Iterative solution of a nonlinear static beam equation
UDC 519.6 The paper deals with a boundary-value problem for the nonlinear integro-differential equation $u''''-m \bigg(\int _0^l {u'}^2\,dx \bigg)u''  = f(x, u, u'),$ $m(z)\geq \alpha>0,$ $0\leq z &am...
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Institute of Mathematics, NAS of Ukraine
2020
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507117977337856 |
|---|---|
| author | Berikelashvili , G. Papukashvili , A. Peradze, J. Berikelashvili , G. Papukashvili , A. Peradze, J. |
| author_facet | Berikelashvili , G. Papukashvili , A. Peradze, J. Berikelashvili , G. Papukashvili , A. Peradze, J. |
| author_sort | Berikelashvili , G. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2022-03-26T11:01:58Z |
| description | UDC 519.6
The paper deals with a boundary-value problem for the nonlinear integro-differential equation $u''''-m \bigg(\int _0^l {u'}^2\,dx \bigg)u''  = f(x, u, u'),$ $m(z)\geq \alpha>0,$ $0\leq z < \infty,$ modeling the static state of the Kirchhoff beam.  The problem is reduced to a nonlinear integral equation, which is solved using the Picard iteration method.  The convergence of the iteration process is established and the error estimate is obtained. |
| doi_str_mv | 10.37863/umzh.v72i8.833 |
| first_indexed | 2026-03-24T02:04:13Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v72i8.833
UDC 519.6
G. Berikelashvili (Georg. Techn. Univ., Tbilisi; A. Razmadze Math. Inst., Tbilisi, Georgia),
A. Papukashvili (I. Javakhishvili Tbilisi State Univ.; I. Vekua Inst. Appl. Math., Tbilisi, Georgia),
J. Peradze (Georg. Techn. Univ., Tbilisi; I. Javakhishvili Tbilisi State Univ., Georgia)
ITERATIVE SOLUTION OF A NONLINEAR STATIC BEAM EQUATION
IТЕРАЦIЙНИЙ РОЗВ’ЯЗОК НЕЛIНIЙНОГО РIВНЯННЯ СТАТИЧНОЇ БАЛКИ
The paper deals with a boundary-value problem for the nonlinear integro-differential equation u\prime \prime \prime \prime - m
\biggl( \int l
0
u\prime 2 dx
\biggr)
u\prime \prime =
= f(x, u, u\prime ), m(z) \geq \alpha > 0, 0 \leq z < \infty , modeling the static state of the Kirchhoff beam. The problem is reduced to a
nonlinear integral equation, which is solved using the Picard iteration method. The convergence of the iteration process is
established and the error estimate is obtained.
Розглядається крайова задача для нелiнiйного iнтегро-диференцiального рiвняння u\prime \prime \prime \prime - m
\biggl( \int l
0
u\prime 2 dx
\biggr)
u\prime \prime =
= f(x, u, u\prime ), m(z) \geq \alpha > 0, 0 \leq z < \infty , що моделює статичний стан балки Кiрхгоффа. Задача зводиться
до нелiнiйного iнтегрального рiвняння, яке розв’язується за допомогою iтерацiйного методу Пiкара. Встановлено
збiжнiсть цього iтерацiйного процесу та отримано оцiнку для похибки.
1. Statement of the problem. Let us consider the nonlinear beam equation
u\prime \prime \prime \prime (x) - m
\left( l\int
0
u\prime
2
(x) dx
\right) u\prime \prime (x) = f(x, u(x), u\prime (x)), x \in (0, l), (1)
with the conditions
u(0) = u(l) = 0, u\prime \prime (0) = u\prime \prime (l) = 0. (2)
Here, u = u(x) is the displacement function of length l of the beam subjected to the action of a
force given by the function f(x, u, u\prime ), the function m(z),
m(z) \geq \alpha > 0, 0 \leq z <\infty , (3)
describes the type of a relation between stress and strain. Namely, if the function m(z) is linear, this
means that the above relation is consistent with Hooke’s linear law, while otherwise we deal with
material nonlinearities.
