An approximate method for solving a boundary value problem in the theory of momentless spherical elastic shells

AN APPROXIMATE METHOD FOR SOLVING A BOUNDARY VALUE PROBLEM IN THE THEORY OF MOMENTLESS SPHERICAL ELASTIC SHELLS *

L.S.KLABUKOVA Moscow (Receiued

SOLUTION

16 February

by the mesh method of a problem

1972)

in the theory of momentless

spherical

elastic shells, given in the displacements at the edge of the shell, is discussed. The unique solvability and stability of the initial and difference problems is proved, zero.

together

with the convergence

of the method as the mesh step tends

to

In this paper we discuss a finite-difference method for solving a boundary value problem for the system of equations of a spherical momentless shell when the displacements on the edge of the shell are given. In Section and stable placements,

1 we show that this boundary

with respect to the right-hand i.e. the problem is correctly

value

problem

sides of the system posed.

is uniquely

solvable

and the edge dis-

In Section 2 we discuss the solution of the problem by the mesh method; we describe how the relevant finite-difference problem is obtained, and show that it is uniquely solvable and stable. With regard to the finite-difference solution, we use the stability of the initial problem, proved in Section 1, to show that it is convergent

to the exact

solution

and we estimate

the error for a given

mesh step.

1. Statement of the problem and investigation to see if it is correctly posed We consider

the boundary

elastic spherical momentless boundary (see 111):

* Zh. vFchis2. Mat. mat. Fiz.,

value shell

problem

for the system

with given displacements

13, 3, 698-711,

207

1973.

of equations on the shell

of an

Solving

a boundary

If the point 0 = 0 belonged the region

problem

to the shell

we should

is sought

is unbounded,

in which the solution

tion to conditions

value

(1.2) we require

that the solution

209

have a = - 00; consequently, and obviously, in addi-

be bounded

region. It is more convenient to deal with bounded regions, out the change of variable .a = eY. The domain of definition z plane

will be denoted

(m + l&connected outer contour

by G, and its boundary

region

with boundary

and the curves

After this change

r

E;:;;ij(T + W-

z

%

lfzi

(1.4)

In the case

(> -

the system

-

di --

In the general

case G is an

+ . . . + rm, where rO is the

(1.3) becomes

(X-iY)+$r+$=O,

when the point z = 0 (or 6 = 0) belongs

(1.4) has a singularity

change

of unknowns:

to the region G, the

at z = 0, which can be eliminated

‘jz

11.5)

(1.6)

so that we carry of the solution in the

i

system

The system

r = rO + rl

the

j = 0, 1, . . . , m, do not intersect.

ri,

of variables,

d(T+iS)

by r.

throughout

(1.4) and boundary

(u + Iv), conditions

by the following

Y (z) = (1.2) then become

W

dY

az=fl,

d-

+,%f,,

2

where f(l-t-o)

o=J

2Eh 2r

(I-+ (1.7)

V(s)

where o(s),

r(s)

1+zz (

az -? ZZ), dz

= N(s)

2 ) '

2

fz(z)=

= o(s)

are given functions

0;

+ PG(S), of the length

of arc s on I?.

The problem with given displacements on the boundary thus amounts to solving the system of equations (1.6) under the boundary conditions (1.7).

210

L. S. Klabukova

Let solvable

us investigate

problem

1. Uniqueness.

Consider

JY (lh’)

c

(1.7).

We shall

show

that

it is uniquely

We then

obtain

r

oIYl’dxdy, JJ G

o > 0, we have

Y(z)

the boundary

of the solution

Existence,

from (1.6’),

(1.7’ ), we obtain

It can be assumed by subtracting

the conditions

(1.7),

to the right-hand

without

from V(z)

we can move the inhomogeneity

side

of system

(1.6).

Consequently,

for V(Z):

equation

JZV

together

1

_Jzdz - 0

f=with

;

1

loss

an arbitrary

or

(leg)

JJG(oY%V~)dxdy=z

z = x + iy.

E 0, so that,

conditions

YVdz+

we have

JV/JY=

V(z) = 0.

The

0 in G. uniqueness

is proved.

otherwise,

the following

+wT==o;

) dzdy=;J

+o$

=

2.

problem

from (1.6’ ), (1.7’ ),

G

since

homogeneous

= 0.

o=JJy(;

Recalling

the corresponding

r we have

V(s)

(1.7’ )

JV _ JZ

= O1

on the boundary

Since

(1.6),

and stable.

Jv __ 7 z J, Jw

=f,

J, Jf2 y- f2+ a, -G, 2

the boundary

conditions

on r:

of generality function

that H(s)

of the boundary from Eqs.

