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Electronic Journal of Differential Equations, Vol. 2010(2010), No. 15, pp. 1–10.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

ENTIRE SOLUTIONS FOR A CLASS OF p-LAPLACE EQUATIONS IN R2

ZHENG ZHOU

Abstract. We study the entire solutions of thep-Laplace equation

div(|∇u|p−2∇u) +a(x, y)W0(u(x, y)) = 0, (x, y)R2

wherea(x, y) is a periodic inxand y, positive function. HereW :RRis a two well potential. Via variational methods, we show that there is layered solution which is heteroclinic inxand periodic inydirection.

1. Introduction

In this paper we consider thep-Laplacian Allen-Cahn equation

−div(|∇u|p−2∇u) +a(x, y)W0(u(x, y)) = 0, (x, y)∈R2

x→±∞lim u(x, y) =±σ uniformly w.r.t. y∈R. (1.1) where we assume 2< p <∞and

(H1) a(x, y) is H¨older continuous onR2, positive and (i) a(x+ 1, y) =a(x, y) =a(x, y+ 1).

(ii) a(x, y) =a(x,−y).

(H2) W ∈C2(R) satisfies

(i) 0 =W(±σ)< W(s) for anys∈R\ {±σ}, andW(s) =O(|s∓σ|p) as s→ ±σ;

(ii) there existsR0> σ such thatW(s)> W(R0) for any|s|> R0. For example, here we may take W(t) = p−1p2−t2|p. This is similar with case p= 2, where the typical examples ofW are given byW(t) =14Qk

i=1(t−zi)2, where zi,i= 1,2, . . . k <∞are zeros ofW(t). The casep= 2 can be viewed as stationary Allen-Cahn equation introduced in 1979 by Allen and Cahn. We recall that the Allen-Cahn equation is a model for phase transitions in binary metallic alloys which corresponds to taking a constant functionaand the double well potentialW(t). The functionuin these models is considered as an order parameter describing pointwise the state of the material. The global minima ofW represent energetically favorite pure phases and different values ofudepict mixed configurations.

2000Mathematics Subject Classification. 35J60, 35B05, 35B40.

Key words and phrases. Entire solution;p-Laplace Allen-Cahn equation;

Variational methods.

c

2010 Texas State University - San Marcos.

Submitted September 15, 2009. Published January 21, 2010.

1

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In 1978, De Giorgi [11] formulated the following question. AssumeN > 1 and consider a solutionu∈C2(RN) of the scalar Ginzburg-Laudau equation:

∆u=u(u2−1) (1.2)

satisfying|u(x)| ≤1, ∂x∂u

N >0 for everyx= (x0, xN)∈RN and lim

xN→±∞u(x0, xN) =

±1. Then the level sets ofu(x) must be hyperplanes; i.e., there existsg ∈C2(R) such thatu(x) =g(ax0−xn) for some fixeda∈ RN−1. This conjecture was first proved forN = 2 by Ghoussoub and Gui in [13] and forN = 3 by Ambrosio and Cabr´e in [5]. For 4≤N ≤8 and assuming an additional limiting condition onu, the conjecture has been proved by Savin in [25] .

Alessio, Jeanjean and Montecchiari [2] studied the equation−4u+a(x)W0(u) = 0 and obtained the existence of layered solutions based on the crucial condition that there is some discrete structure of the solutions to the corresponding ODE.

In [3], when a(x, y) > 0 is periodic in x and y, the authors got the existence of infinite multibump type solutions, where a(x, y) =a(x,−y) takes an important role [3](see also [3, 20, 21, 22, 23, 24]).

Inherited from the above results, I wonder under what condition p-Laplace type equation (1.1) would have two dimensional layered solutions periodical iny. Adapt- ing the renormalized variational introduced in [2, 3] (see also [21, 22]) to the p- Laplace case, we prove

Theorem 1.1. Assume (H1)–(H2). Then there exists entire solution for (1.1), which behaves heteroclinic inxand periodic iny direction.

