Chapter 5 Convergenc e
Chapter 5
Convergenc
e
In Section 3.4 we presented three general problems, Problems A, Band C.
In Section 4.1 we formulated the problems for the Galerkin approximations.
T hese are Problems AG , BG and CG.
5.1 Equilibrium problem
In thi s section we consider the convergence of the s olution of Problem BG to the soluti o n of Problem B.
Assume that
u
h
E
Sh
i s the solution of
b(u h
, v)
=
(j ,
v)
for a ll
v
E
Sh
(5.1.1
) and
u
E
V is the solution of
b ( u,
v)
=
(j,
v )
f o r all
v
E
V.
( 5.1.2)
In the proof of the theorem, we will use the pr o je c tion
P ,
defined in Se c tion 4.6.
Theorem
5.1. 1
1.
II
u h

u"
E
+
0
i
f
h
+ o.
74
CHAPTER
5.
C
ON
VERGENCE
2. If
U
E
H4
n
V ; then and
75
P
roof
If
(5.l.2) is subtracted from (5.l.1), it follow s that
b(
u

u\ v)
=
0 for all
v
E
Sh
This means that uk = Pu. The first part of the theorem follows directly from Lemma 4.6.2
. The estimates in the second part of the theorem follow from Lemma 4.6
.
1 and Lemma 4.6.3. 0
C
H A P
TER
5.
CO
NV
ERGENCE
5.2 Eig
e
nvalu
e
pr
o
bl
e
m
76
Our approach is based on the ideas of [BDSW] , [BF] and [SF] and we fol low the presentation in [SF]. The theory in the book of Strang and Fix ,
[SF, Section 6.3], concerns eigenvalue problems for general symmetric elliptic operators. Most of the presentation is written in a style which encourage abstraction. In collaboration with others, [ZVGV2], we verified that the theory is valid for abstract eigenvalue problems such as Problem
C.
In this thesis we present this abstract version, and also offer a number of modest improvements.
The rate of convergence for eigenvalue problems also depends on the regula rity of the eigenvectors. In the absence of such theory for interface problems, we pose the following assumption which we showed to be true in the one dimensional case. (See Section 3.5.)
Regularit y Ass umption
The eigenvectors of the eigenvalue problem , Pro blem C, are in
Hk
n
V
for
k
=
4 or 6 , and there exists a constant
C b
depending on th bilinear forms band (')' ) such that for e a ch eigenvector
y
5.2.1
T
h
e
Rayl
e
igh
q
uo
t
ient and th
e
Minmax principle
To analyse the convergence of eigenvalues and eigenvectors, some preparation is necessary. First we establish bounds for the approximate eigenvalues using the Rayleigh quotient and the minimax principle .
It is wellknown that the eigenvalues are the stationary values of the Rayleigh quotient. However, the following result gives a more convenient characteri zation of the eigenvalues. See [SF, p 221].
Lemma 5.2.1
Minmax principle
Let T denote the class of subspaces of V having dimens'ion J' , then
Aj
= min max
R(v)
.
SET vES
CHAPTER
5. CO
VERGENCE
W e may assume that the eigenvalues are ordered
77
For some in teg er m, consider the eigenvalues
)'1,
A2, ... , A
m
and corre s ponding eigenvectors Yl, Y2, ... ,
Ym'
Equal eigenvalues are possible but we assume that
Aj
=I
Am
for each
J'
> m.
In the finite dimensional subspace
S
h
we have
A7,
A~,
.. . ,
A~
(a lso ordered) and corresponding eigenvectors
y7,
y~,
... , y~.
( Equal eigenvalues do not matter. In the case of multipliCity,
yJ
is not uniquely determined, but it does not influence any proof.)
5.2.2 Bounds for the approximat
e
eigenvalues
The minimax principle yields lower bounds for the approximate eigenvalues.
L
emma
5.2.2
A]
2::
Aj for
each j.
Proof
The minmax principle is also true for the space
Sh.
Any subspace of
Sh
is a subspace of
V
.
0
Notation
Yi
will be used to denote the normalised eigenvectors, i.e. II Yi l1 = 1.
For
j
=
I, 2, ... , m, let
E j
denote the subspace of
V
spanned by {Yl,
Y 2
, ... ,
Yj}.
Consider t he subspaces
Sj
where
Sj
=
P E j
for
j
=
1 , 2, ... , m.
P
=
Ph
is the projection defined in Section 4.6.
An upper bound for some approximate eigenvalue depends on the construc tion of
Sh.
Clearly the construction of
Sh
must be s uch that dim
Sh
=
N
> m. However, it is st ill possible that dim
Sm
< m .
As
sumption
The construction of
Sh
is such that dim
Sm
= m.
We define a quantity f.L~ to measure the " distortion " of the projection of the unit ball
Em
=
{y
E
Em :
IIYII
=
I} : Se t
fL':n
= inf{IIPyl1
2
:
Y
E
Em}
.
CHAPTER 5. CONVERGENCE 78
Proposi
t
ion 5.2.
1
M':n
>
0
~f
and only if
dim
Sm
= m.
Proof
The function
M':n
IIPyI1
2 ha s a minimum on the compact set
Bm.
Hence
>
0 if and only if
Py
=/:.
0 for each
y
E
Bm·
But this is so if and onl y if the vectors
PYl, PY 2,
.. . ,
pYm
form a linearly independent set. D
P
roposition
5
.2.2
A':n
:::;
m ax{R( Py )
:
y
E
B m}.
P
roof
Since dim
Sm
= m, it follows from the minimax principle that
A':n :::;
max{R(v)
:
v
E
Sm}.
Consider any
v
E
Sm, v=/:.
O. There exists a vector
y
E
Em
such that
Py
=
v.
N ote that
R(P(ay))
Consequently,
=
R(av )
=
R (v).
Choose a such that
ay
E
B m. max{R(P z
) : z
E
B m }
=
m ax{R(v)
:
v
E
Sm}·
D
The following result is crucial.
Lemma
5
.2.3
A':n
:::;
A:.
Mm
Proof
If
y
=
2:::':1
CiYi,
then
b(y, y)
= m
L
i=l orthonormal set. Hence c; \, since
{Yl, Y2,
.
.
. , Ym} is an m i=l
Since
P
is a projection with respect to the inner product
b, b(Py, Py )
:::; Am for each
y
E
Bm.
From the definition of
M':n,
we h a ve
R(P )
y
=
b(Py , Py)
<
Am
IIPyl12 
M':n '
ow use Proposition 5.2.2. D
CHAPTER
5.
CO N VERGENCE
C orollary 5.2.1
Jlr:n :::;
1.
79
This is a direct consequence of Lemmas 5.2.2 and 5.2.3. It i s convenient to formulate error estimates in terms of the quantity 1 
Jl':n.
N otati o n
CJ':n
=
1 
Jl':n.
Corollar y 5.2.2
0:::;
CJ~
<
1
and
A~

