Well-posed Dirichlet problems pertaining to the Duffing equation

In this paper we investigate existence and continuous dependence on a functional parameter of Duffing’s type equation with Dirichlet boundary value conditions. The method applied relies on variational investigation of auxiliary problems and then in order to prove existence, the Banach fixed point theorem is applied. Uniqueness of solutions is also examined.


Introduction
We investigate the classical variational problem for a Duffing type equation.It concerns a nonlinear second order differential equation used to model certain damped and driven oscillators, firstly introduced in [4] by Georg Duffing who was inspired by joint works of O. von Martienssen and J. Biermanns.Variational approach was found successful in proving existence of solution to this problem.The classical variational problem for a Duffing type equation with Dirichlet boundary condition yields whether there exists a function x ∈ H 1 0 (0, 1), such that d 2 dt 2 x(t) + r(t) d dt x(t) + G(t, x(t), u(t)) = 0.
Here r ∈ C 1 (0, 1) stands for the friction term, and G is a nonlinear term, satisfying some suit- able assumptions.In fact G can correspond to a restoring force for a string in string-damper system.The Duffing's equation was also found applicable for some problems concerning current and flux, thus r and G may as well corresponds to its coefficients.The equation is well known for its chaotical behaviour, well described by Holmes [6,7,9,10,11] and jointly by Holmes and Moon [18,8].Recently in [1,2,5,20] some variational approaches were used P. Kowalski in order to receive the existence results for both periodic and Dirichlet type boundary conditions.See for example in [1,2,12,20], where variational approaches are applied such as a direct method, mountain geometry, a min-max theorem due to Manashevich.We note also paper [16] where the topological method is used.Since a Duffing equation serves a mechanical model, it is also important to know whether the solution, once its existence is proved, depends continuously on a functional parameter and also whether this solution is unique.Into the classical variational problem we introduce a control function u ∈ H 1 0 (0, 1) with only function G depending on it.Thus it is of interest to know the conditions which guarantee a) the existence of solutions, b) their uniqueness, c) dependence of solutions on parameters.This is sometimes known as Hadamard's programme and problems satisfying all three conditions are called well-posed.The question of continuous dependence on parameters has a great impact on future applications of any model since it is desirable to know whether the solution to the small deviation from the model would return, in a continuous way, to the solution of the original model.In our investigations we base somehow on [15] however, we use much simpler approach.As concerns the existence of solutions we use generalization of our earlier result [14].
We consider the problem in the following form, under the assumptions that r ∈ L ∞ (0, 1) and some further requirements on f and g.We introduce u ∈ L q (0, 1) to be the functional parameter.Solutions to the above problem are investigated in H 1 0 (0, 1) and these are understood as the weak solutions.We examine problem (DEq) by a kind of a two step method.It can be described as follows.At first we substitute h := dx(t) dt and we consider an auxiliary problem of a form Once the auxiliary problem is solved, we apply the Banach fixed point theorem using condition (H5) to obtain solutions of (DEq).By the fundamental lemma of calculus of variations any weak solutions x to (DEq) is a classical one i.e.
In order to obtain limit solution we assume that for any u ∈ L q (0, 1) there exists a constant L(u) < 1 such that for any x,y ∈ H 1 0 (0, 1), x = y.To obtain continuous dependence on the functional parameter we consider the following (stronger) version of assumptions.Assume there exists d * ∈ (0, 1) such
The inequalities are proved in [14].
for almost every x ∈ I.

Theorem 2.7 ([17]
).Let E be reflexive Banach space and functional f : E → R is s.w.l.s.c. and coercive then there exists a function that minimizes f .Theorem 2.8 (Krasnoselskii's theorem [13]).Let Ω ⊂ R be an interval and let f : for all i ∈ N and t ∈ Ω a.e., then the Nemytskii's operator Well-posed Dirichlet problems pertaining to the Duffing equation 5 is well-defined and sequentially continuous, that is, if

