summary:In this paper we consider a nonlinear hyperbolic boundary value problem. We show that this problem admits weak solutions by using a lifting result for pseudomonotone operators and a surjectivity result concerning coercive and monotone operators

Kandilakis, Dimitrios A.

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Pseudomonotonicity and nonlinear hyperbolic equations

Institute of Mathematics AS CR, v. v. i.

Commentationes Mathematicae Universitatis CarolinaeDimitrios A. KandilakisPseudomonotonicity and nonlinear hyperbolic equationsCommentationes Mathematicae Universitatis Carolinae, Vol. 38 (1997), No. 3, 463--469Persistent URL: http://dml.cz/dmlcz/118945Terms of use:© Charles University in Prague, Faculty of Mathematics and Physics, 1997Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document mustcontain these Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://project.dml.czComment.Math.Univ.Carolin. 38,3 (1997)463–469 463Pseudomonotonicity and nonlinear hyperbolic equationsDimitrios A. KandilakisAbstract. In this paper we consider a nonlinear hyperbolic boundary value problem. Weshow that this problem admits weak solutions by using a lifting result for pseudomono-tone operators and a surjectivity result concerning coercive and monotone operators.Keywords: pseudomonotone operator, demicontinuous operator, maximal monotone op-erator, weak solutionClassification: 35A05, 35L201. IntroductionLet T = [0, b] and Z ⊆ RN be a bounded domain with Lipschitz boundary Γ.Also let Dk =∂∂zk, k ∈ {1, 2, . . . , N}, and D = grad. We consider the followinghyperbolic problem:(1)∂2x∂t2+N∑k=1Dkak(t, z, x, Dx) = g(t, z, x) a.e. on T × Zx(0, z) =∂x∂t(b, z) = 0 a.e. on Z, x |T×Γ= 0.This problem can be equivalently rewritten in abstract form as(1∗){x′′(t) +A(t, x(t)) = g(t, x(t))x(0) = x′(b) = 0}where A(t, .) is a pseudomonotone operator (see Section 3). Existence resultsfor second order evolutions in the form of (1∗) were obtained by Barbu [2, The-orem 1.1, p. 268], Papageorgiou [5, Theorem 3.1] and Zeidler [7, Theorem 33A,p. 924]. All three authors examine the equation (inclusion) x′′(t) + B(t, x′(t)) +A(x(t)) ∋ f(t, x(t)) but they assume conditions on A(.) stronger than pseu-domonotonicity, namely monotonicity, linearity and boundedness. To the bestof the knowledge of the author this is the first time that the notion of pseu-domonotonicity is being used in the theory of hyperbolic problems.Research supported by Grant PENED 678(96)464 D.A.Kandilakis2. PreliminariesIn this section we fix our notation, introduce our assumptions on the data ofthe problem and recall some basic facts from nonlinear analysis that we will needin the sequel.Our hypotheses on the functions ak(t, z, x, y), k ∈ {1, 2, . . . , N}, are the follow-ing:H(a): ak : T × Z × R × RN → R, k ∈ {1, 2, . . . , N}, are