Finite energy solutions and critical conditions of nonlinear equations in R n
aa r X i v : . [ m a t h . A P ] J un Finite energy solutions and critical conditions of nonlinearequations in R n Yutian Lei
Institute of MathematicsSchool of Mathematical SciencesNanjing Normal UniversityNanjing, 210023, Chinaemail: [email protected]
Abstract
This paper is concerned with the critical conditions of nonlinear elliptic equa-tions with weights and the corresponding integral equations with Riesz potentials andBessel potentials. We show that the equations and some energy functionals are invariantunder the scaling transformation if and only if the critical conditions hold. In addition,the Pohozaev identity shows that those critical conditions are the necessary and sufficientconditions for existence of the finite energy positive solutions or weak solutions. Finally,we discuss respectively the existence of the negative solutions of the k -Hessian equationsin the subcritical case, critical case and supercritical case. Here the Serrin exponent andthe critical exponent play key roles. Keywords : critical exponents, finite energy solution, Riesz potential, Bessel potential,k-Hessian equation.
MSC2010
In this paper, we consider the relation between the critical conditions and the finite energysolutions for several semilinear, quasilinear and fully nonlinear elliptic equations.If n ≥
3, and u belongs to the homogeneous Sobolev space D , ( R n ) such that the Sobolevinequality holds k u k L q +1 ( R n ) ≤ C k∇ u k L ( R n ) , (1.1)then q = n +2 n − . In fact, after the scaling transformation u µ ( x ) := µ σ u ( µx ) , µ > , (1.2)by (1.1) we can see that k u µ k L q +1 ( R n ) ≤ Cµ n − − nq +1 k∇ u µ k L ( R n ) . Since u µ also satisfies (1.1), q must be equal to n +2 n − .The Euler-Lagrange equation which the extremal function of (1.1) satisfies is the Lane-Emdenequation − ∆ u = u q , u > in R n . (1.3)Eq. (1.3) and the energy k u k L q +1 ( R n ) are invariant under the scaling transformation if and onlyif q = n +2 n − . In fact, u µ solves (1.3) implies σ = q − . The energy k u µ k L q +1 ( R n ) = k u k L q +1 ( R n ) implies σ = nq +1 . Thus, q = n +2 n − . On the contrary, if q = n +2 n − , (1.3) is invariant under theconformal transformation.The critical exponent n +2 n − plays the key roles on the existence and nonexistence of this Lane-Emden equation. We refer to [8] by Gidas and Sprunk for details.1he solution u is called a finite energy solution if u ∈ L q +1 ( R n ). It is not difficult toverify that u ∈ C ∞ ( R n ) ∩ D , ( R n ) is equivalent to u ∈ C ∞ ( R n ) ∩ L q +1 ( R n ). In addition, k∇ u k L ( R n ) = k u k L q +1 ( R n ) . The classification result by Chen and Li [3] shows that (1.3) has thefinite energy solutions if and only if q = n +2 n − . On the contrary, all the solutions of (1.3) in thecritical case are the finite energy solution.Next, we consider the Lane-Emden system (cid:26) − ∆ u = v q , u, v > in R n , − ∆ v = u q , q , q > . (1.4)Instead of the critical exponent q = n +2 n − , the critical condition which q , q satisfy is1 q + 1 + 1 q + 1 = 1 − n . (1.5)It also comes into play in the study of the existence for (1.4). When q +1 + q +1 ≤ n − n , theexistence of classical positive solutions had been verified by Mitidieri, Serrin and Zou (cf. [19],[23]). Nonexistence of positive solution is still open when q +1 + q +1 > n − n except for the caseof n ≤ u ( x ) = Z R n u q ( y ) dy | x − y | n − α , (1.6)where n ≥ α ∈ (0 , n ), q >
0. It is also invariant under the conformal transformation. Forthe weighted equations, such as the Hardy-Sobolev type, the Caffarelli-Kohn-Nirenberg typeand the weighted Hardy-Littlewood-Sobolev type, the invariant is still true under the scalingtransformation. However, the invariant is absent under the translation. On the other hand, forthe equations involving the Bessel potentials, the invariant is true under the translation, butfalse under the scaling.The following system corresponding (1.6) is related to the study of the extremal functions ofthe Hardy-Littlewood-Sobolev inequality (cf. [17]) u ( x ) = Z R n v q ( y ) dy | x − y | n − α v ( x ) = Z R n u q ( y ) dy | x − y | n − α , (1.7)Recently, [14] shows that the Euler-Lagrange system (1.7) and energy functionals k u k L q ( R n ) and k v k L q ( R n ) are invariant under the scaling transformation u µ ( x ) = µ σ u ( µx ) , v µ ( x ) = µ σ v ( µx ) , (1.8)if and only if the condition q + q = 1 − αn holds. In addition, (1.7) has finite energy solutions ifand only if q , q satisfy such a critical condition. However, it is open whether or not all positivesolutions in the critical case are the finite energy solutions.In this paper, we always assume n ≥ q, q , q >
1. We expect to generalize the argu-ment above to other nonlinear equations, including higher order and fractional order semilinearequations, p-Laplace equation and system, and k-Hessian equations.In Section 2, we point out the relation between finite energy solutions and weak solutions,and prove that the critical conditions are the necessary and sufficient conditions for the existence2f the finite energy solutions of the equations involving the Riesz potentials. For the equationsinvolving the Bessel potentials, we prove that subcritical conditions are the necessary conditionsfor the existence of finite energy solutions. This shows the corresponding energy functional hasno minimizer in critical case. We present the minimum by the least energy whose Euler-Lagrangeequation involves the Riesz potential (cf. Theorem 2.10).In Section 3, we study the Caffarelli-Kohn-Nirenberg type p-Laplacian equation and system,and surprisingly find that the critical condition of the system is degenerate to two simple caseswhen we investigate the invariant of the system and the energy functionals under the scalingtransformation: either p = 2, or the system is reduced to a single equation (cf. Theorem 3.3).Unfortunately, the system has no variational structure, and hence we cannot use the Pohozaevidentity to verify whether or not there exists a nondegenerate critical condition determining theexistence of the finite energy solutions.Finally, in Section 4, we study a k-Hessian equation. We present the nonexistence of negativesolution when the exponent is smaller than the Serrin exponent. In addition, we find a radialsolution with slow decay rate in the supercritical case (cf. Theorem 4.5), and another radialsolution with fast decay rate in the critical case. Based on this result, we prove the criticalcondition is the necessary and sufficient condition for the existence of finite energy solutions (cf.Theorem 4.4). We search the values of q such that the classical Hardy-Sobolev inequality holds( Z R n | x | − t u q +1 dx ) q +1 ≤ C n Z R n |∇ u | dx, (2.1)for all u ∈ D , ( R n ). Here n ≥ t ∈ (0 , u µ (cf. (1.2)), we have( Z R n | x | − t u q +1 µ ( x ) dx ) q +1 = µ σ − n − tq +1 ( Z R n | y | − t u q +1 ( y ) dy ) q +1 ≤ C n µ σ − n − tq +1 ( Z R n |∇ u ( y ) | dy ) / ≤ C n µ n − − n − tq +1 ( Z R n |∇ u µ ( x ) | dx ) / , and hence n − − n − tq +1 = 0, which implies q = n +2 − tn − .The extremal functions in D , ( R n ) \ { } of (2.1) can be obtained by investigating the func-tional E ( u ) = k∇ u k L ( R n ) ( Z R n | x | − t u q +1 dx ) − q +1 . Consider the Euler-Lagrange equation − ∆ u = | x | − t u q , u > in R n . (2.2)In view of − ∆ u µ ( x ) = − µ σ +2 ∆ u ( µx ) = µ σ +2 − t − qσ | x | − t u qµ ( x ) , we can see that σ = − tq − if andonly if u µ solves (2.2). In addition, noting Z R n | x | − t u q +1 µ ( x ) dx = µ σ ( q +1) Z R n | x | − t u q +1 ( µx ) dx = µ σ ( q +1) − n + t Z R n | y | − t u q +1 ( y ) dy, (2.3)3e can see that σ = n − tq +1 if and only if the energy k| x | − tq +1 u k L q +1 ( R n ) is invariant under the scaling(1.2). Eliminating σ we also obtain that q is the critical exponent n +2 − tn − . Theorem 2.1.