Existence result for the auxiliary problem
In order to solve (DEq) we introduce an auxiliary problem The above problem is in a variational form.We consider the following functional We prove that critical points to J u are the weak solutions to (AuxEq).In order to prove that problem (AuxEq) has at least one solution it is sufficient to show that: 1. functional J u is well defined and differentiable in sense of Gâteaux, 2. functional J u is coercive and sequentially weakly lower semicontinuous, 3. and critical points of J u are the solutions of (AuxEq).
In the sequel we shall assume u ∈ L q (0, 1) to be a fixed parameter.In order to simplify proofs we introduce the following functionals.
We start by proving that the functional J u is well defined and admits a Gâteaux derivative.
Proof of this fact is elementary.
Lemma 3.3.Assume (H2) holds.Then The proof of the above properties follows from Lebesgue's dominated convergence theorem.
Lemma 3.4.Assume that (H1) and (H2) hold.Then functional J u is differentiable in sense of Gâteaux and its derivative is equal to for all v ∈ H 1 0 (0, 1).
Proof for this fact is elementary.
We will now focus on proving that problem of finding critical points of functional J u is equivalent to solving problem (AuxEq).Definition 3.5.Every x ∈ H 1 0 (0, 1) which satisfies the equality shall be called a critical point for J u .
The proof follows from the fundamental lemma of calculus of variations.Finally we prove the existence of critical point.Lemma 3.9.Assume (H1) and (H2) holds.Then the functional J u is sequentially weakly lower semicontinuous.
Proof.It is obvious that is s.w.l.s.c.It is easy to show that −J 4 u is s.w.l.s.c.using Lebesgue's domininated convergence theorem.We prove that J 3 u is s.w.l.s.c.Assume x n x in H 1 0 (0, 1).We will prove that lim inf J 3 u (x n ) ≥ J 3 u (x).We reason by contradiction.Suppose there exists such a subsequence that lim J 3 u (x k n ) < J 3 u (x).By the Arzelà-Ascoli theorem this subsequence admits a subsubsequence (x l n ) convergent strongly in C (0, 1).Thus it is bounded in C (0, 1) norm.As (H1) holds, we may reason by using Lebesgue's dominated convergence theorem to obtain the inequality . Thus it contradicts the supposition.Finally J u is s.w.l.s.c.Lemma 3.10.Assume (H1) and (H2) hold.If additionally either (H3) or (H4) holds then J u is coercive.
Proof.We will prove that functional J u is bounded from below by a coercive function depending on x H 1 0 (0,1) .Let x ∈ H 1 0 (0, 1) be taken arbitrarily.We see that J 1 u (x) = 1 2 x 2 H 1 0 (0,1) .By the Cauchy-Schwartz inequality one can prove that We can easily calculate that u s L q (0,1) .
Thus if we assume (H3), functional J u is obviously bounded from below by coercive function.Now we assume (H4).Then and thus functional is obviously coercive since |A| < 1 2 .
We present the following result.
Proof.By Lemmas 3.9 and 3.10, and reflexivity of H 1 0 (0, 1), we see that assumptions of The- orem 2.7 are satisfied.Then there exists a critical point.By Lemma 3.8 this critical point is a classical solution to (AuxEq).

Iterative scheme
In this section we shall prove that using equation (AuxEq) we may produce the solution of (DEq).Since we proved that (AuxEq) for each h ∈ L 2 (0, 1) admits a classical solution it somehow define a solution operator Λ.Up to the section end we will assume u ∈ L q (0, 1) to be a fixed parameter.Proof.By Theorem (3.11) we know that for any function h ∈ L 2 (0, 1) there exists a solution to problem (AuxEq).This means that for any function v ∈ H 1 0 (0, 1) there exists a solution x v to the following problem, Let Ψ : H 1 0 (0, 1) → 2 H 1 0 (0,1) be a multivalued operator which to any v ∈ H 1 0 (0, 1) assigns a set of solutions of (4.1) corresponding to this parameter.Let Λ : H 1 0 (0, 1) → H 1 0 (0, 1) be an arbitrarily chosen single valued selection of operator Ψ, i.e. for any v ∈ H 1 0 (0, 1) Λv ∈ Ψv.We prove that Λ is a contraction mapping.
Well-posed Dirichlet problems pertaining to the Duffing equation 9Let h, v ∈ H 1 0 (0, 1).Denote x v := Λv, x h := Λh.Assume x v = x h .Otherwise condition for contraction mappings holds.Equations (4.1) for h and v are multiplied by (x v − x h ) and then integrated with respect to t ∈ [0, 1].
After subtracting the sides and integrating by parts we get By x h = x v relation, (H5) and by Theorem 2.5 Thus Λ is a contraction mapping.Then the assumptions of Banach's fixed point theorem are satisfied and thus Λ admits a single fixed point in H 1 0 (0, 1), which is a solution of (DEq).However, since Λ was chosen arbitrarily we cannot be sure that there exists a unique solution.We reason by contradiction.Assume that x and y are two distinct solutions of (DEq).We multiply (DEq) by (x(t) − y(t)) and integrate over [0, 1] interval.
Similarly we obtain Thus it contradicts the assumption that there are two distinct solutions.

