Regularity for minimizers for functionals of double phase with variable exponents

Definition 1.1

A function u ∈ W^1,1Ω) is called to be a local minimizer of 𝓕 if F(x, Du) ∈ L¹(Ω) and satisfies

for any φ ∈ Wloc1,p(x) (Ω) with compact support in Ω.

Theorem 1.2

Assume that the conditions (1.5), (1.6) and (1.7) are fulfilled. Let u ∈ W^1,1(Ω) be a local minimizer of 𝓕. Then u ∈ Cloc1,y (Ω) for some y ∈ (0, 1).

Remark 1.3

(About the symbols for Hölder spaces) If we follow the standard textbooks, Dacorogna [49], Evans [50], Gilberg-Trudinger [51], etc., for k ∈ ℕ, 0 < α ≤ 1, C^k,α(Ω) mean the subspaces of C^k(Ω) consisting of functions whose k-th order partial derivatives are locally Hölder continuous. However, recently many authors (especially ones who study regularity problems) write them as Clock,α (Ω), and they use C^k,α(Ω) for C^k,α(Ω̄) (namely, for uniformly Hölder continuous cases). Anyway, with “loc" there is no doubt of misunderstanding. So, in this paper we follow their usage for Hölder spaces.

Theorem 2.1

Let a(x) ∈ C^0,β(Ω) for some β ∈ (0, 1) and 1 < p < q constants satisfying

and let ω ∈ L^∞(ℝⁿ) with ω ≥ 0 and ∫_BR ω dx = 1 for B_R ⊂ Ω with R ∈ (0, 1). Then, there exists a constant C depending only on n, p, q, [a]_0,β, Rⁿ ∥ω∥_L^∞ and ∥Dw∥_L^p(BR) and exponents d₁ > 1 > d₂ depending only on n, p, q, β such that

holds whenever u ∈ W^1,p(B_R), where

Note that for the special choice ω = |B_R|^–1 χ_BR we have

Proof

We can proceed exactly as in the proof of [1, Theorem 1.6] only replacing (3.11) of [1] by

which is shown by [52, Lemma 1.50] (see also the proof of [53, Theorem 7]).□

Corollary 2.2

Assume that all conditions of Theorem 2.1 are satisfied, and let D be a subset of B_R with positive measure. Then, there exists a constant C depending only on n, p, q, [a]_0,β, Rⁿ/|D| and ∥Du∥_L^p(BR) and exponents d₁ > 1 > d₂ depending only on n, p, q, β such that the following inequality holds whenever u ∈ W^1,p(x)(B_R) satifies u ≡ 0 on D:

Proof

Choosing ω so that

and applying Theorem 2.1, we get the assertion.□

Remark 2.3

In [1, Theorem 6.1], and therefore also in the above theorem and corollary, the exponent d₂ ∈ (0, 1) is chosen so that the following conditions hold:

In fact, in [1], they choose a constant y ∈ (1, p) so that

(see [1, (3.6), (3.14)]), and put d₂ = 1/y. Let us mention the that if d₂ satisfies (2.4) and (2.5) for some q = q₀ and p = p₀, then the same d₂ satisfies these inequalities for any q and p with q/p ≤ q₀/p₀.

Proposition 2.4

Let u ∈ Wloc1,p(x) (Ω) be a local minimizer of 𝓕. Then, for any compact subset K ⊂ Ω, F(x, Du) ∈ L^1+δ₀(K) and there exists a positive constant δ₀ and C depending only on the given data and K such that

holds for any B_R(y) ⋐ K.

