On critical exponents of a k -Hessian equation in the whole space
aa r X i v : . [ m a t h . A P ] F e b On critical exponents of a k -Hessian equation in thewhole space Yun Wang and Yutian Lei
Institute of Mathematics, School of Mathematical Sciences,Nanjing Normal University, Nanjing, 210023, China
Abstract
In this paper, we study negative classical solutions and stablesolutions of the following k -Hessian equation F k ( D V ) = ( − V ) p in R n with radial structure, where n ≥
3, 1 < k < n/ p >
1. This equationis related to the extremal functions of the Hessian Sobolev inequality onthe whole space. Several critical exponents including the Serrin type, theSobolev type, and the Joseph-Lundgren type, play key roles in studyingexistence and decay rates. We believe that these critical exponents stillcome into play to research k -Hessian equations without radial structure. Keywords : k -Hessian equation, stable solution, critical exponent, Liou-ville theorem, decay rate MSC2010 : 35B33, 35J60
In 1990, Tso [28] studied the relation between the value of exponent p andthe existence results for the k -Hessian equation F k ( D V ) = ( − V ) p in boundeddomains. The critical exponent p = ( n +2) kn − k plays a key role. Those resultsare associated with the extremal functions of the Hessian Sobolev inequalityfor all k -admissible functions which was introduced by Wang in [32]. Such aninequality with the critical exponent still holds in the whole space R n , and theextremal functions are radially symmetric (cf. [5], [27]).Consider the Euler-Lagrange equation F k ( D V ) = ( − V ) p , V < in R n , (1.1)with a general exponent p >
1, where n ≥
3, 1 < k < n/
2. Here F k [ D V ] = S k ( λ ( D V )), λ ( D V ) = ( λ , λ , · · · , λ n ) with λ i being eigenvalues of the Hes-sian matrix ( D V ), and S k ( · ) is the k -th symmetric function: S k ( λ ) = X ≤ i < ···
V < k -admissible solutions in the coneΦ k := { u ∈ C ( R n ); F s ( D V ) ≥ , s = 1 , , · · · , k } . Such an equation does not only come into play to study the extremal functionsof the Hessian Sobolev inequality, but also is helpful to investigate the global1xistence and blow-up in finite time span for the fully nonlinear parabolic equa-tions (such as the equations studied in [12], [24] and [30]).A special case is F [ D V ] = ∆ V , and (1.1) becomes the Lane-Emden equa-tion − ∆ u = u p , u > in R n . (1.2)The existence results of the solutions of this equation have provided an im-portant ingredient in the study of conformal geometry, such as the extremalfunctions of the Sobolev inequalities and the prescribing scalar curvature prob-lem. It was studied rather extensively. According to Theorem 3.41 in [21], (1.2)has no positive solution even on exterior domains when p is not larger than theSerrin exponent (i.e. p ∈ (1 , nn − )). The Liouville theorem in [9] shows that(1.2) has no positive classical solution in the subcritical case (i.e. p ∈ [1 , n +2 n − )).In the critical case (i.e. p = n +2 n − ), the positive classical solutions of (1.2) mustbe of the form u ( x ) = c ( tt + | x − x ∗ | ) n − (1.3)with constants c, t >
0, and x ∗ ∈ R n (cf. [2]). In supercritical case (i.e. p > n +2 n − ), existence and asymptotic behavior of positive solutions are muchcomplicated and not completely understood. In fact, we can find cylindricalshaped solutions which does not decay along some direction. In addition, thereare radial solutions with the slow decay rates solving (1.2) (cf. [9], [13], [31] andmany others). Furthermore, those radial solutions are of the form u ( x ) = µ p − U ( µ | x | ) , x ∈ R n , where µ = u p − (0), and U ( r ) is the unique solution of (cid:26) − ( U ′′ + n − r U ′ ) = U p , U ( r ) > , r > U ′ (0) = 0 , U (0) = 1 . For the study of ‘stable’ positive solutions of (1.2), the Joseph-Lundgrenexponent p jl ( n ) := 1 + 4 n − − √ n − k -Hessian equation (1.1). As thebeginning of the study, we are concerned about the increasing negative solutionof (1.1) with radial structure as in [5] and [20]. Thus, (1.1) is reduced to thefollowing equation − k C k − n − ( r n − k | u ′ | k − u ′ ) ′ = r n − u p , u ( r ) > as r > . (1.4)Here u ( r ) = u ( | x | ) = − V ( x ), n ≥
3, 1 < k < n/ p >
1. In fact, in thecritical case (i.e. p = ( n +2) kn − k ), the extremal functions of the Hessian Sobolev2nequality are radially symmetric (cf. [5], [27] and [32]). In the noncriticalcase, it is clearer and more concise to study the critical exponents of the radialsolutions. We believe that the ideas are helpful to investigate the correspondingproblems of the solutions with general form, and those critical exponents stillcome into play in the study of k -Hessian equations without radial structure. Clearly, (1.4) has a singular solution u s ( r ) = Ar − kp − k , with A := ( 1 k C k − n − ) p − k ( 2 kp − k ) kp − k ( n − pkp − k ) p − k . (1.5)If write V ( x ) = − u s ( | x | ), then V ( x ) only belong to C ( R n \ { } ) (even it doesnot belong to L ∞ loc ( R n )).We are mainly concerned with the k -admissible solutions of (1.1). Considerthe following boundary values problem ( − k C k − n − ( r n − k | u ′ | k − u ′ ) ′ = r n − u p , u ( r ) > , r > u ′ (0) = 0 , u (0) = ρ (:= µ kp − k ) > . (1.6) Definition 1.1.
