Dirichlet boundary value problem for Duffing’s equation

We use a direct variational method in order to investigate the dependence on parameter for the solution for a Dung type equation with Dirichlet boundary value conditions.


Introduction
Recently the classical variational problem for a Duffing type equation received again some attention.In [1,2,4,8] some variational approaches were used in order to receive the existence of solutions for both periodic and Dirichlet type boundary value problems.Mainly direct method is applied under various conditions pertaining to at most quadratic growth imposed on the nonlinear term given in [1] and further relaxed in [8].Dirichlet problems for such equations could also be considered by some other methods, for example Min-max Theorem due to Manashevich, [5].In [6] the author gives some historical results concerning the Dirichlet problem for Duffing type equations and discusses the methods which are used in reaching the existence results which are different form the ones which we use and comprise the classical variational approach, the topological method.In most sources cited the authors assume the friction term r ∈ C 1 (0, 1); r(τ ) ≥ 0 for τ ∈ [0, 1], and they require some further conditions on r.Mainly a type of monotonicity of r is assumed or else for all t ∈ [0, 1], see [4].The standard procedure to treat Duffing's Equation with control function u ∈ L ∞ (0, 1) and G satisfying some suitable assumptions: EJQTDE, 2013 No. 37, p. 1 Upon putting y = R(t)x boundary problem (ClassicDEq) reads In this paper we are concerned with the variational formulation for the Duffing Equation but we apply the different schema.Namely we consider (DEq) under the assumptions that r ∈ L ∞ (0, 1) and f ∈ L 1 (0, 1).Solutions to above are investigated in H 1 0 (0, 1) and these are the weak solutions.We shall show that by the Fundamental Lemma of the Calculus of Variations, any weak solutions to (DEq) is classical one, i.e.
The equation (DEq) is not in a variational form i.e. there is no suitable functional J for which (DEq) corresponds to its critical points.Then by putting h = dx dt , we may consider the following auxiliary problem We see that weak sense solutions to above are the critical points to J given by following integral (Func) Under the assumptions that h ∈ L ∞ (0, 1), r ∈ L ∞ (0, 1), and some growth requirements on F we can prove that problem (AuxEq) has at least one solution.
To prove this, its sufficient to show that 1. functional J is differentiable in sense of Gâteaux 2. functional J is coercive 3. functional J is weakly lower semi continuous When solutions to (AuxEq) are obtained for any h ∈ L ∞ (0, 1), we will apply the iterative procedure assuming that for any x,y ∈ H 1 0 (0, 1), L < 1 independent of x,y and r 1−L < 1.This will provide solutions to (DEq).EJQTDE, 2013 No. 37, p. 2 Moreover functions F, F x : [0, 1] × R → R will be a Caratheodory functions, satisfying the conditions below: When compared with existing results, see [8], our approach allows for a more general friction term, i.e. it belongs to L ∞ .However in [8] where the friction is continuous the Author obtains results when L = 1 which is not possible with our approach.

Preliminaries
The following two remarks will be essential for our argument. . (1.1) Remark 1.2.Let 1 ≤ p < q and x ∈ L q (0, 1).Then We shall also require Poincarè inequality in following form: .
Since this formulation of Poincarè inequality does not match exactly version presented in [3], we shall present proof for this fact.Proof.Since x ∈ H 1 0 (0, 1), then x(0) = 0 then This leads us to very important estimation, that we will use a lot in this paper: From (1.2) we have that for 2 and n > 2 the following holds .

Variational framework
We shall prove that solving (AuxEq) is equivalent with solving critical points problem for following functional J defined at H 1 0 (0, 1).We shall consider functional described by formula: for each x ∈ H 1 0 (0, 1).We shall consider two versions of assumptions that will provide different results.
We shall prove that this functional is well defined, is Gâteaux differentiable and it's critical points are the weak solutions to (AuxEq).We will also prove that the regularity class of this solution is higher than H 1 0 (0, 1).We would like to compute Gâteaux derivatives, but first we have to ensure that we can differentiate under integration sign.We see the following properties: Lemma 2.1.Under assumption (H1) the following equality holds for any x ∈ H 1 0 (0, 1) and g ∈ H 1 0 (0, 1) In order to prove this, Dominated convergence Theorem is applied.
We see that is well defined.By (H1) for any x we see The differentiability is a consequence of (2.1).
We shall now prove that weak solution (WS) for functional (Func) is a classical solution.Then we shall see that functional critical points to J are the weak solutions to (AuxEq)

Proof.
Since in Theorem 2.2 we have proved that Then the solution of δJ(x, g) = 0, g ∈ H 1 0 (0, 1) is of a class W 2,1 (0, 1) and thus is a classical one.EJQTDE, 2013 No. 37, p. 5 In this section we shall prove the existence of solution to (AuxEq).

