Non-radial solutions to a bi-harmonic equation with negative exponent

Non-radial solutions to a bi-harmonic equation with negative exponent Ali Hyder Department of Mathematics, University of British Columbia, Vancouver BC V6TZ2, Canada ali.hyder@math.ubc.ca Juncheng Wei Department of Mathematics, University of British Columbia, Vancouver BC V6TZ2, Canada jcwei@math.ubc.ca June 22, 208 Abstract We prove the existence of non-radial entire solution to 2 u + u q = 0 in, u > 0, for q >. This answers an open question raised by P. J. McKenna and W. Reichel (E. J. D. E. 37 (2003) -3). Introduction We consider the following bi-harmonic equation with negative exponent 2 u + u q = 0 in, u > 0, () where q > 0. For q = 7, problem () can be seen as a fourth order analog of the Yamabe equation (see [, 5, 2]), namely 2 u = n 4 u n+4 n 4 in R n, u > 0. (2) 2 In the recent past, radial solutions to equation () have been studied by many authors, especially the existence and asymptotic behavior: The author is supported by the Swiss National Science Foundation, Grant No. P2BSP2-72064 The research is partially supported by NSERC

Theorem A ([5, 6, 8,, 6, 20]) i) There is no entire solution to () for 0 < q. ii) If u has exact linear growth at infinity, that is, lim = C > 0, x + x then q > 3. Moreover, for q = 7, u is given by = /5 + x 2, and is unique up to dilation and translations. iii) For q > 3 there exists radial solution with exact linear growth. iv) For q > there exists radial solution with exact quadratic growth, that is, lim x + x 2 = C > 0. v) For < q < 3 there exists a radial solution u such that r 4 q+ u(r) C(q) > 0 as r (the constant C(q) is explicitly known). vi) For q = 3 there exists a radial solution u such that r (log r) 4 u(r) 2 4 as r. It has been shown by Choi-Xu [5] that if u is a solution to () with q > 4, and u has exact linear growth at infinity then u satisfies the integral equation = 8π u q + γ, (3) for some γ R, and γ = 0 if and only if q = 7. In fact, every positive solution u to ( ) n u + u (4n ) = 0 in R 2n, n 2, with exact linear growth at infinity satisfies = c n R 2n u 4n, where c n is a dimensional constant, see [7], [7]. For the classification of solutions to the above integral equation we refer the reader to [2], [20]. In [6] McKenna-Reichel proved the existence of non-radial solution to 2 w + w q = 0 in R n, w > 0 (4) for n 4. This was a simple consequence of their existence results to (4) in lower dimension. More precisely, if u is a radial solution to (4) with n 3 then w(x) := u(x ) is a non-radial solution to 2 w+w q = 0 in R n+, where x = (x, x ) R n R. Then they asked whether in non-radial positive entire solution exist. (See [Open Questions (), [6]].) We answer this question affirmatively. (See Theorem.2 below.) In fact we prove the following theorems. 2

Theorem. Let u be a solution to () for some q >. Assume that β := u q dx < +. (5) 8π Then, up to a rotation and translation, we have = (β + o()) x + i I a i x 2 i + i I 2 b i x i + c, o() x 0, (6) where I, I 2 {, 2, 3}, I I 2 =, a i > 0 for i I, b i < β for i I 2, c > 0. Theorem.2 Let q >. Then for every 0 < κ < κ 2 there exists a non-radial solution u to () such that lim inf x x 2 = κ and lim sup x x 2 = κ 2. (7) Theorem.3 Let q > 7. Then for every κ > 0 there exists a non-radial solution u to () such that lim inf x x (0, ) and lim sup x x 2 = κ. The non-radial solutions constructed in Theorem.2 also satisfies the following integral condition u q dx < +, (8) for q > 3 2. Note that McKenna-Reichel s non-radial example has infinite L bound: R w q dx = +. n+ The existence of infinitely many entire non-radial solutions with different growth rates for the conformally invariant equation 2 u + u 7 = 0 in is in striking contrast to other conformally invariant equations u = u n+2 n 2 in R n, n 3 and ( ) m u = u n+2m n 2m in R n, n > 2m. In both cases all solutions are radially symmetric with respect to some point in R n, see [2], [4], [3] and [8]. Our motivation in the proof of Theorems.2-.3 come from a similar phenomena exhibited in the following equation ( ) n 2 u = e nu in R n, e nu dx < +. (9) R n It has been proved that for n 4 problem (9) admits non-radial entire solutions with polynomial growth at infinity, see [3], [9], [0], [4], [5], [9] and the references therein. It is surprising to see that conformally invariant equations with negative powers share similar phenomena. In the remaining part of the paper we prove Theorems.-.3 respectively. We also give a new proof of iii)-iv) of Theorem A, see sub-section 2.. 3