If a solution u ( r ) of (1.6) satisfies u ( | x | ) ∈ C ( R n ) , then u ( r ) is called a regular solution. Recall two critical exponents: Serrin exponent p se := nkn − k , and Sobolevexponent p so := ( n +2) kn − k .When p is not larger than the Serrin exponent, (1.1) has no negative k -admissible solution (cf. [15], [22] and [23]). Thus, we always assume in thispaper that p is larger than the Serrin exponent p > p se . (1.7)In the critical case (i.e. p = p so ), all the regular solutions of (1.6) can bewritten as the explicit form (cf. Remark 1.4 in [20]) u ρ ( r ) = ( 1 k C k − n − ) p − k ρ (1 + kn /k ( n − k ) ( ρ k +1 n − k r ) ) − n − k k . (1.8)Therefore, we will be concerned with the noncritical cases. Theorem 1.1.
When p < p so , (1.6) has no regular solution. Remark 1.1.
By a direct calculation, when p se < p < p so , besides u s givenby (1.5), (1.4) has other singular solutions U s ( r ) satisfying U s ( r ) /u s ( r ) → r → U s ( r ) r n − kk → λ > r → ∞ . When k = 1, this result can befound in [9], [13], [31]. Theorem 1.2.
When p > p so , all the positive regular solution u µ of (1.6)satisfies u µ ( r ) ≃ r − kp − k for large r . Furthermore, they are the form of u µ ( r ) = µ kp − k u ( µr ) , r ≥ , (1.9)3 here u ( r ) is the solution of (cid:26) − k C k − n − ( r n − k | u ′ | k − u ′ ) ′ = r n − u p , u ( r ) > , r > u ′ (0) = 0 , u (0) = 1 . (1.10)Here, u ( r ) ≃ r − θ means that there exists C > C ≤ u ( r ) r θ ≤ C for large r . Remark 1.2.
Problem (1.10) has a entire solution when p > p so . In fact, bya standard argument of contraction, (1.10) has a unique local positive solution u (cf. Proposition 2.1 in [20]). There holds u ′ < u > u > r >
0. Otherwise, it contradicts with the Liouville theorem in [28].
Definition 1.2.
We say that a positive solution u ∈ C (0 , ∞ ) of (1.4) is stableif Z ∞ [ 1 k C k − n − r n − k | u ′ | k − u ′ ϕ ′ − r n − u p ϕ ] dr = 0; (1.11) Q u ( ϕ ) := C k − n − Z ∞ r n − k | u ′ | k − ( ϕ ′ ) dr − p Z ∞ r n − u p − ϕ dr ≥ for all ϕ ∈ W ∗ , where W ∗ = { ϕ ( r ); ϕ ( r ) = φ ( x ) ∈ C ∞ c ( R n ) , r = | x |} .Similarly, a positive solution u ∈ C (0 , ∞ ) of (1.4) is stable on a set ( R, ∞ ) for some R > , if (1.11) holds for all ϕ ∈ W ∗ , and (1.12) holds for all ϕ ∈ C ∞ c ( R, ∞ ) . Indeed, the fact that the first order Fr´ e chet derivative of the functional J ( u )is equal to zero and the second order Fr´ e chet derivative is nonnegative can leadto this definition, where J ( u ) = C k − n − k ( k + 1) Z ∞ | u ′ | k +1 r n − k dr − p + 1 Z ∞ u p +1 r n − dr. In addition, Q u ( ϕ ) ≥ u ρ given by (1.8) and u µ given by (1.9) satisfy (1.11). For the singular solution u s expressed by (1.5), p > p se implies that 0 is not the singular point in integral terms of (1.11) (seethe proof of Theorem 1.4). Therefore, u s also satisfies (1.11).Recall other two critical exponents: the Joseph-Lundgren exponent p jl = ∞ , if N ≤ k + 8 ,k [ n − k + 3) n + 4 k ] + 4 k p k + 1) n − k ( n − k )( n − k − , if N > k + 8;and p ∗ = k n + 2 kn − k . p se < p so < p jl . In addition, p so < p ∗ by virtue of 1 < k < n/
2. Inview of 2 k ( k + 6 k + 1) / ( k − > k + 8, we can deduce the relation between p ∗ and p jl as follows p ∗ ≥ p jl , if n ≥ k ( k + 6 k + 1) / ( k − ; p ∗ < p jl , if n < k ( k + 6 k + 1) / ( k − . Under the scaling transformation, p = p so ensures that equation (1.1) andenergy k · k p +1 are invariant (cf [15]), and p = p ∗ ensures that equation (1.1)and energy k · k p + k are invariant (cf [17]). In addition, p ∗ is essential to studythe separation property of solutions (see the following Remark). Remark 1.3.
