Volume 11, 2007
15
THE ITERATION PROCESS FOR THE NONLINEAR
TWO-DIMENSIONAL OSCILLATION PROBLEM
Odisharia V.∗) , Peradze J.∗) , Peradze L.∗∗)
∗)
I. Javakhishvili Tbilisi State University
∗∗)
Georgian Technical University
Abstract. The initial boundary value problem for an integro-differential Kirchhoff equation is considered in the case of a square domain. To find an approximate solution, step-by-step discretization is performed with respect to spatial
variables and a time argument. The obtained cubic system is solved by the
iteration method. The method error is estimated.
Key words: Kirchhoff equation, Jacobi nonlinear iteration process, error estimate
AMS subject classification 2000: 65M
Let us consider the following initial boundary value problem
Z
³
´
4
wtt − λ + 2 (wx2 + wy2 )dx dy (wxx + wyy ) = 0,
π Ω
(x, y) ∈ Ω, 0 < t ≤ T,
w(x, y, 0) = w(0) (x, y), wt (x, y, 0) = w(1) (x, y), (x, y) ∈ Ω,
w(x, y, t) = 0, (x, y) ∈ ∂Ω, 0 ≤ t ≤ T,
(1)
(2)
(3)
where Ω = {(x, y) | 0 < x < π, 0 < y < π}, ∂Ω is the boundary of the
domain Ω, w(0) (x, y) and w(1) (x, y) are given functions, λ > 0 and T are given
constants.
Equation (1) is a two-dimensional analogue of the well-known Kirchhoff
equation [1]
Z
³
2 π 2 ´
wtt − λ +
w dx wxx = 0
(4)
π 0 x
for string vibration. The studies of many researchers are devoted to Kirchhoff
type equations (for the bibliography see, e.g., [2], [3]).
We present here one numerical method of the solution of problem (1)–(3).
An approximate solution will be sought for as a finite sum
wn (x, y, t) =
n
X
wnij (t) sin ix sin jy,
i,j=1
where the coefficients wnij (t) are calculated from the Galerkin system
00
(t)
wnij
h
+ λ+
n
X
2
(p + l
p,l=1
2
i
2
(t)
)wnpl
(i2 + j 2 )wnij (t) = 0,
(5)
16
Bulletin of TICMI
wnij (0) =
4
π2
4
0
wnij
(0) = 2
π
i, j = 1, 2, . . . , n,
Z
w(0) (x, y) sin ix sin jy dx dy,
Ω
(6)
Z
w(1) (x, y) sin ix sin jy dx dy, i, j = 1, 2, . . . , n.
Ω
We replace problem (5), (6) by the problem of finding functions w nij (t),
i, j = 1, 2, . . . , n, where the relation between the functions wnij (t) and w nij (t)
is such that
1
w nij (t).
wnij (t) = p
(7)
i2 + j 2
We have to solve the following problem
w00nij (t)
h
+ λ+
n
X
i
w2npl (t) (i2 + j 2 )w nij (t) = 0,
(8)
p,l=1
i, j = 1, 2, . . . , n,
Z
p
4
2
2
w(0) (x, y) sin ix sin jy dx dy,
w nij (0) = 2 i + j
π
Ω
Z
4p
w0nij (0) = 2 i2 + j 2 w(1) (x, y) sin ix sin jy dx dy,
π
Ω
i, j = 1, 2, . . . , n.
(9)
To solve the obtained Cauchy problem (8), (9) we will use a difference
scheme of symmetrical type. To this end, we introduce the grid {tm | 0 = t0 <
t1 < . . . < tM = T } with a constant step τ = tm − tm−1 . The approximate
values w nij (tm ) are denoted by wm
nij , i, j = 1, 2, . . . , n, m = 0, 1, . . . , M .
The scheme has the form
n
m+u−1 2 i
m−1
m−2
2
X
(wm+u
)
wm
1 X nh
nij − 2w nij + w nij
npl ) + (w npl
+
λ
+
2
τ
2 u=−1,0
2
p,l=1
×(i2 + j 2 )
m+u−1 o
wm+u
nij + w nij
= 0,
2
i, j = 1, 2, . . . , n, m = 2, 3, . . . , M,
(10)
w0nij = wnij (0),
τ
w1nij = wnij (0) + w0nij (0)
2 Z
i
2h
4
2τ
(0) 2
(0) 2
((wx ) + (wy ) )dx dy ×
+ 2 λ+ 2
π
π Ω
Z
(0)
(0)
) sin ix sin jy dx dy, i, j = 1, 2, . . . , n.
+ wyy
× (wxx
Ω
(11)
Volume 11, 2007
17
From system (10) we write a subsystem for fixed m, 2 ≤ m ≤ M . Then
n
n
o
X
8
m−1 2
m
m 2
w
+
2λ
+
[(w
)
+
(w
)
]
npl
npl
τ 2 (i2 + j 2 ) nij
p,l=1
m−1
×(wm
nij + w nij ) =
8
f m , i, j = 1, 2, . . . , n,
τ 2 (i2 + j 2 ) nij
(12)
where
m−2
fm
= 2wm−1
nij − w nij −
nij
×(i2 + j 2 )
τ2
2
³
λ+
m−2
wm−1
nij +w nij
2
Pn
m−2 2
2
(wm−1
npl ) +(w npl )
p,l=1
2
´
×
, i, j = 1, 2, . . . , n.