Eq. (2.2) has a weak solution in D , ( R n ) if and only if q = n +2 − tn − .Proof. In fact, if q = n +2 − tn − , the radial function u ( x ) = c ( dd + | x | − t ) n − − t (2.4)belongs to D , ( R n ) and solves (2.2). Here c, d > E ( u ), we havethe Pohozaev identity [ ddµ E ( u ( xµ ))] µ =1 = 0 . Noting E ( u ( xµ )) = µ n − − n − t ) q +1 E ( u ( x )) , we get q = n +2 − tn − .Clearly, if q = n +2 − tn − , then u ∈ D , ( R n ) implies | x | − tq +1 u ∈ L q +1 ( R n ) by the Hardy-Sobolevinequality. A natural question is, for a general exponent q , when the energy k| x | − tq +1 u k L q +1 ( R n ) is finite. Proposition 2.2. (1) If u ∈ C ( R n ) ∩ D , ( R n ) solves (2.2), then k| x | − tq +1 u k q +1 L q +1 ( R n ) < ∞ . Inaddition, k| x | − tq +1 u k q +1 L q +1 ( R n ) = k∇ u k L ( R n ) .(2) Assume u ∈ C ( R n ) solves (2.2) and R R n | x | − t u q +1 dx < ∞ . If u ∈ L nn − ( R n ) , then u ∈ D , ( R n ) , and R R n |∇ u | dx = R R n | x | − t u q +1 dx .Proof. (1) Multiplying (2.2) by u and integrating on B R (0), we have Z B R (0) |∇ u | dx − Z ∂B R (0) u∂ ν uds = Z B R (0) | x | − t u q +1 dx. (2.5)By virtue of u ∈ D , ( R n ), there exists R = R j → ∞ such that R Z ∂B R (0) ( |∇ u | + u nn − ) ds → . (2.6)By this result and the H¨older inequality, we get | Z ∂B R (0) u∂ ν uds | ≤ k ∂ ν u k L ( ∂B R (0)) k u k L nn − ( ∂B R (0)) | ∂B R (0) | − n − n → R = R j → ∞ . Inserting this result into (2.5), we can see R R n u q +1 dx | x | t = R R n |∇ u | dx .(2) On the contrary, take smooth function ζ ( x ) satisfying ζ ( x ) = 1 , f or | x | ≤ ζ ( x ) ∈ [0 , , f or | x | ∈ [1 , ζ ( x ) = 0 , f or | x | ≥ . Define the cut-off function ζ R ( x ) = ζ ( xR ) . (2.7)4ultiplying (2.2) by uζ R and integrating on B R (0), we have Z B R (0) |∇ u | ζ R dx = 2 Z B R (0) uζ R ∇ u ∇ ζ R dx + Z B R (0) | x | − t u q +1 ζ R dx. (2.8)Clearly, there exists C > R , such that | Z B R (0) uζ R ∇ u ∇ ζ R dx | ≤ Z B R (0) |∇ u | ζ R dx + C Z B R (0) u |∇ ζ R | dx. If u ∈ L nn − ( R n ), there holds Z B R (0) u |∇ ζ R | dx ≤ CR ( Z B R (0) u nn − dx ) − n | B R (0) | n ≤ C. Inserting these results into (2.8) and noting R R n | x | − t u q +1 dx < ∞ , we get R B R (0) |∇ u | ζ R dx ≤ C, where C > R . Letting R → ∞ , we have ∇ u ∈ L ( R n ), and hence u ∈D , ( R n ). Thus, (2.6) still holds, and from (2.5) we also deduce R R n |∇ u | dx = R R n | x | − t u q +1 dx .The positive solution u ∈ C ( R n ) is called a finite energy solution of (2.2), if Z R n | x | − t u q +1 dx < ∞ . In the critical case q = n +2 − tn − , (2.4) is a finite energy solution. On the contrary, if (2.2) has afinite energy solution with q ≤ n +2 − tn − , then Proposition 2.2 and Theorem 2.1 imply q = n +2 − tn − .The argument above can be generalized to the higher order system involving two coupledequations (cid:26) ( − ∆) l u = | x | − t v q , u > in R n , ( − ∆) l v = | x | − t u q , v > in R n . (2.9)Here l ∈ [1 , n/
2) is an integer.
Proposition 2.3.
Under the scaling transformation (1.8), the equation (2.9) and the energyfunctionals k| x | − tq u k L q ( R n ) and k| x | − tq v k L q ( R n ) are invariant, if and only if q and q satisfy the critical condition q + 1 + 1 q + 1 = n − ln − t . (2.10) Proof.
Set y = µx . By (1.8) and (2.9), we have( − ∆) l u µ ( x ) = µ σ +2 l ( − ∆) l u ( y ) = µ σ +2 l | y | − t v q ( y ) = µ σ +2 l − t − q σ | x | − t v q µ ( x ) . Eq. (2.9) is invariant under the scaling (1.8) implies σ +2 l − t − q σ = 0 and σ +2 l − t − q σ = 0.By the same derivation of (2.3), we also obtain σ ( q + 1) − n + t = 0 and σ ( q + 1) − n + t = 0by the invariant of k| x | − tq u k L q ( R n ) and k| x | − tq v k L q ( R n ) . Eliminating σ and σ , we cansee q q − q +1)( q +1) = l − tn − t . In view of q q − q + 1)( q + 1) − ( q + 1) − ( q + 1), it follows(2.10).On the contrary, the calculation above still implies the sufficiency.5t seems difficult to generalized this process to the system involving m equations with m ≥ m equations is an interestingproblem.The classical solutions u, v of (2.9) are called finite energy solutions if k| x | − tq u k L q ( R n ) < ∞ , k| x | − tq v k L q ( R n ) < ∞ . Theorem 2.4.
Eq. (2.9) has finite energy solutions if and only if (2.10) holds.Proof.
When q +1 + q +1 = n − l + β + β n , Lieb [17] obtained a pair of extremal functions ( U, V ) ∈ L q +1 ( R n ) × L q +1 ( R n ) of the weighted Hardy-Littlewood-Sobolev inequality, which solves theintegral system U ( x ) = 1 | x | β Z R n V q ( y ) | y | β | x − y | n − l dyV ( x ) = 1 | x | β Z R n U q ( y ) | y | β | x − y | n − l dy. If (2.10) is true, we can choose β and β satisfying β ( q + 1) = β ( q + 1) = t . Taking u ( x ) = | x | β U ( x ) and v ( x ) = | x | β V ( x ), we can see that ( u, v ) solves u ( x ) = Z R n v q ( y ) | y | t | x − y | n − l dyv ( x ) = Z R n u q ( y ) | y | t | x − y | n − l dy. (2.11)In addition, according to the radial symmetry and integrability results (cf. [11], [12]) and theasymptotic behavior of ( U, V ) (cf. [15]), u and v are finite energy solutions. By the propertiesof the Riesz potentials, it follows that ( u, v ) solves (2.9) from (2.11).On the contrary, according to the equivalence results in [4], the classical solutions of (2.9)also satisfy (2.11). In the following, we use the Pohozaev identity of integral forms introducedin [1] to deduce (2.10).By (2.11) we have Z R n u q +1 ( x ) | x | t dx = Z R n u q ( x ) | x | t Z R n v q ( y ) | y | t | x − y | n − l dydx = Z R n v q ( y ) | y | t Z R n u q ( x ) | x − y | n − l dxdy = Z R n v q +1 ( y ) | y | t dy. (2.12)For µ >
0, from (2.11) it follows x · ∇ u ( x ) = ddµ u ( µx ) | µ =1 = (2 l − t ) u ( x ) + Z R n z · ∇ v q ( z ) dz | z | t | x − z | n − l Multiplying by | x | − t u q ( x ) and integrating on R n , we get Z R n u q ( x ) | x | t x · ∇ u ( x ) dx − (2 l − t ) Z R n u q +1 ( x ) | x | t dx = Z R n u q ( x ) | x | t Z R n z · ∇ v q ( z ) dz | z | t | x − z | n − l dx = Z R n z · ∇ v q ( z ) | z | t Z R n u q ( x ) | x | t | z − x | n − l dxdz = Z R n z · ∇ v q ( z ) | z | t v ( z ) dz. (2.13)6ince u, v are finite energy solutions, we can find R = R j → ∞ such that R − t Z ∂B R (0) ( u q +1 + v q +1 ) ds → . Thus, integrating (2.13) by parts yields( t − nq + 1 − l + t ) Z R n u q +1 | x | t dx = q ( t − n ) q + 1 Z R n v q +1 | x | t dx. Combining with (2.12) we obtain (2.10).