= 0,

Vo(z) satisfying (1.6)

conditions we have

Solving

(1.9)

a boundary

value

211

problem

V(s) = 0.

Since

a’V/dza~

can be written equation

where

= 1/,(d’V/dx’

in terms

+ J*V/ay’),

the solution

of problem

function

of the Dirichlet

problem

of Green’s

(1.8),

(1.9)

for Poisson’s

in the form

r = e + iv.

Hence

we obtain,

after

integrating

by parts

in the first

integral:

where

is a function Thus, ponding

with

an integrable

Eq. (2.10)

(1.7)

represents

homogeneous

problem

to the homogeneous The

system

problem

(1.10)

is always We shall

the solution

is thus

at z = 5.

a Fredholm has

system

only the trivial

(1.6),

(1.7),

solvable

which

of equations. solution,

itself

has

for any right-hand

The corres-

since only

side,

i.e.,

it corresponds

the trivial

solution.

the problem

(1.6),

solvable. call

the problem

and the

We shall

singularity

say

right-hand

that

of there sides

the problem

exists

a continuous

and boundary

is correctly

dependence

conditions

posed

between

in certain

if it is uniquely

spaces. solvable

and stable. We shall solvable, 3. side

show

it must

Stability

that then

us first

L,(G)

into

consider

be correctly

of the problem.

and homogeneous

space

the problem

boundary

the space

(1.6),

(1.7)

is stable;

since

it is uniquely

posed. Take

the case

conditions

of solutions

(H(S)

with = 0).

and the space

(/VI(L,(G). From Eq. (1.19),

inhomogeneous We introduce of right-hand

right-hand the norm of sides.

Let

L. S. Klabukova

212

by J the unit

We denote

operator

and by B(V)

c) V(()d(dq

= ~~B(z,

the com-

G pletely

continuous

two operators

of V;

operator

the right-hand

side

of (1.10)

is the sum of

of f, and f2:

Fr (fi) = - 4 &G

(2, Ofi (5) dE dq,

G

Consequently,

+ B) V = F, (f,> + F,(f,).

(1.10’ >

(J

Since

this

equation,

there

exists

function

with

a bounded

G(z,

0

are bounded.

has

completely

inverse a logarithmic

In view

continuous

operator

of all that

singularity, has

operator,

(J + B) -‘.

been

so that

said,

is uniquely

It is well

the operators

we have

IlVll L,(G)

Here

and below,

C denotes

We now consider

VI,

or, since

From

this

JJ

t

av

II -E

Solving

this

IIrIIJI,,(G,.

From

(1.8),

= j-j (+$fz+o$,)vdrdy+ G

I’dxdy

/$-+~;8)dxdy.

and?l.lI),

+

a constant.

= 0,

j-j 1; G +

(iIflilL,(G, + llfzllL,(G,)’

,( c

we have

2 II

G

L?(G)

2x

av

II II -E

~VfilLcGj

+

inequality,

we obtain

av

II II z-

WL2d,

WLdG~

+

Ilf2llL,,GJ+

L?(G) x

=

const.

=z(1 + 1-3 l-tUfAL,(G~+ lifJ,,,,,> .

L,(C)

solvable, that

Green’s

F, and F,

for the solution

(1.10’):

(1.11)

known

of Eq.

Solving

Using

this and (1.6),

Inequalities right-hand

we finally

Take the case

213

value problem

have

s h ow that the solution

(1.11) and (1.12) side.

We now examine boundary conditions.

a boundary

the stability

is stable

with respect

of problem (1.6), (1.7) with respect

fr(z) = 0 and fz(z) = 0 in G.

to its

to the

We move the inhomogeneity

from

the boundary conditions H(z), which is defined

to the right-hand side by introducing an auxiliary function as the solution of the Laplace equation AH = 0 in G and

satisfies

conditions

the boundary

We define

Hi,- = H(s).

U(z) = V (z) - H(z).

Then we have the following

equations

for U(z) and Y(z) in the region G:

(1.13)

z

au

a\y = 0,

and the boundary (1.14)

_)

aZ

az

on r:

U(s) = 0.

