Some existence and nonexistence results for a Schrödinger-Poisson type system
aa r X i v : . [ m a t h . A P ] N ov Some existence and nonexistence results for aSchr¨odinger-Poisson type system
Yutian Lei
Institute of MathematicsSchool of Mathematical SciencesNanjing Normal UniversityNanjing, 210023, ChinaEmail:[email protected]
Abstract
In this paper, we study the Schr¨odinger-Poisson system (cid:26) − ∆ u = √ pu p − v, u > in R n , − ∆ v = √ pu p , v > in R n with n ≥ p >
1. We investigate the existence and the nonexistenceof positive classical solutions with the help of an integral system involvingthe Newton potential u ( x ) = c Z R n u p − ( y ) v ( y ) dy | x − y | n − , u > in R n ,v ( x ) = c Z R n u p ( y ) dy | x − y | n − v > in R n . First, the system has no solution when p ≤ nn − . When p > nn − , thesystem has a singular solution on R n \ { } with slow asymptotic rate p − .When p < n +2 n − , the system has no solution in L n ( p − ( R n ). In fact, if thesystem has solutions in L n ( p − ( R n ), then p = n +2 n − and all the positiveclassical solutions can be classified as u ( x ) = v ( x ) = c ( tt + | x − x ∗ | ) n − ,where c, t are positive constants. When p > n +2 n − , by the shooting methodand the Pohozaev identity, we find another pair of radial solution ( u, v )satisfying u ≡ v and decaying with slow rate p − . Keywords : Schr¨odinger-Poisson equation, positive classical solution, ex-istence and nonexistence, critical and noncritical conditions
MSC2010
Recently, many authors devoted to study the nonlocal stationary Schr¨odingerequation − ∆ u = pu p − ( | x | − n ∗ u p ) , u > in R n , (1.1)where n ≥ p >
1. In particular, Moroz and Van Schaftingen [25] studiedthe existence of the supersolutions and listed several sufficient conditions.1quation (1.1) is defined in many places. One is the example 3.2.8 in thebook [4]. A more general form is the Choquard type equation in the papers [14]and [24]. Another interesting work related to (1.1) is the paper [12] and thereferences therein. Equation (1.1) is also helpful in understanding the blowingup or the global existence and scattering of the solutions of the dynamic Hartreeequation (cf. [18]), which arises in the study of boson stars and other physicalphenomena, and also appears as a continuous-limit model for mesoscopic molec-ular structures in chemistry. Such an equation also arises in the Hartree-Focktheory of the nonlinear Schr¨odinger equations (cf. [23]). More related math-ematical and physical background can be found in [2], [10], [26], [28] and thereferences therein.Since (1.1) has a convolution term, it seems difficult to investigate the exis-tence directly. Write v ( x ) = √ p Z R n u p ( y ) dy | x − y | n − . Then v > R n . Noting the relation between the Newton potential and theconvolution properties of Dirac function, we see that − ∆ v ( x ) = √ p ( − ∆ | x | − n ) ∗ u p = √ pδ x ∗ u p = √ pu p ( x ) , where δ x is the Dirac mass at x . Thus, the positive solution of (1.1) must satisfythe following system (cid:26) − ∆ u = √ pu p − v, u > in R n , − ∆ v = √ pu p , v > in R n . (1.2)It is a simplified model of the Schr¨odinger-Piosson system (cf. [1], [11] andreferences therein).Quittner and Souplet [27] studied positive solutions of another PDE system (cid:26) − ∆ u = v p u r , u > in R n , − ∆ v = v s u q , v > in R n . (1.3)They proved the following results:(R1) If n ≥ p − s = q − r ≥ ≤ r, s ≤ nn − , then positive solutions u, v satisfy u ≡ v .(R2) If n ≥ p − s = q − r ≥
0, then nonnegative solutions u, v satisfy u ≥ v or v ≥ u .In this paper, we study the existence of positive classical solutions to theSchr¨odinger-Piosson type system (1.2). Besides (R1) and (R2), we have furtherexistence and nonexistence results and state them in three cases.First we consider the subcritical case p ∈ (1 , ∗ − ∗ = nn − . Wehave the following nonexistence results. Theorem 1.1. (1)If p ≤ nn − , then (1.2) has no positive solution.(2) If ≤ p < ∗ − , then (1.2) has no positive solution in L n ( p − ( R n ) . emark.
1. When u ≡ v , (1.2) is reduced to the Lane-Emden type equation − ∆ u = √ pu p , u > in R n . (1.4)Its nonexistence results similar to Theorem 1.1 can be found in [9]. Forthe system (1.2), we here introduce an integral system to investigate thenonexistence. By the same argument in [6] and [15], we know that thesolution of (1.2) satisfies the following integral system involving Newtonpotentials u ( x ) = c Z R n u p − ( y ) v ( y ) dy | x − y | n − , u > in R n ,v ( x ) = c Z R n u p ( y ) dy | x − y | n − v > in R n , (1.5)for some constants c , c >
0. By this integral system we can verify thenonexistence of positive solutions when p ≤ nn − . When p > nn − , then(1.2) always has singular solutions on R n \ { } (cf. Theorem 4.1).2. The integral system (1.5) is invariant after translation. Its L n ( p − ( R n )-solutions are radially symmetric about some point x ∗ ∈ R n . In the non-critical case, Kelvin transformation breaks the translation invariant of(1.5). On the other hand, the L n ( p − ( R n )-solutions are still radially sym-metric about the origin in the subcritical case. These facts lead to acontradiction and hence L n ( p − ( R n )-solution does not exist.Next, we consider the critical case and classify the L n ( p − ( R n )-solutions. Theorem 1.2.