Proof

Let K ⊂ Ω be a compact subset and R₀ ∈ (0, dist(K, ∂Ω)) a constant such that

For any x0∈K∘, put

Then, letting x_– ∈ B̄_R0(x₀) be a such that p(x_–) = p₁(x₀, R₀), we have

The above estimate (2.11) implies that

For any B_R(y) ⊂ B_R0(x₀) with 0 < R < 1, and 0 < t ≤ s ≤ R, let η be a cut-off function such that η ≡ 1 on B_t(y), η ≡ 0 outside B_s(y) and |Dη|≤2s−t. Put w := u – η(u – u_R), where u_R = ∫−BR(y) udx. Since

we have

where c₀ is a constant depending only on max_K q(x). On the other hand, since F(x, Du) ∈ L¹, we have

on B_R0(x₀). Thus, mentioning also that w = u outside B_s(y), we see that F(x, Dw) ∈ L¹ (K), namely w is an admissible function. In the following part of the proof, let us abbreviate

Then, we have

We can use hole-filling method. Add c₀ ∫_{Bs(y)∖Bt(y)} F(x, Du) dx to the both side and divide them by c₀ + 1, then we get

Using an iteration lemma [55, Lemma 6.1], we see, for some constant C = C(c₀, p₂, q₂), that

Putting s = R and t = R/2, we have

Since R^p₁–p₂ and R^q₁–q₂ are bounded because of the Hölder continuity of exponents p(x) and q(x), putting

from (2.15), we obtain the estimate

In order to get the boundedness of R^p₁–p₂ and R^q₁–q₂ the so-called “log-Hölder continuity" (see [56, section 4.1]) is sufficient. On the other hand by virtue of the Hölder continuity of q(⋅), we have that ã ∈ C^0,β (β = min{α, σ}). Let d₂ ∈ (0, 1) be a constant satisfying (2.4) and (2.5) for β = min{α, σ}, q = q₂(x₀, R₀) and p = p₁(x₀, R₀). Then, for any B_R(y) ⊂ B_R0(x₀), this d₂ satisfy (2.4) and (2.5) with q = q₂(y, R) and p = p₂(y, R).

By Theorem 2.1, we can estimate II as follows.

As mentioned above, (2.17) holds for for any B_R(y) ⊂ B_R0(x₀) with same d₂. Now, take R > 0 sufficiently small so that

and let θ ∈ (d₂, 1) be a constant satisfying

Then, using Hölder inequality, we can estimate the first term of the right hand side of (2.17) as follows.

Since,

and u locally minimizes 𝓕, ∫_BR(y) |Du|^p(x) dx is bounded. On the other hand, as mentioned after (2.15), R^{–(p₂–p₁)} is bounded. So, there exists a constant c₁ = c₁ (𝓕(u, K), p(x), d₂, n, θ)

where ω_n denotes the volume of a n-dimensional unit ball. Thus, from (2.19) we obtain for some positive constant c₂ = c₂(c₁, θ)

Similarly, we can estimate the second term of the left hand side of (2.17) as follows.

As above, using local minimality of u and the fact that R^{–(q₂–q₁)} is bounded, we have for a positive constant c₃ = c₃(𝓕(u, K), q(x), d₂, n, θ)

Thus, we obtain for some positive constant c₄ = c₄(c₃, θ)

Combining (2.16), (2.17), (2.20) and (2.23), we see that there exists a constant C depending on the given data and 𝓕(u, K) such that

for any B_R(y) ⊂ B_R0 ⊂ K ⋐ Ω. Now, by virtue of the reverse Hölder inequality with increasing domain due to Giaquinta-Modica [57], we get the assertion.□

Proposition 2.5

Let a(x), q and p satisfy the same conditions in Theorem 2.1 and let for A ⊂ BT+

u ∈ W^1,p( BT+ ) be a given function with

for some δ₀ >. Assume that v ∈ W^1,p (B⁺(T)) be a local minimizer of 𝓗 in the class

Then, for any S ∈ (0, T), there exists a constants δ ∈ (0, δ₀) and C > 0 such that for any y ∈ BS+ and R ∈ (0, T–S) we have

Proof

For convenience, we extend u, v, Du, Dv to be zero in B_T ∖ BT+ . Of course, because extended u, v may have discontinuity on Γ_T, they are not always in Wloc1,p (B_T), and therefore Du, Dv do not necessarily coincide with distributional derivatives of u, v on B(T). On the other hand, since u = v on Γ(T), u – v is in the class W^1,p(B(S)) and Du – Dv can be regarded as the weak derivatives of u – v on B(S) for any S < T.