Remark 2.1.
We have two direct corollaries:(1) If l = 1, then (2.9) has finite energy solutions if and only if q +1 + q +1 = n − n − t .(2) If q = q and u = v , then (2.9) has finite energy solution if and only if q = n +2 l − tn − l .Noting the conditions in Theorems 2.1 and 2.4, it is convenient for us to discuss the finiteenergy solutions for integral equations, and the weak solutions in D l, ( R n ) for the differentialequations respectively. Let 1 < r, s < ∞ , 0 < λ < n , β + β ≥ β + β ≤ α . The weighted Hardy-Littlewood-Sobolev (WHLS) inequality states that (cf. [27]) (cid:12)(cid:12)(cid:12)(cid:12)Z R n Z R n f ( x ) g ( y ) | x | β | x − y | n − α | y | β dxdy (cid:12)(cid:12)(cid:12)(cid:12) ≤ C β ,β ,s,α,n k f k r k g k s (2.14)where 1 − r − n − αn < β n < − r . If the inequality (2.14) still holds for the scaling functions f µ and g µ (cf. (1.8)), then we can deduce1 r + 1 s + n − α + β + β n = 2 . (2.15)In order to obtain the sharp constant in the WHLS inequality (2.14), we maximize thefunctional J ( f, g ) = ( k f k r k g k s ) − Z R n Z R n f ( x ) g ( y ) | x | β | x − y | n − α | y | β dxdy. The corresponding Euler-Lagrange equations are the following integral system: λ rf ( x ) r − = 1 | x | β Z R n g ( y ) | y | β | x − y | n − α dy,λ sg ( x ) s − = 1 | x | β Z R n f ( y ) | y | β | x − y | n − α dy, where f, g ≥
0, and λ r = λ s = J ( f, g ).If f ∈ L r ( R n ) and g ∈ L s ( R n ), then the Pohozaev identity dJ ( f ( xµ − ) ,g ( xµ − )) dµ | µ =1 = 0 stillimplies (2.15). In fact, for µ > J ( f ( xµ ) , g ( xµ )) = µ n − ( n − α + β + β ) − nr − ns J ( f ( x ) , g ( x )) . Thus, dJ ( f ( x/µ ) ,g ( x/µ )) dµ | µ =1 = 0 leads to (2.15). 7et u = c f r − , v = c g s − , and q = r − , q = s − . Choosing suitable c and c , we obtainthat the corresponding Euler-Lagrange equations are the following integral system u ( x ) = 1 | x | β Z R n v q ( y ) | y | β | x − y | n − α dyv ( x ) = 1 | x | β Z R n u q ( y ) | y | β | x − y | n − α dy (2.16)where β + β ≤ α , and u, v ≥ , < q , q < ∞ , < α < n, β ≥ , β ≥ ,β n < q + 1 < n − α + β n , β n < q + 1 < n − α + β n . (2.17)The equation (2.16) and the energy functionals k u k L q ( R n ) , k v k L q ( R n ) are invariant underthe scaling transformation (1.8), if and only if1 q + 1 + 1 q + 1 = n − α + β + β n . (2.18)Clearly, (2.18) is equivalent to (2.15).Since ( f, g ) ∈ L r ( R n ) × L s ( R n ) implies ( u, v ) ∈ L q +1 ( R n ) × L q +1 ( R n ), we call such a pairof solutions ( u, v ) the finite energy solutions . In addition, (2.18) is called the critical condition . Theorem 2.5.
Eq. (2.16) has the finite energy positive solutions in C loc ( R n \ { } ) if and onlyif p, q satisfy the critical condition (2.18).Proof. Sufficiency. According to [17], the existence of the extremal functions of the WHLS inequality implies ourconclusion. In fact, those extremal functions are finite energy solutions. By a regularity liftingprocess, the extremal functions also belong to C loc ( R n \ { } ). Necessity.
Denote n − α + β + β by ¯ λ . For x = 0 and µ >
0, we have u ( µx ) = Z R n v q ( y ) dy | µx | β | µx − y | n − α | y | β = µ n − ¯ λ Z R n v q ( µz ) dz | x | β | x − z | n − α | z | β . Differentiating with respect to µ and then letting µ = 1, we get x · ∇ u ( x ) = ( n − ¯ λ ) u + lim d → Z R n \ B d (0) z · ∇ v q ( z ) dz | x | β | x − z | n − α | z | β . (2.19)Multiplying by u q and integrating on R n \ B d (0), we havelim d → Z R n \ B d (0) u q ( x · ∇ u ( x )) dx = ( n − ¯ λ ) Z R n u q +1 ( x ) dx + lim d → Z R n \ B d (0) u q ( x ) Z R n \ B d (0) z · ∇ v q ( z ) dz | x | β | x − z | n − α | z | β dx. (2.20)Integrating by parts, we get K := lim d → Z R n \ B d (0) u q ( x · ∇ u ( x )) dx = lim d → q + 1 Z R n \ B d (0) ( x · ∇ u q +1 ( x )) dx = lim r →∞ rq + 1 Z ∂B r (0) u q +1 ( x ) ds − lim d → dq + 1 Z ∂B d (0) u q +1 ( x ) dx − nq + 1 Z R n u q +1 ( x ) dx.
8n view of u ∈ L q +1 ( R n ), we can find r = r j → ∞ and d = d m → ∞ such thatlim r →∞ r Z ∂B r (0) u q +1 ( x ) ds = lim d → d Z ∂B d (0) u q +1 ( x ) dx = 0 , and hence K = − nq + 1 Z R n u q +1 ( x ) dx. Using the Fubini theorem, we have K := lim d → Z R n \ B d (0) u q ( x ) Z R n \ B d (0) z · ∇ v q ( z ) dz | x | β | x − z | n − α | z | β dx = lim d → Z R n \ B d (0) z · ∇ v q ( z ) Z R n \ B d (0) u q ( x ) dx | z | β | z − x | n − α | x | β dz = lim d → Z R n \ B d (0) ( z · ∇ v q ( z )) v ( z ) dz. Similar to the calculation of K , we also obtain K = − q nq + 1 Z R n v q +1 ( z ) dz. Inserting K and K into (2.20), we have − nq + 1 Z R n u q +1 ( x ) dx = ( n − ¯ λ ) Z R n u q +1 ( x ) dx − q nq + 1 Z R n v q +1 ( z ) dz. By (2.16) and the Fubini theorem, we also have Z R n u q +1 ( x ) dx = Z R n u q ( x ) u ( x ) dx = Z R n u q ( x ) Z R n v q ( y ) | x | β | x − y | n − α | y | β dx = Z R n v q ( y ) Z R n u q ( x ) dx | y | β | y − x | n − α | x | β dy = Z R n v q ( y ) v ( y ) dy = Z R n v q +1 ( y ) dy. Combining two results above yields (2.18).
