In this paper, we are interested in the study of the nonexistence of nontrivial solutions for a class of partial differential equations, in unbounded domains. This leads us to extend these results to systems of m equations. The method used is based on energy type identities

Akrout, K.

Khodja, B.

Electronic Journal of Qualitative Theory of Differential Equations

Kamel Akrout

B. Khodja

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Electronic journal of qualitative theory of differential equations

Absence of nontrivial solutions for a class of partial differential equations and systems in unbounded domains

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Electronic Journal of Qualitative Theory of Differential Equations2009, No. 27, 1-10; http://www.math.u-szeged.hu/ejqtde/Absence of nontrivial solutions for a class ofpartial differential equations and systems inunbounded domainsAKROUT Kamel∗and KHODJA Brahim†Department of Mathematics, Larbi Tebissi University,Tebessa, AlgeriaDepartment of Mathematics, Badji Mokhtar University,Annaba, AlgeriaAbstractIn this paper, we are interested on the study of the nonexistence of non-trivial solutions for a class of partial differential equations, in unboundeddomains. This leads us to extend these results to m-equations systems.The method used is based on energy type identities.Keywords: Differential equations, trivial solution, energy type identities.1 IntroductionThe study of the nonexistence of nontrivial solutions of partial differential equa-tions and systems is the subject of several works of many authors, by usingvarious methods to obtain the necessary and sufficient conditions, so the stud-ied problems admit only the null solutions. The works of Esteban & Lions [2],Pohozaev [6] and Van Der Vorst [7], contains results concerning the semilin-ear elliptic equations and systems. A semilar result can be found in [4], wherestudied equations one of the form{λ∂2u∂t2− ∂∂x(p (x, y) ∂u∂x)− ∂∂y(q (x, y) ∂u∂y) + f (x, y, u) = 0 in R× ω,u+ ε ∂u∂n= 0 on R×∂ω,considered in H2(R × ω) ∩ L∞(R × ω), where ω = ]a1, b1[ × ]a2, b2[ and thisequation does not admit nontrivial solutions if the following conditions holdsf(0) = 0, 2F (u)− uf(u) ≤ 0.∗akroutkamel@gmail.com†bmkhodja@yahoo.frEJQTDE, 2009 No. 27, p. 1In this work similar results for a class of the partial differential equations andsystems were also obtained.Let us consider the following problem inH2(R×Ω)∩L∞(R×Ω), Ω a boundeddomain of Rn, for a function λ : R→ R, not changing sign and p : R× Ω → Ralso had not changing sign. −∂∂t(λ (t) ∂u∂t)−n∑i=1∂∂xi(p (t, x) ∂u∂xi)+ f (x, u) = 0 in R×Ω,u+ ε∂u∂ν= 0 on R× ∂Ω.(1.1)We use the notationsH = L2 (Ω) ,‖u (t, x)‖ =(∫Ω|u(t, x)|2dx) 12, the norm of u in H,‖∇u (t, x)‖2 =∫Ωn∑i=1∣∣∣ ∂u∂xi∣∣∣2 dx,F (x, u) =u∫0f(x, σ)dσ, ∀ x ∈ Ω, u ∈ R.Let L be the operator defined byLu (t, x) = −n∑i=1∂∂xi(p (t, x) ∂u∂xi), (t, x) ∈ R× Ω,and f : Ω×R→ R a real continuous function, locally Lipschitz in u, such thatf(x, 0) = 0, ∀ x ∈ Ω.We assume thatu ∈ H2(R;H) ∩ L∞(R;L∞(Ω)),satisfies the equation− ∂∂t(λ (t) ∂u∂t(t))+ Lu (t, x) + f(x, u) = 0 , (t, x) ∈ R× Ω, (1.2)under the boundary conditions(u+ ε ∂u∂n) (t, σ) = 0, (t, σ) ∈ R× ∂Ω, Robin condition, (1.3)u(t, σ) = 0, (t, σ) ∈ R× ∂Ω, Dirichlet condition, (1.4)∂u(t,σ)∂n= 0, (t, σ) ∈ R× ∂Ω, Neumann condition. (1.5)We extend the above result of (1.1) to the system of m-equations of the form −∂∂t(λ (t) ∂uk∂t)−n∑i=1∂∂xi(pk (t, x)∂uk∂xi)+ fk (x, u1, ..., um) = 0 in R×Ω,uk + ε∂uk∂ν= 0 on R× ∂Ω.