Radial symmetry of positive solutions for semilinear elliptic equations in $Bbb R^n$

RADIAL SYMMETRY OF POSITIVE SOLUTIONS FOR SEMILINEAR ELLIPTIC EQUATIONS IN jRn

YUKI NAITO

ABSTRACT. Symmetry properties of positive solutions for semilin- ear elliptic problems in ]Rn are considered. We give a symmetry result for the problem in the general case, and then derive various results for certain classes of semilinear elliptic equations. We em- ploy the moving plane method based on the maximum principle on unbounded domains to obtain the result on symmetry.

1. Introduction

In this paper, we study the radial symmetry of positive solutions for semilinear elliptic equations in jRn. We consider the problem of the form

{

~u + f(lxl, u) = ° ^{in jRn,}

(1.1)

u(x) ^-+ ° ^{as Ixl -+ 00,}

where n ::::: 3. We establish a symmetry result for the problem (1.1) in the general case, and then derive various results for certain classes of semilinear elliptic equations. Typical equations we shall be interested in are as follows:

flu - u + uP = 0, p > 1;

A 1 P _ 0

uu + (1 + Ixl)e u - 0, to::::: 0, p > 1;

flu + _{(1 +} /3 _Ixl)/L uP - _{(1 +} 1 _Ixl)v u ^q = ° _' ^{/3 >} ° _{, ,."} ^{11. 1/ > 2} _{" - '} ^{P q > 1}

Some of those results are already known, but our aim is to treat those results from a unified point of view.

Received August 18, 1999.

2000 Mathematics Subject Classification: 35J60.

Key words and phrases: radial symmetry, semilinear elliptic equation, moving

plane method.

752 YUki Naito

Our arguments are based on the moving plane method. The method was first developed by Serrin [13] in PDE theory, and later extended and generalized by Gidas, Ni, and Nirenberg [2, 3]. In this paper, we employ the moving plane method based on the maximum principle on unbounded domains to obtain the result on symmetry.

In Section 2, we state the main theorem, and give some corollaries of the theorem. In Section 3, we prove the theorem by using the method . of moving planes.

2. Statement of the results

In (1.1), we assume that f(r, u) is continuous and Cl in u ~ 0, and that f(r, u) is nonincreasing in r > 0 for each fixed u ~ o. Our main result is the following:

THEOREM 1. Let u E C ² (]Rn) be a positive solution of (1.1). Define U and <I> as

(2.1)

U(r) = ^{sup{u(x) :} Ixl ~ r} and <I>(r) = ^sup{fu(r, s) : 0:::; s:::; U(r)}, respectively. Assume tbat tbere exists a positive function W on Ixl ~ Ho for some Ho > 0 satisfying

(2.2) (2.3)

~W + <I>(lxl)W :::; 0 in Ixl > R o and lim U(lxl) = O.

Ixl-+oo W(x)

Tben u must be radially symmetric about some point Xo E Rn and UT < 0 for r = Ix - ^xol > o.

We give some corollaries of the theorem. Some of the results presented here are already known, but our aim is to treat these results from a unified point of view.

COROLLARY 1. Assume tbat

fu(r,u):::; 0 forr _~ ro, 0:::; u:::; uo,

witb some constants ro ~ 0 and Uo > O. Let u be a positive solution of (1.1). Tben u must be radia1ly symmetric about some point Xo E ]Rn

and UT < 0 for r = Ix - ^xol > o.

REMARK. This result has been obtained by [9]. Related results have been obtained by [3, 5].

Proof. We see that the function U defined by (2.1) satisfies U(r) -+ 0 as r ^{-+ 00.} Take Ro 2: ro so large that U(r) ::; Uo for r 2: _Ro. Define W as W(x) == 1 on Ixl 2: Ro. Then W satisfies (2.3). Since <P(r) = max{fu(r, u) : 0 ::; u ::; U(r)} ::; 0 for r 2: _Ro, we have (2.2). Therefore, we can apply Theorem 1 to conclude the assertion.

