Bulletin of TICMI
Vol. 16, No. 1, 2012, 27–33
An Iteration Method for the Kirchhoff Static Beam
Peradze Jemal
I. Javakhishvili Tbilisi State University, Georgian Technical University
(Received December 26, 2011; Accepted March 5, 2012)
RL
00
u02
k−1 dx)uk = f , k = 1, 2, . . . , is used to solve the
R
boundary value problem for the nonlinear differential equation uiv −(λ+2/L 0L u02 dx)u00 = f .
The approximation uk is expressed as well-defined integrals of the functions uk−1 and f . The
method error uk − u is estimated.
The iteration method uiv
k − (λ + 2/L
0
Keywords: Kirchhoff equation, Static beam, Iteration method, Algorithm error.
AMS Subject Classification: 65L10, 65L70.
1.
Statement of the problem
We consider the following boundary value problem
¶
µ
Z
2 L 02
u (x) dx u00 (x) = f (x),
u (x) − λ +
L 0
ıv
0 < x < L,
u(0) = u(L) = 0,
(1.1)
λ = const > 0,
u00 (0) = u00 (L) = 0,
(1.2)
where f (x) is a given continuous function, and u(x) is the sought solution.
Equation (1.1) is the stationary problem related to the equation
µ
Z L µ ¶2 ¶ 2
∂2u
∂u
∂4u
∂ u
+
α
−
α
+
α
= 0,
dx
0
1
2
∂t2
∂x4
∂x
∂x2
0
(1.3)
which was proposed by Woinowsky–Krieger [9] in 1950 as a model for the deflection of an extensible dynamic beam with hinged ends. The nonlinear term of this
equation was for the first time used by Kirchhoff [4] who generalized D’Alembert’s
classical model. Therefore equations (1.1) and (1.3) are frequently called a Kirchhoff type equation for a dynamic and a static beam, respectively. The results of
one of the initial mathematical studies of equations of (1.3) type are presented in
[1] and [2].
For equation (1.1) and its generalizations, as well as for equations similar to
(1.1), the problem of construction of numerical algorithms and estimation of their
accuracy is studied in [3], [5]–[8]. Each of the algorithms used in these papers is
a combination of two approximate methods, one of which reduces the problem to
ISSN: 1512-0082 print
c 2012 Tbilisi University Press
°
28
Bulletin of TICMI
the finite-dimensional one and the other is some iterative process of solution of
the discrete system. In the present paper, a technique somewhat different from
the above-mentioned one is proposed to solve problem (1.1),(1.2). The differential
equation (1.1) is solved by an iteration method. At each iteration step, a boundary value problem is obtained for a linear differential equation whose solution is
written in integrals. The algorithm accuracy is estimated by the method of a priori
inequalities.
2.
The algorithm
On choosing a function u0 (x), 0 ≤ x ≤ L, that together with its second derivative
vanishes for x = 0 and x = L, we will seek for a solution of problem (1.1),(1.2)
using the iteration process
uıv
k (x)
¶
µ
Z
2 L 02
u (x) dx u00k (x) = f (x),
− λ+
L 0 k−1
(2.4)
0 < x < L,
u00k (0) = u00k (L) = 0,
uk (0) = uk (L) = 0,
(2.5)
k = 1, 2, . . . ,
where uk (x) is the k-th approximation of the solution of problem (1.1),(1.2), k =
0, 1, . . . .
The considered algorithm makes it possible to express uk (x) through the preceding approximation in the integral form. Indeed, on denoting
2
αk = λ +
L
Z
0
L
u02
k (x) dx,
we introduce the function vk (x) = u00k (x), k = 0, 1, . . . .
Now, (2.4),(2.5) can be rewritten as relations
u00k (x) = vk (x),
0 < x < L,
uk (0) = uk (L) = 0
and
vk00 (x) − αk−1 vk (x) = f (x),
0 < x < L,
vk (0) = vk (L) = 0.