(1.6)EJQTDE, 2009 No. 27, p. 21 ≤ k ≤ m, where fk : Ω × Rm→ R,are real continuous functions, locallyLipschitz in ui, verifingfk(x, u1, ..., 0, ..., um) = 0, ∀ x ∈ Ω,∃Fm : Ω× Rm → R such that∂Fm∂sj= fj (x, s1, ..., sm) , 1 ≤ j ≤ m,Let Lk be the operators defined byLku (t, x) = −n∑i=1∂∂xi(pk (t, x)∂u∂xi), (t, x) ∈ R× Ω,we assume thatuk ∈ H2(R;H) ∩ L∞(R;L∞(Ω)),are solutions of the system− ∂∂t(λ (t) ∂uk∂t(t))+m∑k=1Lkuk (t, x) + f(x, u1, ..., um) = 0 , (t, x) ∈ R× Ω,(1.7)1 ≤ k ≤ m, with boundary conditions(uk + ε∂uk∂n) (t, σ) = 0, (t, σ) ∈ R× ∂Ω, Robin condition, (1.8)uk(t, σ) = 0, (t, σ) ∈ R× ∂Ω, Dirichlet condition, (1.9)∂uk(t,σ)∂n= 0, (t, σ) ∈ R× ∂Ω, Neumann condition. (1.10)According to the sign of λ, this type of problems comprises equations of bothhyperbolic or elliptic type.Our proof is based on energy type identities established in section 2, whichmake it possible to obtain the main nonexistence result in section 3. In section4 we apply the results to some examples.2 Identities of energy typeIn this section, we give essential lemmas for showing the main result of thispaper.Lemma 1 Let λ and p satisfyλ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ R,∂p∂t(t, x) ≤ 0 (resp ≥ 0) , ∀ (t, x) ∈ R× Ω.(2.1)Then the following energ identity,− 12λ (t)∥∥∂u∂t(t, x)∥∥2 + 12 ∫Ωp(t, x) |∇u|2dx+∫ΩF (x, u)dx+12ε∫∂Ωp(t, s)u2 (t, s) ds = 0.(2.2)holds for any solution of the Robin problem of (1.2)− (1.3).EJQTDE, 2009 No. 27, p. 3Proof. The assumptions f ∈ W 1,∞loc (Ω× R) , p ∈ L∞(R × Ω) and f(x, 0) =0, ∀x ∈ Ω, allow us to deduce the existence of two positive constants C1and C2,such that|p(t, x)| ≤ C1, |F (x, u)| ≤ C2 |u (t, x)|2.In addition, consider the functionsΨ (t) =12∫Ωp(t, x) |∇u|2 dx, Φ (t) =∫ΩF (x, u)dx, t ∈ R,where Φ and Ψ are of class C1,and|Ψ(t)| ≤ C1 ‖∇u(t, x)‖2, |Φ (t)| ≤ C2 ‖u(t, x)‖2, ∀t ∈ R.Then,Φ′ (t) =∫Ωf(x, u)∂u∂tdx, ∀t ∈ R,andΨ′ (t) =∫Ω(n∑i=1p(t, x) ∂u∂xi∂2u∂xi∂t+ 12∂p∂t(t, x) |∇u|2)dx= −∫Ωn∑i=1∂∂xi(p(t, x) ∂u∂xi)∂u∂tdx+ 12∫Ω∂p∂t(t, x) |∇u|2dx+∫∂Ωp(t, s)∂u∂ν∂u∂t(t, s) ds.Define the function K : R→ R byK (t) = − 12λ (t)∥∥∂u∂t(t, x)∥∥2 +Ψ(t) + Φ(t).The function K is absolutely continuous and differentiable in R, andK ′(t) = − 12λ′ (t)∥∥∂u∂t(t, x)∥∥2 − λ (t) ∫Ω∂u∂t∂2u∂t2dx+ Ψ′(t) + Φ′(t)= 12λ′ (t)∥∥∂u∂t(t)∥∥2 + 12 ∫Ω∂p∂t(t, x) |∇u|2 dx+∫∂Ωp(t, s)∂u∂ν∂u∂t(t, s) ds+∫Ω(− ∂∂t(λ (t) ∂u∂t)−n∑i=1∂∂xi(p(t, x) ∂u∂xi)+ f(x, u))∂u∂tdx.Because u is solution of (1.2)− (1.3), we deduce thatK ′(x) = 12λ′ (t)∥∥∂u∂t(t, x)∥∥2 + 12 ∫Ω∂p∂t|∇u|2dx+∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds,while on the boundary,∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds = −1ε∫∂Ωp(t, s)∂u∂tu (t, s) ds.AlsoK ′(t) = 12λ′ (t)∥∥∂u∂t(t, x)∥∥2 + 12 ∫Ω∂p∂t(t, x) |∇u|2dx−1ε∫∂Ωp(t, s)∂u∂tu (t, s)ds=12λ′ (t)∥∥∂u∂t(t, x)∥∥2 + 12 ∫Ω∂p∂t(t, x) |∇u|2 dx− 12ε∂∂t( ∫∂Ωp(t, s)u2 (t, s) ds)+ 12ε∫∂Ω∂p∂t(t, s)u2 (t, s) ds,EJQTDE, 2009 No. 27, p. 4i.eddt(K(t) + 12ε∫∂Ωp(t, s)u2 (t, s) ds)= 12λ′ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ω∂p∂t(t, x) |∇u|2 dx + 12ε∫∂Ω∂p∂t(t, s)u2 (t, s) ds.We setM (t) = K(t) +12ε∫∂Ωp(t, s)u2 (t, s)ds.Conditions (2.1) imply thatM ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ R,i.e H is monotonous. But also this function verifieslim|t|→+∞M (t) = 0,becauseM ∈ L2 (R) . Hence M (t) = 0, ∀t ∈ R, and this gives the desired result.Lemma 2 Let λ and p verify (2.1). The solution of the Dirichlet problem (1.2) , (1.4)or the Neumann problem (1.2) , (1.5) , satisfies the following energ identity− 12λ (t)∥∥∂u∂t(t, x)∥∥2 + 12 ∫Ωp(t, x) |∇u|2 dx+∫ΩF (x, u)dx = 0. (2.3)Proof. For the problem (1.2) , (1.4), the fact that u = 0 on the boundaryimplies that∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds = ddt( ∫∂Ωp(t, x)∂u∂νu (t, s) ds)−∫∂Ω∂p∂t∂u∂νu (t, s) ds−∫∂Ωp(t, x) ∂2u∂t∂νu (t, s)ds = 0.For the problem (1.2) , (1.5), the fact that ∂u∂ν= 0 on the boundary implies that∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds = 0.The remainder of the proof is similar to that of Lemma 1.Lemma 3 Let λ and pk satisfyλ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ R,∂pk∂t(t, x) ≤ 0 (resp ≥ 0) , 1 ≤ k ≤ m, ∀ (t, x) ∈ R× Ω.