We consider for U(z):

IIUIlL2iG, and

aw (1.15)

of

aH

:+wuI=-

conditions

system

-

azaz

and the boundary

-

1 -

o

aw --

From (1.13)) we have the equation l[dU/d~(lL2cG,.

au

az a2

1 =-

o

aa

--

aH

a2 aF

conditions

u,/, =o. We denote equation;

by G(z, 5) Green’s function of the Dirichlet we then easily find that U(z) is the solution

problem for Poisson’s of the equation

L. S. Klabukova

214

The homogeneous continuous

equation

operator,

homogeneous

corresponding

has only a trivial

problem (1.13,

is therefore

solvable

(l.l4),

whatever

We now consider

the function

jldliid~lL2,,,

Solving

to this last equation

with completely

solution, since it corresponds to the which itself has only a trivial solution. H(z) and we have

. From Eq. (1.15),

recalling

that U lr = 0,

where

x = con&.

where

and I/YI\L2(G,, we C=(li-')'2)~.T o obtain final bounds for //V\j,,zrG~

have to find bounds We consider the equation (1.18)

conjugate

the auxiliary

aQLl/aT= Re@*(z)/,-

-

function

0, and is defined

k

-a,(s)

+ck,

Qo(z)

= II, + iu,,

by the following

which satisfies conditions

in G

on I’:

k = 1, 2, . . . . m,

and ck are constants.

Hence

Gg(2)

is the Solution

of

for the ck by writing the Dirichlet pkrobiem. We find expressions for problem (1.18) to be solvable (See [21). We take the homogeneous problem

(I. 19)

we get

for IIH/IL2(c) and ~~d?A’~3~i~,~c;~.

where ak (s) = (r(s) /,a modified conditions

this inequality,

It

= 0 in the region G, a.2

Re (-2 @,*) = 0 on I’.

Solving

The index of problem

n* = -1 + m; the problem linearly

a boundary

215

value problem

(1.18) is n = 0, while the index of problem (1.19) is

number

of linearly

independent

solutions

(1.18) is equal to t = 1 (@,, = ic, c = const), so that the number of independent solutions of problem (1.19) is equal to P = I - (n - n*> = m.

We denote these solutions by a=,*, tPcz*, . . . , Qft,,*, these solutions can be normalized by the conditions

JIm (z’(I),,,*) ds = &,

is the Kronecker The conditions m

ZJ ,

Consequently,

delta). for solvability

of problem (1.18) are

(6 (s) + cd Im (z’Ool*) ds = 0,

I=

1,2,. . . , m.

the cl are given by

Cl =

-

C J ok(s)Im(z’Q,,*)ds, h=O

Hence

It can be shown that

. , , E-2

k, I!= 1,2,.

=r

%l

of the homogeneous

we obtain

the following

We now consider

bounds,

IIu~]I~,~~,

2

IIdz IIL,,,=’ du,

I = 1,2, _. . , m.

I’,

JJ

required

later:

and jja~,_Jd~l,~~~).

We have

[( $)‘f (%)‘I dxdy.

0

Since Au= = 0 in the region G, we have

o=

Hence

du, z II II dZ

f$u,ds-JJ [ I%)‘+ ($“)‘Idxdy=

ffAu,g,dxdy= G

LItc,= -4

1 -du, uo J

r ds

ds.

It can be shown that there exists the solution ip,(z) = ucr + iv, of the modified Dirichlet problem (1.18), for which we have the bound (1.21)

L. S. Klabukova

216

This

can be proved by reducing

geneous

solvable

integral

terms of the resolvent,

the Dirichlet

equation

problem

13, 41, expressing

and obtaining

the tequisite

to the equivalent the solution

bounds.

Using

homo-

of this in (1.21),

we get

iv,., defined

by the

and hence

We can treat conditions

similarly

the analytic

=

k-1,

‘k(S) + d,,

k, and d, are constants,

where ‘k(S) = r(s)lr

(1.22’)

dZ

We introduce

the further

the Laplace

Pk (Tj = ‘,j

and obtain

the inequalities

h(G)

functions pk (x, y), k = 1, 2, . . . , m, bpk = 0 in G and the boundary conditions

auxiliary

equation

by M, and N, the constants

IIapr II

Ilpkllt,cG) s M*,

-z

We turn to lIHl/L,(G, Q,(Z),

Re@,(z)IrO~TO(S),

on IT*

We denote (1.23)

2, . . ..m.

dU, II II

(drl G Clldm,;

satisfying

@,,(z) = u,+

on I?

Re@Jz>1~k

(1.20’)

function

@,(z),

pk(x,

and

La(G)

k = I, 2,


By definition IIJH/J~II~,(GI~

y), we easily

. . . , m.

