Research Article Entire Solutions of an Integral Equation in R 5

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1 ISRN Mathematical Analysis Volume 03, Article ID 3839, 7 pages Research Article Entire Solutions of an Integral Equation in Xin Feng and Xingwang Xu Department of Mathematics, National University of Singapore, Block S7 (SOC), 0 Lower Kent Ridge Road, Singapore 9076 Correspondence should be addressed to Xingwang Xu; matxuxw@nus.edu.sg Received 9 May 03; Accepted 7 June 03 Academic Editors: G. Gripenberg, M. McKibben, and M. Winter Copyright 03 X. Feng and X. Xu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We will study the entire positive C 0 solution of the geometrically and analytically interesting integral equation: u(x) = / x y u q (y)dy with 0<qin. We will show that only when q=, there are positive entire solutions which are given by the closed form u(x) = c( + x ) / up to dilation and translation. The paper consists of two parts. The first part is devoted to showing that q must be equal to if there exists a positive entire solution to the integral equation. The tool to reach this conclusion is the wellknown Pohozev identity. The amazing cancelation occurred in Pohozev s identity helps us to conclude the claim. It is this exponent which makes the moving sphere method work. In the second part, as normal, we adopt the moving sphere method based on the integral form to solve the integral equation.. Introduction In this paper, we will study a very special type of the integral equation. From an analytical point of view, such equation is interesting to be studied. We should point out that even forradiuscase,theanalysisoftheequationhasbeenalready difficult. The method of moving planes has become a very powerful tool in the study of nonlinear elliptic equations; see Alexandrov [], Serrin [], Gidas et al. [3], and others. The moving plane method can be applied to prove the radial symmetry of solutions,andthenoneonlyneedstoclassifyradialsolutions. The method of moving planes in the integral form has been developed by Chen and Li []. This technique requires not only to prove that the solutions are radially symmetric, but also to take care of a possible singularity at origin. We note that in [], Li and Zhang have given different proofs to some previous established Liouville type theorems based on the method of moving sphere. It is this method that we applied to the integral equation to capture the solutions directly. As usual, in order for such a method to work, Pohozev s identity is a must. In our argument, we derive and use this powerful identity to conclude that the negative exponent q =. Then it is the method of moving spheres in the integral form which helps us to deduce the exact solution to the integral equation in this paper, instead of only getting radius solutions orprovingradialsymmetryofsolutions. The organization of the paper is as follows. In Section, we prove that q=. In this part, we use integration by parts to derive Pohozev s identity in the form Δ 3 u=u q.itiseasy to observe that if u solves our integral equation, then u also solves the differential equation Δ 3 u=u q.thuswecanput the integral form into Pohozev s identity to calculate every term. It turns out that the boundary integral term tends to zero which forces q=unless u + which is not allowed. In the third section, we adopt the moving sphere method in the integral form to solve the integral equation with q=.. q= Themainpurposeinthissectionistoshowthefollowing theorem. It is this exact exponent that makes the moving sphere workable for the integral equation. Notice that it is not hard to see that every C 0 positive solution of the integral equation is C 6. Thus in the following, we mainly assume that the solution is in C 6 ( ). Theorem. Suppose that u is a C 6 entire positive solution of the integral equation u (x)= x y u q (y) dy in, =ω with q>0,thenq=. ()

2 ISRN Mathematical Analysis Lemma. If u(x) satisfies (),thenδ 3 u(x) = u q (x). Proof. By direct calculation, we have Δ x u (x) = Δ x ( x y )u q (y) dy = x y u q (y) dy; Δ x u (x) = Δ C x ( x y )u q (y) dy = 8 x y 3 u q (y) dy. Since K(r) = (/ 3ω )r 3 is the fundamental solution of Laplace equation in,itfollowsthat Δ 3 8 xu (x) = ( 3ω ) Δ x ( 3ω x y 3 )u q (y) dy = ω u q (x) =u q (x). Thusthelemmaistrue. () (3) I = (x Br (0) u)q(x)u q (x)dx; I = Br (x u) (0) Δ 3 x u(x). First, rewrite I as I = [x ( u q (x) )] Q (x) dx. (8) q Using integration by parts, we have I = q u q (x) Q (x) rdδ q (Q + x Q) u q (x) dx. Now, the Second Green identity implies that I = [(x u) Δ u (Δ u) r + Δ (x u) Δ udx = (x u) ]dδ r (9) [r u Δ u (Δ u) r r r (r u r )]dδ+i 3 +I, (0) I 3 = Br (0) ΔuΔ udx; I = Br (0) x (Δu)Δ udx. As for I,sinceΔ(x u) = Δu + x Δu and by the Second Green identity again, we obtain Lemma 3. If u>0is the solution of the equation Δ 3 x u (x) =Q(x) u q (x) in, () then the following Pohozev identity holds: ( + q ) Qu q dx + P (u, r) = q rqu q dδ q (x Q) u q dx, () P (u, r)= [r u Δ u Δ u r r r (r u r )+ r( Δu r ) Δu r (r Δu r )+ rδuδ u + u r Δ u u u Δ r ] dδ. (6) Proof. Multiplying x uon the both sides of () and integrating over the ball,onehas 0= (x u)[δ 3 x u (x) Q(x) u q (x)]dx=i I, (7) I = Δ [x Δu] Δu dx + [x Δu Δu r = [Δ u+x Δ u] Δu dx Hence, + [r( Δu r ) I =I 3 + x Δ uδu dx + [r( Δu r ) Observe that I + x Δ uδu dx = x (ΔuΔ u) dx (x Δu) Δu] dδ r r (r Δu )Δu] dδ. r r (r Δu )Δu] dδ. r = rδuδ udδ ΔuΔ udx = rδuδ udδ I 3 ; () () (3)