Same as (1.3), the fractional order equation( − ∆) α/ u = u q , u > in R n , (2.21)is still invariant under the conformal transformation as long as q = n + αn − α . Here the fractionalorder differential operator ( − ∆) α/ can be defined via the properties of the Riesz potential (cf.[26]). According to the results in [4] and [5], it is equivalent to the integral equation (1.6). Inaddition, the fact δ − α/ = c α R ∞ exp ( − tδ ) t α/ dtt shows that the kernel of the Riesz potential canbe written as a static heat kernel. Namely, besides (1.6), we obtain another integral equationwhich is equivalent to (2.21): u ( x ) = Z R n u q ( y ) Z ∞ (4 πt ) α − n exp( − | x − y | t ) dtt dy. (2.22)9f replacing the static heat kernel H ( x ) = c α R ∞ (4 πt ) α − n exp( − | x − y | t ) dtt by the Bessel kernel g α ( x ) = c α R ∞ (4 πt ) α − n exp( − | x − y | t − t π ) dtt , then we have a new integral equation u ( x ) = Z R n g α ( x − y ) u q ( y ) dy, u > in R n , (2.23)which is equivalent to the fractional order equation (cf. [9])( id − ∆) α/ u = u q , u > in R n . (2.24)This equation is not invariant under the scaling (1.2).When α = 2, (2.24) becomes a semilinear equation − ∆ u + u = u q , u > in R n . (2.25)Here q >
1. It can be used to describe the solitary wave of the Schr¨odinger equation. A knownresult implied in Chapter 8 of [2] is q < n +2 n − (namely q is subcritical) if u ∈ H ( R n ) is a weaksolution of (2.25).Next, we investigate the relation between weak solutions and finite energy solutions of (2.25). Proposition 2.6.
Assume u is a positive solution of (2.25). Then u ∈ C ( R n ) ∩ L q +1 ( R n ) ifand only if u ∈ H ( R n ) . In addition, k u k H ( R n ) = k u k q +1 L q +1 ( R n ) .Proof. Step 1. If u ∈ H ( R n ) is a weak solution, then u ∈ C ( R n ) (cf. [2]). Testing by uζ R yields Z B R (0) ∇ u ∇ ( uζ R ) dx + Z B R (0) u ζ R dx = Z B R (0) u q +1 ζ R dx. (2.26)Therefore, by the H¨older inequality, from u ∈ H ( R n ) we deduce that R B R (0) u q +1 ζ R dx ≤ C, where C > R . Letting R → ∞ , we get u ∈ L q +1 ( R n ). Step 2.
On the contrary, if u ∈ C ( R n ) ∩ L q +1 ( R n ), multiplying by uζ R and integrating on B R (0), we also have (2.26). In view of u ∈ L q +1 ( R n ), it follows Z B R (0) u |∇ ζ R | dx ≤ k u k L q +1 ( R n ) ( Z B R (0) |∇ ζ R | q +1) q − dx ) q − q +1 ≤ CR n ( q − q +1 − . Since q is subcritical, lim R →∞ R B R (0) u |∇ ζ R | dx = 0. Thus, we can easily see u ∈ H ( R n ). Step 3.
We claim k u k H ( R n ) = k u k q +1 L q +1 ( R n ) . In fact, under each assumption, u ∈ C ( R n ).Multiplying (2.25) by u , we get Z B R (0) ( |∇ u | + | u | ) dx = Z B R (0) u q +1 dx + Z ∂B R (0) u∂ ν uds. (2.27)By virtue of u ∈ H ( R n ) ∩ L q +1 ( R n ), we can find R = R j → ∞ such that R Z ∂B R (0) ( |∇ u | + u q +1 ) ds → . Therefore, by the H¨older inequality, we have | Z ∂B R (0) u∂ ν uds | ≤ ( R Z ∂B R (0) |∇ u | dx ) ( R Z ∂B R (0) u q +1 dx ) q +1 | ∂B R | − q +1 ≤ CR ( n − − q +1 ) − ( + q +1 ) . q is subcritical, lim R →∞ | R ∂B R (0) u∂ ν uds | = 0. Inserting this into (2.27), we obtain Z R n ( |∇ u | + | u | ) dx = Z R n u q +1 dx. Theorem 2.6 is proved.
Proposition 2.7.
Assume u solves (2.23), then u ∈ H α/ ( R n ) if and only if u ∈ L q +1 ( R n ) . Inaddition, k u k H α/ ( R n ) = k u k q +1 L q +1 ( R n ) .Proof. From (2.23), we have ˆ u ( ξ ) = ˆ g α ( ξ )( u q ) ∧ ( ξ ), or (1+4 π | ξ | ) α ˆ u ( ξ ) = ( u q ) ∧ ( ξ ). Multiplyingby ¯ˆ u and using the Parseval identity, we get Z R n (1 + 4 π | ξ | ) α | ˆ u | dξ = Z R n ( u q ) ∧ ¯ˆ udξ = Z R n u q +1 dx. Therefore, the proof is easy to complete.
Theorem 2.8. (1) If q < n + αn − α , (2.23) has a positive solution in L q +1 ( R n ) . Moreover, if α > ,then u ∈ C ( R n ) .(2) If (2.23) has a positive solution u ∈ C ( R n ) ∩ L q +1 ( R n ) , then q < n + αn − α .Proof. (1) By the analogous argument of the existence of ground state in [7], we can find acritical point of E ( u ) = 12 Z R n (1 + 4 π | ξ | ) α/ | ˆ u ( ξ ) | dξ − Z R n u q +1 ( x ) q + 1 dx on the Nihari manifold { u ∈ H α/ ( R n ) \ { } ; E ′ ( u ) u = 0 } . According to Proposition 2.7, u ∈ L q +1 ( R n ). This implies the existence of the weak solution of (2.24). According to theequivalence, (2.23) also has a finite energy solution. In addition, u is radially symmetric anddecreasing about some point in R n (cf. [18]). In the same way to lift regularity process in [13],we can also deduce the regularity of the solution from H α/ ( R n ) to C ( R n ) by virtue of α > u ( µx ) = µ α Z R n u q ( µy ) Z ∞ (4 πt ) α − n exp( − | x − y | t − µ t π ) dtt dy. Thus, x · ∇ u ( x ) = [ du ( µx ) dµ ] µ =1 = αu + Z R n y · ∇ u q ( y ) g α ( x − y ) dy − Z R n u q ( y ) Z ∞ (4 πt ) α − n exp( − | x − y | t − t π ) t π dtt dy. Multiplying by u q ( x ) and integrating, we get1 q + 1 Z R n x · ∇ u q +1 ( x ) dx = α Z R n u q +1 ( x ) dx + qq + 1 Z R n y · ∇ u q +1 ( y ) dy − Z R n Z R n u q ( x ) u q ( y ) Z ∞ (4 πt ) α − n exp( − | x − y | t − t π ) t π dtt dxdy. u ∈ L q +1 ( R n ), we can find R = R j → ∞ such that R R ∂B R (0) u q +1 ds →
0. Thus, the resultabove leads to ( q − q + 1 n − α ) Z R n u q +1 dx = − Z R n Z R n u q ( x ) u q ( y ) Z ∞ (4 πt ) α − n exp( − | x − y | t − t π ) t π dtt dxdy. Since the right hand side is positive, we can deduce that q − q +1 n − α >
0, which implies q < n + αn − α .Consider the system (cid:26) ( id − ∆) α/ u = v q , u > in R n , ( id − ∆) α/ v = u q , v > in R n , (2.28)where n ≥ α ∈ (0 , n ), q , q > u, v of (2.28) in H α/ ( R n ), for all φ ∈ H α/ ( R n ),there hold Re Z R n (1 + 4 π | ξ | ) α/ ˆ u ( ξ ) ¯ˆ φ ( ξ ) dξ = Z R n v q ( x ) φ ( x ) dx,Re Z R n (1 + 4 π | ξ | ) α/ ˆ v ( ξ ) ¯ˆ φ ( ξ ) dξ = Z R n u q ( x ) φ ( x ) dx. Here ˆ u is the Fourier transformation of u . Theorem 2.9.
If (2.28) has weak positive solutions in H α/ ( R n ) . Then q + 1 + 1 q + 1 > n − αn . (2.29) Proof.
