On Second-Order Differential Equations with Nonsmooth Second Member

Abstract

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).

1. 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

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 [1]).

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 [3] 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 [4] (cf. also Jörgens [5]).

The physical motivation of (8) with can be seen in Lourêdo et al. [6].

Problem (4) with of the form (5) generalizes (8) when . 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 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

Remark 2. When it is possible to obtain a solution of Problem (14) by using the Theory of Semigroups (cf. Pazy [7]).

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

Theorem 3. Assume that , , and satisfy the hypotheses of Theorem 1 and . Then there exists a function in the class (16) such that is solution of the problem

Remark 4. Note that if belongs to class (16) then (cf. Lions and Magenes [8]).

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 in Theorem 1. Then the solution gives by this theorem when satisfies

Theorem 6. Let 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 .

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 [9] and Komornik [10]).

Let be , . Then the linear operator is continuous, bijective, and Also, is given by These results can be found in Lions [1] and Medeiros and Milla Miranda [11].

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 , and where .

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.

Proof of Theorem 1. We use the Galerkin method (cf. Lions [12] and Vicente and Frota [13]). Thus consider an approximate solution of Problem (14); that is, where and .

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

Substituting (69)–(71) into (68), we find where Applying Gronwall inequality in (72), we deduce

With this inequality, we determine a subsequence of , still denoted by , and a function such that By (13) we have that (cf. [1] and [11]). Then convergences (75) and Aubin-Lions Theorem (cf. [14]) 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 [12]). Thus, we consider the orthogonal projection where is the subspace generated by .

By similar arguments employed to obtain (67) and by (54) or (30), we find

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 .

We solve (88). Consider an approximate solution of (88) given by

By similar arguments used to obtain (74), we derive where

The preceding inequality gives By the projection method, we obtain, as in (83), This and estimate (91) provide

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 [14]) 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 .

Let . Consider . By (94) and (90)₂ we obtain Use the notation . Then the preceding problems give

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 [14] 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).

By (I)–(III), we deduce that there exists 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.

5. Proof of Theorem 6

Let and be solutions of Problem (26) with and in class (25). Consider . Then by (26) we have

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 Combining (140) and (142), we deduce

The preceding inequality with gives On the other hand, Hence, where Thus estimate (146) gives This expressions, (142), and provide Combining inequalities (138), (144), and (149) with inequality (134), we obtain This implies

We will prove that . In fact as belongs to class (25) we have that ; that is, is continuous on for all . Consider . Then by (128)₁ we obtain Integrating this equality on and using (151), we derive This, (146), and (151) give Therefore, for all . By density we obtain for all . Thus . Note that the constants ,  , and given, respectively, by (139) and (147) are independent of .

We apply similar arguments to the problem and we obtain for . After a finite number of steps we proof that for .

When , that is, , for all , expressions (144), (149), and similar arguments used to obtain the preceding result, allow us to deduce the uniqueness of solutions in this case.

6. Applications

Let be an open bounded set of with boundary of class . Let be the self-adjoint operator defined by the triplet , where denotes the usual scalar product of . The norm of is denoted by .

Lemma 11. Let be a real number. Then, is contained in and

Proof. First we prove (156) when , a natural number. More precisely, we prove that if , then and To prove (157), we use the method of mathematical induction. Consider and . Then, by the regularity of solutions of elliptic problems, we obtain and Assume that (157) holds for . Consider . Then, . By induction hypothesis it follows that and Consider the problem where . As , by the regularity of solutions of elliptic problems, we have and Inequalities (159) and (161) provide This and (158) prove (157).
Next we use the interpolation of Hilbert spaces. Consider a natural number such that . By results of intermediate spaces, we have (cf. Lions and Magenes [8]). We have by (157) that the injections are continuous. Then, by interpolation of Hilbert spaces, we have that the injection is continuous. We choose . Then, by (163), we obtain Also These last two equalities and (165) give the lemma.

Consider the operator , . For ; we have that This embedding and (156) imply that

Define By (169) we have that . Thus, for Theorems 1 and 3 give, respectively, solutions of problems (14) and (22).

Consider (171) for the particular case . Then Theorems 1 and 6 provide a unique solution of the problem On the other hand, we have if ( defined in (21)). In this case , that is, . The two restrictions and give, respectively, for , and the variations , , and . In all three cases, Corollary 5 gives a solution of the problem where is an open bounded set of , .

Conflict of Interests

The authors report that there is no conflict of interests in the publication of this paper.