3 ISRN Mathematical Analysis 3 that is, I = rδuδ udδ I 3. () Combining ()and(), we have I = 3 I 3 + [r( Δu r ) r (r Δu r )Δu+rΔuΔ u] dδ. () Thus I = I 3+ [r u Δ u Δ u r r r (r u r )+ r( Δu r ) Δu r (r Δu r )+ rδuδ u] dδ. (6) To deal with I 3, we observe that if multiplying u on both sides of () and integrating over the ball,onehas 0= u[δ 3 u Q(x) u q ]dx. (7) Thus, with the help of the Second Green identity, one obtains Q (x) u q dx = uδ 3 udx = ΔuΔ udx+ [u Δ u u r r Δ u] dδ. Therefore, we get I 3 = ΔuΔ udx (8) = Q (x) u q dx + [ u r Δ u u Δ u r ] dδ. (9) Put it into (6)toconcludethat I = Qu q dx + [r u Δ u Δ u r r r (r u r )+ r( Δu r ) Δu r (r Δu r )+ rδuδ u + u r Δ u u u Δ r ] dδ. (0) Put I and I into the equation I I =0and rearrange the terms; then we get the desired identity. Thus the lemma holds. Lemma. If u is an entire solution of the integral equation (), then q=. Proof. Define d = (/ ) u q (y)dy, d = (/ ) y u q (y)dy.thenitfollowsthat u (x) = x y u q (y) dy x u q (y) dy + y u q (y) dy = x d+d. If 0<q 6,then d =u(0) = y u q (y) dy y ( y d+d ) q (y) dy = ω + r (rd + d C ) q dr 0 =, () () which contradicts the assumption that u is an entire solution. Therefore, we only consider the case q>6. Rewrite the equation as u (x) =d x +β (x), (3) β (x) = [ x y x ]u q (y) dy. () Clearly, we have β (x) y u q (y) dy = d. () And, by (), one has u r = x (x y) r x y u q (y) dy = u q (y) dy+ := d + β (x), r [ (x y) x y x r ]u q (y) dy (6) β (x) = (x y) x r x y x y u q (y) dy. (7)

4 ISRN Mathematical Analysis Note that β (x) canbecontrolledasfollows: β (x) = [ x y x ] x y +y (x y) x y u q (y) dy y u q (y) dy =d. (8) Clearly, with the definition of β (x), thefollowingholds true: r u r =dr+β (x). (9) Then differentiate it to get r (r u r ) =d+ [ (x y) x y x (x y)(x y) x y 3 x x ] := d + β 3 (x), x r u q (y) dy β 3 (x) = [( x x y ) One observes that x β 3 (x) = [ x ( x x y M 3, u q (y) dy x y (30) x (x y)y (x y) r x y 3 ]u q (y) dy. (3) x y u q (y) dy ) x (x y)y (x y) x y 3 ] (3) for a constant independent of x and for x sufficiently large. Here we have used the assumption that q > 6 and u(x) = d x + β (x), β (x) is bounded by a constant. The integral over canbecontrolledasfollows: x y x y u q (y) dy = + ) x y (A A x y u q (y) dy =S +S, (33) A ={y : x y x /}, A ={y : x y x /}. Then S = A x y x y u q (y) dy A y u q (y) dy d. (3) As for S, y A implies that y x + x y (3/) x and y x x y (/) x. Therefore S = A x y x y u q (y) dy q [ min u(y)] [ x / y / A x y dy] q =[ min u(y)] x / y / [ min u(y)] qω 6 x / y / 7. x / [ω r 3 dr] 0 (3) Since u(x) = O( x ) and q>6,soi is bounded for x R 0 when R 0 is large enough. It follows that x β 3 (x) is bounded for x R 0 when R 0 is large enough. Now we deal with the higher order derivatives. First Δu (x) = x y u q (y) dy = x d+β (x), (36) β (x) = [ x y x ]u q (y) dy. (37) For a constant M independent of x, when x is large enough, it is easy to see that x β (x) = x [ x x y x x y ]u q (y) dy M, by the estimation in β 3. x y x y u q (y) dy (38)