Testing (2.28) by u and v respectively, we have Re Z R n (1 + 4 π | ξ | ) α/ ˆ u ( ξ )¯ˆ v ( ξ ) dξ = Z R n v q +1 ( x ) dx,Re Z R n (1 + 4 π | ξ | ) α/ ˆ v ( ξ )¯ˆ u ( ξ ) dξ = Z R n u q +1 ( x ) dx. Since the left hand sides of two equalities above are equal, it follows Re Z R n (1 + 4 π | ξ | ) α/ ˆ u ( ξ )¯ˆ v ( ξ ) dξ = Z R n v q +1 ( x ) dx = Z R n u q +1 ( x ) dx. (2.30)On the other hand, the positive weak solutions u, v are the critical points of the functional E ( u, v ) = Re Z R n (1 + 4 π | ξ | ) α/ ˆ u ( ξ )¯ˆ v ( ξ ) dξ − Z R n ( u q +1 q + 1 + v q +1 q + 1 ) dx. Thus, the Pohozaev identity [ ddµ E ( u ( xµ ) , v ( xµ ))] µ =1 = 0 holds. By virtue of E ( u ( xµ ) , v ( xµ )) = µ n − α Re Z R n (1 + 4 π | ζ | ) α/ ˆ u ( ζ )¯ˆ v ( ζ ) dζ − µ n Z R n ( u q +1 q + 1 + v q +1 q + 1 ) dy, n − α ) Re Z R n (1 + 4 π | ζ | ) α/ ˆ u ( ζ )¯ˆ v ( ζ ) dζ + αRe Z R n (1 + 4 π | ζ | ) ( α − / ˆ u ( ζ )¯ˆ v ( ζ ) dζ = n Z R n ( u q +1 q + 1 + v q +1 q + 1 ) dy. Combining with (2.30), we get αRe Z R n (1 + 4 π | ζ | ) α − ˆ u ( ζ )¯ˆ v ( ζ ) dζ = [ n ( 1 q + 1 + 1 q + 1 ) − ( n − α )] Z R n u q +1 ( x ) dx. (2.31)We claim that the left hand side of (2.31) is positive. In fact, set w ( x ) = Z R n g ( x − y ) v ( y ) dy. Then, w > H α/ ( R n ), and ˆ w = (1 + 4 π | ξ | ) − ˆ v. Testing (2.28) by w yields Re Z R n (1 + 4 π | ξ | ) α/ ˆ u ¯ˆ wdξ = Z R n v q wdx, which implies Re Z R n (1 + 4 π | ξ | ) ( α − / ˆ u ¯ˆ vdξ > . Combining this result with (2.31), we see the subcritical condition (2.29). Theorem 2.9 is proved.
Remark 2.2.
If we prove the second conclusion of Theorem 2.8 by the same way of Theorem2.9, the assumption of u ∈ C ( R n ) can be removed. Consider the minimum of the following energy functional in H α/ ( R n ) \ { } E ( u ) = 12 Z R n (1 + 4 π | ξ | ) α/ | ˆ u ( ξ ) | dξ − α ∗ Z R n u α ∗ ( x ) dx, where α ∗ = nn − α . Clearly, α ∗ − § E ( u ) has no minimizer in H α/ ( R n ) \ { } in thecritical case. However, the radial function U ∗ ( x ) = a ( bb + | x − x | ) ( n − α ) / , a, b > and x ∈ R n is the extremal the Hardy-Littlewood-Sobolev inequality (cf. [17]). Furthermore, according tothe classification results in [5] and [16], the radial function U ∗ is the unique solution of (1.6). Inaddition, it is also the extremal function in D α/ , ( R n ) \ { } of the functional E ∗ ( u ) = (cid:20)Z R n | ( − ∆) α/ u | dx (cid:21) (cid:20)Z R n | u | n/ ( n − α ) dx (cid:21) ( α − n ) /n . D α/ , ( R n ) to L n/ ( n − α ) ( R n ): c ( Z R n | u | n/ ( n − α ) dx ) ( n − α ) /n ≤ Z R n | ( − ∆) α/ u | dx. The following result shows the relation between the energy functionals involving the Rieszpotential and the Bessel potential in the critical case.
Theorem 2.10. inf { E ( u ); u ∈ H α/ ( R n ) \ { }} = α n [ E ∗ ( U ∗ )] n/α .Proof. The ideas in [2] and [10] are used here.Write the scaling function u λt,s ( x ) = e tλ u ( e − sλ x ) , where λ ≥ t ≥ t + s > µ := 2 t + ( n − s ≥
0, and ν := 2 t + ns ≥
0. Set ¯ µ = max { µ, ν } .By a simply calculation, we have K ( u ) := dE ( u λt,s ) dλ | λ =0 = µ Z R n (1 + 4 π | ξ | ) α | ˆ u ( ξ ) | dξ + sα Z R n (1 + 4 π | ξ | ) α − | ˆ u ( ξ ) | dξ − µ Z R n u α ∗ ( x ) dx = ν Z R n (1 + 4 π | ξ | ) α | ˆ u ( ξ ) | dξ − sα Z R n (1 + 4 π | ξ | ) α − π | ξ | | ˆ u ( ξ ) | dξ − µ Z R n u α ∗ ( x ) dx. Similarly, if we set E ( u ) = 12 Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ − α ∗ Z R n u α ∗ ( x ) dx, then K ( u ) := dE ( u λt,s ) dλ | λ =0 = µ Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ − µ Z R n u α ∗ ( x ) dx. Write L ( u ) := E ( u ) − K ( u )¯ µ = sα Z R n (1 + 4 π | ξ | ) α − π | ξ | | ˆ u ( ξ ) | dξ + ( µ ν − α ∗ ) Z R n u α ∗ dx, µ < ν ; − sα µ Z R n (1 + 4 π | ξ | ) α − | ˆ u ( ξ ) | dξ + ( 12 − α ∗ ) Z R n u α ∗ dx, µ ≥ ν, and L ( u ) := E ( u ) − K ( u )¯ µ = ( 12 − µ ν ) Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ + ( µ ν − α ∗ ) Z R n u α ∗ dx, µ < ν ;( 12 − α ∗ ) Z R n u α ∗ dx, µ ≥ ν. µ < ν , s > µ ν > α ∗ ; when µ ≥ ν , s ≤
0. Thus, L ( u ) , L ( u ) ≥ m = inf { E ( u ); K ( u ) = 0 , u ∈ H α/ ( R n ) \ { }} ;¯ m = inf { L ( u ); K ( u ) ≤ , u ∈ H α/ ( R n ) \ { }} ; m = inf { E ( u ); K ( u ) = 0 , u ∈ D α/ , ( R n ) \ { }} ;¯ m = inf { L ( u ); K ( u ) < , u ∈ D α/ , ( R n ) \ { }} . Clearly, m = ¯ m . Set F = { u ∈ D α/ , ( R n ) \ { } ; K ( u ) < } , ˜ F = { u ∈ D α/ , ( R n ) \ { } ; K ( u ) ≤ } , ¯ F = { u ∈ D α/ , ( R n ) \ { } ; K ( u ) = 0 } . We claim F = ∪ λ> { u ∈ D α/ , ( R n ) \ { } ; K ( u λt,s ) = 0 } . Once it holds, then m = ¯ m .In fact, for any λ >
0, if K ( u λt ′ ,s ′ ) = 0, then e [2 t ′ +( n − α ) s ′ ] λ Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ = e [2 t ′ +( n − α ) s ′ ] nλn − α Z R n u α ∗ dx. This leads to K ( u ) <
