c
Applied Mathematics E-Notes, 7(2007), 159-166 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Unilateral Monotone Iteration Scheme For A Forced
Duffing Equation With Periodic Boundary Conditions∗
Ahmed Alsaedi†
Received 12 June 2006
Abstract
In this paper, we apply the generalized quasilinearization technique to a forced
Duffing equation with periodic boundary conditions and obtain a monotone sequence of approximate solutions converging quadratically to the unique solution
of the problem.
1
Introduction
An interesting and fruitful technique for proving existence results for nonlinear problems is the method of upper and lower solutions. This method coupled with the monotone iterative technique, known as quasilinearization technique [1], manifests itself as
an effective and flexible mechanism that offers theoretical as well as constructive existence results in a closed set, generated by the lower and upper solutions. However, the
concavity/convexity assumption that is demanded by the method of quasilinearization,
proved to be a stumbling block for further development of the theory. The nineties
brought new dimensions to this technique. The most interesting new idea was introduced by Lakshmikantham [2-3] who generalized the method of quasilinearizaion by
relaxing the convexity assumption. This development was so significant that it attracted the attention of many researchers and the method was extensively developed
and applied to a wide range of initial and boundary value problems for different types
of differential equations, see [4-14].
In this paper, we consider a nonlinear periodic boundary value problem involving
a forced Duffing equation and develop the generalized quasilinearization method for
the problem at hand. In fact, we obtain a sequence of lower solutions (approximate
solutions) converging monotonically and quadratically to the unique solution of the
problem. Duffing equation is a well known nonlinear equation of applied science which is
used as a powerful tool to discuss some important practical phenomena such as periodic
orbit extraction, nonuniformity caused by an infinite domain, nonlinear mechanical
oscillators, etc. Another important application of Duffing equation is in the field of the
prediction of diseases. In Section 2, we discuss some basic results while Section 3 is
devoted to the main result.
∗ Mathematics
Subject Classifications: 34B10, 34B15.
of Mathematics, Faculty of science, King Abdul Aziz University, P.O. Box. 80257,
Jeddah 21589, Saudi Arabia
† Department
159
160
2
Monotone Iteration for a Forced Duffing Equation
Preliminary Notes
We know that the homogeneous periodic boundary value problem
−u00(x) − ku0(x) − λu(x) = 0, x ∈ [0, π],
u(0) = u(π), u0(0) = u0 (π),
has only the trivial solution if and only if λ 6= 4n2 for k = 0 and λ 6= 0 for k 6= 0 for all
n ∈ {0, 1, 2, ...}. Consequently, for these values of λ and for any σ(x) ∈ C([0, π]), the
non homogeneous problem
−u00(x) − ku0 (x) − λu(x) = σ(x), x ∈ [0, π]
u(0) = u(π), u0(0) = u0 (π),
has a unique solution
u(x) =
Z
π
Gλ(x, y)σ(y)dy,
0
where Gλ(x, y) is the Green’s function given by

√
√
−k
k 2 −4λ
k 2 −4λ
 ξ1 sinh
2 π sinh
(π
−
(y
−
x))
+
e
(y
−
x)
, 0 ≤ x ≤ y,
√ 2
√ 2 Gλ(x, y) =
k
2
2
k −4λ
k −4λ
 ξ1 sinh
(π − (x − y)) + e 2 π sinh
(x − y) , y ≤ x ≤ π,
2
2
for λ <
k2
4
for λ >
k2
,
4
and

√
√
−k
4λ−k 2
4λ−k 2
 ξ2 sin
2 π sin
(π
−
(y
−
x))
+
e
(y
−
x)
, 0 ≤ x ≤ y,
√ 2
√ 2 Gλ(x, y) =
k
2
2
4λ−k
4λ−k
 ξ2 sin
(π − (x − y)) + e 2 π sin
(x − y) , y ≤ x ≤ π,
2
2
where
ξ1 = √
2e
k2 − 4λ(2e
−k
2 π
−k
2 (x+π−y)
√
cosh(
k 2 −4λ
π)
2
− 1 − e−kπ )
,
−k
−2e 2 (x+π−y)
√
ξ2 = √
.
