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Applied Mathematics E-Notes, 9(2009), 160-167 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Approximate Solutions Of The Forced Duffing
Equation With Mixed Nonlinearities∗
Bashir Ahmad†, Badra Salah Alghamdi‡
Received 1 February 2008
Abstract
In this paper, an improved form of the generalized quasilinearization technique
is developed to obtain a monotone sequence of approximate solutions converging
uniformly and quadratically to a unique solution of the forced Duffing equation
involving mixed nonlinearities with periodic boundary conditions.
1
Introduction
The monotone iterative technique coupled with the method of upper and lower solutions
[1-7] 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. In general, the convergence of the sequence of approximate solutions given
by the monotone iterative technique is at most linear [8-9]. To obtain a sequence of approximate solutions converging quadratically, we use the method of quasilinearization
[10]. This method has been developed for a variety of problems [11-20]. In reference
[21], a generalized quasilinearization technique was developed for Duffing equation with
periodic boundary conditions. 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.
The purpose of this paper is to study an extended form of the generalized quasilinearization method for a periodic boundary value problem involving the forced Duffing
equation with mixed type of nonlinearities. Precisely, we obtain a sequence of approximate solutions converging monotonically and quadratically to a unique solution of the
following problem
−u00 (x) − ku0 (x) = f(x, u(x)), x ∈ [0, π],
∗ Mathematics
(1)
Subject Classifications: 34B10, 34B15.
of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box. 80203,
Jeddah 21589, Saudi Arabia
‡ Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box. 80203,
Jeddah 21589, Saudi Arabia
† Department
160
161
B. Ahmad and B. S. Alghamdi
u0 (0) = u0 (π),
u(0) = u(π),
(2)
where f ∈ [0, π] × R → R is continuous and is allowed to admit a decomposition of the
form
f(x, u) = h(x, u) + g(x, u).
Here, we assume that (h(x, u) + M1 u2 ) is convex for some M1 > 0 and instead of
requiring the convexity/concavity assumption on g(x, u), we demand a less restrictive
condition on g(x, u) namely, [g(x, u) + M2 u+1 ] satisfies a nondecreasing condition for
some M2 > 0 and > 0.
2
Preliminaries
We observe that the homogeneous periodic boundary value problem
−u00 (x) − ku0 (x) − λu(x) = 0, x ∈ [0, π],
u0 (0) = u0 (π),
u(0) = u(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, the solution u(x) of (1)-(2) can be
written by Green’s function method as
Z π
u(x) =
Gλ (x, y)f(y, u(y))dy,
(3)
0
where Gλ (x, y) is the Green’s function given by

h
√ 2
√ 2
i
−k
k −4λ(π−(y−x))
k −4λ(y−x)
 γ1 sinh
2 π sinh
+
e
, 0 ≤ x ≤ y,
h
√ 2 2
√ 2 2
i
Gλ (x, y) =
k
k
−4λ(π−(x−y)
k
−4λ(x−y)
π
 γ1 sinh
+ e 2 sinh
, y ≤ x ≤ π,
2
2
for λ <
k2
4
for λ >
k2
,
4
and

√
h √
i
−k
4λ−k 2(π−(y−x)
4λ−k 2(y−x)
 γ2 sin
+ e 2 π sin
, 0 ≤ x ≤ y,
2
2
√
h √
i
Gλ (x, y) =
k
4λ−k 2(π−(x−y)
4λ−k 2(x−y)
π
 γ2 sin
+ e 2 sin
, y ≤ x ≤ π,
2
2
where
γ1 = √
2e
k 2 − 4λ 2e
γ2 = √
−k
2
π
−k
2
cosh
−2e
(x+π−y)
√
−k
2
k 2−4λ
π
2
− 1 − e−kπ
,
(x+π−y)
4λ − k 2 1 + e−kπ − 2e
−k
2
π
cos
√
We note that Gλ (x, y) > 0 for λ < 0 and Gλ (x, y) ≤ 0 for
4λ−k 2
π
2
k2
4
.
<λ≤
k2
4
+ 1.
162
Forced Duffing Equations
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).
We state the following results to prove the main result. We do not provide the proof
of these results as it is based on the standard arguments [1, 11, 12].