For simplicity we consider the equation of the form (2.4) D.u + </>(Ixl)f(u) = 0

In equation (2.4), we assume that </> E e[O, (0) satisfies

</>(r) 2: 0 for r 2: 0 and </>(r) is nonincreasing in r> 0, and that f E G ¹ [0, (0) with f(u) > 0 for u > o. 0

COROLLARY 2. Assume, furthermore, that

</>(r) ⁼ O(r ^{f )} If'(u)1 = O(u ^{p - 1 )}

as r -+ 00 for some f E [-2,0]" and as u ^-+ 0 for some p > _~~;.

Let u be a positive solution of (2.4) in !Rn satisfying

as Ixl -+ ⁰⁰ if - 2 < f ::; 0

aslxl-+oo iff=-2.

Then u must be radia1ly symmetric about some point Xo E Rn and U r < 0 for r = Ix - xol > O.

REMARK. Related results have been obtained by [3, 5, 7,8]. Li[6] and Zou[14, 15] have obtained the symmetry results for positive solutions which have slow decay at infinity.

Proof. We see that the function U defined by (2.1) satisfies

as r -+ 00 if - 2 < f ::; 0,

as r ^-+ 00 if f = -2.

754 Yilki Naito

(i) The case where -2 < f ::; O. Let W(x) = Ixl-~=~ for Ixl > o. Then W satisfies (2.3) and

.6.W+ m (n-2-m)W=0 Ixl>O,

Ixl ^{2 '}

where m = (f + 2)j(n - 2). We note that n - 2 - m > 0 from p >

(n + f)j(n - 2). We find that

<P(r) = max{et>(r)J'(u) : 0 ::; u ::; U(r)} = O(lxj-2) as r ^{-+ 00.}

Then we have (2.2) for sufficiently large Ro > o. Therefore, we can apply Theorem 1 to conclude the assertion.

(ii) The case where f = -2. Let W(x) = (log Ixl)-P-l for Ixl 1 > l.

Then W satisfies (2.3) and

.6.W + (; =~ - ^{(p -l~lOg Ix') IX12l~g Ixl W = 0, Ixl > l.}

We find that

<P(r) = 0 (lxl-2(log Ixlt ^{1 )} as r -+ 00.

Then we have (2.2) for sufficiently large Ro > o. Therefore, we can apply

Theorem 1 to conclude the assertion. 0

COROLLARY 3. In equation (2.4), assume furthermore that et> t= ⁰

satisfies

(2.5) 100 ret>(r)dr < 00.

Let u be a positive solution of (2.4) in lR ⁿ satisfyingu(x) -+ cas lxl -+ 00

for some constant c 2: O. Then u must be radia1ly symmetric about the origin and ^UT < 0 for r = Ixl > O.

REMARK. Related results have been obtained by [7, 8, 11]. If et>

is locally Holder continuous, then every bounded positive solution u satisfies u(x) -+ c as Ixl -+ 00 for some constant c 2: O. (See [11].)

Proof.

(2.6)

Let v(x) = u(x) - c. Then v satisfies

{

.6.v + et>(lxl)h(v) = 0 in lR ^{n ,}

v(x) -+ 0 as Ixl -+ 00,

where h(v) = g(v + c). Since _-~v = cjJh :::: 0, we have v > 0 in lR. ⁿ by the maximum principle. We apply Theorem 1 to the problem (Z.6). We define U and <I> as

U(r) = sup{v(x) : Ixl :::: r} and <I>(r) = sup{cjJ(r)h'(s) : 0:::; s:::; U(r)}, respectively. Since <I>(r) :::; McjJ(r) for some constant M > 0 and (2.5) holds, we have

100 r<I>(r)dr < 00.

Then there exists a positive function W on Ixl :::: Ra for some Ra > 0 satisfying

~W + <I>(lxl)W = 0 and liminf W(x) > o.