−k
2
4λ − k2(1 + e−kπ − 2e 2 π cos( 4λ−k
)π)
2
2
2
We note that Gλ(x, y) > 0 for λ < 0 and Gλ(x, y) ≤ 0 for k4 < λ ≤ k4 + 1.
Now, we consider the following nonlinear periodic boundary value problem (PBVP)
−u00(x) − ku0(x) = f(x, u(x)), x ∈ [0, π],
(1)
u(0) = u(π), u0(0) = u0 (π),
(2)
where f ∈ [0, π] × R → R is continuous.
A. Alsaedi
161
We say that α ∈ C 2([0, π]) is a lower solution of (1)-(2) if
−α00 (x) − kα0 (x) ≤ f(x, α(x)), x ∈ [0, π],
α(0) = α(π), α0(0) ≥ α0 (π).
Similarly, β ∈ C 2([0, π]) is an upper solution of (1)-(2) if
−β 00 (x) − kβ 0 (x) ≥ f(x, β(x)), x ∈ [0, π],
β(0) = β(π), β 0 (0) ≤ β 0 (π).
In order to prove the main result, we need the following theorem. We omit the proof
of this theorem as it is based on the standard arguments [5].
THEOREM 1. Suppose that α, β ∈ C 2([0, π], R) are lower and upper solutions of
(1)-(2) respectively such that α(x) ≤ β(x), ∀x ∈ [0, π]. Then there exists at least one
solution u(x) of (1)-(2) such that α(x) ≤ u(x) ≤ β(x) on [0, π].
3
Generalized Quasilinearization Method
We have the following result.
THEOREM 2. Assume that
(A1 ) α, β ∈ C 2 ([0, π], R) are lower and upper solutions of (1)-(2) such that α(x) ≤ β(x)
on [0, π].
(A2 ) f ∈ C 2 ([0, π] × R) and
∂f
∂u (x, u)
< 0 for every (x, u) ∈ S, where
S = {(x, u) ∈ R2 : x ∈ [0, π] and u ∈ [α(x), β(x)]}.
Then there exists a monotone sequence {qn} which converges uniformly and quadratically to a unique solution of (1)-(2).
PROOF. In view of the assumption (A2 ) and the mean value theorem, we have
f(x, u) ≥ f(x, v) + [
∂
f(x, v) + 2mv](u − v) − m(u2 − v2 ), m > 0,
∂u
for every x ∈ [0, π] and u, v ∈ R such that α(x) ≤ v ≤ u ≤ β(x) on [0, π]. Here, we have
∂ 2f
used ∂u
(x, u) ∈ S, which follows from (A2 ). We define the function
2 (x, u) ≥ −2m,
g(x, u, v) as
g(x, u, v) = f(x, v) + [
∂
f(x, v) + 2mv](u − v) − m(u2 − v2 ).
∂u
Observe that
g(x, u, v) ≤ f(x, u), g(x, u, u) = f(x, u).
(3)
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Monotone Iteration for a Forced Duffing Equation
It follows from (A2 ) and (3) that g(x, u, v) is strictly decreasing in u for each fixed
(x, v) ∈ [0, π] × R and satisfies one sided Lipschitz condition
g(x, u1, v) − g(x, u2, v) ≤ L(u1 − u2), L > 0.
(4)
Now, we set α = q0 and consider the PBVP
−u(x) − ku0(x) = g(x, u(x), q0(x)), x ∈ [0, π]
(5)
u(0) = u(π), u0(0) = u0 (π).
