On absence of solutions of a semi-linear elliptic equation with biharmonic operator in the exterior of a ball

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1 Tansactions of NAS of Azebaijan, Issue Mathematics, 36, Seies of Physical-Technical and Mathematical Sciences. On absence of solutions of a semi-linea elliptic euation with bihamonic opeato in the exteio of a ball Shimayil H. Bagiov Mushfi J. Aliyev Received: / Revised: / Accepted: Abstact. We study absence of global solutions of a semi-linea elliptic euation with a bihamonic opeato u c u x σ u = 0 in the exteio of a ball. Sufficient condition absence of global x solutions is obtained.the poof is based on the method of a test functions. Keywods. Semi-linea elliptic euation, bihamonic opeato, global solution, citical exponent, method of test functions. Mathematics Subject Classification 010: 35A01, 35J61, 35J91 1 Intoduction Intoduce the following denotations. Let R > 0, B R = {x; x < R}, BR c = {x; x > R}, B R1, R = {x; R 1 < x < R }, B R = {x; x = R}, whee x = x 1,..., x n R n, = x = x 1... x. In BR c conside the euation u c x u x σ u = 0, 1.1 whee 0 c n, > 1, σ >, u = u, is ndimensional Laplace opeato. Denote α ± = n ± D, D = n c. We will study the existence of global solutions of euation 1.1 in BR c satisfying the condition udx B R Unde the global solution of poblem 1.1, 1. we undestand the function u x C BR c,satisfying condition 1. on the bounday and euation 1.1 at evey point of BR c. Shimayil H.Bagiov sh bagiov@yahoo.com Baku State Univesity Institute of Mathematics and Mechanics, Baku, Azebaijan Mushfi J.Aliyev a.mushfi@amble.u Institute of Mathematics and Mechanics, Baku, Azebaijan

2 6 On absence of solutions of a semi-linea elliptic euation... The issue on the existence of global solutions of weakly nonlinea elliptic euations occupies a significant place in theoy of such euations, and a lot of papes have been devoted to it. Fo eview of such papes see the monogaph []. Weakly nonlinea euations with a bihamonic opeato wee consideed by many authos. In the pape [1] fo c = 0 euation 1.1 is consideed in the ball B R with the bounday conditions B R udx 0, B R udx 0 and it is poved that if σ, > 1, then the solution is absent. In the pape [5] fo σ = 0, c = 0 euation 1.1 is consideed in BR c with diffeent bounday conditions on B R and citical exponent of absence of solution of the poblems unde consideation ae found. In the pape [6] fo σ 0, simila poblems ae consideed and it is shown how the esults of the pape [5] may genealized fo an abitay σ 0. In this pape we conside poblem 1.1, 1. fo 0 c n, σ > and also in the papes [1],[5],[6] using the techniue of test function, woked out Mitidiei and Pohozaev in the papes []-[] find an exact exponent of absence of a global solutions. Fomulation of the main esult and poof The main esult of this pape is the following theoem: Theoem.1 Let > 1, 0 c n, 1 α σ 0. If u x is the solution of poblem 1.1, 1. in BR c, then u 0. Poof. Fo simplicity of notation, we conside the euation in B1 c. Multiply euation 1.1 by the function ϕ C0 B 1, ρ and integate by pats. We get the following: u x σ c ϕdx = u ϕdx B 1, ρ B 1, ρ x uϕdx u = B 1, ρ n = ϕds B 1, ρ u ϕ n ds B 1, ρ c u, ϕ dx B 1, ρ B 1, ρ x uϕdx c u ϕdx uϕdx..1 B 1, ρ x Take ϕ x = ξ x ψ x, 1, fo 1 x ρ, β ψ x = ρ, fo ρ x ρ, 0, fo x, ξ x = x α x α, whee β is a athe lage positive numbe. It is easy to veify that ξ x is the solution of the euation ξ c x ξ = 0, and ξ / x =1 = 0. As fo x = 1, ϕ n = ξ n = ξ = α α 1 α α 1 = α α = D, then B 1 u ϕ n ds = D B 1 udx 0.

