Lecture Notes of TICMI, Vol.9, 2008
N. Chinchaladze
ON SOME NONCLASSICAL PROBLEMS FOR
DIFFERENTIAL EQUATIONS AND THEIR
APPLICATIONS TO THE THEORY OF CUSPED
PRISMATIC SHELLS
© Tbilisi University Press
Tbilisi
2
Preface
The present Lecture Notes contains extended material mainly based on the lectures presented at the Workshop on Mathematical Methods for Elastic Cusped
Plates and Bars (Tbilisi, September 27–28, 2001).
The work consists of the list of notation, introduction, three chapters and references.
The Introduction contains a survey of results related to the subject and a brief
presentation of results of the present work.
In Chapter 1 some auxiliary materials are given which are used in Chapters 2
and 3.
Chapter 2 deals with the problems of cylindrical bending and bending vibration
of a cusped plate. Bending problems of cusped plates fall outside of the limits of
classical bending theory. The aim of this chapter is to study the problem of wellpossedness of boundary value problems and initial boundary value problems in case
of cylindrical bending of shells with two cusped edges and in some cases to solve
these problems in explicit forms.
Chapter 3 is dedicated to the interface problem of the interaction of a plate with
two cusped edges and a flow of an incompressible fluid.
Acknowledgments. The author is very grateful to Prof. G. Jaiani, Prof.
S. Kharibegashvili, and Prof. D. Natroshvili for their useful discussions.
Author
Contents
List of Notation
4
Introduction
6
1 Preliminary Materials
23
1.1. Cusped Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.2. Hilbert-Schmidt Theorems . . . . . . . . . . . . . . . . . . . . . . . . 24
1.3. Singular and Supersingular Integral Equations . . . . . . . . . . . . . 26
2 Bending of a Cusped Plate
2.1. Cylindrical Bending of a Cusped Plate . . . . . . . . . . . . . . . . .
2.2. Vibration of the Plate with Two Cusped Edges . . . . . . . . . . . .
2.3. Harmonic Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
30
46
59
3 A Cusped Elastic Plate-Fluid Interaction Problem
69
3.1. Case of an Ideal Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.2. Case of a Viscous Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . 81
References
86
3
List of Notations
N := {1, 2, · · ·},
N := {1, 2, · · ·},
Rn n−dimensional Euclidean space (n ∈ N)
Ω := {(x1 , x2 , x3 ) : −∞ < x1 < ∞, 0 < x2 < l, x3 = 0} - the projection of a plate
on the plane x3 = 0
I := {[0, l] × {0}}
Ωf := {x1 , x2 , x3 : x1 = 0, x2 := (x2 , x3 ) ∈ R2 \I} - space which occupies the fluid
(+)
(−)
2h(x) := h (x) − h (x) - thickness of a plate at point x
ω - oscillation frequency
D(x2 ) - flexural rigidity
ρ - density of a plate
w(x2 , t) - deflection of a plate
q(x2 , t) - lateral load
M2 (x2 , t) - bending moment
Q2 (x2 , t) - intersecting force
E - Young’s modulus
σ - Poisson’s ratio
F := (F2 , F3 ) - plane volume forces
δij - Kroneker Delta
ρf - density of a fluid
u := (u1 , u2 , u3 ) - displacement vector of a fluid
v := (v1 , v2 , v3 ) - velocity vector of a fluid
p - pressure of a fluid
(+)
(−)
p(x2 , h (x2 ), t) (p(x2 , h (x2 ), t)) - the value of the pressure on the upper (lower)
surface of the plate
v3∞ (t), p∞ (t) - values
of the ¶
velocity vector component and pressure at infinity
µ
∂vk
∂vj
f
+
- stress tensor of a fluid
σjk
= −pδjk + µ
∂xk ∂xj
ν, µ - coefficients of viscosity
∂2
∂2
+
∆ :=
∂x22 ∂x23
∂w
w,t :=
∂t
4
LIST OF NOTATIONS
5
∂w
, i = 1, 2, 3
∂xi
C n (]0, l[) (C n ([0, l])) - n-times continuously differentiable functions in ]0, l[ (on [0, l])
C n (Ωf ) - n-times continuously differentiable functions in Ωf with respect to x2 and
x3
C(t > 0) - continuous functions with respect to t for t > 0
H([0, l]) - class of Hölder continuous functions
L2 ([0, l]) - class of square integrable functions on [0, l]
w,i :=
Introduction
In 1955 I.Vekua [95]-[97] raised the problem of investigation of cusped plates, i.e.
such ones whose thickness on the part of the plate boundary or on the whole one
vanishes. The problem mathematically leads to the question of setting and solving of
boundary value problems for even order equations and systems of elliptic type with
the order degeneration in the statical case and of initial boundary value problems
for even order equations and systems of hyperbolic type with the order degeneration in the dynamical case (for corresponding investigations see the survey [35] and
also I. Vekua’s comments in [97, p.86]). There exists a wide literature devoted to
the theory of degenerate and mixed type equations (see, e.g., [5], [30]), which was
developed intensively in the period from early 50-ies till early 70-ies but it could not
cover the above equations and systems because of distinct peculiarities of the latter
caused by the geometry of the mechanical problem.
The first work concerning classical bending of cusped elastic plates was done by
E. Makhover [67], [68] and S. Mikhlin [71].
In 1957 E. Makhover [67], [68], by using the results of S. Mikhlin [71], had
considered such a cusped plate with the stiffness D(x1 , x2 ) satisfying
D1 xκ2 1 ≤ D(x1 , x2 ) ≤ D2 xκ2 1 , D1 , D2 , κ1 = const > 0,
(1)
within the framework of classical bending theory. She particularly studied in which
cases the deflection (κ1 < 2) or its normal derivative (κ1 < 1) on the cusped edge
of the plate can be given. In 1971, A. Khvoles [62] represented the forth order Airy
stress function operator as the product of two second order operators in the case
when the plate thickness 2h is given by
2h = h0 xκ2 2 , h0 , κ2 = const > 0, x2 ≥ 0,
(2)
and investigated the general representation of corresponding solutions. Since 1972
the work of G. Jaiani in [36]–[51] is also devoted to these problems. By using more
natural spaces than E. Makhover, G. Jaiani in [48] has analyzed in which cases
the cusped edge can be freed (κ1 > 0) or freely supported (κ1 < 2). Moreover,
he established well–posedness and the correct formulation of all admissible principal boundary value problems (BVPs). In [41], [42], [47] he also investigated the
tension–compression problem of cusped plates, based on I. Vekua’s model of shallow
prismatic shells. G. Jaiani’s results can be summarized as follows.
6
7
INTRODACTION
Let n be the inward normal of the plate boundary. In the case of the tensioncompression problem on the cusped edge, where
0≤
∂h
< +∞(in the case (2) this means κ2 ≥ 1) ,
∂n
which will be called a sharp cusped edge, one can not prescribe the displacement
vector; while on the cusped edge, where
∂h
= +∞(in the case (2) this means κ2 < 1) ,
∂n
called a blunt cusped edge, the displacement vector can be prescribed. In the case
of the classical bending problem with a cusped edge, where
∂h
= O(dκ−1 )as d → 0, κ = const > 0
∂n
(3)
and where d is the distance between an interior reference point of the plate projection
and the cusped edge, the edge can not be fixed if κ ≥ 13 , but it can be fixed if
0 < κ < 31 ; it can not be freely supported if κ ≥ 23 , and it can be freely supported
if 0 < κ < 23 ; it can be free or arbitrarily loaded by a shear force and a bending
moment if κ > 0. Note that in the case (2), the condition (3) implies that d2 = x2
and κ = κ2 = κ31 .
For the specific cases of cusped cylindrical and conical shell bending, the above
results remain valid as it has been shown by G. Tsiskarishvili and N. Khomasuridse
[89]-[92]. These results also remain valid in the case of classical bending of orthotropic cusped plates (see [51]). However, for general cusped shells and also for
general anisotropic cusped plates, the corresponding analysis is done.
The problems involving cusped plates lead to correct mathematical formulations
of BVPs for even order elliptic equations and systems whose orders degenerate at
the boundary (see [47], [52]-[53]).
Applying the functional–analytic method developed by G. Fichera in [28], [29]
(see also [21], [22]), in [47] the particular case of Vekua’s system for general cusped
plates has been investigated.
The classical bending of plates with the stiffness (1) in energetic and in weighted
Sobolev spaces has been studied by G. Jaiani in [48], [50]. In the energetic space
some restrictions on the lateral load has been relaxed by G. Devdariani in [20].
G. Tsiskarishvili [90] characterized completely the classical axial symmetric bending
of specific circular cusped plates without or with a hole.
In the case (2), the basic BVPs have been explicitly solved in [43] and [53] with
the help of singular solutions depending only on the polar angle.
If we consider the cylindrical bending of a plate, in particular of a cusped one,
with rectangular projection a ≤ x1 ≤ b, 0 ≤ x2 ≤ `, then we actually get the
corresponding results also for cusped beams (see [49], [43], [93], [73]-[77], [12], [13],
[54], [55]).
8
INTRODACTION
In 1999-2001 two contact problems were considered by N. Shavlakadze [86], [87],
namely, the contact problem for an unbounded elastic medium composed of two
half-planes x1 > 0 and x1 < 0 having different elastic constants and strengthened
on the semi-axis x2 > 0 by an inclusion of variable thickness (cusped beam) with
constant Young’s modulus and Poisson’s ratio. It was assumed that the plate is
subjected to plane deformation, the flexural rigidity D had the form
D = D0 xκ2 , D0 , κ = const > 0,
and the cusped end x2 = 0 of the beam was free.
At the same time (in the fifties of the twentieth century), I.Vekua [95] introduced
a new mathematical model for elastic prismatic shells (i.e., of plates of variable
thickness) which was based on expansions of the three–dimensional displacement
vector fields and the strain and stress tensors in linear elasticity into orthogonal
Fourier-Legendre series with respect to the variable plate thickness. By taking
only the first N + 1 terms of the expansions, he introduced the so–called N –th
approximation. Each of these approximations for N = 0, 1, ... can be considered as
an independent mathematical model of plates. In particular, the approximation for
N = 1 corresponds to the classical Kirchhoff plate model. In the sixties, I. Vekua
[96] developed the analogous mathematical model for thin shallow shells. All his
results concerning plates and shells are collected in his monograph [97]. Works
of I. Babuška, D. Gordeziani, V. Guliaev, I. Khoma, A. Khvoles, T. Meunargia,
C. Schwab, T. Vashakmadze, V. Zhgenti, and others (see [2], [31], [33], [61], [62],
[69], [84], [85], [94], [100] and the references therein) are devoted to further analysis
of I.Vekua’s models (rigorous estimation of the modeling error, numerical solutions,
etc.) and their generalizations (to non-shallow shells, to the anisotropic case, etc.).
In [56] variational hierarchical two–dimensional models for cusped elastic plates
are constructed. With the help of variational methods, existence and uniqueness theorems for the corresponding two–dimensional boundary value problems are proved
in appropriate weighted functional spaces. By means of the solutions of these two–
dimensional boundary value problems, a sequence of approximate solutions in the
corresponding three-dimensional region is constructed. This sequence converges in
the Sobolev space H 1 to the solution of the original three-dimensional boundary
value problem. The systems of differential equations corresponding to the twodimensional variational hierarchical models are explicitly given for a general orthogonal system and for Legendre polynomials, in particular.
Recently N.Chinchaladze, R. Gilbert, G. Jaiani, S. Kharibegashvili and D. Natroshvili have studied the well posedness of boundary value problems for elastic
cusped prismatic shells in the N th approximation of I. Vekua’s hierarchical models
under (all reasonable) boundary conditions at the cusped edge and given displacements at the non-cusped edge and stresses at the upper and lower faces of the shell
[19].
For the last decades the direct and inverse problems connected with the interaction between difference vector fields have received much attention in the mathe-
9
INTRODACTION
matical and engineering scientific literature and have been intensively investigated.
They arise in many physical and mechanical models describing the interaction of
two different media where the whole process is characterized by a vector-function of
dimension k in one medium and by a vector-function of dimension n in the other (for
example, fluid-structure interaction where a streamlined body is an elastic obstacle,
scattering of acoustic and electromagnetic waves by an elastic obstacle, interaction
between an elastic body and seismic waves, etc.).
A lot of authors have considered and studied in detail the direct problems of
interaction between an elastic isotropic body occupying a bounded region Ω with a
three-dimensional elastic vector field to be defined, and some isotropic medium (say
fluid) occupying the unbounded exterior region, the compliment of Ω with respect
to the whole space, where a scalar field is to be defined. The time-harmonic dependent unknown vector and scalar fields are coupled by some kinematic and dynamic