2. The periodic problem To prove Theorem 1.1, we first consider the equation

−div(|∇u|p−2∇u) +a(x, y)W0(u(x, y)) = 0, (x, y)∈R2 u(x, y) =u(x, y+ 1)

x→±∞lim u(x, y) =±σ uniformly w.r.t. y∈R.

(2.1)

The main feature of this problem is that it has mixed boundary conditions, requiring the solution to be periodic in the y variable and of the heteroclinic type in the x variable.

LettingS0=R×[0,1], we look for minima of the Euler-Lagrange functional I(u) =

Z

S0

1

p|∇u(x, y)|p+a(x, y)W(u(x, y))dx dy on the class

Γ ={u∈Wloc1,p(S0) :ku(x,·)∓σkLp(0,1)→0. x→ ±∞}

whereku(x1,·)−u(x2,·)kpLp(0,1)=R1

0 |u(x1, y)−u(x2, y)|pdy. Setting Γp={u∈Γ :u(x,0) =u(x,1)for a.e. x∈R}

cp= inf

Γp

I and Kp={u∈Γp:I(u) =cp}

Then we use the reversibility assumption (H1)-(ii) to show that the minimacon Γ equals minimacp on Γp, and so solutions of (2.1).

Note the assumptions onaandW are sufficient to prove thatIis lower semicon- tinuous with respect to the weak convergence inWloc1,p(S0); i.e., if un →u weakly

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in Wloc1,p(Ω) for any Ω relatively compact in S0, then I(u) ≤ lim infn→∞I(un).

Moreover we have

Lemma 2.1. If (un) ⊂ Wloc1,p(S0) is such that un → u weakly in Wloc1,p(S0) and I(un)→I(u), thenI(u)≤lim infn→∞un and

Z

S0

a(x, y)W(un)dx dy→ Z

S0

a(x, y)W(u)dx dy Z

S0

|∇un|pdx dy→ Z

S0

|∇u|pdx dy

Proof. Since un → uweakly in Wloc1,p(S0), k∇ukLp(S0) ≤lim infn→∞k∇unkLp(S0)

by the lower semicontinuous of the norm. By compact embedding theorem, we have un →uin Lploc(S0), using pointwise convergence and Fatou lemma, we have R

S0a(x, y)W(u)dx dy≤lim infn→∞R

S0a(x, y)W(un)dx dy, then Z

S0

a(x, y)W(u)dx dy≤lim sup

n→∞

Z

S0

a(x, y)W(un)dx dy

= lim sup

n→∞

h

I(un)− Z

S0

1

p|∇un|pdx dyi

=I(u)−lim inf

n→∞

Z

S0

1

p|∇un|pdx dy

≤ Z

S0

a(x, y)W(u)dx dy.

Thus,R

S0a(x, y)W(un)dx dy→R

S0a(x, y)W(u)dx dy, and sinceI(un)→I(u), we haveR

S0|∇un|pdx dy→R

S0|∇u|pdx dy.

By Fubini’s Theorem, if u ∈ Wloc1,p(S0), then u(x,·) ∈ W1,p(0,1), and for all x1, x2∈R, we have

Z 1 0

|u(x1, y)−u(x2, y)|pdy= Z 1

0

| Z x2

x1

xu(x, y)dx|pdy

≤ |x1−x2|p−1 Z 1

0

Z x2 x1

|∂xu(x, y)dx|pdx dy

≤pI(u)|x1−x2|p−1.

If I(u)<+∞, the functionx→u(x,·) is H¨older continuous from a dense subset ofRwith values inLp(0,1) and so it can be extended to a continuous function on R. Thus, any functionu∈Wloc1,p(S0)∩ {I <+∞}defines a continuous trajectory inLp(0,1) verifying

d(u(x1,·), u(x2,·))p= Z 1

0

|u(x1, y)−u(x2, y)|pdy

≤pI(u)|x1−x2|p−1,∀x1, x2∈R.