(6)
In view of (A1 ) and (3), we have
−q000 (x) − kq00 (x) ≤ f(x, q0(x)) = g(x, q0(x), q0(x)), x ∈ [0, π],
q0(0) = q0(π), q00 (0) ≥ q00 (π),
and
−β 00 (x) − kβ 0 (x) ≥ f(x, β(x)) ≥ g(x, β(x), q0(x)), x ∈ [0, π],
β(0) = β(π), β 0 (0) ≤ β 0 (π),
which imply that q0(x) and β(x) are lower and upper solutions of (5)-(6) respectively.
Hence, by Theorem 2 and (4), there exists a unique solution q1(x) of (5)-(6) such that
q0 (x) ≤ q1(x) ≤ β(x), on [0, π].
Next, consider the PBVP
−u(x) − ku0 (x) = g(x, u(x), q1(x)), x ∈ [0, π],
(7)
u(0) = u(π), u0(0) = u0 (π).
(8)
using (A1 ) and employing the fact that q1(x) is a solution of (5)-(6), we obtain
−q100 (x) − kq10 (x) = g(x, q1(x), q0(x)) ≤ g(x, q1(x), q1(x)), x ∈ [0, π],
(9)
q1(0) = q1(π), q10 (0) ≥ q10 (π),
(10)
−β 00 (x) − kβ 0 (x) ≥ f(x, β(x)) ≥ g(x, β(x), q1(x)), x ∈ [0, π],
(11)
and
0
0
β(0) = β(π), β (0) ≤ β (π),
(12)
From (9)-(10) and (11)-(12), we find that q1(x) and β(x) are lower and upper solutions
of (7)-(8) respectively. Again,by Theorem 2 and (4), there exists a unique solution
q2(x) of (7)-(8) such that
q1(x) ≤ q2(x) ≤ β(x) on [0, π].
This process can be continued successively to obtain a monotone sequence {qn(x)}
satisfying
q0(x) ≤ q1 (x) ≤ .... ≤ qn−1(x) ≤ qn(x) ≤ β(x), on [0, π],
A. Alsaedi
163
where the element qn(x) of the sequence {qn(x)} is a solution of the problem
−u(x) − ku0(x) = g(x, u(x), qn−1(x)), x ∈ [0, π]
u(0) = u(π), u0(0) = u0 (π).
Since the sequence {qn} is monotone, it follows that it has a pointwise limit q(x). To
show that q(x) is in fact a solution of (1)-(2), we note that qn(x) is a solution of the
problem
−u00(x) − ku0(x) − λu(x) = Ψn (x), x ∈ [0, π],
(13)
u(0) = u(π), u0(0) = u0 (π),
(14)
where Ψn (x) = g(x, qn(x), qn−1(x)) − λqn (x) for every x ∈ [0, π]. Since g(x, u, v) is
continuous on S and α(x) ≤ qn(x) ≤ β(x) on [0, π], it follows that {Ψn (x}) is bounded
in C[0, π]. Thus, qn(x), the solution of (13)-(14) can be written as
Z π
qn(x) =
Gλ (x, y)Ψn (x)dy.
(15)
0
This implies that {qn(x)} is bounded in C 2([0, π]) and hence {qn(x)} % q(x) uniformly
on [0, π]. Consequently, taking limit n → ∞ of (15) yields
Z π
q(x) =
Gλ (x, y)[f(y, q(y)) − λq(y)]dy, x ∈ [0, π].
0
Thus, we have shown that q(x) is a solution of (1)-(2).
Now, we prove that the convergence of the sequence is quadratic. For that, we
define
F (x, u) = f(x, u) + mu2 .
In view of (A2 ), we can find a constant C such that
0≤
∂2
F (x, u) ≤ C.