conditions on the boundary ∂Ω, which lead to various type of non-classical interface
problems of steady oscillations for a piecewise homogeneous isotropic medium. An
exhaustive information in this direction concerning theoretical and numerical results
can be found in [4], [6], [7], [24], [25], [59], [60], [32], [34] [26], [27], [78], [84].
Some particular cases where the elastic body under consideration is anisotropic
have been treated in [57], [58], [79].
Various authors dedicated their works to the solid-fluid (see e.g. [79], [83], [98][99], [80]-[82], [9]-[11]), [14]-[18] contact problems. The present work is devoted to
the interaction problems when profile of an elastic part is cusped on some part
boundary.
Bending problems of cusped plates fall outside of the limits of classical bending
theory. The aim of the dissertation is to study the problem of well-possedness of
boundary value problems and initial boundary value problems in case of cylindrical
bending of shells with two cusped edges and in some cases to solve these problems
in explicit forms.
The work consists of the list of notations, introduction, three chapters and bibliography.
The Introduction contains a survey of results related to the subject and a brief
presentation of results of the present work.
In Chapter 1 some auxiliary materials are given used in Chapters 2 and 3.
Chapter 2 deals with the problems of cylindrical bending and bending vibration
of a plate.
Let us consider the plate whose projection on x3 = 0 occupies the domain Ω
Ω = {(x1 , x2 , x3 ) : −∞ < x1 < ∞, 0 < x2 < l, x3 = 0},
and where the thickness of the plate are given by the equation
α/3
2h(x2 ) = h0 x2 (l − x2 )β/3 ,
h0 , l, α, β = const, h0 , l > 0, α, β ≥ 0.
When α2 + β 2 > 0 a plate is called a cusped plate. A profile of the plate under
consideration has one of the forms shown in Figures 4-12.
10
INTRODACTION
The equation of cylindrical bending of the plate has the form (see, e.g., [88])
(D(x2 )w, 22 (x2 )), 22 = q(x2 ),
0 < x2 < l,
(4)
where w(x2 ) is a deflection of the plate, q(x2 ) is a load, D(x2 ) is a flexural rigidity
∂w
of the plate, and by w,i we denote w,i :=
.
∂xi
In general,
2Eh3 (x2 )
D(x2 ) :=
,
3(1 − σ 2 )
where E is a Young’s modulus, σ is a Poisson’s ratio. Let E =const, σ =const, and
D(x2 ) = D0 xα2 (l − x2 )β , D0 = const > 0.
In the case of cylindrical bending of an isotropic plate, the bending moment
M2 (x2 ) and the intersection force Q2 (x2 ) are given by the formulae (see [88])
M2 (x2 ) := −D(x2 )w,22 (x2 ), Q2 (x2 ) := M2,2 (x2 ).
(5)
Section 2.1 is devoted to the investigation of properties of equation (4) and
formulation of all admissible classical bending boundary value problems (BVPs).
If q(x2 ) ∈ C([0, l]) then
M2 (x2 ), Q2 (x2 ) ∈ C([0, l]),
the behaviour of the w,2 (x2 ) and w(x2 ) when x2 → 0+ and x2 → l− depends on α
and β. As a result of the corresponding analysis we obtain that, e.g., at the point
x2 = 0 the following classical bending boundary conditions are admissible
1. w(0) = w0 (0) = 0 iff(if and only if) α < 1;
2. w0 (0) = Q2 (0) = 0 iff α < 1;
3. w(0) = M2 (0) = 0 iff α < 3;
4. M2 (0) = Q2 (0) = 0 for any α.
(6)
(7)
(8)
(9)
Similar conditions we have at the point x2 = l, under the same restrictions on
β. All BVPs are solved in the explicit integral forms. Using these integral representations and the difference equation corresponding to (4) by means of MATLAB we
get numerical results for the deflection, the bending moment and the intersecting
force for different materials (see Figures 13-16).
In Section 2.2 a dynamical problem is investigated for the above cusped plate.
The corresponding equation has the following form
(D(x2 )w, 22 (x2 , t)), 22 = q(x2 , t) − 2ρh(x2 )
where ρ is a density of the plate.
∂ 2 w(x2 , t)
,
∂t2
0 < x2 < l,
(10)
11
INTRODACTION
We solve equation (10) under the following initial conditions (IC)
w(x2 , 0) = ϕ1 (x2 ), w,t (x2 , 0) = ϕ2 (x2 ), x2 ∈ [0, l],
(11)
where ϕ1 (x2 ), ϕ1 (x2 ) ∈ C([0, l]) are given functions.
In this case the bending moment and the intersecting force are given by the
expressions
M2 (x2 , t) := −D(x2 )w,22 (x2 , t),
Q2 (x2 , t) := M2,2 (x2 , t).
(12)
(13)
Since of (10) is not degenerate equation with respect to t = 0, taking into account
(6)-(9), the following initial boundary value problems (IBVPs) are admissible
Problem 11 Let 0 ≤ α < 3, 0 ≤ β < 1. Find a function w(x2 , t), which satisfies
the following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C([0, l]) ∩ C 1 (]0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0),
w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the BCs
w(0, t) = M2 (0, t) = w,2 (l, t) = Q2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]) ∩ C([0, l]),
ϕi (0) = −D(x2 )ϕ00i (x2 )|x2 =0+ = ϕ0i (l)
0
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 12 Let 0 ≤ α, β < 1. Find a function w(x2 , t), which satisfies the following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the boundary conditions (BCs)
w(0, t) = w,2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
ϕi (0) = ϕ0i (0) = ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
12
INTRODACTION
Problem 13 Let 0 ≤ α, β < 1. Find a function w(x2 , t), which satisfies the following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the BCs
w(0, t) = w,2 (0, t) = w,2 (l, t) = Q2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
0
ϕi (0) = ϕ0i (0) = ϕ0i (l) = (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 14 Let 0 ≤ α, < 1, 0 ≤ β < 3. Find a function w(x2 , t), which satisfies
the following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the BCs
w(0, t) = w,2 (0, t) = w(l, t) = M2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]),
ϕi (0) = ϕ0i (0) = ϕi (l) = (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 15 Let 0 ≤ α < 1, β ≥ 0. Find a function w(x2 , t), which satisfies the
following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 < l, t ≥ 0),
equation (10), the BCs
w(0, t) = w,2 (0, t) = M2 (l, t) = Q2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[),
ϕi (0) = ϕ0i (0) = (−D(x2 )ϕ00i (x2 )) |x2 =l−
0
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
13
INTRODACTION
Problem 16 Let 0 ≤ α, β < 1. Find a function w(x2 , t), which satisfies the following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the BCs
w,2 (0, t) = Q2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
0
ϕ0i (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
Problem 17 Let 0 ≤ α < 1, 0 ≤ β < 3. Find a function w(x2 , t), which satisfies
the following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the BCs
w,2 (0, t) = Q2 (0, t) = w(l, t) = M2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]),
0
ϕ0i (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l)
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 18 Let 0 ≤ α < 3, 0 ≤ β < 1. Find a function w(x2 , t), which satisfies
the following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the BCs
w(0, t) = M2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]) ∩ C([0, l]),
ϕi (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
14
INTRODACTION
Problem 19 Let 0 ≤ α, β < 3. Find a function w(x2 , t), which the satisfies following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (10), the BCs
w(0, t) = M2 (0, t) = w(l, t) = M2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C([0, l]),
ϕi (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l)
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 20 Let α ≥ 0, 0 < β < 1. Find a function w(x2 , t), which satisfies the
following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 < x2 ≤ l, t ≥ 0),
equation (10), the BCs
M2 (0, t) = Q2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (11), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]),
0
(−D(x2 )ϕ00i (x2 )) = (−D(x2 )ϕ00i (x2 )) |x2 =0+
= ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
Let q ≡ 0. Using the Fourier method, we look for w(x2 , t) in the following form
w(x2 , t) = X(x2 )T (t),
where T (t) and X(x2 ) are satisfying the following equations
T 00 (t) + λT (t) = 0,
and
Zl
X(x2 ) = λ
g(ξ)K(x2 , ξ)X(ξ)dξ, g(x2 ) := 2ρh(x2 ),
(14)
0
where K(x2 , ξ) ∈ C([0, l] × [0, l]) is constructed explicitly and it depends on the
coefficients of equation (10) and the type of boundary conditions in Problems 11-20.
We denote by λn and Xn the corresponding eigenvalues and eigenfunctions of
(14).
The following propositions hold.
15
INTRODACTION
Proposition 2.2 K(x2 , ξ) is symmetric with respect to x2 and ξ.
Proposition 2.3 Number of eigenvalues λn of (14) is not finite.
Proposition 2.4 All λn are positive.
The solution of equation (10) under the initial conditions (11) and one of the
boundary conditions (see Problems 11-20) can be written as follows [15]
w(x2 , t) =
∞
X
´
³
p
p
Xn (x2 ) bn1 sin( λn t) + bn2 cos( λn t) ,
(15)
n=1
where
1
bn1 = √
λn
Zl
Zl
g(x2 )Xn (x2 )ϕ2 (x2 )dx2 , bn2 =
g(x2 )Xn (x2 )ϕ1 (x2 )dx2 .
(16)
0
0
Let us consider one of the IBVP. For the sake of simplicity we consider Problem
11.
00 00
(Dϕ )
Further, if we suppose that ψi (x2 ) := √ i ∈ C([0, l]) (i = 1, 2), we can prove
g(x2 )
the following theorems [15]
Theorem 2.5 The series (15) converges absolutely and uniformly on [0, l]. Moreover, the series
∞
´
X
p
p
p ³
n
n
w,t (x2 , t) =
Xn (x2 ) λn b1 cos( λn t) − b2 sin( λn t)
n=1
and
∞
³
´
X
p
p
w,tt (x2 , t) = − Xn (x2 )λn bn1 sin( λn t) + bn2 cos( λn t)
n=1
converge absolutely and uniformly on any [a, b] ⊂]0, l[ if the functions
ψi (x2 )
Ψi (x2 ) := p
fori = 1, 2satisfyBCsgiveninProblem11
g(x2 )
and the functions
p
00
χi (x2 ) g(x2 ) := (D(x2 )Ψ00i (x2 )) , i = 1, 2, are integrable on ]0, l[
(For this, e.g., it is sufficient that
x2 → 0+ ,
j = 2, 8).
dj
ϕ (x )
dxj2 i 2
dj
ϕ (x )
dxj2 i 2
γ
(17)
(18)
= O(x2ij ), γij = const > 7 − j −
= O((l − x2 )δij ), δij = const > 7 − j −
5β
,
3
5α
,
3
x2 → l− , i = 1, 2;
16
INTRODACTION
Theorem 2.6 The series
∞
´
³
X
p
p
∂i
di
n
n
w(x2 , t) =
Xn (x2 ) b1 sin( λn t) + b2 cos( λn t) , i = 1, 2, 3, 4,
∂xi2
dxi2
n=1
are convergent absolutely and uniformly on any [a, b] ⊂]0, l[, while the series
³
X di−1
p
∂ i−1
00
n
(D(x
)w,
(x
,
t))
=
(D(x
)X
(x
))
b
sin(
λn t)+
2
x
x
2
2
2
2 2
n
1
∂xi−1
dxi−1
2
2
n=1
∞
¢
√
+bn2 cos( λn t) , i = 1, 2
are convergent absolutely and uniformly on [0, l].
Thus, (15) is the solution of the Problem 11 for q(x2 , t) ≡ 0.
Let us consider the case when q(x2 , t) 6≡ 0, ϕi = 0, and let √qg (·, t) ∈ L2 (0, l).
Then q(x2 , t) can be represented as a convergent series in L2 (0, l):
Z
∞
X
q(x2 , t) =
g(x2 )Xn (x2 )qn (t), qn (t) := q(x2 , t)Xn (x2 )dx2 .
l
n=1
0
Further, we look for the solution in the form w(x2 , t) =
∞
P
wn (x2 , t), where
n=1
wn (x2 , t) is a solution of the equation (10) under the homogeneous initial conditions
and under the boundary conditions given in Problem 11 with q(x2 , t) replaced by
g(x2 )Xn (x2 )qn (t). Now, using the method of separation of variables we can write
wn (x2 , t) = Xn (x2 )T1n (t),
where
00
T1n
(t) + λn T1n (t) = qn (t).
Therefore, w(x2 , t) can be expressed as follows
Z
∞
X
p
1
√ Xn sin( λn (t − τ ))qn (τ )dτ.
w(x2 , t) =
λn
n=1
t
(19)
0
Similarly to Theorems 2.5 and 2.6, if the following conditions are fulfilled
Ã
!
¶
µ
1
q(x2 , t)
τ (x2 , t) := p
D(x2 )
∈ C[0, l],
g(x2 ) ,x2 x2
g(x2 )
,x x
2 2
τ (x2 , t)
satisfies the BCs given in Problem 11
and p
g(x2 )
(20)
17
INTRODACTION
(For this, e.g., it is sufficient that
∂j
q(x2 , t)
∂xj2
∂j
q(x2 , t)
∂xj2
γ
2α
,
3
2β
,
3
j = 0, 8) we get the absolute
= O(x2j ) x2 → 0+ , γj > 7 − j −
= O((l − x2 )δj ) x2 → l− , γj > 7 − j −
and uniform convergence of the series (19) and
∞
X di
∂i
(x
,
t))
=
(D(x
)w
(D(x2 )Xn00 )T1n (t), i = 0, 1,
2
,x
x
2
2 2
i
∂xi2
dx
2
n=1
on [0, l], and absolute and uniform convergence of
∞
X di
∂i
w
(x
,
t)
=
Xn (x2 )T1n (t), i = 1, ..., 4,
x
2
i
∂xi2 2
dx
2
n=1
∞
X
di
∂i
w(x
,
t)
=
X
(x
)
T1n (t), i = 1, 2,
2
n
2
∂ti
dti
n=1
on any [a, b] ⊂]0, l[.
Now, let q(x2 , t) 6≡ 0, ϕi (x2 ) 6≡ 0. If conditions (20), (17), and (18) are satisfied
then the solution of Problem 11 can be expressed as follows
∞
X
w(x2 , t) =
wn (x2 , t),
n=1
where
wn (x2 , t) = Xn (x2 )(T1n (t) + Tn (t)),
w1 (x2 , t) := Xn T1n (t) is given by the formula (19) and w2 (x2 , t) := Xn Tn (t) is given
by the formula (15).
Remark 1 Similarly are solved IBVPs corresponding to the Problems 12-20.
We can avoid the restrictions (20) if we consider harmonic vibration. In this case
w(x2 , t) = eiω t w0 (x2 ), q(x2 , t) = eiω t q0 (x2 ),
where ω = const is an oscillation frequency, q0 (x2 ) ∈ C([0, l]) is a given function.
E.g., in the case of Problem 11, for w0 (x2 ) we get the following problem
00
(D(x2 )w000 (x2 )) = q0 (x2 ) + 2ω 2 ρh(x2 )w0 (x2 ),
w0 (0) = M2 (0) = w0 (l) = Q2 (l) = 0, 0 ≤ α < 2, 0 ≤ β < 1,
w0 (x2 ) ∈ C 4 (]0, l[) ∩ C([0, l]) ∩ C 1 (]0, l]).
(21)
This problem is equivalent to the integral equation
Zl
w0 (x2 ) − ω
2
K(x2 , ξ) g(ξ) w0 (ξ)dξ = F (x2 ),
0
(22)
18
INTRODACTION
where
Zl
F (x2 ) :=
K(x2 , ξ) q0 (ξ)dξ,
0
K(x2 , ξ) has the same form as in integral equation (14).
If ω 2 6= λn , the unique solution of (22) can be written as follows (see, e.g., [66])
p
w1 (x2 ) = F (x2 ) g(x2 )


