Holder convergence of the weak solution to an evolution equation of $p$-Ginzburg-Landau type

J. Korean Math. Soc. 44 (2007), No. 3, pp. 585–603

H ¨ OLDER CONVERGENCE OF THE WEAK SOLUTION TO

AN EVOLUTION EQUATION OF p-GINZBURG-LANDAU TYPE

Yutian Lei

Reprinted from the

Journal of the Korean Mathematical Society Vol. 44, No. 3, May 2007

c

°2007 The Korean Mathematical Society

H ¨ OLDER CONVERGENCE OF THE WEAK SOLUTION TO AN EVOLUTION EQUATION OF p-GINZBURG-LANDAU

TYPE

Yutian Lei

Abstract. The author studies the local H¨ older convergence of the so- lution to an evolution equation of p-Ginzburg-Landau type, to the heat flow of the p-harmonic map, when the parameter tends to zero. The con- vergence is derived by establishing a uniform gradient estimation for the solution of the regularized equation.

1. Introduction

Let G ⊂ R ² be a bounded and simply connected domain with smooth bound- ary ∂G, g ∈ C ^∞ (∂G, S ¹ ), where S ¹ = {x ∈ R ² ; |x| = 1}, and deg(g, ∂G)

= d = 0, u ₀ ∈ C ^∞ (G, S ¹ ) and u ₀ (x) = g(x) as x ∈ ∂G. Denote G _T = G×(0, T ] with T ∈ (0, ∞). We are concerned with the asymptotic behavior of the weak solution u ε of the following problem when ε → 0,

(1.1) u t = div(|∇u| ^p−2 ∇u) + 1

ε ^p u(1 − |u| ² ), a.e. on G T

(1.2) u| _∂G×R

⁺

= g(x),

(1.3) u(x, 0) = u 0 (x), x ∈ G.

Recall that u ∈ C loc (0, T ; L ² (G, R ² )) ∩ L ^p _loc (0, T ; W ^1,p (G, R ² )) is a weak solu- tion of (1.1), if u satisfies

(1.4) R

G

T

uφ t dxdt = R

G

T

|∇u| ^p−2 ∇u∇φdxdt

− _ε ¹

p

R

G

T

uφ(1 − |u| ² )dxdt, ∀φ ∈ C ₀ ^∞ (G T ).

For the fixed ε > 0, the weak solution of (1.1)-(1.3) exists. In fact, it can be derived as the limit of the solution u ⁿ _ε of the regularized problem

(1.5) u t = div[(|∇u| ² + 1

n ) ^(p−2)/2 ∇u] + 1

ε ^p u(1 − |u| ² ), on G T Received October 6, 2005.

2000 Mathematics Subject Classification. 35B25, 35K55, 35K65.

Key words and phrases. H¨ older convergence, p-Ginzburg-Landau equations, heat flow of p-harmonic map.

c

°2007 The Korean Mathematical Society

585

u| _∂G×R

⁺

= g(x), u(x, 0) = u 0 (x), x ∈ G,

by letting n → ∞ (cf. [9]). The uniqueness of the weak solution will be proved in §2.

When p = 2, F. Bethuel, H. Brezis and F. Helein discussed the stationary case. They obtained the asymptotic behavior of the minimizer of the simplified Ginzburg-Landau functional

E ε (u) = 1 2

Z

G

|∇u| ² dx + 1 4ε ²

Z

G

(1 − |u| ² ) ² dx

under the assumption of deg(g, ∂G) = 0 (see [1]). In particular, it was proved that the Ginzburg-Landau minimizer converges to the harmonic map with boundary data g on G.

In the evolutionary case and under some general assumption on u(x, 0), F. H. Lin, R. Jerrard and H. Soner studied the dynamics of the zeros of the solution u _ε , i.e., the dynamical properties of Ginzburg-Landau vortices (see [6]

and [8]). Under the special assumption of u 0 (x) ∈ C ^∞ (G, S ¹ ), some uniform estimates and the H¨older convergence of u ε is derived in [5]. The work shows that the limit of the solution u ε is a heat flow of the harmonic map.

In this paper, under the assumption of u 0 (x) ∈ C ^∞ (G, S ¹ ) as in [5], we will present the local H¨older convergence of the p-Ginzburg-Landau solution (the weak solution of (1.1)-(1.3)) when p > 2, by establishing a series of uniform estimations of the solution u ⁿ _ε of (1.5), (1.2) and (1.3). The reason to consider the estimation of u ⁿ _ε is that, comparing with the case of p = 2, the system (1.1) is degenerate and nonlinear in the main part. It leads to the difficulty when setting up the regularity estimation of u ε .

Besides the results in [5], the study is also motivated by the research work in [3]. The authors proved that the heat flow of the p-harmonic map u p can be approached by the weak solution u ε of (1.1)-(1.3) in some weak sense, when ε → 0. This work proffered a new idea to study the p-harmonic maps. Namely, we can firstly research the properties of weak solutions u ε to some p-Laplace system with a penalization term, such as p-Ginzburg-Landau system (1.1).

Next, letting ε → 0, and investigating the asymptotic behavior of u ε to obtain the corresponding properties of the limit p-harmonic map. We hope the con- vergence of u ε to the p-harmonic map u p can be improved in more strong sense (cf. Theorems 5.2 and 5.3). Thus, the relation between u ε and u p will be more clear.

The limit function u p of u ε when ε → 0 can be introduced as follows. If u _p ∈ L ^∞ _loc (0, T ; L ² (G, S ¹ )) ∩ L ^p _loc (0, T ; W ^1,p (G, S ¹ )) is a weak solution of the system

(1.6) u t = div(|∇u| ^p−2 ∇u) + u|∇u| ^p ,

then it is called the heat flow of the p-harmonic map. By deg(g, ∂G) = 0,

there exists θ ∈ L ^∞ _loc (0, T ; L ² (G, R)) ∩ L ^p _loc (0, T ; W ^1,p (G, R)), θ 1 ∈ C ^∞ (∂G, R)

and θ 2 ∈ C ^∞ (G, R) such that u p = (cos θ, sin θ), g = (cos θ 1 , sin θ 1 ), u 0 = (cos θ 2 , sin θ 2 ) and

(1.7) θ| _∂G×R

⁺

= θ 1 ,

(1.8) θ| _G×{t=0} = θ 2 .

