε (u ε , K) which implies the conclusion of Theorem 1.2.By virtue of the uniform estimate we can also derive the proof of Theorem 1.3 in §4.For the regularized minimizer, we will give the proofs of Theorems 1.4 and 1.5 in §5 and §6, respectively.
By the embedding theorem we derive, from |u ε | = max{1, b} and proposition 2.1, the following Proposition 2.2 Let u ε be a symmetric minimizer of E ε (u, B).Then there exists a constant C independent of ε ∈ (0, 1) such that As a corollary of Proposition 2.1 we have EJQTDE, 2003 No. 22, p. 5 with some constant C > 0 independent of ε ∈ (0, 1).
Proposition 2.4 Let u ε be a symmetric minimizer of E ε (u, B).Then for any γ ∈ (0, γ 0 ) with γ 0 < b sufficiently small, there exist positive constants λ, µ independent of ε ∈ (0, 1) such that if where B 2lε is some disc of radius 2lε with l ≥ λ, then Proof.First we observe that there exists a constant β > 0 such that for any x ∈ B and 0 Suppose that there is a point Then applying Proposition 2.2 we have which contradicts (2.2) and thus the proposition is proved.

EJQTDE, 2003 No. 22, p. 6
To find the points where u 2 ε3 = b 2 based on Proposition 2.4, we may take (2.2) as the ruler to distinguish the discs of radius λε which contain these points.
Let u ε be a symmetric minimizer of E ε (u, B).Given γ ∈ (0, 1).Let λ, µ be constants in Proposition 2.4 corresponding to γ.If Then, one has Proposition 2.5 There exists a positive integer N (independent of ε) such that the number of bad discs Card J ε ≤ N.
Proof.Since (2.6) implies that every point in B can be covered by finite, say m (independent of ε) discs, from Proposition 2.3 and the definition of bad discs,we have and hence Card J ε ≤ mC µ ≤ N .Applying TheoremIV.1 in [1], we may modify the family of bad discs such that the new one, denoted by {B(x ε i , hε); i ∈ J}, satisfies The last condition implies that every two discs in the new family are not intersected.From Proposition 2.4 it is deduced that all the points where |u ε3 | = b are contained in these finite, disintersected bad discs.

EJQTDE, 2003 No. 22, p. 7
Proof of Theorem 1.1.Suppose there exists a point x 0 ∈ Z ε such that x 0 ∈B(0, hε).Then all points on the circle By virtue of Proposition 2.4 we can see that all points on S 0 are contained in bad discs.However, since |x 0 | ≥ hε, S 0 can not be covered by a single bad disc.As a result, S 0 has to be covered by at least two bad disintersected discs.This is impossible.
Proof.For j = 2, the inequality (3.3) is just the one in Proposition 2.1.Suppose that (3.3) holds for all j ≤ n.Then we have, in particular If n = N then we are done.Suppose n < N .We want to prove (3.3) for j = n + 1. Obviously (3.4) implies from which we see by integral mean value theorem that there exists Consider the functional It is easy to prove that the minimizer ρ 1 of E(ρ, η n+1 ) in W 1,p fε ((η n+1 , 1), R + ) exists and satisfies ) where v = ρ 2 r + 1.It follows from the maximum principle that ρ 1 ≤ π/2 and the last inequality of which is implied by Theorem 1.
Proof.Similar to the derivation of (3.5) we may obtain from Proposition 3.1 for j = N that there exists Also similarly, consider the functional whose minimizer ρ 2 in W 1,p fε ((η N +1 , 1), R + ) exists and satisfies where v = ρ 2 r + 1.From (3.4) for n = N it follows immediately that Similar to the proof of (3.13) and (3.14), we get, from Proposition 3.1 and (3.16), Now we define EJQTDE, 2003 No. 22, p. 11 and then we have Hence Using (3.17) we have This is my conclusion.
4 Proof of Theorem 1.3 Firstly, it follows from Jensen's inequality that Combining this with (3.15) yields Using (4.1) and the integral mean value theorem we can see that there exists EJQTDE, 2003 No. 22, p. 13 Consider the functional It is easy to prove that the minimizer ρ 3 of E(ρ, η 1 ) in W 1,p fε ((η 1 , 1), R + ) exists.By the same way to proof of (3.14), using (3.2) and (4.2) we have Hence, similar to the derivation of (3.15), we obtain Thus (4.1) may be rewritten as Let Proceeding in the way above (whose idea is improving the exponent of ε from step by step), we can get that for any m ∈ N , Letting m → ∞, we derive (1.2).From (1.2) we can see that On the other hand, for any x 0 ∈ K, we have Substituting this into (4.3)we obtain By the method in the calculus of variations we can see the following Proposition 5.1 The minimizer f ε ∈ V of the functional E ε (f, (0, 1)) satisfies the following equality .
Assume u τ ε = (e idθ sin f τ ε , cos f τ ε ) is the minimizer of the regularized functional E τ ε (u, B).It is easy to prove that the minimizer f τ ε is a classical solution of the equation ) where A = v + τ .By the same argument of Theorem 1.1 and Proposition 3.2, we can also see that for any compact subset K ∈ (0, 1], there exist constants η ∈ (0, 1/2) and C > 0 which are independent of ε and τ , such that where Proposition 5.2 Denote f τ ε = f .Then for any closed subset K ⊂ (0, 1), there exists C > 0 which is independent of ε, τ such that Proof.Without loss of the generality, we assume d = 1.Take R > 0 sufficiently small such that on (0, 1).Differentiating (5.1), multiplying with f r ζ 2 and integrating, we have where The second term of the right hand side of the inequality above can be handled similar to the proof of Proposition 5.2.Computing the first term of the right hand side yields with any δ ∈ (0, 1) by using Young inequality.In view of (5.1), we have I(r) = Hence, we may also obtain the result as (5.5) Now, if we take φ = ζ 2/q |f r | (p+2)/q , then the interpolation inequality (5.6) is invalid since φ = 0 near r = 1.Thus, we apply a new interpolation inequality [6, (2.19) in Chapter 2] Then it still follows the same result as (5.7).The rest of the proof is similar to the proof of Proposition 5.2.
Noting the limit (e idθ , 0) is unique, we can see that the convergence (5.11) holds not only for some subsequence but for all ũε .Theorem is proved.
6 Proof of Theorem 1.5 Without loss of the generality, we assume d = 1.Denote f = f τ ε .Set ψ = cos f ε p .Multiplying (5.1) by sin f we obtain Substituting ψ r = − sin f ε p f r into (6.1)we have Suppose ψ(r) achieves its maximum at the point r 0 in K, where K is an arbitrary open interval in any compact subset of (0, 1).Then ψ r (r 0 ) = 0, ψ rr (r 0 ) ≤ 0. And (sin f ) 2 ≥ C 1 > 0 with the constant C 1 independent of ε and τ which is implied by (5.2).Thus, it is deduced that, from Proposition 5.2, which implies sup K | cos f | ≤ Cε p with the constant C > 0 independent of ε and τ , where K is any compact subset of (0, 1).Letting τ → 0 and using (5.9) we may see the conclusion Substituting this and (6.3),(6.4)into (6.2) yields By virtue of the arbitrary of the point T , it is not difficult to get our Theorem.