Local Gradient Estimates for Degenerate Elliptic Equations

Theorem 1.1 ([3, Theorem 1], [12, Theorem 1])

Assume (H1)–(H3), and that

hold for every (x,z,η)∈B3×𝕂×ℝn. If u is a bounded weak solution of (1.1), then u∈Cloc1,α⁢(B3) and there exists a constant M>0 depending only on n,p,γ0,γ1 and ∥u∥L∞⁢(B3) such that

Assume that (H1)–(H5) hold. Let u be a weak solution of (1.1) that satisfies

Then there exists C>0 depending only on n, p, γ0, γ1 and M0 such that

Assume (H2)–(H3), and

Then there exists C=C⁢(n,p,γ0,γ1)>0 such that for any weak solution u of (1.1), estimate (1.4) holds true.

Assume that (H2)–(H5) hold. There exists a constant C>0 depending only on n, γ0 and γ1 such that

for any nonnegative function ξ∈C0∞⁢(B3) and any β∈Liploc⁡([0,∞)) satisfying β,β′≥0.

Proof.

Using the difference-quotient argument as indicated in [13] (or [12, Proposition 1]) or using the approximation procedure as in [3], we may assume that u∈C2⁢(B3) and |∇⁡u⁢(x)|>0 for every x∈B3. For each i=1,2,…,n, define

By differentiating equation (1.1) with respect to xi, we have

div ⁡ [ ∑ j = 1 n u x i ⁢ x j ⁢ ∂ ⁡ 𝐚 ∂ ⁡ η j ⁢ ( x , u , ∇ ⁡ u ) + 𝐛 i ⁢ ( x , u , ∇ ⁡ u ) ] + d d ⁢ x i ⁢ b ⁢ ( x , u , ∇ ⁡ u ) = 0   in ⁢ B 3

in the weak sense. Using φ=uxi⁢β⁢(w)⁢ξ2 as a test function in the weak formulation and summing over i=1,2,…,n, we obtain

Dealing with the left-hand side of (2.3), we have from assumptions (H2) and (H3) that

Therefore, the left-hand side of (2.3) is greater than or equal to

For the right-hand side of (2.3), note that

∑ i = 1 n | ∇ ⁡ φ i | ≤ C n ⁢ ( | ∇ 2 ⁡ u | ⁢ β ⁢ ( w ) ⁢ ξ 2 + | ∇ ⁡ u | ⁢ β ′ ⁢ ( w ) ⁢ | ∇ ⁡ w | ⁢ ξ 2 + | ∇ ⁡ u | ⁢ | ∇ ⁡ ξ | ⁢ β ⁢ ( w ) ⁢ ξ ) ,

and from (H4)–(H5) that

Therefore, there exists a constant C=C⁢(n,γ1)>0 such that the right-hand side of (2.3) is no greater than

C ⁢ ∫ B 3 ( w p - 1 2 + w p 2 ) ⁢ ( | ∇ 2 ⁡ u | ⁢ β ⁢ ( w ) ⁢ ξ 2 + w 1 2 ⁢ β ′ ⁢ ( w ) ⁢ | ∇ ⁡ w | ⁢ ξ 2 + w 1 2 ⁢ | ∇ ⁡ ξ | ⁢ β ⁢ ( w ) ⁢ ξ ) ⁢ 𝑑 x .

We then estimate for ϵ>0 that

Consequently, the right-hand side of (2.3) is no greater than

The lemma then follows from the bounds (2.4) and (2.6) by taking ϵ=γ04. ∎

Assume that (H2)–(H5) hold. Let v=wp/2=|∇⁡u|p. Then there exists C=C⁢(n,p,γ0,γ1)>0 such that

∫ B 3 | ∇ ( v - k ) + | 2 ξ 2 d x ≤ C ∫ B 3 [ ( v - k ) + ] 2 | ∇ ξ | 2 d x + C ∫ B 3 ( w p + w p + 1 ) χ v > k ( x ) ξ 2 d x

for every constant k>0 and every nonnegative function ξ∈C0∞⁢(B3).

Proof.

We apply Lemma 2.1 with β⁢(s)=(sp/2-k)+. Then by dropping the first term in (2.1) and using β⁢(w)+w⁢β′⁢(w)≤(1+p2)⁢wp/2⁢χv>k, we obtain

The lemma then follows from Cauchy–Schwarz’s inequality and the fact that

∫ B 3 w p - 2 | ∇ w | 2 χ v > k ( x ) ξ 2 d x = 4 p 2 ∫ B 3 | ∇ ( v - k ) + | 2 ξ 2 d x . ∎

If we assume (1.5) in place of (H4)–(H5), then (2.5) becomes

| 𝐛 i ⁢ ( x , u , ∇ ⁡ u ) | + | b ⁢ ( x , u , ∇ ⁡ u ) | ≤ γ 1 ⁢ w p - 1 2 .

Then by inspecting the proof we see that (2.1) holds without the terms wp+12 and wp+22. As a consequence, instead of Lemma 2.2 we now obtain

for every constant k>0 and every nonnegative function ξ∈C0∞⁢(B3).

Assume that (H1)–(H3) and (H5) hold. There exists a constant C>0 depending only on p, n, γ0 and γ1 such that

Proof.

We follow the arguments in the proof of [6, p. 247, Lemma 1.1]. Let M=∥u∥L∞⁢(Br⁢(x0)). Since (2.8) is trivial if M=∞, we can assume that M<∞. Let ξ∈C0∞⁢(Br⁢(x0)) be the standard cut-off function with ξ=1 on Br/2⁢(x0) and |∇⁡ξ|≤cr. Then for any λ>0, by taking eλ⁢u⁢ξp as a test function we obtain

∫ B r ⁢ ( x 0 ) e λ ⁢ u ⁢ [ λ ⁢ ( 𝐚 ⋅ ∇ ⁡ u ) ⁢ ξ p + p ⁢ ( 𝐚 ⋅ ∇ ⁡ ξ ) ⁢ ξ p - 1 ] ⁢ 𝑑 x = ∫ B r ⁢ ( x 0 ) b ⁢ ( x , u , ∇ ⁡ u ) ⁢ e λ ⁢ u ⁢ ξ p ⁢ 𝑑 x .