(2.2)

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Lemma 2.2. For allr >0, there existsµr>0, such that ifu∈Wloc1,p(S0)satisfies minku(x,·)±σkW1,p(0,1)≥r for a.e. x∈(x1, x2), then

Z x2 x1

hZ 1 0

1

p|∇u|p+a(x, y)W(u(x, y))dyi dx

≥ 1

p(x2−x1)p−1d(u(x1,·), u(x2,·))p+p−1 p µ

p p−1

r (x2−x1)

≥µrd(u(x1,·), u(x2,·))

(2.3)

Proof. We define the functional F(u(x,·)) =

Z 1 0

1

p|∂yu(x, y)|p+aW(u(x, y))dy

onW1,p(0,1), wherea= minR2a(x, y)>0. To prove the lemma, we first to claim that:

For any r >0, there exists µr >0, such that if q(y)∈ W1,p(0,1) is such that minkq(y)±σkW1,p(0,1)≥r, thenF(q(y))≥ p−1p µ

p p−1

r . Namely, ifqn(·)∈W1,p(0,1) andF(qn)→0, then minkqn±σkW1,p(0,1)→0.

Assume by contradiction that if F(qn)→0 and minkqn±σkL(0,1)≥ε0 >0.

Then there exists a sequence (y1n)⊂[0,1] such that min|qn(y1n)±σ| ≥ε0. Since R1

0 aW(qn)dy→0 there exists a sequence (yn2)⊂[0,1] such that|qn(y2n)±σ|< ε20. Then

ε0

2 ≤ |qn(yn2)−qn(y1n)|

≤ | Z y2n

yn1

|q˙n(t)|dt|

≤ |y2n−y1n|1−1phZ 1 0

|q˙n(t)|pdti1/p

≤p1p(F(qn))1/p →0.

It is a contradiction.

Since minkqn±σkL(0,1) → 0 as F(qn) → 0, then R1

0 |q˙n(y)|pdy → 0, and it follows thatkqn−σkW1,p(0,1)→0 asF(qn)→0.

Observe that if (x1, x2) ⊂ R and u ∈ Wloc1,p(S0) are such that F(u(x,·)) ≥

p−1 p µ

p p−1

r for a.e. x∈(x1, x2), by H¨older’s and Yung’s inequalities we have Z x2

x1

hZ 1 0

1

p|∇u|p+a(x, y)W(u(x, y))dyi dx

≥ Z x2

x1

Z 1 0

1

p|∂xu|pdy dx+ Z x2

x1

Z 1 0

1

p|∂yu|p+aW(u)dy dx

=1 p

Z 1 0

Z x2

x1

|∂xu|pdx dy+ Z x2

x1

F(u(x,·))dx

≥ 1

p(x2−x1)p−1d(u(x1,·), u(x2,·))p+p−1 p µ

p p−1

r (x2−x1)

≥µrd(u(x1,·), u(x2,·)).

The proof is complete.

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As a direct consequence of Lemma 2.2, we have the following result.

Lemma 2.3. If u∈Wloc1,p(S0)∩ {I <+∞}, thend u(x,·),±σ

→0 asx→ ±∞.

Proof. Note that since I(u) =

Z

S0

1

p|∇u|p+a(x, y)W(u(x, y))dx dy <+∞,

W(u(x, y)) →0 as |x| →+∞. Then by Lemma 2.2, lim infx→+∞d u(x,·), σ

= 0. Next we show that lim supx→+∞d u(x,·), σ

= 0 by contradiction. We as- sume that there exists r ∈ (0, σ/4) such that lim supx→+∞d(u(x,·), σ) > 2r, by (2.2) there exists infinite intervals (pi, si), i ∈ N such that d u(pi,·), σ

= r, d u(si,·), σ

= 2r and r≤ d u(x,·), σ

≤ 2r for x∈ ∪i(pi, si), i ∈N by Lemma 2.2, this implies I(u) = +∞, it’s a contradiction. Similarly, we can prove that limx→−∞d u(x,·),−σ

= 0.

Now we consider the functional on the class

Γ ={u∈Wloc1,p(S0) :I(u)<+∞, d u(x,·),±σ

→0 as x→ ±∞}

Let

c= inf

Γ I and K={u∈Γ :I(u) =c} (2.4) We will show that K is not empty, and we start noting that the trajectory in Γ with action close to the minima has some concentration properties.