(2.4)Then any solutions of the system (1.7)−(1.8) satisfies for all t ∈ R, the followingenergetic identity− 12∑mk=1 λ (t)∥∥∂uk∂t(t, x)∥∥2 + 12 ∑mk=1 ∫Ωpk(t, x) |∇uk|2dx+∫ΩFm(x, u1, ..., um)dx +12ε∑mk=1∫∂Ωpk(t, s)u2k (t, s) ds = 0.(2.5)EJQTDE, 2009 No. 27, p. 5Lemma 4 Let λ and pk verify (2.5). Then the solutions of the systems (1.7)−(1.9) or (1.7)− (1.10) , satisfies for all t ∈ R, the following estimate− 12∑mk=1 λ (t)∥∥∂uk∂t(t, x)∥∥2 + 12 ∑mk=1 ∫Ωpk(t, x) |∇uk|2dx +∫ΩFm(x, u1, ..., um)dx = 0.(2.6)Proof. Let us define the function Km : R→ R byKm (t) = −12∑mk=1 λ (t)∥∥∂uk∂t(t, x)∥∥2 +Ψm(t) + Φm(t),where the functions Ψm and Φm are defined as followsΨm (t) =12∫Ω∑mk=1 pk(t, x) |∇uk|2dx, t ∈ R,Φm (t) =∫ΩFm(x, u1, ..., um)dx, t ∈ R,the rest of the proof is similar to the proofs of the preceding lemmas.3 The main ResultTheorem 1 Let us suppose that λ, F and f verifyλ (t) > 0 (resp < 0) , ∀t ∈ R,2F (x, u)− uf (x, u) ≤ 0 (resp ≥ 0) ,(3.1)and (2.1) holds. Then the problem (1.2)−(1.3) admit only the null solution.Proof. Let us define the function E byE (t) = ‖u (x, t)‖2.Multiplying equation (1.1) by u and integrating the new equation on Ω, weobtains∫Ω[− ∂∂t(λ (t) ∂u∂t)−n∑i=1∂∂xi(p (t, x) ∂u∂xi)− f (x, u)]udx=∫Ω[− 12(λ′ (t)∂(u2)∂t+ λ (t)∂2(u2)∂t2)+ λ (t)(∂u∂t)2+p (t, x) |∇u|2 + uf (x, u)]dx −n∑i=1∫∂Ωp (t, s) ∂u∂xiu (t, s) νids= − 12 (λ (t)E′′ (t) + λ′ (t)E′ (t)) + λ (t)∥∥∂u∂t(t, x)∥∥2 + ∫Ωp (t, x) |∇u|2dx+∫Ωuf (x, u) dx −∫∂Ωp (t, s) ∂u∂νu (t, s) ds = 0.Using identity (2.2), we haveddt(λ (t)E′ (t)) = λ (t)E′′ (t) + λ′ (t)E′ (t)= 2λ (t)∥∥∂u∂t(t)∥∥2 + 2 ∫Ωp (t, x) |∇u (t, x)|2 dx+2∫Ωu (t, x) f (x, u (t, x)) dx+ 2ε∫∂Ωp (t, s)u2 (t, s) ds= 4λ (t)∥∥∂u∂t(t)∥∥2 − 2 ∫Ω(2F (x, u)− uf (x, u)) dx.EJQTDE, 2009 No. 27, p. 6If λ (t) > 0, the assumption (3.1) implies thatddt(λ (t)E′ (t)) = λ (t)E′′ (t) + λ′ (t)E′ (t) ≥ 0 , ∀t ∈ R. (3.2)We conclude thatE′ (t) ≤ 0,otherwise,∃t1 ≥ 0, E′ (t1) ≥ 0. (3.3)Equation (3.2) implies that λ (t)E′ (t) is an increasing functionλ (t1)E′ (t1) ≤ λ (t)E′ (t) , ∀t ≥ t1,but, one haslim|t|→+∞E′ (t) = 0,because E′ ∈ L2 (R), thenλ (t1)E′ (t1) ≤ 0 and λ (t1) > 0⇒ E′ (t1) ≤ 0,which contradicts relation (3.3). Hence,E′ (t) ≤ 0, ∀t ∈ Ri.e E is monotonous. But, this function verifieslim|t|→+∞E (t) = 0,witch implies thatE (t) = 0, ∀t ∈ R,Thus u = 0 in R×.If λ (t) < 0, we deduce by the same manner that u = 0 in R×.Theorem 2 Let λ, F and f verify (3.1) and (2.1) holds. Then the only solutionof the problems (1.2)− (1.4) or (1.2)− (1.5) is the null solution.Proof. Identical to that of Theorem 1.Theorem 3 Let us suppose that λ, Fm and fk, 1 ≤ k ≤ m, satisfyλ (t) > 0 (resp < 0) , ∀t ∈ R,Fm (x, u1, ..., um)−∑mk=1 ukfk (x, u1, ..., um) ≤ 0 (resp ≥ 0) ,(3.4)and (2.5) holds. Then the system (1.7)− (1.8) admit only the null solutions.Proof. Multiplying equation (1.6) by uk and integrating the new equation onΩ, one obtains− 12(λ (t) d2dt2‖uk (t, x)‖2+ λ′ (t) ddt‖uk (t, x)‖2)+ λ (t)∥∥∂uk∂t(t, x)∥∥2 + ∫Ωpk (t, x) |∇uk|2dx+∫Ωukfk (x, u1, ..., um) dx−∫∂Ωpk (t, s)∂uk∂νuk (t, s) ds = 0.EJQTDE, 2009 No. 27, p. 7The sum on k from 1 to m gives− 12 (λ (t)E′′m (t) + λ′ (t)E′m (t)) + λ (t)∑mk=1∥∥∂uk∂t(t, x)∥∥2 +∑mk=1 ∫Ωpk (t, x) |∇uk|2dx+∑mk=1∫Ωuk (t, x) fk (x, u1, ..., um) dx−∑mk=1∫∂Ωpk (t, s)∂uk∂νuk (t, s) ds = 0.By using identity (2.6), we deduce thatddt(λ (t)E′m (t)) = λ (t)E′′m (t) + λ′ (t)E′m (t)= 4λ (t)∑mk=1∥∥ ∂uk∂t(t)∥∥2 − 2 ∫Ω(2Fm (x, u1, ..., um)−∑mk=1 uk (t, x) fk (x, u1, ..., um)) dx.Then the assumption (3.4) gives the result.Theorem 4 Let λ, p and F verifyλ (t) > 0, p (t, x) < 0 and F (x, u) ≤ 0, ∀ (t, x) ∈ R× Ω,orλ (t) < 0, p (t, x) > 0 and F (x, u) ≥ 0, ∀ (t, x) ∈ R× Ω,(3.5)and (2.1) holds. Then the problems (1.2)− (1.3), (1.2)− (1.4) and (1.2)− (1.5)admit only the null solution.Proof. Assumptions (3.4) and equality (2.3) allow is− 12λ (t)∥∥ ∂u∂t(t, x)∥∥2 + 12 ∫Ωp(t, x) |∇u|2dx+∫ΩF (x, u)dx = 0,which implies that∂u∂t(t, x) = 0, ∀ (t, x) ∈ R× Ω,i.e.u (t, x) = u (x) .But the following condition is necessary∫R×Ω|u (t, x)|2dtdx =∫R×Ω|u (x)|2dtdx < +∞,then u ≡ 0.Theorem 5 Let λ, pk (1 ≤ k ≤ m)and f satisfyλ (t) > 0, pk (t, x) < 0 and Fm (x, u1, ..., um) ≤ 0, ∀ (t, x) ∈ R× Ω,orλ (t) < 0, pk (t, x) > 0 and Fm (x, u1, ..., um) ≥ 0, ∀ (t, x) ∈ R× Ω.(3.6)and (2.4) holds. Then the systems (1.7)− (1.8) , (1.7)− (1.9) and (1.7)− (1.10)admit only the null solutions.Proof. Similar to that of Theorem 2.Remark 1 Note that one can apply these results in the field R+ × Ω, with theconditionu (0, x) = 0, ∀x ∈ Ω.EJQTDE, 2009 No. 27, p. 84 ApplicationsExample 1 Letθ, θ1, θ2 : Ω→ R,be a nonnegative functions of class C (R) , p, q ≥ 1 and m ∈ R, such thatf(x, u) = mu+ θ1 (x) |u|p−1u+ θ2 (x) |u|q−1u.Then the problem defined by∂2u∂t2−n∑i=1∂∂xi(θ (x) ∂u∂xi)+ f(x, u) = 0 in R× Ω,(u+ ε ∂u∂n)(x, σ) = 0 on R× ∂Ω,(4.1)admits anly the null solution.In this case it suffice to check that2F (x, u)− uf(x, u) =θ1 (x) (2p+1 − 1) |u|p+1 + θ2 (x) (2q+1 − 1) |u|q+1 ≤ 0,and apply Theorem 1.Example 2 Let Ω be a bounded open of set Rn. Then, problem∂∂t(e−t2 ∂u∂t)−∆u = θ (x) |u|p−1u in R+ × Ω,(u+ ε ∂u∂n)(t, σ) = 0 on R+ × ∂Ω,u (0, x) = 0, ∀x ∈ Ω,(4.2)wherep ≥ 1, θ : Ω→ R, is nonnegative,admits only the trivial solution, u ≡ 0.Indeed,λ (t) = −e−t2< 0, λ′ (t) = 2te−t2≥ 0, ∀t ≥ 0,2F (x, u)− uf(x, u) = θ (x) ( 2p+1 − 1) |u|p+1≥ 0.Theorem 1 gives the result.Example 3 Let Ω be a bounded open of set Rn, p, q ≥ 1,Then, the system− ∂∂t(λ (t) ∂u∂t)−∆u + (p+ 1) θ (x) u |u|p−1|v|q+1= 0 in R× Ω,− ∂∂t(λ (t) ∂v∂t)−∆v + (q + 1) θ (x) v |v|q−1|u|p+1= 0 in R× Ω,(u+ ε ∂u∂n)(t, σ) =(v + ε ∂v∂n)(t, σ) = 0 on R× ∂Ω,(4.3)whereθ : Ω→ R, is nonnegative,λ (t) > 0 of a class L∞ (R) ,admit only the trivial solutions, u ≡ v ≡ 0.EJQTDE, 2009 No. 27, p. 9Indeed, there exist a function F defined as followsF (x, u, v) = θ (x) |u|p+1|v|q+1,witch satisfies∂F∂u= f1(x, u, v) = (p+ 1) θ (x) u |u|p−1|v|q+1,∂F∂v= f2(x, u, v) = (q + 1) θ (x) v |v|q−1|u|p+1,F (x, u, v)− uf1(x, u, v)− vf2 (x, u, v) = −θ (x) (p+ q + 1) |u|p+1|v|q+1≤ 0.Theorem 3 gives the result.References[1] H. Brezis, Analyse fonctionnelle, The´orie et Applications, Masson, Paris(1983).[2] M. J. Esteban & P. Lions, Existence and non-existence results for semilinear elliptic problems in unbounded domains, Proc.Roy.Soc.Edimburgh. 93-A(1982),1-14.[3] N. Kawarno, W. Ni & Syotsutani, Generalised Pohozaev identity andits applications. J. Math. Soc. Japan. Vol. 42 N◦3 (1990), 541-563.[4] B. Khodja, Nonexistence of solutions for semilinear equations and systemsin cylindrical domains. Comm. Appl. Nonlinear Anal. (2000), 19-30.[5] W. Ni & J. Serrin, Nonexistence theorems for quasilinear partial differ-ential equations. Red. Circ. Mat. Palermo, suppl. Math. 8 (1985), 171-185.[6] S. I. Pohozaev, Eeigenfunctions of the equation ∆u + λf (u) = 0, So-viet.Math.Dokl.(1965), 1408-1411.[7] R. C. A. M. Van der Vorst, Variational identities and applications todifferential systems, Arch.Rational; Mech.Anal.116 (1991) 375-398.[8] C. Yarur, Nonexistence of positive singular solutions for a class of semilin-ear elliptic systems. Electronic Journal of Diff. Equations, 8 (1996), 1-22.(Received January 6, 2009)EJQTDE, 2009 No. 27, p. 10