Let u be a classical solution of (1.2). Then the following itemsare equivalent to each other(1) u ∈ L n ( p − ( R n ) ;(2) u is bounded and decaying with the fast rate n − ;(3) u belongs to the homogeneous Sobolev space D , ( R n ) ;(4) u ∈ L ∗ ( R n ) and p = 2 ∗ − ;(5) u ( x ) = v ( x ) = c ( tt + | x − x ∗ | ) n − with constants c, t > . In the special case u ≡ v , we recall the Lane-Emden equation (1.4). Theclassification of the solutions of this single equation has provided an importantingredient in the study of the conformal geometry, such as the prescribing scalarcurvature problem and the extremal functions of the Sobolev inequalities. It wasstudied rather extensively (cf. [3], [8], [9], [21] and the reference therein). Inparticular, Chen and Li [5] proved that all the positive solutions of (1.4) withthe critical exponent p = 2 ∗ − u ( x ) = c ( tt + | x − x ∗ | ) n − c, t >
0. For the system (1.2), we expect to prove both u ≡ v and p = 2 ∗ −
1. Once these results are verified, we can use the result in [5] toclassify the positive solutions of (1.2).At last, we consider the supercritical case p > ∗ −
1. We use the shootingmethod and Pohozaev identity to prove the following result.
Theorem 1.3.
When p > ∗ − and p ≥ , we can find radial solutions u, v decaying with the slow rate p − when | x | → ∞ . Not all the solutions in the supercritical case are radially symmetric. Insection 4 we introduce an example to show that some bounded solutions areneither radial nor decaying when | x | → ∞ .Another integral system as (1.5) is the following Lane-Emden type equations u ( x ) = c Z R n v q ( y ) dy | x − y | λ , u > in R n ,v ( x ) = c Z R n u p ( y ) dy | x − y | λ , v > in R n . (1.6)It is essential in studying the extremal functions of the Hardy-Littlewood-Sobolev inequality (cf. [22]) Z R n Z R n f ( x ) g ( y ) | x − y | λ dxdy ≤ C ( n, s, λ ) || f || r || g || s with 0 < λ < n , 1 < s, r < ∞ , f ∈ L r ( R n ) and g ∈ L s ( R n ), r + s + λn = 2.Define T g ( x ) = Z R n g ( y ) | x − y | n − α dy with α = n − λ . The Hardy-Littlewood-Sobolev inequality becomes || T g || p ≤ C ( n, s, α ) || g || npn + αp , where nn − α < p < ∞ , and 1 < s < n/α . This inequality will be used in thispaper to research the radial symmetry and the integrability of the solutions of(1.5).In the critical case, the classification results for the single equation of (1.6)can be found in [6] and [19]. In particular, the method of moving planes ofintegral forms was introduced in [6]. It has become a powerful tool to handle thequalitative properties including the existence and the nonexistence, the radialsymmetry, and the priori estimates. Jin and Li [13] applied a regularity liftinglemma by the contraction maps to obtain the optimal integrability of positiveregular solutions. Based on this result, [16] estimated the fast decay rates when | x | → ∞ . 4 Subcritical case
In this section, we always assume that p < ∗ − Theorem 2.1. (Liouville theorem) If p ≤ nn − , then there does not exist anypositive solution of (1.5).Proof. The idea in [15] can be used here.If u, v are positive solutions, we can deduce a contradiction.(1) When p < nn − , it follows n − − p − < a = n − a j +1 = (2 p − a j − b j = pa j − j = 0 , , , · · · .By (1.5), we have u ( x ) ≥ c Z B (0) u p − ( y ) v ( y ) dy | x − y | n − ≥ c | x | a , (2.1) v ( x ) ≥ c Z B ( x, | x | / u p ( y ) dy | x − y | n − ≥ c | x | b ,u ( x ) ≥ c Z B ( x, | x | / u p − ( y ) v ( y ) dy | x − y | n − ≥ c | x | a , · · · · · · By induction, we have u ( x ) ≥ c | x | aj as long as a j >
0. Noting p > n − − p − <
0, from a j = a (2 p − j − p − j − + (2 p − j − + · · · + (2 p − ]= ( n − − p − )(2 p − j + p − , we can find some j > a j ≤
0. This leads to u ( x ) = ∞ and yieldsa contradiction.(2) When p = nn − , from u ( x ) ≥ c Z B R (0) u p − ( y ) v ( y ) dy | x − y | n − ≥ c ( R + | x | ) n − Z B R (0) u p − ( y ) v ( y ) dy, we deduce Z B R (0) u p ( x ) dx ≥ Z B R (0) cdx ( R + | x | ) n ( Z B R (0) u p − ( y ) v ( y ) dy ) p ≥ c ( Z B R (0) u p − ( y ) v ( y ) dy ) p . (2.2)Here c is independent of R . Similarly, from v ( x ) ≥ c ( R + | x | ) n − Z B R (0) u p ( y ) dy, (2.3)5e also deduce Z B R (0) u p − ( x ) v ( x ) dx ≥ Z B R (0) cu p − ( x ) dx ( R + | x | ) n − Z B R (0) u p ( y ) dy. Using (2.1), (2.2) and noting p = nn − , we get Z B R (0) u p − ( x ) v ( x ) dx ≥ c ( Z B R (0) u p − ( y ) v ( y ) dy ) p , which implies u p − v ∈ L ( R n ) if we let R → ∞ .Multiplying (2.3) by u p − and integrating on A R := B R (0) \ B R/ (0), westill have Z A R u p − ( x ) v ( x ) dx ≥ c ( Z B R (0) u p − ( y ) v ( y ) dy ) p . Letting R → ∞ and noting u p − v ∈ L ( R n ), we obtain k u p − v k L ( R n ) = 0. Itis impossible.By using the method of moving planes in integral forms, which was estab-lished by Chen-Li-Ou (cf. [6] and [7]), we prove a radial symmetry result. Theorem 2.2.