For anyδ >0, we set λδ =1

p+ max

R2

a(x, y)· max

|s±σ|≤p1/pδ

W(s). (2.5)

Lemma 2.4. There exists ¯δ0 ∈ (0, σ/2) such that for anyδ ∈(0,δ¯0)there exists ρδ>0 andlδ>0, for which, ifu∈Γ andI(u)≤c+λδ, then

(i) minku(x,·)±σkW1,p(0,1)≥δ for a.e. x∈(s, p)thenp−s≤lδ.

(ii) if ku(x,·) +σkW1,p(0,1) ≤ δ, then d(u(x,·),−σ)≤ ρδ for any x≤ x, and if ku(x+,·)−σkW1,p(0,1)≤δ, thend(u(,·), σ)≤ρδ for anyx≥x+. Proof. By Lemma 2.2, as in this case, there existsµδ >0 such that

Z p s

Z 1 0

1

p|∇u|p+a(x, y)W(u)dx dy≥µδ(p−s).

SinceI(u)≤c+λδ there existslδ <+∞such thatp−s < lδ.

To prove (ii), we first do some preparation, µrδp−1p λδ, ρδ = max{δ, rδ}+ 3(p−1

)p−1p λδ. Let ¯δ0 ∈ (0, σ/2) be such that ρδ < σ/2 for all δ ∈ (0,¯δ0). Let δ∈(0,δ¯0), u∈Γ, I(u)≤+∞andx∈Rbe such thatku(x,·) +σkW1,p(0,1)≤δ.

Define

u(x, y) =





−1 ifx < x−1,

x−x+ (x−x+ 1)u(x, y) ifx−1≤x,

u(x, y) ifx≥x.

and note thatu ∈Γ andI(u)≥c, thenku+σkW1,p(0,1)=|x−x+1|·ku(x,·)+

σkW1,p(0,1) ≤δ whenx−1≤x≤x. Recall thatkqkL(0,1) ≤p1/pkqkW1,p(0,1)

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for anyq∈W1,p(0,1), then ku+σkL(0,1)≤p1/pku+σkW1,p(0,1)≤p1/pδ, by definition (2.5) ofλδ, we have

Z x

x−1

[ Z 1

0

1

p|∇u|p+a(x, y)W(u)dy]dx≤λδ. Since

I(u) =I(u)− Z x

−∞

Z 1 0

1

p|∇u|p+a(x, y)W(u)dy dx +

Z x

x−1

Z 1 0

1

p|∇u|p+a(x, y)W(u)dy dx we obtain

Z x

−∞

Z 1 0

1

p|∇u|p+a(x, y)W(u)dy dx≤2λδ. (2.6) Now, assume by contradiction that there existsx1< xsuch that d(u(x1,·),−σ)≥ ρδ, by (2.2) there exists x2 ∈ (x1, x) such that d(u(x,·),−σ) ≥ max{δ, rδ} for x∈(x1, x2) and d(u(x1,·), u(x1,·))≥ρδ−max{δ, rδ}. By Lemma 2.2, we have

Z x

−∞

Z 1 0

1

p|∇u|p+a(x, y)W(u)dy dx≥(pµrδ

p−1)p−1p ρδ−max{δ, rδ}

≥3λδ which contradicts (2.6). Thus d(u(x,·),−σ)≤ρδ for anyx≤x. Analogously, we can prove ifku(x+,·)−σkW1,p(0,1)≤δ, then d(u(x,·), σ)≤ρδ asx≥x+. To exploit the compactness of I on Γ, we set the function X : Wloc1,p(S0) → R∪ {+∞}given by

X(u) = sup{x: d(u(x,·), σ)} ≥σ/2.

Settingχ(s) = min|s±σ|, by(H3), there exist 0< w1< w2 such that

w1χp(s)≤W(s)≤w2χp(s) whenχ(s)≤σ/2. (2.7) Now, we can get the compactness of the minimizing sequence ofIin Γ.