P. Kowalski
We note that however we obtain the uniqueness of the weak solution, the classical one is also unique since Lemma 3.8 holds in this case as well.We can also prove similar property in the limit case with p = 2.In fact one can similarly prove that in case (H5) holds, the operator Λ is actually single-valued.Lemma 4.2.If 1 − u s L q (0,1) a L q q−s (0,1) > 0, (H1), (H2), (H3) and (H5) are satisfied then problem (DEq) has at least one solution.
Proofs follow the steps from proof of Theorem 4.1.In the next section we will investigate the impact of functional parameter, which until now was considered as fixed.

Continuous dependence on functional parameter
We will prove that sequence of solutions corresponding to sequence of parameters is bounded.Theorem 5.1.Let (u k ) k∈N ⊂ L q (0, 1) be a bounded sequence of functional parameters.Assume (H1c), (H2), (H5) are satisfied and either (H3) or (H4) holds.Then there exists a sequence (x k ) k∈N of solutions to (DEq), such that each x k corresponds to a parameter u k and that sequence (x k ) k∈N is bounded in H 1 0 (0, 1).Proof.Let (u k ) k∈N be a bounded sequence of functional parameters.By Theorem 4.1 for any u k there exists x k ∈ H 1 0 (0, 1) ∩ W 2,1 (0, 1) being a solution of (DEq).We may equivalently consider the following problem.For all k ∈ N, and for all v ∈ H 1 0 (0, 1), we have that: We shall test against v := x k function.Then we have that We integrate by parts By Lemma 2.5 we obtain that If x k H 1 0 (0,1) = 0, the assertion is trivial.We may assume that x k H 1 0 (0,1) > 0. Then: Well-posed Dirichlet problems pertaining to the Duffing equation 11 Suppose the sequence x k is unbounded in H 1 0 (0, 1).By (H1c) and (H2) for sufficiently large k we obtain The above is equivalent to u s L q (0,1) a L q q−s (0,1) ≤ f L 1 (0,1) . (5.1) Since the left-hand side is a coercive functional, its values would go up to infinity as k → ∞.This contradicts (5.1).
Now we focus on dependence on functional parameter.
Theorem 5.2.Let (u k ) ⊂ L q (0, 1), k ∈ N be a bounded sequence of functional parameters.Assume (H1c), (H2), (H5) are satisfied and either (H3) holds or (H4) does.Then there exists a sequence of solutions x k of (DEq) corresponding to u k .Moreover, • If u k → u strongly in L q (0, 1) then (x k ) x in H 1 0 (0, 1) and x is a solution to (DEq) corresponding to u.
• If g(t, x, u) = g(t, x)u, and if assumption (H2) takes a form |g(t, x)| ≤ |x| p−1 a(t), a ∈ L q q−s (0, 1) , (H2c) then for any sequence of parameters u k u converging weakly in L q (0, 1) there exists a sequence of solutions to (DEq) such x k x converging weakly in H 1 0 (0, 1) and x is a solution to (DEq) corresponding to u. Proof.By Theorem 4.1 there exists a sequence of solutions x k of (DEq) corresponding to each u k .By Theorem 5.1 this sequence is bounded in H 1 0 (0, 1).By the Rellich-Kondrachov theorem this sequence admits a subsequence x n k convergent strongly in L 2 (0, 1) and in C ([0, 1]) and also weakly in H 1 0 (0, 1).Let x denote the element such that x n k x in H 1 0 (0, 1).By the fundamental lemma of calculus of variations we can use equivalently the weak formulation.Let v ∈ H 1 0 (0, 1).with L = 0.71 < 1.Since r L ∞ (0,1) = 0.25 • e −t 2 /2 L ∞ (0,1) = 0.25 and r L ∞ (0,1) 1−L = 0.25 1−0.71 < 0.87 < 1 then by Proposition 4.2 we conclude that problem (6.1) has at least one solution to each u n .Then the solution to u such that u n → u is x n x. Thus x is a solution to

Theorem 2 . 9 (
Duality pairing convergence[3, prop.3.5.(iv)]).Let E be a Banach space.If x n x in E and if f n → f strongly in E * then f n , x n → f , x strongly.