The equation (1) is the stationary problem associated with the equation
utt + uxxxx - m
\left( l\int
0
u2x dx
\right) uxx = f(x, t, u, ux),
m(z) \geq \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t} > 0,
which for the case where m(z) = m0 +m1z, m0,m1 > 0, and f(x, t, u, ux) = 0, was proposed
by Woinowsky-Krieger [10] as a model of deflection of an extensible dynamic beam with hinged
c\bigcirc G. BERIKELASHVILI, A. PAPUKASHVILI, J. PERADZE, 2020
1024 ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
ITERATIVE SOLUTION OF A NONLINEAR STATIC BEAM EQUATION 1025
ends. The nonlinear term
\int l
0
u2x dx was for the first time used by Kirchhoff [3] who generalized
D’Alembert’s classical linear model. Therefore (1) is frequently called a Kirchhoff type equation for
a static beam.
The problem of construction of numerical algorithms and estimation of their accuracy for equa-
tions of type (1) is investigated in [1, 5, 8, 9]. In [4], the existence of a solution of the problem (1),
(2) is proved when the right-hand part of the equation is written in the form q(x)f(x, u, u\prime ), where
f \in C([0, l]\times [0,\infty )\times \BbbR ) is a nonnegative function and q \in C[0, l] is a positive function.
In the present paper, in order to obtain an approximate solution of the problem (1), (2), an
approach is used, which differs from those applied in the above-mentioned references. It consists in
reducing the problem (1), (2) by means of Green’s function to a nonlinear integral equation, to solve
which we use the iterative process. The condition for the convergence of the method is established
and its accuracy is estimated.
The Green’s function method with a further iteration procedure has been applied by us previously
also to a nonlinear problem for the axially symmetric Timoshenko plate [6].
2. Assumptions. Let us assume that besides (3) the function m(z) also satisfies the Lipschitz
condition \bigm| \bigm| m(z1) - m(z2)
\bigm| \bigm| \leq l1| z2 - z1| , 0 \leq z1, z2 <\infty , l1 = \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t} > 0.
Suppose that f(x, u, v) \in L2 ((0, l),\BbbR ,\BbbR ) and, additionally, that the inequalities\bigm| \bigm| f(x, u, v)\bigm| \bigm| \leq \sigma 1(x) + \sigma 2(x) | u| + \sigma 3(x) | v| , (4)\bigm| \bigm| f(x, u2, v2) - f(x, u1, v1)
\bigm| \bigm| \leq l2(x) | u2 - u1| + l3(x) | v2 - v1| , (5)
where
0 < x < l, u, v, ui, vi \in \BbbR , i = 1, 2, \sigma 1(x) \in L2(0, l), \sigma i(x), li(x) \in L\infty (0, l), i = 2, 3,
\sigma 1(x) \geq \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t} > 0, \sigma i(x) \geq 0, li(x) > 0, i = 2, 3,
are fulfilled.
We impose one more restriction on the beam length l and the parameters \alpha and \sigma 2(x), \sigma 3(x)
from the conditions (3) and (4), (5)
\omega = \alpha +
\Bigl( \pi
l
\Bigr) 2
- l
\pi
\biggl(
2
\pi
\bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty +
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty \biggr) > 0. (6)
Let us assume that there exists a solution of the problem (1), (2) and u \in W 2,2
0 (0, l) [2].
3. The method. We will need the Green function for the problem
v\prime \prime \prime \prime (x) - av\prime \prime (x) = \psi (x),
0 < x < l, a = \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t} > 0,
v(0) = v(l) = 0, v\prime \prime (0) = v\prime \prime (l) = 0.
(7)
In order to obtain this function, we split the problem (7) into two problems
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
1026 G. BERIKELASHVILI, A. PAPUKASHVILI, J. PERADZE
w\prime \prime (x) - aw(x) = \psi (x),
w(0) = w(l) = 0,
and
v\prime \prime (x) = w(x),
v(0) = v(l) = 0.
Calculations convince us that
w(x) = - 1
\surd
a \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
al
\bigr)
\left( x\int
0
\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\bigl( \surd
a(x - l)
\bigr)
\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\bigl( \surd
a\xi
\bigr)
\psi (\xi )d\xi +
+
l\int
x
\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\bigl( \surd
ax
\bigr)
\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}
\bigl( \surd
a(\xi - l)
\bigr)
\psi (\xi ) d\xi
\right) ,
v(x) =
1
l
\left( x\int
0
(x - l)\xi w(\xi )d\xi +
l\int
x
x(\xi - l)w(\xi )d\xi
\right) .