3 Sh.H. Bagiov, M.J. Aliyev 65 Fom.1 we get u x σ c ξψdx u ξψ dx B 1, ρ B 1, ρ B 1, ρ x uξψdx = uψ ξ c B 1, ρ x ξ dx u ξ, ψ ξ ψ dx B 1, ρ u = u ξ, ψ ξ ψ dx = ξ, ψ ξ ψ ds B 1, ρ B 1, ρ v u, ξ, ψ ξ ψ dx B 1, ρ = v ξ, ψ v ξ ψ ds B 1, ρ u u ξ, ψ ξ ψ dx B 1, ρ = u ξ, ψ ξ ψ dx B ρ, ρ 1 u x σ ξ, ψ ξ ψ ξψdx B ρ, ρ B ρ, ρ x σ 1 ξ 1 ψ 1 whee 1 1 = 1. Hee we used the fact that on B 1, ρ all deivatives ψ eual zeo. As a esult we have u x σ ξ, ψ ξ ψ ξψdx B 1, ρ B ρ, ρ x σ 1 ξ 1 ψ 1 dx 1,. dx..3 Estimate the last integal. Denote J ξ, ψ = ξ, ψ ξ ψ. At fist we calculate each addend of J ξ, ψ sepaately ξ, ψ = ξ n 1 = ξ 3 ψ 3 ξ ψ ξ 3 3 n 1 ξ n 1 ξ ξ ξ ψ = ξ ψ n 1 ξ ψ = ξ ψ ξ ψ ξ ψ n 1 n 1 ξ ψ = ξ ψ n 1 ψ,. ξ ξ ψ

4 66 On absence of solutions of a semi-linea elliptic euation... ξ ψ n 1 ξ ψ n 1 ξ ψ n 1 ξ n 1 = ξ ψ n 1 ξ n 1 ξ n 1 ξ 3 ψ ξ ξ ψ n 1 ξ ψ n 1 ξ ξ n 1 ψ 3 ψ 3 ξ n 1 n 1 ψ 3 n 1 3 ξ n 1 n 1 n 1 ξ ψ = ξ ψ ξ 3 ψ ξ ψ 3n 1 Fom. and.5 we get ξ ψ n 1 n 1 ξ n 1 n 1 3 ξ J ξ, ψ = ξ ψ ξ 3 ψ ξ ξ 3 ψ 3 n 1 ξ ψ ξ ψ 3 3 n 1 ξ 3 ψ 3 n 1 n 1 ξ ξ ψ ξ n 1 3 ξ..5 3 n 1 ξ 3 ψ 3 ξ ψ 5 3 ξ 3 n 1 ξ ψ n 1 ξ ψ n 1 ξ ψ n 1 ξ n 1 n 3 ξ n 1 n 3 3 ξ. If in the ight side of.3 we pass to pola coodinates, we get u x σ J ξ, ψ ξψdx c 1 B 1, ρ ρ ρ σ 1 ξ 1 ψ 1 n1 d..6 In the ight integal we make a substitution: t = ρ, = tρ, ψ t = ψ tρ = ψ, ξ t = ξ tρ = ξ. Then we get ρ ρ J ξ, ψ n1 σ 1 ξ 1 ψ 1 d = 1 t ρ α J ξ, ψ t n1 ρ n dt ρ σα 1 t σα 1 ρ D t D 1 ψ 1 = ρ α nσ 1 A ξ, ψ,.7