0, and hence F ⊃ ∪ λ> { u ∈ D α/ , ( R n ) \ { } ; K ( u λt,s ) = 0 } .On the other hand, for any u ∈ F , there holds R R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ < R R n u α ∗ dx . Thus, wecan find λ ∗ > e [2 t ′ +( n − α ) s ′ ] λ ∗ Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ = e [2 t ′ +( n − α ) s ′ ] nλ ∗ n − α Z R n u α ∗ dx. This shows u ∈ { u ∈ D α/ , ( R n ) \ { } ; K ( u λ ∗ t,s ) = 0 } .In addition, it is easy to see that F is dense in ˜ F , which implies¯ m = inf { L ( u ); K ( u ) ≤ , u ∈ D α/ , ( R n ) \ { }} . (2.32)Set G = { u ∈ H α/ ( R n ) \ { } ; K ( u ) ≤ } . Clearly, G ⊂ ˜ F .Noting K ( u λt ′ ,s ′ ) = µ e λ (2 t ′ +( n − α ) s ′ ) Z R n ( e sλ + 4 π | ξ | ) α | ˆ u ( ξ ) | dξ + sα e λ (2 t ′ +( n − α +2) s ′ ) Z R n ( e sλ + 4 π | ξ | ) α − | ˆ u ( ξ ) | dξ − µ e λ ( α ∗ t ′ + ns ′ ) Z R n u α ∗ ( x ) dx. we can deduce by taking t ′ = n − α and s ′ = − λ → + ∞ K ( u λ ( n − α ) / , − ) = K ( u ) , (2.33)Similarly, we also get lim λ → + ∞ L ( u λ ( n − α ) / , − ) = L ( u ) , (2.34)15learly, (2.33) shows that G is dense in ˜ F . Combining with (2.32) yields¯ m = inf { L ( u ); K ( u ) ≤ , u ∈ D α/ , ( R n ) \ { }} . In addition, (2.34) impliesinf { L ( u ); K ( u ) ≤ , u ∈ D α/ , ( R n ) \ { }} = ¯ m. Therefore, ¯ m = ¯ m .The argument above shows that m = ¯ m = m = ¯ m .Take t = 0, then α ∗ µ = 2 ν . Thus, m = m = inf (cid:26) α n Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ ; Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ = Z R n u α ∗ ( x ) dx (cid:27) = inf α n Z R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ (cid:20) R R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ R R n u α ∗ ( x ) dx (cid:21) n − αα ; u ∈ D α/ , ( R n ) \ { } = α n inf ((cid:20) R R n (2 π | ξ | ) α | ˆ u ( ξ ) | dξ ( R R n u α ∗ ( x ) dx ) ( n − α ) /n (cid:21) nα ; u ∈ D α/ , ( R n ) \ { } ) = α n c n/α ∗ . Here c ∗ is the sharp constant of the inequality c ( Z R n u α ∗ ( x ) dx ) ( n − α ) /n ≤ Z R n | ( − ∆) α/ u ( x ) | dx. According to the classification result in [5], we know that the corresponding minimizer in D α/ , ( R n ) \ { } is U ∗ . Consider the Caffarelli-Kohn-Nirenberg inequality( Z R n | u | q +1 | x | b ( q +1) dx ) p/ ( q +1) ≤ C a,b Z R n |∇ u | p | x | ap dx, where n ≥ p >
1, 0 ≤ a < n − pp , and a ≤ b ≤ a + 1. Since the scaling function u µ ( x ) alsosatisfies this inequality, by a simple calculation we can see q = npn − p + p ( b − a ) − D ,pa ( R n ) \ { } can be obtained by investigating the functional E ( u ) = Z R n |∇ u | p | x | ap dx ( Z R n | u | q +1 | x | b ( q +1) dx ) − p/ ( q +1) . Here D ,pa ( R n ) is the completion of C ∞ ( R n ) with respect to the norm k| x | − a ∇ u k L p ( R n ) . Clearly,the extremal function satisfies the Pohozaev identity ddµ E ( u ( xµ )) | µ =1 = 0 . Noting E ( u ( xµ )) = µ n − p ( a +1) − pnq +1 + pb E ( u ( x )) , we also obtain q = npn − p + p ( b − a ) − E ( u ) satisfies: − div ( 1 | x | ap |∇ u | p − ∇ u ) = 1 | x | b ( q +1) u q , u > in R n . (3.1)By a direct calculation we also deduce that (3.1) and the energy R R n | u | q +1 | x | b ( q +1) dx are invariantunder the scaling (1.2) if and only if q = npn − p + p ( b − a ) −
1. If a = b = 0, this exponent q = npn − p − Theorem 3.1. (1) If u ∈ D ,pa ( R n ) is a weak solution of (3.1), then | x | − b u ∈ L q +1 ( R n ) .Moreover, if u ∈ C ( R n ) , then k| x | − a ∇ u k pL p ( R n ) = k| x | − b u k q +1 L q +1 ( R n ) . (3.2) (2) On the contrary, assume u ∈ C ( R n ) solves (3.1), and | x | − b u ∈ L q +1 ( R n ) . If Z R n ( u | x | b ) npn + p ( b − a − dx < ∞ . then u ∈ D ,pa ( R n ) and (3.2) still holds.Proof. Step 1. If u ∈ D ,pa ( R n ) is a weak solution of (3.1), testing by uζ pR yields Z R n | x | − ap |∇ u | p − ∇ u ∇ ( uζ pR ) dx = Z R n | x | − b ( q +1) u q +1 ζ pR dx. (3.3)By the Young inequality, it follows Z R n | x | − b ( q +1) u q +1 ζ pR dx ≤ C Z R n | x | − ap |∇ u | p ζ pR dx + C Z R n | x | − ap u p |∇ ζ R | p dx. Using the H¨older inequality and the Caffarelli-Kohn-Nirenberg inequality, we get Z R n u p | x | ap |∇ ζ R | p dx ≤ CR p [ Z R n ( u | x | b ) npn + p ( b − a − dx ] − p (1+ a − b ) n ( Z B R (0) | x | ( b − a ) na +1 − b dx ) p (1+ a − b ) n ≤ C [ Z R n ( u | x | b ) npn + p ( b − a − dx ] − p (1+ a − b ) n ≤ C Z R n | x | − ap |∇ u | p dx. Combining two results above and letting R → ∞ , we obtain | x | − b u ∈ L q +1 ( R n ).In addition, if u ∈ C ( R n ) solves (3.1), multiplying by u and integrating on B R (0), we have Z B R (0) | x | − ap |∇ u | p dx − Z ∂B R (0) | x | − ap u |∇ u | p − ∂ ν uds = Z B R (0) | x | − b ( q +1) u q +1 dx. (3.4)Using the H¨older inequality, we get | Z ∂B R (0) | x | − ap u |∇ u | p − ∂ ν uds |≤ ( Z ∂B R (0) | x | − ap |∇ u | p ds ) − p [ Z ∂B R ( u | x | b ) npn − p ( a +1 − b ) ds ] p − (1+ a − b ) n · ( Z ∂B R (0) | x | ( b − a ) na +1 − b ds ) a +1 − bn ≤ C ( R Z ∂B R (0) | x | − ap |∇ u | p ds ) − p [ R Z ∂B R ( u | x | b ) npn − p ( a +1 − b ) ds ] p − (1+ a − b ) n · R [ n − ( b − a ) na +1 − b ] a +1 − bn − p − p + a +1 − bn . (3.5)17n view of u ∈ D ,pa ( R n ), by the Caffarelli-Kohn-Nirenberg inequality, there holds Z R n ( u | x | b ) npn − p ( a +1 − b ) dx < ∞ , and hence we can find R = R j → ∞ such that R Z ∂B R (0) | x | − ap |∇ u | p ds + R Z ∂B R (0) ( u | x | b ) npn − p ( a +1 − b ) ds → . Therefore, it follows from (3.5) that | Z ∂B R (0) | x | − ap u |∇ u | p − ∂ ν uds | → R → ∞ . Inserting this into (3.4) and letting R = R j → ∞ , we get (3.2). Step 2. If u ∈ C ( R n ), multiplying (3.1) by uζ pR and integrating, we also obtain (3.3). Usingthe Young inequality, we get Z R n |∇ u | p | x | ap ζ pR dx ≤ Z R n u q +1 | x | b ( q +1) ζ pR dx + 12 Z R n |∇ u | p | x | ap ζ pR dx + C Z R n u p | x | ap |∇ ζ R | p dx. This result, together with | x | − b u ∈ L q +1 ( R n ), implies Z R n |∇ u | p | x | ap ζ pR dx ≤ C + C Z R n u p | x | ap |∇ ζ R | p dx. (3.6)If R R n ( u | x | b ) npn + p ( b − a − dx < ∞ , Z R n u p | x | ap |∇ ζ R | p dx ≤ R p ( Z R n ( u | x | b ) npn − p ( a +1 − b ) dx ) − p ( a +1 − b ) n ( Z B R | x | n ( b − a ) a +1 − b dx ) p ( a +1 − b ) n ≤ C. Thus, we can see R R n u p | x | ap |∇ ζ R | p dx < ∞ . Inserting this into (3.6), we have | x | − a ∇ u ∈ L p ( R n ),and hence u ∈ D ,pa ( R n ). Similar to Step 1, we also obtain (3.2). Theorem 3.1 is proved. Theorem 3.2.