Let p ≥ . If h ≥ satisfies Z R n | y | − nh/ u n ( p − / ( y ) dy < ∞ , Z R n | y | − nh/ v n ( p − / ( y ) dy < ∞ . (2.4) Then the positive continuous solutions of u ( x ) = c Z R n u p − ( y ) v ( y ) dy | y | h | x − y | n − , u > in R n ,v ( x ) = c Z R n u p ( y ) dy | y | h | x − y | n − v > in R n , (2.5) are radially symmetric and decreasing about x ∗ ∈ R n . Moreover, x ∗ = 0 as longas h > .Proof. For some real number λ , define Σ λ := { x = ( x , . . . , x n ); x > λ } , x λ =(2 λ − x , x , . . . , x n ), u λ ( x ) = u ( x λ ), Σ uλ = { x ∈ Σ λ | u ( x ) ≤ u λ ( x ) } , Σ vλ = { x ∈ Σ λ | v ( x ) ≤ v λ ( x ) } . It is not difficult to see that u λ ( x ) − u ( x )= c Z Σ λ (cid:18) | x − y | n − − | x λ − y | n − (cid:19) | y λ | h ( v λ u p − λ − vu p − ) dy − c Z Σ λ (cid:18) | x − y | n − − | x λ − y | n − (cid:19) (cid:18) | y | h − | y λ | h (cid:19) vu p − dy. (2.6)6ince the second term of the right hand side is nonpositive, from the definitionof Σ uλ and Σ vλ , it follows u λ ( x ) − u ( x ) ≤ c Z Σ λ (cid:18) | x − y | n − − | x λ − y | n − (cid:19) | y λ | h ( v λ u p − λ − vu p − ) dy ≤ c Z Σ vλ | x − y | n − | y λ | h u p − λ ( v λ − v )( y ) dy + c Z Σ uλ | x − y | n − | y λ | h v ( u p − λ − u p − )( y ) dy. Using the Hardy-Littlewood-Sobolev inequality and the H¨older inequality, wehave k u λ − u k L s (Σ uλ ) ≤ C k| y | − h u p − λ ( v λ − v ) k L nsn +2 s (Σ vλ ) + C k| y | − h vu p − λ ( u λ − u ) k L nsn +2 s (Σ uλ ) ≤ C k| y | − h u p − λ k L n (Σ vλ ) k ( v λ − v ) k L s (Σ vλ ) + C k| y | − h u p − λ v k L n (Σ uλ ) k ( u λ − u ) k L s (Σ uλ ) . Similarly, we also obtain k v λ − v k L s (Σ vλ ) ≤ C k| y | − h u p − λ k L n (Σ uλ ) k ( u λ − u ) k L s (Σ uλ ) . By (2.4), as λ → −∞ , C k| y | − h u p − λ k L n (Σ vλ ) ≤ , k| y | − h u p − λ v k L n (Σ uλ ) ≤ . Combining these results, we can see that Σ uλ and Σ vλ are empty set as long as λ is near −∞ .Suppose that at λ <
0, we have u ( x ) ≥ u λ ( x ) and v ( x ) ≥ v λ ( x ) but u ( x ) u λ ( x ) and v ( x ) v λ ( x ) on Σ λ . By the same argument above, we canprove that there exists an ǫ >
0, such that u ( x ) ≥ u λ ( x ) and v ( x ) ≥ v λ ( x ) onΣ λ for all λ ∈ [ λ , λ + ǫ ). Therefore, we can move plane x = λ to the right aslong as u ( x ) ≥ u λ ( x ) and v ( x ) ≥ v λ ( x )hold on Σ λ . If the plane stops at x = λ for some λ <
0, then u ( x ) and v ( x )must be radially symmetric and decreasing about the plane x = λ . Other-wise, we can move the plane all the way to x = 0. Since the direction of x can be chosen arbitrarily, we obtain that u ( x ) , v ( x ) are radially symmetric anddecreasing about some x ∗ ∈ R n . 7f h = 0, we claim x ∗ = 0. Otherwise, we can find λ < x = λ is the stopped plane. From (2.6), we get0 = u λ ( x ) − u ( x )= − c Z Σ λ (cid:18) | x − y | n − − | x λ − y | n − (cid:19) (cid:18) | y | h − | y λ | h (cid:19) vu p − dy = 0 . It is impossible.