5 Sh.H. Bagiov, M.J. Aliyev 67 whee A ξ, ψ = 1 t t σα 1 J ξ, ψ t n1 dt,.8 1 ρ D t D 1 ψ 1 J ξ, ψ = t α 1 tρ D ψ tα 1 α α tρ D 3 ψ 3 n 1 t α 1 1 tρ D 3 ψ 3 5t α α α 1 α α 1 tρ D ψ t α 3 α α 1 α α α 1 α tρ D ψ n 1 t α α α tρ D ψ n 1 tα 1 tρ D ψ n 1 t α 1 tρ D ψ n 1 t α 3 α α 1 α α 1 tρ D ψ n 1 n 3 t α 3 α α tρ D ψ n 1 n 3 t α 3 1 tρ D ψ. Fom hee J ξ, ψ t α ψ c t α 1 3 ψ 3 c 3t α ψ c t α 3 ψ, whee c, c 3, c ae dependent on n, α, α. Taking all these into account in.8, we get that fo lage ρ t A ξ, α ψ ψ c t α 1 3 ψ 3 c3 t α 3 ψ c t α 3 1 t t σα 1 ψ 1 It easy to see that fo lage β A ξ, ψ < c 5 <, whee c 5 is independent of ρ. Then fom.6,.7 and.8 it follows that u x σ ξψdx c 1 ρ α nσ 1 A ξ, ψ B 1, ρ ψ dt. c 1 c 5 ρ 1 α nσ 1 1 = c 6 ρ 1α σ 1..9

6 68 On absence of solutions of a semi-linea elliptic euation... If 1 α σ > 0, then tending ρ fom.9, we get u x σ ξdx 0. Conseuently u 0 in B1 c. Let now 1 α σ = 0. Then fom.9 it follows u x σ ξdx c 6. B c 1 B c 1 Then B ρ, ρ u x σ ξdx 0 as ρ. Fom. and.7 it follow that 1 u x σ ξψdx u x σ ξψdx B 1, ρ B ρ, ρ c 1 5 as ρ, i.e. in the limit B ρ, ρ u x σ ξψdx c 5 B c 1 u x σ ξdx = 0. So, in this case u 0 B1 c. This poves the theoem. B ρ, ρ B ρ, ρ u x σ ξdx J ξ, ψ 1 x σ 1 ξ 1 ψ dx 1 1 0, Now let us show that estimation on absence of a global solution is exact, i.e., if 1 α σ < 0, then thee exists a global solution of poblem 1.1, 1..We will look fo this solution in the fom A x µ. Substituting this function in the euation, we get µ µ µ n µ n x µ cµ µ n x µ Hence µ = σ 1, As A 1 = µ µ n A 1 x σµ = 0. µ µ n c = µ µ n µ α µ α. udx = Aµ µ n dx = µ µ n c 6 0, B 1 B 1 then fom condition 1. As 1 α σ < 0, then µ n 0. σ 1 < n n c,

7 i.e. n Sh.H. Bagiov, M.J. Aliyev 69 µ < n n c. Obviously, then µ n 0. Coming back the expession A, we get that fo n n the function u x = A x σ 1 Hence A = c < σ 1 < n c will be a positive solution of poblem 1.1, 1., whee [ ] 1 σ σ σ σ 1 1 n 1 α 1 α 1. We can wite the estimation 1 α σ 0 as follows: 1 σ α = 1 σ 1 α. σ n n c =. The value is said to the an exact citical exponent on absence of global solution of poblem 1.1, 1.. Refeences 1. Laptev, G.G.: On absence of solutions of a class of singula semilinea diffeential ineualites, Poc.Steklov Inst. Math., 3, in Russian.. Mitidiei, E., Pohozaev, S.: Absence of global positive solutions fo uasilinea elliptic ineualites. Dokl. Russ. Acad. Sci., 359, Mitidiei, E., Pohozaev, S.: Absence of positive solutions fo uasilinea elliptic poblems in RN. Poc. Steklov Inst. Math. 7, in Russian.. Mitidiei, E., Pohozaev, S.: A pioi estimates and blow-up of solutions to nonlinea patial diffeential euations and ineualities. Poc. Steklov Inst. Math. 3, Volodin, Yu.V.: On citical exponents of some semilinea bounday value poblems with bihamonic opeatos in the exteio of a sphee. Matem. zametki, 79 issue, in Russian. 6. Volodin, Yu.V.: Citical exponents of semilinea bounday value poblems with bihamonic opeatos in the exteio of a sphee with bounday conditions of fist type, Ucheniye zapiski RGSU 13, in Russian.

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