It follows from (1.6) that

(1.9) θ t = div(|∇θ| ^p−2 ∇θ).

From Lemma 1.2 in [3] it is easy to see that there exists a unique solution Θ ∈ L ^∞ _loc (0, T ; L ² (G, R)) ∩ L ^p _loc (0, T ; W ^1,p (G, R))

satisfying the problem (1.7)-(1.9) in the weak sense. Hence the heat flow of the p-harmonic map u p = (cos Θ, sin Θ) is also unique.

We shall prove a series of preliminary propositions in Section 2. By setting up the gradient estimations of u ⁿ _ε locally in Section 3 and 4, we will present the H¨older convergence of the weak solution u ε and its gradient ∇u ε in §5 (Theorem 5.2 and 5.3).

2. Preliminaries

Proposition 2.1. The weak solution of (1.1)-(1.3) is unique.

Proof. Set 0 < t 1 < t 2 < T . By a limit process we see that the test function φ in (1.4) can be taken in W ₀ ^1,p (G, R ² ). Then the weak solution satisfies

(2.1)

Z

G

(u(x, t 2 ) − u(x, t 1 ))φ(x)dx + Z _t

₂

t

1

Z

G

|∇u| ^p−2 ∇u∇φdxdt

= 1 ε ^p

Z _t

₂

t

1

Z

G

uφ(1 − |u| ² )dxdt.

Fix τ ∈ (0, T ). Choose h satisfying 0 < τ < τ + h < T . Taking t 1 = τ , t 2 = τ + h in (2.1) and multiplying with h ⁻¹ we obtain

(2.2)

Z

G

(u h (x, τ )) τ φ(x)dx + Z

G

(|∇u| ^p−2 ∇u) h (x, τ )∇φ(x)dx

= 1 ε ^p

Z

G

φ[u(1 − |u| ² )] _h dx

for any φ ∈ W ₀ ^1,p (G, R ² ), where u h is the Steklov’s mean value of u, i.e., u h = _h ¹ R _t+h

t u(x, τ )dτ as t ∈ (0, T − h); u h = 0 as t > T − h.

Assume that both u 1 and u 2 are weak solutions of (1.1)-(1.3), and write

w = u 1 − u 2 . In virtue of (2.2), we see that for any φ ∈ W ₀ ^1,p (G, R ² ) and any

given τ ∈ (0, T ), Z

G

w hτ φdx + Z

G

(|∇u 1 | ^p−2 ∇u 1 − |∇u 2 | ^p−2 ∇u 2 ) h ∇φdxdt

= 1 ε ^p

Z

G

T

φ[u 1 (1 − |u 1 | ² ) − u 2 (1 − |u 2 | ² )] h dxdt.

Taking φ(x) = w h (x, τ ) and integrating on (0, t), we see that Z

G

w ² _h dx

= − Z _t

0

Z

G

(|∇u ₁ | ^p−2 ∇u ₁ − |∇u ₂ | ^p−2 ∇u ₂ ) _h (∇u ₁ − ∇u ₂ ) _h dxdτ

+ 1 ε ^p

Z _t

0

Z

G

w ² _h dxdτ − 1 ε ^p

Z _t

0

Z

G

(u 1 |u 1 | ² − u 2 |u 2 | ² ) h w h dxdτ in view of R

G w ² _h (x, 0)dx = 0. Letting h → 0, we have Z

G

w ² (x, t)dx

= − Z _t

0

Z

G

(|∇u 1 | ^p−2 ∇u 1 − |∇u 2 | ^p−2 ∇u 2 )(∇u 1 − ∇u 2 )dxdτ

+ 1 ε ^p

Z _t

0

Z

G

w ² dxdτ − 1 ε ^p

Z _t

0

Z

G

(u 1 |u 1 | ² − u 2 |u 2 | ² )(u 1 − u 2 )dxdτ.

Applying Lemma 1.2 in [3], we know that the first and the third terms of the right hand side of the equality above are not positive. Using Gronwall’s inequality, we obtain

Z

G

|u 1 − u 2 | ² dx = 0, ∀t ∈ (0, T ).

This implies our conclusion. ¤

Proposition 2.2. Assume u ⁿ _ε solves the problem (1.5), (1.2) and (1.3). Then

|u ⁿ _ε | ≤ 1 on G T .

Proof. Denote u = u ⁿ _ε . Suppose ψ = |u| ² achieves its maximum on ∂G × (0, T ] or G × {0}, then the conclusion can be seen easily. Suppose ψ achieves its maximum at (x 0 , t 0 ) in G T . Then

(2.3) ∇ψ(x 0 , t 0 ) = 0, ∆ψ(x 0 , t 0 ) ≤ 0, and ψ t (x 0 , t 0 ) ≥ 0.

We claim that ψ ≤ 1 over G T . Otherwise, multiplying (1.5) with u, we have 1

2 (|u| ² ) t = div(v ^(p−2)/2 u∇u) − v ^(p−2)/2 |∇u| ² + 1

ε ^p |u| ² (1 − |u| ² ),

where v = |∇u| ² + _n ¹ . It follows from (2.3) that at (x 0 , t 0 ), 0 ≤ 1

2 ψ t = 1

2 div(v ^(p−2)/2 )∇ψ + 1

2 v ^(p−2)/2 ∆ψ − v ^(p−2)/2 |∇u| ² + 1

ε ^p |u| ² (1 − |u| ² ) < 0,

which is impossible. The proof is complete. ¤

Proposition 2.3. Assume u ⁿ _ε solves (1.5), (1.2) and (1.3). Then, there exists a constant C 0 > 0, which is independent of ε and n, such that

(2.4) sup

t∈(0,T ]

[ Z _t

0

Z

G

| ∂

∂t u ⁿ _ε (x, t)| ² dxdτ + E _ε ⁿ (u ⁿ _ε (x, t))] ≤ C 0 . Here

E ⁿ _ε (u) = 1 p Z

G

(|∇u| ² + 1

n ) ^p/2 dx + 1 4ε ^p

Z

G

(1 − |u| ² ) ² dx.

