On Second-Order Differential Equations with Nonsmooth Second Member

In an abstract framework, we consider the following initial value problem: u + μAu + F(u)u = f in (0,T), u(0) = u0, u󸀠(0) = u, where μ is a positive function and f a nonsmooth function. Given u, u, and f we determine F(u) in order to have a solution u of the previous equation. We analyze two cases of F(u). In our approach, we use theTheory of Linear Operators in Hilbert Spaces, the compactness Aubin-LionsTheorem, and an argument of Fixed Point. One of our two results provides an answer in a certain sense to an open question formulated by Lions in (1981, Page 284).


Introduction
Let  and  be two real separable Hilbert spaces with  dense in  and  continuously embedding in .The scalar product and norms of  and  are represented, respectively, by (, V) , || , ((, V)) , ‖‖ .

ISRN Applied Mathematics
He say also that one of the difficulties in the study of existence of solutions of the nonlinear equations lies in the difficulty in defining weak solutions, since the transposition method is essentially a linear method.This is ultimately connected to the fact that one cannot multiply distributions.
Problem (4) with () of the form ( 6) is an abstract formulation of Problem (7) with a slight modification of the nonlinear term.Theorem 3 gives the existence of solutions of this problem.In applications we give examples of Problem (7), with the modification of the nonlinear term, for Ω an open bounded set of R  ,  = 1, 2, 3.
In Grotta Ragazzo [3] the following equation is studied: This equation is considered as a first approximation of the Klein-Gordon equation Observe that (8) with  = 0 and  = 1 is the meson equation of Schiff [4] (cf. also Jörgens [5]).
The physical motivation of (8) with  = 1 can be seen in Lourêdo et al. [6].
Problem (4) with () of the form ( 5) generalizes ( 8) when  = 0.The existence of solutions of this problem is studied in Theorem 1.
In Louredo et al., loc.cit., is analyzed the equation with nonlinear boundary condition.The () given in (5) is different from the (∫ Ω  2 ) of this equation.The term () is related to the nonsmoothness of .

Main Results
We use the notation (  )  = ( − ),  ∈ R,  ≥ 0. Identifying  with   , we have  (  ) →  →  ( − ) .(11) Here and in what follows the notation  →  means that the space  in dense in the space  and the embedding of  in  are continuous.Note that ( − )  = (  ).Also, if ,  ∈ R with  ≥ , we have Assume that the embedding of  in  is compact. ( First we analyze Problem (2) with () = | −/2 | 2 , that is, the problem Theorem 1. Assume condition (13).Let  and  be real numbers with  ≥ 1.Consider Then there exists a function  in the class such that  is solution of the equation and satisfies the initial conditions Remark 2. When  = 1 it is possible to obtain a solution  of Problem ( 14) by using the Theory of Semigroups (cf.Pazy [7]).
Corollary 5.Under the same hypotheses of Theorem 1, there exists a function  in the class (16) such that  is solution of the problem We analyze the uniqueness of solutions.Consider  = 0 in Theorem 1. Then the solution  gives by this theorem when Theorem 6.Let  ≥ 1 be a real number.Consider Then there exists a unique solution  of Problem (26) in the class (25).
We do not know if there is uniqueness of solutions for Theorem 3, even when  = 0.
In what follows we prove the above results.
Introduce the adjoint operator (  ) * of   , that is, where X = (  ).Note that  is identified with   .By the properties of   , we have that where X = (  ).

ISRN Applied Mathematics
On the other hand, by (30) we derive The last two expressions give This and (38) provide that So (i) 1 is proved.Taking the limit in (41) and observing (33) 2 , we obtain (36).We prove (ii).We have that there exists a unique  ∈  such that By (33) 2 and (30) 1 , we have (44) This implies that Thus (ii) 2 is proved.By (43) and (45), we obtain This concludes the proof of the proposition.
Motived by (37), we equip the space (  )  with the scalar product where X = (  ).This scalar product on X  yields a norm which is equivalent to the usual norm of (  )  .By similarity between expressions (30) and (36) and between (32) and (37), respectively, we introduce the notations Also we use the notation With these considerations and expressions ( 29) and (33), we obtain and by expressions ( 35) and (31), Also by ( 37) and ( 48), ( 49), respectively, we find Proposition 8. Consider ,  ∈ R. Then the linear operator defined by is continuous.
Proof.We obtain where Y = ( − ).Then which proves the proposition.

Proof of Theorem 3
The idea is to apply a fixed point argument to the problem where  ∈ R,  ≥ 0.
We solve (88).Consider an approximate solution   of (88) given by By similar arguments used to obtain (74), we derive The preceding inequality gives By the projection method, we obtain, as in (83),  93) and (95) allow us to find a subsequence of (  ), still denoted by (  ), and a function   such that, by passing to limit in (94), we obtain This, initial conditions (90) 2 , and estimates (95) imply By taking the lim inf in both side of (91), we obtain As  →  →  and the embedding  in  are compact ( < (1 − )/2), it follows from of Aubin-Lions Theorem (see Simon [14]) that           0 ([0,];) ≤  1 , ∀ ≥ 0.
Define the map where   is the solution of Problem (97).We will prove that  has a fixed point.Consider only the case  ̸ = 0.The case  = 0 is outside of our attention.We will prove the following results.
Let  0 > 0. Consider  > 0. By (94) and (90) 2 we obtain Use the notation   =   −   0  .Then the preceding problems give Taking the scalar product of  of both sides of this equation with  −    , we find We obtain As (107) Also, where the constant  1 > 0 is independent of  and .Taking the lim inf in both sides of this inequality, we find By Simon [14] and noting that the embedding of  in  is compact, we derive Thus.
which proves the continuity of  at  0 > 0. In similar way we prove the continuity of  at  0 = 0.
Let (  ) be a sequence of positive numbers with   → ∞.It follows from (98) and the compactness of the embedding of  in  that there exists a subsequence of (  ), still denoted by (  ), and a function  such that This implies By estimate (91), we obtain Then Convergences ( 118) and (120) provide Thus by (117) we find As the sequence (  ) was arbitrary and the limit of (   ) is always the same, we conclude that Thus           0 ([0,];) → 0,  → ∞ which proves part (III).
By (I)-(III), we deduce that there exists  > 0,  ∈ R such that Considering this  in (97), we obtain a solution  of ( 22) that satisfies all conditions of the theorem.The proof of Corollary 5 follows by defining the map where   is the solution of the problem and applying similar arguments to those used in the proof of Theorem 3.
When   = 0, that is, () =  0 , for all  ∈ [0,], expressions (144), (149), and similar arguments used to obtain the preceding result, allow us to deduce the uniqueness of solutions in this case.