Let us agree that for the solution of problem (10), (11) we will use iteration
m−1
layerwise, more exactly, knowing the approximate solutions wm−2
nij and w nij ,
m
i, j = 1, 2, . . . , n, from (12) we find w nij , i, j = 1, 2, . . . , n, by iteration. As to
the initial values, i.e. values at the zero and the first layer, formulas (11) make
it possible to find w0nij and w1nij , i, j = 1, 2, . . . , n.
m
Denote the k-th iteration approximation wm
nij by w nij,k , i, j = 1, 2, . . . , n,
k = 0, 1, . . ..
From our algorithmic approach it follows that (12) is a system of equations
with respect to wm
nij , i, j = 1, 2, . . . , n. To solve this system we use the nonlinear Jacobi iteration process. From (12) it follows that the process has the
form
w3nij,k+1 + aij w2nij,k+1 + bij w nij,k+1 + cij = 0,
(13)
where
m−1 2
aij = wm−1
nij , bij = dij + (w nij ) +
m−1
2
cij = (dij + (wm−1
nij ) )w nij −
dij = 2λ +
n
X
8
,
+ j 2)
τ 2 (i2
8
fm ,
+ j 2 ) nij
τ 2 (i2
(14)
m−1 2
2
[(wm
npl,k ) + (w npl ) ], k = 0, 1, . . .
p, l = 1
p 6= i
l 6= j
In addition to (14) we need the values
2
8
2
,
rij = dij + (wm−1
nij ) + 2 2
3
τ (i + j 2 )
2³
10 m−1 2 ´ m−1
sij =
dij +
(w nij ) w nij −
3
9
³1
´
8
m
m−1
w
, i, j = 1, 2, . . . , n.
− 2 2
+
f
nij
τ (i + j 2 ) 3 nij
(15)
18
Bulletin of TICMI
Using the Cardano formulas [4], by (13) we write wm
nij,k+1 in the explicit
form
wm
(16)
nij,k+1 = ψij,k ,
where
aij
ψij,k = −
+ σij,1 − σij,2 , i, j = 1, 2, . . . , n,
(17)
3
3 ´1 i1
h
³ s2
rij
2 3
ij
v sij
σij,v = (−1)
, v = 1, 2, i, j = 1, 2, . . . , n. (18)
+
+
2
4
27
To establish the convergence conditions for the iteration process used, we
consider the Jacobi matrix
n ∂ψ
o
ij,k
J=
, i, j = 1, 2, . . . , n, i1 , j1 = 1, 2, . . . , n.
(19)
∂w ni1 j1 ,k
Here ij and i1 j1 are respectively the numbers of a row and a column of matrix
(19).
By (14)–(18) on the principal diagonal of the matrix J we find zeros. As
to nondiagonal elements, i 6= i1 , j 6= j1 , we find
2
X
∂ψij,k
1 m
1 h m−1
= − w ni1 j1 ,k
2w nij
∂wm
9
σ2
ni1 j1 ,k
v=1 ij,v
3 ´− 1 i
³
1 2 ´³ s2ij rij
2
+(−1)v wm−1
s
+
r
+
.
ij
ij
nij
3
4
27
(20)
From (18) we obtain the formulas
σij,1 σij,2 =
3 ´1
3
3
³ s2
rij
σij,1
+ σij,2
rij
2
ij
3
3
, σij,2
− σij,1
= sij ,
+
=
,
3
4
27
2
which together with (20) give
∂ψij,k
(1)
(2)
= ψiji1 j1 + ψiji1 j1 ,
∂wm
ni1 j1 ,k
(21)
where
³ 2
´
´2−v
m−1
v−1
−
wm
(s
)
w
ij
3 ni1 j1 ,k
3 nij
´−1
³
³ r ´v
ij
2v
2v
+ σij,2
, v = 1, 2.
× σij,1
+ −
3
³2
(v)
ψiji1 j1 =
(v)
(22)
Let us estimate |ψiji1 j1 |, v = 1, 2. To this end, we consider the functions
ψ (u) (z) = (−r)u +
2
X
1
2u
[z + (−1)v (z 2 + r3 ) 2 ] 3 ,
v=1
−∞ < z < ∞, r = const > 0,
Volume 11, 2007
19
u = 1, 2. Each function ψ (u) (z), u = 1, 2, is even and increasing for any z ≥ 0
and therefore
min
−∞<z<∞
|ψ (u) (z)| = ψ (u) (0) = (2u − 1)ru ,
u = 1, 2.