Ixl->oo

(See, e.g., [11, Lemma Bol].) Then W satisfies (2.2) and (2.3). Therefore, Theorem 1 can be applied to conclude the assertion. 0

Finally, we consider the semilinear elliptic equations of the form

(2 7) ^A f3 p 1 q - 0

o

uU + (1 + Ixl)/l U - (1 + Ix!)v U - III m.. ,

where f3 > 0, J-l, v > 2 and p, q :::: 1 are real constants. The problem of existence of positive solutions to (2.7) has been studied by [4, 10, 1], ^and the solutions structures have been investigated by Chern [1]

Combining the result [1, Theorem 1.4] and Theorem 1, we obtain the following:

COROLLARY 4. Assume that v :::: J-l > 2 and q > p :::: 1. Let U be a positive solution of (2.7) satisfying u(x) :::; (f3/iJv)l/(q-p) in lR. ⁿ . Then U must be radially symmetric about the origin and ^U _r < 0 for r = Ixl > o.

Furthermore, for the case v = J-l > 2, all bounded positive solutions of (2.7) are radially symmetric about the origin and ^U r < 0 for r = Ixl > o.

Proof. Let

Since

f ( rr,u --(I+r)/l+l u ) - ^{f3J-l p} + (l+r)V+l u :::; (l+r)/l+l - ^{v q 1 (f3} J-lu +VU , ^{p q)}

we have fr(r,u) :::; 0 for 0:::; u:::; (f3/iJV)l/(q-P).

756 Yuki Naito

Let u be a positive solution of (2.7) satisfying u:::; (j3f-£/II)I!(q-P). By the similar arguments as in [1], we obtain limlxl->oo u(x) = c for some .constant c 2: O. Let v(x) = u(x) - c. Then v satisfies v(x) ---. 0 as

Ixl ---. ⁰⁰ and

b.v + g(lxl, v) = 0 in !Rn,

where g(r, v) = f(r, v + c). Therefore, we can apply Theorem 1 to con- clude the assertion by the similar argument as in the proof of Corollary 3.

Let us consider the case where 11 = f-£ > 2. Chern [1, Theorem 1.4]

has shown that every bounded solution u satisfies

(2.8) u(x) ---. c as Ixl ---. ⁰⁰

for some constant c 2: 0, and that u must be radially symmetric provided u(x) 2: j31!(q-p) in !Rn.

Let u be a positive solution of (2.7) satisfying (2.8). We claim that u(x) 2: /31!(q-P) in !Rn if c 2: j31!(q-p), and u(x) :::; j31!(q-p) in !Rn if c :::; ^j31!(q-P). We show the former case. The latter case can be shown similarly. Assume to the contrary that ^u(xo) < ^j31!(q-p) for some ^{Xo E}

!Rn. Define D = {x E !Rn : u(x) < j31!(q-P)}. Since f(r,u) 2: 0 for 0:::; u :::; ^j31!(q-p\ we have -b.u 2: 0 in D and u = ^j31!(q-p) on oD. By the maximum principle, we obtain ^u 2: j31!(q-p) in D. This contradicts the assumption. Therefore u(x) 2: j31!(q-P) in !Rn.

Let u be a positive solution of (2.7) satisfying (2.8). Assume that c 2: j31!(q-p). Then u(x) 2: j31!(q-p) in !Rn, and u must be radial symmetric by Chern's result. On the other hand, assume that c :::; ^j31!(q-p). Then

u :::; j31!(q-P) in !Rn, and ^u must be radial symmetric by the first part of

this corollary. This completes the proof. 0

3. Proof of Theorem 1

To prove Theorem 1, we introduce a few notations. For.\ E !R, we define TA and ~A as

TA = {x = (Xl, •.. ,x n ) E!R n

: Xl =.\} and ~A = {x E!R n

: Xl < .\}.