IJ’

of the functions

find that

H(z)=u.(z)+irc&g

(Ck+ic&)p*(X,y). k=l

From

this,

in

conjunction

with (1.20),

(1.20’ ), (l-22),

(l-22’ ), and (l.23)~

H(Z),

Solving

From these

inequalities

a boundary

and (1.16),

value

(1.17),

217

problem

and recalling

the equations

lav V(Z) = U(z) + H(z) and Y(z) = - _ 0 we

_ a;

get

This

proves

the stability

Obviously, problem

(1.6),

with respect

from (l.ll), (1.7),

IIVII L,(C)

(1.12),

to the boundary

and (1.24),

conditions.

we can obtain

for the general

the bounds G

C(

llfiIIL,cc, + IlfzLcc, + ll~llL*(~~),

(1.25) II~IIW, G C( Ilfi IIIW, + IlfzllL,cG,:+ IwIIL2cr, + IlH’IlLncr,), which in fact show that the initial

problem

is correctly

posed.

2. An approximate (difference) method for solving problem Cl.@, (1.7) 1. We first take the case

of homogeneous

boundary

We shall show that in this case the difference scheme devised by reducing the initial problem to a variational we obtained

(2.1)

the following

-f

equation

conditions

H(s)

for V(z):

G)=F(z) in

the

region

(,$

where

On the boundary

(2.2)

q-

r we have the condition

=o.

We shall show that the problem variational problem.

(2.9,

f 0.

can be conveniently problem. In Section

(2.2) can be reduced

to an equivalent

1

L. S. Klabukova

218

Consider

the operator

Let us show that, the boundary

r,

in the class

L(V) is a symmetric

We introduce

Symmetry

of functions

the scalar

of L(V).

satisfying

and positive

the condition

definite

V(s) = 0 on

operator.

product

We can easily

The fact that L(V) is positive

definite

where A, is the minimum eigenvalue

see that,

for VI,

= 0 and Wlr = 0,

is shown by putting

W = V, and obtaining

of the problem

L(V) = AV in the region G, (2.4) V(s)=Oonr. Clearly, all the A > 0, since (L(V), V) > 0. I n addition, X >,A, > 0. This follows from the inequality (l.ll), which holds for the solution of $e problem (2.1), (2.2). In the case which we obtain

llvllL,(G) It follows Having the problem

of problem (2.4),

we have to set f, z 0, f, = hV in (l.ll),

after

4 C*IIVIIL,IG,’

from this that X >, l/C. shown that L is symmetric (2.1), (2.2) reduces

and positive

to a variational

definite,

problem

it follows

that

on the minimum of the

functional

Z(V) = (L(V), in the class the function

V) - (F, V) - (FV)

of functions VI,- = 0. Hence the initial V(a) minimizing the functional

problem

reduces

to finding

Solving

in the class

of functions

VI,

a boundary

value

219

problem

= 0, and to the construction

of the function

Y(z)

from the expression

The following

difference

method may be used to find the approximate

solu-

tion of this problem. We cover the region G with a mesh of (for simplicity) with the leg h and isolate wholly in G;

we denote

We consider

it by G,,

the functions

linear in every triangle vertices of a triangle. by identical

the region

which consists

and its boundary

V(z),

defined

of G,, and hence We consider the

right-angled

of triangles

triangles

which lie

by rh.

and continuous

in G,, which are

are defined by their values at the which vanish on r, and are continued

v(2)

zero into the region G \ G,.

We shall

minimize

the functional

(2.5) by means

of the functions

V(z).

minimization conditions for the functional yield a system of linear algebraic equations in the values of V(z) at the mesh base-points. Since L is positive definite, this system of equations is uniquely solvable; hence there exists a 6 (z), minimizing the difference

solution

After finding

(2.7)

I(?>;

this function

will be denoted

of the problem (2.1),

V,(z),

we find Y,(z)

by V, (z) and called

(2.2).

from the expression

%=-+Z).

The functions (1.6), (1.7).

V, (z), YJh(z) represent

the difference

solution

of the problem

Let us show that the difference solution is convergent to the solution of problem (1.6), (1.7) as h -10; for this we examine the errors

(IvIt is easily

VhIIL,tG)

seen that

and IP - y~~\L,(GI.

The

220

t.

(2.8)

Z(C)-

We denote

by zP the base-points

at the points

S. Klabukova

tQ- V)).

Z(V) = (L(C - VI,

zp by V (zP).

of our mesh region_Gh,

A function

of the class

at the points .zP, will be denoted by V,(z). It is clear that, for smooth functions V(Z),

V(Z,>

max G

I

at,

-

from Eqs.

(2.9)

z(&-r(v)=(L(~a-

(2.8)

JJCd I

Since V, minimizes

which

Consequently,

takes

?,(z,)

of V(z) values

= V (z,>.

= Q(h),

and (2.3), V),

a&-v>

1

0

(2.10)

I

$2 --FE

-

and the values

ilv

and hence,

=

V(2),

I2

az

the functional

(@0-V))=

dxdy = O(P).

i(c),

we have

I (V,> 4 I (Q.