Proof. Multiplying (1.5) with u t and integrating over G, we have Z

G

u ² _t dx = Z

G

u t div(v ^(p−2)/2 ∇u)dx + 1 ε ^p

Z

G

uu t (1 − |u| ² )dx, where v = |∇u| ² + _n ¹ . Noting u t = g t = 0 on ∂G, we obtain

Z

G

u ² _t dx = − ∂

∂t E _ε ⁿ (u) and thus for any t > 0,

Z _t

0

Z

G

| ∂

∂t u ⁿ _ε (x, t)| ² dxdτ + E _ε ⁿ (u ⁿ _ε (x, t)) ≤ E ⁿ _ε (u 0 ) = C 0 .

This means that (2.4) holds. ¤

Proposition 2.4. For any ε ∈ (0, ε 0 ) with some fixed small constant ε 0 , if n → ∞, then

u ⁿ _ε → u ε , weakly star in L ^∞ ((0, T ); W ^1,p (G, R ² ));

u ⁿ _ε → u _ε , strongly in L ^p ((0, T ); L ^p (G, R ² ));

∂

∂t u ⁿ _ε → ∂

∂t u ε , weakly in L ² (0, T ; L ² (G, R ² )).

In addition, the limit u ε is just the unique weak solution of (1.1)-(1.3).

Proof. From Proposition 2.2 and (2.4), we see that there exists C = C(T ) > 0 (independent of ε and n), such that

(2.5) ku ⁿ _ε k L

^∞

(0,T ;W

^1,p

(G,R

²

)) ≤ C(T ),

(2.6) k ∂

∂t u ⁿ _ε k L

²

(0,T ;L

²

(G,R

²

)) ≤ C(T ).

So for any ε ∈ (0, ε 0 ), there exists a subsequence u ⁿ _ε

^k

of u ⁿ _ε such that as k → ∞, (2.7) u ⁿ _ε

^k

→ w ε , weakly star in L ^∞ ((0, T ); W ^1,p (G, R ² ));

(2.8) u ⁿ _ε

^k

→ w ε , strongly in L ^p ((0, T ); L ^p (G, R ² ));

(2.9) ∂

∂t u ⁿ _ε

^k

→ ∂

∂t w ε , weakly in L ² (0, T ; L ² (G, R ² ))

by applying Lemma 1.4 and the proof of Theorem 2.1 in [3]. Assume K is any compact subset of G T . Set ζ ∈ C ₀ ^∞ (G T , R) and ζ = 1 on K. Multiplying (1.5) by uζ and integrating over G T we obtain

1 ε ^p

Z

K

(1 − |u| ² )|u| ² dxdt ≤ C(1 + Z

G

T

v ^p/2 dxdt + Z

G

T

|u _t | ² dxdt) ≤ C(T ), by applying the H¨older inequality, (2.5) and (2.6). Here v = |∇u| ² + _n ¹ . Com- bining this with _ε ¹

p

R

G

T

(1 − |u| ² ) ² dxdt ≤ C(T ) (which is implied by (2.4)) yields

1 ε ^p

Z

K

(1 − |u| ² )dxdt ≤ C.

Hence, using Theorem 2.1 in [3] yields that for any ε ∈ (0, ε 0 ), as n k → ∞, (2.10) v ^(p−2)/2 ∇u ⁿ _ε

^k

→ |∇w ε | ^p−2 ∇w ε , in L ¹ (K).

Obviously, u ⁿ _ε satisfies R

G

T

uφ t dxdt = R

G

T

v ^(p−2)/2 ∇u∇φdxdt

− _ε ¹

p

R

G

T

uφ(1 − |u| ² )dxdt, ∀φ ∈ C ₀ ^∞ (G T ).

Letting n k → ∞, and using (2.7)-(2.10), we get R

G

T

w ε φ t dxdt = R

G

T

|∇w ε | ^p−2 ∇w ε ∇φdxdt

− _ε ¹

p

R

G

T

w ε φ(1 − |w ε | ² )dxdt, ∀φ ∈ C ₀ ^∞ (G T ).

This, together with the uniqueness of the weak solution u ε of (1.1)-(1.3), implies w ε = u ε . Our proposition can be completed by (2.7), (2.8), (2.9) and the

uniqueness of u ε . ¤

Proposition 2.5. Assume u ε is the weak solution of (1.1)-(1.3). Then as ε → 0,

u ε → u p , weakly star in L ^∞ (0, T ; W ^1,p (G, R ² )),

∂

∂t u ε → ∂

∂t u p , weakly in L ² (0, T ; L ² (G, R ² )),

u ε → u p , strongly in L ^p (0, T ; L ^p (G, R ² )).

Proof. By (2.5), (2.6) and Proposition 2.4, we can find a positive constant C (independent of ε), such that

(2.11) ku ε k L

^∞

(0,T ;W

^1,p

(G,R

²

)) ≤ C,

(2.12) k ∂

∂t u ε k L

²

(0,T ;L

²

(G,R

²

)) ≤ C.

Hence, there is a subsequence ε k such that as ε k → 0,

u ε

k

→ u p , weakly in L ^∞ (0, T ; W ^1,p (G, R ² ))

∂

∂t u _ε

_k

→ ∂

∂t u _p , weakly in L ² (0, T ; L ² (G, R ² )) u _ε

_k

→ u _p , strongly in L ^p (0, T ; L ^p (G, R ² )).

Using Lemma 1.4 and Theorem 2.2 in [3], we see that u p is just the unique heat flow of the p-harmonic map. Thus, the convergence above can be generalized

to all ε instead of the subsequence ε k . ¤

Proposition 2.6. Assume u ε is the weak solution of (1.1)-(1.3). Then for any compact subset K ⊂ G, there exists a constant C > 0, which is independent of ε, such that for any x 1 , x 2 ∈ K and t ∈ (0, T ],

|u(x 1 , t) − u(x 2 , t)| ≤ C|x 1 − x 2 |.