5 ISRN Mathematical Analysis Δu r Then, by computation, one gets = [ (x y) x x y 3 x ]u q (y) dy = x d+ [ (x y) x x y 3 x + x ]u q (y) dy = x d+β (x), r (39) β (x) = (x y) x [ x y 3 + x ]u q (y) dy. (0) And as usual, we observe that x β (x) = [ (x y) x x x y 3 + x ]u q (y) dy = x x y 3 x (x y) x x y 3 u q (y) dy = x [ x y (x y)(x y) x x (x y)] x y 3 u q (y) dy = x [( x y x )x (x y) x y y (x y)] x y 3 u q (y) dy A ={y : x y x /}, A ={y : x y x /}. Then S 3 = A x y x y u q (y) dy A y u q (y) dy d. (3) As for S, y A implies that y x + x y (3/) x and y x x y (/) x. Therefore S = A x y x y u q (y) dy q 3 [ min u(y)] [ x / y / A x y dy] q 3 =[ min u(y)] x / y / x [ min u(y)] qω 6 x / y /. x / [ω r dr] 0 () Since u(x) = O( x ) and q>6,sos is bounded for x R 0 when R 0 is large enough. It follows that x β (x) is bounded for x R 0 when R 0 is large enough. Since r( Δu/ r) = (/ x )d + β (x),weobtain r (r Δu r )= x d+β 6 (x), () [ x y x y + x y x y ]u q (y) dy. Similarly, we have the following estimation: x y x y u q (y) dy + ) x y =(A A x y u q (y) dy =S 3 +S, () () β 6 (x) = r β (x) = (x y) x [ x y 3 + x ] x x u q (y) dy = [ (x y) x y 3(x y) x(x y) x y x x 3 ] x x u q (y) dy

6 6 ISRN Mathematical Analysis = { x x y [3 x y (x y) x u q (y) dy +3y (x y)(x y) x x (x y) x y x x y ] x } = [ (x y) x x x y 3 + 3y (x y)(x y) x x x y x x y 3 x ]u q (y) dy. As before, we can estimate β 6 (x) as x 3 β 6 (x) = [ (x y) x x x y 3 x x y 3 x ]u q (y) dy = [( (x y) x x x y 3 x x y 3 ) [ x x y x y 3 (x y) x x +( x y 3 x ) (6) + 3y (x y)(x y) x x x y + 3y (x y)(x y) x x x y ]u q (y) dy +(( x x (x y) x x y x (x y) + x x y y (x y)) ( x y 3 ) ) + 3y (x y)(x y) x x x y ]u q (y) dy x 3 y x y 3 u q (y) dy+ x y x y u q (y) dy + M 6, x y x y u q (y) dy+ x 3 y x y 3 u q (y) dy (7) for a constant independent of x and for x sufficiently large. We have proved that the first integral is bounded in the previous estimation. The second integral over can be controlled as follows: x 3 y x y 3 u q (y) dy + ) x 3 y =(A A x y 3 u q (y) dy =S +S 6, (8) A ={y : x y x /}, A ={y : x y x /}. Then S = A x 3 y x y 3 u q (y) dy 8A y u q (y) dy 8d. (9) As for S 6, y A implies that y x + x y (3/) x and y x x y (/) x. Therefore S 6 = A x 3 y x y 3 u q (y) dy q [ min u(y)] [ x / y / A x y 3 dy] q =[ min u(y)] x / y / [ min u(y)] qω 6 x / y /. x / [ω rdr] 0 (0) Since u(x) = O( x ) and q>6,sos 6 is bounded for x R 0 when R 0 is large enough. It follows that x 3 β 6 (x) is bounded for x R 0 when R 0 is large enough. Once again, calculation shows that Δ u (x) = 8 x y 3 u q (y) dy = 8 x 3 d+β 7 (x), () β 7 (x) = 8 [ x 3 x y 3 ]u q (y) dy. ()