Substituting the first of these formulas into the second and performing integration by parts, we
obtain
v(x) =
1
a
\left( x\int
0
\Bigl(
k1(l - x)\xi + k2 \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} (
\surd
a(x - l)) \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
a\xi
\bigr)
\psi (\xi )
\Bigr)
d\xi +
+
l\int
x
\bigl(
k1x(l - \xi ) + k2 \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
ax
\bigr)
\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
a(\xi - l)
\bigr)
\psi (\xi )
\bigr)
d\xi
\right) ,
k1 =
1
l
, k2 =
1
\surd
a \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
al
\bigr) .
The application of (7) to the problem (1), (2) makes it possible to replace the latter problem by
the integral equation
u(x) =
l\int
0
G(x, \xi )f
\bigl(
\xi , u(\xi ), u\prime (\xi )
\bigr)
d\xi , 0 < x < l, (8)
where
G(x, \xi ) =
1
\tau
\left\{
1
l
(x - l)\xi +
1
\surd
\tau \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
\tau l
\bigr) \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} \bigl( \surd \tau (x - l)
\bigr)
\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
\tau \xi
\bigr)
, 0 < \xi \leq x < l,
1
l
x(\xi - l) +
1
\surd
\tau \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
\tau l
\bigr) \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} \bigl( \surd \tau x\bigr) \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} (\surd \tau (\xi - l)), 0 < x \leq \xi < l,
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
ITERATIVE SOLUTION OF A NONLINEAR STATIC BEAM EQUATION 1027
\tau = m
\left( l\int
0
u\prime
2
(x) dx
\right) .
The equation (8) is solved by the Picard iteration method. After choosing a function u0(x),
0 \leq x \leq l, which together with its second derivative vanish for x = 0 and x = l, we find subsequent
approximations by the formula
uk+1(x) =
l\int
0
Gk(x, \xi )f
\bigl(
\xi , uk(\xi ), u
\prime
k(\xi )
\bigr)
d\xi , 0 < x < l, k = 0, 1, . . . , (9)
where
Gk(x, \xi ) =
1
\tau k
\left\{
1
l
(x - l)\xi +
1
\surd
\tau k \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
\tau k l
\bigr) \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} \bigl( \surd \tau k(x - l)
\bigr)
\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
\tau k\xi
\bigr)
, 0 < \xi \leq x < l,
1
l
x(\xi - l) +
1
\surd
\tau k \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}
\bigl( \surd
\tau k l
\bigr) \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} \bigl( \surd \tau kx\bigr) \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h} \bigl( \surd \tau k(\xi - l)
\bigr)
, 0 < x \leq \xi < l,
\tau k = m
\left( l\int
0
u\prime k
2
(x) dx
\right) ,
and uk(x) is the kth approximation of the solution of the equation (8).
4. The problem for the error. Our aim is to estimate the error of the method, by which we
understand the difference between the approximate and exact solutions
\delta uk(x) = uk(x) - u(x), k = 0, 1, . . . . (10)
For this, it is advisable to use not the formula (9), but the system of equalities
u\prime \prime \prime \prime k+1(x) - m
\left( l\int
0
u\prime k
2
(x) dx
\right) u\prime \prime k+1(x) = f
\bigl(
x, uk(x), u
\prime
k(x)
\bigr)
, (11)
uk(0) = uk(l) = 0, u\prime \prime k(0) = u\prime \prime k(l) = 0, (12)
which follows from (9).
If we subtract the respective equalities in (1) and (2) from (11) and (12), then we get
\delta u\prime \prime \prime \prime k (x) - 1
2
\left( \left[ m
\left( l\int
0
u\prime
2
k - 1(x) dx
\right) +m
\left( l\int
0
u\prime
2
(x) dx
\right) \right] \delta u\prime \prime k(x) +
+
\left[ m
\left( l\int
0
u\prime
2
k - 1(x) dx
\right) - m
\left( l\int
0
u\prime
2
(x) dx
\right) \right] (u\prime \prime k(x) + u\prime \prime (x))
\right) =
= f
\bigl(
x, uk - 1(x), u
\prime
k - 1(x)
\bigr)
- f
\bigl(
x, u(x), u\prime (x)
\bigr)
, (13)
\delta uk(0) = \delta uk(l) = 0, \delta u\prime \prime (0) = \delta u\prime \prime (l) = 0, k = 1, 2, . . . . (14)
We will come back to (13), (14) to estimate the error of the method (9). In meantime we have to
derive several a priori estimates.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
1028 G. BERIKELASHVILI, A. PAPUKASHVILI, J. PERADZE
5. Auxiliary inequalities. Let
\bigm\| \bigm\| u(x)\bigm\| \bigm\|
p
=
\left( l\int
0
\biggl(
dpu
dxp
(x)
\biggr) 2
dx
\right) 1/2
, p = 0, 1, 2,
\bigm\| \bigm\| u(x)\bigm\| \bigm\| =
\bigm\| \bigm\| u(x)\bigm\| \bigm\|
0
. (15)
The symbol (\cdot , \cdot ) is understood as a scalar product in L2(0, l).