functions such that(i) (t, z)→ ak(t, z, x, y) is measurable;(ii) (x, y)→ ak(t, z, x, y) is continuous;(iii) | ak(t, z, x, y) |≤ β1(t, z) + c1(| x | +‖y‖) a.e. on T × Z for all x ∈ R,y ∈ RN , with β1 ∈ L2(T × Z) and c1 > 0;(iv)N∑k=1(ak(t, z, x, y)− ak(t, z, x, y′))(yk − y′k) > 0 a.e. on T ×Z, for all x ∈ Rand all y, y′ ∈ RN , y 6= y′; and(v)N∑k=1ak(t, z, x, y)yk ≥ c2‖y‖2 − β2(t, z) a.e. on T × Z, for all x ∈ R, ally ∈ RN , with β2 ∈ L1(T × Z) and c2 > 0.Because of hypothesis H(a), we can introduce the semilinear forma : L2(T, H10 (Z))× L2(T, H10 (Z))→ R defined bya(x, y) =b∫0∫ZN∑k=1ak(t, z, x, Dx)Dky(t, z) dz dt.In what follows by ((., .)) we will denote the duality brackets between L2(T, H10 (Z))and L2(T, H−1(Z)).Definition 2.1. A function x ∈ L2(T, H10 (Z)) is said to be a (weak) solution ofproblem (1) if((∂2x∂t2, u))− a(x, u) =b∫0∫Zg(t, z, x)u(t, z) dz dtfor all u ∈ L2(T, H−1(Z)).Finally our hypotheses on the function g(t, z, x) are the following:H(g): g : T × Z × R → R is a function such that(i) (t, z)→ g(t, z, x) is measurable;(ii) x → g(t, z, x) is continuous; and(iii) | g(t, z, x) |≤ β3(t, z) for almost all (t, z) ∈ T × Z and x ∈ R, withβ3 ∈ L2(T × Z).Suppose now that X is a separable Hilbert space with inner product (., .). Werecall the following generalization of the notion of a maximal monotone operator(see Zeidler [7, p. 585]).Pseudomonotonicity and hyperbolic equations 465Definition 2.2. An operator A : X → X is said to be “pseudomonotone”if xnw→ x in X as n → ∞ and lim supn→∞(A(xn), xn − x) ≤ 0, imply that (A(x), x −y) ≤ lim infn→∞(A(xn), xn − y) for all y ∈ X .Remark 2.1. A monotone hemicontinuous operator or a completely continuousoperator A : X → X is pseudomonotone. Pseudomonotonicity is preserved byaddition and it is easy to see that it implies property (M) (i.e. if xnw→ x in X ,A(xn)w→ u in X as n → ∞ and lim supn→∞(A(xn), xn − x) ≤ 0, then A(x) = u). Fordetails we refer to Zeidler [7, pp. 583–588].We will also need the following notion:Definition 2.3. Suppose that L : D(L) ⊆ X → X is a linear maximal monotoneoperator and V : X → X is a bounded nonlinear operator. We say that V (.) is“pseudomonotone with respect to D(L)” if for {xn}n≥1 ⊆ D(L) such that xnw→ xin X and L(xn)w→ L(x) in X as n → ∞ and also lim supn→∞(V (xn), xn − x) ≤ 0,then we have V (xn)w→ V (x) in X and (V (xn), xn)→ (V (x), x) as n → ∞.The following proposition will be used in order to prove the maximal mono-tonicity of the differential operator (see Zeidler [7, Theorem 32L, p. 897]).Proposition 2.1. A linear operator L : D(L) ⊆ X → X is maximal monotoneif and only if L is densely defined, closed and both L and L∗ are monotone.Now let L : D(L) ⊆ L2(T, X) → L2(T, X) be defined by Lx = −x′′ for allx ∈ D(L) = {y ∈ L2(T, X) : y′′ ∈ L2(T, X), y(0) = y′(T ) = 0} (here thetime derivatives of x(.) and y(.) are understood