Theorem 2.3. If ≤ p < ∗ − , then there does not exist any positive solutionof (1.5) in L n ( p − / ( R n ) .Proof. Step 1. Suppose u, v are the L n ( p − ( R n )-solutions of (1.5). Accordingto Theorem 2.2 with h = 0, we see that they are radially symmetric about x ∗ ∈ R n . Since (1.5) is invariant after translation, x ∗ can be chosen arbitrarily. Step 2.
Take the Kelvin transformation of u, v ˜ u ( x ) = 1 | x | n − u ( x | x | ) , ˜ u ( x ) = 1 | x | n − u ( x | x | ) . By (1.5), we see that ˜ u, ˜ v solve (2.5) with h = n + 2 − p ( n − p < ∗ −
1, it follows h >
0. In addition, from u, v ∈ L n ( p − ( R n ), we see that(2.4) for ˜ u, ˜ v is true. According to Theorem 2.2, ˜ u, ˜ v are also radially symmetricbut the center point x ∗ must be the origin. So the translation invariant isabsent. By the same argument of Theorem 3 in [7], we can also deduce acontradiction. Theorem 2.1 implies that p > nn − u ( x ) = v ( x ) = c ( tt + | x − x ∗ | ) n − is a positive solution of (1.2) in L n ( p − / ( R n ). However, it is nontrivial to showthat all solutions of (1.2) in L n ( p − / ( R n ) are the form above. In this section,we prove this conclusion. Theorem 3.1. If u ∈ L n ( p − ( R n ) solves (1.5), then(R1) u, v ∈ L s ( R n ) for all s > nn − ; u, v L s ( R n ) for all s ≤ nn − ; R2) u, v are bounded and lim | x |→∞ u ( x ) = lim | x |→∞ v ( x ) = 0; (3.2) (R3) u, v are radially symmetric and decreasing about some point x ∗ ∈ R n .Proof. (1) By the Hardy-Littlewood-Sobolev inequality, from u ∈ L n ( p − ( R n )we can deduce v ∈ L n ( p − ( R n ).Write w = u + v . Then w ∈ L n ( p − ( R n ). From (1.5), it follows that w satisfies w ( x ) = K ( x ) Z R n w p ( y ) dy | x − y | n − . (3.3)Here K ( x ) > w A = w as | x | > A or w > A ; w = 0 as | x | ≤ A and w ≤ A . For f ∈ L s ( R n ) with s > nn − , define T f ( x ) := K ( x ) Z R n w p − A ( y ) f ( y ) dy | x − y | n − ,F ( x ) := K ( x ) Z R n ( w − w A ) p ( y ) dy | x − y | n − . Therefore, w solves the operator equation f = T f + F. By the Hardy-Littlewood-Sobolev inequality, we get k T f k s ≤ C k w p − A f k nsn +2 s ≤ C k w A k p − n ( p − k f k s , and k F k s ≤ C k w − w A k p npsn +2 s < ∞ . Take A suitably large such that C k w A k p − n ( p − <
1. Thus, T is a contractionmap from L s ( R n ) to itself. In view of n ( p − > nn − (which is implied by (3.1)), T is also a contraction map from L n ( p − / ( R n ) to itself. By the lifting lemmaon the regularity (cf. Lemma 2.1 in [13]), we obtain w ∈ L s ( R n ) for s > nn − .Thus, u, v ∈ L s ( R n ) for s > nn − .On the other hand, if s ≤ nn − , by (2.1) we have k u k s ≥ c Z ∞ R r n − s ( n − drr = ∞ . Similarly, we also deduce v L s ( R n ) for all s ≤ nn − . (R1) is proved.(2) For r >
0, write w ( x ) = K ( x ) Z B r ( x ) w p ( y ) dy | x − y | n − + K ( x ) Z R n \ B r ( x ) w p ( y ) dy | x − y | n − := K + K . ǫ >
0, from (R1) we deduce K ≤ C k w k p /ǫ ( Z d r n − n − − pǫ drr ) − pǫ ≤ C. On the other hand, by virtue of (3.1) and (R1), we get w ∈ L p ( R n ). Thus K ≤ Cr − n k w k pp ≤ C. Combining the estimates of K and K , we know that w is bounded. Thus, u, v are bounded.Next, we show that w is decaying. Take x ∈ R n . By exchanging the orderof the integral variables, we have w ( x ) = ( n − α ) K ( x ) Z ∞ ( R B t ( x ) w p ( y ) dyt n − α ) dtt . Since w ∈ L ∞ ( R n ), ∀ ε >
0, there exists δ ∈ (0 , /
2) such that Z δ [ R B t ( x ) w p ( z ) dzt n − α ] dtt ≤ C k w k p ∞ Z δ t α dtt < ε. (3.4)As | x − x | < δ , Z ∞ δ [ R B t ( x ) w p ( z ) dzt n − α ] dtt ≤ Z ∞ δ [ R B t + δ ( x ) w p ( z ) dz ( t + δ ) n − α ]( t + δt ) n − α +1 d ( t + δ ) t + δ ≤ Cw ( x ) . Combining this result with (3.4), we get w ( x ) < Cε + Cw ( x ) , f or | x − x | < δ. Let s = n ( p − , then w s ( x ) = | B δ ( x ) | − Z B δ ( x ) w s ( x ) dx ≤ Cε s + C | B δ ( x ) | − k w k sL s ( B δ ( x )) . (3.5)Since w ∈ L s ( R n ), lim | x |→∞ k w k L s ( B δ ( x )) = 0. Inserting this result into (3.5),we have lim | x |→∞ w s ( x ) = 0 . This result means u, v converge to zero when | x | → ∞ . (R2) is proved.(3) By the same proof of Theorem 2.2 with h = 0, we can see the conclusion(R3). Theorem 3.2. If u ∈ L n ( p − ( R n ) solves (1.2) in the classical sense, then u p v ∈ L ( R n ) and u ∈ D , ( R n ) . Moreover, Z R n |∇ u | dx = Z R n |∇ v | dx = √ p Z R n u p vdx. (3.6)10 roof. Step 1. By the H¨older inequality, from Theorem 3.1 (R1) and (3.1), wecan deduce that u p v ∈ L ( R n ). Step 2.