Lemma 1. The following estimates are true:\bigm\| \bigm\| u(x)\bigm\| \bigm\| \leq l
\pi
\bigm\| \bigm\| u(x)\bigm\| \bigm\|
1
,
\bigm\| \bigm\| u(x)\bigm\| \bigm\|
1
\leq l
\pi
\bigm\| \bigm\| u(x)\bigm\| \bigm\|
2
, (16)
respectively, for u(x) \in W 1,2
0 (0, l) and u(x) \in W 2,2(0, l) \cap W 1,2
0 (0, l).
Proof. Indeed, the first estimate of (16) is Friedrich’s inequality (see, e.g., [7, p. 192]). Applying
this inequality and taking into account that\bigm\| \bigm\| u(x)\bigm\| \bigm\| 2
1
= u(x)u\prime (x)| l0 - (u(x), u\prime \prime (x)) = - (u(x), u\prime \prime (x)) \leq
\bigm\| \bigm\| u(x)\bigm\| \bigm\| \bigm\| \bigm\| u(x)\bigm\| \bigm\|
2
,
we get the second inequality of (16).
Lemma 2. The inequality
\| f(x, u(x), u\prime (x))\| \leq
\bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\| + \biggl( l
\pi
\bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty +
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty \biggr) \bigm\| \bigm\| u(x)\bigm\| \bigm\| 1 (17)
is fulfilled for u(x) \in W 1,2
0 (0, l).
Proof. By (4) we write\bigm\| \bigm\| f(x, u(x), u\prime (x))\bigm\| \bigm\| \leq
\bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\| + \bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty \bigm\| \bigm\| u(x)\bigm\| \bigm\| + \bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty \| u\prime (x)\| .
Recall also (16). The result is (17).
Lemma 3. For the solution of the problem (1), (2) we have the inequality\bigm\| \bigm\| u(x)\bigm\| \bigm\|
1
\leq c1, (18)
where
c1 =
l
\omega \pi
\bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\| . (19)
Proof. We multiply the equation (1) by u(x) and integrate the resulting equality with respect
to x from 0 to l. By using (2), we get\bigm\| \bigm\| u(x)\bigm\| \bigm\| 2
2
+m
\Bigl( \bigm\| \bigm\| u(x)\bigm\| \bigm\| 2
1
\Bigr) \bigm\| \bigm\| u(x)\bigm\| \bigm\| 2
1
=
\bigl(
f(x, u(x), u\prime (x)), u(x)
\bigr)
.
By (16) and (3), we obtain\biggl(
\alpha +
\Bigl( \pi
l
\Bigr) 2\biggr) \bigm\| \bigm\| u(x)\bigm\| \bigm\| 2
1
\leq l
\pi
\bigm\| \bigm\| f(x, u(x), u\prime (x))\bigm\| \bigm\| \bigm\| \bigm\| u(x)\bigm\| \bigm\|
1
.
Therefore, by (17),\Biggl(
\alpha +
\Bigl( \pi
l
\Bigr) 2
-
\biggl(
l
\pi
\biggr) 2 \bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty - l
\pi
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty
\Biggr) \bigm\| \bigm\| u(x)\bigm\| \bigm\|
1
\leq l
\pi
\bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\| .
From this relation and (6) follows (18).