7 ISRN Mathematical Analysis 7 We still need to control β 7 (x) which can be done as follows: x β 7 (x) = 8 [ x x x y 3 ]u q (y) dy = 8 x R x y ( x x y+ y ) x x y 3 u q (y) dy = 8 x 3 [ R x y x ]+ x x y ( x y + y ) x y 3 u q (y) dy 8 x 3 y x y 3 u q (y) dy + 8 x y x y u q (y) dy + 8 x y x y u q (y) dy M 7, (3) for a constant independent of x and for x sufficiently large. Now, the third integral over can be controlled as follows: x y x y u q (y) dy = x y y x y x y u q (y) dy x y x y [ x y + x x y ]u q (y) dy x y x y u q (y) dy + x y x y u q (y) dy. () The two integrals are bounded as we have shown before. It follows that x β 7 (x) is bounded for x R 0 when R 0 is large enough. Finally, by computation, one obtains Δ u r = (x y) x x y x u q (y) dy = x d+β 8 (x), () β 8 (x) = [ (x y) x x y x x ] u q (y) dy. (6) The estimate on β 8 (x) can proceed as follows: x β 8 (x) = [ (x y) x x x y x ]u q (y) dy = x (x y) x x x y x y u q (y) dy = (([ x 3 R x y 3 ] x x (x y)+ x x y 3 y (x y))( x y 3 y (x y)) ) u q (y) dy [ x 3 x y ( x x y+ y )] x R x y + u q (y) dy x y x y u q (y) dy [ x y 3 x y + x y x y 3 + x y x y 3 ]u q (y) dy + M 8, x y x y u q (y) dy (7) for a constant independent of x and for x sufficiently large. As before, the aforementioned first integral can be controlled as follows: x y x y u q (y) dy =(A + A ) x y x y u q (y) dy =S 7 +S 8, (8) A ={y : x y x /}, A ={y : x y x /}.

8 8 ISRN Mathematical Analysis Then Also we have S 7 = A x y x y u q (y) dy 6A y u q (y) dy (9) [r u Δ u Δ u r r r (r u r ) + r( Δu r ) Δu r (r Δu r ) 6d. As for S 8, y A implies that y x + x y (3/) x and y x x y (/) x. Therefore S 8 = A x y x y u q (y) dy q [ min u(y)] dy] x / y / [A q =[ min u(y)] x / y / [ min u(y)] qω 6 x / y /, x / [ω dr] 0 (60) which is bounded, since u(x) = O( x ) and q>6,soi 8 is bounded for x R 0 when R 0 is large enough. Besides, y x x y 3 u q (y) dy = x y x y y x y u q (y) dy x y x y u q (y) dy + x 3 y x y 3 u q (y) dy, (6) andthetwointegralshavebeenshowntobeboundedbefore. It follows that x β 8 (x) is bounded for x R 0 when R 0 is large enough. Combining these estimates, we can see that () x β (x)β 8 (x) 0 as x ; () x β 3 (x)β 7 (x) 0 as x ; (3) x 3 β (x) 0as x ; + rδuδ u+ u r Δ u u u Δ r ] =r[d+ β (x) ][ r x d+β 8 (x)] [d+β 3 (x)][ 8 x 3 d+β 7 (x)] + r[ x d+β (x) r ] [ x d+β 6 (x)][ x d+β (x)] + r[ x d+β (x)][ 8 x 3 d+β 7 (x)] + [d + β (x) ][ 8 r x 3 d+β 7 (x)] [d x +β (x)][ x d+β 8 (x)] =d[ x β (x) + 0 x β (x) + 8 x 3 β 3 (x) + 6 x β (x) + x β (x) + x β 6 (x) + 3 β 7 (x) + x β 8 (x)] +[β (x) β 8 (x) β 3 (x) β 7 (x) + r β (x) β (x) β 6 (x) + rβ (x) β 7 (x)+ r β (x) β 7 (x) β (x) β 8 (x)]. (6) Observe that x [β (x) β 8 (x) β 3 (x) β 7 (x)+ r β (x) () x β (x)β 6 (x) 0 as x ; () x β (x)β 7 (x) 0 as x ; (6) x 3 β (x)β 7 (x) 0 as x ; (7) x β (x)β 8 (x) 0 as x. β (x) β 6 (x) + rβ (x) β 7 (x) + r β (x) β 7 (x) β (x) β 8 (x)] 0 as x. (63)

9 ISRN Mathematical Analysis 9 The coefficient of d is given by [ x β (x) + 0 x β (x) + 8 x 3 β 3 (x) + 6 x β (x) + x β (x) + x β 6 (x) + 3 β 7 (x) + x β 8 (x)] =[ x ( 3β (x) +β (x))+ 8 x 3 β 3 (x) x (3β (x) +β (x)) x β 6 (x) + 3 β 7 (x) + x β 8 (x)] = [( x y x +(8 x x y 8 x 3 (x y) x +0 x x y 8 x 3 ) 8 x (x y)y (x y) x x y 3 ) +( x x y +8 x 3 +6(x y) x x y 3 x ) +( 6 (x y) x x y 3 x y (x y)(x y) x x x y +8 x y 3 +8 x 3 ) +( x 3 x y 3 ) +( (x y) x x y x 3 )] u q (y) dy = { x x y [ 3 x y 6 +(x y) x x y x x y x (x y)y (x y) x y 6y (x y)(x y) x x x x y +3 (x y) x x ]} S = [ 3 x y 6 +(x y) x x y x x y x (x y)y (x y) x y 6y (x y)(x y) x x x x y +3(x y) x x ] =[x (x y) x y +3y (x y) x y x x y x (x y)+ x x y y (x y) x (x y)y (x y) x y 6y (x y)(x y) x x + x x (x y)+ x y (x y)] =x (x y)[ x y x x y + x ] +y (x y)[3 x y + x x y x And directly calculation gives (x y) x y 6x (x y) x + x ]. (6) [3 x y + x x y x (x y) x y 6x (x y) x + x ] =x (x y) x y + x x y 6x (x y) x + x 3y (x y) x y =(x y) +y (x y)[x (x y)+ x 3y (x y)]. (66) By substitution, we have S = x (x y) [y (x y) y (x y) + (x y) y (x y)+(x y) ] +y (x y)(x y) [y (x y)] [x (x y)+ x 3y (x y)] u q (y) dy = { x x y S} u q (y) dy, (6) = x (x y)[y (x y)] +(x y)x (x y)y (x y) x x[y (x y)] +(x y) x (x y) +y (x y)(x y) +3[y (x y)] 3