Zl
∞
X
p
1

F (ξ) g(x2 ) Yn (ξ)dξ  Yn (x2 ),
+ ω2
(23)
2
λ
−
ω
n
n=1
0
It is shown that the series in the right hand side of (23) is absolutely and uniformly
convergent on [0, l], because of q0 ∈ C([0, l]).
Using the difference equation corresponding to (21), by means of MATLAB we
get numerical and graphical results for harmonic vibration problems.
Chapter 3 is dedicated to the interface problem of the interaction of a plate with
two cusped edges and a flow of a fluid.
We assume that the flow is independent of x1 , parallel to the plane 0x2 x3 , i.e.
v1 ≡ 0, and generates a bending of the plate. Let at infinity, for pressure we have
p(x2 , x3 , t) → p∞ (t), when |x| → ∞,
(24)
and let for the velocity components conditions at infinity be either
v2 (x2 , x3 , t) = O(1), v3 (x2 , x3 , t) → v3∞ (t),
(25)
vj (x2 , x3 , t) = O(1), j = 2, 3,
(26)
or
where v := (v2 , v3 ) is a velocity vector of the fluid, p(x2 , x3 , t) is a pressure, and
v3∞ (t), p∞ (t) are given functions.
Let us introduce the following notations
I := {[0, l] × 0},
©
ª
Ωf := x1 , x2 , x3 : x1 = 0, x := (x2 , x3 ) ∈ R2 \I .
If the middle plane of the plate lies in the plane 0x1 x2 and the flow of moving
fluid involves bending of the plate then transmission conditions could have the form:
µ
¶
µ
¶
(+)
(−)
f
f
σN 3 x1 , x2 , h (x1 , x2 ), t − σN 3 x1 , x2 , h (x1 , x2 ), t = q(x1 , x2 , t),
(27)
µ
(+)
(+)
(+)
v3 x1 − h (x1 , x2 )w,1 (x1 , x2 , t), x2 − h (x1 , x2 )w,2 (x1 , x2 , t), h (x1 , x2 )
19
INTRODACTION
¶
µ
(−)
+w(x1 , x2 , t), t
= v3 x1 − h (x1 , x2 )w,1 (x1 , x2 , t), x2
¶
(−)
(−)
∂w(x1 , x2 , t)
− h (x1 , x2 )w,2 (x1 , x2 , t), h (x1 , x2 ) + w(x1 , x2 , t), t =
,
∂t
(28)
(the first of the last pair of equalities is valid since deflection of plate w is independent
of x3 ).
After corresponding analysis we arrive at the conclusion that, for the normal
component of the velocity vector and the pressure in the case of an ideal fluid, we
have the following transmission conditions (compare with [65], [99], [83])
v3 (x2 , 0, t) =
(−)
∂w(x2 , t)
, x2 ∈]0, l[, t ≥ 0.
∂t
(29)
(−)
→
− p(x2 , h (x2 ), t) cos(−
n (x2 , h (x2 )), x3 )
(+)
(+)
→
− p(x2 , h (x2 ), t) cos(−
n (x2 , h (x2 )), x3 ) = q(x2 , t), x2 ∈]0, l[.
(30)
In the case of a viscous fluid we add to (29) the transmission condition for the
tangential component of the velocity vector
v2 (x2 , 0, t) = 0, x2 ∈]0, l[, t ≥ 0.
(31)
In Section 3.1 the solution of the interaction problem in the case of an ideal fluid
is given [2].
For the potential motion of the flow there exists a complex function Φ = −ψ +iϕ
such that
∂ϕ(x2 , x3 , t)
∂ψ(x2 , x3 , t)
=
= v2 (x2 , x3 , t),
∂x2
∂x3
(32)
∂ϕ(x2 , x3 , t)
∂ψ(x2 , x3 , t)
=−
= v3 (x2 , x3 , t).
∂x3
∂x2
The pressure is given by the formula
·
f
p(x2 , x3 , t) = ρ
¸
2
v∞
p∞ ∂ϕ∞ ∂ϕ 1 2
2
+ f +
−
− (v2 + v3 ) .
2
ρ
∂t
∂t
2
(33)
We calculate w(x2 , t) from the equation (10).
Problem 21 Find a function w(·, t) ∈ C 4 (]0, l[) (and additional smoothness conditions indicated in Problems 11-20), also the functions v2 (x2 , x3 , t) ∈ C 2 (Ωf )∪C 1 (t >
0), v3 (x2 , x3 , t) ∈ C 2 (Ωf ) ∪ C 1 (t > 0) and p(x2 , x3 , t) ∈ C(Ωf ) ∪ C(t > 0) which
satisfy the system of equations (10), (32), (33), transmission conditions (29), (30),
conditions at infinity (24), (25) and one of the BCs given in Problems 11-20.
20
INTRODACTION
For Φ,2 (x2 , x3 , t) = v3 + iv2 , in view of (25) and (29), we get the following
expression [72]
Φ,2 = −
πi
Zl p
p
1
(x2 + ix3 )(x2 + ix3 − l)
0
(ξ2 + ix3 )(ξ2 + ix3 − l)
w,t (ξ2 , t)dξ2
(ξ2 − x2 ) − ix3
x2 + ix3 − l/2
+v3∞ p
(x2 + ix3 )(x2 + ix3 − l)
.
(34)
Let
w(x2 , t) = eiωt w0 (x2 ), q(x2 , t) = eiωt q0 (x2 ),
p(x2 , x3 , t) = eiωt p0 (x2 , x3 ),
u2 (x2 , x3 , t) = eiωt u02 (x2 , x3 ), u3 (x2 , x3 , t) = eiωt u03 (x2 , x3 ),
ϕ(x2 , x3 , t) = ieiωt ϕ0 (x2 , x3 ), ψ(x2 , x3 , t) = ieiωt ψ0 (x2 , x3 ),
v2 (x2 , x3 , t) = ieiωt v20 (x2 , x3 ), v3 (x2 , x3 , t) = ieiωt v30 (x2 , x3 ),
0
0
p∞ (t) = eiωt p0∞ , v3∞ (t) = ieiωt v3∞
, p0∞ , v3∞
= const,
where ω = const > 0 is an oscillation frequency, v2 = u2,t (v3 = u3,t ).
After separating real and imaginary parts of (34), we obtain the expressions for
v2 and v3 . By means of the latter, in view of (32), we can calculate ϕ and then
substitute it into (33). Then substituting the obtained expression for p(x2 , x3 , t)
into (30), we get the expression for q(x2 ). Therefore, all the mechanical quantities
in the fluid part and the lateral load are calculated by means of deflection. In the
case of harmonic vibration for deflection we get the second order Fredholm type
linear integral equation [2]
Zl
w0 (x2 ) − ω 2
K1 (x2 , ξ)w0 (ξ)dξ = f1 (x2 ),
(35)
0
where K1 (x2 , ξ2 ) ∈ C([0, l] × [0, l]) and f1 (x2 ) ∈ C([0, l]) are defined explicitly.
They depend on the coefficients of the equation (21), on the type of the boundary
conditions in Problems 11-20, and the conditions at infinity (24)-(25).
The following proposition is valid
Proposition 3.2 Problem of the harmonic vibration corresponding to the Problem
17 has a unique solution when
1
ω2 <
,
Ml
where
M := max {|K1 (x2 , ξ)|} .
x2 ,ξ∈[0,l]
21
INTRODACTION
Remark 2 If the plate thickness is sufficiently small, we can assume that:
1. the fluid occupies R2 \I;
2. the plate occupies I (its geometry depending on the thickness is taken into account
in the coefficient of the bending equation);
(±)
3. h can be neglected. Since the normals of I are (0, 0, 1) and (0, 0, −1), (30) can
be rewritten as follows
−p(x2 , 0+ , t) + p(x2 , 0− , t) = q(x2 , t), x2 ∈]0, l[.
In Section 3.2 an interaction problem for the case of an incompressible viscous
fluid is solved [3]. We consider the case when the motion of the fluid is sufficiently
slow, i.e., vj and vj,k (j, k = 2, 3) are so small that linearized Navier-Stokes equations
can be applied
∂v2
1 ∂p
+ ν∆v2 ,
=− f
∂t
ρ ∂x2
(36)
∂v3
1 ∂p
=− f
+ ν∆v3 ,
∂t
ρ ∂x3
where ν = µ/ρf is a coefficient of viscosity, ∆ =
We add to the (36) the following equation
∂2
∂2
+
.
∂x22 ∂x23
div v(x2 , x3 , t) = 0, (x2 , x3 ) ∈ Ωf , t ≥ 0.
Let us consider the problem of harmonic vibration.
Problem 22 Find a function w0 (x2 ) on I, which satisfies the equation (21), one
of the boundary conditions given in the Problems 11-20, and also find functions
u0i (x2 , x3 ), p0 (x2 , x3 ), q0 (x2 ) on Ωf , which satisfy the following system of equations
∆p0 (x2 , x3 ) = 0,
1 ∂p0
−ω 2 u0j = − f
+ νiω∆u0j , j = 2, 3,
ρ ∂xj
smoothness conditions
u0i ∈ C 2 (Ωf ) ∩ C(R2 ) ∩ C(t > 0), i = 2, 3;
p0 ∈ C 2 (Ωf );
q0 ,2 (·, t) ∈ C γ ([0, l]), 0 < γ ≤ 1,
following conditions at infinity
¯
p0 ||x|→∞ = O(1), u0j ¯|x|→∞ = O(1), j = 2, 3,
22
INTRODACTION
and transmissions conditions as follows
−p0 (x2 , 0+ ) + p0 (x2 , 0− ) = q0 (x2 ), x2 ∈]0, l[,
u03 (x2 , 0) = w0 (x2 ), u02 = 0, x2 ∈]0, l[,
where q(x2 , t) = eiωt q0 (x2 ).
In this case all the mechanical quantities in question are calculated by means
of the lateral load q0 , for which we obtain a second order supersingular integral
equation
Zl
0
q0 (ξ2 )
dξ2 + 2πω 2 ρf
(ξ2 − x2 )2
Zl
K2 (x2 , ξ2 )q0 (ξ2 )dξ2 = f2 (x2 ),
0
where the supersingular integral is defined in H’adamard’s finite part sense, K2 (x2 , ξ2 )
∈ C([0, l] × [0, l]) and f2 (x2 ) ∈ C([0, l]) are quite definite functions. If q0 ,2 (x2 ) ∈
C γ ([0, l]) 0 < γ ≤ 1, this equation is solved by a method developed by Boikov, I.V.,
Dobrynin, N.F., Domnin, L. in [9] (see also Chapter 1, section 1.3).
Chapter 1
Preliminary Materials
1.1.
Cusped Plates
Let 0x1 x2 x3 be the Cartesian coordinate system, and Ω be a domain in the plane
0x1 x2 with a piecewise smooth boundary.
The body bounded upper by the surface
(+)
x3 = h (x1 , x2 ) ≥ 0, (x1 , x2 ) ∈ Ω,
lower by the surface
(−)
x3 = h (x1 , x2 ) ≥ 0, (x1 , x2 ) ∈ Ω,
and from the side by a cylindrical surface parallel to the x3 -axis, will be called a
cusped plate.
(+)
(−)
2h(x1 , x2 ) := h (x1 , x2 ) − h (x1 , x2 ), (x1 , x2 ) ∈ Ω,
is the thickness of the plate.
The points P ∈ ∂Ω, at which plate thickness 2h(x1 , x2 ) = 0, will be called plate
cusps. If h ∈ C 1 (Ω), obviously,
0 ≤ L := lim
Q→P
∂2h(Q)
≤ +∞, Q ∈ Ω, P ∈ ∂Ω,
∂n
provided that the finite or infinite limit L exists; if P is an angular point of the
boundary ∂Ω under the inward to ∂Ω normal n we mean bisector of angle between
unilateral tangents to ∂Ω at P . Ω will be called a projection of the plate. ∂Ω will
be called a plate boundary. On the figures 1-3 are represented the possible normal
(+)
(−)
sections (profiles) of a symmetric plate ( h (x1 , x2 ) = − h (x1 , x2 )) at the point P in
its neighborhood.
23
24
PRELIMINARY MATERIALS
6
6
6
(+)
(+)
x3 = h
x3 = h
3́
´
´
6
P
(+)
x3 = h
-
n
?
´
´
-
P QQ
n
Q
-
P
-
n
Q
Q
s
(−)
(−)
x3 = h
L = +∞
fig.1
x3 = h
0 < L < +∞
fig.2
(−)
x3 = h
L=0
fig.3
Let us now consider an isotropic cusped plate.
The equation of the classical bending theory for an isotropic plate has the following form (see [88], p.364)
(Dw,11 ),11 + (Dw,22 ),22 +ν(Dw,22 ),11 +ν(Dw,11 ),22
+ 2(1 − ν)(Dw,12 ),12 = q(x1 , x2 ),
(1.1)
where D ∈ C 2 (Ω), and
2Eh3
,
3(1 − ν 2 )
where E is Young’s modulus and ν is Poisson’s ratio.
We recall (see [88]) that
D :=
Mα
M12
Qα
Q∗α
=
=
=
=
−(Dα ,αα +D3 w,ββ ), α 6= β, α, β = 1, 2,
−M21 = 2D4 w,12 ,
Mα,α + M12,β , α 6= β, α, β = 1, 2,
Qα + M21,β , α 6= β, α, β = 1, 2,
(1.2)
where Mα are bending moments, Mαβ , α 6= β, are twisting moments, Qα are shearing
forces and Q∗α are generalized shearing forces (bars under repeated indices mean that
we do not sum with respect to these indices).
At the points of the boundary, where the thickness vanishes, all quantities will
be defined as limits from inside of Ω.
1.2.
Hilbert-Schmidt Theorems
Recall the following three Hilbert-Schmidt theorems (see [1], [66], [70])
Theorem 1.1 If u(x2 ) has the form
Zl
u(x2 ) = λ
R(x2 , ξ)f (ξ)dξ,
0
25
1.2. HILBERT-SCHMIDT THEOREMS
with f (x2 ) piece-wise continuous on [0, l], and a symmetric kernel R(x2 , ξ) ∈ C([0, l]×
[0, l]), then
∞
X
u(x2 ) =
(u, Yn )Yn (x2 ),
(1.3)
n=1
where
Zl
(u, Yn ) :=
u(x2 )Yn (x2 )dx2 ,
0
Yn is an eigenfunction of R(x2 , ξ), and the series on the right-hand side of (1.3) is
convergent absolutely and uniformly on [0, l].
Theorem 1.2 If the number of eigenvalues λn of the symmetric kernel is finite then
R(x2 , ξ) =
N
X
Yn (x2 )Yn (ξ)
n=1
λn
.
Theorem 1.3 If f (x2 ) ∈ C([0, l]), then
Zl
R(x2 , ξ)f (ξ)dξ =
∞
X
(f, Yn )
n=1
0
λn
Yn ,
and the series is convergence absolutely and uniformly, here R(x2 , ξ) is a symmetric
kernel with respect to x2 , ξ, Yn are eigenfunctions of R corresponding to eigenvalues
λn .
Definition 1.4 Kernel R(x2 , ξ) ∈ C([0, l] × [0, l]) is called positive (negative) definite if for any f (x2 ) piecewise continuous function the following integral form
Zl Zl
J =:
R(x2 , ξ)f (x2 )f (ξ)dξdx2
0
0
is positive (negative).
Theorem 1.5 R(x2 , ξ) is a positive definite if and only if all eigenvalues λn of
R(x2 , ξ) are positive.
Theorem 1.6 If R(x2 , ξ) ∈ C([0, l] × [0, l]) is positive definite kernel, then it can
be represented as follows
R(x2 , ξ) =
∞
X
Yn (x2 )Yn (ξ)
n=1
λn
,
(1.4)
where Yn are eigenfunctions of R(x2 , ξ), and λn are eigenvalues of R(x2 , ξ). The
series in the right-hand side of (1.4) is convergent uniformly on [0, l].
26
PRELIMINARY MATERIALS
Let us consider the following integral equation
Zl
ϕ(x2 ) − λ
R(x2 , ξ)ϕ(ξ)dξ = f (x2 ),
(1.5)
0
where R(x2 , ξ) ∈ C([0, l] × [0, l]) and f (x2 ) ∈ C([0, l]).
The solution of (1.5) has the following form
ϕ(x2 ) = f (x2 ) + λ
∞
X
n=1
where
fn
Yn (x2 ),
λn − λ
Zl
f (ξ)Yn (ξ)dξ.
fn =
0
Proposition 1.7 If R(x2 , ξ) is a positive definite kernel then
Γ(x2 , ξ; λ) =
∞
X
Yn (x2 )Yn (ξ)
n=1
λn − λ
,
where Γ(x2 , ξ; λ) is a resolvent of the equation (1.5)
1.3.
Singular and Supersingular Integral Equations
Definition 1.8 We say that a function ϕ(x) satisfies the condition H(µ) (Hölder
continuous function) on [0, l], if for arbitrary points x1 , x2 ∈ [0, l] we have
|ϕ(x1 ) − ϕ(x2 )| ≤ A|x1 − x2 |µ ,
where A, µ = const > 0, 0 < µ ≤ 1. A is a coefficient, µ is an exponent of the
condition H(µ).
Let us denote
S := R2 \L,
where L = ∪Lj := aj bj , aj bj ∈ 0x2 , j = 1, p, a1 < b1 < a2 < b2 < . . . .
Problem (Dirichlet Problem). (see [72]) Find such a harmonic function u(x2 , x3 ),
which is a bounded function everywhere on S and satisfying the following conditions
u+ = f + (x2 ), u− = f − (x2 ), on L,
1.3. SINGULAR AND SUPERSINGULAR INTEGRAL EQUATIONS
27
where f + (x2 ) and f − (x2 ) are given real functions and f + (x2 ), f − (x2 ) ∈ H.
Solution. We assume that derivations of f + (x2 ) and f − (x2 ) exist and we denote
by Φ(z) the analytic function whose real part is u(x2 , x3 ),
Φ(z) := u(x2 , x3 ) + iv(x2 , x3 ),
∂u(x2 , x3 )
∂v(x2 , x3 )
Φ0 (z) :=
+i
.
∂x2
∂x2
The solution of the Problem is given by the following formulae
Z p
Z 0
R(ξ)f 0 (ξ)dξ
1
1
g (ξ)dξ c1 z p−1 + . . . + cp
0
p
Φ = p
,
+
−
ξ−ζ
πi
ξ−ζ
πi R(z)
R(z)
L
(1.6)
L
where
R(z) =
q
Y
(z − cj ), cj = {aj bj }, q = 2p,
j=1
2f (x2 ) := f + (x2 ) + f − (x2 ), 2g(x2 ) := f + (x2 ) − f − (x2 ).
Thus, from (1.6) we get
Zz
Φ0 (z)dz + c.
Φ(z) =
(1.7)
0
c1 , ..., cp , c are real constants. We define c1 from the conditions at infinity, and
c2 , ..., cp , c from the following conditions (see [72])
Zak
Φ0 (z)dz + ck = f (ak ).
Re
0
the last system is uniquely solvable.
Let ϕ0 (x) ∈ H([0, l]) and let us consider the following integral
Zl
I(x0 ) =
0
ϕ(x)dx
, n = const ≥ 2,
(x − x0 )n
which we define as H’adamard integral as follows (see [8], [3])


Z
ϕ(x)dx
2ϕ(x0 ) 
I(x0 ) = lim 
+
,
n
ε→0
(x − x0 )
ε
L∗
where L∗ := [0, l]\Q(ε, x0 ), Q(ε, x0 ) := (x0 − ε, x0 + ε).
28
PRELIMINARY MATERIALS
Let us consider the following supersingular integral equation
Z1
Z1
X(τ )
dτ +
(τ − x)2
−1
K(x, τ )X(τ )dτ = f (x),
(1.8)
−1
where K(x, τ ) ∈ C([−1, 1] × [−1, 1]), f (x) ∈ C([−1, 1]).
The approximate solution of (1.8) is given in book [8] for X 0 (x) := (dX(x)/dx) ∈
H([−1, 1]). Below we briefly state this result.
Let us divide interval [−1, 1] into N parts as follows
yk0 := −1 +
2k + 1
2k
, k = 0, N , yk := −1 +
, k = 0, N − 1,
N
2N
XN := (X(y0 ), ..., X(yN1 )),
we will call XN an approximate solution of (1.8). For XN we get the following
system of linear equations
·
¸
N
−1
X
1
1
−2N X(yi ) −
−
X(yj ) 0
yj+i − yi yj0 − yi
j=0
j6=i
+
2
N
N
−1
X
K(yi , yj )X(yj ) = f (yi ),
i = 0, N − 1.
(1.9)
j=0
The system (1.9) is uniquely solvable [8].
Let us denote by X ∗ the solution of (1.8), by XN∗ the solution of (1.9) and let X̂N∗
be a projection of X0∗ on yk . In order to get the error estimate of the approximate
solution of the equation (1.8), we consider
¯
¯
¯
¯
¯
¾¯
−1 n
o½
³
´ N
X
¯
¯
1
1
∗
∗
¯
¯−2N XN∗ (yi ) − X̂N∗ (yi ) −
−
X
(y
)
−
X̂
(y
)
j
j
N
N
0
0
¯
¯
y
−
y
y
−
y
i
i
j+i
j
¯
¯
j=0
¯
¯
j6=i
¯
¯
¯ l
¯
¯
¯Z
½
¾
N
−1
∗
X
¯
¯
1
1
X
(ξ)
∗
∗
¯
X̂
(y
)
dξ
−
2N
X̂
(y
)
+
−
= ¯¯
i
j
N
N
0
0
¯
2
y
−
y
y
−
y
i
i
j+i
j
¯
¯ (ξ − yi )
j=0
¯0
¯
j6=i
¯ 0
¯
¯
¯ 0
y
¯y
¯
¯
¯Z
−1 ¯ Zi+1 ∗
¯
¯ i+1 X ∗ (ξ) − X ∗ (y ) ¯ N
∗
X
X
(ξ)
−
X
(y
)
¯
¯
¯
¯
j
i
+
dξ
dξ
≤¯
¯
¯
¯ =: I1 + I2 .
2
¯
¯
¯
¯
(ξ − yi )2
(ξ
−
y
)
j
¯ j=0 ¯ y0
¯
¯ y0
i
j6=i
i
Therefore, since X 0 (x) ∈ H([0, l]), we have that there exist A = const > 0, and
α1 = const 0 < α1 < 1 such that
|X 0 (y1 ) − X 0 (y2 )| ≤ A|y1 − y2 |α1 .
29
1.3. SINGULAR AND SUPERSINGULAR INTEGRAL EQUATIONS
Using the following expression
0
yi+1
Z
yi0
l/(2N )
dξ
y0
= ln|ξ − yi |yi+1
= ln
= 0,
0
i
ξ − yi
l/(2N )
we obtain
I1
¯ 0
n ∗
o ¯¯
yi+1
¯Z
dX (ξ)
∗
∗
¯
¯
X (ξ) − X (yi ) − (ξ − yi )
|ξ=yi
dξ
¯
¯
= ¯
dξ
¯
2
¯
¯
(ξ − yi )
¯ y0
¯
i
¯ 0
¯
y
¯Z
¯
µ ¶−α1
¯ i+1 dX ∗ (ξ) − dX ∗ (ξ) |ξ=y ¯
N
i
¯
¯
dξ
dξ
= ¯
dξ ¯ ≤ A
,
¯
¯
ξ − yi
2
¯ y0
¯
(1.10)
i
Analogously, we get
µ
I2 ≤ A(N − 1)
N
2
¶−α1
.
(1.11)
From (1.10) and (1.11) we obtain that the error of this method might be too
large. For getting the most better results instead of the system (1.9) we consider
the following system
aii X(yi ) −
+
·
N −1
X
0
1
1
X(yj ) 0
− 0
yj+i − yi yj − yi
j=0
¸
N −1
2 X
K(yi , yj )X(yj ) = f (yi ), i = 0, N − 1.
N j=0
(1.12)
where
Z
aii := −2N
∆ii
h
dξ2
n 0
ni
0
,
∆
:=
[−1,
1]
∩
y
−
,
y
+
,
ii
i
(ξ2 − yi )2
N i+1 N
n :=
√
N
X0
:=
N
−1
X
.
j=0
j6=i−1, i, i+1
In this case, after repeating the above calculations, the error of the approximate
solution of (1.8) will be
|X ∗ − XN∗ | ≤ An−α1 ,
where X ∗ and XN∗ are the solutions of equations (1.8) and (1.12), respectively.
Chapter 2
Bending of a Cusped Plate
2.1.
Cylindrical Bending of a Cusped Plate
In this chapter we will consider a plate, whose projection on x3 = 0 occupies the
domain Ω
Ω = {(x1 , x2 , x3 ) : −∞ < x1 < ∞, 0 < x2 < l, x3 = 0}.
The equation of the cylindrical bending of plates has the following form [see,
Chapter 1, equation (1.1)]
(D(x2 )w, 22 (x2 )), 22 = q(x2 ),
0 < x2 < l.
(2.1)
In general D(x2 ) is given by the equation
2Eh3 (x2 )
D(x2 ) :=
,
3(1 − ν 2 )
(2.2)
We consider the case when E =const, ν =const, and
D(x2 ) = D0 xα2 (l − x2 )β ,
Then
D0 , α, β = const, D0 > 0, α, β ≥ 0.
(2.3)
α/3
2h(x2 ) = h0 x2 (l − x2 )β/3 , h0 = const > 0.
If α2 + β 2 > 0, equation (2.1) becomes degenerate one. Such plates are called cusped
plates.
The profile of the plate under consideration has one of the forms shown in Figures
4-12.
In case under consideration [see Chapter 1, formulas (1.2)]
M2 (x2 ) := −D(x2 )w,22 (x2 ),
Q2 (x2 ) := M2,2 (x2 ),
30
(2.4)
(2.5)
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
31
32
CHAPTER 2. BENDING OF A CUSPED PLATE
where M2 (x2 ) is a bending moment, Q2 (x2 ) is a shearing force.
Obviously, if we assume that q(x2 ) ∈ C([0, l]), w(x2 ) ∈ C 4 (]0, l[), for Q2,2 (x2 ),
M2,2 (x2 ), w,2 (x2 ), and w(x2 ) we have
Zx2
Q2 (x2 ) := −
q(ξ)dξ − c1 ,
(2.6)
x02
Zx2
M2 (x2 ) := − (x2 − ξ)q(ξ)dξ − c1 x2 − c2 ,
(2.7)
x02