The latter relation, (14), (15) and (22) imply
(u)
|ψiji1 j1 | ≤
[τ 2 (i2 + j 2 )]u
2−u
|sij |u−1 |wm−1
|wm
ni1 j1 ,k |, u = 1, 2.
nij |
26u − 20
From formulas (21), (23) and (14), (15) it follows that
¯ ∂ψ
¯ 1
n1
h
¯
ij,k ¯
m−1
2 2
2
m
2 2
2
≤
|
|
λ+
τ
(i
+
j
)|w
τ
(i
+
j
)|w
¯
¯
ni1 j1 ,k
nij
∂w ni1 j1 ,k
4
6
n ³
o
X
1 m
5 m−1 2 ´i
2
m
m−1
+
(w
) + (w npl )
+ |w nij | + |f nij | .
2 npl,k
9
p,l=1
(23)
(24)
Let us introduce the vectors
n
n
n
m
m
m
= (wm−1
wm−1
n
nij )i,j=1 , w n,k = (w nij,k )i,j=1 , f n = (f nij )i,j=1 .
Besides, we also need vector and matrix norms. For the vector µ = (µs )N
s=1 and
N
N
P
P
|grs |.
the matrix G = (grs )N
|µs | and kGk1 = max
r,s=1 we define kµk1 =
Let us consider the sums
n
P
s=1
1≤s≤N r=1
(i2 + j 2 )u , u = 1, 2. Taking into account
i,j=1
that [5]
n
X
l2u =
l=1
n(n + 1)(2n + 1) ³ 3n2 + 3n − 1 ´u−1
, u = 1, 2,
6
5
we obtain
n
X
n2 (n + 1)(2n + 1) ³ 6n2 + 6n − 2 ´u−1
(i2 + j 2 )u ≤
, u = 1, 2.
3
5
i,j=1
(25)
Note that for u = 1 in (25) we have the equality.
Let us estimate the norm of the Jacobi matrix (19). By virtue of (24) and
(25)
kJk1
τ 2 n2 (n + 1)(2n + 1)
12
n τ 2 (3n(n + 1) − 1)
× max |wm
|
kwm−1
k1 ×
ni1 j1 ,k
n
i1 ,j1
15
n ³
o
h
X
1 m 2 5 m−1 2 ´i
m
+kwm−1
k
+kf
k
, (26)
× λ+
(w nij,k ) + (w nij )
1
1
n
n
2
9
i,j=1
≤
i1 , j1 = 1, 2, . . . , n.
20
Bulletin of TICMI
Applying the principle of compressed mapping [6], we assume that
kJk1 ≤ q,
0 < q < 1,
1
(27)
m
m
kwm
kwm
n,k − w n,0 k1 ≤
n,1 − w n,0 k1 , k = 1, 2, . . . ,
1−q
is fulfilled. From (26) it follows that for (27) to be hold, it is sufficient that
the following biquadratic inequality
α τ4 + β τ2 − γ ≤ 0
(28)
be fulfilled with respect to the step τ . Here
1³ m
3n(n + 1) − 1 m−1 h
α =
kw n,0 k1 +
kwn k1 λ +
15
2
n
´2 5 X
i
1
m−1 2
m
m
+
kw − w n,0 k1 +
|w
| ,
1 − q n,1
9 i,j=1 nij
β = kwm−1
k1 + kf m
k,
n
n 1
³
´−1
12q
1
m
m
m
γ =
kw n,0 k1 +
kw − w n,0 k1
.
n2 (n + 1)(2n + 1)
1 − q n,1
Thus we come to a conclusion that if the parameter q and the step τ
satisfy inequality (28), then system (12) has a unique solution with respect to
m
m n
the unknowns wm
nij , i, j = 1, 2, . . . , n. The vector w n = (w nij )i,j=1 consisting
m
of the components of solutions wm
nij is the limit of a sequence of vectors w n,k ,
as k → ∞. Moreover, the estimate
m
kwm
n,k − w n k1 ≤
qk
kwm − wm
n,0 k1 , k = 0, 1, . . . ,
1 − q n,1
is true.
Formula (7) enables us to construct with the aid of wm
nij,k approximate
solutions of the function wnij (t) at the grid points.
The problem considered here for the one-dimensional equation (4) is studied
in [7].
References
[1] Kirchhoff, G., Vorlesungen über Mechanik, Teubner, Leipzig, 1883.
[2] Medeiros, L., Limaco, I., Menezes, S., Vibrations of an elastic string, Mathematical
Aspects, I and II., J. Comput. Anal. Appl., 4, 2(2002), 211–263.
[3] Peradze, J., A numerical algorithm for the nonlinear Kirchhoff string equation, Numer. Math., 102, 2(2005), 311–342.
[4] Kurosh, A., A course on higher algebra, Nauka, Moscow, 1975 (in Russian).
[5] Gradstein, I., Rizhik, I., Tables of integrals, sums, series and products, Nauka,
Moscow, 1971 (in Russian).
[6] Trenogin, V. Functional analysis, Nauka, Moscow, 1980 (in Russian).
[7] Peradze, J., The Jacobi nonlinear iteration method for a discrete Kirchhoff system,
AMIM, I. Vekua Appl. Math. Inst. Tbilisi State Univ., 6, 1(2001), 81–89.
Received April, 16, 2007; revised October, 15, 2007; accepted November, 7, 2007.