For uk (x) we have
µ
¶
Z x
Z L
1
uk (x) =
(x − L)
ξvk (ξ) dξ + x
(ξ − L)vk (ξ) dξ
L
0
x
(2.6)
Vol. 16, No. 1, 2012
29
and vk (x) is representable in the form
1
vk (x) = √
√
αk−1 sinh( αk−1 L)
µ
Z
¡√
¢
× sinh αk−1 (x − L)
¡√
¢
+ sinh αk−1 x
Z
x
sinh
0
x
L
¡√
¢
αk−1 ξ f (ξ) dξ
¶
¡√
¢
sinh αk−1 (ξ − L) f (ξ) dξ . (2.7)
Substituting (2.7) into (2.6) and applying the well-known equalities for hyperbolic
functions, we obtain the desired formula
uk (x) = −
1
αk−1 L
µ
Z
(x − L)
Z
x
ξf (ξ) dξ + x
0
¶
L
(ξ − L)f (ξ) dξ
x
µ
Z
¡√
¢ x
¡√
¢
1
+
sinh αk−1 (x−L)
sinh αk−1 ξ f (ξ) dξ
√
√
αk−1 αk−1 sinh( αk−1 L)
0
¶
Z L
¡√
¢
¡√
¢
+ sinh αk−1 x
sinh αk−1 (ξ − L) f (ξ) dξ , k = 1, 2, . . . , (2.8)
x
which makes it possible to obtain approximations uk (x) evading the differential
problem (2.4),(2.5). Therefore approximations uk (x) are found by means of (2.8).
3.
Error of the algorithm and the equation for it
Let us define the algorithm error as a difference
∆uk (x) = uk (x) − u(x),
k = 0, 1, . . . ,
between an approximate and an exact solutions.
Subtracting the respective relations in (1.1) and (1.2) from (2.4) and (2.5), we
obtain
∆uıv
k (x)
¶
µ
Z
¢
1 L ¡ 02
02
uk−1 (x) + u (x) dx ∆u00k (x)
− λ+
L 0
µZ L
¶
¡ 0
¢
1
0
0
−
uk−1 (x) + u (x) ∆uk−1 (x) dx (u00k (x) + u00 (x)) = 0,
L
0
∆uk (0) = ∆uk (L) = 0,
∆u00k (0) = ∆u00k (L) = 0.
(3.9)
(3.10)
We use (3.9), (10) to estimate the algorithm accuracy. For this, we need some a
priori relations which are derived in the next paragraph.
30
4.
Bulletin of TICMI
Auxiliary inequalities
For the well-defined functions, sufficiently smooth for 0 ≤ x ≤ L, we introduce the
notation
Z
L
(u(x), v(x)) =
µZ
ku(x)kp =
¶
Lµ p
d u(x) 2
dxp
0
u(x)v(x) dx,
0
¶1
2
dx ,
p = 0, 1, 2,
ku(x)k = ku(x)k0 .
Lemma 4.1: For a twice differentiable function u(x), 0 ≤ x ≤ L, that vanishes
at x = 0 and x = L the inequalities
√
2
L
ku(x)k ≤ ku(x)k1 ≤ √ ku(x)k2
L
2
(4.11)
are valid.
Proof : We have
Z
x
u(x) =
u0 (ξ) dξ.
0
Hence
µZ
|u(x)| ≤
¶1 µ Z
x
2
dξ
0
x
¶1
02
u (ξ) dξ
0
2
1
≤ x 2 kuk1 .
Therefore
ku(x)k2 ≤
L2
kuk21 ,
2
which implies the left inequality of (4.11). Using the latter and taking into account
that
¯L
ku(x)k21 = u(x)u0 (x)¯0 − (u(x), u00 (x)) = −(u(x), u00 (x)),
we complete the proof.
¤
Lemma 4.2: For the solution of problem (1.1), (1.2) we have the inequality
ku(x)k1 ≤ c1 ,
(4.12)
where c1 is the constant calculated by the formula
L
c1 =
2
µ
¶− 1
4
1
2
kf (x)k 2 .