Let u µ ( r ) be a regular solution of (1.6). Corollary 1.7 in [20]implies that, when p ≥ max { p ∗ , p jl } , u µ ( r ) < u s ( r ) for r >
0, and u µ ( r ) µ < µ .The exponent p ∗ also appears in the study of γ -Laplace equations (cf. [16]and [20]) and integral equations involving Wolff potentials (cf. [3], [19], [26] and[29]). In particular, it plays an important role to investigate integrability, decayrates and intersection properties of the positive entire solutions. In addition,this exponent ensures that equation and energy k · k p + γ − are invariant underthe scaling transformation (cf [17]).In particular, for the γ -Laplace equation − div ( |∇ u | γ − ∇ u ) = K ( x ) u p , u > in R n , (1.13)we write p se ( γ ) = n ( γ − n − γ , p so ( γ ) = nγn − γ − p ∗ ( γ ) = n + γn − γ ( γ − p jl = γ − γ [ n − γ − − p ( n − / ( γ − − as n > γ ( γ +3) γ − , and p jl = ∞ as n ≤ γ ( γ +3) γ − .If γ ∈ (1 , p se ( γ ) < p ∗ ( γ ) < p so ( γ ). When K ( x ) ≡
1, according to theLiouville theorem in [25], (1.13) has no positive solution as p < p so ( γ ), and p ∗ ( γ )does not make sense. When K ( x ) is a double bounded function, according tothe result in [17], (1.13) has positive radial solutions as long as p > p se ( γ ).Now, p ∗ comes into play in studying integrability and decay rates of positivesolutions.Now, we state the results about the stable solutions. Theorem 1.3.
When p < p jl , (1.4) has no stable solution. Theorem 1.4.
When p ≥ p jl , the singular solution u s given by (1.5) is a stablesolution of (1.4). Theorem 1.5.
When p = p so or p ≥ max { p ∗ , p jl } , all the regular solutions of(1.6) are stable solutions of (1.4) on ( R, ∞ ) for some R > . When p se < p
. Remark 1.4.
Theorem 1.4 shows that u s is also a stable solution of (1.4)on ( R, ∞ ) for some R > p ≥ p jl . Combining with Theorem 1.5, weknow that (1.4) has stable solutions on ( R, ∞ ) for some R > p ∈ ( p se , p so ] ∪ [ p jl , ∞ ). To our knowledge, it is unknown whether (1.4) has nostable solution on ( R, ∞ ) for some R > p belongs to the gap ( p so , p jl ).5 Regular solutions
Lemma 2.1.
Let u be a regular solution of (1.6). Then, u ′ < for r > , and u ( r ) → as r → ∞ . Moreover, there are positive constants C , C such that forlarge r , C r − n − kk ≤ u ( r ) ≤ C r − kp − k . (2.1) Proof. Step 1.
Since u is a positive solution of (1.4), − k C k − n − ( r n − k | u ′ | k − u ′ ) ′ > , r > . Integrating from 0 to R with R >
0, we obtain R n − k | u ′ ( R ) | k − u ′ ( R ) < u ′ < Step 2. By u > u ′ < r >
0, we know that lim r →∞ u ( r ) exists andhence is nonnegative. Suppose that lim r →∞ u ( r ) >
0, then there exists a constant c > u ≥ c , and hence − k C k − n − ( r n − k | u ′ | k − u ′ ) ′ ≥ c p r n − . Integrating from 0 to R , we obtain R n − k | u ′ ( R ) | k − u ′ ( R ) ≤ − CR n . Here
C > R . This result, together with u ′ <
0, implies u ′ ( R ) ≤ − CR . Integrating again yields u ( r ) ≤ u (0) − Cr . Letting r → ∞ , we see a contradiction with u >
0. This shows that u ( r ) → r → ∞ . Step 3.
According to the results in [14] or [23], the regular solution of (1.6)satisfies c W kk +1 ,k +1 ( u p )( x ) ≤ u ( | x | ) ≤ c [ inf x ∈ R n u ( | x | ) + W kk +1 ,k +1 ( u p )( x )] , (2.2)where c , c are positive constant, and W kk +1 ,k +1 ( u p ) is the Wolff potential of u p . Namely, W kk +1 ,k +1 ( u p )( x ) = Z ∞ ( R B t ( x ) u p ( | y | ) dyt n − k ) k dtt . Therefore, for large | x | , u ( | x | ) ≥ c Z ∞| x | +1 ( R B (0) u p ( | y | ) dyt n − k ) k dtt ≥ c Z ∞| x | +1 t k − nk dtt = c | x | k − nk . Since u is radially symmetric and decreasing, we can also get u ( | x | ) ≥ c Z | x | / ( R B | x | (0) ∩ B t ( x ) u p ( | y | ) dyt n − k ) k dtt ≥ cu pk ( | x | ) | x | , which implies that u ( | x | ) ≤ c | x | kk − p for large | x | .The proof is complete. 6 .1 Proof of Theorem 1.1 Let p < p so . Assume that (1.6) has a positive regular solution u , we will deducea contradiction. Step 1.