Eq. (3.1) has a solution in C ( R n ) ∩ D ,pa ( R n ) if and only if q = npn − p + p ( b − a ) − .Proof. If q = npn − p + p ( b − a ) −
1, we know that the following extremal function of the Caffarelli-Kohn-Nirenberg inequality is a finite energy solution U a,b ( x ) = c ( n − p − pa | x | p ( n − p − pa )(1+ a − b )( p − n − (1+ a − b ) p ] ) n − p (1+ a − b ) p (1+ a − b ) with c = [ n ( p − − p ( n − p (1 + a − b )) − ] n − p (1+ a − b ) p a − b ) . We verify the sufficiency.Next, we prove the necessity. Multiplying (3.1) by ( x · ∇ u ) and integrating on B R (0), we get Z B R (0) |∇ u | p − | x | ap ∇ u ∇ ( x · ∇ u ) dx − Z ∂B R (0) |∇ u | p − | x | ap | ∂ ν u | ds = Z B R (0) u q ( x · ∇ u ) | x | b ( q +1) dx. (3.7)Integrating by parts, we obtain Z B R (0) |∇ u | p − | x | ap ∇ u ∇ ( x · ∇ u ) dx = Z B R (0) | x | − ap [ |∇ u | p + 1 p x · ∇ ( |∇ u | p )] dx = Z B R (0) |∇ u | p | x | ap + Rp Z ∂B R (0) |∇ u | p | x | ap ds − n − app Z B R (0) |∇ u | p | x | ap dx, (3.8)18nd Z B R (0) u q ( x · ∇ u ) | x | b ( q +1) dx = 1 q + 1 Z B R (0) x · ∇ u q +1 | x | b ( q +1) dx = Rq + 1 Z ∂B R (0) u q +1 | x | b ( q +1) ds − n − b ( q + 1) q + 1 Z B R (0) u q +1 | x | b ( q +1) dx. (3.9)According to the first conclusion of Theorem 3.1, ∇ u | x | a ∈ L p ( R n ) implies u | x | b ∈ L q +1 ( R n ). Wecan find R = R j → ∞ such that R Z ∂B R (0) |∇ u | p | x | ap ds + R Z ∂B R (0) u q +1 | x | b ( q +1) ds → . (3.10)Inserting (3.8) and (3.9) into (3.7), and using (3.10), we have(1 − n − app ) Z R n |∇ u | p | x | ap dx = − n − b ( q + 1) q + 1 Z R n u q +1 | x | b ( q +1) dx. Combining with (3.2) yields 1 − n − app = − n − b ( q +1) q +1 , which implies q = npn − p + p ( b − a ) − (cid:26) − div ( | x | − ap |∇ u | p − ∇ u ) = | x | − b ( q +1) v q , u > in R n ; − div ( | x | − ap |∇ v | p − ∇ v ) = | x | − b ( q +1) u q , v > in R n . (3.11) Theorem 3.3.
Under the scaling transformation (1.8), the system (3.11) and the energy func-tionals Z R n u q +1 | x | b ( q +1) dx and Z R n v q +1 | x | b ( q +1) dx, are invariant if and only if one of the two degenerate conditions holds: p = 2 and q = q .Moreover, if p = 2 , the critical condition is q + 1 + 1 q + 1 = n + 2( b − a − n . (3.12) If q = q , the critical condition is q = q = npn − p + p ( b − a ) − . In addition, (3.11) is reduced to asingle equation (3.1) in the weak sense (i.e. u, v ∈ D ,pa ( R n ) ).Proof. Step 1. By calculation, we have − div [ | x | − ap |∇ u µ ( x ) | p − ∇ u µ ( x )] = − µ ( p − σ +( a +1) p − b ( q +1) − σ q | x | − b ( q +1) v q µ ( x ) . If (3.11) is invariant under the scaling (1.8), there hold ( p − σ + ( a + 1) p − b ( q + 1) = σ q and ( p − σ + ( a + 1) p − b ( q + 1) − σ q . Thus, σ = p ( q + p − a + 1 − b ) q q − ( p − − b, σ = p ( q + p − a + 1 − b ) q q − ( p − − b. In addition, the energy functionals R R n u q | x | b ( q dx and R R n v q | x | b ( q dx are invariant implies σ = nq +1 − b and σ = nq +1 − b . Therefore, n [ q q − ( p − ] p ( a + 1 − b ) = ( q + 1)( q + p −
1) = ( q + 1)( q + p − . (3.13)19he latter equality implies p = 2 or q = q .On the contrary, the argument above also shows that if p = 2 or q = q holds, then systemand the energy functionals are invariant under the scaling (1.8). Step 2. If p = 2, (3.11) becomes a Laplace system. Thus, the former equality of (3.13) implies(3.12). In particular, if a = 0, it is identical with (1) of Remark 2.1.If q = q , we denote them by q . The former equality of (3.13) implies q = npn − p + p ( b − a ) − φ ∈ C ∞ ( R n ), there holds Z R n | x | − ap ( |∇ u | p − ∇ u − |∇ u | p − ∇ u ) ∇ φdx = Z R n | x | − b ( q +1) ( v q − u q ) φdx. Since C ∞ ( R n ) is dense in D ,pa ( R n ), we can take φ = u − v . Noting the monotonicity inequality( |∇ u | p − ∇ u − |∇ u | p − ∇ u ) · ∇ ( u − v ) ≥ , we get Z R n | x | − b ( q +1) ( u q − v q )( u − v ) dx ≤ . By the integral mean value theorem we have u = v a.e. on R n . Namely, (3.11) is reduced to(3.1). Remark 3.1.
Different from the cases p = 2 and q = q , (3.11) has no variational structure. Tso [28] obtained the critical exponent and the existence/nonexistence results for the k-Hessian equation on the bounded domain. Other related work can be seen in [6] and the referencestherein. Here we consider the following k-Hessian equation on R n F k ( D u ) = ( − u ) q , u < in R n . (4.1)Here F k [ D u ] = S k ( λ ( D u )), λ ( D u ) = ( λ , λ , · · · , λ n ) with λ i being eigenvalues of the Hessianmatrix ( D u ), and S k ( · ) is the k -th symmetric function: S k ( λ ) = X ≤ i < ··· C − ≤ R ( x ) ≤ C , we can see the analogous result still holds. Theorem 4.1. If q ≤ nkn − k , then F k ( D u ) = R ( x )( − u ) q , u < in R n , (4.2) has no negative solution satisfying inf R n ( − u ) = 0 for any double bounded coefficient R ( x ) .Proof. If (4.2) has some negative solution u satisfying inf R n ( − u ) = 0 for some double bounded R ( x ), then − u solves an integral equation − u ( x ) = K ( x ) Z ∞ " R B t ( x ) ( − u ) q ( y ) dyt n − k k dtt , u < in R n , K ( x ) is also double bounded. However, by the Wolff potential estimates (cf. [14]), weknow this integral equation has no positive solution for any double bounded fonction.Hereafter, we always assume that q is larger than the Serrin exponent nkn − k : q > nkn − k . (4.3) Theorem 4.2.