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 ) . (3.7)Multiplying the first equation of (1.2) by uζ R and integrating on D := B R (0), we have − Z D ζ R u ∆ udx = √ p Z D u p vζ R dx. Integrating by parts, we obtain Z D |∇ u | ζ R dx + 2 Z D uζ R ∇ u ∇ ζ R dx = √ p Z D u p vζ R dx. (3.8)Applying the Cauchy inequality, we get | Z D uζ R ∇ u ∇ ζ R dx | ≤ δ Z D |∇ u | ζ R dx + C Z D u |∇ ζ R | dx (3.9)for any δ ∈ (0 , / u, v ∈ L ∗ ( R n ).By the H¨older inequality, we obtain Z D u |∇ ζ R | dx ≤ ( Z D u ∗ dx ) − /n ( Z D |∇ ζ R | n dx ) /n ≤ C. (3.10)Noting u p v ∈ L ( R n ), from (3.8)-(3.10) we deduce Z D |∇ u | ζ R dx ≤ C. Letting R → ∞ yields Z R n |∇ u | dx < ∞ . Similarly, we also obtain Z R n |∇ v | dx < ∞ . Combining the results above, we can see Z R n ( u ∗ + v ∗ + u p v + |∇ u | + |∇ v | ) dx < ∞ . Therefore, we can find R j such thatlim R j →∞ R j Z ∂B Rj ( u ∗ + v ∗ + u p v + |∇ u | + |∇ v | ) ds = 0 . (3.11)11 tep 3. Multiplying the first equation of (1.2) by u and integrating on D ,we get Z D |∇ u | dx − Z ∂D u∂ ν uds = √ p Z D u p vdx. (3.12)Here ν is the unit outward norm vector on ∂D . By the H¨older inequality, from(3.11) we deduce | Z ∂D u∂ ν uds | ≤ CR n − n − / − / ∗ ( R Z ∂D |∇ u | ds ) / ( R Z ∂D u ∗ ds ) / ∗ → R = R j → ∞ . Letting R = R j → ∞ in (3.12), we have k∇ u k = √ p k u p v k . Similarly, we can also obtain k∇ v k = √ p k u p v k .The following result shows that there does not exist L n ( p − ( R n )-solution if p is not equal to the critical exponent 2 ∗ − Theorem 3.3.
If a classical solution u belongs to L n ( p − ( R n ) , then p = 2 ∗ − ,and hence L n ( p − / ( R n ) = L ∗ ( R n ) .Proof. Write B = B R (0). Multiply (1.2) by x · ∇ u and x · ∇ v , respectively.Integrating on B , we get − Z B ( x · ∇ u )∆ udx = 1 √ p Z B v ( x · ∇ u p ) dx, − Z B ( x · ∇ v )∆ vdx = √ p Z B u p ( x · ∇ v ) dx. Integrating by parts yields − p Z ∂B | x || ∂ ν u | ds + p Z ∂B | x ||∇ u | ds − n − p Z B |∇ u | dx = √ p Z B v ( x · ∇ u p ) dx, and − Z ∂B | x || ∂ ν v | ds + 12 Z ∂B | x ||∇ v | ds − n − Z B |∇ v | dx = √ p Z B u p ( x · ∇ v ) dx. Adding two results together and integrating by parts again, we obtain − Z ∂B | x | ( p | ∂ ν u | + | ∂ ν v | ) ds + 12 Z ∂B | x | ( p |∇ u | + |∇ v | ) ds − n − Z B ( p |∇ u | + |∇ v | ) dx = √ p Z B x · ∇ ( u p v ) dx = √ p Z ∂B | x | u p vds − n √ p Z B u p vdx. R = R j → ∞ and using (3.11), we have n − Z B ( p |∇ u | + |∇ v | ) dx = n √ p Z B u p vdx. By (3.6) we see p = 2 ∗ − Theorem 3.4.
If a classical solution u of (1.2) belongs to L n ( p − ( R n ) , then u ( x ) = v ( x ) = c ( tt + | x − x ∗ | ) n − . (3.13) with some constant c = c ( n ) and for some t > .Proof. Step 1. We claim u ≡ v .Let W = u − v . By Theorems 3.1 and 3.2, we see Z R n ( | W | ∗ + |∇ W | ) dx < ∞ . Thus, when R = R j → ∞ , R Z ∂B R (0) ( | W | ∗ + | ∂ ν W | ) ds → . From (1.2), it follows ∆ W = √ pu p − W . Therefore, Z B |∇ W | dx + √ p Z B u p − W dx = Z ∂B W ∂ ν W ds. (3.14)Here B = B R (0). By the H¨older inequality, as R = R j → ∞ , | Z ∂B W ∂ ν W ds |≤ C ( R Z ∂B | W | ∗ ds ) / ∗ ( R Z ∂B | ∂ ν W | ds ) / R ( n − / − / ∗ ) − / − / ∗ → . Inserting this into (3.14) with R = R j → ∞ , we get Z R n ( |∇ W | + √ pu p − W ) dx = 0 , which implies u ≡ v . Step 2.