THEOREM 1. Assume that α, β ∈ C 2 [0, π] are respectively lower and upper solutions of the boundary value problem (1)-(2). If f(x, u) is strictly decreasing in u for
each x ∈ [0, π], then α(x) ≤ β(x) for x ∈ [0, π].
THEOREM 2. Suppose that α, β ∈ C 2 ([0, π]) are lower and upper solutions of
(1)-(2) respectively such that α(x) ≤ β(x) for all x ∈ [0, π]. Then there exists at least
one solution u(x) of (1)-(2) such that α(x) ≤ u(x) ≤ β(x) for x ∈ [0, π].
3
Main Result
THEOREM 3. Assume that
(A1 ) α, β ∈ C 2 ([0, π]) are lower and upper solutions of (1)-(2) respectively such that
α(x) ≤ β(x) for all x ∈ [0, π];
(A2 ) hu (x, u), huu(x, u) exist, are continuous with huu (x, u) + 2M1 ≥ 0 for all (x, u) ∈
Ω, where Ω = {(x, u) ∈ R2 : x ∈ [0, π], α(x) ≤ u(x) ≤ β(x)};
(A3 ) gu (x, u) exists and satisfy
{[gu(x, u) + (1 + )M2 u ] − [gu (x, v) + (1 + )M2 v ]}(u − v) ≥ 0, > 0;
(A4 ) hu (x, u) + gu (x, u) < 0 for all x ∈ [0, π].
Then there exists a monotone sequence {αn } which converges uniformly and quadratically to a unique solution of (1)-(2).
PROOF. For any u ≥ v, it follows from (A2 ) and (A3 ) that
h(x, u) + g(x, u) ≥ G(x, u, v), G(x, u, u) = h(x, u) + g(x, u)
(4)
where
G(x, u, v) =
h(x, v) + [hu (x, v) + 2M1 v](u − v) − M1 (u2 − v2 )
+g(x, v) + [gu (x, v) + (1 + )M2 v ](u − v) − M2 (u(1+) − v(1+) ).(5)
163
B. Ahmad and B. S. Alghamdi
Further, in view of (A4 ), it follows that Gu (x, u, v) < 0 for each fixed (x, v) ∈ [0, π] × R.
Now, we set α0 (x) = α(x) and consider the problem
−u00 (x) − ku0 (x) = G(x, u(x), α0(x)), x ∈ [0, π],
u(0) = u(π),
(6)
u0 (0) = u0 (π).
(7)
In view of (A1 ), (4) and (5), we have
−α000 (x) − kα00 (x) ≤ h(x, α0 (x)) + g(x, α0(x)) = G(x, α0(x), α0 (x)),
α0 (0) = α0 (π),
x ∈ [0, π],
α00 (0) ≥ α00 (π),
−β 00 (x) − kβ 0 (x) ≥ h(x, β(x)) + g(x, β(x)) ≥ G(x, β(x), α0(x)), x ∈ [0, π],
β(0) = β(π),
β 0 (0) ≤ β 0 (π),
which imply that α0 (x) and β(x) are lower and upper solutions of (6)-(7) respectively.
Hence, by Theorems 1 and 2, there exists a unique solution α1 (x) of (6)-(7) such that
α0 (x) ≤ α1 (x) ≤ β(x) for x ∈ [0, π].
Next, consider the problem
−u00 (x) − ku0 (x) = G(x, u(x), α1(x)),
u(0) = u(π),
x ∈ [0, π],
u0 (0) = u0 (π).
(8)
(9)
Using (4) and (5) together with the fact that α1 (x) is a solution of (6)-(7), we obtain
≤
−α001 (x) − kα01 (x) = G(x, α1(x), α0 (x))
h(x, α1(x)) + g(x, α1 (x)) = G(x, α1 (x), α1(x)),
x ∈ [0, π],
0
0
α1 (0) = α1 (π),
α1 (0) = α1 (π),
−β 00 (x) − kβ 0 (x) ≥ h(x, β(x)) + g(x, β(x)) ≥ G(x, β(x), α1(x)),
β(0) = β(π),
β 0 (0) ≤ β 0 (π).
x ∈ [0, π],
Thus it follows that α1 (x) and β(x) are respectively lower and upper solutions of (8)(9). Again, by Theorems 1 and 2, there exists a unique solution α2 (x) of (8)-(9) such
that
α1 (x) ≤ α2 (x) ≤ β(x),
x ∈ [0, π].