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
ITERATIVE SOLUTION OF A NONLINEAR STATIC BEAM EQUATION 1029
Lemma 4. Suppose we are given some numbers vk \geq 0, k = 0, 1, . . . , for which the inequality
vk \leq avk - 1 + b, k = 1, 2, . . . , (20)
where 0 \leq a < 1, b > 0, holds. Then we have the following uniform estimate with respect to the
index k:
vk \leq b
1 - a
+ a\mathrm{m}\mathrm{a}\mathrm{x}
\biggl(
0, v0 -
b
1 - a
\biggr)
, k = 1, 2, . . . . (21)
Proof. By virtue of (20), by the method of mathematical induction, we have vk \leq akv0 +
+ (ak - 1 + ak - 2 + . . .+ 1)b, k = 1, 2, . . . , which implies
vk \leq akv0 +
1 - ak
1 - a
b =
b
1 - a
+ ak
\biggl(
v0 -
b
1 - a
\biggr)
. (22)
Let us denote \nu k = ak
\biggl(
v0 -
b
1 - a
\biggr)
and consider two cases v0 \leq b
1 - a
and v0 >
b
1 - a
. In the
first case \nu k \leq 0 and by virtue of (22) vk \leq b
1 - a
, k = 1, 2, . . . . In the second case \nu k > 0,
\mathrm{m}\mathrm{a}\mathrm{x} \nu k = \nu 1 = a
\biggl(
v0 -
b
1 - a
\biggr)
, which, by virtue of (22), yields vk =
b
1 - a
+ a
\biggl(
v0 -
b
1 - a
\biggr)
,
k = 1, 2, . . . . From this conclusions the validity of the estimate (21) follows.
Lemma 5. Approximations of the iteration method (9) satisfy the inequality\bigm\| \bigm\| uk(x)\bigm\| \bigm\| 1 \leq c2, k = 1, 2, . . . , (23)
where
c2 =
\left\{ c1, if
\bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty +
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty = 0,
c1 + c0\mathrm{m}\mathrm{a}\mathrm{x}
\bigl(
0, \| u0(x)\| 1 - c1
\bigr)
, if
\bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty +
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty \not = 0,
(24)
c0 =
\left( 1 + \omega
\Biggl( \biggl(
l
\pi
\biggr) 2 \bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty +
l
\pi
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty
\Biggr) - 1
\right) - 1
.
Proof. Replace k by the index k - 1 in the equation (11), multiply the resulting relation by
uk(x) and integrate over x from 0 to l. Taking (12) into account, we get\bigm\| \bigm\| uk(x)\bigm\| \bigm\| 22 +m
\Bigl( \bigm\| \bigm\| uk - 1(x)
\bigm\| \bigm\| 2
1
\Bigr) \bigm\| \bigm\| uk(x)\bigm\| \bigm\| 21 = \Bigl( f\bigl( x, uk - 1(x), u
\prime
k - 1(x)
\bigr)
, uk(x)
\Bigr)
, k = 1, 2, . . . .
Applying (3) and (15), we have\biggl(
\alpha +
\Bigl( \pi
l
\Bigr) 2\biggr) \bigm\| \bigm\| uk(x)\bigm\| \bigm\| 21 \leq l
\pi
\bigm\| \bigm\| f\bigl( x, uk - 1(x), u
\prime
k - 1(x)
\bigr) \bigm\| \bigm\| \bigm\| \bigm\| uk(x)\bigm\| \bigm\| 1,
which implies \biggl(
\alpha +
\Bigl( \pi
l
\Bigr) 2\biggr)
\| uk(x)\| 1 \leq
l
\pi
\bigm\| \bigm\| f(x, uk - 1(x)u
\prime
k - 1(x))
\bigm\| \bigm\| .
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
1030 G. BERIKELASHVILI, A. PAPUKASHVILI, J. PERADZE
Hence, by using (17), we conclude that\bigm\| \bigm\| uk(x)\bigm\| \bigm\| 1 \leq 1
\alpha +
\biggl(
\pi
l
\biggr) 2
l
\pi
\biggl( \bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\| + \biggl( \bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty l
\pi
+
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty \biggr) \bigm\| \bigm\| uk - 1(x)
\bigm\| \bigm\|
1
\biggr)
.
This relation is an inequality of type (20), where
vk =
\bigm\| \bigm\| uk(x)\bigm\| \bigm\| 1,
a =
1
\alpha +
\biggl(
\pi
l
\biggr) 2 l\pi \Bigl( \bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty l
\pi
+
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty \Bigr) , b =
1
\alpha +
\biggl(
\pi
l
\biggr) 2 l\pi \bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\| .