Brahim Khodja

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University of Szeged

Electronic Journal of Qualitative Theory of Differential Equations2009, No. 27, 1-10; http://www.math.u-szeged.hu/ejqtde/Absence of nontrivial solutions for a class ofpartial differential equations and systems inunbounded domainsAKROUT Kamel∗and KHODJA Brahim†Department of Mathematics, Larbi Tebissi University,Tebessa, AlgeriaDepartment of Mathematics, Badji Mokhtar University,Annaba, AlgeriaAbstractIn this paper, we are interested on the study of the nonexistence of non-trivial solutions for a class of partial differential equations, in unboundeddomains. This leads us to extend these results to m-equations systems.The method used is based on energy type identities.Keywords: Differential equations, trivial solution, energy type identities.1 IntroductionThe study of the nonexistence of nontrivial solutions of partial differential equa-tions and systems is the subject of several works of many authors, by usingvarious methods to obtain the necessary and sufficient conditions, so the stud-ied problems admit only the null solutions. The works of Esteban & Lions [2],Pohozaev [6] and Van Der Vorst [7], contains results concerning the semilin-ear elliptic equations and systems. A semilar result can be found in [4], wherestudied equations one of the form{λ∂2u∂t2− ∂∂x(p (x, y) ∂u∂x)− ∂∂y(q (x, y) ∂u∂y) + f (x, y, u) = 0 in R × ω,u + ε ∂u∂n= 0 on R×∂ω,considered in H2(R × ω) ∩ L∞(R × ω), where ω = ]a1, b1[ × ]a2, b2[ and thisequation does not admit nontrivial solutions if the following conditions holdsf(0) = 0, 2F (u) − uf(u) ≤ 0.∗akroutkamel@gmail.com†bmkhodja@yahoo.frEJQTDE, 2009 No. 27, p. 1In this work similar results for a class of the partial differential equations andsystems were also obtained.Let us consider the following problem in H2(R×Ω)∩L∞(R×Ω), Ω a boundeddomain of Rn, for a function λ : R → R, not changing sign and p : R × Ω → Ralso had not changing sign.− ∂∂t(λ (t) ∂u∂t)−n∑i=1∂∂xi(p (t, x) ∂u∂xi)+ f (x, u) = 0 in R×Ω,u + ε∂u∂ν= 0 on R × ∂Ω.(1.1)We use the notationsH = L2 (Ω) ,‖u (t, x)‖ =(∫Ω|u(t, x)|2dx)12, the norm of u in H,‖∇u (t, x)‖2 =∫Ωn∑i=1∣∣∣∂u∂xi∣∣∣2dx,F (x, u) =u∫0f(x, σ)dσ, ∀ x ∈ Ω, u ∈ R.Let L be the operator defined byLu (t, x) = −n∑i=1∂∂xi(p (t, x) ∂u∂xi), (t, x) ∈ R × Ω,and f : Ω × R → R a real continuous function, locally Lipschitz in u, such thatf(x, 0) = 0, ∀ x ∈ Ω.We assume thatu ∈ H2(R; H) ∩ L∞(R; L∞(Ω)),satisfies the equation− ∂∂t(λ (t) ∂u∂t(t))+ Lu (t, x) + f(x, u) = 0 , (t, x) ∈ R × Ω, (1.2)under the boundary conditions(u + ε ∂u∂n) (t, σ) = 0, (t, σ) ∈ R × ∂Ω, Robin condition, (1.3)u(t, σ) = 0, (t, σ) ∈ R × ∂Ω, Dirichlet condition, (1.4)∂u(t,σ)∂n= 0, (t, σ) ∈ R × ∂Ω, Neumann condition. (1.5)We extend the above result of (1.1) to the system of m-equations of the form− ∂∂t(λ (t) ∂uk∂t)−n∑i=1∂∂xi(pk (t, x)∂uk∂xi)+ fk (x, u1, ..., um) = 0 in R×Ω,uk + ε∂uk∂ν= 0 on R × ∂Ω.(1.6)EJQTDE, 2009 No. 27, p. 21 ≤ k ≤ m, where fk : Ω × Rm→ R,are real continuous functions, locallyLipschitz in ui, verifingfk(x, u1, ..., 0, ..., um) = 0, ∀ x ∈ Ω,∃Fm : Ω × Rm → R such that∂Fm∂sj= fj (x, s1, ..., sm) , 1 ≤ j ≤ m,Let Lk be the operators defined byLku (t, x) = −n∑i=1∂∂xi(pk (t, x)∂u∂xi), (t, x) ∈ R × Ω,we assume thatuk ∈ H2(R; H) ∩ L∞(R; L∞(Ω)),are solutions of the system− ∂∂t(λ (t) ∂uk∂t(t))+m∑k=1Lkuk (t, x) + f(x, u1, ..., um) = 0 , (t, x) ∈ R × Ω,(1.7)1 ≤ k ≤ m, with boundary conditions(uk + ε∂uk∂n) (t, σ) = 0, (t, σ) ∈ R × ∂Ω, Robin condition, (1.8)uk(t, σ) = 0, (t, σ) ∈ R × ∂Ω, Dirichlet