By virtue of u ≡ v and Theorem 3.3, (1.2) is reduced to the singleequation − ∆ u = √ pu ∗ − , u > in R n . According to the classification results in [5], the positive solutions must be theform as (3.13) in the critical case. 13he argument above implies u ∈ L n ( p − ( R n ) is equivalent to (3.13).At last, we complete the proof of Theorem 1.2. Theorem 3.5.
If the classical solution u of (1.2) belongs to L n ( p − ( R n ) , if andonly if one of the following items holds(C1) u ∈ L ∗ ( R n ) and p = 2 ∗ − ;(C2) u ∈ L ∞ ( R n ) and u ( x ) = O ( | x | − n ) as | x | → ∞ ;(C3) u ∈ D , ( R n ) .Proof. The necessity can be seen in Theorems 3.1-3.4. Next, we state the suffi-ciency.Both (C1) and (C2) can lead easily to u ∈ L n ( p − ( R n ). So, u ∈ L n ( p − ( R n )is equivalent to (C1) and (C2).Finally, (C3) implies (C1). In fact, by the Sobolev inequality we get u ∈ L ∗ ( R n ). On the other hand, using (3.8)-(3.10) we can deduce u p v ∈ L ( R n )from u ∈ D , ( R n ). Thus, (3.11) holds, and the proof of Theorem 3.3 still works.This shows p = 2 ∗ − Remark. u ( x ) = O ( | x | − n ) as | x | → ∞ in condition (C2) implies 0 < c ≤ u ( x ) | x | n − ≤ c as | x | → ∞ . It is led to by (2.1). If (3.1) holds, then (1.2) has a singular solution.
Theorem 4.1.
Let u ( x ) = c | x | t and v ( x ) = c | x | t solve (1.2) on R n \ { } , where c , c , t , t are positive constants. Then such singular solutions must be u ( x ) = v ( x ) = c | x | p − with c = [ n √ p ( p − − √ p ( p − ] p − .Proof. Write U ( r ) = U ( | x | ) = u ( x ) and V ( r ) = V ( | x | ) = v ( x ). Thus, − U ′′ − n − r U ′ = c t r t ( n − t − , − V ′′ − n − r V ′ = c t r t ( n − t − . Since u, v solve (1.2), it is easy to get (cid:26) t + 2 = t ( p −
1) + t ,t + 2 = t p, and (cid:26) c t ( n − t −
2) = √ pc p − c ,c t ( n − t −
2) = √ pc p as long as n − > max { t , t } . Therefore, t = t = p − and c = c =[ n √ p ( p − − √ p ( p − ] p − . In view of (3.1), the condition n − > max { t , t } is trueobviously. 14 emark. The fact u ≡ v is natural by an analogous argument in Step 4 ofthe proof of Theorem 4.4.Next, we search for bounded solutions.An example is the pair of cylinder-shaped solutions ( u ∗ , v ∗ ) (cf. [7]). Let u ∗ ( x ) = c ( tt + | x − x ∗ | ) n − . (4.1)According to Theorem 1.2, u ∗ solves (1.2) in the whole space R n in the criticalcase p = n +2 n − . Thus, it is not difficult to see that u ∗ ( x, x n +1 ) = u ∗ ( x ) and v ∗ ( x , x ) = u ∗ ( x ) still solves (1.2) in R n +1 . In view of p > n +3 n − , the problem(1.2) which u ∗ satisfies in R n +1 is equipped with the supercritical exponent.Clearly, this pair of positive solution ( u ∗ , v ∗ ) is neither radial nor decaying when | x | → ∞ . We also see u ∗ = v ∗ because the generating lines of the cylinders aredifferent.At last, we search for a radial bounded solution decaying with the slow rate p − .In order to find the existence of entire solutions in R n , we need the followingnonexistence result on a bounded domain. It can be verified by the Pohozaevidentity. Theorem 4.2.
Let D ⊂ R n be a ball centered at the origin. If p ≥ ∗ − , (4.2) then the following boundary value problem has no positive radial solution in C ( D ) ∩ C ( ¯ D ) − ∆ u = √ pu p − v, in D, (4.3) − ∆ v = √ pu p , in D, (4.4) u = v = 0 on ∂D. (4.5) Proof.