10 0 ISRN Mathematical Analysis =y (x y) [ x (x y)+ y (x y) y (x y)] +(x y) x (x y)+y (x y) (x y) +3[y (x y)] 3 = x (x y)y (x y)+(x y) x (x y) +y (x y)[6 x y (x y) +3[y (x y)] ] y (x y)y (x y) =8 x y (x y) 6 x y 0 y (x y) + y (x y) 3 y 6. Hence [ x ( 3β (x)+β (x))+ 8 x 3 β 3 (x) x (3β (x)+β (x)) x β 6 (x)+ 3 β 7 (x)+ x β 8 (x)] = { x x y [8 x y (x y) 6 x y (67) 0 y (x y) + y (x y) 3 y 6 ]u q (y)}dy. (68) Now, integrate it over with r= x,andweobtain that A:= { x x y [8 x y (x y) 6 x y 0 y (x y) + y (x y) 3 y 6 ] u q (y)}dydδ = ( + ) y < x y > x { x x y [8 x y (x y) 8 x y + x y 0 y (x y) + y (x y) 3 y 6 ] u q (y) dδ} dy := I +I +I 3 +II +II +II 3, (69) I = y < x x x y [8 x y (x y) 8 x y ] u q (y) dy dδ 3 = y y < x x u q (y) [ (x y) y x y dδ] dy. (70) It is clear that Δ x [(x y) y ]=0, by Poisson representation formula for harmonic function, the previous integral equals 3 I = y y < x x u q (y) [ x ω x y (y y y )] dy =0. (7) Similarly, calculation gives Δ x [ x y 0 y (x y) ]= 0 y 0 y =0, and then Poisson representation formula implies that I = y < x x x y = y < x x u q (y) [ x y 0 y (x y) ]u q (y) dy dδ x y 0 y (x y) x y dδ dy = y < x x u q (y) [ x ω x y ( y 6 0 y 6 )] dy = y < x x u q (y) [ x ω x y ( 8 y 6 )] dy. (7)

11 ISRN Mathematical Analysis Once again, Δ x [ y (x y) 3 y 6 ]=0,andPoisson formula leads to I 3 = y < x x x y [ y (x y) 3 y 6 ]u q (y) dδ dy = y < x x u q (y) y (x y) 3 y 6 x y dδ dy = y < x x u q (y) [ x ω x y (8 y 6 )] dy. Notice that I +I 3 =0. Next, consider the following integral: II = y > x x x y [8 x y (x y) + y (x y)] u q (y) dδ dy = y > x x u q (y) [8 x y + y ] x y (73) x y dδ dy. (7) Setting z=( x y)/ y, the previous integral is equal to II = y > x x u q (y) [8 x y + y ] x ( x z/ z ) x ( x z/ z ) dδ dy = y > x x u q (y) [8 x y + y ] ( x / z )x z ( x / z ) dδ dy x z = y > x x u q (y) [8 x y + y ]( x 3 z ) x ω x z z zdy = u q (y) y > x Also, we have x [ 3ω [ y ( y x ) x +ω y y x ]dy. (7) II = y > x x x y [ 8 x y 3 y 6 ] u q (y) dδ dy = y > x x u q (y) [ 8 x y 3 y 6 ] x y dδ dy = y > x x u q (y) [ 8 x y 3 y 6 ] x ( x z/ z ) dδ dy = y > x x u q (y) [ 8 x y 3 y 6 ] ( x z ) dδ dy x z = y > x x u q (y) [ 8 x y 3 y 6 ] = y > x ( x z ) x ω x z dy u q (y) [ 3ω y x y x ω y 3 y x ]dy. (76) Finally, it follows from variable change and Poisson formula for the harmonic function ( 0 y (x y) + x y ) that II 3 = y > x x x y [ 0 y (x y) + x y ] u q (y) dδ dy