2 Proof of the theorems We begin by proving Theorem.. Proof of Theorem. Let u be a solution to ()-(5). We set v(x) := 8π u q, w := u v. (0) Fixing ε > 0 and R = R(ε) > 0 so that dx u q (x) < 8πε, one gets v(x) 8π x 2 y u q B c R 8π BR c Using that x, form (0), we obtain v(x) β x in. Combining these estimates we deduce that It follows that w satisfies v(x) lim = β. x x 2 w = 0 in, w(x) β x, x u q (β 2ε) x C(R). and hence, w is a polynomial of degree at most 2, see for instance [4, Theorem 5]. Indeed, up to a rotation and translation, we can write w(x) = i I a i x 2 i + i I 2 b i x i + c 0, where I, I 2 are two disjoint (possibly empty) subsets of {, 2, 3}, a i 0 for i I, b i 0 for i I 2 and c 0 R. Therefore, up to a rotation and translation, we have = 8π u q + a i x 2 i + b i x i + c. i I i I 2 Now u > 0 and v(x) β x lead to a i > 0 for i I, b i β for i I 2 and c = u(0) > 0. In order to prove that b i < β we assume by contradiction that b i0 = β for some i 0 I 2. Up to relabelling we may assume that i 0 =. Then C + b x + b x on C := {x = (x, x) R R 2 : x }, 4

a contradiction to (5). We conclude the proof. Now we move on to the existence results. We look for solutions to () of the form u = v + P where P is a polynomial of degree 2. Notice that u = v + P satisfies () if and only if v satisfies 2 v = (v + P ) q, v + P > 0. () In particular, if P 0, and v satisfies the integral equation v(x) =, (2) 8π (P + v) q then v satisfies (). Thus, we only need to find solutions to (2) (or a variant of it), and we shall do that by a fixed point argument. Let us first define the spaces on which we shall work: X := { v C 0 ( ) : v X < }, v(x) v X := sup x + x, X ev := { v X : v(x) = v( x) x }, v Xev := v X, X rad := {v X : v is radially symmetric}, v Xrad := v X. The following proposition is crucial in proving Theorem.2. Proposition 2. Let P be a positive function on such that P ( x) = P (x) and for some q > 0 x dx <. (P (x)) q Then there exists a function v X ev satisfying min v = v(0) = 0, v(x) :=, (3) 8π (P + v) q and v(x) lim = α P,v := x x 8π (P + v) q. Moreover, if P is radially symmetric then there exists a solution to (3) in X rad. Proof. Let us define an operator T : X ev X ev, v v, (In case P is radial we restrict the operator T on X rad. Notice that T (X rad ) X rad.) where v(x) :=. (4) 8π (P + v ) q We proceed by steps. Step T is compact. 5

Using that x we bound v X C 8π (P ) q for every v X. (5) Differentiating under the integral sign one gets v(x) 8π (P ) q C for every v X, x R3. We let (v k ) be a sequence in X ev. Then v k := T (v k ) is bounded in C loc (R3 ). Moreover, up to a subsequence, for some c i 0 with i = 0,, we have R3 y i k c 8π (P + v k ) q i. We rewrite (4) (with v = v k and v = v k ) as v k (x) = x y 8π (P + v k ) q + x 8π (P + v k ) q =: I,k(x)+I 2,k (x). It follows that Using that x y we bound I 2,k (x) 8π This implies that Since up to a subsequence, I,k (x) c 0 x c in X. y C. (P ) q lim sup I 2,k (x) sup R k x \ + x = 0. sup k sup I 2,k (x) <, x I 2,k I in X ev, for some I X ev. This proves Step as T is continuous. Step 2 T has a fixed point in X ev. It follows form (5) that there exists M > 0 such that T (X ev ) B M X ev. In particular, T ( B M ) B M. Hence, by Schauder fixed point theorem there exists a fixed point of T in B M. v(x) Step 3 lim x x = 8π (P + v ) =: α(p, v). q Step 3 follows from v(x) α(p, v) x 8π x (P + v ) q 4π 6 y C. (P ) q

Step 4 If v is a fixed point of T then v 0. Differentiating under the integral sign, from (4) one can show that the hessian D 2 v is strictly positive definite, and hence v is strictly convex. Moreover, using that (P + v ) is an even function, one obtains v(0) = 0. This leads to min v(x) = v(0) = 0. x R3 We conclude the proposition. In the same spirit one can prove the following proposition. Proposition 2.2 Let P be a positive even function on such that for some q > 0 x dx <. (P (x)) q Then there exists a positive function v X ev satisfying v(x) :=, min v = v(0). (6) 8π (P + v) q R3 Proof of Theorem.2 Let q > and 0 < κ < κ 2 be fixed. For every ε > 0 let v ε X ev be a solution of (3), that is, v ε (x) =, (7) 8π (P ε + v ε ) q where P ε (x) := + κ x 2 + κ 2 (x 2 2 + x 2 3) + ε x 4, x = (x, x 2, x 3 ). We claim that for every multi-index β N 3 with β = 2 D β v ε (x) C on B 2 and D β v ε (x) Cf q (x) on B c 2, (8) where x if q > 3/2 f q (x) := x log x if q = 3/2 x 2 2q if q < 3/2. For β = 2, differentiating under the integral sign, from (7), we obtain D β v ε ((x)) C (P ε + v ε ) q C ( + κ y 2 ) q 3 = C I i (x), i= 7