Let R be a positive constant satisfying R ≤ (T – S)/2. For x₀ ∈ BS+ , we treat the two cases x0n≤34R and x0n>34R separately.

1. Suppose that x0n≤34R . Take radii s, t so that 0 < R/2 ≤ t < s ≤ R and choose a η ∈ C0∞ (B_T) such that 0 ≤ η ≤ 1, η ≡ 1 on B_t, supp η ⊂ B_s and |Dη| ≤ 2/(s – t). Defining

   φ:=η(v−u),

   we see that φ∈W01,1(BT+) with supp φ ⊂ B_s, and that

   D(v−φ)=(1−η)Dv−(v−u)Dη+ηDu.

   Then, by virtue of the minimality of v, for a positive constant c₄ depending only on q, we have

   ∫ΩtH(x,Dv)dx≤∫ΩsH(x,Dv)dx≤∫ΩsH(x,D(v−φ))dx=∫Ωs|D(v−φ)|p+a(x)|D(v−φ)|qdx≤c4∫Ωs∖Ωt|Dv|p+a(x)|Dv|qdx+c4∫Ωs|Du|p+a(x)|Du|qdx+c4∫Ωs2s−tp|v−u|p+a(x)2s−tq|v−u|qdx≤c4∫Ωs∖Ωt|Dv|p+a(x)|Dv|qdx+c4∫Ωs|Du|p+a(x)|Du|qdx+c42s−tp∫Ωs|v−u|pdx+c42s−tq∫Ωsa(x)|v−u|qdx.

   Now, we use the hole filling method as in the proof of Proposition 2.4. Namely, adding

   c4∫Ωt|Dv|p+a(x)|Dv|qdx

   and dividing both side by c₄ + 1, we obtain

   ∫ΩtH(x,Dv)dx≤c4c4+1∫ΩsH(x,Dv)dx+∫ΩsH(x,Du)dx+1(s−t)p∫Ωs|v−u|pdx+1(s−t)q∫Ωsa(x)|v−u|qdx,

   Using the iteration lemma [55, Lemma 6.1], we get for some constant C = C(c₄, p, q)

   ∫ΩtH(x,Dv)dx≤C∫ΩsH(x,Du)dx+C(s−t)p∫Ωs|v−u|pdx+C(s−t)q∫Ωsa(x)|v−u|qdx.

   Putting t = R/2 and s = R, we have

   ∫ΩR/2H(x,Dv)dx≤C∫ΩRHx,v−uRdx+C∫ΩRH(x,Du)dx.

   Let us now consider the mean integral in all the terms, we obtain

   ∫−ΩR/2H(x,Dv)dx≤C∫−ΩRH(x,Du)dx+C∫−ΩRHx,v−uRdx.

   Since we are assuming that x0n≤34R we can apply Corollary 2.2 with a constant independent on R for the last term in the right hand side and get

   ∫−ΩR/2H(x,Dv)dx≤C∫−ΩRH(x,Du)dx+C∫−ΩR(H(x,D(v−u)))d2dx1d2.

   Taking into consideration that d₂ < 1 we share in the last term Dv and Du, apply Hölder inequality for the integral of H(x, Du)^d₂, and obtain

   ∫−ΩR/2H(x,Dv)dx≤C∫−ΩRH(x,Du)dx+C∫−ΩRH(x,Dv)d2dx1d2. (2.26)

2. Let us deal with the case that x0n>34R . In this case, since B_3R/4(x₀) ⋐ BT+ , we can proceed as in [1, 9. Proof of Theorem 1.1:(1.8)], slightly modifying the radii, to get

   ∫−ΩR/2H(x,Dv)dx=∫−BR/2H(x,Dv)dx≤C∫−B3R/4H(x,Dv)d2dx1d2≤C′∫−ΩRH(x,Dv)d2dx1d2. (2.27)

   Thus, we see that (2.26) holds for every 0 < R < (S – T)/2. Now, the reverse Hölder inequality allows us to obtain

   ∫−ΩRH(x,Dv)1+δdx11+δ≤C∫−ΩR2H(x,Dv)dx+C∫−ΩRH(x,Du)1+δdx11+δ.

   □