condition, (1.9)∂uk(t,σ)∂n= 0, (t, σ) ∈ R × ∂Ω, Neumann condition. (1.10)According to the sign of λ, this type of problems comprises equations of bothhyperbolic or elliptic type.Our proof is based on energy type identities established in section 2, whichmake it possible to obtain the main nonexistence result in section 3. In section4 we apply the results to some examples.2 Identities of energy typeIn this section, we give essential lemmas for showing the main result of thispaper.Lemma 1 Let λ and p satisfyλ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ R,∂p∂t(t, x) ≤ 0 (resp ≥ 0) , ∀ (t, x) ∈ R × Ω.(2.1)Then the following energ identity,− 12λ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ωp(t, x) |∇u|2dx+∫ΩF (x, u)dx +12ε∫∂Ωp(t, s)u2 (t, s) ds = 0.(2.2)holds for any solution of the Robin problem of (1.2) − (1.3).EJQTDE, 2009 No. 27, p. 3Proof. The assumptions f ∈ W 1,∞loc (Ω × R) , p ∈ L∞(R × Ω) and f(x, 0) =0, ∀x ∈ Ω, allow us to deduce the existence of two positive constants C1and C2,such that|p(t, x)| ≤ C1, |F (x, u)| ≤ C2 |u (t, x)|2.In addition, consider the functionsΨ (t) =12∫Ωp(t, x) |∇u|2 dx, Φ (t) =∫ΩF (x, u)dx, t ∈ R,where Φ and Ψ are of class C1,and|Ψ (t)| ≤ C1 ‖∇u(t, x)‖2, |Φ (t)| ≤ C2 ‖u(t, x)‖2, ∀t ∈ R.Then,Φ′ (t) =∫Ωf(x, u)∂u∂tdx, ∀t ∈ R,andΨ′ (t) =∫Ω(n∑i=1p(t, x) ∂u∂xi∂2u∂xi∂t+ 12∂p∂t(t, x) |∇u|2)dx= −∫Ωn∑i=1∂∂xi(p(t, x) ∂u∂xi)∂u∂tdx + 12∫Ω∂p∂t(t, x) |∇u|2dx +∫∂Ωp(t, s)∂u∂ν∂u∂t(t, s) ds.Define the function K : R → R byK (t) = − 12λ (t)∥∥∂u∂t(t, x)∥∥2+ Ψ(t) + Φ(t).The function K is absolutely continuous and differentiable in R, andK ′(t) = − 12λ′ (t)∥∥∂u∂t(t, x)∥∥2− λ (t)∫Ω∂u∂t∂2u∂t2dx + Ψ′(t) + Φ′(t)= 12λ′ (t)∥∥∂u∂t(t)∥∥2+ 12∫Ω∂p∂t(t, x) |∇u|2 dx +∫∂Ωp(t, s)∂u∂ν∂u∂t(t, s) ds+∫Ω(− ∂∂t(λ (t) ∂u∂t)−n∑i=1∂∂xi(p(t, x) ∂u∂xi)+ f(x, u))∂u∂tdx.Because u is solution of (1.2) − (1.3), we deduce thatK ′(x) = 12λ′ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ω∂p∂t|∇u|2dx +∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds,while on the boundary,∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds = −1ε∫∂Ωp(t, s)∂u∂tu (t, s) ds.AlsoK ′(t) = 12λ′ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ω∂p∂t(t, x) |∇u|2dx −1ε∫∂Ωp(t, s)∂u∂tu (t, s)ds=12λ′ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ω∂p∂t(t, x) |∇u|2 dx− 12ε∂∂t(∫∂Ωp(t, s)u2 (t, s) ds)+ 12ε∫∂Ω∂p∂t(t, s)u2 (t, s) ds,EJQTDE, 2009 No. 27, p. 4i.eddt(K(t) + 12ε∫∂Ωp(t, s)u2 (t, s) ds)= 12λ′ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ω∂p∂t(t, x) |∇u|2 dx + 12ε∫∂Ω∂p∂t(t, s)u2 (t, s) ds.We setM (t) = K(t) +12ε∫∂Ωp(t, s)u2 (t, s)ds.Conditions (2.1) imply thatM ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ R,i.e H is monotonous. But also this function verifieslim|t|→+∞M (t) = 0,because M ∈ L2 (R) . Hence M (t) = 0, ∀t ∈ R, and this gives the desired result.Lemma 2 Let λ and p verify (2.1). The solution of the Dirichlet problem (1.2) , (1.4)or the Neumann problem (1.2) , (1.5) , satisfies the following energ identity− 12λ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ωp(t, x) |∇u|2 dx +∫ΩF (x, u)dx = 0. (2.3)Proof. For the problem (1.2) , (1.4), the fact that u = 0 on the boundaryimplies that∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds = ddt(∫∂Ωp(t, x)∂u∂νu (t, s) ds)−∫∂Ω∂p∂t∂u∂νu (t, s) ds −∫∂Ωp(t, x) ∂2u∂t∂νu (t, s)ds = 0.For the problem (1.2) , (1.5), the fact that ∂u∂ν= 0 on the boundary implies that∫∂Ωp(t, x)∂u∂ν∂u∂t(t, s) ds = 0.The remainder of the proof is similar to that of Lemma 1.Lemma 3 Let λ and pk satisfyλ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ R,∂pk∂t(t, x) ≤ 0 (resp ≥ 0) , 1 ≤ k ≤ m, ∀ (t, x) ∈ R × Ω.