where I i (x) := ( + κ y 2 ) q, A := B x, A 2 := B 2 x \ A, A 3 := \ B 2 x. 2 A i Since q > we have D β v ε C on B 2. For x 2 we bound I (x) 2 x A ( + κ y 2 ) q Cf q(x), I 2 (x) C x 2q A 2 C x 2q y 3 x y C x 2 2q, I 3 (x) 2 y ( + κ y 2 ) q C x 2 2q. A 3 This proves (8). Since v ε (0) = v ε (0) = 0, by (8), we have ( + x ) log(2 + x ) if q > 3/2 v ε (x) C ( + x )(log(2 + x )) 2 if q = 3/2 ( + x ) 4 2q if q < 3/2. (9) Therefore, for some ε k 0 we must have v εk v in C 3 loc (R3 ) for some v in, where v satisfies 2 v = (v + P 0 ) q in, v 0 in, P 0 (x) := + κ x 2 + κ 2 (x 2 2 + x 2 3). Hence, u = v + P 0 is a solution to (). Moreover, as v satisfies (9), we have lim inf x x 2 This completes the proof. P 0 (x) = lim inf x x 2 = κ, lim sup x x 2 P 0 (x) = lim sup x x 2 = κ 2. Proof of Theorem.3 Let q > 7 be fixed. Then for every ε > 0 there exists a positive solution v ε to (6) with P (x) = P ε (x) := + εx 2 + κ(x 2 2 + x 2 3). Setting u ε := v ε + P ε one gets u ε (x) = 8π u q ε + P ε(x), Since c q := 2 3 q > 0 for q > 7, from (23), one obtains 0 = c q = 2 u q ε (x) dx + 2 3P ε (x) + 2c q u ε (x) 4 u q ε(x) 8 min u ε = u ε (0). (20) 2x P ε (x) P ε (x) u q (x) dx, dx

which implies that 2c q u ε (0) < 4, that is, u ε (0) C. Therefore, by (20) y 8π u q ε = u ε(0) C. (2) Hence, differentiating under the integral sign, from (20) (u ε (x) P ε (x)) 8π u q ε C. Thus, (u ε ) 0<ε is bounded in Cloc (R3 ). This yields u ε (x) 8π B u q δ x for x 2, ε for some δ > 0. Using this, and recalling that q > 4, we deduce lim sup y R u q = 0. ε 0<ε y R Therefore, for some ε k 0, we have u εk u, where u satisfies = 8π u q + + κ(x2 2 + x 2 3). We conclude the proof. 2. A new proof of iii)-iv) of Theorem A Proof of iii) Let q > 3 be fixed. Then by Proposition 2., for every ε > 0, there exists a radial function u ε satisfying u ε (x) = 8π u q + + ε x 2, min ε u ε = u ε (0) =. Since u ε is radially symmetric, one has (see Eq. (3.3) in [5]) for some δ > 0. Therefore, as q > 3 dx u q ε(x) C which gives As u ε (0) =, one would get u ε (r) δ( + r 4 ) q+, dx ( + x 4 ) q q+ C, u ε (x) 8π u q + 2ε x C + 2ε x. ε u ε (x) + C x + Cε x 2. 9

Thus, the family (u ε ) 0<ε is bounded in C loc (R3 ). Hence, for some ε k 0 we have u εk u where u satisfies Finally, as before, we have = 8π u q +, This completes the proof of iii). lim = x x 8π min u = u(0) =. R3 u q. Proof of iv) Let q > be fixed. Then by Proposition 2., for every ε > 0, there exists a non-negative radial function v ε satisfying v ε (x) = 8π ( + y 2 + ε y 4 + v ε ) q. The rest of the proof is similar to that of Theorem.2. In the spirit of [5, Lemma 4.9] we prove the following Pohozaev type identity. Lemma 2.3 (Pohozaev identity) Let u be a positive solution to = 8π u q + P (x), (22) for some non-negative polynomial P of degree at most 2 and q > 4. Then ( 2 3 ) q u q (x) dx + 2x P (x) P (x) 2 u q dx = 0 (23) (x) Proof. Differentiating under the integral sign, from (22) x = x (x y) 8π u q + x P (x). Multiplying the above identity by u q (x) and integrating on x u q dx = x (x y) (x) 8π u q (x)u q dx + x P (x) u q dx. (x) (24) Integration by parts yields x u q dx = x (u q (x))dx (x) q = 3 u q dx + R u q dσ. q q 0

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