For x = ^{(Xl, .. .} ,x n ) E !Rn and .\ E !R, let x ^A be the reflection of x with respect to the hyperplane TA, i.e., x ^A = (2.\ - ^Xl, X2, . .. ,x n ). It is easy to see that, if .\ > 0,

(3.1)

We define vA(x) = u(x) - u(xA) for x E 1: A•

LEMMA 1. Let>. > o. Then VA satisfies (3.2)

where

Proof. Since f(r, u) is nonincreasing in rand (3.1) holds, we have

o ~u(x) + f(lxl, u(x)) - _~U(XA) - f(lxAI, u(xA))

> ~ (u(x) - u(xA)) + f(lxJ,u(x)) - f(lxl,u(x A))

~VA(X) + CA(X)VA(x)

for x E 1: _{A ,} where cA(x) is the function defined by (3.3). D

Define the set A as A = {>. E (0,00) : VA(X) > 0 in ~A}. Put B _o = {x E lR. ⁿ : Ixl < Ho}, where R o is the constant in the statement of

Theorem 1.

LEMMA 2. Let>. > o. If VA > 0 on 1: A n B o then>' ^E A.

Proof. From Lemma 1 and the assumption, we obtain

~VA + CA(X)VA::; ⁰ in 1: A \ B o, VA 2: 0 on 8(1: A \ B o).

Since U(r) is nonincreasing, we have 0::; u(xA)+t(u(x)-u(x ^A )) ::; U(lxl) for 0 ::; t ::; 1. Then, by (3.3), we have cA(x) ::; <p(lxl) in 1: A• From (2.2) and (2.3), the positive function W satisfies

~W + CA(X)W ::; 0 in 1: _A \ B _{o and} VA(X) < U(lxl) ---.0 W(x) - W(x) as Ixl ---. 00, x E 1: A \ B o.

Hence the maximum principle (Lemma A in Appendix) implies that VA > 0 in 1: A \ Bo. Then vA(x) > 0 in 1: A by the assumption, which

implies that >. E A. D

758 YUki Naito

By Lemma 2, we obtain the following:

LEMMA 3. Let A > O. If A tJ. A, then there exists Xo E _~>. n B o such that v>.(xo) ::; O.

LEMMA 4. Let AEA. Then 8u/8xl < 0 on T>..

Proof. By Lemma 1, we have (3.2) and v>. > 0 in ~>.. Since v>. = 0 on T>., we obtain 8v>./8xl < 0 on T>. by the Hopf boundary lemma.

Therefore, we have

8u 18v>.

-8 = -2 -8 < 0 on T>.. 0

Xl Xl

Proof of Theorem 1. Since u(x) is positive and limlxl->oou(x) = 0, there exists RI > Ho such that

(3.4) max{u(x) : Ixl ~ RI} < min{u(x) : Ixl ::; Ho},

where Ho is the constant in the statement of Theorem 1. We decompose the proof of Theorem 1 into several steps.

Step 1. We have [RI, (0) c A.

Let A ~ RI. We note that _Bo c ~>.. From (3.4), we have v>. > 0 in B o. Then, by Lemma 2, we have ^A E A, which implies that [RI, (0) cA.

Step 2. Let Ao EA. Then thereexists € > 0 such that (Ao-c, Ao] cA.

Assume to the contrary that there exists an increasing sequence {Ai}, i = 1,2, ... , such that Ai tJ. A and Ai ~ Ao as i ^---7 00. By Lemma 3, we have a sequence {Xi}' i = 1,2, ... , such that Xi E _~>.; n B o and V>'i(Xi) ::; O. A subsequence, which we call again {Xi}, converges to some point Xo E ~>'o n _Bo· Then v>.o(xo) ::; O. Since v>'o > 0 in ~>.o, we must have Xo E T>.o' By the mean value theorem we observe that there exists a point Yi satisfying (8u/8xl)(Yi) ~ 0 on the straight segment joining Xi to X;i for each ⁱ = 1,2,.... Since Yi ^---7 Xo as ^{i ---7 00,} we have (8u/8xI)(XO) ~ O. On the other hand, since Xo E T>.o' we have (8u/8xl)U(XO) < 0 by Lemma 4. This is a contradiction and Step 2 is established.