in the sense of vector-valueddistributions). With the use of the above proposition we easily see that L(.) ismaximal monotone.Consider now an operator A : T ×X → X satisfying the following hypotheses:H(A): A : T × X → X is an operator such that(i) t → A(t, x) is measurable;(ii) x → A(t, x) is demicontinuous and pseudomonotone (recall that demicon-tinuity means that if xn → x in X as n → ∞, then A(t, xn)w→ A(t, x) in X asn → ∞);(iii) ‖A(t, x)‖ ≤ β4(t) + c3‖x‖ a.e. on T with c3 > 0, β4 ∈ L2(T )+ ={f ∈ L2(T ) : f(t) ≥ 0 a.e.}; and(iv) (A(t, x), x) ≥ c4‖x‖2 − β5(t)‖x‖ − β6(t) for almost all t ∈ T , allx ∈ X , with c4 > 0, β6 ∈ L1(T ), β5 ∈ L2(T )+.Let Â : L2(T, X) → L2(T, X) be the Nemitsky (superposition) operator cor-responding to A(t, x) i.e. Â(x)(.) = A(., x(.)). We will show that, in some sense,the pseudomonotonicity of A(t, .) can be lifted to Â(.).466 D.A.KandilakisProposition 2.2. Let A : T ×X → X be an operator satisfying hypothesis H(A)and L : D(L) ⊆ L2(T, X)→ L2(T, X) be the linear maximal monotone operatordefined by L(x) = −x′′ for all x ∈ D(L) = {y ∈ L2(T, X) : y′′ ∈ L2(T, X),y(0) = y′(T ) = 0}. Then the Nemitsky operator Â : L2(T, X) → L2(T, X) isdemicontinuous and pseudomonotone with respect to D(L).Proof: We will first prove the demicontinuity of Â(.). So let xn → x in L2(T, X)as n → ∞. By passing to a subsequence if necessary, we may assume thatxn(t) → x(t) in X a.e. on T as n → ∞. Then hypothesis H(A)(ii) implies thatfor every y ∈ L2(T, X) we have that (A(t, xn(t)), y(t)) → (A(t, x(t)), y(t)) a.e.on T . Because of hypothesis H(A)(iii) we can apply the extended dominatedconvergence theorem (see Ash [1, Theorem 7.52, p. 295]) and get that((Â(xn), y)) =b∫0(A(t, xn(t)), y(t)) dt→b∫0(A(t, x(t)), y(t)) dt = ((Â(x), y))(here ((., .)) denotes the inner product of L2(T, X)). Since y ∈ L2(T, X) wasarbitrary, we conclude that Â(xn)w→ Â(x) in L2(T, X) as n → ∞, which showsthat Â(.) is demicontinuous.Next we will show that Â(.) is pseudomonotone with respect to D(L). Soassume that x, xn ∈ D(L), n ≥ 1, xnw→ x in L2(T, X), x′′nw→ x′′ in L2(T, X) asn → ∞ andlim supn→∞((Â(xn), xn − x)) = lim supn→∞b∫0(A(t, xn(t)), xn(t)− x(t)) dt ≤ 0.Since x′′nw→ x′′ in L2(T, X) as n → ∞ we havex′n(t) = −T∫tx′′n(s) dsw→ −T∫tx′′(s) ds = x′(t)for every t ∈ T , i.e. x′n(t)w→ x′(t) in X for every t ∈ T . Similarly we can showthat xn(t)w→ x(t) in X for every t ∈ T .Now let ξn(t) = (A(t, xn(t)), xn(t)−x(t)). Also let B ⊆ T be the Lebesgue-nullset outside of which hypotheses H(A)(iii) and (iv) hold. Then we have:(2)ξn(t) ≥ ϑn(t) = c4‖xn(t)‖p − β5(t)‖xn(t)‖ − β6(t)−(β4(t) + c3‖xn(t)‖)‖x(t)‖ for all t ∈ T \B.Pseudomonotonicity and hyperbolic equations 467Let also ϑn(t) = 0 for t ∈ B. Set C = {t ∈ T : lim infn→∞ξn(t) < 0}. This is aLebesgue measurable subset of T . We will show that C ⊆ B and therefore λ(C) =0 (here λ(.) is the Lebesgue measure on T ). Indeed, if t ∈ C∩(T \B), then considera subsequence {nm}m≥1 of N such that lim infn→∞ξn(t) = limn→∞ξnm(t) < 0. Exploit-ing the fact that A(t, .) is pseudomonotone we obtain that (A(t, xnm(t)), xnm(t)−x(t)) → 0, contradicting the fact that t ∈ C. So λ(C) = 0, which means that0 ≤ lim infn→∞ξn(t) a.e. on T . Then from the extended Fatou’s lemma (see Ash [1,Theorem 7.5.2, p. 295]), we obtain0 ≤b∫0lim infn→∞ξn(t) dt ≤ lim infn→∞b∫0ξn(t) dt ≤ lim supn→∞b∫0ξn(t) dt ≤ 0.Henceb∫0ξn(t) dt → 0 as n → ∞. Since 0 ≤ lim infn→∞ξn(t) a.e. on T , we deducethat ξ−n (t) → 0 a.e. on T as n → ∞. Moreover from (2) we see that {ϑn}n≥1is uniformly integrable in L1(T ). Then 0 ≤ ξ−n (t) ≤ ϑ−n (t) a.e. on T and ofcourse {ϑ−n }n≥1 is also uniformly integrable. By applying the extended dominatedconvergence theorem we see thatb∫0ξ−n (t) dt → 0 as n → ∞. Therefore we havelimn→∞b∫0| ξn(t) | dt = limn→∞b∫0(ξn(t) + 2ξ−n (t)) dt = 0,so by passing to a subsequence if necessary, we may assume that ξn(t)→ 0 a.e. onT as n → ∞. Since A(t, .) is pseudomonotone, we obtain that (A(t, xn(t)), xn(t))→ (A(t, x(t)), x(t)) a.e. on T and A(t, xn(t))w→ A(t, x(t)) a.e. on T as n → ∞.A final application of the extended dominated convergence theorem gives thatÂ(xn)w→ Â(x) in L2(T, X) and ((Â(xn), xn))→ ((Â(x), x)) as n → ∞. ThereforeÂ(.) is pseudomonotone with respect to D(L). In the proof of our main theorem in the next section we will need the followingwell known surjectivity result (see B.-A. Ton [6] or Lions [4, Theorem 1.2, p. 319]):Proposition 2.3. If Y is a reflexive Banach space, L : D(L) ⊆ Y → Y ∗ isa linear maximal monotone operator and G : Y → Y ∗ is a bounded, demicon-tinuous operator which is pseudomonotone with respect to D(L) and satisfies(G(y),y)Y ∗,Y‖y‖Y→ +∞ as ‖y‖Y → ∞ (i.e. G(.) is coercive) then L+G is surjective;i.e. R(L+G) = Y ∗.468 D.A.Kandilakis3. Main theoremTheorem 3.1. If hypotheses H(a) and H(g) hold, then problem (1) has a solu-tion.Proof: In what follows, for notational simplicity, we write X = H10 (Z) andX∗ = H−1(Z). Note that X is a separable Hilbert space. Let â : T ×X×X → Rbe the following time dependent semilinear form:â(t, x, y) =∫ZN∑k=1ak(t, z, x(t, z), Dx(t, z))Dky(t, z) dz.Because of hypothesis H(a)(iii), y → â(t, x, y) is a continuous linear functionaldefined on X . Therefore the equation(A1(t, x), y) = â(t, x, y)defines uniquely a function A1 : T×X → X∗. Using hypothesis H(a) we can verifythat t → A1(t, x) is measurable, x → A1(t, x) is demicontinuous, ‖A1(t, x)‖∗ ≤β′1(t)+ c′1‖x‖ a.e. on T for all x ∈ X , with β′1 ∈ L2(T ), c′1 > 0 and (A1(t, x), x) ≥c′2‖x‖2 − β′2(t) a.e. on T for all x ∈ X , with β′2 ∈ L1(T ) and c′2 > 0. Now letf : T ×X → L2(T ×Z) be defined by f(t, x)(z) = −g(t, z, x(z)). From hypothesesH(g)(i) and (ii) we see that the function t → f(t, x) is measurable, the functionx → f(t, x) is continuous and|(f(t, x), x)| ≤∫Z|g(t, z, x(z))x(z)| dz≤ c(t)‖x‖2 a.e. on T with c(.) ∈ L2(T ).Also from hypothesis H(g)(iii) we have‖f(t, x)‖2 ≤ ‖β3(t, .)