12 ISRN Mathematical Analysis = y y > x x u q (y) x ( x z/ z ) = y y > x x u q (y) [ 0(x x z z ) + x 6 z ]dδdy ( x / z ) x z = y y > x x [ 0( (x z) x z ) + x 6 z ]dδdy x y u q (y) 0(x z) + x z x z dδ dy = y x ω y > x x 3 u q (y) x z [8 z ]dy = 3ω u q (y) y > x x ( y x ) y dy. (77) Combining those previous calculations, we can conclude that A=I +I +I 3 +II +II +II 3 = u q (y) y > x + y > x [ x 3ω [ y ( y x ) +ω x y ] y dy x ] u q (y) [ 3ω y x y x ω y 3 y x ]dy 3ω u q (y) y > x = ω u q (y) y > x y dy 0, x ( y x ) y dy as x,sinceu(0) = u q (y) y dy < +. (78) What we have shown so far is that, if u is regular, that is, q>6,then P (u, r) 0 as r. (79) Setting Q=,whenq>6,wehave ru q dδ 0, (80) as x,sinceu(x) = O( x ) when x is large enough. Hence, let x, and we conclude that q=from (). 3. Moving Sphere Method This section is to run the method of moving sphere for the integral equation with the exact exponent q=. First, define V(x) = x u(x/ x ) and V (x) = u(x/ ); it is rather easy to see that V (x) = ( x /)υ(( x)/ x ). Lemma. If one sets ω (x) = υ(x) + υ (x), thenω (x) = B [G(x, z)ψ(z)ω (z)]dz, ψ = (/(υ υ )) ] 0; G(x, z) = [( z /) x (( z)/ z ) x z ] 0when x and z. Inparticular, ω (0) = B [ z ]ψ(z)ω (z)dz,whichbelongsto(0, + ). [ 0 i=0 υi υ 0 i Proof. Since u(x) = (/ ) x y u (y)dy, V (x) = x u( x x ) = x x x y u (y) dy = R y x = (I +I ), y y u (y) dy (8) I = B/ [ y x (y/ y ) u (y)]dy, I = R [ y x (y/ y ) u (y)]dy. /B / As for I,sety=z/ to obtain I = [ z B x z/ z/ u ( z )] d z = [ B z x ( z z ) u ( z )] dz. (8)

13 ISRN Mathematical Analysis 3 As for I,ifwesety = z/ z,thenweobtain I = [ z B z [ x z/ z z/ z u ( z z ) ] ] = B [ z x z u ( z z )]dz. Similarly, one can write V (x) as V (x) =u( x ) d z z = x + )[ (B / /B / y u (y)] dy = (I 3 +I ). As for I 3,bysettingy=z/, one concludes that (83) (8) Now, since z x z z x z = z [ x x, z + z z ] [ x x, z + z ] = x z + z x = x ( z )+( z ) =( x )( z ) = ( x )( z ), (88) G(x, z) = [( z /) x ( z/ z ) x z ] 0when x and z. Put x=0to obtain I 3 = B [ x z u ( z )] dz. (8) For I,asbefore,sety = z/ z to get I = [ z 0 x z B z u ( z )]dz. (86) z Therefore, combining these previous calculations, we get ω (x) = [(I I )+(I I 3 )] = [ z x z B z = [ z B x z z [ υ (z) υ ω (z) x z ][V (z) V (z)]dz x z ] = B [G (x, z) ψ (z) ω (z)]dz, (z) ]ω (z) dz (87) ψ = [(υ (z) υ (z))/ω (z)] = (/υ υ ) ] 0and G = [( z /) x ( z/ z ) x z ]. [ 0 i=0 υi υ 0 i ω (0) = B [G (0, z) ψ (z) ω (z)]dz which belongs to (0, + ). Thuswecompletetheproof. = B [ z ] ψ (z) ω (z) dz, (89) Proposition 6. If >0is large enough, then ω (x) 0 for all x such that x. Proof. We prove that ω (x) 0 when >0is large enough by three steps. Step. There exists R 0 >0such that for R 0 x /, we have ω 0. It is clear to see that x/ x = / x /R 0 ; x/ / /R 0. When R 0 is large enough, ω (x) = x u( x x )+u(x ) = x [u (0) +O( )]+[u(0) +O( x x )] x u (0) >0. (90) Step. There exists an R >R 0 such that for R / x,wehaveδω 0and ω 0.