Lemma 2.5. If(un)⊂Γis such thatI(un)→candX(un)→X0∈R, then there existsu0∈ K such that, along a sequence, un→u0 weakly inW1,p(S0).

Proof. We now show that (un) is bounded in Wloc1,p(S0), i.e., (un) is bounded in Lploc(S0), (∇un) is bounded in Lploc(S0). Since I(un)→ cand R

S0|∇un|pdx dy ≤ pI(un), we have that (∇un) is bounded inLploc(S0). If we can prove thatun(x,·) is bounded inLp(0,1) for a.e. x∈R, then (un) is bounded inLploc(S0).

LetBr={q∈Lp(0,1)/kqkLp(0,1)≤r}, we assume by contradiction that for any R >2σ, there exists ¯x∈Rsuch thatu(¯x,·)∈/BRforu∈Γ∩ {I(u)≤c+λ}, λ >0, such that ku(¯x,·)kLp(0,1) ≥R, then d(u(¯x,·), σ) ≥ ku(¯x,·)kLp(0,1)− kσkLp(0,1) ≥ R−σ. Since d(u(x,·),±σ)→0 asx→ ±∞, by continuity there existsx1>x¯such that d(u(x1,·), σ)≤σ/2 and d(u(x,·), σ)≥σ/2 for x∈(¯x, x1). Using Lemma 2.2, we get

c+λ≥I(u)≥µσ/2d(u(x1,·), u(¯x,·))≥µσ/2(R−3σ/2).

which is a contradiction forRlarge enough. We conclude that (un) is bounded in Wloc1,p(S0), thus there exists u0 ∈ Wloc1,p(S0) such that up to a sequence, un → u0

weakly in Wloc1,p(S0). We shall prove that u0 ∈ Γ; i.e., d(u0(x,·),±σ) → 0 as x→ ±∞. First we claim that:

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For any smallε > 0, there existsλ(ε)∈(0, λ¯δ) and l(ε)> lδ¯such that ifu∈Γ∩ {I(u)≤c+λ(ε)} then

Z

|x−X(u)|≥l(ε)

Z 1 0

W(u(x, y))dy dx≤ε. (2.8) Indeed, let δ < δ¯be such that 3λδ ≤ aw1ε where a = minR2a(x, y). Given any u ∈ Γ∩ {I(u) ≤ c+λδ}, by Lemma 2.4, there exists x ∈ (X(u)−lδ, X(u)) andx+∈(X(u), X(u) +lδ) such thatku(x,·) +σ)kW1,p(0,1)≤δandku(x+,·)− σkW1,p(0,1)≤δ. We define the function

˜

u(x, y) =













−σ ifx < x−1,

σ(x−x) + (x−x+ 1)u(x, y) ifx−1≤x,

u(x, y) ifx≤x≤x+,

(x+−x+ 1)u(x+, y) +σ(x−x+) ifx+≤x < x++ 1,

σ ifx > x++ 1

which belongs to Γ, andI(˜u)≥c, Z

|x−X(u)|≥lδ

Z 1 0

1

p|∇u|p+a(x, y)W(u)dy dx

≤I−∞x (u) +Ix+∞+ (u)

=I(u)−I(˜u) +Ixx−1(˜u) +Ixx+++1(˜u)

≤3λδ

then (2.8) follows settingl(ε) =l¯δ andλ(ε) =λδ.

From (2.8) it is easy to see thatu(x, y)→σas x→+∞. Combining (2.8) and (2.7) we obtain

Z

|x−X(u)|≥l(ε)

Z 1 0

w1|u(x, y)−σ|pdx dy≤ Z

|x−X(u)|≥l(ε)

Z 1 0

W(u(x, y))dy dx≤ε;

i.e., d u(x,·), σ

→0 asx→+∞. Analogously, we can get that d u(x,·),−σ

→0

asx→ −∞, it follows thatu0∈Γ.

As a consequence, we get the following existence result.

Proposition 2.6. K 6= ∅ and any u∈ K satisfies u ∈C1,α(R2) is a solution of

−div(|∇u|p−2∇u) +a(x, y)W0(u(x, y)) = 0 on S0 with ∂yu(x,0) = ∂yu(x,1) = 0 for allx∈R, andkukL(S0)≤R0. Finally, u(x, y)→ ±σ asx→ ±∞uniformly iny∈[0,1].