Continuing this process successively, we obtain a monotone sequence {αn (x)} satisfying
α0 (x) ≤ α1 (x) ≤ ... ≤ αn (x) ≤ β(x), x ∈ [0, π],
where αn (x) is the unique solution of the problem
u00 (x) − ku0 (x) = G(x, u(x), αn−1(x)),
u(0) = u(π),
x ∈ [0, π],
u0 (0) = u0 (π).
Since the sequence {αn (x)} is monotone, it follows that it has a pointwise limit u(x).
164
Forced Duffing Equations
To show that u(x) is in fact a solution of (1)-(2), we note that αn (x) is a solution
of the following problem
−u00 (x) − ku0 (x) − λu(x) = ψn (x),
u(0) = u(π),
x ∈ [0, π],
u0 (0) = u0 (π),
(10)
(11)
where ψn (x) = G(x, αn(x), αn−1 (x)) − λαn (x) for every x ∈ [0, π]. Since G(x, u, v) is
continuous on Ω and α0 (x) ≤ αn (x) ≤ β(x) on [0, π], it follows that ψn (x) is bounded
in C[0, π]. Thus, αn (x), the solution of (10)-(11) can be written as
αn (x) =
Z
π
Gλ (x, y)ψn (y)dy.
(12)
0
Since [0, π] is compact and the monotone convergence of the sequence {αn (x)} is pointwise, it follows by the standard arguments (Arzela Ascoli convergence criterion, Dini’s
theorem [1, 12]) that the convergence of the sequence is uniform. Thus, taking the
limit n → ∞ in (12) yields
Z π
u(x) =
Gλ (x, y)[h(y, u(y)) + g(y, u(y)) − λu(y)]dy, x ∈ [0, π].
0
Thus, u(x) is a solution of (1)-(2). Now, we prove that the convergence of the sequence
is quadratic. For that, setting en (x) = u(x)−αn (x) ≥ 0, n = 1, 2, 3, ... and P (x, u(x)) =
h(x, u(x))+g(x, u(x))+M1 u2 (x)+M2 u1+(x), we obtain en (0) = en (π), e0n (0) = e0n (π)
and
=
=
=
−e00n (x) − ke0n (x)
α00n (x) + kα0n (x) − u00 (x) − ku0 (x)
−G(x, αn(x), αn−1 (x)) + h(x, u(x)) + g(x, u(x))
−[h(x, αn−1(x)) + {hu (x, αn−1(x)) + 2M1 αn−1(x)}(αn (x) − αn−1 (x))
−M1 (α2n (x) − α2n−1 (x)) + g(x, αn−1(x))
+{gu (x, αn−1(x)) + (1 + )M2 αn−1 (x)}(αn (x) − αn−1 (x))
=
=
=
=
1+
−M2 (α1+
n (x) − αn−1(x))] + h(x, u(x)) + g(x, u(x))
−P (x, αn−1(x)) + P (x, u(x)) − M1 u2 (x) − M2 u1+ (x)
−Pu (x, αn−1(x))(αn (x) − αn−1 (x)) + M1 α2n (x) + M2 α1+
n (x)
2
2
Pu (x, ξ)(u(x) − αn−1(x)) − M1 (u (x) − αn (x)) − M2 (u1+ (x) − α1+
n (x))
+Pu (x, αn−1(x))(u(x) − αn (x)) − Pu (x, αn−1(x))(u(x) − αn−1(x))
[Pu (x, ξ) − Pu (x, αn−1(x))](u(x) − αn−1 (x)) − M1 (u(x) + αn (x))(u(x) − αn (x))
−M2 (u(x) − αn (x))η(u(x), αn (x)) + Pu (x, αn−1(x))(u(x) − αn (x))
[Pu (x, ξ) − Pu (x, αn−1(x))](u(x) − αn−1 (x))
+[Pu (x, αn−1(x)) − M1 (u(x) + αn (x)) − M2 η(u(x), αn(x))](u(x) − αn (x))
165
B. Ahmad and B. S. Alghamdi
=
=
=
[Pu (x, ξ) − Pu (x, αn−1(x))]en−1 (x) + [Pu (x, αn−1(x)) − M1 (u(x) + αn (x))