14 ISRN Mathematical Analysis First, we calculate Δω. In fact Δω = Δ(u(x/ )) + Δ( x u(x/ x )), Δ(u(x/ )) = (/ )(Δu)(x/ ) and Δ x u(x/ x ) = (Δ x )u(x/ x ) + (Δu(x/ x )) x + x u(x/ x ). To carry out the detail, forn =one observes that Δ x = x, x u( x x )= x 3 u i ( x x ) x i, Δu ( x )= x (Δu) ( x )+( 6) x x u i ( x x i= x ) x i. (9) Thus Δω = x [ x 3 (Δu) (x ) x (Δu) ( x x ) +8 x i= i= (9) u i ( x x ) x i V (x)]. x If R >,then x/ x =/ x /R if x R.Let C=max B u. Then 8 i= u i(x/ x ) (x i / x ) 8C/ x when R is large enough. Now let B=max B Δu. Hencewe have x 3 (Δu) ( x ) x (Δu) ( x x ) 3 B+ x B 3 B. (93) Notice that if x is sufficiently large, then V(x) (u(0)/ ) x. Therefore if >0is large enough, then Δω (x) 0 for R / x. Besides, ω (x) x = =0by definition, and ω (x) x =/ 0 by Step. Thus the maximal principle implies that ω (x) 0 on the domain R / x. Step 3. There exists an R R such that ω (x) 0 for x R 0 for large enough. On the one hand, when x is small, V(x) = x u(x/ x ) x c x/ x =c,andv is continuous on the rest of ball, so V is bounded for x R 0. On the other hand, the definition V (x) = u(x/ ) implies that V (x) (u(0)/) when is large enough. Therefore, ω (x) > 0 for 0< x R 0 when is large enough. Combining Steps to 3, we have ω (x) 0. Thuswecompletetheproofofthisproposition. Now, for any b, we denote u(x + b) by u b (x), V b (x) by x u b (x/ x ),andsetω,b (x) = V b (x) + (V b ) (x). By Proposition 6,if is large enough, then ω,b 0.Thus, we define b = inf{ > 0 : ω μ,b (x) 0 in B μ (0) for <μ< + }. Proposition 7. There exists b such that b >0. Proof. Suppose that the proposition is not true. Hence b =0 for all b; thatis,forallb and all >0,wealwayshave ω,b (x) 0 for all x B (0). Note that is, ω x,b (x) = V b (x) + x V b ( x x ) =0, = x ω x,b (rx) = V b (rx) +rv b ( x r ) 0 when 0<r<. (9) Therefore, it follows that (d/(dr))ω x,b (rx) r= 0;that V b (rx) x + V b ( x r )+r V b ( x r )( x r ) 0. (9) r= Simplifying this, we get ( V b ) x+v b (x) 0, (96) V b (x) = x u((x/ x )+b). Setting y = x/ x,bycalculationweget V b x= x u( x x +b)+ x ( u) ( x x +b) x x = x u(y+b) x ( u) (y + b) y. (97) Substituting (97)into(96)anddoingvariablechange,we obtain u(y)+( u) (y) (y b) 0. (98) Then both sides being divided by b and sending b to +,letb = b ω b,weget ( u) (y) (ω b ) 0. (99) Since the previous inequality holds for all b,so ( u) = 0. It implies that u is a constant which contradicts u(x) = (/ ) x y u (y)dy. Thus this proposition holds. Proposition 8. Suppose that there exists a point b such that b >0;thenω b,b 0for all x. Proof. Without loss of generality, we assume that b = 0. Suppose that the proposition is not true; then ω 0,0(x) 0. In order to complete the argument, we need to do some preparations. First of all, setting C 0 = min ω x B0 / 0,0(x) > 0,thenitis clear that ω 0 (x) C 0 >0 x B 0 /. (00)

15 ISRN Mathematical Analysis Next, calculate ( G(x, z)/ η x ) x =0 as follows. Sincewehavetheformula which is because thus, ( z G (x, z) = z z x z z x z z = z = z ) = x z x x x z z,x z z [ x, x x, z + z z ] = z x x, z + = x =( x x z z [ z, z x, z + x x ] z x x ), x z = x Hence, by direct calculation, one obtains G(x, z) η x x =0 = ( x G (x, z) i i x i x ) x = 0 = = = i= i= i= [ z [( z [( z x x i ( z z x i x ] ) x i x i ( z i / z ) x ( z/ z ) x ] x i ( z i / z ) i= ( x / z ) z ( x/ x ) x = 0, (0) z x x (0) x z. (03) [( x i x z ) x i x ] x i z i x z ) x i x ] x i z i x z ) x = 0 = z 0 0 x z. (0) It clearly implies that whenever z. Therefore, by Lemma, ω 0 (x) η x B0 (0) G (x, z) 0 η (0) x B0 (0) = z 0 B 0 (0) 0 x z ψ (z) ω 0 (z) dz. (06) By the definition of 0 and our assumption that ω 0,0(x) 0,fromLemma, weseethatω 0 (x) > 0 inside the ball B 0 (0). Then, at every boundary point x,wehave ω 0 (x) η x B0 (0) <0. (07) Besides, the definition of 0 implies that there is a sequence k such that k 0, k < 0, inf ω k (x) <0. B k (0) (08) It follows from (08) that there exists x k B k (0) such that ω k (x k )= min (x) <0. B k (0) (09) Since ω k (x) 0 for x = k,wehave x k < k < 0.Clearly, x k has a convergent subsequence, still denoted by x k,witha limit point x 0.Also, ω k (x k )=0.Thenx 0 satisfies ω 0 (x 0 ) =0, ω 0 (x 0 ) 0. (0) Now consider the following two cases. Case ( x 0 < 0 ).Theconclusionω 0 (x 0 ) 0contradicts the fact that ω 0 (x 0 ) is strictly positive inside the ball B 0 (0). Case ( x 0 = 0 ). Then the fact that ω 0 (x 0 )=0contradicts (07). Hence, the proposition follows. Proposition 9. For all b,onehas b >0. Proof. It follows from Propositions 7 and 8 that there exists some b such that b >0and ω b (x) = 0;thatis,V b (x) = (V b (x)) ;thatis, x u b ( x x ) = b u b ( x ). () b Setting y = x/ x,weget y u b (y) = b u b ( y b y ). ()