Proof. By Lemma 2.5, the setKis not empty. By (H2),kukL(S0)≤R0. Indeed,

˜

u= max{−R0,min{R0, u}} is a fortiori minimizer. Letη ∈ C0(S0) andτ ∈ R, then u+τ η ∈ Γ and since u ∈ K, I(u+τ η) is a C1 function of τ with a local minima atτ = 0. Therefore,

I0(u)η= Z

S0

|∇u|p−2∇u∇η+aW0(u)η dx dy= 0

for all such η, namely uis a weak solution of the equation −div(|∇u|p−2∇u) + a(x, y)W0(u(x, y)) = 0 on S0. Standard regularity arguments show that u ∈ C1,α(S0) for some α∈ (0,1) and satisfies the Neumann boundary condition (see

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[14][17][27]). SincekukL(S0)≤R0, there existsC >0 such thatkukC1,α(S0)≤C, which guarantees thatusatisfies the boundary conditions. Indeed, assume by con- tradiction thatudoes not verifyu(x, y)→ −σas x→ −∞uniformly with respect toy∈[0,1]. Then there existsδ >0 and a sequence (xn, yn)∈S0withxn→ −∞

and |u(xn, yn) +σ| ≥2δfor all n∈N. The C1,α estimate of uimplies that there exists ρ >0 such that |u(x, y) +σ| ≥ δ for∀(x, y)∈ Bρ(xn, yn), n∈N. Along a subsequencexn→ −∞, yn →y0∈[0,1], |u(x, y) +σ| ≥δfor (x, y)∈Bρ/2(xn, y0), which contradicts with the fact that d(u(x,·),−σ)→ 0 asx→ −∞ sinceu∈Γ.

The other case is similar.

We shall explore the reversibility condition of (H1)-(ii), and we will prove that the minimizer on Γ is in fact a solution of (2.1).

Lemma 2.7. cp=c.

Proof. Since Γp ⊂ Γ, cp ≥ c. Assume by contradiction that cp > c, then there existsu∈Γ such thatI(u)< cp. Writing

I(u) = Z

R

hZ 1/2 0

1

p|∇u|p+aW(u)dyi dx+

Z

R

hZ 1 1/2

1

p|∇u|p+aW(u)dyi dx

=I1+I2

it follows that min{I1, I2}<c2p. Suppose for exampleI1< cp/2, define v(x, y) =

(u(x, y) ifx∈Rand 0≤y≤12, u(x,1−y) ifx∈Rand 12 ≤y≤1.

Thenv∈Γp, by condition (H1)-(ii),I(v) = 2I1< cp, this is a contradiction.

We shall prove that anyu∈ K is periodic iny.

Lemma 2.8. If u∈ K thenu(x,0) =u(x,1) for allx∈R.

Proof. Supposeu∈ Kand v as above, then v(x, y) =u(x, y) for y ∈[0,1/2]. By (H1)-(ii),I(u) =c=cp=I(v), sov∈ K. Thenuandv are solutions of

−div(|∇u|p−2∇u) +aW0(u(x, y)) = 0, onS0,

yu(x,0) =∂yu(x,1) = 0 for allx∈R. (2.9) Since u=v for y ∈ [0,1/2], by the principle of unique continuation (see [8]), we

haveu=v inR×[0,1]. i.e. u(x,0) =u(x,1).

Remark 2.9. It is an open problem for the principle for p-harmonic functions in casen≥3 andp6= 2. Whenp=∞, the principle of unique continuation does not hold.

Proof of Theorem 1.1. We now extend u periodically in y direction to the entire space R2, and write it as U(x, y). As a consequence of the above lemmas and proposition 2.6,U(x, y) is an entire solution of (1.1), which is heteroclinic inxand

1-periodic iny direction.

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Zheng Zhou

College of Mathematics and Econometrics, Hunan University, Changsha, China E-mail address:[email protected]

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