By (2.1), there exists
R > u ( r ) ≤ Cr − kp − k for r > R .Thus, Z ∞ r n − u p +1 dr ≤ C ( R ) + Z ∞ R r n − p +1) kp − k drr < ∞ . (2.3) Step 2.
Let ϕ ∈ C ∞ (0 , ∞ ) satisfy ϕ ( r ) = 1 when r ∈ (0 , ϕ ( r ) = 0 when r ∈ [2 , ∞ ), and 0 ≤ ϕ ≤
1. Write ϕ R ( r ) = ϕ ( rR ). Multiply (1.4) by uϕ k +1 R andintegrate on (0 , ∞ ). By the initial value condition in (1.6), we get Z ∞ r n − k | u ′ | k +1 ϕ k +1 R dr = kC k − n − Z ∞ r n − u p +1 ϕ k +1 R dr − ( k + 1) Z ∞ r n − k uϕ kR | u ′ | k − u ′ ϕ ′ R dr. (2.4)By the Young inequality and the H¨older inequality, for a small ǫ >
0, thereholds that | Z ∞ r n − k uϕ kR | u ′ | k − u ′ ϕ ′ R dr | ≤ ǫ Z ∞ r n − k | u ′ | k +1 ϕ k +1 R dr + C ǫ R k +1 ( Z ∞ r n − u p +1 dr ) k +1 p +1 ( Z RR r θ +1 drr ) p − kp +1 , (2.5)where p − kp +1 θ = n − k − ( n − k +1 p +1 . Therefore, ( θ +1) p − kp +1 − ( k +1) = n p − kp +1 − k < p < p so . Letting R → ∞ , we deduce from (2.3), (2.4) and (2.5)that Z ∞ r n − k | u ′ | k +1 dr < ∞ . (2.6) Step 3.
Multiplying (1.4) by u and integrating on (0 , R ), we obtain that Z R r n − k | u ′ | k +1 dr − R n − k u ( R ) | u ′ ( R ) | k − u ′ ( R ) = kC k − n − Z R r n − u p +1 dr. (2.7)By (2.6) and (2.3), there exists R j → ∞ such that R n − k +1 j | u ′ ( R j ) | k +1 + R nj u p +1 ( R j ) → . (2.8)Therefore, by p < p so , R n − kj u ( R j ) | u ′ ( R j ) | k − u ′ ( R j ) → , as R j → ∞ . Inserting this result into (2.7) and letting R = R j → ∞ , we obtain Z ∞ r n − k | u ′ | k +1 dr = kC k − n − Z ∞ r n − u p +1 dr (2.9) Step 4.
Multiplying (1.4) by ru ′ and integrating on (0 , R ), we have thePohozaev type equality − n − kk + 1 Z R r n − k | u ′ | k +1 dr + kC k − n − np + 1 Z R r n − u p +1 dr = kk + 1 R n − k +1 | u ′ ( R ) | k +1 + k ( p + 1) C k − n − R n u p +1 ( R ) . (2.10)7y (2.8), the right hand side of (2.10) converges to zero when R = R j → ∞ .Letting R = R j → ∞ in (2.10) and using (2.9), we can see n − kk +1 = np +1 , whichcontradicts with p < p so . When k = 1, the proof of the slow decay is based on the comparison principle(cf. Lemma 2.20 and Theorem 2.25 in [18]). For the quasilinear equation (1.4),we use the monotony inequality replacing the comparison principle. Lemma 2.2.
Let u ( r ) be a regular solution of (1.6). If u ( r ) = O ( r − kp − k − ε ) with some ε ∈ (0 , n − kk − kp − k ) for large r , then u ( r ) = O ( r (2 k − n ) /k ) for large r .Proof. If u ( r ) = O ( r − kp − k − ε ) for large r , we can find a large R > r > R , u ( r ) ≤ Cr − kp − k − ε . (2.11)By Lemma 2.1, inf [0 , ∞ ) u ( r ) = 0. Using (2.2) we have u ( | x | ) ≤ C ( I + I + I ) , where I = Z | x | ( R B t ( x ) u p ( | y | ) dyt n − k ) k dtt ,I = Z ∞ | x | ( R B t ( x ) ∩ B R (0) u p ( | y | ) dyt n − k ) k dtt ,I = Z ∞ | x | ( R B t ( x ) \ B R (0) u p ( | y | ) dyt n − k ) k dtt . For sufficiently large | x | , we can deduce from (2.11) that I ≤ C | x | − pk ( kp − k + ε ) Z | x | ( R B t ( x ) dyt n − k ) k dtt ≤ C | x | − kp − k − ε pk ,I ≤ C ( | B R (0) | u p (0)) /k Z ∞ | x | t k − nk dtt ≤ C | x | − n − kk ,I ≤ Z ∞ | x | ( R B t + | x | (0) \ B R (0) u p ( | y | ) dyt n − k ) k dtt ≤ C | x | − kp − k − ε pk . These estimates show that u ( r ) ≤ C ( r − n − kk + r − kp − k − ε pk ) ≤ Cr − kp − k − ε pk . Re-placing (2.11) by this result to estimate I , I and I as we have done above, weget u ( r ) ≤ C ( r − n − kk + r − kp − k − ε ( pk ) ) ≤ Cr − kp − k − ε ( pk ) . By iterating m times, we can obtain u ( r ) ≤ C ( r − n − kk + r − kp − k − ε ( pk ) m ) . Clearly, there exists a sufficiently large m such that n − kk ≤ kp − k + ε ( pk ) m .Thus, after m steps, we derive that, u ( r ) ≤ Cr − n − kk f or large r. Lemma 2.2 is proved. 8 emma 2.3.