Let us apply (6), (19) to these formulas and carry out some calculations. As a result, for
\bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty +
+
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty = 0, we obtain a = 0 and
b
1 - a
= c1, while for
\bigm\| \bigm\| \sigma 2(x)\bigm\| \bigm\| \infty +
\bigm\| \bigm\| \sigma 3(x)\bigm\| \bigm\| \infty \not = 0 we have
a = c0 and
b
1 - a
= c1. By considering these two cases with the estimate (21), we get convinced
that (23) is valid.
By Lemmas 3 and 5 it will be natural to require that the initial approximation u0(x) in (9) satisfy
the condition \bigm\| \bigm\| u0(x)\bigm\| \bigm\| 1 \leq c1. (25)
Then, by virtue of (24) and (23), we have
\bigm\| \bigm\| uk(x)\bigm\| \bigm\| 1 \leq c1, which, with (19) taken into account,
implies \bigm\| \bigm\| uk(x)\bigm\| \bigm\| 1 \leq l
\omega \pi
\bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\| , k = 0, 1, . . . . (26)
6. Convergence of the method. Multiplying (13) by \delta uk(x), integrating the resulting equality
with respect to x from 0 to l and using (14), we come to the relation\bigm\| \bigm\| \delta uk(x)\bigm\| \bigm\| 22 + 1
2
\Bigl( \Bigl(
m
\Bigl( \bigm\| \bigm\| uk - 1(x)
\bigm\| \bigm\| 2
1
\Bigr)
+m
\Bigl( \bigm\| \bigm\| u(x)\bigm\| \bigm\| 2
1
\Bigr) \Bigr) \bigm\| \bigm\| \delta uk(x)\bigm\| \bigm\| 21+
+
\Bigl(
m
\Bigl( \bigm\| \bigm\| uk - 1(x)
\bigm\| \bigm\| 2
1
\Bigr)
- m
\Bigl( \bigm\| \bigm\| u(x)\bigm\| \bigm\| 2
1
\Bigr) \Bigr) \bigl(
u\prime k(x) + u\prime (x), \delta u\prime k(x)
\bigr) \Bigr)
=
=
\Bigl(
f
\bigl(
x, uk - 1(x), u
\prime
k - 1(x)
\bigr)
- f
\bigl(
x, u(x), u\prime (x)
\bigr)
, \delta uk(x)
\Bigr)
, k = 1, 2, . . . .
Applying (3) – (5) and (16), we first obtain
\bigm\| \bigm\| \delta uk(x)\bigm\| \bigm\| 22 + \alpha
\bigm\| \bigm\| \delta uk(x)\bigm\| \bigm\| 21 \leq 1
2
l1
1\prod
p=0
\bigm| \bigm| \bigl( u\prime k - p(x) + u\prime , \delta u\prime k - p(x)
\bigr) \bigm| \bigm| +
+
\Bigl(
\| l2(x)\| \infty \| \delta uk - 1(x)\| + \| l3(x)\| \infty \| \delta u\prime k - 1(x)\|
\Bigr) \bigm\| \bigm\| \delta uk(x)\bigm\| \bigm\| \leq
\leq 1
2
l1
1\prod
p=0
\Bigl(
\| uk - p(x)\| 1 +
\bigm\| \bigm\| u(x)\bigm\| \bigm\|
1
\Bigr) \bigm\| \bigm\| \delta uk - p(x)
\bigm\| \bigm\|
1
+
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
ITERATIVE SOLUTION OF A NONLINEAR STATIC BEAM EQUATION 1031
+
l
\pi
\biggl(
l
\pi
\| l2(x)\| \infty + \| l3(x)\| \infty
\biggr) 1\prod
p=0
\| \delta uk - p(x)\| 1,
and after that, by virtue of (18) and (26), we have
\bigm\| \bigm\| \delta uk(x)\bigm\| \bigm\| 1 \leq \biggl( \alpha +
\biggl(
\pi
l
\biggr) 2\biggr) - 1
\left( 1
2
l1
1\prod
p=0
\bigl(
\| uk - p(x)\| 1 + \| u\| 1(x)
\bigr)
+
+
\biggl(
l
\pi
\biggr) 2
\| l2(x)\| \infty +
l
\pi
\| l3(x)\| \infty
\right) \| \delta uk - 1\| 1 \leq q\| \delta uk - 1\| 1, k = 1, 2, . . . ,
where
q =
\Biggl(
\alpha +
\biggl(
\pi
l
\biggr) 2\Biggr) - 1\biggl(
2c21l1 + \| l2(x)\| \infty
\biggl(
l
\pi
\biggr) 2
+
\bigm\| \bigm\| l3(x)\bigm\| \bigm\| \infty l
\pi
\biggr)
.