ξ
ξ

Z
Zx2 
Z






w,2 (x2 ) :=
−
ηq(η)dη
+
c
+
ξ
q(η)dη
+
c
D−1 (ξ)dξ


2
1




0
0
0
x2
(2.8)
x2
x2
+ c3 ,




ξ
ξ


Z
Z
Zx2






w(x2 ) :=
(x2 − ξ) − ηq(η)dη + c2  + ξ  q(η)dη + c1  D−1 (ξ)dξ




0
0
0
x2
x2
+ c3 x2 + c4 ,
x02
x2
∈]0, l[.
(2.9)
At points 0, l all above quantities are defined as the corresponding limits when
x2 → 0+ and x2 → l− .
Obviously,
Q2 (x2 ), M2 (x2 ) ∈ C([0, l]),
w(x2 ), w,2 (x2 ) ∈ C(]0, l[),
the behavior of the w,2 (x2 ) and w(x2 ) when x2 → 0+ and x2 → l− depends, in view
of (2.8), (2.9), on α, β.
As a result of the corresponding analysis we arrive at the admissible classical
bending BVPs:
Problem 1 Let α < 1, β < 1. Find w ∈ C 4 (]0, l[) ∩ C 1 ([0, l]) satisfying (2.1) and
the following boundary conditions (BCs):
w(0) = g11 , w,2 (0) = g21 , w(l) = g12 , w,2 (l) = g22 ;
(2.10)
Problem 2 Let α < 1, β < 1. Find w ∈ C 4 (]0, l[) ∩ C 1 ([0, l]) satisfying (2.1) and
BCs:
w(0) = g11 , w,2 (0) = g21 w,2 (l) = g22 Q2 (l) = h22 ;
Problem 3 Let 0 ≤ α < 1,
satisfying (2.1) and BCs:
0 ≤ β < 2. Find w ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l])
w(0) = g11 , w,2 (0) = g21 , w(l) = g12 , M2 (l) = h12 ;
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
33
Problem 4 Let 0 ≤ α < 1, β ≥ 0. Find w ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) satisfying (2.1)
and the following BCs:
w(0) = g11 , w,2 (0) = g21 M2 (l) = h12 , Q2 (l) = h22 ;
Problem 5 Let 0 ≤ α, β < 1. Find w ∈ C 4 (]0, l[) ∩ C 1 ([0, l]) satisfying (2.1) and
the following BCs:
w,2 (0) = g21 Q2 (0) = h21 , w(l) = g12 , w,2 (l) = g22 ;
Problem 6 Let 0 ≤ α < 1, 0 ≤ β < 2. Find w ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l])
satisfying (2.1) and the following BCs:
w,2 (0) = g21 , Q2 (0) = h21 , w(l) = g12 , M2 (l) = h12 ;
Problem 7 Let 0 ≤ α < 2, 0 ≤ β < 1. Find w ∈ C 4 (]0, l[) ∩ C 1 (]0, l]) ∩ C([0, l])
satisfying (2.1) and the following BCs:
w(0) = g11 , M2 (0) = h11 , w(l) = g12 , w,2 (l) = g22 ;
Problem 8 Let 0 ≤ α < 2, 0 ≤ β < 1. Find w ∈ C 4 (]0, l[) ∩ C([0, l]) ∩ C 1 (]0, l])
satisfying (2.1) and the following BCs:
w(0) = g11 , M2 (0) = h11 , w,2 (l) = g22 , Q2 (l) = h22 ;
(2.11)
Problem 9 Let 0 ≤ α, β < 2. Find w ∈ C 4 (]0, l[) ∩ C([0, l]) satisfying (2.1) and
the following BCs:
w(0) = g11 , M2 (0) = h11 w(l) = g12 , M2 (l) = h12 ;
Problem 10 Let α ≥ 0, 0 ≤ β < 1. Find w ∈ C 4 (]0, l[) ∩ C 1 (]0, l]) satisfying
(2.1) and the following BCs:
M2 (0) = h11 , Q2 (0) = h22 w(l) = g12 , w,2 (l) = g22 .
In all these problems gij , hij (i, j = 1, 2) are given constants.
All above problems are solved explicitly. Let us solve typical ones. For the sake
of simplicity we consider homogeneous BCs.
Solution of Problem 1:
By virtue of (2.8) and homogeneous boundary conditions for w,2 we have


0
Zx2 Zξ
 −1

(2.12)
c3 =
 (η)q(η)dη + c2 + ξc1  D (ξ)dξ,
0
c3
x02

Zx2 Zξ
 −1

=
 (ξ − η)q(η)dη + c2 + ξc1  D (ξ)dξ.
l

0
x02
(2.13)
34
CHAPTER 2. BENDING OF A CUSPED PLATE
Taking into account of (2.9) and homogeneous conditions (2.10) for w, we obtain


0
Zx2 Zξ


c4 = − ξ  (ξ − η)q(η)dη + c2 + ξc1  D−1 (ξ)dξ,
(2.14)
0
x02

0
Zx2
c4 = −

ξ
l
Zξ


(ξ − η)q(η)dη + c2 + ξc1  D−1 (ξ)dξ.
(2.15)
x02
Obviously, from (2.12)-(2.15), for c1 and c2 we have the following system
Zl
Zl
−1
ξD (ξ)dξ + c2
c1
Zl
−1
D (ξ)dξ = −
0
0
Z
0
(ξ − η)D−1 (ξ)dξ =: d1 ,
(2.16)
0
Zl
ξ 2 D−1 (ξ)dξ + c2
Zl
Zl
ξD−1 (ξ)dξ = −
0
0
0
ξ(ξ − η)D−1 (ξ)dξ
q(η)dη
η
x02
Zx2
+
η
Zη
q(η)dη
Zl
c1
(ξ − η)D−1 (ξ)dξ
q(η)dη
x02
x02
+
Zl
Zη
ξ(ξ − η)D−1 (ξ)dξ =: d2 .
q(η)dη
0
(2.17)
0
The determinant of this system is equal to
2
 l
Zl
Zl

Z
−1
−1
ξD (ξ)dξ
− D (ξ)dξ ξ 2 D−1 (ξ)dξ < 0.
∆1 =


0
0
(2.18)
0
1
The last assertion follows from the Hölder inequality which is strict since ξD− 2 (ξ)
1
and D− 2 (ξ) are positive on ]0, l[, and ξ 2 D−1 (ξ) and D−1 (ξ) differ from each other
by a nonconstant factor ξ 2 .
Further,
d1
0
c1 =
ξ
dξ
ξ α (l−ξ)β
− d2
Rl
0
1
dξ
ξ α (l−ξ)β
,
∆1
d2
c2 =
Rl
Rl
0
ξ
ξ α (l−ξ)β
dξ − d1
Rl
0
∆1
ξ2
dξ
ξ α (l−ξ)β
.
35
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
After substitution the c1 and c2 into (2.12) and (2.14) we get the expressions for c3
and c4 . It is obvious, that the last integral of the expression c1 exists if and only if
α < 1, β < 1.
Solution of Problem 8: From (2.6), (2.7) and BC we get
0
Zx2
Zl
c1 = −
q(ξ)dξ, c2 = −
ξq(ξ)dξ,
(2.19)
0
x02
hence
Zl
Zx2
q(ξ)dξ,
Q2 (x2 ) =
M2 (x2 ) =
Zl
ξq(ξ)dξ + x2
x2
0
q(ξ)dξ,
x2
Substituting (2.19) in (2.8) and (2.9) and taking into account BCs, after using
Dirichlet formula we have


Z l Zξ
Zl
Zl
Zl
−1
c3 =  ηq(η)dη + ξ q(η)dη  D (ξ)dξ + ξq(ξ) ηD−1 (η)dηdξ
0
x02
ξ
0
Zx2
Zl
Zl
−1
q(ξ)
−
x02
(η − ξ)D (η)dηdξ +
0
Zx2
c4 =
x02
0
0
0
Zx2
Zl
η(η − ξ)D−1 (η)dηdξ +
q(ξ)
D−1 (η)dηdξ,
q(ξ)
Zξ
0
η 2 D−1 (η)dηdξ
q(ξ)
x02
0
0
Zx2
+
Zl
0
ξ
x02
x02
Zx2
ηD−1 (η)dηdξ,
ξq(ξ)
0
0
and
Zl
Rξ
ηq(η)dη + ξ
0
w,2 (x2 ) =
Rl
q(η)dη
ξ
dξ,
D0 ξ α (l − ξ)β
x2
w(x2 ) =
(2.20)


Zx2
Zx2
Zl
Zl
dη
dη
dη
q(ξ) 
 dξ
−x2
+
+ x2 ξ
α−1
β
α−2
β
α
D0
η (l − η)
η (l − η)
η (l − x)β
x2
Zx2
+
l

q(ξ) 
ξ
D0
ξ
0
ξ
Zx2
Zξ
Zl
dη
+
− η)β
η α−1 (l
ξ
dη
+ x2 ξ
− η)β
η α−2 (l
0

dη
 dξ.
− x)β
η α (l
ξ
36
CHAPTER 2. BENDING OF A CUSPED PLATE
It is easy to see that w(x2 ) and w,2 (x2 ) belong to C([0, l]), since
Rξ
lim
ηq(η)dη
ξq(ξ)
− ξ)β − βξ α (l − ξ)β−1
q(ξ)
= lim α−2
.
ξ→0+ ξ
[α(l − ξ)β − βξ(l − ξ)β−1 ]
0
ξ→0+ ξ α (l
−
=
ξ)β
lim
ξ→0+ αξ α−1 (l
Solution of Problem 9:
Zl
Zl
1
Q2 (x2 ) = q(ξ)dξ −
ξq(ξ)dξ,
l
x2
0
Zl
Zx2
M2 (x2 ) = x2
q(ξ)dξ +
x2
x2
ξq(ξ)dξ −
l
0
Zl
w,2 (x2 ) = −
1
R1 (ξ)
dξ +
β
− ξ)
l
Zl
w(x2 ) = −x2
R1 (ξ)
dξ −
α
ξ (l − ξ)β
Zx2
x2
x2
l
R1 (ξ)
dξ,
α−1
ξ (l − ξ)β
0
x2
+
ξq(ξ)dξ,
0
Zl
ξ α (l
Zl
Zl
R1 (ξ)
dξ
α−1
ξ (l − ξ)β
0
R1 (ξ)
dξ,
α−1
ξ (l − ξ)β
0
where
1
R1 (ξ) := − (l − ξ)
l
Zξ
ξ
ηq(η)dη −
l
0
Zl
(l − η)q(η)dη.
ξ
Solution of Problem 10:
Zx2
Q2 (x2 ) = − q(ξ)dξ,
Zx2
M2 (x2 ) = − (x2 − ξ)q(ξ)dξ,
0
Zl
0
Rξ
(ξ − η)q(η)dη
0
w,2 (x2 ) =
ξ α (l − ξ)β
dξ,
x2
Rξ
Zl
w(x2 ) = −
(x2 − ξ)
x2
(ξ − η)q(η)dη
0
ξ α (l − ξ)β
dξ,
37
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
w,2 and w are bounded as x2 → 0+ if
∃q ([α]−2) , such that lim q ([α]−2) (ξ) 6= ∞, lim q (i−3) = 0, i = 3, [α], α ≥ 3,
ξ→0+
ξ→0+
is fullfiled since
Rξ
ηq(η)dη
0
lim
ξ→0+ ξ α (l
−
ξ)β
= lim
ξ→0+ ξ α−2
q(ξ)
.
[α(l − ξ)β − βξ(l − ξ)β−1 ]
For homogeneous BCs solutions of all the above problems can be represented as
follows [14]
Zl
(2.21)
w(x2 ) = K(x2 , ξ)q(ξ)dξ,
0
where
½
K(x2 , ξ) =
K3 (ξ, x2 ), 0 ≤ ξ ≤ x2 ,
K3 (x2 , ξ), x2 ≤ ξ ≤ l.
(2.22)
K3 (x2 , ξ) has different forms for different problems, e.g.,
Problem 1.
Zx2
K3 (x2 , ξ) = (η − x2 )(η − ξ)D−1 (η)dη
 ξ
Z
+

0
Zx2
(ξ − η)D−1 dη (x2 − η)ηD−1 (η)dη
0
Zξ
+
0
Zξ
−
0
Zξ
+
0
0
 Rl
−1
Zx2
 ηD (η)dη
η(ξ − η)D−1 (η)dη (x2 − η)D−1 (η)dη 0

∆1
0
Rl −1
D (η)dη
Zx2
0
−1
−1
(ξ − η)ηD (η)dη (x2 − η)ηD (η)dη
∆1
0
Rl 2 −1
η D (η)dη
Zx2
0
−1
−1
,
(ξ − η)D (η)dη (x2 − η)D (η)dη
∆1
where ∆1 is given by (2.18).
0
(2.23)
38
CHAPTER 2. BENDING OF A CUSPED PLATE
Problem 2.
Zx2
K3 (x2 , ξ) =
(x2 − η)(ξ − η)D−1 dη
0
Zξ
1
−
Rl
(ξ − η)D−1 (η)dη
(l − η)D−1 (η)dη
0
0
Zx2
(x2 − η)D−1 (η)dη.
×
(2.24)
0
Problem 3.
Zx2
K3 (x2 , ξ) =
(x2 − η)(ξ − η)D−1 (η)dη
0
−
Zx2
(x2 − η)(l − η)D−1 (η)dη
1
Rl
(l − η)2 D−1 (η)dη
0
0
Zξ
(ξ − η)(l − η)D−1 (η)dη.
×
(2.25)
0
Problem 4.
Zx2
K3 (x2 , ξ) =
(ξ − η)(x2 − η)D−1 (η)dη.
(2.26)
0
Problem 5.
Zl
(x2 − η)(ξ − η)D−1 (η)dη
K3 (x2 , ξ) =
x2
−
1
Rl
Zl
(x2 − η)D−1 (η)dη
ηD−1 (η)dη x2
0
Zl
(ξ − η)D−1 (η)dη.
×
ξ
(2.27)
39
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
Problem 6.
Zx2
Zx2
−1
η 2 D−1 (η)dη
ηD (η)dη +
K3 (x2 , ξ) = −(l − x2 )
ξ
l
Z0
D−1 (η)dη.
+ (l − x2 )ξ
(2.28)
ξ
Problem 7.
Zl
(x2 − η)(ξ − η)D−1 (η)dη
K3 (x2 , ξ) =
x2
−
Zl
1
Rl
(x2 − η)ηD−1 (η)dη
η 2 D−1 (η)dη x2
0
Zl
(ξ − η)ηD−1 (η)dη.
×
(2.29)
ξ
Problem 8.
Zx2
Zx2
ηD−1 (η)dη +
K3 (x2 , ξ) := −x2
η 2 D−1 (η)dη
0
ξ
Zl
D−1 (η)dη.
+ x2 ξ
(2.30)
ξ
Problem 9.
x2 ξ
K3 (x2 , ξ) =
l2
Zl
x2 (l − ξ)
(l − η)D (η)dη +
l2
−1
ξ
+
Zx2
(l − η)ηD−1 (η)dη
ξ
(l − x2 )(l − ξ)
l2
Zx2
η 2 D−1 (η)dη.
(2.31)
0
Problem 10.
Zl
(x2 − η)(η − ξ)D−1 (η)dη.
K3 (x2 , ξ) = −
ξ
(2.32)
40
CHAPTER 2. BENDING OF A CUSPED PLATE
Obviously, taking into account (2.23)-(2.32), we have (see (2.22))

 C([0, l] × [0, l]), in case of Problems 1 − 3, 5 − 9;
C(]0, l]×]0, l]), in case of Problems 10;
K(x2 , ξ) ∈

C([0, l[×[0, l[), in case of Problems 4,
(2.33)
and

 C([0, l] × [0, l]), in case of Problems 1, 2, 5;
C(]0, l]×]0, l]), in case of Problems 7, 8, 10;
K 0 ,2 (x2 , ξ) ∈