Suppose that u, v are positive radial solutions, we will deduce a contra-diction.Multiply (4.3) and (4.4) by u and v , respectively. Integrating on D and using(4.5), we have Z D |∇ u | dx = Z D |∇ v | dx = √ p Z D u p vdx. (4.6)Since u has the radial structure, |∇ u | = | ∂ ν u | on ∂D . Here ν is the unitoutward norm vector on ∂D . 15ultiplying (4.3) by ( x · ∇ u ) and integrating on D , we get − Z ∂D | x || ∂ ν u | ds + Z D |∇ u | dx + 12 Z D x · ∇ ( |∇ u | ) dx = 1 √ p Z D v ( x · ∇ u p ) dx. Integrating by parts and noting (4.5), we obtain12 Z ∂D | x || ∂ ν u | ds + n − Z D |∇ u | dx = n √ p Z D u p vdx + 1 √ p Z D u p ( x · ∇ v ) dx. Similarly, from (4.4) we also deduce that − Z ∂D | x || ∂ ν v | ds − n − Z D |∇ v | dx = √ p Z D u p ( x · ∇ v ) dx. Combining two results above with (4.6) yields − Z ∂D | x | ( | ∂ ν v | + 1 p | ∂ ν v | ) ds = n −
22 ( √ p + 1 √ p ) Z D u p vdx − n √ p Z D u p vdx. Therefore, n −
22 ( √ p + 1 √ p ) − n √ p < , which contradicts with (4.2).Based on the Liouville type result above, we can search for positive solutionsof (1.2) with radial structures. Let u, v be radially symmetric about x ∗ ∈ R n .We can write U ( r ) = U ( | x − x ∗ | ) = u ( x − x ∗ ) ,V ( r ) = V ( | x − x ∗ | ) = v ( x − x ∗ ) . Theorem 4.3.
Let p ≥ max { , ∗ − } . Then the following ODE system (cid:26) − ( U ′′ + n − r U ′ ) = √ pU p − V, − ( V ′′ + n − r V ′ ) = √ pU p , r > U ′ (0) = V ′ (0) = 0 , U (0) = 1 , V (0) = a, (4.7) has entire solutions for some positive constant a .Proof. Here we use the shooting method.
Step 1.
By the standard contraction argument, by p ≥ u a ( r ) , v a ( r ). Step 2.
We claim that either (4.7) has entire solutions for all a >
1, or forsome a ∗ >
1, there exists R ∈ (0 ,
1] such that u a ∗ ( r ) , v a ∗ ( r ) > r ∈ [0 , R )and u a ∗ ( R ) = 0.In fact, integrating (4.7) twice yields v a ( r ) = v a (0) − √ p Z r τ − n Z τ s n − u pa ( s ) dsdτ, u a ( r ) = u a (0) − √ p Z r τ − n Z τ s n − u p − a ( s ) v a ( s ) dsdτ. Thus, v a ( r ) − u a ( r ) = ( a −
1) + √ p Z r τ − n Z τ s n − u p − a ( s )( v a ( s ) − u ( s )) dsdτ. (4.8)Let a >
1. So we can find δ > v a ( r ) > u a ( r ) for r ∈ [0 , δ ) by thecontinuity of u a , v a .We claim v a ( r ) > u a ( r ) for all r ≥
0. Otherwise, there exists r ≥ δ suchthat v ( r ) = u ( r ) and v ( r ) > u ( r ) as r ∈ [0 , r ). From (4.8) with r = r wecan deduce a contradiction easily.Therefore, if u a ( r ) > r ≥
0, then (4.7) has entire solutions and theproof is complete. Otherwise, we can find
R > u a ( r ) , v a ( r ) > r ∈ (0 , R ) and u a ( R ) = 0. We denote the a in this state by a ∗ . Step 3.
We claim that for a < ε = n p √ p , there exists R ∈ (0 , u a ( r ) , v a ( r ) > r ∈ [0 , R ) and v a ( R ) = 0.In fact, from (4.7) we obtain u ( r ) > v ( r ) for r ≥ u ′ a , v ′ a < r > v a ( r ) ≤ a < ε and u a ( r ) ≤ u a (0) = 1. Thus, u a ( r ) = u a (0) − √ p Z r τ − n Z τ s n − u p − a ( s ) v a ( s ) dsdτ ≥ − ε √ pr n ≥ , for r ∈ (0 , v a ( r ) = v a (0) − √ p Z r τ − n Z τ s n − u pa ( s ) dsdτ < ε − √ p p r n . This proves that for a < ε , we can find R ∈ (0 ,
1] such that u a ( r ) , v a ( r ) > r ∈ (0 , R ) and v a ( R ) = 0. Step 4.
Let a = sup S , where S := { ε ; when a ∈ (0 , ε ) , ∃ R a > , such that u a ( r ) > , v a ( r ) ≥ , for r ∈ [0 , R a ] , v a ( R a ) = 0 } . Clearly, S = ∅ by virtue of ε ∈ S . From Step 2, it follows ε ≤ a ∗ for ε ∈ S .Namely, S is upper bounded, and hence we see the existence of a . Step 5.
Write ¯ u ( r ) = u a ( r ) and ¯ v ( r ) = v a ( r ). We claim that ¯ u ( r ) , ¯ v ( r ) > r ∈ [0 , ∞ ), and hence they are entire positive solutions of (4.7).Otherwise, there exists ¯ R > u ( r ) , ¯ v ( r ) > r ∈ (0 , ¯ R ) and oneof the following consequences holds:(1) ¯ u ( ¯ R ) = 0 , ¯ v ( ¯ R ) > v ( ¯ R ) = 0 , ¯ u ( ¯ R ) > u ( ¯ R ) = 0 , ¯ v ( ¯ R ) = 0 .