(2.4)Then any solutions of the system (1.7)−(1.8) satisfies for all t ∈ R, the followingenergetic identity− 12∑mk=1 λ (t)∥∥∂uk∂t(t, x)∥∥2+ 12∑mk=1∫Ωpk(t, x) |∇uk|2dx+∫ΩFm(x, u1, ..., um)dx +12ε∑mk=1∫∂Ωpk(t, s)u2k (t, s) ds = 0.(2.5)EJQTDE, 2009 No. 27, p. 5Lemma 4 Let λ and pk verify (2.5). Then the solutions of the systems (1.7)−(1.9) or (1.7)− (1.10) , satisfies for all t ∈ R, the following estimate− 12∑mk=1 λ (t)∥∥∂uk∂t(t, x)∥∥2+ 12∑mk=1∫Ωpk(t, x) |∇uk|2dx +∫ΩFm(x, u1, ..., um)dx = 0.(2.6)Proof. Let us define the function Km : R → R byKm (t) = −12∑mk=1 λ (t)∥∥∂uk∂t(t, x)∥∥2+ Ψm(t) + Φm(t),where the functions Ψm and Φm are defined as followsΨm (t) =12∫Ω∑mk=1 pk(t, x) |∇uk|2dx, t ∈ R,Φm (t) =∫ΩFm(x, u1, ..., um)dx, t ∈ R,the rest of the proof is similar to the proofs of the preceding lemmas.3 The main ResultTheorem 1 Let us suppose that λ, F and f verifyλ (t) > 0 (resp < 0) , ∀t ∈ R,2F (x, u) − uf (x, u) ≤ 0 (resp ≥ 0) ,(3.1)and (2.1) holds. Then the problem (1.2)−(1.3) admit only the null solution.Proof. Let us define the function E byE (t) = ‖u (x, t)‖2.Multiplying equation (1.1) by u and integrating the new equation on Ω, weobtains∫Ω[− ∂∂t(λ (t) ∂u∂t)−n∑i=1∂∂xi(p (t, x) ∂u∂xi)− f (x, u)]udx=∫Ω[− 12(λ′ (t)∂(u2)∂t+ λ (t)∂2(u2)∂t2)+ λ (t)(∂u∂t)2+p (t, x) |∇u|2 + uf (x, u)]dx −n∑i=1∫∂Ωp (t, s) ∂u∂xiu (t, s) νids= − 12 (λ (t)E′′ (t) + λ′ (t)E′ (t)) + λ (t)∥∥∂u∂t(t, x)∥∥2+∫Ωp (t, x) |∇u|2dx+∫Ωuf (x, u) dx −∫∂Ωp (t, s) ∂u∂νu (t, s) ds = 0.Using identity (2.2), we haveddt(λ (t)E′ (t)) = λ (t)E′′ (t) + λ′ (t)E′ (t)= 2λ (t)∥∥∂u∂t(t)∥∥2+ 2∫Ωp (t, x) |∇u (t, x)|2 dx+2∫Ωu (t, x) f (x, u (t, x)) dx + 2ε∫∂Ωp (t, s)u2 (t, s) ds= 4λ (t)∥∥∂u∂t(t)∥∥2− 2∫Ω(2F (x, u) − uf (x, u)) dx.EJQTDE, 2009 No. 27, p. 6If λ (t) > 0, the assumption (3.1) implies thatddt(λ (t)E′ (t)) = λ (t)E′′ (t) + λ′ (t)E′ (t) ≥ 0 , ∀t ∈ R. (3.2)We conclude thatE′ (t) ≤ 0,otherwise,∃t1 ≥ 0, E′ (t1) ≥ 0. (3.3)Equation (3.2) implies that λ (t)E′ (t) is an increasing functionλ (t1)E′ (t1) ≤ λ (t)E′ (t) , ∀t ≥ t1,but, one haslim|t|→+∞E′ (t) = 0,because E′ ∈ L2 (R), thenλ (t1)E′ (t1) ≤ 0 and λ (t1) > 0 ⇒ E′ (t1) ≤ 0,which contradicts relation (3.3). Hence,E′ (t) ≤ 0, ∀t ∈ Ri.e E is monotonous. But, this function verifieslim|t|→+∞E (t) = 0,witch implies thatE (t) = 0, ∀t ∈ R,Thus u = 0 in R × .If λ (t) < 0, we deduce by the same manner that u = 0 in R × .Theorem 2 Let λ, F and f verify (3.1) and (2.1) holds. Then the only solutionof the problems (1.2) − (1.4) or (1.2) − (1.5) is the null solution.Proof. Identical to that of Theorem 1.Theorem 3 Let us suppose that λ, Fm and fk, 1 ≤ k ≤ m, satisfyλ (t) > 0 (resp < 0) , ∀t ∈ R,Fm (x, u1, ..., um) −∑mk=1 ukfk (x, u1, ..., um) ≤ 0 (resp ≥ 0) ,(3.4)and (2.5) holds. Then the system (1.7)− (1.8) admit only the null solutions.Proof. Multiplying equation (1.6) by uk and integrating the new equation onΩ, one obtains− 12(λ (t) d2dt2‖uk (t, x)‖2+ λ′ (t) ddt‖uk (t, x)‖2)+ λ (t)∥∥∂uk∂t(t, x)∥∥2+∫Ωpk (t, x) |∇uk|2dx+∫Ωukfk (x, u1, ..., um) dx −∫∂Ωpk (t, s)∂uk∂νuk (t, s) ds = 0.EJQTDE, 2009 No. 27, p. 7The sum on k from 1 to m gives− 12 (λ (t)E′′m (t) + λ′ (t)E′m (t)) + λ (t)∑mk=1∥∥∂uk∂t(t, x)∥∥2+∑mk=1∫Ωpk (t, x) |∇uk|2dx+∑mk=1∫Ωuk (t, x) fk (x, u1, ..., um) dx −∑mk=1∫∂Ωpk (t, s)∂uk∂νuk (t, s) ds = 0.By using identity (2.6), we deduce thatddt(λ (t) E′m (t)) = λ (t)E′′m (t) + λ′ (t)E′m (t)= 4λ (t)∑mk=1∥∥∂uk∂t(t)∥∥2− 2∫Ω(2Fm (x, u1, ..., um) −∑mk=1 uk (t, x) fk (x, u1, ..., um)) dx.Then the assumption (3.4) gives the result.Theorem 4 Let λ, p and F verifyλ (t) > 0, p (t, x) < 0 and F (x, u) ≤ 0, ∀ (t, x) ∈ R × Ω,orλ (t) < 0, p (t, x) > 0 and F (x, u) ≥ 0, ∀ (t, x) ∈ R × Ω,(3.5)and (2.1) holds. Then the problems (1.2) − (1.3), (1.2) − (1.4) and (1.2) − (1.5)admit only the null solution.Proof. Assumptions (3.4) and equality (2.3) allow is− 12λ (t)∥∥∂u∂t(t, x)∥∥2+ 12∫Ωp(t, x) |∇u|2dx +∫ΩF (x, u)dx = 0,which implies that∂u∂t(t, x) = 0, ∀ (t, x) ∈ R × Ω,i.e.u (t, x) = u (x) .But the following condition is necessary∫R×Ω|u (t, x)|2dtdx =∫R×Ω|u (x)|2dtdx < +∞,then u ≡ 0.Theorem 5 Let λ, pk (1 ≤ k ≤ m)and f satisfyλ (t) > 0, pk (t, x) < 0 and Fm (x, u1, ..., um) ≤ 0, ∀ (t, x) ∈ R × Ω,orλ (t) < 0, pk (t, x) > 0 and Fm (x, u1, ..., um) ≥ 0, ∀ (t, x) ∈ R × Ω.