C([0, l[×[0, l[), in case of Problems 3, 4, 6;
(2.34)
Remark 2.1 Problems 1-10 are not correct for the different from the indicated in
Problems 1-10 values of α and β. It is evident from the fact that in the above cases,
in general, the limits of w and w,2 as x2 → 0+ , l− do not exist. The last assertions
easily follow from the general representations (2.9) and (2.8) of w and w,2 with
(2.3).
Using integral representations and the difference equation corresponding to (2.1)
by means of MATLAB we get numerical results and corresponding graphical results
for deflection, bending moment and intersecting force for different materials. These
numerical results coincide up to 10−3 . In Figures 13-17 is shown graphical results
taking into difference equation corresponding to (2.1), In Figure 18 is shown the
graphical results for deflection using integral representation corresponding to Figures
13-17.
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
Fig. 13
.
41
42
CHAPTER 2. BENDING OF A CUSPED PLATE
Fig. 14
.
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
Fig. 15
.
43
44
CHAPTER 2. BENDING OF A CUSPED PLATE
Fig. 16
.
2.1. CYLINDRICAL BENDING OF A CUSPED PLATE
Fig. 17
.
45
46
CHAPTER 2. BENDING OF A CUSPED PLATE
2.2.
Vibration of the Plate with Two Cusped Edges
The equation of bending vibration has the following form
∂ 2 w(x2 , t)
(D(x2 )w, 22 (x2 , t)), 22 = q(x2 , t) − 2ρh(x2 )
,
∂t2
0 < x2 < l,
(2.35)
where ρ is a density of the shell.
In this case we have to add to the BCs of Problems 1-10 the initial conditions
w(x2 , 0) = ϕ1 (x2 ), w,t (x2 , 0) = ϕ2 (x2 ),
(2.36)
where ϕi (x2 ) ∈ C 4 (]0, l[), i = 1, 2, are given functions..
Let us consider the following initial boundary value problem (IBVP):
Problem 11 Let 0 ≤ α < 2, 0 ≤ β < 1. Find
w(·, t) ∈ C 4 (]0, l[) ∩ C([0, l]) ∩ C 1 (]0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0)
(2.37)
satisfying equation (2.35), the BCs
w(0, t) = M2 (0, t) = w,2 (l, t) = Q2 (l, t) = 0,
(2.38)
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C([0, l]) ∩ C 1 (]0, l]), i = 1, 2.
(2.39)
ϕi (0) = −D(x2 )ϕ00i (x2 )|x2 =0+ = ϕ0i (l) =
0
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
(2.40)
Solution. In this section all quantities, in particular, in (2.4), (2.5) depend on x2
and t.
Using the Fourier method, we will look for w(x2 , t) in the following form
w(x2 , t) = X(x2 )T (t).
(2.41)
Let firstly q(x2 , t) ≡ 0. Then from (2.35) we get
T 00 (t)
(D(x2 )X 00 (x2 ))00
=−
= λ = const.
g(x2 )X(x2 )
T (t)
Hence,
T 00 (t) + λT (t) = 0,
and
00
(D(x2 )X 00 (x2 )) = λg(x2 )X(x2 ),
(2.42)
(2.43)
2.2. VIBRATION OF THE PLATE WITH TWO CUSPED EDGES
47
where g(x2 ) := 2ρh(x2 ).
From (2.38) for X(x2 ) we obtain the following BCs
X(0) = −D(x2 )X 00 (x2 )|x2 =0 = X 0 (l) = (−D(x2 )X 00 (x2 ))0 |x2 =l = 0.
Now, in view of (2.37), we have to solve the following BVP:
Find
X(x2 ) ∈ C 4 (]0, l[) ∩ C([0, l]) ∩ C 1 (]0, l]),
(2.44)
(2.45)
which satisfies equation (2.43) and BCs (2.44).
If in (2.21) we replace w(x2 ) and q(x2 ) by X(x2 ) and λg(x2 )X(x2 ), respectively,
then, similarly to Section 2.1, for X(x2 ) we obtain
Zl
X(x2 ) = λ
g(ξ)K(x2 , ξ)X(ξ)dξ,
(2.46)
0
where
½
K(x2 , ξ) =
K3 (ξ, x2 ), 0 ≤ ξ ≤ x2 ,
K3 (x2 , ξ), x2 ≤ ξ ≤ l.
Zx2
Zx2
−1
K3 (x2 , ξ) := −x2
Zl
2
ηD (η)dη +
η D (η)dη + x2 ξ
0
ξ
D−1 (η)dη.
−1
(2.47)
ξ
Proposition 2.2 K(x2 , ξ) is symmetric with respect to x2 and ξ.
Proof For z1 and z2 , such that 0 < z1 , z2 < l we get
½
K3 (z2 , z1 ), 0 ≤ z2 ≤ z1 ,
K(z1 , z2 ) =
K3 (z1 , z2 ), z1 ≤ z2 ≤ l,
½
K(z2 , z1 ) =
K3 (z1 , z2 ), z1 ≤ z2 ≤ l,
K3 (z2 , z1 ), 0 ≤ z2 ≤ z1 ,
i.e.,
K(z1 , z2 ) = K(z2 , z1 ),
for any z1 , z2 ∈ [0, l].
¤
(2.46) can be rewritten as follows
Zl
Y (x2 ) = λ
R(x2 , ξ)Y (ξ)dξ,
(2.48)
0
where
Y (x2 ) =
p
g(x2 )X(x2 ),
R(x2 , ξ) =
p
p
g(x2 )K(x2 , ξ) g(ξ).
(2.48) is an integral equation with a symmetric kernel.
(2.49)
48
CHAPTER 2. BENDING OF A CUSPED PLATE
Proposition 2.3 Number of eigenvalues λn of (2.48) is not finite.
Proof Let it be finite, and n = 1, m. Then we can express R(x2 , ξ) as follows (see
Theorem 1.2)
m
X
Yn (x2 )Yn (ξ)
R(x2 , ξ) =
,
λ
n
n=1
where Yn (x2 ) ∈ C 4 (]0, l[), i.e.,
R(x2 , ξ) ∈ C 4 (]0, l[×]0, l[) .
(2.50)
On the other hand, by virtue of (2.47),
Kx0002 (x2 , ξ)|ξ→x2 − − Kx0002 (x2 , ξ)|ξ→x2 + =
1
,
D(x2 )
then kernel
R(x2 , ξ) 6∈ C 4 (]0, l[×]0, l[) .
(2.51)
But, (2.50) and (2.51) contradict to each other, thus the number of λn is not finite.
¤
Proposition 2.4 All of λn are positive.
Proof Obviously, if we denote by Yn orthonormalized eigenfunctions (it can be
assumed without loss of generality) of (2.48), then
Yn (x2 )
Xn (x2 ) = p
g(x2 )
are eigenfunctions of (2.46) (i.e., of (2.43)). Let us multiply both sides of the following equation
(D(x2 )Xn00 (x2 ))00 = λn g(x2 )Xn (x2 ),
(2.52)
by Xn (x2 ) and integrate it from 0 to l. Taking into account the first expression of
(2.49), we obtain
Zl
Zl
Xn (x2 )(D(x2 )Xn00 (x2 ))00 dx2
g(x2 )Xn (x2 )Xn (x2 )dx2
= λn
0
0
Zl
Yn (x2 )Yn (x2 )dx2 = λn .
= λn
0
49
2.2. VIBRATION OF THE PLATE WITH TWO CUSPED EDGES
Further,
Zl
λn =
0
¯l
¯
¯
Xn (x2 )(D(x2 )Xn00 (x2 ))00 dx2 = Xn (x2 )(D(x2 )X 00 (x2 ))0 ¯¯
¯
0
Zl
Xn0 (x2 )(D(x2 )Xn00 (x2 ))0 dx2 =
−
0
(by virtue of the BCs (2.44))
¯l
¯
Zl
¯
00
0
0
00
0
= − Xn (x2 )(D(x2 )Xn (x2 )) dx2 = Xn (x2 )(D(x2 )X (x2 ))¯¯
¯
0
0
Zl
Zl
+
D(x2 )(Xn00 )2 (x2 )dx2 = D(x2 )(Xn00 )2 (x2 )dx2 ≥ 0.
0
0
Hence, λn > 0 for any n, since in non trivial case Xn 6≡ 0. ¤
The solution of (2.42) can be written as follows
³p ´
³p ´
n
n
Tn (t) = b1 sin
λn t + b2 cos
λn t , bni = const, i = 1, 2.
Now, we can find a solution of the Problem 11 in the form as follows
∞
³p ´
³p ´´
X
Yn (x2 ) ³ n
n
p
b1 sin
λn t + b2 cos
λn t
w(x2 , t) =
g(x2 )
n=1
(2.53)
or, taking into account (2.49), in the following form
w(x2 , t) =
∞
X
³p ´´
³
³p ´
λn t + bn2 cos
λn t .
Xn (x2 ) bn1 sin
(2.54)
n=1
In view of initial conditions (2.36), we formally have
∞
X
∞ p
X
p
p
Yn (x2 )bn2 = ϕ1 (x2 ) g(x2 ),
λn Yn (x2 )bn1 = ϕ2 (x2 ) g(x2 ).
n=1
(2.55)
n=1
00 00
(Dϕ )
If ψi (x2 ) := √ i ∈ C[0, l], (i = 1, 2), then after integration of the last expresg(x2 )
p
sion, g(x2 )ϕi (x2 ) can be expressed as follows
p
g(x2 )ϕi (x2 ) =
Zl p
0
g(x2 )g(ξ)K(x2 , ξ)ψi (ξ)dξ,
50
CHAPTER 2. BENDING OF A CUSPED PLATE
i.e.,
p
Zl
g(x2 )ϕi (x2 ) =
R(x2 , ξ)ψi (ξ)dξ.
0
Hence, by virtue of Theorem 1.1, since ψi (ξ) ∈ C([0, l]) and symmetric R(x2 , ξ) ∈
C([0, l] × [0, l]), we get absolutely and uniformly convergence of the series
∞ Z
X
p
l
p
g(x2 )ϕi (x2 ) =
g(ξ)ϕi (ξ)Yn (ξ)dξ · Yn (x2 ),
n=1 0
i.e., of (2.55) on [0, l], and
bn1
1
=√
λn
Zl
Zl
g(x2 )Xn (x2 )ϕ2 (x2 )dx2 ,
bn2
g(x2 )Xn (x2 )ϕ1 (x2 )dx2 .
=
(2.56)
0
0
Further, taking into account (2.45), X(x2 ) ∈ C([0, l]). Then, by virtue of (2.49),
we can rewrite (2.55) as follows
ϕ1 (x2 ) =
∞
X
n=1
Xn (x2 )bn2 ,
ϕ2 (x2 ) =
∞ p
X
λn Xn (x2 )bn1 .
(2.57)
n=1
Evidently, last series will be absolutely and uniformly nt on ]0, l[. Since there
exists positive minimum of eigenvalues, from the convergence of the second series folN
P
Xn (x2 )bn1 . Therefore,
lows absolute and uniform convergence on ]0, l[ of the series
n=1
the series (2.54) is absolutely and uniformly convergent on ]0, l[.
After formal differentiation of (2.54) with respect to t we get
w,t (x2 , t) =
∞
´
X
p ³
p
p
Xn (x2 ) λn bn1 cos( λn t) − bn2 sin( λn t) ,
(2.58)
n=1
∞
´
³
X
p
p
w,tt (x2 , t) = − Xn (x2 )λn bn1 sin( λn t) + bn2 cos( λn t) .
(2.59)
n=1
Theorem 2.5 (2.57) and (2.54) converge absolutely and uniformly on [0, l], and
(2.58) - (2.59) converge absolutely and uniformly on any [a, b] ∈]0, l[ if
ψi (x2 )
Ψi (x2 ) := p
for i = 1, 2, are satisfying BCs 2.40
g(x2 )
(2.60)
p
00
χi (x2 ) g(x2 ) := (D(x2 )Ψ00i (x2 )) , i = 1, 2, are an integrable ones on ]0, l[ (2.61)
51
2.2. VIBRATION OF THE PLATE WITH TWO CUSPED EDGES
dj
5α
γij
,
j ϕi (x2 ) = O(x2 ), γij = const > 7 − j −
3
dx2
dj
5β
δij
x2 → 0+ ,
, x2 → l− , i = 1, 2;
j ϕi (x2 ) = O((l − x2 ) ), δij = const > 7 − j −
3
dx2
j = 2, 8).
(for this, e.g. it is sufficient that
Proof Substituting in (2.56) the function g(x2 )Xn (x2 ) found from (2.52), we get
bn1 =
1
√
λn λn
Zl
(D(x2 )Xn00 (x2 ))00 ϕ2 (x2 )dx2
0
(after integrating by parts 4-times, taking into account BCs, (2.40), (2.44), and
(2.49))


Zl


1
= √
(D(x2 )Xn00 (x2 ))0 ϕ2 (x2 )|l0 − (D(x2 )Xn00 (x2 ))0 ϕ02 (x2 )dx2

λn λn 
0


Zl


1
l
= √
−D(x2 )Xn00 (x2 )ϕ02 (x2 )|0 + D(x2 )Xn00 (x2 )ϕ002 (x2 )dx2

λn λn 
0
=
1
√
λn λn
Zl
Xn00 (x2 )D(x2 )ϕ002 (x2 )dx2
0
Zl
Xn0 (x2 )(D(x2 )ϕ002 (x2 ))0 dx2
−
0
Zl





Xn (x2 )(D(x2 )ϕ002 (x2 ))00 dx2
+

0
=
1
√
1
bn2 =
λn
=
l
Xn0 (x2 )D(x2 )ϕ00 (x2 )|0
λn λn
1
√
½
− Xn (x2 )(D(x2 )ϕ002 (x2 ))0 |l0
λn λn
1
√
λn λn
Zl
Xn (x2 )(D(x2 )ϕ002 (x2 ))00 dx2
0
Zl
λn λn
Analogously,
=
=
½
1
√
Yn (x2 )ψ2 (x2 )dx2 .
0
Zl
Yn (x2 )ψ1 (x2 )dx2 .
0
In view of (2.61), Ψi (x2 ) can be expressed as follows
Zl
Ψi (x2 ) =
0
(2.62)
p
K(x2 , ξ) g(ξ)χi (ξ)dξ, i = 1, 2,
(2.63)
52
CHAPTER 2. BENDING OF A CUSPED PLATE
and by virtue of (2.60), (2.49) we obtain
Zl
ψi (x2 ) =
R(x2 , ξ)χi (ξ)dξ, i = 1, 2.
0
According to the Theorem 1.1, the following series
∞
X
βin Yn (x2 ),
n=1
where
Zl
βin
Yn (x2 )ψi (x2 )dx2 , i = 1, 2,
=
(2.64)
0
is convergent absolutely and uniformly on ]0, l[, i.e.,
∞
X
|βin ||Yn (x2 )| < +∞, i = 1, 2.
n=1
p
By view of K(x2 , ξ) g(ξ) ∈ C([0, l] × [0, l]), there exists such M that
¯
¯
p
¯
¯
M := max ¯K(x2 , ξ) g(ξ)¯ < +∞.
0≤x2 , ξ≤l
Using (2.48), (2.63), (2.64) we have
¯
¯
¯ Zl
¯
p
¯
¯
n
n
¯
|Xn (x2 )b2 | = ¯λn K(x2 , ξ) g(ξ)Yn (ξ)b2 dξ ¯¯
¯
¯
0
¯
¯ l
¯
¯Z
p
¯
¯
n
¯
= ¯ K(x2 , ξ) g(ξ)Yn (ξ)β2 dξ ¯¯
¯
¯
0
Zl
≤
p
|K(x2 , ξ) g(ξ)||Yn (ξ)||β2n |dξ := c1n
0
On the other hand in virtue of (2.65) we obtain
∞
X
n=1
c1n
=
∞
X
Zl
c1n
n=1
Zl
≤ M
0
p
|K(x2 , ξ) g(ξ)||Yn (ξ)||β2n |dξ
0
∞
X
n=1
|Yn (ξ)||β2n |dξ ≤ M M1 l < ∞.
(2.65)
2.2. VIBRATION OF THE PLATE WITH TWO CUSPED EDGES
53
From the last two uniquality we get
|ϕ1 | ≤
∞
X
|Xn (x2 )bn2 |
≤
n=1
∞
X
c1n < +∞.
n=1
Which means that ϕ1 can be expressed as absolutely and uniformly convergent
series. Analoguously, we can prove that ϕ2 converges absolutely and uniformly on
[0, l].
Let, now consider (2.54) series. It is obviously that
|w(x2 , t)| ≤
∞
X
|Xn (x2 )bn1 |
+
∞
X
|Xn (x2 )bn2 |,
n=1
n=1
and from the convergent of ϕ1 and ϕ2 we obtain that (2.54) converges absolutely
and uniformly on [0, l].
Further, from (2.58)
¯
¯
∞
¯X
³
´¯
p
p
p
¯
¯
|w,t (x2 , t)| = ¯ Xn (x2 ) λn bn1 cos( λn t) − bn2 sin( λn t) ¯
¯ n=1
¯
¯
¯
∞
¯X
¯
p
p
¯
¯
≤ ¯ Xn (x2 ) λn bn1 cos( λn t)¯
¯ n=1
¯
¯
¯
∞
¯
¯X
p
p
¯
¯
+ ¯ Xn (x2 ) λn bn2 sin( λn t)¯
¯
¯ n=1
∞ ¯
∞ ¯
¯ X
¯
X
p
p
¯
¯
¯
n¯
≤
(2.66)
¯Xn (x2 ) λn b1 ¯ +
¯Xn (x2 ) λn bn2 ¯ .
n=1
n=1
According to Proposition 2.4, all of λn are positive. Therefore, we can find λ0 such
that λ0 ≤ min {λi }, and by virtue of (2.49), (2.62)-(2.65), we obtain
1≤i≤∞
¯
∞ ¯
∞ ¯
¯
X
p
1 X ¯¯ p 1 n ¯¯
¯
n¯
¯Xn (x2 ) λn b2 ¯ = p
¯Yn λn λn β1 ¯
g(x
)
2
n=1
n=1
∞
1
1 X
≤ √ p
|Yn ||β1n | < ∞,
λ0 g(x2 ) n=1
¯
∞ ¯
∞ ¯
¯
X
p
¯
1 X ¯¯ p
1
¯
n¯
n¯
Yn λn √ β2 ¯
¯Xn (x2 ) λn b1 ¯ = p
¯
λ λ
g(x )
2 n=1
n=1
≤
n
n
∞
X
1
1
p
|Yn ||β2n | < ∞, x2 ∈]0, l[.
λ0 g(x2 ) n=1
Hence, the series in (2.66) are convergent. Thus, (2.58) is convergent absolutely and
uniformly on ]0, l[. Similarly, we get the absolute and uniform convergence of (2.59)
on ]0, l[. ¤
54
CHAPTER 2. BENDING OF A CUSPED PLATE
Let us now differentiate (2.54) formally i-times with respect to x2 and consider
the following expressions
∞
´
³
X
p
p
∂i
di
n
n
w(x
,
t)
=
X
(x
)
b
sin(
λ
t)
+
b
cos(
λ
t)
,
n
n
2
n 2
1
2
i
∂xi2
dx
2
n=1
i = 1, 2, 3, 4,
³
X di−1
p
∂ i−1
00
n
(D(x
)w,
(x
,
t))
=
(D(x
)X
(x
))
b
sin(
λn t)
2
x2 x2
2
2
2
n
1
i−1
∂xi−1
dx
2
2
n=1
(2.67i )
∞
(2.68i )
¢
√
+bn2 cos( λn t) , i = 1, 2.
Theorem 2.6 The series (2.67i ) (i = 1, ..., 4) are convergent absolutely and uniformly on any [a, b] ∈]0, l[. The series (2.68i ) (i = 1, 2) are convergent absolutely
and uniformly on [0, l].
Proof Obviously, in view of (2.44), after integration of (2.52), we get
Zl
Xn0 (x2 ) = λn
R1 (x2 , ξ)Xn (ξ)dξ,
(2.69)
0
where