Proof. Assume R > 0 and t 0 ∈ (0, T ). Let J η be the mollification operator, such that

u η (x, t) = J η u(x, t) = Z _T

0

Z

R

²

j η (x − y, t − τ )u(y, τ )dydτ,

where 0 < η < t 0 < t < T − η. Denote u ε by u. For any x 1 , x 2 ∈ B R , there holds

u η (x 1 , t) − u η (x 2 , t)

= R _T

0

R

R

²

R ₁

0 d

ds j η (sx 1 + (1 − s)x 2 − y, t − τ )u(y, τ )dsdydτ

= (x 1 − x 2 ) R _T

0

R

R

²

R ₁

0 ∇ x j η (sx 1 + (1 − s)x 2 − y, t − τ )u(y, τ )dsdydτ

= −(x 1 − x 2 ) R ₁

0

R

R

²

R _T

0 ∇ y j η (sx 1 + (1 − s)x 2 − y, t − τ )u(y, τ )dτ dyds

= (x ₁ − x ₂ ) R ₁

0

R

R

²

R _T

0 j _η (sx ₁ + (1 − s)x ₂ − y, t − τ )∇ _y u(y, τ )dτ dyds.

Hence, applying H¨older inequality and (2.11) we obtain (2.13)

|u η (x 1 , t) − u η (x 2 , t)|

≤ |x 1 − x 2 | R ₁

0

R

R

²

R _T

0 |j η (sx 1 + (1 − s)x 2 − y, t − τ )||∇ y u(y, τ )|dτ dyds

≤ C|x 1 − x 2 |.

Letting η → 0 we derive

(2.14) |u(x 1 , t) − u(x 2 , t)| ≤ C|x 1 − x 2 |,

which implies our consequence. ¤

Proposition 2.7. For any n > 0, as ε → 0, there holds |u ⁿ _ε | → 1 in C loc (G T ).

Proof. Assume K is a compact subset of G T . For any (x 0 , t 0 ) ∈ K, denote γ = |u ε (x 0 , t 0 )|. Similar to the proof of (2.14), for u ⁿ _ε we also have

|u ⁿ _ε (x, t 0 ) − u ⁿ _ε (x 0 , t 0 )| ≤ C 0 |x − x 0 |

by using (2.5). Here C 0 is independent of ε and n. Hence |u ⁿ _ε (x, t 0 )| ≤ γ + ε as x ∈ B(x 0 , C ₀ ⁻¹ ε) and

Z

B(x

0

,C

⁻¹₀

ε)

(1 − |u ⁿ _ε (x, t 0 )| ² ) ² dx ≥ C ₀ ⁻² π(1 − γ − ε) ² ε ² .

Combining this with (2.4) we obtain (1−γ −ε) ² ≤ Cε ^p−2 or 1−γ ≤ Cε ^(p−2)/2 + ε. Noting the arbitrariness of (x 0 , t 0 ), we see our consequence. ¤ Proposition 2.7 implies that, there exists ε 0 > 0 sufficiently small, such that as ε ∈ (0, ε 0 ),

(2.15) |u ⁿ _ε | ≥ 1/2

on any compact subset of G T .

3. Estimate for k∇u ⁿ _ε k _L

^l

loc

Proposition 3.1. Assume u ⁿ _ε is a solution of (1.5), (1.2) and (1.3). If p ≥ 4, then for any compact subset K ⊂ G, t ∈ (0, T ) and l > 1, there exists a constant C > 0, which is independent of n and ε, such that

sup

τ ∈[t,T ]

Z

K

|∇u ⁿ _ε | ^l dx + k∇u ⁿ _ε k _L

l

(K

T

,R

²

) ≤ C = C(K T , l), where K _T = K × [t, T ].

Proof. Obviously u = u ε satisfies (1.5). Denote v = |∇u| ² + ¹ _n . Differentiate (1.5) with respect to x j ,

(3.1) ∂u x

j

∂t = (v ^(p−2)/2 u x

i

) x

i

x

j

+ 1

ε ^p [u(1 − |u| ² )] x

j

. Let ζ ∈ C ^∞ (B 2ρ × [0, T ], [0, 1]) satisfy

0 ≤ ζ(x, τ ) ≤ 1, (x, τ ) ∈ B 2ρ × [0, T ];

ζ(x, τ ) = 1, (x, τ ) ∈ B ρ × (t, T );

ζ(x, τ ) = 0, (x, τ ) ∈ ∂B 2ρ × [0, T ] or τ ≤ t/2;

|∇ζ| ≤ Cρ ⁻¹ , |ζ t | ≤ Ct ⁻¹ .

Multiplying (3.1) by ζ ² v ^b u x

j

(b ≥ 0) and integrating over B 2ρ ×[0, t 0 ], we obtain (3.2)

1 2(b+1)

R

B

2ρ

ζ ² v ^b+1 (x, t 0 )dx + R _t

₀

0

R

B

2ρ

ζ ² (v ^b u x

j

) x

i

(v ^(p−2)/2 u x

i

) x

j

dxdτ

= _b+1 ¹ R _t

₀

0

R

B

2ρ

ζζ t v ^b+1 dxdτ − 2 R _t

₀

0

R

B

2ρ

ζζ x

i

v ^b u x

j

(v ^(p−2)/2 u x

i

) x

j

dxdτ + _ε ¹

p

R _t

₀

0

R

B

2ρ

(1 − |u| ² )ζ ² v ^b u ² _x

_j

dxdτ − _2ε ¹

p

R _t

₀

0

R

B

2ρ

[(|u| ² ) x

j

] ² ζ ² v ^b dxdτ.

Applying (1.5) and (2.15) we see that

1 ε

^p

R _t

₀

0

R

B

2ρ

(1 − |u| ² )ζ ² v ^b u ² _x

_j

dxdτ

≤ R _t

₀

0

R

B

2ρ

|u| ⁻¹ [|u _t | + | div(v ^(p−2)/2 ∇u)|]ζ ² v ^b+1 dxdτ.