+ λL
L
(4.13)
Vol. 16, No. 1, 2012
31
Proof : We multiply equation (1.1) by u(x) and then integrate the obtained equality with respect to x from 0 to L. Using (1.2), we get the relation
ku(x)k22
¶
µ
2
2
+ λ + ku(x)k1 ku(x)k21 = (f (x), u(x)) ,
L
which, together with (4.11), yields
µ
¶
2
2
L
2
λ + 2 + ku(x)k1 ku(x)k21 ≤ √ kf (x)k ku(x)k1 .
L
L
2
Thus we obtain the inequality
2
L2
ku(x)k41 ≤
L
8
µ
¶−1
2
+λ
kf (x)k2 ,
L2
which implies estimate (4.12).
¤
Lemma 4.3: The solution of problem (2.4), (2.5) satisfies the relation
kuk (x)k1 ≤ c2 ,
k = 1, 2, . . . ,
(4.14)
where c2 is a constant independent of k and defined by
L2
c2 = √
2
µ
¶−1
2
+ λL
kf (x)k.
L
(4.15)
Proof : We multiply equation (2.4) by uk (x) and integrate with respect to x from
0 to L. Taking (2.5) into account, we see
kuk (x)k22
¶
µ
2
2
+ λ + kuk−1 (x)k1 kuk (x)k21 = (f (x), uk (x)) .
L
By applying (4.11) we find
¶
µ
2
L
2
2
λ + 2 + kuk−1 (x)k1 kuk (x)k21 ≤ √ kf (x)k kuk (x)k1 .
L
L
2
This implies (4.14).
¤
32
5.
Bulletin of TICMI
Estimation of the algorithm error
We multiply equation (3.9) by ∆uk (x) and integrate the obtained equality with
respect to x from 0 to L. Applying (3.10) we obtain
k∆uk (x)k22
¶
µ
¢
1¡
2
2
kuk−1 (x)k1 + ku(x)k1 k∆uk (x)k21
+ λ+
L
+
1
¢
1 Y¡ 0
uk−p (x) + u0 (x), ∆u0k−p (x) = 0.
L
p=0
By (4.11)
µ
¶
¢
2
1¡
2
2
kuk−1 (x)k1 + ku(x)k1 k∆uk (x)k21
λ+ 2 +
L
L
≤
1
1 Y
(kuk−p (x)k1 + ku(x)k1 ) k∆uk−p (x)k1 .
L
p=0
Therefore
µ ¶1 µ
¶1
¢ 2
2 2 1¡
2
2
k∆uk (x)k1 ≤
kuk−1 (x)k1 + ku(x)k1
L
L
¶
µ
¢ −1
1¡
2
2
2
kuk−1 (x)k1 + ku(x)k1
× λ+ 2 +
L
L
× (kuk (x)k1 + ku(x)k1 ) k∆uk−1 (x)k1 .
y
2
0≤y<∞ α+y
Since max
=
1
2
α−1/2 holds for α > 0, we have
µ µ
¶¶− 1
2
2
(kuk (x)k1 + ku(x)k1 ) k∆uk−1 (x)k1 .
k∆uk (x)k1 ≤ 2
+ λL
L
(5.16)
¡ ¡
¢¢− 1
2
Let the condition q = 2 L2 + λL
(c1 + c2 ) < 1 be fulfilled, which, as follows
from (4.13) and (4.15), is equivalent to the requirement
Ã
!p
µ
¶− 3
2
√
4
1
1X
2
< 1.
q=
L 2
+ λL
kf (x)k 2
4
L
p=1
Then, by virtue of (5.16), (4.12), (4.14) and (4.11), we come to a conclusion that
the iteration method (2.4),(2.5) or, which is the same, (2.8) reduces to the solution
Vol. 16, No. 1, 2012
33
of problem (1.1),(1.2) and the estimate
µ
k∆uk (x)kp ≤
L
√
2
¶1−p
k = 1, 2, . . . ,
q k k∆u0 (x)k1 ,
p = 0, 1,
holds for the error of the method.
Acknowledgement
The author acknowledges the support under grant 7353 of the Estonian Science
Foundation.
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