(3.6)and (2.4) holds. Then the systems (1.7)− (1.8) , (1.7)− (1.9) and (1.7)− (1.10)admit only the null solutions.Proof. Similar to that of Theorem 2.Remark 1 Note that one can apply these results in the field R+ × Ω, with theconditionu (0, x) = 0, ∀x ∈ Ω.EJQTDE, 2009 No. 27, p. 84 ApplicationsExample 1 Letθ, θ1, θ2 : Ω → R,be a nonnegative functions of class C (R) , p, q ≥ 1 and m ∈ R, such thatf(x, u) = mu + θ1 (x) |u|p−1u + θ2 (x) |u|q−1u.Then the problem defined by∂2u∂t2−n∑i=1∂∂xi(θ (x) ∂u∂xi)+ f(x, u) = 0 in R × Ω,(u + ε ∂u∂n)(x, σ) = 0 on R × ∂Ω,(4.1)admits anly the null solution.In this case it suffice to check that2F (x, u) − uf(x, u) =θ1 (x) (2p+1 − 1) |u|p+1 + θ2 (x) (2q+1 − 1) |u|q+1 ≤ 0,and apply Theorem 1.Example 2 Let Ω be a bounded open of set Rn. Then, problem∂∂t(e−t2 ∂u∂t)− ∆u = θ (x) |u|p−1u in R+ × Ω,(u + ε ∂u∂n)(t, σ) = 0 on R+ × ∂Ω,u (0, x) = 0, ∀x ∈ Ω,(4.2)wherep ≥ 1, θ : Ω → R, is nonnegative,admits only the trivial solution, u ≡ 0.Indeed,λ (t) = −e−t2< 0, λ′ (t) = 2te−t2≥ 0, ∀t ≥ 0,2F (x, u) − uf(x, u) = θ (x) ( 2p+1 − 1) |u|p+1≥ 0.Theorem 1 gives the result.Example 3 Let Ω be a bounded open of set Rn, p, q ≥ 1,Then, the system− ∂∂t(λ (t) ∂u∂t)− ∆u + (p + 1) θ (x) u |u|p−1|v|q+1= 0 in R × Ω,− ∂∂t(λ (t) ∂v∂t)− ∆v + (q + 1) θ (x) v |v|q−1|u|p+1= 0 in R × Ω,(u + ε ∂u∂n)(t, σ) =(v + ε ∂v∂n)(t, σ) = 0 on R × ∂Ω,(4.3)whereθ : Ω → R, is nonnegative,λ (t) > 0 of a class L∞ (R) ,admit only the trivial solutions, u ≡ v ≡ 0.EJQTDE, 2009 No. 27, p. 9Indeed, there exist a function F defined as followsF (x, u, v) = θ (x) |u|p+1|v|q+1,witch satisfies∂F∂u= f1(x, u, v) = (p + 1) θ (x) u |u|p−1|v|q+1,∂F∂v= f2(x, u, v) = (q + 1) θ (x) v |v|q−1|u|p+1,F (x, u, v) − uf1(x, u, v) − vf2 (x, u, v) = −θ (x) (p + q + 1) |u|p+1|v|q+1≤ 0.Theorem 3 gives the result.References[1] H. Brezis, Analyse fonctionnelle, Théorie et Applications, Masson, Paris(1983).[2] M. J. Esteban & P. Lions, Existence and non-existence results for semilinear elliptic problems in unbounded domains, Proc.Roy.Soc.Edimburgh. 93-A(1982),1-14.[3] N. Kawarno, W. Ni & Syotsutani, Generalised Pohozaev identity andits applications. J. Math. Soc. Japan. Vol. 42 N◦3 (1990), 541-563.[4] B. Khodja, Nonexistence of solutions for semilinear equations and systemsin cylindrical domains. Comm. Appl. Nonlinear Anal. (2000), 19-30.[5] W. Ni & J. Serrin, Nonexistence theorems for quasilinear partial differ-ential equations. Red. Circ. Mat. Palermo, suppl. Math. 8 (1985), 171-185.[6] S. I. Pohozaev, Eeigenfunctions of the equation ∆u + λf (u) = 0, So-viet.Math.Dokl.(1965), 1408-1411.[7] R. C. A. M. Van der Vorst, Variational identities and applications todifferential systems, Arch.Rational; Mech.Anal.116 (1991) 375-398.[8] C. Yarur, Nonexistence of positive singular solutions for a class of semilin-ear elliptic systems. Electronic Journal of Diff. Equations, 8 (1996), 1-22.(Received January 6, 2009)EJQTDE, 2009 No. 27, p. 10

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Absence of nontrivial solutions for a class of partial differential equations and systems in unbounded domains

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