Zl





ξ D−1 (η)dη, 0 ≤ ξ ≤ x2 ,



R1 (x2 , ξ) =
x2
Zl
Zx2



−1

− ηD (η)dη + ξ D−1 (η)dη, x2 ≤ ξ ≤ l,




ξ
ξ
and
R1 (x2 , ξ) ∈ C([0, l] × [0, l]),
(2.70)
because of 0 ≤ α < 2, 0 ≤ β < 1.
Substituting (2.69) into (2.711 ) for i = 1, we obtain
∞
X
∂
w(x2 , t) =
λn
∂x2
n=1
Zl
=
R1 (x2 , ξ)
0
Zl
³
´
p
p
R1 (x2 , ξ)Xn (ξ)dξ bn1 sin( λn t) + bn2 cos( λn t) =
0
"∞
X
n=1
³
Xn (ξ)λn
bn1
sin(
p
λn t) +
bn2
cos(
#
´
λn t) dξ,
p
(2.71)
55
2.2. VIBRATION OF THE PLATE WITH TWO CUSPED EDGES
since (2.59) is absolutely and uniformly convergent on ]0, l[ and in view of (2.70) and
Xn (x2 ) ∈ C([0, l]) we conclude that the corresponding integral in (2.71) is absolutely
convergent on ]0, l[. Similarly, we can prove the convergence of the series (2.672 ),
(2.673 ), (2.674 ), on ]0, l[ and (2.68i ) (i = 1, 2) on [0, l]. ¤
Thus, (2.53) is the solution of the Problem 11 for q(x2 , t) ≡ 0.
Now, let us consider Problem 11 when q(x2 , t) 6≡ 0, ϕi = 0, i = 1, 2, and let
q
√ (·, t) ∈ L2 (0, l). Then q(x2 , t) can be represented as a convergent series in L2 (0, l):
g
∞
X
q(x , t)
p 2
=
g(x2 ) n=1
hence,
Ã
q(x , t)
p 2 , Yn
g(x2 )
!
∞
X
√
Yn =
(q, Xn )Xn g,
n=1
Z
∞
X
q(x2 , t) =
g(x2 )Xn (x2 )qn (t), qn (t) := q(x2 , t)Xn (x2 )dx2 .
l
n=1
0
Further, we look for the solution in the form
∞
X
w(x2 , t) =
wn (x2 , t),
n=1
where wn (x2 , t) is a solution of the Problem 11 with q(x2 , t) replaced by g(x2 )Xn (x2 )qn (t).
Using the method of separation of variables, we can write
wn (x2 , t) = Xn (x2 )T1n (t),
where
00
T1n
(t) + λn T1n (t) = qn (t)
and Xn (x2 ) satisfies (2.46).
Therefore, w(x2 , t) can be expressed as follows
Z
∞
X
p
1
√ Xn sin( λn (t − τ ))qn (τ )dτ.
w(x2 , t) =
λn
n=1
t
(2.72)
0
Now, similarly to the proofs of Theorems 2.5 and 2.6, if the following conditions
are fulfilled
Ã
!
µ
¶
q(x2 , t)
1
D(x2 )
τ (x2 , t) := p
∈ C[0, l],
g(x2 ) ,x2 x2
g(x2 )
,x2 x2
τ
√ (0, t) = −D(x2 )
g
Ã
¯
¯
¯
τ (x2 , t)
¯
p
g(x2 ) ,x x ¯¯
2 2
Ã
!
=
x2 =0+
! ¯¯
¯
τ (x , t)
¯
p 2
g(x2 ) ,x ¯¯
2
x2 =l
(2.73)
56
CHAPTER 2. BENDING OF A CUSPED PLATE

 ¯
¯
¯
τ (x2 , t)
 ¯
p
¯
g(x2 ) ,x x
¯
2 2
Ã
!
= −D(x2 )
= 0,
,x2 x2 =l−
(for this, e.g., it is sufficient that
∂j
q(x2 , t)
∂xj2
∂j
q(x2 , t)
∂xj2
γ
= O(x2j ) x2 → 0+ , γj > 7 − j −
= O((l − x2 )δj ) x2 → l− , γj > 7 − j −
2β
,
3
2α
,
3
j = 0, 8) we have the absolute
and uniform convergence of the series (2.72) and
∞
X di
∂i
(D(x
)w
(x
,
t))
=
(D(x2 )Xn00 )(x2 )T1n (t), i = 0, 1,
2
,x
x
2
2 2
i
∂xi2
dx
2
n=1
on [0, l], and the absolute and uniform convergence of the series
∞
X di
∂i
w(x
,
t)
=
Xn (x2 )T1n (t), i = 1, ..., 4,
2
∂xi2
dxi2
n=1
∞
X
∂i
di
w(x2 , t) =
Xn (x2 ) i T1n (t), i = 1, 2,
∂ti
dt
n=1
on any [a, b] ∈]0, l[.
Remark 2.7 Let q(x2 , t), ϕi (x2 ) 6≡ 0. If conditions (2.60), (2.61) and (2.73) are
satisfying then the solution of the Problems 1-10 can be expressed as follows
w(x2 , t) =
∞
X
wn (x2 , t),
n=1
where
wn (x2 , t) = Xn (x2 )(T1n (t) + Tn (t)),
Xn (x2 )T1n (t) is given by the formula (2.72) and Xn (x2 )Tn (t) is given by the formula
(2.54).
Remark 2.8 Similarly, we can solve the following initial boundary value problems
which correspond to the Problems 1-7, 9, 10.
Problem 12 Let 0 ≤ α, β < 1. Find a function w(x2 , t), which satisfies following
smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (2.35), the boundary conditions (BCs)
w(0, t) = w,2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
ϕi (0) = ϕ0i (0) = ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
2.2. VIBRATION OF THE PLATE WITH TWO CUSPED EDGES
57
Problem 13 Let 0 ≤ α, β < 1. Find a function w(x2 , t), which satisfies following
smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (2.35), the BCs
w(0, t) = w,2 (0, t) = w,2 (l, t) = Q2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
0
ϕi (0) = ϕ0i (0) = ϕ0i (l) = (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 14 Let 0 ≤ α, < 1, 0 ≤ β < 3. Find a function w(x2 , t), which satisfies
following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (2.35), the BCs
w(0, t) = w,2 (0, t) = w(l, t) = M2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]),
ϕi (0) = ϕ0i (0) = ϕi (l) = (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 15 Let 0 ≤ α < 1, β ≥ 0. Find a function w(x2 , t), which satisfies
following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 < l, t ≥ 0),
equation (2.35), the BCs
w(0, t) = w,2 (0, t) = M2 (l, t) = Q2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[),
ϕi (0) = ϕ0i (0) = (−D(x2 )ϕ00i (x2 )) |x2 =l−
0
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
58
CHAPTER 2. BENDING OF A CUSPED PLATE
Problem 16 Let 0 ≤ α, β < 1. Find a function w(x2 , t), which satisfies following
smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (2.35), the BCs
w,2 (0, t) = Q2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l]),
0
ϕ0i (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
Problem 17 Let 0 ≤ α < 1, 0 ≤ β < 3. Find a function w(x2 , t), which satisfies
following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (2.35), the BCs
w,2 (0, t) = Q2 (0, t) = w(l, t) = M2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 ([0, l[) ∩ C([0, l]),
0
ϕ0i (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l)
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 18 Let 0 ≤ α < 3, 0 ≤ β < 1. Find a function w(x2 , t), which satisfies
following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (2.35), the BCs
w(0, t) = M2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]) ∩ C([0, l]),
ϕi (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
59
2.3. HARMONIC VIBRATION
Problem 19 Let 0 ≤ α, β < 3. Find a function w(x2 , t), which satisfies following
smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C([0, l]), M2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 ≤ x2 ≤ l, t ≥ 0),
equation (2.35), the BCs
w(0, t) = M2 (0, t) = w(l, t) = M2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C([0, l]),
ϕi (0) = (−D(x2 )ϕ00i (x2 )) |x2 =0+ = ϕi (l)
= (−D(x2 )ϕ00i (x2 )) |x2 =l− = 0, i = 1, 2.
Problem 20 Let α ≥ 0, 0 < β < 1. Find a function w(x2 , t), which satisfies
following smoothness conditions
w(·, t) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]), M2 (·, t) ∈ C([0, l]), Q2 (·, t) ∈ C([0, l]),
w(x2 , ·) ∈ C 1 (t ≥ 0) ∩ C 2 (t > 0), w(x2 , t) ∈ C(0 < x2 ≤ l, t ≥ 0),
equation (2.35), the BCs
M2 (0, t) = Q2 (0, t) = w(l, t) = w,2 (l, t) = 0, t > 0,
and ICs (2.36), where
ϕi (x2 ) ∈ C 4 (]0, l[) ∩ C 1 (]0, l]),
0
(−D(x2 )ϕ00i (x2 )) = (−D(x2 )ϕ00i (x2 )) |x2 =0+
= ϕi (l) = ϕ0i (l) = 0, i = 1, 2.
In all these cases we get integral equations with symmetric kernels.
2.3.
Harmonic Vibration
We can avoid the restrictions (2.73) if we consider the problem of harmonic vibration.
In this case
w(x2 , t) = eiω t w0 (x2 ), q(x2 , t) = eiω t q0 (x2 ),
where ω = const is an oscillation frequency, q0 (x2 ) ∈ C([0, l]) is a given function.
Now, for w0 (x2 ) from (2.35) we get the following equation
(D(x2 )w000 (x2 ))
00
= q0 (x2 ) + 2ω 2 ρh(x2 )w0 (x2 ),
(2.74)
60
CHAPTER 2. BENDING OF A CUSPED PLATE
which we solve under the above BVPs (see problems 1-10), where we replace w(x2 )
and w0 (x2 ) by w0 (x2 ) and w00 (x2 ). For bending moment and intersecting force we
obtain
M2 (x2 ) = −D(x2 )w0 ,22 , Q2 (x2 ) = M2 ,2 (x2 ).
All these problems are equivalent to the following integral equation (which we get
from (2.21) after replacing w(x2 ) and q(x2 ) by w0 (x2 ) and q0 (x2 ) + 2ω 2 ρh(x2 )w0 (x2 )
respectively),
Zl
2
w0 (x2 ) − ω
K(x2 , ξ) g(ξ) w0 (ξ)dξ = Φ(x2 ),
(2.75)
0
where
Zl
Φ(x2 ) :=
K(x2 , ξ) q0 (ξ)dξ.
(2.76)
0
Introducing a new unknown function
p
w1 (x2 ) = w0 (x2 ) g(x2 )
(2.77)
we can reduce (2.75) to the following integral equation
Zl
w1 (x2 ) − ω 2
p
R(x2 , ξ) w1 (ξ)dξ = Φ(x2 ) g(x2 )
(2.78)
0
where R(x2 , ξ) is given by (2.49)
Further, in view of (2.52) we have
Zl
Xn (x2 ) = λn
g(ξ)K(x2 , ξ)Xn (ξ)dξ.
(2.79)
0
If ω 2 6= λn , the unique solution of (2.78) can be written as follows (see, e.g., [66],
Theorem XVIII, p.157)
p
w1 (x2 ) = Φ(x2 ) g(x2 )


Zl
∞
X
p
1

+ ω2
Φ(ξ) g(ξ) Yn (ξ)dξ  Yn (x2 ),
2
λ
−
ω
n
n=1
(2.80)
0
where the series in the right hand side of (2.80) is absolutely and uniformly convergent on [0, l].
61
2.3. HARMONIC VIBRATION
After substituting (2.80) into (2.77) we formally have
w0 (x2 ) = Φ(x2 )
+ ω2
∞
X
n=1


1
λn − ω 2

Zl
Φ(ξ)
p
g(ξ) Yn (ξ)dξ  Xn (x2 ).
(2.81)
0
We have to prove that (2.81) is a solution of (2.74) under homogeneous BCs.
Let differentiate (2.81) formally i-times with respect to x2 and consider the following expressions
di
di
w
(x
)
=
Φ(x2 )
0 2
dxi2
dxi2


Zl
∞
X
p
1
di
2


+ω
Φ(ξ) g(ξ) Yn (ξ)dξ
Xn (x2 ),
λn − ω 2
dxi2
n=1
(2.82)
0
i = 1, ..., 4
Proposition 2.9 The series on the right hand side of (2.81) and (2.82) are absolutely and uniformly convergent on ]0, l[.
Proof Let denote by
1
Sn (x2 ) :=
λn − ω 2
Zl
Φ(ξ)
p
g(ξ) Yn (ξ)dξXn (x2 ).
0
Taking into account (2.79), (2.49) we have
1
1
Sn (x2 ) = p
g(x2 ) λn − ω 2
Zl
Φ(ξ)
0
Rl
1
λn
= p
g(x2 ) λn − ω 2
p
g(ξ) Yn (ξ)dξ · Yn (x2 )
Φ(ξ)
p
g(ξ) Yn (ξ)dξ · Yn (x2 )
0
λn
(2.83)
According to Proposition 2.3, the number of eigenvalues is not finite, that means
λn → ∞ when n → ∞, and further
1
λn
=
2 → 1.
2
λn − ω
1 − λωn
(2.84)
62
CHAPTER 2. BENDING OF A CUSPED PLATE
Zl
In view of Φ(x2 ) :=
K(x2 , ξ) q0 (ξ)dξ and (2.33) we obtain, that the following
0
series
Rl
∞
X
Φ(ξ)
p
g(ξ) Yn (ξ)dξ · Yn (x2 )
0
;
λn
n=1
(2.85)
is absolutely and uniformly convergent on ]0, l[.
Further, in view of (2.83)-(2.85) and (2.33) we get, that
∞
X
n=1
Sn (x2 ) =
∞
X
n=1
λn
λn − ω 2
Zl p
g(η)K(x2 , η)
0
 l

Z
p
×  Φ(ξ) g(ξ) Yn (ξ)dξ  Yn (η)dη
0
=
Zl p
∞
X
0
n=1
g(η)K(x2 , η)
λn
λn − ω 2

 l
Z
p
×  Φ(ξ) g(ξ) Yn (ξ)dξ  Yn (η)dη
(2.86)
0
is also absolutely and uniformly convergent on ]0, l[, i.e., w0 (x2 ) ∈ C(]0, l[).
Analogously, we obtain
Zl p
∞
X
di
di
∂i
λn
w (x ) =
Φ+
g(η) i K(x2 , η)
i 0 2
i
dx2
dx2
∂x2
λ − ω2
n=1 n
0
 l

Z
p
×  Φ(ξ) g(ξ) Yn (ξ)dξ  Yn (η)dη, i = 1, ..., 4.
0
Because of
p
∂i
g(η) i K(x2 , η) ∈ C(]0, l[×]0, l[),
∂x2
we get Φ(i) ∈ C(]0, l[) and w(i) ∈ C(]0, l[), i = 1, ..., 4. ¤
(2.87)
63
2.3. HARMONIC VIBRATION
Proposition 2.10
 1
C ([0, l]),




C 1 ([0, l[) ∩ C([0, l]),


 1
C (]0, l]) ∩ C([0, l]),
w0 (x2 ) ∈
C([0, l]),




C(]0, l]),



C([0, l[),
in
in
in
in
in
in
case
case
case
case
case
case
of
of
of
of
of
of
Problems 1, 2, 5;
Problems 3, 6;
Problems 7, 8;
Problem 9;
Problem 10;
Problem 4,
(2.88)
because of q0 (x2 ) ∈ C([0, l]).
Proof. (2.81) can be rewritten as follows
Rl
ω
w0 (x2 ) = Φ(x2 ) + p
∞
X
2
g(x2 )
n=1
λn
λn − ω 2
p
Φ(ξ) g(ξ)Yn (ξ)dξ
0
λn
Yn (x2 ).
(2.89)
Taking into account of (2.84), and Theorem 1.3 we have
Rl
1
p
g(x2 )
∞
X
n=1
p
Φ(ξ) g(ξ)Yn (ξ)dξ · Yn (x2 )
0
√
λn
= (because of Φ g is a continious function on [0, l])
Zl
Zl
p
1
= p
R(x2 , ξ)Φ(ξ) g(ξ)dξ = K(x2 , ξ)Φ(ξ)g(ξ)dξ
g(x2 )
0
 1 0
C
([0,
l]),
in
case
of
Problems 1, 2, 5;

 1


C ([0, l[) ∩ C([0, l]), in case of Problems 3, 6;


 1
C (]0, l]) ∩ C([0, l]), in case of Problems 7, 8;
∈
C([0, l[),
in case of Problem 10;




C(]0, l]),
in case of Problem 4;