Using Young’s inequality and (2.4) to estimate the right hand side of the in- equality above, we may obtain that for any δ ∈ (0, 1),

1 ε

^p

R _t

₀

0

R

B

2ρ

(1 − |u| ² )ζ ² v ^b u ² _x

_j

dxdτ

≤ δ R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b−4)/2 |∇v| ² dxdτ +C(δ)( R _t

₀

0

R

B

2ρ

v ^(p+2b)/2 dxdτ + R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b+2)/2 dxdτ ) + CI 0 , where I 0 = R _t

₀

0

R

B

2ρ

ζv ^2b+2 dxdτ , and R _t

₀

0

R

B

2ρ

ζζ x

i

v ^b u x

j

(v ^(p−2)/2 u x

i

) x

j

dxdτ

≤ C(δ) R _t

₀

0

R

B

2ρ

v ^(p+2b)/2 dxdτ +δ( R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b−2)/2 P ₂

j=1 |∇u x

j

| ² dxdτ + R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b−4)/2 |∇v| ² dxdτ ).

Substituting these (with δ sufficiently small) into (3.2), and calculating its left hand side, we deduce that

(3.3) 1 2(b + 1)

Z

B

2ρ

ζ ² v ^b+1 (x, t 0 )dx + Z _t

₀

0

Z

B

2ρ

ζ ² v ^(p+2b−2)/2 X 2 j=1

|∇u x

j

| ² dxdτ

+ p + 2b − 2 4

Z _t

₀

0

Z

B

2ρ

ζ ² v

^p+2b−42

|∇v| ² dxdτ

+ b(p − 2) 2

Z _t

₀

0

Z

B

_2ρ

ζ ² v

^p+2b−62

(∇u∇v) ² dxdτ

≤ C(

Z _t

₀

0

Z

B

2ρ

v ^(p+2b)/2 (|∇ζ| ² + 1)dxdτ + Z _t

₀

0

Z

B

2ρ

ζ ² v ^(p+2b+2)/2 dxdτ ) + CI 0 . To estimate the last term of the right hand side of (3.3), we take

f = ζ ^2/q v (p+2b+2)/(2q)

in the embedding inequality (3.4)

Z

G

T

|f (x, t)| ^q dxdt ≤ C(

Z

G

T

|∇f (x, t)| ^h

¹

dxdt)(sup

t

Z

G

|f (x, t)| ^h

²

dx) ^h

¹

^/2

for f ∈ L ^∞ (0, T ; L ^q (G, R ² )) ∩ L ^p (0, T ; W ₀ ^1,p (G, R ² )), where h 1 , h 2 ≥ 1, q =

h

1

(h

2

+2)

2 , C only depends on h 1 and h 2 . Set h 1 = 4/3, h 2 = 1, thus q = 2.

Using H¨older inequality we obtain

(3.5)

R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b+2)/2 dxdτ

≤ C(sup 0<t<t

0

R

B

2ρ

ζv ^(p+2b+2)/4 dx) ^2/3

·[ R _t

₀

0

R

B

_2ρ

|∇ζ| ^4/3 v ^(p+2b+2)/3 dxdτ +( R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b−4)/2 |∇v| ² dxdτ ) ^2/3 ( R _t

₀

0

R

B

2ρ

v ² dxdτ ) ^1/3 ].

From (2.4) it is led to R

G v ^p/2 dx ≤ pC 0 . By using the reverse H¨older in- equality, we see that there is a small constant κ ∈ (0, 1), such that

Z

B

2ρ

v ^(p+2κ)/2 dx ≤ [C(

Z

G

v ^p/2 dx) ^2/p + (pC 0 ) ^1/p ] ^(p+2κ)/2 ≤ C(p, C 0 , κ).

Substituting this into Z

B

2ρ

ζv ^(p+2b+2)/4 dx ≤ ( Z

B

2ρ

ζ ²

^b+1−κb+1

v ^b+1−κ dx)

¹²

( Z

B

2ρ

v

^2κ+p2

ζ

^b+12κ

dx)

¹²

,

which is implied by H¨older inequality, we have Z

B

2ρ

ζv ^(p+2b+2)/4 dx ≤ C(

Z

B

2ρ

ζ ²

^b+1−κb+1

v ^b+1−κ dx)

¹²

.

Substitute this into (3.5). Since p ≥ 4, we see 2 ≤ p/2 and ^p+2b+2 ₃ ≤ ^p+2b ₂ . Using R

G v ^p/2 dx ≤ pC 0 , we derive that for any δ 1 ∈ (0, 1), R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b+2)/2 dxdτ

≤ C(sup _t R

B

2ρ

ζ ²

^b+1−κb+1

v ^b+1−κ dx)

¹³

[( R _t

₀

0

R

B

2ρ

v ^(p+2b)/2 dxdτ )

^{2(p+2b+2)3(p+2b)}

+C( R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b−4)/2 |∇v| ² dxdτ ) ^2/3 ]

≤ δ 1

R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b−4)/2 |∇v| ² dxdτ +C(δ 1 ) sup _t R

B

2ρ

ζ ²

^b+1−κb+1

v ^b+1−κ dx +C(δ 1 )( R _t

₀

0

R

B

2ρ

v ^(p+2b)/2 dxdτ ) ^C

with C independent of n and ε. Using H¨older inequality again, we see that for any δ 2 ∈ (0, 1),

R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b+2)/2 dxdτ

≤ δ 1

R _t

₀

0

R

B

2ρ

ζ ² v ^(p+2b−4)/2 |∇v| ² dxdτ +δ 2 sup _t R

B

2ρ

ζ ² v ^b+1 dx + C(δ 1 , δ 2 )[1 + ( R _t

₀

0

R

B

2ρ

v ^(p+2b)/2 dxdτ ) ^C ].