Proof.
By norm continuity, the first part of (Func) is weakly lower semicontinuous.
is linear and continuous thus it is w.l.s.c.For F function we need to apply some additional theory in order to prove its w.l.s.c.
Lets consider weakly converged sequence in H 1 0 (0, 1), x n x 0 .By Arzela-Ascoli Theorem there exists such subsequence that converge uniformly in C (0, 1).Then for sufficiently large d, the below condition holds: for sufficiently large n.By Lebesgue's dominated convergence Theorem we obtain: Then it is proved that functional (Func) is w.l.s.c.Lemma 3.2.Functional J (Func) is coercive if (H1) and one of the assumptions holds 1. F satisfies (H2) 2. F satisfies (H3) with A satisfying: |A| < 1 2 . Proof.
First we observe that We see that: x H 1 0 (0,1) .
Which proves the lemma in first case.If the other condition (H3) holds so there exists such A ∈ R \ {0}, B, C ∈ R for which the following holds Integrating the both sides of (3.4) we get We should consider two cases: 1.If sequence of norms x n diverges in H 1 0 (0, 1) it may still converge in L 2 (0, 1).In such case 2. In opposite case, the same inequality holds since Poincarè inequality is applicable. Thus Then functional (Func) is coercive since |A| < 1 2 for unbounded case.Together with bonded case this proves lemma in second case.
Above means that we have 2 cases in which we proved coerciveness.
The theorem below proves the existence of solution.
Theorem 3.3.Let E be reflexive Banach space and let the functional f : E → R be w.l.s.c. and coercive.Then there exist a function that minimizes f .[7] Then we have the following Theorem 3.4.There exists at least one solution to (AuxEq) if (H1) is satisfied and one of the following holds: 1. F satisfies (H2) 2. F satisfies (H3) with A satisfying |A| < 1 2 .
EJQTDE, 2013 No. 37, p. 7 Proof.By Lemmas 3.1 and 3.2, and reflexiveness of H 1 0 (0, 1) we see that assumptions of Theorem 3.3 are satisfied.Then there exists solutions in functional critical points problem.By Lemma 2.5 this solution is a classical solution to (AuxEq).

Iterative scheme framework
In this section we shall prove that using equation (AuxEq) we may provide the solution of (DEq).2. F is bounded (H3) and |A| < 1 2 .and moreover for any x,y and L < 1 independent of x,y.Then if 1−L < 1 then problem (DEq) has at least one solution

Proof.
Let h be an arbitrary taken function h ∈ H 1 (0, 1).Lets then define a sequence (x n ) ⊂ H 1 0 (0, 1) ∩ W 2,1 (0, 1) , n ∈ N. We consider following formula By using Theorem 3.4 and induction with respect to n, it is easy to prove that such sequence is well defined.We shall prove that (x n ) is Cauchy sequence in H 1 0 (0, 1) with respect to norm.Since the solution in understood in weak sense, we do the following.Let n, m ∈ N. Then the (4.2) for n and m is multiplied by (x n − x m ) and then integrated with respect to t ∈ [0, 1].
After subtracting the sides and integrating by parts EJQTDE, 2013 No. 37, p. 8 Thus by (1.3) for 0 = x n − x m ∈ H 1 0 (0, 1), we have that By (4.1) we have that ) .Thus we have that: we have that (x n ) is Cauchy sequence with respect to H 1 0 (0, 1) norm.

Example
Example 5.1.The above schema can be applied for the following equation Also (H2) is satisfied since F (•, x) := 1 2 e −(•) x 2 is convex with respect to its second variable.

Theorem 4 . 1 .
If (H1) is satisfied and if one of below conditions holds 1. F is convex (H2)