16 6 ISRN Mathematical Analysis It follows that lim [ y y u b (y)] = b u b (0). (3) Suppose that there exists a point b such that b =0. Then ω,b (x) = V b (x) + (V b (x)) 0for all >0and x B (0)/{0}. That is, u b ( x / ) x u b (x/ x ) for all >0and x B (0)\{0}. Fixing > 0 and letting x 0, wehaveu b (0) b u b (0). Therefore, b u (b) u(b). () Letting go to 0,wehave0 u(b),whichcontradictsthe fact that u isapositiveentiresolution. Hence, this proposition holds. Proposition 0. If u > 0 satisfies u(x) = (/ ) x y u (y)dy, then there exist some x 0 and constants a, d such that u(x)=(a x x 0 +d) /. Proof. From Propositions 8 and 9,weknowthatforallb, we have ω b,b 0;thatis,V b (x) = ( x / b )V b ( b x/ x ). Using V(x) = x u(x/ x ),settingy=x/ b and μ b = b, we can rewrite it as follows: u(y+b)= y μ b u(μ b y y +b). () Now setting x=y+b,thenwehave u (x) =μ b x b u( μ b (x b) x b +b). (6) Itfollowsthatthelimitlim x x u(x) exists and is equal to μ b u(b) for all b R. Then set A=lim x x u(x) = μ b u(b). Case. If A=0,thenu 0whichisimpossibletobeapositive solution. Case. IfA =0, the definition of A implies that A > 0. Without loss of generality, we can assume that A=.Now for x large, u (x) = μ 0 u {u (0) + (0) μ 0 x i x x i x +o( )}, (7) x so the coefficient of x i is (μ 0 / x )( u/ x i )(0),and u (x) = μ b u {u (b) + (b) μ b (x i b i ) x b x i x b +o( x )} =μ x b [ x + i x ( b i ) +o( x )] [u(b) + u x i (b) μ b x x i +o( x )], (8) so the coefficient of x i is (/ x ) + ( u/ x i )(b)(μ b / x ). Comparing these two coefficients and A=,we obtain Then Thus, x + u (b) μ b x i x = μ 0 u (0). (9) x x i u b u x i (b) =u 0 u x i (0) +b i. (0) u = u + x. () x i (x)x=b x i (x)x=0 x i (x)x=b It gives u (x) x =C x i. () i x=b By solving the previous equation, we obtain u (x) = x +Cx+D= x C for some x 0,equivalentto Acknowledgments +E= x x 0 +d, (3) u (x) =a( x x 0 +d) /. () The paper is mainly taken from first author s master thesis at National University of Singapore. She is sincerely grateful to her family, friends, and all other people who helped her during the completion of the project. The research of the second author is supported through his NUS research Grant R References [] A. D. Alexandrov, Uniqueness theorems for surfaces in the large, American Mathematical Society Translations, vol.,pp. 6, 96. [] J. Serrin, A symmetry problem in potential theory, Archive for Rational Mechanics and Analysis,vol.3,pp.30 38,97.

17 ISRN Mathematical Analysis 7 [3] B.Gidas,W.M.Ni,andL.Nirenberg, Symmetryandrelated properties via the maximum principle, Communications in Mathematical Physics,vol.68,no.3,pp.09 3,979. [] W. X. Chen and C. Li, A necessary and sufficient condition for the Nirenberg problem, Communications on Pure and Applied Mathematics,vol.8,no.6,pp ,99. [] Y. Li and L. Zhang, Liouville-type theorems and Harnack-type inequalities for semilinear elliptic equations, Journal d Analyse Mathématique,vol.90,pp.7 87,003.

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