Let u ( r ) be a regular solution of (1.6). If u ( r ) = o ( r − kp − k ) forlarge r , then u ( r ) = O ( r (2 k − n ) /k ) for large r .Proof. Step 1. Let ϕ ( r ) ∈ C (0 , ∞ ) satisfy lim r →∞ r n − pkp − k ϕ ( r ) = 0. Integrat-ing (1.4) from 0 to r , we have | u ′ | k − u ′ = − kC k − n − r k − n Z r s n − u p ( s ) ds. (2.12)Thus, by u ( r ) = o ( r − kp − k ) when r → ∞ , it follows that r n − k | u ′ ( r ) | k ϕ ( r ) → . (2.13)Multiply (1.4) by ϕ and integrate from R to ∞ . By (2.13), we obtain that Z ∞ R r n − k | u ′ | k − ( u ′ ) ϕ ′ dr = − R n − k | u ′ ( R ) | k − u ′ ( R ) ϕ ( R ) + kC k − n − Z ∞ R r n − u p ϕdr. (2.14)Write h ( r ) := c ∗ r − θ , where c ∗ is a positive constant determined later, and θ := kp − k + ǫ with suitably small ǫ >
0. By simply calculating and integratingby parts, we get Z ∞ R r n − k | h ′ | k − h ′ ϕ ′ dr = − ( c ∗ θ ) k Z ∞ R r n − k ( θ +2) ϕ ′ dr = ( c ∗ θ ) k [ n − k ( θ + 2)] Z ∞ R r n − − k ( θ +2) ϕdr + ( c ∗ θ ) k R n − k ( θ +2) ϕ ( R ) . Subtracting this result from (2.14) yields Z ∞ R r n − k [ | u ′ | k − u ′ − | h ′ | k − h ′ ] ϕ ′ dr = [( c ∗ θ ) k R n − k ( θ +2) + R n − k | u ′ ( R ) | k − u ′ ( R )] ϕ ( R )+ Z ∞ R r n − [ ku p C k − n − − ( c ∗ θ ) k [ n − k ( θ + 2)] r k ( θ +2) ] ϕdr. (2.15) Step 2.
In view of k ( θ + 2) = pkp − k + kǫ , we can find η ∈ (0 , kǫ /p ) suchthat k ( θ + 2) > pkp − k + pη . (2.16)Since u ∈ C is decreasing and u ( r ) = o ( r − kp − k ) for large r , then either thereexist positive constants c , c such that c r − kp − k ≥ u ( r ) ≥ c r − kp − k − η (2.17)when r is suitably large, or lim r →∞ u ( r ) r kp − k + η = 0, which implies that thereexists η ∈ (0 , η ) such that for large r , u ( r ) ≤ cr − kp − k − ( η − η ) . (2.18)9f (2.18) is true, Lemma 2.3 can be proved easily by Lemma 2.2.In the following, we assume that (2.17) is true. Take ϕ = r − m ( u − h ) + in(2.15), where m > n − pkp − k is sufficiently large. Then, Z ∞ R r n − k − m [ | u ′ | k − u ′ − | h ′ | k − h ′ ][( u − h ) + ] ′ dr = [( c ∗ θ ) k R n − k ( θ +2) − m + R n − k − m | u ′ ( R ) | k − u ′ ( R )][ u ( R ) − h ( R )] + + Z ∞ R r n − m − [ ku p C k − n − − ( c ∗ θ ) k [ n − k ( θ + 2)] r k ( θ +2) ]( u − h ) + dr + m Z ∞ R r n − k − m − [ | u ′ | k − u ′ − | h ′ | k − h ′ ]( u − h ) + dr. (2.19)By (2.12), (2.17) and (2.16), for any δ ∈ (0 , R > r ≥ R , | h ′ | k ≤ δ | u ′ | k . Therefore, the last term of the right hand side of(2.19) with R = R is not larger than m (1 − δ ) R ∞ R r n − k − m − | u ′ | k − u ′ ( u − h ) + dr .Choose c ∗ = u ( R ) R θ to ensure u ( R ) = h ( R ). Therefore, the first term of theright hand side of (2.19) with R = R is zero. Thus, from (2.19) with R = R it follows that Z ∞ R r n − k − m [ | u ′ | k − u ′ − | h ′ | k − h ′ ][( u − h ) + ] ′ dr ≤ Z ∞ R r n − m − [ ku p C k − n − − m (1 − δ ) r − k | u ′ | k ]( u − h ) + dr. (2.20)By (2.12) and the monotonicity of u ( r ), there holds r − k | u ′ ( r ) | k ≥ kC k − n − r − n u p ( r ) Z r s n − ds ≥ ku p ( r ) nC k − n − . Taking m suitably large, we obtain that the right hand side of (2.20) is notlarger than zero. In view of the monotony inequality ( | a | k − a − | b | k − b )( a − b ) ≥ k − | a − b | k +1 , we obtain from (2.20) that Z ∞ R r n − k ([( u − h ) + ] ′ ) k +1 dr ≤ , which implies [ u ( r ) − h ( r )] + ≡ Constant for r ≥ R . In view of u ( R ) = h ( R ),it follows Constant = 0, which implies u ( r ) ≤ h ( r ) for r ≥ R . ApplyingLemma 2.2, we can also see the conclusion of Lemma 2.3. Proof of Theorem 1.2.