Step 3. We have either (3.5)

u(x) == U(X>'I) for some Al > 0

or

(3.6) in "Eo and. OU < 0 on TA for .A > O.

OXI

Let )\1 = inf{.A > 0 : (.A, (0) c A}. We distinguish the following two cases: (i) .AI > 0; (ii) .AI = O.

(i) The case where .AI > O. Let V A1 (x) = u(x) - U(X A1 ). From the continuity of u, we have V A1 (x) 2: 0 in "E _{A1 •} It follows from Lemma 1 that 6.v _A1 + CA1 (X)V _{A1 :::;} 0 in "E _{A1 •} Hence, by the strong maximum principle, we have that either V _A1 > 0 in "E _A1 or V _A1 == 0 in "E _{A1 •} Assume that

V A1 > 0 in "E _{A1 .} Then.AI E A. From Step 2, there exists c > 0 such that (.AI -c,.Ad cA. This contradicts the definition of .Al. Therefore, ^V A1 == 0 in "E A1 • Since (.AI, (0) c A, we have OU/OXI < 0 on TA for .A > .AI by Lemma 4. Thus we obtain (3.5).

(ii) The case where .AI = O. From the continuity of u, we have u(x) 2:

u(xO) in "Eo. By Lemma 4 we have OU/OXI < 0 on TA for .A > O. Thus, (3.6) holds.

If (3.6) occurs in Step 3, we can repeat the previous Steps 1-3 for the negative xrdirection to conclude that either ^U is symmetric in the Xl-direction about some plane Xl = .AI < 0 or

(3.7) u(x) :::; u(XO) in "Eo.

If (3.7) occurs, then u(x) == u(xO) in "Eo. Therefore, U must be symmetric in Xl-direction about some plane, and strictly decreasing away from the plane. Since the equation (1.1) is invariant under rotation, we may take any direction as the Xl-direction and conclude that u is symmetric in every direction about some plane. Therefore, u is radially symmetric about some point Xo E ffin and U _r < 0 for r = Ix - xol. 0

Appendix

Let D be an unbounded domain in ^]Rn, and let L denote a uniformly elliptic differential operator of the form

Lu == aij(x)oijU + bi(x)OiU + c(x)u,

where aij , bi, c E LOO(D). (We use the notation Oi = O/OXi and

Oij = 02/0XiOXj, together with the summation convention for repeated

indices.)

760 Yiili Naito

LEMMA A. Suppose that u "¥= 0 satisfies Lu ::; 0 in n and u ;:::: 0 on an. Suppose, furthermore, that there exists a function w such that

·w > 0 on n u an ^and Lw ::; 0 in n. ^If

(A.I) u(x) ---+ 0 I I ^En

( \ as x ---+ 00, X H, W xJ

then u > 0 in n.

REMARK. If n is bounded, we do not require the condition (A.I). See [12, Chap. 2, Theorem 10J.

Proof. First we show that u ;:::: 0 in n. Assume to the contrary that u(xo) < 0 for some Xo E n. Choose <5 > 0 so that

(A.2) u(xo) + <5w(xo) = O.

From (A.I), there exists R > Ixol satisfying u(x) + <5w(x) ;:::: 0 on {Ixl = R} n n. Define BR = {x E jRn : Ixl < R}. Then u + <5w satisfies L(u+<5w)::; 0 on nnB R and u+6w;:::: 0 on a(nnB R ). By [12, Chap.2, Theorem 10], (u + 6w)/w cannot attain a nonpositive minimum at an interior point of n n ^BR unless it is a constant. This contradicts (A.2).

Therefore, u ;:::: 0 in n. By the strong maximum principle, we conclude

that u > 0 in n. 0

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Department of Applied Mathematics Faculty of Engineering

Kobe University

Rokkodai Kobe, 657-8501 Japan

E-mail: [email protected]