If (4.3) holds, then (4.2) has radial solutions with the fast and the slow decayrates respectively for some double bounded functions R ( x ) .Proof. Clearly, if the following ODE has solution U ( r ) C k − n − ( U r r ) k − U rr + C kn − ( U r r ) k = R ( r )( − U ) q , (4.4)then u ( x ) = U ( r ) solves (4.2). Here C k − n − and C kn − are combinatorial constants, and R ( r ) is adouble bounded function.We search the radial solution as the form u ( x ) = U ( r ) = − (1 + r ) − θ , r = | x | , (4.5)where θ > U r = 2 θr (1 + r ) θ +1 , U rr = 2 θ (1 + r ) θ +1 (cid:20) − θ + 1) r r (cid:21) . Thus, the left hand side of (4.4) C k − n − ( U r r ) k − U rr + C kn − ( U r r ) k = (cid:20) θ (1 + r ) θ +1 (cid:21) k " C k − n − + C kn − + [ C kn − − C k − n − (2 θ + 1)] r r . (4.6)In view of (4.3), it follows 2 kq − k < n − kk . We next determine that the decay rate 2 θ is either the fast rate n − kk or the slow rate kq − k .In fact, if C kn − − C k − n − (2 θ + 1) >
0, then 2 θ < n − kk . We choose θ such that 2 θ = kq − k , andfrom (4.6) we can see that C k − n − ( U r r ) k − U rr + C kn − ( U r r ) k = R ( r )(1 + r ) − ( θ +1) k = R ( r )(1 + r ) − θq . This result shows that U ( r ) as the form (4.5) with 2 θ = k +1 q − k solves (4.4).In addition, if C kn − − C k − n − (2 θ + 1) = 0, then 2 θ = n − kk . Let q = ( n +2) kn − k . Thus, from (4.6)we can also deduce C k − n − ( U r ) k − U rr + C kn − r − ( U r ) k = R ( r )(1 + r ) − ( θ +1) k − = R ( r )(1 + r ) − θq , (4.7)and hence U ( r ) as the form (4.5) with 2 θ = n − kk also solves (4.4).21ome back the equation (4.1). We consider the critical and the supercritical cases. Theorem 4.3.
Under the scaling transformation (1.2), the equation (4.1) and the energy func-tional k ( − u ) k L q +1 ( R n ) are invariant if and only if q = ( n +2) kn − k .Proof. Clearly, if we denote µx by y , then F k ( D u µ ( x )) = µ − k ( σ +2) F k ( D u ( y )) = µ − k ( σ +2) ( − u ) q ( y ) = µ qσ − k ( σ +2) ( − u µ ) q ( x )implies σ = kq − k . On the other hand, R R n ( − u µ ) q +1 dx = µ ( q +1) σ − n R R n ( − u ) q +1 dx implies σ = nq +1 . Eliminating σ yields q = ( n +2) kn − k . The necessity is complete.On the contrary, the argument above still works for the sufficiency.The following result shows that (4.1) has no finite energy solution when q is not the criticalexponent. Theorem 4.4.
Eq. (4.1) has negative solution u belonging to C ( R n ) ∩ L q +1 ( R n ) if and only if q = ( n +2) kn − k .Proof. Sufficiency. If q = ( n +2) kn − k , Theorem 4.2 implies (4.4) has a radial solution as the form(4.5) with the fast decay rate 2 θ = n − kk . In addition, (4.6) shows that R ( r ) is a constant R ( x ) = C ∗ := ( n − kk ) k ( C k − n − + C kn − ) . Thus, setting L := C q − k ∗ , and V ( r ) := LU ( r ) = − L (1 + r ) − n − k k , we know that v ( x ) = V ( | x | ) is a radial solution of (4.1) in C ( R n ) ∩ L q +1 ( R n ). Necessity.
Multiply (4.1) by − uζ R and integrate on B R (0). Noting u ∈ L q +1 ( R n ) and thenletting R → ∞ , we obtain uF k ( D u ) ∈ L ( R n ), and hence − Z R n uF k ( D u ) dx = Z R n ( − u ) q +1 dx. (4.8)In addition, (4.1) is the Euler-Lagrange equation for the functional E k ( u ) = − k + 1 Z R n uF k ( D u ) dx − Z R n ( − u ) q +1 q + 1 dx. By virtue of E k ( u ( xµ )) = − µ n − k k + 1 Z R n uF k ( D u ) dx − µ n q + 1 Z R n ( − u ) q +1 dx, the Pohozaev identity ddµ E k ( u ( xµ )) | µ =1 = 0 shows n − kk + 1 Z R n uF k ( D u ) dx + nq + 1 Z R n ( − u ) q +1 dx = 0 . Inserting (4.8) into this result, we obtain n − kk +1 = nq +1 , which implies q = ( n +2) kn − k .Theorem 4.4 shows (4.1) has a radial solution with fast decay rate n − kk when q is a criticalexponent.In the supercritical case, Theorem 4.4 implies that the solution of (4.1) is not the finite energysolution, and hence the decay rate should be slower than n − kk .22 heorem 4.5. If q > ( n +2) kn − k , then (4.1) has radial solutions with slow decay rate kq − k .Proof. Consider the problem k C k − n − [( f r ) k r n − k ] r = r n − ( − f ) q , r > ,f ( r ) < and f ′ ( r ) > , r > ,f (0) = − A, f ′ (0) = 0 , (4.9)where A > f ( r ). Then u ( x ) = f ( | x | ) is a solution of (4.1).Let t >
0. Integrating (4.9) from 0 to t yields f r ( t ) = ( 1 k C k − n − ) − k [ t k − n Z t r n ( − f ( r )) q drr ] k . For s ≥
0, integrating from s to R again, we get f ( R ) = f ( s ) + ( 1 k C k − n − ) − k Z Rs [ t k − n Z t r n ( − f ( r )) q drr ] k dt. (4.10)We claim that inf r ≥ [ − f ( r )] = 0 as long as f is an entire positive solution of (4.9). Otherwise,there exists c ∗ > − f ( r ) ≥ c ∗ for r ≥
0. Therefore, (4.10) with s = 0 shows f ( R ) ≥ ( kc q ∗ C k − n − ) k R − A. Thus, we can find some R A such that f ( R A ) = 0. This shows that (4.9) has no entire positivesolution.By virtue of f ′ ( r ) > r ≥ [ − f ( r )] = 0, we know that the global solution f ( r ) shouldbe bounded and increasing to zero when r → ∞ . In the following, we construct the function f ( r ) with the slow decay rate kq − k .First, (4.9) admits a local negative solution by the standard argument. Namely, for each A >
0, we can find R A > f ( r ) < r ∈ (0 , R A ). We claim f ( R A ) < f ( R A ) = 0. Thus, f ( r ) solves the two point boundary value problem (cid:26) k C k − n − [( f r ) k r n − k ] r = r n − ( − f ) q , r ∈ (0 , R A ) ,f (0) = − A, f ( R A ) = 0 . Namely, u ( x ) = f ( | x | ) is a classical solution of (cid:26) F k ( D u ) = ( − u ) q on B R A (0) ,u | ∂B RA (0) = 0 , u < on B R A (0) . It contradicts with the nonexistence result in the supercritical case (cf. [28]).Thus, we can extend the solution towards right in succession and hence obtain an entiresolution for each A .Next, by the shooting method, it is easy to find suitable A > f A ( r ) < ,
1] satisfying f A (0) = − A and f A (1) = − C A . Here C A = [ C k − n − k ( n − qkq − k )] q − k ( q − k k ) kk − q .
23y virtue of (4.3), we see C A > f ( R ) = − C A R − kq − k solves (4.9) as r >
1. It is sufficient to verify (4.10)with s = 1: f ( R ) = f (1) + ( 1 k C k − n − ) − k Z R [ t k − n Z t r n ( − f ( r )) q drr ] k dt. In fact, (4.3) shows that n − qkq − k >
0. By a calculation, we get the right hand side of theintegral equation above ( kC qA C k − n − ) k Z R ( t k − n Z t s n − qkq − k dss ) k dt − C A = ( kC qA C k − n − ) k ( n − qkq − k ) − k Z R t − qq − k dtt − C A = ( kC qA C k − n − ) k ( n − qkq − k ) − k q − k k (1 − R − kq − k ) − C A = − C A R − kq − k . It is the left hand side.Thus, we find a radial solution of (4.1) in the supercritical case u ( x ) = (cid:26) f A ( | x | ) , f or r ∈ [0 , − C A | x | − kq − k , f or r ≥ . Theorem 4.5 is proved.
Acknowledgements
The author is very grateful to Dr. Xingdong Tang for providing thereference [10], and many fruitful discussions.
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