−M2 η(u(x), αn(x))]en (x)
[hu (x, ξ) − hu (x, αn−1(x)) + gu (x, ξ) − gu (x, αn−1(x)) + 2M1 (ξ − αn−1 (x))
+(1 + )M2 (ξ − αn−1 (x))]en−1 (x) + [Pu (x, αn−1(x)) − M1 (u(x) + αn (x))
−M2 η(u(x), αn(x))]en (x)
[huu(x, σ)(ξ − αn−1 (x)) + gu (x, ξ) − gu (x, αn−1(x)) + 2M1 (ξ − αn−1 (x))
+(1 + )M2 (ξ − αn−1 (x))]en−1 (x) + [Pu (x, αn−1(x)) − M1 (u(x) + αn (x))
−M2 η(u(x), αn(x))]en (x),
(13)
where αn−1 (x) ≤ σ ≤ ξ ≤ u(x) on [0, π] and η(a, b) > 0 for all a, b. Substituting
Pu (x, αn−1(x)) − M1 (u(x) + αn (x)) − M2 η(u(x), αn (x)) = an (x)
and using the estimate
(huu (x, σ) + 2M1 )(ξ − αn−1 (x)) + gu (x, ξ) − gu (x, αn−1(x))
+(1 + )M2 (ξ − αn−1 (x)) ≤ V (ξ − αn−1 (x)) ≤ V en−1 (x)
in (13) gives
−e00n (x) − ke0n (x) − en (x)an (x) = V e2n−1 (x) + bn (x), x ∈ [0, π],
e0n (0) = e0n (π),
en (0) = en (π),
where bn (x) ≤ 0 on [0, π]. Since limn→∞ an (x) = Pu (x, u(x)) − 2M1 u(x) − (1 +
)M2 u(x) = hu (x, u(x)) + gu (x, u(x)) and hu (x, u(x)) + gu (x, u(x)) < 0, therefore,
for λ < 0, there exists n0 ∈ N such that an (x) − λ < 0, x ∈ [0, π] for n ≥ n0 . Therefore,
the error function en (x) satisfies the following problem
−e00n (x) − ke0n (x) − λen (x) = (an (x) − λ)en (x) + V e2n−1 (x) + bn (x),
x ∈ [0, π],
whose solution is
en (x) =
Z
0
π
Gλ (x, y)[(an (y) − λ)en (y) + V e2n−1 (y) + bn (y)]dy.
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) + V e2n−1 (y) + bn (y)] < Gλ (x, y)V e2n−1 (y).
Thus, we obtain
0 ≤ en (x) ≤ V
Z
0
π
Gλ (x, y)e2n−1 (y)dy,
which can be expressed as
ken k ≤ V ∗ ken−1 k2 ,
166
Forced Duffing Equations
Rπ
where V ∗ = V max 0 Gλ (x, y)dy and ken k = max{|en | : x ∈ [0, π]} is the usual
uniform norm. This completes the proof of the theorem.
REMARK. It is interesting to note that the generalized quasilinearization technique
for the PBVP (1)-(2) [21] follows as a special case if we take g(x, u) = 0 in the forcing
term of the Duffing equation.
EXAMPLE. Consider the boundary value problem
−u00 (x) − ku0 (x) = ea(1−u(x)) − π + |u(x)|2, a > 2, x ∈ [0, π],
u(0) = u(π),
u0 (0) = u0 (π).
(14)
(15)
Here, h(x, u(x)) = ea(1−u(x)) − π, g(x, u(x)) = |u(x)|2 . Let α(x) = −1 and β(x) = 1 be
lower and upper solutions of (14)-(15) respectively. By choosing M1 > 0, 0 < M2 ≤
1, = 1, it can easily be verified that the assumptions (A1 ) − (A4 ) of Theorem 3 are
satisfied. Thus, the conclusion of Theorem 3 applies to the problem (14)-(15).
ACKNOWLEDGMENT. The authors thank the reviewer for his/her valuable comments and suggestions.
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