Taking (10), (19) and (16) into consideration we come to the following result.
Theorem 1. Let assumptions (3) – (6) and (25) be fulfilled. Suppose besides
q =
1
\alpha +
\biggl(
\pi
l
\biggr) 2 \biggl( l\pi
\biggr) 2\biggl(
2l1
\biggl( \bigm\| \bigm\| \sigma 1(x)\bigm\| \bigm\|
\omega
\biggr) 2
+ \| l2(x)\| \infty +
\pi
l
\| l3(x)\| \infty
\biggr)
< 1.
Then the approximations of the iteration method (9) converge to the exact solution of the problem
(1), (2) and for the error the estimate
\bigm\| \bigm\| uk(x) - u(x)
\bigm\| \bigm\|
p
\leq
\biggl(
l
\pi
\biggr) 1 - p
qk
\bigm\| \bigm\| u0(x) - u(x)
\bigm\| \bigm\|
1
, k = 1, 2, . . . , p = 0, 1,
is true.
7. Numerical experiment. The results on the convergence of the iteration process (9) to the
sought function (8) were confirmed by numerical experiments. For illustration, we present here the
result of numerical computations of one of the test problem.
We consider a special case, where m(z) = m0+m1z, m0 = 1, m1 =
1
2
, the beam length l = 1,
exact solution u(x) = x(x - 1)
\bigl(
x2 - x - 1
\bigr)
, i.e., u(x) = x4 - 2x3 + x, the right-hand side of the
equation (1)
f
\bigl(
x, u(x), u\prime (x)
\bigr)
=
1
35
\Bigl(
43.5u\prime
2
(x) - 348x3u\prime (x) -
- 1566u(x) + 696x6 - 3132x3 + 2088x+ 796.5
\Bigr)
.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
1032 G. BERIKELASHVILI, A. PAPUKASHVILI, J. PERADZE
Fig. 1. Case of five iterations, n = 10. Fig. 2. Case of five iterations, n = 20.
We carried out nine iterations. In the case of division of the interval [0, 1] into n = 10, 20 parts
(with h = 0.1, 0.05, respectively) the integrals were computed using the generalized trapezoid rule.
Here we applied the following definition of the kth iteration error:
error k = \mathrm{m}\mathrm{a}\mathrm{x}
i=0,1,...,n
\bigl\{
abs
\bigl(
uk(xi) - uk - 1(xi)
\bigr) \bigr\}
, xi = ih, k = 1, 2, . . . , 9.
The error of numerical values are given in the table below.
n
error k
1 2 3 4 5 7 9
10 0.43203 0.16734 0.06405 0.02473 0.00953 0.00142 0.00021
20 0.43328 0.16715 0.06365 0.02446 0.00938 0.00138 0.00020
Initial approximation of approximate solution u0(x) = 0. In the case of five iterations for n =
= 10, 20 the exact and approximate solutions are graphically illustrated (Figs. 1, 2).
In Figures 1 and 2 the numbers 1, 2, 3, 4, 5 corespond to the numbers of iterations, respectively.
And the graph of the fifth iteration u5(x) actually coincides with the graph of the exact solution
u(x).
Numerical experiments clearly show the convergence of approximate solutions with the increase
of the number of iterations and the influence of n number of the interval divisions on the rate of
convergence.
References
1. C. Bernardi, M. I. M. Copetti, Finite element discretization of a thermoelastic beam, Archive ouverte HAL-UPMC,
29/05/2013, 23 p.
2. S. Fučik, A. Kufner, Nonlinear differential equations, Stud. Appl. Mech., 2. Elsevier Sci. Publ. Co., Amsterdam etc.
(1980).
3. G. Kirchhoff, Vorlesungen über mathematische Physik, I. Mechanik, Teubner, Leipzig (1876).
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
ITERATIVE SOLUTION OF A NONLINEAR STATIC BEAM EQUATION 1033
4. T. F. Ma, Positive solutions for a nonlocal fourth order equation of Kirchhoff type, Discrete Contin. Dyn. Syst.,
694 – 703 (2007).