Using

Eqs.

(2.3)

h,(V,

and (2.8)-~2.1~),

we get

- v, Vh - v>,( (L(Vh - VI,

tv, - VI) = UQ - I(V) ,<

Z(O,>- Z(V) = O(h3. Hence (2.11)

/IV, - VI/j&I&

and since,

from Eq. (2.3,

we obtain,

recalling

(2.12)

IlY’h - YIIL*(c) =

The

estimates

2. (1.7),

Eqs.

(2. ll),

Consider we construct

= o(h)’

(2.6)

(2.12)

the general

and (2.7):

1 d(Vh_V) II 0 dz guarantee

case.

the functional

that

= O(h). 11LZ(G) the method

To find the solution

is convergent

as h -+ 0.

of the problem (1.61,

Soluing

221

a boundary value problem

where (dv/ds) I,- = (aV/d z ) z’ + (&‘/dz) z’ , and Z’ is the derivative with respect to s on r. We shall solve the problem (1.6), (1.7) by a difference method. For this,

we cover the region

denote

the region

c by a mesh of right-angled

of triangles

triangles

with leg h and

by G,.

We take a covering such that G E gh, and furthermore, such that the intersection of any triangle of G, with the region G is non-empty. We consider the function 9, and \?, defined and continuous in the region G,, which are linear in every triangle of G,, i.e., are defined by their values at the vertices of a triangle. To form the corresponding

difference

problem,

we write the conditions

for the

minimization of the functional J(V, v) in the class of functions @(.z), P(Z). We obtain a system of linear algebraic equations, the unknowns in which are the A values of V(z) and s(z) at the mesh base-points. The resulting system of equations

is uniquely

problem (1.6), (1.7), difference geneous for f,(z)

system

solvable. proved

This

in Section

of equations,

follows I.

from the unique

For, to examine

solvability

of the

the solvability

of the

we only have to look at the corresponding

homo-

system, which is obtained by minimizing the functional J(v^, \ii, taken 3 0, f2(z) = 0, H(S) 3 0. It can easily be seen that the functions c and

9, achieving

this minimization,

give J(O, 9

= 0, i.e.,

Hence it foliois that the homogeneous problem (1.6), (1.7) holds for 9 and 9, and this problem only has a trivial solution. Hence the homogeneous difference system only has a trivial solution, and the unique solvability of the difference problem now follows. Let us estimate tion is convergent

J@,

the error of the difference to the solution

solution

of problem (1.6)-(1.7)

We denote by vh(z) and ‘?,(z) the piecewise plane ‘&. These functions will be called the difference

and show that this soluas h -+ 0. functions, minimizing solution of the problem.

L. S. Klabukova

222

We denote by V,(Z) and *,
a(v,-V)

I

mas fi

dz

I=O(h),

max G

=0(h),

m:xI$(tio-V)I

F

We now obtain

Y,) < J(V,, from (2.13)

atv,dZ

+j{ P Returning we obtain

I

dZ

= O(h),

Vl=O(h2),

maxleo-

r

Since V,(z) and Y,(z) minimize the functional functions in the given region G,, we have J(V,,

I

Y)

Y I= O(Wv

maxl~o-

(2.14)

d (k-

--

V)

in the class

of piecewise

plane

‘I’01= O(h3.

and (2.14):

+w(Y,,-Y)

dxdy+

IV,-Vlz+‘&V,,-V)

I’)ds=O(h’)<

to the condition (1.25) under which the problem the required estimates

is correctly

posed,

(2.15) IIVh- vlILtcr, = 0 These

expressions

estimate

time show that the difference problem (1.6)-(1.7) as h + 0.

(h),

l&Vh

- v> II,,;,,

the error of the difference solution

converges

= 0 (h).

solution

and at the same

in the mean to the solution

Translated by D. E. Brown

of

Solving

223

a boundary value problem REFERENCES

1.

GOL’DENVEIZER, obolochek).

2.

VEKUA,

I. N.,

Fizmatgiz, 3.

MIKHLIN,

A. L.

Analytic Moscow,

S. G.

MUSKHELISHVILI, uravneniya),

of thin elastic

Moscow,

distributions

shells

(Teoriya

uprugikh tonkikh

1953. (Obobshchennye

analitichesksie

funktsii),

1959.

Variational

v matematicheskoi 4.

Theory

Gostekhizdat,

methods

fizike), N. I.

Fizmatgiz,

in mathematical

Gostekhizdat.

Singular Moscow,

integral 1962.

physics

(Variatsionnye

1957.

equations

(Singulyarnye

integral’nye

metody