C([0, l]),
in case of Problem 9.
According to (2.90), (2.89) we have (2.88). ¤
Similarly, it can be proved that the following series
di
di
00
(x
))
=
(D(x
)w
(D(x2 )F 00 (x2 ))
2
2
0
dxi2
dxi2


l
Z
∞
X
p
1

+ ω2
Φ(ξ)
g(ξ)Yn (ξ)dξ 
2
λ
−
ω
n
n=1
0
di
(D(x2 )Xn00 (x2 )), i = 1, 2
×
i
dx2
(2.90)
64
CHAPTER 2. BENDING OF A CUSPED PLATE
are absolutely and uniformly convergent in [0, l].
So, we obtaine that the formal solution (2.81) is a solution of (2.74) under BCs
1-10.
Using the difference equation corresponding to (2.74) by means of MATLAB we
get numerical and graphic (Figures 18-21) results for harmonic vibration problems.
65
2.3. HARMONIC VIBRATION
Problem 10, Aluminium,
α = 0.3, β = 0.3
l = 2π, q(x2 , t) = cos(x2 ) ∗ cos(t), t ∈ [0, 2π]
w0 (x2 )cos(t)
Fig. 18
66
CHAPTER 2. BENDING OF A CUSPED PLATE
M2 (x2 )cos(t)
Fig. 19
67
2.3. HARMONIC VIBRATION
Problem 8, Iron,
α = 1, β = 0.3
l = 2, q(x2 , t) = const ∗ cos(t), t ∈ [0, 2π]
w0 (x2 )cos(t)
Fig. 20
68
CHAPTER 2. BENDING OF A CUSPED PLATE
M2 (x2 )cos(t)
Fig. 21
Chapter 3
A Cusped Elastic Plate-Fluid
Interaction Problem
Let us consider the interface problem of the interaction of a plate, whose variable
flexural rigidity is given by the equation (2.3), and of a flow of fluid. Let the flow
be independent of x1 , parallel to the plane 0x2 x3 , i.e. v1 ≡ 0, and generates a
bending of the plate. Let at infinity, for pressure we have
p(x2 , x3 , t) → p∞ (t), when |x| → ∞,
(3.1)
and let for the velocity components conditions at infinity be either
v2 (x2 , x3 , t) = O(1), v3 (x2 , x3 , t) → v3∞ (t), when |x| → ∞
(3.2)
or
vj (x2 , x3 , t) = O(1), j = 2, 3, when |x| → ∞
(3.3)
where v := (v2 , v3 ) is a velocity vector of the fluid, p(x2 , x3 , t) is a pressure, and
v3∞ (t), p∞ (t) are given functions.
In what follows we suppose that the plate is so thin that, we can assume: the fluid
occupies the whole space R3 but the middle plane Ω of the plate.
Let,
I := {[0, l] × 0},
©
ª
Ωf := x1 , x2 , x3 : x1 = 0, x := (x2 , x3 ) ∈ R2 \I .
If the middle plane of the plate lies in the plane 0x1 x2 and the flow of moving fluid
involves bending of the plate then transmission conditions have the form:
µ
¶
µ
¶
(+)
(−)
f
f
σN 3 x1 , x2 , h (x1 , x2 ), t − σN 3 x1 , x2 , h (x1 , x2 ), t = q(x1 , x2 , t),
69
70 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
µ
(+)
(+)
(+)
v3 x1 − h (x1 , x2 )w,1 (x1 , x2 , t), x2 − h (x1 , x2 )w,2 (x1 , x2 , t), h (x1 , x2 )
¶
µ
(−)
+w(x1 , x2 , t), t = v3 x1 − h (x1 , x2 )w,1 (x1 , x2 , t), x2
¶
(−)
(−)
∂w(x1 , x2 , t)
− h (x1 , x2 )w,2 (x1 , x2 , t), h (x1 , x2 ) + w(x1 , x2 , t), t =
,
∂t
(3.4)
(the first of the last pair of equalities is valid since deflection of plate w is
independent of x3 ),
Because of incompressibility we have
div v(x2 , x3 , t) = 0, (x2 , x3 ) ∈ Ωf , t ≥ 0,
f
σjk
and (see e.g., [23], p.5)
µ
¶
∂vj
∂vk
= −pδjk + µ
+
, j, k = const = 2, 3,
∂xk ∂xj
(3.5)
(3.6)
f
where σjk
is a stress tensor, µ is a coefficient of viscosity, δjk is Kroneker delta. In
case of ideal fluid µ = 0.
From (3.5) and (3.6) we obtain
∂v3 (x2 , x3 , t)
∂x3
∂v2 (x2 , x3 , t)
.
= −p(x2 , x3 , t) − 2µ
∂x2
f
σ33
(x2 , x3 , t) = −p(x2 , x3 , t) + 2µ
(3.7)
In case of ideal fluid in virtue of (3.7) we get
(±)
±
f
(x2 , h (x2 ), t) = p(x2 , h(x2 ), t).
σ33
Therefore, the transmission condition for p has the following form
(−)
(−)
→
− p(x2 , h (x2 ), t) cos(−
n (x2 , h (x2 )), x3 )
(+)
(+)
→
− p(x2 , h (x2 ), t) cos(−
n (x2 , h (x2 )), x3 ) = q(x2 , t), x2 ∈]0, l[.
(3.8)
Remark. If the plate thickness is sufficiently small, we can assume that:
1. the fluid occupies R2 \I;
2. the plate occupies I (its geometry depending on the thickness is taken into
account in the coefficient of the bending equation);
(±)
3. h can be neglected. Since the normals of I are (0, 0, 1) and (0, 0, −1), (3.8) can
be rewritten as follows
−p(x2 , 0+ , t) + p(x2 , 0− , t) = q(x2 , t), x2 ∈]0, l[.
(3.9)
71
3.1. CASE OF AN IDEAL FLUID
4.Further, using (3.4), we can write transmission conditions for v3 (x2 , x3 , t) in the
following form (see [65], [98], [82])
v3 (x2 , 0, t) =
∂w(x2 , t)
, x2 ∈]0, l[, t ≥ 0.
∂t
(3.10)
In case of a viscous fluid we add to (3.10) the transmission condition for
v2 (x2 , x3 , t)
v2 (x2 , 0, t) = 0, x2 ∈]0, l[, t ≥ 0.
(3.11)
In virtue of (3.7) and (3.11)
f
σ33
(±)
(x2 , 0, t) = p± (x2 , 0, t).
Further, taking into account of smallness of the thickness, in case of viscous fluid
we rewrite transmission conditions for p can be (3.9)
3.1.
Case of an Ideal Fluid
For the potential motion of the flow there exists a complex function Φ = −ψ + iϕ
such that
∂ϕ(x2 , x3 , t)
∂ψ(x2 , x3 , t)
=
= v2 (x2 , x3 , t),
∂x2
∂x3
(3.12)
∂ϕ(x2 , x3 , t)
∂ψ(x2 , x3 , t)
=−
= v3 (x2 , x3 , t).
∂x3
∂x2
The pressure is given by the formula
¸
· 2
p∞ ∂ϕ∞ ∂ϕ 1 2
f v∞
2
+ f +
−
− (v2 + v3 ) .
p(x2 , x3 , t) = ρ
2
ρ
∂t
∂t
2
(3.13)
In case under consideration w(x2 , t) is given by the equation (2.35).
Taking into account transmission condition (3.8), we have
¡ α
¢
2h(x2 )ρs
w,tt (x2 , t)+
x2 (l − x2 )β w,22 (x2 , t) ,22 = −
D0
(−)
(−)
(+)
(3.14)
(+)
→
→
p(x2 , h (x2 ), t) cos(−
n (x2 , h (x2 )), x3 ) + p(x2 , h (x2 ), t) cos(−
n (x2 , h (x2 )), x3 )
.
−
D0
72 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
Problem 21 Find a function w(·, t) ∈ C 4 (]0, l[) (satisfying additional smoothness
conditions indicated in Problems 11-20), also the functions v2 (x2 , x3 , t) ∈ C 2 (Ωf ) ∪
C 1 (t > 0), v3 (x2 , x3 , t) ∈ C 2 (Ωf ) ∪ C 1 (t > 0) and p(x2 , x3 , t) ∈ C(Ωf ) ∪ C(t > 0)
which satisfy the system of equations (3.12), (3.13), (3.14), transmission conditions
(3.10), (3.8), conditions at infinity (3.1), (3.2) and one of the BCs given in Problems
11-20.
For Φ,2 (x2 , x3 , t) = v3 + iv2 , in view of (3.10) and (3.2), we get the following
expression (see [72])
Φ,2 = −
πi
Zl p
1
p
(x2 + ix3 )(x2 + ix3 − l)
+v3∞ (t) p
(ξ2 + ix3 )(ξ2 + ix3 − l)
w,t (ξ2 , t)dξ2
(ξ2 − x2 ) − ix3
0
x2 + ix3 − l/2
(x2 + ix3 )(x2 + ix3 − l)
.
(3.15)
Let
w(x2 , t) = eiωt w0 (x2 ), q(x2 , t) = eiωt q0 (x2 ),
(3.16)
p(x2 , x3 , t) = eiωt p0 (x2 , x3 ),
(3.17)
u2 (x2 , x3 , t) =
eiωt u02 (x2 , x3 ),
u3 (x2 , x3 , t) =
eiωt u03 (x2 , x3 ),
where ω = const > 0, v2 = u2,t (v3 = u3,t ). Further,
ϕ(x2 , x3 , t) = ieiωt ϕ0 (x2 , x3 ), ψ(x2 , x3 , t) = ieiωt ψ0 (x2 , x3 ),
v2 (x2 , x3 , t) = ieiωt v20 (x2 , x3 ), v3 (x2 , x3 , t) = ieiωt v30 (x2 , x3 ),
0
0
p∞ (t) = eiωt p0∞ , v3∞ (t) = ieiωt v3∞
, p0∞ , v3∞
= const.
From (3.15), we have expressions for v2 and v3 as follows
1
v2 (x2 , x3 , t) = −
π
Zl
R1 (ξ, x2 , x3 )w,t (ξ, t)dξ + v3∞ (t)R3 (x2 , x3 )
0
v3 (x2 , x3 , t) =
1
π
Zl
R2 (ξ, x2 , x3 )w,t (ξ, t)dξ + v3∞ (t)R4 (x2 , x3 ),
0
73
3.1. CASE OF AN IDEAL FLUID
where
p
r(ξ, x3 )
R1 (ξ, x2 , x3 ) = p
r(x2 , x3 )
(x2 − ξ) cos[(φ(ξ, x3 ) − φ(x2 , x3 ))/2] + x3 sin[(φ(ξ, x3 ) − φ(x2 , x3 ))/2]
,
×
(ξ − x2 )2 + x23
p
r(ξ, x3 )
R2 (ξ, x2 , x3 ) = p
r(x2 , x3 )
(x2 − ξ) sin[(φ(ξ, x3 ) − φ(x2 , x3 ))/2] + x3 cos[(φ(ξ, x3 ) − φ(x2 , x3 ))/2]
×
,
(ξ − x2 )2 + x23
½
¾
φ(x2 , x3 )
φ(x2 , x3 )
1
p
R3 (x2 , x3 ) = (x2 − l/2) cos
,
+ x3 sin
2
2
r(x2 , x3 )
½
¾
φ(x2 , x3 )
φ(x2 , x3 )
1
p
R4 (x2 , x3 ) = (x2 − l/2) sin
+ x3 cos
,
2
2
r(x2 , x3 )
here φ(x2 , x3 ) is defined by either
cosφ(x2 , x3 ) = (x22 − x23 − lx2 )/r(x2 , x3 )
or
sinφ(x2 , x3 ) = (2x2 − l)x3 /r(x2 , x3 )
q
r(x2 , x3 ) =
and
(x22 − x23 − lx2 )2 + ((2x2 − l)x3 )2 .
By means of the latter, in view of (3.12), we can calculate ϕ which we have to
substitute in (3.13)
ρf
p(x2 , x3 , t) =
π
Zl
Zx2
Zx2
f
w,tt (ξ, t) R1 (ξ, x2 , x3 )dx3 dξ + v3∞ (t)ρ R3 (x2 , x3 )dx3
·0
0
0
¸
2
p
(t)
∂ϕ
(t)
v
(t)
∞
∞
+
+ ρf ∞ +
2
ρf
∂t

2

Zl
f 
ρ
 1 R1 (ξ, x2 , x3 )w,t (ξ, t)dξ + v3∞ (t)R3 (x2 , x3 )
−
2 
 π
0

2 

Zl

1


+
R2 (ξ, x2 , x3 )w,t (ξ, t)dξ + v3∞ (t)R4 (x2 , x3 )
.

π

0
74 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
Then substituting the obtained expression of p(x2 , x3 , t) in (3.8), by virtue of
(3.16), we get the following expression for q0 (x2 )
(−)
Zl
2 f
ω ρ
q0 (x2 ) =
π
hZ(x2 )
0
0
(+)
Zl
+
(−)
−
R1 (ξ, x2 , x3 ) · cos(→
n (x2 , h (x2 )), x3 )dx3 dξ
w0 (ξ)
hZ(x2 )
0
(+)
→
R1 (ξ, x2 , x3 ) · cos(−
n (x2 , h (x2 )), x3 )dx3 dξ
w0 (ξ)
0
 (−)


 hZ(x2 )
(−)
0
2 f
−
R3 (x2 , x3 ) · cos(→
n (x2 , h (x2 )), x3 )dx3
− v3∞ ω ρ


 0

(−)

(x
)
hZ 2


(−)
−
→
+
R3 (x2 , x3 ) · cos( n (x2 , h (x2 )), x3 )dx3



0
(3.18)
Taking into account (3.16), (3.17), (3.18) from (3.14) after four times integration
with respect to x2 we get the following relation
Zx2
Zx2
w0 (x2 ) − 2ρ ω h(ξ)K(x2 , ξ)w0 (ξ)dξ = (c1 ξ + c2 ) (x2 − ξ)D−1 (ξ)dξ
s
2
x02
x02
Zx2
+ c3 x2 + c4 + K(x2 , ξ)q0 (ξ)dξ,
(3.19)
x02
where
Zx2
x02 ∈]0, l[, K(x2 , ξ) = − (x2 − η)(ξ − η)D−1 (η)dη.
ξ
Constants ci (i = 1, ..., 4) should be defined from the admissible boundary
conditions (see Section 2 Problems 1-10).
75
3.1. CASE OF AN IDEAL FLUID
Let us consider, e.g., Problem 8. Then for w0 (x2 ) we get the following equation
Zl
w0 (x2 ) − ω 2
K1 (x2 , ξ)w0 (ξ)dξ


Zl
Zx2
s 2
− 2ρ ω
h(ξ)K(x2 , ξ)w0 (ξ)dξ + h(ξ)Kl (x2 , ξ)w0 (ξ)dξ

0
x2
x02

0
x

Z2

+
h(ξ)K0 (x2 , ξ)w0 (ξ)dξ


0
(3.20)
0
= f (x2 ),
where
x

Zx2
Z 2

−1
−1
K0 (x2 , ξ) = ξ
x2 D (η)dη − ηD (η)dη − K(0, ξ),


0
l
x
Z2
Zx2
ηD−1 (η)dη −
Kl (x2 , ξ) = x2
η 2 D−1 (η)dη + x2
0
l



Z l
ρf
Kl (x2 , ζ)
K1 (x2 , ξ) =
π 

x0
2
Zl
(η − ξ)D−1 (η)dη,
ξ
(−)
Zh (ζ)
(−)
→
R1 (ξ, ζ, x3 ) · cos(−
n (ζ, h (ζ)), x3 )dx3 dζ
0
(+)
Zl
Zh (ζ)
(+)
→
+ Kl (x2 , ζ)
R1 (ξ, ζ, x3 ) · cos(−
n (ζ, h (ζ)), x3 )dx3 dζ
x02
0
(−)
Z0
Zh (ζ)
(−)
−
+ K0 (x2 , ζ)
R1 (ξ, ζ, x3 ) · cos(→
n (ζ, h (ζ)), x3 )dx3 dζ
x02
0
(+)
Z0
Zh (ζ)
(+)
→
+ K0 (x2 , ζ)
R1 (ξ, ζ, x3 ) · cos(−
n (ζ, h (ζ)), x3 )dx3 dζ
x02
0
(−)
Zx2
Zh (ζ)
(−)
→
+ K(x2 , ζ)
R1 (ξ, ζ, x3 ) · cos(−
n (ζ, h (ζ)), x3 )dx3 dζ
x02
0
76 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
(+)
Zx2
Zh (ζ)
(+)
→
+ K(x2 , ζ)
R1 (ξ, ζ, x3 ) · cos(−
n (ζ, h (ζ)), x3 )dx3 dζ ,
0
x02


f (x2 ) = x2 g22 + h22
Zl
Zl
ξD−1 (ξ)dξ + h11
x02

0
Zx2

−1
D (ξ)dξ  + g11 + h22 ξ 2 D−1 (ξ)dξ
x02
0
0
Zx2
Zx2
−1
−h11 ξD (ξ)dξ − (h22 ξ + h11 )(x2 − ξ)D−1 (ξ)dξ
0
−ω 2 ρf


Z l


Kl (x2 , ξ)
x02
 (−)
Zh (ξ)
(−)

→

R3 (ξ, x3 ) · cos(−
n (ξ, h (ξ)), x3 )dx0

0
x02

(+)
Zh (ξ)
(+)

→
R3 (ξ, x3 ) · cos(−
n (ξ, h (ξ)), x3 )dx3 
 dξ
+
0
Z0
−
K0 (x2 , ξ)
x02
 (−)
Zh (ξ)
(−)

−
→

R
(ξ,
x
)
·
cos(
n
(ξ,
h (ξ)), x3 )dx3
3
3

0

Zh (ξ)
(+)

→
+
R3 (ξ, x3 ) · cos(−
n (ξ, h (ξ)), x3 )dx3 
 dξ
(+)
0
Zx2
− K0 (x2 , ξ)
 (−)
Zh (ξ)
(−)

−
→

R
(ξ,
x
)
·
cos(
n
(ξ,
h (ξ)), x3 )dx3
3
3

0
x02
Zh (ξ)




0



(+)

(+)

−
R3 (ξ, x3 ) · cos(→
n (ξ, h (ξ)), x3 )dx3 
 dξ
+
.
It is easy to show that 2ρs h(ξ)K(x2 , ξ), 2ρs h(ξ)K0 (x2 , ξ), 2ρs h(ξ)Kl (x2 , ξ),
K1 (x2 , ξ) ∈ C([0, l]) (in our case 0 ≤ α < 2, 0 ≤ β < 1).
77
3.1. CASE OF AN IDEAL FLUID
The integral equation (3.20) can be solved by method of successive approximations
when
ω2 <
1
,
Ml
where
M := max {|2ρs h(ξ)K(x2 , ξ)|, |2ρs h(ξ)K0 (x2 , ξ)|, |2ρs h(ξ)Kl (x2 , ξ)|, |K1 (x2 , ξ)|} .
x2 ,ξ∈[0,l]
Remark 3.1 In case of the other boundary conditions (see problems 1-7, 9, 10), the
problem under consideration is solved analogously and in all cases we get integral
equations of type (3.20).
Below we give expressions for kernels K0 and Kl under BCs of Problem 1-7, 9, 10.
Problem 1.
!
Ã
l
l
R