Note that as a consequence of (H1)–(H3), we have

𝐚 ⁢ ( x , u , ∇ ⁡ u ) ⋅ ∇ ⁡ u ≥ γ 0 ⁢ | ∇ ⁡ u | p   and   | 𝐚 ⁢ ( x , u , ∇ ⁡ u ) ⋅ ∇ ⁡ ξ | ⁢ ξ p - 1 ≤ γ 1 ⁢ | ∇ ⁡ u | p - 1 ⁢ | ∇ ⁡ ξ | ⁢ ξ p - 1 .

These together with condition (H5) give

where C depends only on p and γ1. Choosing λ=(γ1+2⁢C)/γ0, we then get

∫ B r ⁢ ( x 0 ) | ∇ ⁡ u | p ⁢ ξ p ⁢ 𝑑 x ≤ e 2 ⁢ ( γ 1 + 2 ⁢ C ) ⁢ M γ 0 ⁢ ∫ B r ⁢ ( x 0 ) ( | ∇ ⁡ ξ | p + ξ p ) ⁢ 𝑑 x ≤ e 2 ⁢ ( γ 1 + 2 ⁢ C ) ⁢ M γ 0 ⁢ ( c p ⁢ r - p + 1 ) ⁢ | B r ⁢ ( x 0 ) | .

This yields (2.8) as desired since ξ=1 on Br/2⁢(x0). ∎

Theorem 2.5 ([6, p. 251, Theorem 1.1])

Assume that (H1)–(H3) and (H5) hold. Let u be a weak solution of (1.1) that satisfies (1.3). Then there exist constants C0>0 and α∈(0,1) depending only on n, p, γ0, γ1 and M0 such that

| u ⁢ ( x ) - u ⁢ ( y ) | ≤ C 0 ⁢ | x - y | α   for every ⁢ x , y ∈ B 21 / 8 .

Lemma 3.1 ([6, p. 63, Lemma 4.5] and [3, Lemma 2.4])

Let p>1, ρ>0, and let f∈C2⁢(Bρ⁢(x0)¯) satisfy |∇⁡f|>0. Then for any ξ∈C01⁢(Bρ⁢(x0)), we have

∫ B ρ ⁢ ( x 0 ) | ∇ ⁡ f | p + 2 ⁢ ξ 2 ⁢ 𝑑 x ≤ 2 ⁢ ( n + p ) 2 ⁢ ( osc B ρ ⁢ ( x 0 ) ⁡ f ) 2 ⁢ ∫ B ρ ⁢ ( x 0 ) [ | ∇ ⁡ f | p - 2 ⁢ | ∇ 2 ⁡ f | 2 ⁢ ξ 2 + | ∇ ⁡ f | p ⁢ | ∇ ⁡ ξ | 2 ] ⁢ 𝑑 x ,

where oscBρ⁢(x0)⁡f=supx∈Bρ⁢(x0)⁡|f⁢(x)-f⁢(x0)|.

Proof.

We include a proof for the sake of completeness. Let v=|∇⁡f|2. Then

∫ B ρ ⁢ ( x 0 ) v p + 2 2 ⁢ ξ 2 ⁢ 𝑑 x = ∫ B ρ ⁢ ( x 0 ) v p 2 ⁢ | ∇ ⁡ f | 2 ⁢ ξ 2 ⁢ 𝑑 x = ∫ B ρ ⁢ ( x 0 ) v p 2 ⁢ f x i ⁢ [ f ⁢ ( x ) - f ⁢ ( x 0 ) ] x i ⁢ ξ 2 ⁢ 𝑑 x .

Therefore, the integration by parts yields

The lemma then follows. ∎

Let f∈L∞⁢(BR) with R>0. Assume that there exist constants q>p>0 and γ>0 such that

for every r∈(0,R) and every σ∈(0,1). Then we have

∥ f ∥ L ∞ ⁢ ( B R / 2 ) ≤ γ ′ R n p ⁢ ( ∫ B R | f | p ⁢ 𝑑 x ) 1 p   𝑤𝑖𝑡ℎ   γ ′ = p q - p ⁢ 2 n p ⁢ ( 2 n p + 1 ⁢ q - p q ⁢ γ ) q p .

In particular, γ′=8np⁢γ2 if q=2⁢p.

Proof.

The proof of this lemma for particular q=p+2 is in [1, p. 55]. For the sake of completeness, we include the same arguments for all q>p here.

Let G=(∫BR|f|p⁢𝑑x)1p, and for s=0,1,…,

r s = R 2 ⁢ ∑ i = 0 s 2 - i , F s = ∥ f ∥ L ∞ ⁢ ( B r s ) .

Then by applying (3.1) to r=rs+1 and σ⁢r=rs+1-rs=R/2s+2, we obtain that

F s ≤ 2 n ⁢ s q ⁢ ( 4 R ) n q ⁢ γ ⁢ ( ∫ B r s + 1 | f | q ⁢ 𝑑 x ) 1 q ≤ 2 n ⁢ s q ⁢ ( 4 R ) n q ⁢ γ ⁢ F s + 1 q - p q ⁢ G p q .

Using Young’s inequality, it follows for any δ>0 and s=0,1,…,

Thus by iterating the inequality in (3.2), we get for any s=1,2,…,