Similarly, we also have the same upper bound for I 0 in (3.3). Substitute this into (3.3) and choose δ ₁ and δ ₂ sufficiently small. Noting that it always true for all t 0 ∈ (0, T ), we obtain

(3.7)

sup 0<τ <T

R

B

2ρ

ζ ² v ^b+1 (x, τ )dx + R _T

0

R

B

2ρ

ζ ² v ^(p+2b−4)/2 |∇v| ² dxdτ

≤ C + C( R _T

0

R

B

2ρ

v ^(p+2b)/2 dxdτ ) ^C . First, take b = 0. Using (2.4) and (2.5), we have

sup

0<τ <T

Z

B

2ρ

(ζ ^2/λ w) ^λ (x, τ )dx + Z _T

0

Z

B

2ρ

(ζ ^2/λ |∇w|) ² dxdτ ≤ C,

where w = v ^(p+2b)/4 and λ = ^4(b+1) _p+2b . For ζ ^2/λ w, using the embedding inequality (3.4), we have R _T

0

R

B

2ρ

(ζ ^2/λ w) ^r dxdτ ≤ C. It implies R _T

t

R

B

ρ

w ^r dxdτ ≤ C, where r = 2 + ⁴ _p . This means |∇u| ∈ L ^p+2 (B ρ × (t, T )) and

(3.8)

Z _T

t

Z

B

ρ

|∇u| ^p+2 dxdτ ≤ C.

Next, take b = 1. Applying (3.8) and proceeding in the same way above, we can see that |∇u| ∈ L ^p+6 (B ρ × (t, T )) and

Z _T

t

Z

B

ρ

|∇u| ^p+6 dxdτ ≤ C.

For a given l > 0, by discussing inductively, we may find s k (for some k) such that p + s k > l, and deduce that |∇u| ∈ L ^p+s

^k

(B ρ × (t, T )),

Z _T

t

Z

B

ρ

|∇u| ^p+s

^k

dxdτ ≤ C.

Substituting this into (3.7) with some sufficiently large b, we see that sup

t<τ <T

Z

B

2ρ

|∇u| ^l dx ≤ C.

Thus Proposition 3.1 is proved. ¤

4. Estimate for k∇u _ε k L

^∞_loc

By means of the Moser iteration, from the estimate Proposition 3.1, we can further prove the following:

Proposition 4.1. Assume p ≥ 4. Then for any compact subset K ⊂ G T , there exists a constant C > 0 (independent of n and ε), such that

k∇u ⁿ _ε k _L

^∞

_(K,R

²

₎ ≤ C = C(K).

Proof. Assume R > 0, t ₁ ∈ (0, T ). Let t ₀ ∈ (t ₁ , T ), t ₀ > 4t ₁ . Define Q(ρ, t 0 ) = B ρ (x 0 ) × (t 0 , T ), T m = ^t ₂

⁰

− ₂

m+2

^t

⁰

, ρ m = ρ + ₂ ^ρ

m

, ρ _m = ¹ ₂ (ρ m + ρ m+1 ) = ρ + ₂

m+2

^3ρ , B m = B(x 0 , ρ m ), B _m ⁰ = B(x 0 , ρ _m ),

Q m = B m × (T m , T ), Q ⁰ _m = B _m ⁰ × (T m+1 , T ), and assume ζ m (x, t) ∈ C ₀ ¹ (Q m ) satisfying

ζ m = 0, ∀t ≤ T m ; ζ m = 1, ∀(x, t) ∈ Q ⁰ _m

|∇ζ m | ≤ ρ ⁻¹ 2 ^m+2 , 0 ≤ ζ mt ≤ t ⁻¹ ₀ 2 ^m+3 .

Without loss of generality, we suppose x 0 = 0. Take ρ > 0 sufficiently small such that B 2ρ ⊂ B R and |Q m | ≤ 1. Multiplying (3.1) with ζ _m ² v ^b u x

j

(b ≥ 1) and integrating over B _2ρ × (t ₀ /4, t), we obtain

1 2(b + 1)

Z

B

2ρ

ζ _m ² v ^b+1 (x, t)dx

+ Z _t

t

0

/4

Z

B

2ρ

ζ _m ² (v ^b u x

j

) x

i

(v ^(p−2)/2 u x

i

) x

j

dxdτ

= 1

b + 1 Z _t

t

0

/4

Z

B

2ρ

ζ m ζ mτ v ^b+1 dxdτ

− 2 Z _t

t

0

/4

Z

B

2ρ

ζ _m ζ _mx

_i

v ^b u _x

_j

(v ^(p−2)/2 u _x

_i

) _x

_j

dxdτ

+ 1 ε ^p

Z _t

t

0

/4

Z

B

2ρ

(1 − |u| ² )ζ _m ² v ^b (u x

j

) ² dxdτ

− 1 2ε ^p

Z _t

t

0

/4

Z

B

2ρ

ζ _m ² v ^b ((|u| ² ) x

j

) ² dxdτ.

Similar to the derivation of (3.3), we may also get

(4.1)

1 2(b + 1)

Z

B

2ρ

ζ _m ² v ^b+1 (x, t 0 )dx

+ Z _t

t

0

/4

Z

B

2ρ

ζ _m ² v ^(p+2b−2)/2 X 2 j=1

|∇u x

j

| ² dxdτ

+ p + 2b − 2 4

Z _t

t

0

/4

Z

B

2ρ

ζ _m ² v

^p+2b−42

|∇v| ² dxdτ

+ b(p − 2) 2

Z _t

t

0

/4

Z

B

2ρ

ζ _m ² v

^p+2b−62

(∇u∇u) ² dxdτ

≤ C ÃZ _t

t

0

/4

Z

B

2ρ

|ζ mτ |v ^b+1 dxdτ + Z _t

t

0

/4

Z

B

2ρ

v ^(p+2b)/2 |∇ζ m | ² dxdτ

+ p + 2b − 2 2

Z _t

t

0

/4

Z

B

2ρ

ζ _m ² (v ^(p+2b+2)/2 + v ^2b+2 )dxdτ

!

with the constant C independent of n, ε, b and m. To estimate Z _t

t

₀

/4

Z

B

_2ρ

ζ _m ² v ^(p+2b+2)/2 dxdτ,

we take f = ζ m ^2/q v (p+2b+2)/(2q) in (3.4) with h 1 = 1, h 2 = 2(q−1) and q ∈ ( ³ ₂ , 2).

