- About this Journal ·
- Abstracting and Indexing ·
- Aims and Scope ·
- Article Processing Charges ·
- Articles in Press ·
- Author Guidelines ·
- Bibliographic Information ·
- Citations to this Journal ·
- Contact Information ·
- Editorial Board ·
- Editorial Workflow ·
- Free eTOC Alerts ·
- Publication Ethics ·
- Reviewers Acknowledgment ·
- Submit a Manuscript ·
- Subscription Information ·
- Table of Contents
ISRN Applied Mathematics
Volume 2014 (2014), Article ID 305718, 13 pages
On Second-Order Differential Equations with Nonsmooth Second Member
1DM, UEPB, Campina Grande-PB, Brazil
2IM, UFRJ, Rio de Janeiro, RJ, Brazil
Received 15 January 2014; Accepted 27 February 2014; Published 24 March 2014
Academic Editors: A. Bellouquid and C. Join
Copyright © 2014 M. Milla Miranda et al. 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.
In an abstract framework, we consider the following initial value problem: u′′ + μAu + F(u)u = f in (0,T), , where is a positive function and f a nonsmooth function. Given u0, u1, and f we determine in order to have a solution u of the previous equation. We analyze two cases of . In our approach, we use the Theory of Linear Operators in Hilbert Spaces, the compactness Aubin-Lions Theorem, 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).
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
Let be the self-adjoint operator of defined by the triplet . Consider , . We denote by the Hilbert space equipped with the scalar product (cf. Lions ).
Consider the following initial value problem: where is a positive function and a nonsmooth function.
The objective of this work is to study the following inverse problem: given , , and ( dual space of ) to determine such that Problem (4) has a solution . We analyze two cases of , more precisely, the cases where is an appropriate Hilbert space.
In Lions [2, Page 284], the following problem is formulated where is an open bounded set of with boundary , , and being the Dirac mass supported at . He says not to know if this problem admits a solution. 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 an open bounded set of , .
In Grotta Ragazzo  the following equation is studied: This equation is considered as a first approximation of the Klein-Gordon equation Observe that (8) with and is the meson equation of Schiff  (cf. also Jörgens ).
In Louredo et al., loc.cit., is analyzed the equation with nonlinear boundary condition. The given in (5) is different from the of this equation. The term is related to the nonsmoothness of .
2. Main Results
We use the notation , , . Identifying with , we have 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 , with , we have Assume that
First we analyze Problem (2) with , that is, the problem
Theorem 1. Assume condition (13). Let and be real numbers with . Consider Then there exists a function in the class such that is solution of the equation and satisfies the initial conditions
To formulate the second problem, we introduce some notations. In fact, let us define where By (12) we have
Consider Problem (4) with and , that is, the problem
We analyze the uniqueness of solutions. Consider in Theorem 1. Then the solution gives by this theorem when satisfies
We do not know if there is uniqueness of solutions for Theorem 3, even when .
In what follows we prove the above results.
3. Proof of Theorem 1
Before proving the theorem, we make some considerations on the operator . Recall hypothesis (13). By solving the spectral problem , for all , we determine the eigenfunctions and eigenvalues, respectively, and of the operator , that is, Note that is a Hilbert basis of (cf. Brezis  and Komornik ).
Introduce the adjoint operator of , that is, where . Note that is identified with . By the properties of , we have that Thus, the linear operator is continuous and bijective.
Proposition 7. Let and . Then one has the following.(i) converges in , and(ii) converges in , andwhere .
Proof. We prove . As , we have
Consider . Then noting that , we obtain
On the other hand, by (30) we derive
The last two expressions give
This and (38) provide that
So is proved. Taking the limit in (41) and observing (33)2, we obtain (36).
We prove . We have that there exists a unique such that By (33)2 and (30)1, we have Then This implies that Thus 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 . This scalar product on 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 , . Then the linear operator defined by is continuous.
Proof. We obtain where . Then which proves the proposition.
Remark 9. Note that
Remark 10. Observe that if then
Multiply both sides of (61)1 by and add from up to . We obtain where Then, By Proposition 8 with and and noting that , we obtain Substituting this equality into (66) and integrating on , we obtain Note that Also
With this inequality, we determine a subsequence of , still denoted by , and a function such that By (13) we have that (cf.  and ). Then convergences (75) and Aubin-Lions Theorem (cf. ) imply Therefore, This convergence and convergence (75)1 provide which implies since , defined in (65).
In order to obtain an estimate for , we apply the method of projections to the approximate equation (61) (cf. Lions ). Thus, we consider the orthogonal projection where is the subspace generated by .
Multiply both members of (61)1 by and add from up to . Then, applying to this result, expression (82), affirmation (54), or (30) and noting that , , belong to , we obtain which gives Then estimates (74) and (80) provide Thus, there exists a subsequence of , still denoted by , such that
Expressions (75) and (86) tell us that belongs to class (16). Convergences (75)1, (80), and (86) allow us to pass to limit in (83) and to obtain which provides (17). Initial conditions (18) follow from convergences (75) and (86).
4. Proof of Theorem 3
The idea is to apply a fixed point argument to the problem where .
By similar arguments used to obtain (74), we derive where
Estimates (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 , it follows from of Aubin-Lions Theorem (see Simon ) that
Define the map where is the solution of Problem (97). We will prove that has a fixed point. Consider only the case . The case is outside of our attention. We will prove the following results.(I) One has.
In fact if , we have that is a solution of (97) with , but this a contradiction since .(II) One has is continuous on .
Taking the scalar product of of both sides of this equation with , we find We have Integrating on both sides of the last two expressions, we derive We obtain As , we have Also,
Taking into account estimate (91) in (107) and (109), we find Substituting the last two inequalities into (106), we obtain Combining this inequality with (105), we derive Considering and using the Gronwall inequality, this expression gives where the constant is independent of and . Taking the lim inf in both sides of this inequality, we find By Simon  and noting that the embedding of in is compact, we derive Thus. which proves the continuity of at . In similar way we prove the continuity of at .(III) One has as .
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 which proves part (III).
5. Proof of Theorem 6
Fix . Consider . Introduce the function We have Apply the operator given by the first member of (128)1 to . We obtain where . By (130)2, we find Also by (130)3, Integrate (131) on and use (128)1, (130), (132), and (133). We deduce We modify each term of the second member of (134). We have Then First, we assume that . In this case, we choose . We have Combining this last inequality with (136), we find We introduce the notations (Note that since belongs to class (25)). We have By (130)4, we obtain Therefore