∂u2
Letting en (x) = q(x) − qn(x), n = 1, 2, 3, ..., we have
−e00n (x) − ke0n (x)
= qn00 (x) + kqn0 (x) − q00(x) − kq0 (x)
= f(x, q(x)) − f(x, qn−1(x))
∂
−
f(x, qn−1(x)) + 2mqn−1(x) (qn(x) − qn−1(x))
∂u
2
+m(qn2 (x) − qn−1
(x))
2
= F (x, q(x)) − mq (x) − F (x, qn−1(x))
∂
2
+mqn−1
(x) −
F (x, qn−1(x))(qn(x) − qn−1(x))
∂u
2
+m(qn2 (x) − qn−1
(x))
= F (x, q(x)) − F (x, qn−1(x))
∂
− F (x, qn−1(x))(qn (x) − qn−1(x)) − m(q2 (x) − qn2 (x)),
∂u
164
Monotone Iteration for a Forced Duffing Equation
en (0) = en (π), e0n (0) = e0n (π).
Using the mean value theorem repeatedly, we obtain
∂
∂
F (x, ξ)(q(x) − qn−1(x)) −
F (x, qn−1(x))(qn(x) − qn−1(x))
∂u
∂u
−m(q2 (x) − qn2 (x))
∂
∂
F (x, ξ)(q(x) − qn−1(x)) −
F (x, qn−1(x))(q(x) − qn−1(x))
∂u
∂u
∂
− F (x, qn−1(x))(qn(x) − q(x)) − m(q2 (x) − qn2 (x))
∂u
∂
∂
F (x, ξ(x)) −
F (x, qn−1(x)) (q(x) − qn−1(x))
∂u
∂u
∂
+
F (x, qn−1(x)) − m(q(x) + qn(x)) (q − qn(x))
∂u
2
∂
F (x, η)(ξ(x) − qn−1(x))en−1 (x)
∂u2
∂
+
F (x, qn−1(x)) − m(q(x) + qn(x)) en (x),
∂u
−e00n (x) − ke0n (x) =
=
=
=
en (0) = en (π), e0n (0) = e0n (π).
where qn−1(x) ≤ η(x) ≤ ξ(x) ≤ q(x) on [0, π] (η and ξ also depend on qn−1(x) and
q(x)). Substituting
an (x) =
∂
F (x, qn−1(x)) − m(q(x) + qn(x)),
∂u
∂2
F (x, η(x))en−1(x)(ξ − qn−1(x)),
∂u2
in the above expression gives bn (x) ≤ 0 on [0, π] and
bn(x) + Ce2n−1(x) =
−e00n (x) − ke0n (x) − en (x)an (x) = Ce2n−1(x) + bn(x), x ∈ [0, π],
en (0) = en (π), e0n (0) = e0n (π).
∂f
∂f
Since limn→∞ an (x) = ∂F
∂u (x, q(x)) − 2mq(x) = ∂u (x, q(x)) and ∂u (x, q(x)) < 0, therefore, for λ < 0, there exists n0 ∈ N such that for n ≥ n0 , we have an (x) < λ < 0, x ∈
[0, π]. Therefore,the error function en(x) satisfies the following problem
−e00n (x) − ke0n (x) − λen (x) = (an(x) − λ)en (x) + Ce2n−1(x) + bn(x), x ∈ [0, π],
whose solution is
en (x) =
Z
π
Gλ (x, y)[(an(y) − λ)en (y) + Ce2n−1(y) + bn(y)]dy.
0
Since an (y) − λ < 0, bn(y) ≤ 0, and Gλ (x, y) > 0 for λ < 0, therefore, it follows that
Gλ (x, y)[(an(y) − λ)en (y) + Ce2n−1(y) + bn(y)] < Gλ (x, y)Ce2n−1(y).
A. Alsaedi
165
Thus, we obtain
0 ≤ en (x) ≤ C
Z
π
Gλ(x, y)e2n−1 (y)dy,
0
which can be expressed as
ken k ≤ C1ken−1 k2,
where C1 = C max
norm.
Rπ
0
Gλ (x, y)dy and ken k = max{|en | : x ∈ [0, π]} is the usual uniform
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