Then, using H¨older inequality, and noticing |Q m | ≤ 1, we obtain

(4.2) Z

Q

m

ζ _m ² v ^(p+2b+2)/2 dxdτ ≤ CI ₃

¹²

· 2 ^m ρ I

p+2b+2 q(p+2b)

4 + p + 2b + 2 2q I ₁

¹²

I

p+2b+2 q(p+2b)

−

¹₂

4

¸

(in which we replace t 0 by t ∈ (T m , T ), since (4.2) still holds for all t 0 ∈ (T m , T )), where

I 1 = R

Q

m

ζ _m ² v ^(p+2b−4)/2 |∇v| ² dxdτ, I 3 = sup _T

_m

_<t<T R

B

m

ζ m ^4(q−1)/q v (1−1/q)(p+2b+2) dx, I ₄ = R

Q

m

v ^(p+2b)/2 dxdτ.

Substituting (4.2) into (4.1) and similar to the derivation of (3.7), we obtain I 1 + I 2

≤ C

½ 2 ^m t 0 I

2(b+1) p+2b

4

+ µ 2 ^m

ρ

¶ ₂

I 4 + p + 2b − 2 2 I ₃

¹²

· 2 ^m ρ I

p+2b+2 q(p+2b)

4 + p + 2b + 2 2q I ₁

¹²

I

p+2b+2 q(p+2b)

−

¹₂

4

¸¾ , where I 2 = sup _T

_m

_<t<T R

B

m

ζ _m ² v ^b+1 dx. Applying H¨older inequality and (2.4) we have

I 3 ≤ I ₂ ^2(q−1)/q (sup

t

Z

Q

m

v

^{p(q−1)2−q}

dx) ^2/q−1 ≤ CI ₂ ^2(q−1)/q . Hence, by using Young’s inequality, we can see that

I 1 + I 2 ≤ C

·

2

^m

t

0

I

2(b+1) p+2b

4 + ( ² _ρ

^m

) ² I 4 + ( ^p+2b−2 ₂ ) ^λ ( ² _ρ

^m

) ^q I

p+2b+2 p+2b

4

+( ^p+2b+2 _2q ) ^λ I ⁽

p+2b+2

q(p+2b)

−

¹₂

)(

_2−q^2q

) 4

¸ , where λ > 0 only depends on q. Since

2(b + 1)

p + 2b ≤ 1 ≤ p + 2b + 2

p + 2b ≤ ( p + 2b + 2 q(p + 2b) − 1

2 )( 2q 2 − q ), from Young’s inequality it follows that

I 1 + I 2 ≤ C 1 C ₂ ^m µ

1 + I ¹⁺

C3 (p−2)/2+2m−1

4

¶ ,

where C 1 > 0; C 2 depends on t 0 , ρ, q and λ; 2 ^m−1 = b + 1 and C 3 = _2−q ² . Denote w = v ^(p+2b)/4 , σ = ^4(b+1) _p+2b . Then

sup

T

m

<τ <T

Z

B

m

(ζ _m ^2/σ w) ^σ (x, τ )dx + Z

Q

m

(ζ _m ^2/σ |∇w|) ² dxdτ ≤ I ₁ + I ₂ . Applying the embedding inequality (3.4) we have

Z

Q

m+1

w ^2+σ dxdτ ≤ (I 1 + I 2 ) ² . Since b = 2 ^m−1 − 1, the inequality above implies

(4.3)

Z

Q

m+1

v ^(p−2)/2+2

^m

dxdτ

≤(C 1 C ₂ ^m ) ² µ

1 + Z

Q

m

v ^(p−2)/2+2

^m−1

dxdτ

¶ _2(1+C

₃

/((p−2)/2+2

^m−1

))

. If there exists a subsequence of positive integers {m i } with m i → ∞, such that

I 4 = Z

Q

_mi

v ^(p−2)/2+2

^mi

dxdτ < 1,

then letting m i → ∞ yields immediately

(4.4) kvk _L

^∞

_(Q

_∞

_,R) ≤ C.

Otherwise, there must be a positive integer m ₀ such that I ₄ =

Z

Q

m

v

^p−22

⁺²

^m

dxdτ ≥ 1 for m ≥ m ₀ . This, together with (4.3), implies

Z

Q

m+1

v ^(p−2)/2+2

^m

dxdτ

≤(C 1 C ₂ ^m ) ² µZ

Q

m

v ^(p−2)/2+2

^m−1

dxdτ

¶ _2(1+C

₃

/((p−2)/2+2

^m−1

))

.

Using Proposition 3.1 and an iteration lemma (Proposition 2.3 in [7]), we can also deduce the estimation (4.4). Thus the proof of Proposition 4.1 is complete.

¤ 5. H¨ older convergence

Once the uniform estimate k∇u _ε k _L

^∞

_(Ω

_R

₎ is established, it is not difficult to prove

Proposition 5.1. Assume p ≥ 4. Let ψ ε = _ε ¹

p

(1 − |u ε | ² ). Then for any compact subset K of G T , there exists a constant C > 0 (independent of ε), such that

(5.1) kψ ε k _L

^∞

_(K,R) ≤ C.

Proof. Consider the inner product of the both sides of (1.5) with u = u ⁿ _ε , uu t − div(v ^(p−2)/2 ∇u)u = 1

ε ^p |u| ² (1 − |u| ² ) = |u| ² ψ, where ψ = _ε ¹

p

(1 − |u ⁿ _ε | ² ). Combining this and ∇ψ = − _ε ²

p

u · ∇u with

div(v ^(p−2)/2 ∇u)u = div(v ^(p−2)/2 u · ∇u) − v ^(p−2)/2 |∇u| ² yields

|u| ² ψ = v ^(p−2)/2 |∇u| ² + ε ^p

2 div(v ^(p−2)/2 ∇ψ) − ε ^p 2 ψ _t . Using (2.15) we further obtain

(5.2) 1

4 ψ ≤ v ^(p−2)/2 |∇u| ² + ε ^p

2 div(v ^(p−2)/2 ∇ψ) − ε ^p

2 ψ t , ∀ε ∈ (0, ε 0 ).