Am
:::;
A~P~,
Since the eigenvalue error is bounded by and p rove that
CJ!
>
O.
CJ!, it is sufficient to estimate
CJ!
5.2.3 Estimates
Proposition 5.2.
3
CJ!
= max{2(y,
y

Py) II Y
Pyl12 : y
E
Em}.
P roof
IIPyll2
+ lIyll 2 
2(y, Py)
IIPyll2
+
2(y, y)

2(y,
Py) 
1 (since lIyll2
=
1 ).
As a consequence 1 
IIPyll2
=
2(y , y
 Py
)

Ily 
p Y 11 2 .
The result follows from the definition of
CJ!.
0
Remark
In [SF , Section 6 .
3]
CJ! is defined by
CJ~
= max{12(y,
y
 Py
)
lIy pYll21 : y
E
Em}.
The absolute value is not necessary since max{2(y,
y
Py) 
Ily
Pyl12 : y
E
Em}
~
O.
The assumption is then made that
CJ!
<
1 , and they prove tha t dim
Sm
= m.
W e proved the fact that dim
Sm
= m is equivalent to
CJ!
<
1 and we believe that it is important to take note of this equivalence.
CHAPTER 5. CO
N
VERGE
NC
E
Proposition
5.2
.4
For any y
E
Em !
(y,y
 Py
)
=
b(y*  Py*,y  Py
) .
Proof
b(Yi  PY
i, Y
 Py)
=
b(Yi, Y
 Py
)
since
b(y
 Py, PY
i)
=
O.
H ence,
Multiply by c i A;l and sum over i. We have
(y,
y  Py)
= m
L
i=l
Ci\lb(Yi
 P
i,
Y  Py)
80 o
The following result also differ from [SF].
L
emma
5.2
.4
(J~ ~ max{2 1Iy * 
pY*IIE IIY
 pYllE :
Y
E
E m }·
P
roof
Consider the result of Proposition 5.2.3. We have demonstrated that the quantity
2(y, Y
_ py) _
Ily 
pyll2
must have a non negative maximum (Corollary 5.2.2) . Cons e quently
(J~ ~ max{2(y,
y
 Py ) :
y
E
Em}.
Use Proposition 5.2.4 and the Schwartz inequality for the inner product
b.
0
Prop
o
si
ti
on
5.2.5
For
any c >
0
there exists a
<5
>
0
such that for h
<
<5!
Ily* 
Ily 
Py *I IE pYllE
< c
for each y
E
E m,
< c
for each y
E
E m·
CHAPTER
5.
C O
VERGENCE
81
P roof
From Lemma 4.6.2 there exist po s itive numb e rs
0 1) 02 )
.
.
. )
On
such that for each i
N ow, suppose
h
< min i
Oi,
then m i =l
The same arguments are valid for
Ily* 
PY*//E'
Lemma
5.2.5
For any
c
>
0
there exists a 0
>
0
such that
(J~
< c
i f h
<
O.
Proof
For any c
>
0 there exists a
0
>
0 su c h that if
h
<
0 ,
then
Ilu