Let p > p so . Step 1.
By Lemma 2.1, we see that u ( r ) ≤ Cr − kp − k for large r . We claimthat there exists c > u ( r ) ≥ cr − kp − k for large r .Otherwise, lim r →∞ u ( r ) r kp − k = 0. By Lemma 2.3 it follows that u ( r ) = O ( r k − nk ) for large r . Thus, V ( x ) ∈ L p +1 ( R n ) ∩ C ( R n ) (here V ( x ) = − u ( | x | )).According to Theorem 4.4 in [15], we know p = p so , which contradicts with p > p so . Step 2.
We define by scaling a new function w ( r ) = µ kp − k u ( µr ) , µ > .
10y a direct calculation, we see that w still satisfies (1.4). Applying the initialvalue conditions, we can obtain the second conclusion of Theorem 1.2. Remark 2.1.
Let u µ ( r ) be a regular solution of (1.6) with p > p so . When p ≥ p ∗ , Miyamoto used the technique of phase plane analysis to show that u µ ( r ) /u s ( r ) → r → ∞ (cf. Lemma 2.5 in [20]). When p ≥ p so , Theorem1.2 shows that the decay rate of u µ is the same as that of u s . Furthermore, iflim r →∞ u ( r ) r − kp − k exists, then it must be A which is introduced in (1.5). Infact, integrating (1.4) twice yields u ( r ) = u (0) − ( kC k − n − ) /k Z r [ t k − n Z t s n − u p ( s ) ds ] /k dt. Write B := lim r →∞ u ( r ) r kp − k . Using the L’Hospital principle twice, we get B k = kC k − n − ( 2 kp − k ) − k R r s n − u p ( s ) dsr n − pkp − k = kC k − n − ( 2 kp − k ) − k ( n − pkp − k ) − B p , which implies B = A . Step 1.
We claim that for every γ ∈ [1 , p +2 √ p ( p − k ) − kk ) and any integer m ≥ max { p + γp − k , } , there exists a constant C > ψ ∈ W ∗ , thereholds Z R r n − u p + γ ψ m ( k +1) dr ≤ C Z R ( r ( n − k )( p + γ ) − ( n − γ + k ) p + γ | ψ ′ | k +1 ) p + γp − k dr. (3.1) Proof of (3.1).
Let ψ ∈ W ∗ be a cut-off function such that 0 ≤ ψ ≤ ψ ( r ) = (cid:26) , if r ≤ R/ , , if r ≥ R. Clearly, there exists a constant
C > | ψ ′ | ≤ CR .Taking ϕ = u γ ψ m ( k +1) in (1.11), we get γk C k − n − Z R r n − k | u ′ | k +1 u γ − ψ m ( k +1) dr ≤ k + 1 k C k − n − Z R r n − k | u ′ | k u γ ψ mk | ( ψ m ) ′ | dr + Z R r n − u p + γ ψ m ( k +1) dr. Using the Young inequality to the first term of the right hand side, we canobtain that for any small ε > γk C k − n − − ε ) Z R r n − k u γ − | u ′ | k +1 ψ m ( k +1) dr ≤ C ε Z R r n − k u γ + k | ( ψ m ) ′ | k +1 dr + Z R r n − u p + γ ψ m ( k +1) dr. (3.2)11aking ϕ = u γ +12 ψ m ( k +1)2 in (1.12), we have p Z R r n − u p + γ ψ m ( k +1) dr ≤ C k − n − ( γ + 1) Z R r n − k u γ − | u ′ | k +1 ψ m ( k +1) dr + C k − n − ( k + 1) Z R r n − k | u ′ | k − u γ +1 ψ m ( k − | ( ψ m ) ′ | dr + C k − n − ( γ + 1)( k + 1)2 Z R r n − k | u ′ | k u γ ψ mk | ( ψ m ) ′ | dr. (3.3)Using the Young inequality to the second and the third terms of the right handside of (3.3), we get p Z R r n − u p + γ ψ m ( k +1) dr ≤ ( C k − n − ( γ + 1) ε ) Z R r n − k | u ′ | k +1 u γ − ψ m ( k +1) dr + C ε Z R r n − k | ( ψ m ) ′ | k +1 u γ + k dr. (3.4)Combining (3.2) and (3.4), we obtain by the H¨older inequality that[ p − ( C k − n − ( γ + 1) ε ) 1 γk C k − n − − ε ] Z R r n − u p + γ ψ m ( k +1) dr ≤ C Z R r n − k | ( ψ m ) ′ | k +1 u γ + k dr ≤ C [ Z R ( r ( n − γ + kp + γ u γ + k ψ ( m − k +1) ) p + γγ + k dr ] γ + kp + γ · [ Z R ( r ( n − k )( p + γ ) − ( n − γ + k ) p + γ | ψ ′ | k +1 ) p + γp − k dr ] p − kp + γ . (3.5)In view of γ ∈ [1 , p +2 √ p ( p − k ) − kk ), lim ε → [ p − ( C k − n − ( γ +1) + ε ) γk C k − n − − ε ] = p − k ( γ +1) γ >
0. Therefore, the coefficient of the left hand side of (3.5) is positive aslong as ε is sufficiently small. Therefore, noting ( m − k + 1) p + γγ + k ≥ m ( k + 1)which is implied by m ≥ max { p + γp − k , } , we can deduce (3.1) from (3.5) by theYoung inequality. Step 2.