‖2 a.e. on T.Now let A : T × X → X∗ be defined by A(t, x) = A1(t, x) + f(t, x). It is easy tosee that t → A(t, x) is measurable, x → A(t, x) is demicontinuous, ‖A(t, x)‖∗ ≤β(t) + c‖x‖ a.e. on T with β ∈ L2(T ), c > 0, and | (A(t, x), x) |≥ c′2‖x‖2 −β′2(t)‖x‖ − β′3(t) a.e. on T with c′2 > 0, β′2 ∈ L2(T ) and β′3 ∈ L1(T ). Moreoverfrom Theorem 3.1 of Gossez-Mustonen [3] we know thatA(t, .) is pseudomonotone.Thus if Â : L2(T, X) → L2(T, X∗) is the Nemitsky operator corresponding toA(t, x), then by virtue of Proposition 2.2 we see that Â is pseudomonotone withrespect to D(L).Let f̂ : L2(T, X) → L2(T × Z) be the Nemitsky operator corresponding tof(t, x). We claim that the operator x → Â1(x)+ f̂(x) = Â(x) is coercive. Indeed,from hypothesis H(a)(iv) we know that(3) ((Â1(x), x)) ≥ c2‖x‖2L2(T,X) − ‖β2‖L1(T×Z).Pseudomonotonicity and hyperbolic equations 469Also using Young’s inequality with ε > 0 we have((f̂(x), x)) = (f̂(x), x)L2(T ;L2(Z)) ≤ ‖f̂(x)‖L2(T×Z)‖x‖L2(T×Z)≤ε22‖f̂(x)‖2L2(T×Z) +12ε2‖x‖2L2(T×Z).But by virtue of hypothesis H(g)(iii) and Minkowski’s inequality we have‖f̂(x)‖2L2(T×Z) ≤ γ1 + γ2‖x‖2L2(T,X) for some γ1, γ2 > 0.Hence it follows that(4) ((f̂ (x), x)) ≤ε22γ1 +ε22γ2‖x‖2L2(T,X) +12ε2‖x‖2L2(T×Z).From (3) and (4) we get(5)((Â1(x) + f̂(x), x)) ≥ (c −12ε2γ2)‖x‖2L2(T,X),−12ε2‖x‖2L2(T×Z) −ε22γ1.Choose ε > 0 so that c2 >12ε2γ2. Then from (5) we deduce that x → Â1(x)+ f̂(x)is coercive.Proposition 2.3 implies that the operator L+ Â is surjective. Therefore thereexists x ∈ D(L) such that(6) Lx+ Â(x) = 0.But (6) is equivalent to (1), therefore x solves (1). References[1] Ash R., Analysis and Probability, Academic Press, NY, 1972.[2] Barbu V., Nonlinear Semigroups and Differential Equations in Banach Spaces, NoordhoffInter. Pub. Leyden, The Netherlands.[3] Gossez J.-P., Mustonen V., Pseudomonotonicity and the Leray-Lions condition, Diff. andIntegral Equations 6 (1993), 37–45.[4] Lions J.-L., Quelques Methodes de Resolution des Problemes aux Limites Non-Lineaires,Dunod, Paris, 1969.[5] Papageorgiou N.S., Existence of solutions for second order evolution inclusions, J. Appl.Math and Stoch. Anal. 4, vol. 7 (1994), pp. 525–535.[6] Ton B.-A., Nonlinear evolution equations in Banach spaces, J. Diff. Equations 9 (1971),608–618.[7] Zeidler E., Nonlinear Functional Analysis and its Applications, Springer Verlag, NY, 1990.Department of Mathematics, University of the Aegean, 83200 Karlovassi Samos,Greece(Received October 2, 1996, revised October 24, 1996)

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Pseudomonotonicity and nonlinear hyperbolic equations

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