Since at the point (x 0 , t 0 ), where ψ achieves its maximum, ∇ψ = 0, ∆ψ ≤ 0, ψ t ≥ 0 and

div(v ^(p−2)/2 ∇ψ) = v ^(p−2)/2 ∆ψ + p − 2

2 v ^(p−4)/2 ∇v∇ψ ≤ 0

we derive that, from (5.2),

(5.3) kψ _ε ⁿ k L

^∞

(K,R) ≤ C

by using the local estimate Proposition 4.1. Letting n → ∞ and applying

proposition 2.4, we can see that (5.1) holds. ¤

Theorem 5.2. Assume p ≥ 4. For any α ∈ (0, 1), there holds

ε→0 lim u ε = u p , in C _loc ^α,

^α2

(G T , R ² ).

Proof. Assume 0 < η < t 0 < t 1 < t 2 < T . Set B(∆t) = B(x 0 , (∆t) ^1/2 ), where x 0 ∈ B R , ∆t = t 2 − t 1 . Take φ ∈ C ₀ ¹ (B(∆t)), and denote u ⁿ _ε by u. Then,

(5.4)

R

B(∆t) φ(x)(u η (x, t 2 ) − u η (x, t 1 ))dx

= R

B(∆t) φ(x) R ₁

0 d

ds u η (x, st 2 + (1 − s)t 1 )dsdx

= ∆t R

B(∆t) φ(x) R ₁

0

R _T

0

R

R

²

j ηt (x − y, st 2 + (1 − s)t 1 − τ ) u(y, τ )dydτ dsdx

= −∆t R

B(∆t) φ(x) R ₁

0

R _T

0

R

R

²

j ητ (x − y, st 2 + (1 − s)t 1 − τ ) u(y, τ )dydτ dsdx.

For given (x, t) ∈ G T with t satisfying 0 < η < t 0 < t < T − η, we have J η (x − y, t − τ ) ∈ C ₀ ¹ (G T ). Since u is the solution of (1.5), it must satisfy

R _T

0

R

R

²

j ητ (x − y, st 2 + (1 − s)t 1 − τ )u(y, τ )dydτ

= R _T

0

R

R

²

v y ^(p−2)/2 ∇ _y u∇ _y j _η (x − y, st ₂ + (1 − s)t ₁ − τ )dydτ

− _ε ¹

p

R _T

0

R

R

²

u(y, τ )(1 − |u(y, τ )| ² )j η (x − y, st 2 + (1 − s)t 1 − τ )dydτ, where v y = |∇ y u| ² + _n ¹ . Substituting this into (5.4) yields

R

B(∆t) φ(x)(u η (x, t 2 ) − u η (x, t 1 ))dx

= −∆t R

B(∆t) φ(x) R ₁

0

£ R _T

0

R

R

²

v y ^(p−2)/2 ∇ y u∇ y j η (x − y, st 2 + (1 − s)t 1 − τ )dydτ

− _ε ¹

p

R _T

0

R

R

²

u(y, τ )(1 − |u(y, τ )| ² )j η (x − y, st 2 + (1 − s)t 1 − τ )dydτ ¤

dsdx

(5.5)

= −∆t R ₁

0

R _T

0

R

R

²

v y ^(p−2)/2 ∇ y u( R

B(∆t) ∇ x φj η (x − y, st ₂ + (1 − s)t ₁ − τ )dx)dydτ ds

+∆t R

B(∆) φ(x) R ₁

0 J η ( _ε ¹

p

u(1 − |u| ² ))(x, st 2 + (1 − s)t 1 )dsdx

= −∆t R ₁

0

R

B(∆t) ∇ x φ( R _T

0

R

R

²

j η (x − y, st 2 + (1 − s)t 1 − τ ) v ^(p−2)/2 y ∇ y udydτ )dxds

+∆t R

B(∆) φ(x) R ₁

0 J η ( _ε ¹

p

u(1 − |u| ² ))(x, st 2 + (1 − s)t 1 )dsdx

= −∆t R ₁

0

R

B(∆t) ∇ x φJ η (v ^(p−2)/2 ∇u)(x, st 2 + (1 − s)t 1 )dxds +∆t R ₁

0

R

B(∆) φ(x)J η ( _ε ¹

p

u(1 − |u| ² ))(x, st 2 + (1 − s)t 1 )dxds.

Take δ ∈ C ₀ ¹ (R, R ⁺ ) satisfying δ(s) = 0 as |s| ≥ 1 and R

R δ(s)ds = 1. For h > 0 define δ h (s) = _h ¹ δ( _h ^s ). By a limit process, we see that (5.5) still holds for φ ∈ W ₀ ^1,1 (B(∆t)). Hence, taking

φ = φ h (x) =

Z _(∆t)

^1/2

_−|x−x

₀

_|−2h

−h

δ h (s)ds in (5.5), we have

R

B(∆t) φ h (x)(u η (x, t 2 ) − u η (x, t 1 ))dx

= ∆t R ₁

0

R

B(∆t) δ h ((∆t) ^1/2 − |x − x 0 | − 2h) ^x _|x−x

ⁱ

^−x

⁰ⁱ

0

| J η (v ^(p−2)/2 u x

i

) (x, st 2 + (1 − s)t 1 )dxds

+∆t R ₁

0

R

B(∆) φ h (x)J η ( _ε ¹

p

u(1 − |u| ² ))(x, st 2 + (1 − s)t 1 )dxds.

Clearly, lim h→0 φ h (x) = 1 as x ∈ B(∆t); δ((∆t) ^1/2 − |x − x 0 | − 2h) = 0 as

|x − x 0 | < (∆t) ^1/2 − h; δ h ≤ Ch ⁻¹ and |B(∆) \ B(x 0 , (∆) ^1/2 − h)| ≤ Ch(∆t) ^1/2 . Using Proposition 4.1 and (5.3), we derive

| Z

B(∆t)

φ h (x)(u η (x, t 2 ) − u η (x, t 1 ))dx| ≤ C(∆t) ^3/2 . Letting h → 0 yields for all n > 0,

| Z

B(∆t)

(u η (x, t 2 ) − u η (x, t 1 ))dx| ≤ C(∆t) ^3/2 .

Letting n → ∞, and using Proposition 2.4, we see that

| Z

B(∆t)

(u εη (x, t 2 ) − u εη (x, t 1 ))dx| ≤ C(∆t) ^3/2 .