By the definition of ψ , from (3.1) we can deduce that Z R r n − u p + γ ψ m ( k +1) dr ≤ CR n +1 − (2 k +1)( p + γ ) − ( γ + k ) p − k . (3.6)When n +1 − (2 k +1)( p + γ ) − ( γ + k ) p − k <
0, the desired claim follows by letting R → ∞ .Consider a real-valued function f ( t ) = (2 k + 1)( t + γ ( t )) − ( γ ( t ) + k ) t − k , t ∈ ( k, ∞ ) , γ ( t ) = t +2 √ t ( t − k ) − kk . Clearly, we know f ( t ) is a strictly decreasingfunction (by virtue of f ′ ( t ) < k, ∞ )), satisfying lim t → k f ( t ) = ∞ andlim t →∞ f ( t ) = 2 k + 9. Therefore, we consider separately two cases: n ≤ k + 8,and n ≥ k + 9.Case I: n ≤ k + 8. In view of p > p se , there exists γ ∈ [1 , p +2 √ p ( p − k ) − kk )such that n + 1 − (2 k +1)( p + γ ) − ( γ + k ) p − k < n ≥ k + 9. In view of p > p se , there exists a unique p > k suchthat n + 1 = f ( p ) since f ( t ) is decreasing in ( k, ∞ ). Therefore, p satisfies( n − k )( n − k − p − k [ n − k + 3) n + 4 k ] p + k ( n − = 0 , (3.7)and ( n − k − p − ( n − k > p − k ) . (3.8)The roots of equation (3.7) are p = k [ n − k + 3) n + 4 k ] + 4 k p k + 1) n − k ( n − k )( n − k − , (3.9) p = k [ n − k + 3) n + 4 k ] − k p k + 1) n − k ( n − k )( n − k − . (3.10)Inequality (3.8) implies p > p , and hence we take p = p (it equals exactly p jl ). Thus, when p < p jl , there exists γ ∈ [1 , p +2 √ p ( p − k ) − kk ) satisfying n + 1 − (2 k +1)( p + γ ) − ( γ + k ) p − k < R → ∞ in (3.6), we can deduce R R r n − u p + γ dr →
0. This contradiction shows that (1.4) has no positive stablesolution as long as p < p jl . Let u s be the singular solution of (1.4) given by (1.5). We will prove that thesingular solution u s ( r ) is stable when n ≥ k + 9 and p ≥ p jl .First, we claim that u s satisfies (1.11). In fact, by (1.7), the improper integral R ∞ r n − u ps ϕdr ≤ C R R r n − − pkp − k dr < ∞ . Similarly, the left hand side of (1.11)also makes sense. In addition, u s solves (1.4). Multiply by the test function ϕ ∈ W ∗ and integrate from 0 to ∞ . Noting r n − k | u ′ s ( r ) | k → r →
0, we knowthat the claim is true.To prove that u s satisfies (1.12), we observe firstly that p ( 2 p − k )( n − pkp − k ) ≤ ( n − − p ( k − p − k ) ⇔ n ( p − kp ) − kp ≤ ( n − ( p − kp + k )+4( k − p − k − n − p − kp ) ⇔ ( n − k )( n − k − p − k ( n − k + 3) n + 4 k ) p + k ( n − ≥ ⇔ p ∈ ( −∞ , p ] S [ p jl , + ∞ ) (3.11)13here p is defined in (3.10). On the other hand, by Definition 1.2, we havethat for any φ ∈ C ∞ c ( R n ), C k − n − Z R n | x | k − | u ′ s | k − |∇ φ | dx − p Z R n u p − s φ dx = C k − n − Z R n ( 1 k C k − n − ) k − p − k ( 2 kp − k ) ( k − pp − k ( n − pkp − k ) k − p − k | x | p ( k − p − k |∇ ϕ | dx − p Z R n ( 1 k C k − n − ) p − p − k ( 2 kp − k ) ( p − kp − k ( n − pkp − k ) p − p − k | x | p − kp − k ϕ dx = C ( Z R n | x | p ( k − p − k |∇ φ | dx − p ( 2 p − k )( n − pkp − k ) Z R n | x | p − kp − k φ ) dx, where C = C k − n − ( 1 k C k − n − ) k − p − k ( 2 kp − k ) ( k − pp − k ( n − pkp − k ) k − p − k . (3.12)By p ≥ p jl , (3.11) implies that Z R n | x | p ( k − p − k |∇ φ | dx − p ( 2 p − k )( n − pkp − k ) Z R n | x | p − kp − k φ dx ≥ Z R n | x | p ( k − p − k |∇ φ | dx − ( n − − p ( k − p − k ) Z R n | x | p − kp − k φ dx. It follows that Q u s ( ϕ ) > , ∀ ϕ ∈ W ∗ (3.13)by the Caffarelli-Kohn-Nirenberg inequality (cf. [4]) Z R n |∇ φ | | x | a dx > C a,b Z R n φ | x | b dx, ∀ φ ∈ D , a ( R n ) , (3.14)where n ≥
3, 0 ≤ a < n − and a ≤ b ≤ a + 1, the best constant C a,b is givenby C a,b = ( n − − a ) a = p ( k − p − k and b = a + 1. This resultshows that u s is a stable solution of (1.4) when n ≥ k + 9 and p ≥ p jl . Theproof of Theorem 1.4 is complete. Step 1.