17e deduce the contradictions from three consequences above.(1) By C -continuous dependence of u a , v a in a , and the fact ¯ u ′ ( ¯ R ) <
0, wesee that for all | a − a | small, there exists R a > u ( r ) , ¯ v ( r ) > , f or r ∈ (0 , R a );¯ u ( R a ) = 0 , ¯ v ( R a ) > . This contradicts with the definition of a .(2) Similarly, for | a − a | small, there exists R a > u ( r ) , ¯ v ( r ) > , f or r ∈ (0 , R a );¯ u ( R a ) > , ¯ v ( R a ) = 0 . This implies that a + δ ∈ S for some δ >
0, which contradicts with the definitionof a .(3) The consequence implies that u ( x ) = ¯ u ( | x | ) and v ( x ) = ¯ v ( | x | ) are solu-tions of the system (cid:26) − ∆ u = √ pu p − v, − ∆ v = √ pu p , in B R ,u, v > in B R , u = v = 0 on ∂B R . (4.9)It is impossible by Theorem 4.2.All the contradictions show that our claim is true. Thus, the entire positivesolutions exist. Theorem 4.4.
The entire solutions obtained in Theorem 4.3 satisfy lim | x |→∞ u ( x ) = lim | x |→∞ v ( x ) = 0 (4.10) and u ≡ v . Moreover, either p = 2 ∗ − and u = u ∗ , or p > ∗ − and c ≤ u ( x ) | x | p − ≤ c when | x | → ∞ , where c , c are positive constants.Proof. Step 1. We claim lim r →∞ ¯ u ( r ) = 0.Eq. (4.7) implies ¯ u ′ < v ′ < r >
0. So ¯ u and ¯ v are decreasingpositive solutions, and lim r →∞ ¯ u ( r ), lim r →∞ ¯ v ( r ) exist.If there exists c > u ( r ) ≥ c for r >
0, then (4.7) shows that ¯ v satisfies ¯ v ′′ + n − r ¯ v ′ ≤ −√ pc p . Integrating twice yields ¯ v ( r ) ≤ ¯ v (0) − √ pc p r n r >
0. It is impossible since ¯ v is an entire positive solution. This shows that¯ u → r → ∞ . Step 2.
Furthermore, we claim u ( x ) ≤ c | x | − p − when | x | → ∞ .In fact, since u ( x ) , v ( x ) are the positive entire solutions of (1.2), they satisfy(1.5) according to the Remark in section 1. Moreover, u ( x ) , v ( x ) are radiallysymmetric and decreasing, we can deduce v ( x ) ≥ cu p ( x ) Z | x | r drr ≥ c | x | u p ( x ) , and hence u ( x ) ≥ cu p − ( x ) v ( x ) Z | x | r drr ≥ c | x | u p − ( x ) v ( x ) ≥ c | x | u p − ( x ) . This implies our claim.
Step 3.
We claim lim | x |→∞ v ( x ) = 0.Clearly, for large R >
0, we obtain from (1.5) that v ( x ) = c ( Z B R (0) u p ( y ) dy | x − y | n − + Z B | x | / ( x ) u p ( y ) dy | x − y | n − + Z B cR (0) \ B | x | / ( x ) u p ( y ) dy | x − y | n − ):= c ( I + I + I ) . First, u is bounded. So, as | x | → ∞ , I ≤ C Z B R (0) dy | x − y | n − ≤ C | x | − n → . Next, for large | x | , Step 2 implies u ( y ) ≤ c | x | − p − as y ∈ B | x | / ( x ). Therefore, I ≤ c | x | − pp − Z B | x | / ( x ) dy | x − y | n − ≤ C | x | − pp − → . Finally, using the Young inequality and letting R → ∞ , we have I ≤ C Z B cR (0) \ B | x | / ( x ) dy | x − y | n − | y | pp − ≤ C Z ∞ R r − pp − drr → . Combining the estimates of I , I and I , we getlim | x |→∞ v ( x ) = 0 . This result, together with Step 1, implies (4.10).
Step 4.
We claim u ≡ v . The argument in Step 1 of the proof of Theorem3.4 does not work, since the boundary integral is difficult to handle. We use19he idea in [17] to prove this uniqueness. This idea is universally valid, whichimplies the uniqueness as long as (4.10) holds. In addition, this idea is alsoindependent whether the value of p is critical or not.For any r ≥
0, we prove U ( r ) = V ( r ). Otherwise, either U ( r ) < V ( r ) , (4.11)or U ( r ) > V ( r ) . (4.12)If (4.11) holds, by the continuity of U and V , we can find R > U ( r ) < V ( r ) as r ∈ [ r , R ) . (4.13)Set R = sup { R ; (4 .
13) is true } . Thus, when R < ∞ , U ( R ) = V ( R ) . (4.14)In view of (4.10), this conclusion still holds even if R = ∞ .By (1.2), we see that for r > − ( r n − U ′ ) ′ = √ pr n − U p − V, − ( r n − V ′ ) ′ = √ pr n − U p . Integrating twice, we get U ( R ) = U ( r ) − √ p Z R r r − n Z r s n − U p − ( s ) V ( s ) dsdr,V ( R ) = V ( r ) − √ p Z R r r − n Z r s n − U p ( s ) dsdr. Using (4.11), (4.13) and (4.14), we obtain0 > U ( r ) − V ( r ) = √ p Z R r r − n Z r s n − U p − ( s )( V ( s ) − U ( s )) dsdr ≥ , which is impossible. So (4.11) is not true. Similarly, we also see that (4.12) isnot true. Since r is arbitrary, we know u ≡ v . Step 5.
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