 −K(0, ξ) + l (ξ − η)D−1 (η)dη R η 2 D−1 (η)dη



0
ξ
K0 (x2 , ξ) = −K(0, ξ) +
¶2
µ

Rl
Rl
Rl



ηD−1 (η)dη
η 2 D−1 (η)dη D−1 (η)dη −

0
R0
−
0
Rl
η 2 D−1 (η)dη D−1 (η)dη −
0
+
µ
Rl
R0
Rl
(ξ − η)D−1 (η)dη η 2 D−1 (η)dη
0
ξ
−
−1
¶2  η(x2 − η)D (η)dη

0
ηD−1 (η)dη 
0
0









 Zl
Rl
(ξ − η)D−1 (η)dη ηD−1 (η)dη
ξ
Rl
0
0
¶2
µl

R
Rl
Rl


−1
2
−1
−1

ηD (η)dη
η D (η)dη D (η)dη −
0
0
0




 Zx2
0
−1
+
¶2  (x2 − η)D (η)dη,
µl
l
l
R
R
R

0
ηD−1 (η)dη 
η 2 D−1 (η)dη D−1 (η)dη −
Rl
K(0, ξ) ηD−1 (η)dη
0
0
0
!
Ã
l
R
Rl −1


−1

K(l,
ξ)
+
l
(ξ
−
η)D
(η)dη
D (η)dη


0
ξ
Kl (x2 , ξ) = −
µl
¶2

Rl
Rl
R


2
−1
−1
−1

ηD (η)dη
 η D (η)dη D (η)dη −
0
0
0
78 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
Rl
Rl
−1
(ξ − η)D (η)dη ηD (η)dη
0
ξ
−
Rl





−1
Rl
µ
η 2 D−1 (η)dη D−1 (η)dη −
0
Zl
0
Rl
¶2 


ηD−1 (η)dη 
0
η(x2 − η)D−1 (η)dη
×
0


+

Zl
(ξ − η)D−1 (η)dη
K(l, ξ) + l
 l
Z

(x2 − η)D−1 (η)dη.
0
ξ
Problem 2.
Z0
Z0
−1
K0 (x2 , ξ) = −K(0, ξ) +
(ξ − η)D−1 (η)dη
x2 (ξ − η)D (η)dη −
ξ
ξ
Rx2
(x2 − η)D−1 (η)dη
0
×
Rl
,
D−1 (η)dη
0
Rx2
(x2 − η)D−1 (η)dη Z l
Zx2
η(x2 − η)D−1 (η)dη −
Kl (x2 , ξ) =
0
Rl
0
(ξ − η)D−1 (η)dη
D−1 (η)dη
ξ
0
Rx2 −1
Rl
ηD (η)dη ηD−1 (η)dη
−
0
0
Rl
.
D−1 (η)dη
0
Problem 3.
R0
Z0
(ξ − η)D−1 (η)dη −
K0 (x2 , ξ) = x2
η(x2 − η)D−1 (η)dη
x2
Rl
η 2 D−1 (η)dη
0


0
Z


×
−K(0, ξ) + l (ξ − η)D−1 (η)dη ,


ξ
ξ
79
3.1. CASE OF AN IDEAL FLUID
R0
K(l, ξ) η(x2 − η)D−1 (η)dη
Z0
ηD−1 (η)dη −
Kl (x2 , ξ) = −K(l, ξ) + ξ
x2
Rl
x2
η 2 D−1 (η)dη
0
R0
(l − ξ) ηD−1 (η)dη Z l
x2
−
Rl
η(x2 − η)D−1 (η)dη.
η 2 D−1 (η)dη
0
0
Problem 4.
Z0
(ξ − η)D−1 (η)dη,
K0 (x2 , ξ) = −K(0, ξ) + x2
ξ
Z0
(x2 − η)(ξ − η)D−1 (η)dη.
Kl (x2 , ξ) =
x2
Problem 5.
Rl
Zl
η(x2 − η)D−1 (η)dη −
K0 (x2 , ξ) =
Rl
(x2 − η)D−1 (η)dη D−1 (η)dη
x2
x2
Rl
x2
D−1 (η)dη
0
Rl
−
(x2 − η)D−1 (η)dη Z0
x2
Rl
(ξ − η)D−1 (η)dη,
D−1 (η)dη
ξ
0
Zl
Kl (x2 , ξ) = K(l, ξ) − (x2 − l) (ξ − η)D−1 (η)dη
ξ
Rl
Zl
(ξ − η)D−1 (η)dη ×
−
(x2 − η)D−1 (η)dη
x2
Rl
ξ
.
D−1 (η)dη
0
Problem 6.
Zx2
Zx2
Z0
K0 (x2 , ξ) = (x2 − l) ηD−1 (η)dη − η 2 D−1 (η)dη + (x2 − l) (η − ξ)D−1 (η)dη,
0
l
ξ
80 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
x

Zx2
Z 2

−1
−1
Kl (x2 , ξ) = ξ
(x2 − l)D (η)dη − ηD (η)dη − K(l, ξ)


0
l
Problem 7.
R0
Z0
(ξ − η)D−1 (η)dη −
K0 (x2 , ξ) = x2
η(x2 − η)D−1 (η)dη
x2
Rl
η 2 D−1 (η)dη
0


0
Z


−K(0, ξ) + l (ξ − η)D−1 (η)dη ,
×


ξ
ξ
R0
K(l, ξ) η(x2 − η)D−1 (η)dη
Z0
Kl (x2 , ξ) = −K(l, ξ) + ξ
ηD−1 (η)dη −
x2
Rl
x2
η 2 D−1 (η)dη
0
R0
(l − ξ) ηD−1 (η)dη Z0
x2
−
Rl
η(x2 − η)D−1 (η)dη.
η 2 D−1 (η)dη
x2
0
Problem 9.
ξ
K0 (x2 , ξ) =
l
Zx2
Z0
ξ(l − x2 )
−1
(l − η)(x2 − η)D dη +
η(l − η)D−1 (η)dη
2
l
x02
x2 ξ
−
l
x02
Zl
(l − η)D−1 (η)dη −
x2
K(0, ξ),
l
x02
ξ−l
Kl (x2 , ξ) =
l
Z0
Zx2
(ξ − l)(l − x2 )
−1
η 2 D−1 (η)dη
η(x2 − η)D dη −
2
l
x02
x02
x2 (ξ − l)
−
l
Zl
ηD−1 (η)dη +
x02
x2
K(l, ξ).
l
81
3.2. CASE OF A VISCOUS FLUID
Problem 10.
Zl
K0 (x2 , ξ) = −
(x2 − η)(ξ − η)D−1 (η)dη,
x2
Zl
Kl (x2 , ξ) = K(l, ξ) − (x2 − ξ) (ξ − η)D−1 (η)dη.
ξ
Thus, the following Preposition holds true.
Proposition 3.2 Problem of the harmonic vibration corresponding to Problem 21
has a unique solution when
1
ω2 <
,
Ml
where
M := max {|2ρs h(ξ)K(x2 , ξ)|, |2ρs h(ξ)K0 (x2 , ξ)|, |2ρs h(ξ)Kl (x2 , ξ)|, |K1 (x2 , ξ)|} .
x2 ,ξ∈[0,l]
3.2.
Case of a Viscous Fluid
Let the motion of the fluid is sufficiently slow, i.e., vj and vj,k (j, k = 2, 3) be so
small that linearized Navier-Stokes equations (see [65], [82], [98]) can be applied.
Hence,
∂v2
1 ∂p
=− f
+ ν∆v2 ,
∂t
ρ ∂x2
(3.21)
∂v3
1 ∂p
=− f
+ ν∆v3 ,
∂t
ρ ∂x3
where ν = µ/ρf , ∆ =
∂2
∂x22
+
∂2
.
∂x23
Let
vi ∈ C 2 (Ωf ) ∩ C(R2 ) ∩ C(t > 0), i = 2, 3;
p ∈ C 2 (Ωf );
q,2 (·, t) ∈ H([0, l]).
(3.22)
After differentiation of the first equation of (3.21) with respect to x2 , of the second
equation of (3.21) with respect to x3 and termwise summation, by virtue of (3.5),
we obtain that p(x2 , x3 , t) is satisfying the following equation
∆p(x2 , x3 , t) = 0.
(3.23)
82 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
In case of harmonic vibration in the fluid part, from (3.17), (3.21), (3.23) we
obtain the following system
∆p0 (x2 , x3 ) = 0,
1 ∂p0
−ω 2 u0j = − f
+ νiω∆u0j , j = 2, 3.
ρ ∂xj
(3.24)
(3.25)
Problem 22 Find a function w0 (x2 ) on I, which satisfies the equation
(D(x2 )w, 22 (x2 , t)), 22 = q(x2 , t) + F3 (x2 ) − 2ρh(x2 )
∂ 2 w(x2 , t)
,
∂t2
0 < x2 < l (3.26)
(where F := (0, F3 ) is a plane volume forces), one of the boundary conditions given in
the Problems 1-10, and also find functions u0i (x2 , x3 ), p0 (x2 , x3 ), q0 (x2 ) on Ωf , which
satisfy system of equations (3.24)-(3.25), smoothness conditions (3.22), following
conditions at infinity
¯
(3.27)
p0 ||x|→∞ = O(1), u0j ¯|x|→∞ = O(1), j = 2, 3,
and transmissions conditions as follows
−p0 (x2 , 0+ ) + p0 (x2 , 0− ) = q0 (x2 ), x2 ∈]0, l[,
u03 (x2 , 0) = w0 (x2 ), u02 = 0, x2 ∈]0, l[,
(3.28)
(3.29)
where q(x2 , t) = eiωt q0 (x2 ), p∞
0 is a given constant.
Solution. After separating real and imaginer parts from (3.25) we have
u0j =
1 ∂p0
, j = 2, 3,
∂xj
ω 2 ρf
∆u0j = 0, j = 2, 3.
(3.30)
(3.31)
From the last equality, taking into account u02 = 0, x2 ∈]0, l[ we get
∂p0
= 0, x2 ∈]0, l[.
∂x2
(3.32)
The solution of the equation (3.24) under condition (3.27), (3.29), and (3.32), has
the following form (see [72])
x3
p0 (x2 , x3 ) = −
2π
Zl
0
q0 (ξ2 )dξ2
.
(ξ2 − x2 )2 + x23
(3.33)
Substituting (3.33) into (3.30), for u02 and u03 we get
u02 (x2 , x3 )
x3
=
πω 2 ρf
Zl
0
q0 (ξ2 )(ξ2 − x2 )dξ2
,
[(ξ2 − x2 )2 + x23 ]2
(3.34)
83
3.2. CASE OF A VISCOUS FLUID
u03
1
=
2πω 2 ρf
Zl
0
q0 (ξ2 )[x23 − (ξ2 − x2 )2 ]
dξ2 .
[(ξ2 − x2 )2 + x23 ]2
(3.35)
Let now consider the following limit, when x2 ∈]0, l[,
½
q0 (ξ2 )[x23 − (ξ2 − x2 )2 ]
l − x2
lim
dξ2 = lim q0 (l)
2
2
2
x3 →0
x3 →0
[(ξ2 − x2 ) + x3 ]
(l − x2 )2 + x23
0

½
Zl

q0 (ξ2 )(ξ2 − x2 )
l − x2
x2
−
dξ2 = lim q0 (l)
+q0 (0) 2
2
2
2
x2 + x3
(ξ2 − x2 ) + x3  x3 →0
(l − x2 )2 + x23
Zl
0
Zl
[q00 (ξ2 ) − q00 (x2 )] (ξ2 − x2 )
dξ2
(ξ2 − x2 )2 + x23
0

l
½
Z

0
¤ª
l − x2
q0 (x2 ) © £
2
2
−
ln (ξ2 − x2 ) + x3 ,ξ2 dξ2 = lim q0 (l)
 x3 →0
2
(l − x2 )2 + x23
x2
+q0 (0) 2
−
x2 + x23
0
+q0 (0)
x22
x2
−
+ x23
q00 (x2 )
2
2
ln
(l − x2 ) +
x22 + x23
x23
Zl
−
[q00 (ξ2 )
− q00 (x2 )] (ξ2 −
(ξ2 − x2 )2 + x23
0
x2 )


dξ2

(because of q00 ∈ H([0, l]))
q0 (l)
q0 (0)
l − x2
=
+
− q00 (x2 )ln
−
l − x2
x2
x2
Zl
0
q00 (ξ2 ) − q00 (x2 )
dξ2 .
ξ2 − x2
On the other hand if we define the following supersingular integral in H’adamard’s
finite part sense, we analogously obtain

 x −ε
Zl
Zl
Z2
q0 (ξ2 )
q0 (ξ2 )
q0 (ξ2 )
2q0 (x2 ) 
dξ2 = lim 
dξ2 +
dξ2 +
2
2
2
ε→0
(ξ2 − x2 )
(ξ2 − x2 )
(ξ2 − x2 )
ε
0
=
0
x2 +ε
q0 (l)
q0 (0)
l − x2
+
− q00 (x2 )ln
−
l − x2
x2
x2
Zl
0
q00 (ξ2 ) − q00 (x2 )
dξ2 .
ξ 2 − x2
Hence, using transmission condition (3.28) for u03 , we get the following expression
1
w0 (x2 ) = −
2πω 2 ρf
Zl
0
q0 (ξ2 )
dξ2 , x2 ∈]0, l[,
(ξ2 − x2 )2
where the supersingular integral on the right hand side we define in H’adamard’s
finite part sense (see [8], [3]).
84 CHAPTER 3. A CUSPED ELASTIC PLATE-FLUID INTERACTION PROBLEM
Taking into account of Proposition 2.4 and Theorem 1.5 R(x2 , ξ) is a positive
definite kernel. Further, by view of Proposition 1.7 and (2.76), we can rewrite
(2.81) as follows (see, also formula (3.26))
Zl
Zl
w0 (x2 ) =
K(x2 , ξ)F3 (ξ)dξ +
0
Zl

0
0
Γ(x2 , η, ω 2 )g(η)K(η, ξ)dη  q0 (ξ)dξ
0
Zl
Zl
K(x2 , ξ)F3 (ξ)dξ +
:=

Zl

+ ω2
K(x2 , ξ)q0 (ξ)dξ
0
K1 (x2 , ξ)q0 (ξ)dξ,
(3.36)
0
p
where Γ(x2 , ξ, ω 2 ) is a resolvent of the symmetric kernel K(x2 , η) g(x2 )g(η).
Substituting (3.36) into (3.36), for q0 (x2 ) we obtain the following supersingular
integral equation
Zl
0
q0 (ξ2 )
dξ2 + 2πω 2 ρf
(ξ2 − x2 )2
Zl
K1 (x2 , ξ2 )q0 (ξ2 )dξ2
0
Zl
K(x2 , ξ)F3 (ξ)dξ =: f (x2 ).
=
(3.37)
0
We will find approximate solution of (3.37) using the method of solving given in
Section 1.3 (see equation (1.8), where interval [-1,1] should be replaced by [0, l]) for
q00 (x2 ) := (dq0 (x2 )/dx2 ) ∈ H([0, l]).
Let divide interval [0, l] into N parts as follows
yk0 :=
lk
lk
l
, k = 0, N , yk :=
+
, k = 0, N − 1,
N
N
2N
q0N := (q0 (y0 ), ..., q0 (yN1 )),
For q0N
we will call q0N approximate solution of (3.37).
we get the following system of linear equations [see Chapter 1, system
(1.12)]
N −1
X
0
·
1
1
− 0
aii q0 (yi ) −
q0 (yj ) 0
yj+i − yi yj − yi
j=0
N −1
2πω 2 ρf l X
+
K1 (yi , yj )q0 (yj ) = f (yi ),
N
j=0
¸
i = 0, N − 1.
(3.38)
85
3.2. CASE OF A VISCOUS FLUID
aii := −
Z
4N
l
∆ii
where
h
dξ2
n 0
ni
0
,
∆
:=
[0,
l]
∩
y
−
,
y
+
,
ii
i
(ξ2 − yi )2
N i+1 N
n :=
√
N
X0
N
−1
X
:=
.
j=0
j6=i−1, i, i+1
After repeating the calculation given in Section 1.3, we get
µ ¶−α1
2n
∗
∗
|q0 − q0N | ≤ A
,
l
∗
where q0∗ and q0N
are the solutions of the equations (3.37) and (3.38) respectively.
After calculating q0N , from (3.33) and (3.36) we get approximate expressions for
p0 (x2 , x3 ) and w0 (x2 ), as follows
N −1
x3 l X
q0 (yj )
p0 (x2 , x3 ) = −
, (x2 , x3 ) ∈ Ωf ;
2
2
2πN j=0 (yj − x2 ) + x3
1
w0 (yi ) = −
2πω 2 ρf
(
N −1
X
0
·
1
1
aii q0 (yi ) −
− 0
q0 (yj ) 0
yj+i − yi yj − yi
j=0
¸)
, x2 ∈]0, l[,
Let us denote by w̄0 (yi ) the projection of w0 on yi and let estimate the error of the
approximate solution of deflection. If we repeat the above calculation we get
µ ¶−α1
A
2n
|w̄0 (yi ) − w0 (yi )| ≤ f 2
.
2ρ πω
l
Further, after substituting p0 (x2 , x3 ) in (3.28) we obtain u0j (x2 , x3 ).
u02 (x2 , x3 )
N −1
x3 l X q0 (yj )(yj − x2 )
=
,
πN ω 2 ρf j=0 [(yj − x2 )2 + x23 ]2
u03 (x2 , x3 ) = −
N
−1
X
q0 (yj )(x23 − (yj − x2 )2 )
1
, (x2 , x3 ) ∈ Ωf .
2πN ω 2 ρf j=0
[(yj − x2 )2 + x23 ]2
Proposition 3.3 In case of the harmonic vibration of the plate with two cusped
edges under action of the incompressible viscous fluid all quantities can be expressed
by lateral load (q0 (x2 )) (see formulas (3.33)-(3.34)) and for the calculating of q0 (x2 )
we get (3.37) type supersingular integral equation, where supersingular integral is
defined in H’adamard’s finite part sense. This equation has solution in class q00 ∈
H([0, l]).
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