When p = p so , all regular solutions u ρ of (1.6) can be written as theform given by (1.8). When r is suitably large, u ρ ( r ) ≤ D r − n − kk , | u ′ ρ | ≥ D r − n − kk , (3.15)where D , D are positive constants independent of r . Thus, pu p − ( r ) = O ( r − ( k − nk − ) , as r → ∞ . Therefore, we can find some
R > | x | > R and φ ∈ C ∞ c ( R n \ B R (0)), there holds pu p − ρ ( | x | ) φ ( x ) < C ∗ | x | − ( k − nk − φ ( x ) , C ∗ = ( n − − k − k n ) D k − C k − n − ( n − kk ) k − . Thus, C k − n − Z R n | x | k − | u ′ ρ | k − |∇ φ | dx − p Z R n u p − ρ φ dx ≥ D k − C k − n − ( n − kk ) k − Z R n | x | k − k n |∇ φ | dx − C ∗ Z R n | x | k − k n +2 φ dx = D k − C k − n − ( n − kk ) k − ( Z R n | x | k − k n |∇ φ | dx − ( n − − k − k n ) Z R n | x | k − k n +2 φ ) dx, (3.16)and the right hand side is nonnegative by the Caffarelli-Kohn-Nirenberg in-equality (3.14) with a = k − k n and b = a + 1. Therefore, Q u ρ ( ϕ ) ≥ ϕ ∈ C ∞ c ( R, ∞ ). In addition, u ρ also satisfies (1.11). So the regular solution u ρ is stable on ( R, ∞ ). Step 2.
Let u µ (see (1.9)) be a regular solution of (1.6) with p ≥ max { p ∗ , p jl } .We claim that u µ is stable on ( R, ∞ ) for some R > r →∞ u ′ µ ( r ) /u ′ s ( r ) = 1 when p ≥ max { p ∗ , p jl } .Clearly, u ′ s = − ( k C k − n − ) p − k ( kp − k ) pp − k ( n − pkp − k ) p − k r − p + kp − k .Combining with (2.12) and using the L’Hospital principle, we getlim r →∞ ( u ′ µ u ′ s ) k = lim r →∞ Z r s n − u pµ ( s ) ds ( k C k − n − ) pp − k ( kp − k ) pkp − k ( n − pkp − k ) kp − k r n − pkp − k = lim r →∞ r n − u pµ ( r )( k C k − n − ) pp − k ( kp − k ) pkp − k ( n − pkp − k ) pp − k r n − pkp − k − = lim r →∞ u pµ ( r ) u ps ( r ) . By Remark 2.1, there holds lim r →∞ u ′ µ ( r ) /u ′ s ( r ) = 1 when p ≥ max { p ∗ , p jl } . Thus,there exists sufficiently large R > r > R , | u ′ µ ( r ) | k − = | u ′ s ( r ) | k − + o (1) r − ( k − p + k ) p − k . Therefore, by the strict inequality (3.13), we can find a suitably small δ > ψ ∈ C ∞ c ( R n \ B R (0)), C k − n − Z R n | u ′ µ ( | x | ) | k − | x | k − |∇ φ | dx = C k − n − Z R n | u ′ s ( | x | ) | k − + o (1) | x | − ( k − p + k ) p − k | x | k − |∇ φ | dx ≥ C [ p ( 2 p − k )( n − pkp − k ) + δ + o (1)] Z R n | x | p − kp − k φ dx ≥ p Z R n u s ( | x | ) p − φ dx. C is the constant in (3.12). In view of u s ( r ) > u µ ( r ) for r > R (seeRemark 1.3), we can see Q u µ ( ϕ ) ≥ ϕ ∈ C ∞ c ( R, ∞ ). In addition, u µ satisfies (1.11). Thus, u µ is stable on ( R, ∞ ) for some R > Step 3.
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