5. J. Peradze, A numerical algorithm for a Kirchhoff-type nonlinear static beam, J. Appl. Math., 2009, Article ID 818269
(2009), 12 p.
6. J. Peradze, On an iteration method of finding a solution of a nonlinear equilibrium problem for the Timoshenko plate,
Z. angew. Math. und Mech., 91, № 12, 993 – 1001 (2011).
7. K. Rektorys, Variational methods in mathematics, science and engineering, Springer Science & Business Media
(2012).
8. H. Temimi, A. R. Ansari, A. M. Siddiqui, An approximate solution for the static beam problem and nonlinear
integro-differential equations, Comput. Math. Appl., 62, № 8, 3132 – 3139 (2011).
9. S. Y. Tsai, Numerical computation for nonlinear beam problems, M. S. Thesis, Nat. Sun Yat-Sen Univ., Kaohsiung,
Taiwan (2005).
10. S. Woinowsky-Krieger, The effect of an axial force on the vibration of hinged bars, J. Appl. Mech., 17, 35 – 36
(1950).
Received 12.06.17
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
|
| id | umjimathkievua-article-833 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:04:13Z |
| publishDate | 2020 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/c5/390db5ba1667ea91a554dd0e274220c5.pdf |
| spelling | umjimathkievua-article-8332022-03-26T11:01:58Z Iterative solution of a nonlinear static beam equation Iterative solution of a nonlinear static beam equation Berikelashvili , G. Papukashvili , A. Peradze, J. Berikelashvili , G. Papukashvili , A. Peradze, J. UDC 519.6 The paper deals with a boundary-value problem for the nonlinear integro-differential equation $u''''-m \bigg(\int _0^l {u'}^2\,dx \bigg)u'' &nbsp;= f(x, u, u'),$ $m(z)\geq \alpha&gt;0,$ $0\leq z &lt; \infty,$ modeling the static state of the Kirchhoff beam.&nbsp;&nbsp;The problem is reduced to a nonlinear integral equation, which is solved using the Picard iteration method.&nbsp;&nbsp;The convergence of the iteration process is established and the error estimate is obtained. УДК 519.6 Ітерацiйний розв’язок нелiнiйного рiвняння статичної балки Розглядається крайова задача для нелінійного інтегро-диференціального рівняння $u''''-m \bigg(\int _0^l {u'}^2\,dx \bigg)u'' = f(x, u, u'),$ $m(z)\geq \alpha&gt;0,$ $0\leq z &lt; \infty,$ що моделює статичний стан балки Кірхгоффа.&nbsp;&nbsp;Задача зводиться до нелінійного інтегрального рівняння, яке розв'язується за допомогою ітераційного методу Пікара.&nbsp;&nbsp;Встановлено збіжність цього ітераційного процесу та отримано оцінку для похибки. Institute of Mathematics, NAS of Ukraine 2020-08-18 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/833 10.37863/umzh.v72i8.833 Ukrains’kyi Matematychnyi Zhurnal; Vol. 72 No. 8 (2020); 1024-1033 Український математичний журнал; Том 72 № 8 (2020); 1024-1033 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/833/8736 |
| spellingShingle | Berikelashvili , G. Papukashvili , A. Peradze, J. Berikelashvili , G. Papukashvili , A. Peradze, J. Iterative solution of a nonlinear static beam equation |
| title | Iterative solution of a nonlinear static beam equation |
| title_alt | Iterative solution of a nonlinear static beam equation |
| title_full | Iterative solution of a nonlinear static beam equation |
| title_fullStr | Iterative solution of a nonlinear static beam equation |
| title_full_unstemmed | Iterative solution of a nonlinear static beam equation |
| title_short | Iterative solution of a nonlinear static beam equation |
| title_sort | iterative solution of a nonlinear static beam equation |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/833 |
| work_keys_str_mv | AT berikelashvilig iterativesolutionofanonlinearstaticbeamequation AT papukashvilia iterativesolutionofanonlinearstaticbeamequation AT peradzej iterativesolutionofanonlinearstaticbeamequation AT berikelashvilig iterativesolutionofanonlinearstaticbeamequation AT papukashvilia iterativesolutionofanonlinearstaticbeamequation AT peradzej iterativesolutionofanonlinearstaticbeamequation |