PUIlE
< c for each
u
E
Em
·
The r e sult follows from Lemma 5.2.4 and Proposition 5 .
2.5. o o
Proposition
5.2.6
If Problem
C
satisfies the regularity assumption, then for any
Y
E
Em and
Proof
W e may assum e t hat C i
~
0 for each i.
First estimate:
IIY* 
Py*IIE
<
<
< m
L
c) , .
;lIlYi 
pYiliE
i = l m c
c)\i
1
IYil k · h
P

2 i=l
c/ ]
' m
L
Ci A f
1
I1Yill hk
*2 i =l
<
CCbAr:n1 h k '
2.
CHAPTER
5.
C O NVERGENCE
82
Second estimate: m
<
i = l
CG b
A
f
hY2,
using the same arguments as for the first estimate.
L e mma 5 .2.6
If Problem
C satisfies
the
regularity assumption) then
CT~
:::;
CG
A;::lh 2(kO2) b
P roof
Use Lemma 5.2.4 and Proposition 5.2.6. o o
5.2
.4 C onv e rg e n ce of e i g e n v a l ues
W e may now use the results of the previous subsection to establish the con vergence of
A~ to
Am.
L e mma 5.2.7
There e.'Eists a
6
>
0
such
that for h
<
6,
A~
 Am :::;
2AmCJ~.
Proof
Choose 6 such that
CJ~
<
~.
Consequently
A ~
<
2A m.
o
T he o rem 5.
2 .1
1.
A~

Am
t
0
as
h
t
O.
2. If Probl em
C satisfies the regularity assumptwn
,
then
.
A~

Am :::;
15G b
A;:
h
2
(k "
2) .
Proof
1. This is a direct consequence of Lemmas 5.2.5 and 5.2
.
7.
2. Use Lemmas 5.2.6 and 5.2.7. o
CHAPTER
5.
CO N
VERGENCE
83
5.2.5 Convergence o f eig e nv e c t ors
To estimate the error
"ymy~ll, we need to estimate the difference
IIPYmy~ll·
It is necessary to consider the possibility that
Am
has multiplicity more than one. Suppose that the multiplicity of
A
=
1, 2, ... , m 
r,
m
Am
is
r
and let
+
1, ... ,
N.
From Theorem 5.2.1 it follows that there exist real numbers
p
>
0 and
0
>
0 such that if
h
<
0 ,
then lAm 
AJI
>
p
for each
j
E
A.
As sumption
Assume that
h
is sufficiently small for
(5.2.1) to hold.
(5.2.1)
Suppose
{y~r+l' y~r+2 '
... , corresponding to
A~ r+l' A~r+2 ' y~}
... is
, an orthonormal set
A~.
The strategy of eigenvectors now is to estimate the distance between y~r+ i and some (uniquely defined) vector in
E
Am' the eigenspace corresponding to
Am.
We define a projection
Pm
with domain
P(EArn ):
PmW
= m
L
(w, yJ)yJ
for each
w
E
P(EArn)'
j
=
mr+l
This projection enables us to deal with the case of a repeated eigenvalue.
Here we differ from [SF]. Although most of the computations are the same , we believe that our construction of the projection
Pm
is a worthwhile contri bution. W e will show that
PmP
(and hence
Pm)
is invertible for
h
sufficiently small.
Propo s ition 5.2.7
For
each
j
E
A
and each
y
E
E
AmJ
(AJ
 Am)(Py,
yJ)
= Am(Y  Py
, yJ).
Proof
It is only necessary to prove that
AJ(Py, yJ)
=
Am(Y, yJ)
since the term 
Am
(Py,
yJ)
appears on both sides of the equation.
Since
yJ
and yare eigenvectors, it follows that
A7(Py, yJ)
=
b(Py,
yJ)
and
Am(Y, yJ)
=
b(y, yJ).
But
b(Py  y, yJ)
=
0 for each
j,
thus ( 5.2.2) follows.
(5.2.2) o
CHAPTER
5 .
CO
N
VERGENCE
Lemma 5.2.8
84
P roof
From the assumption we have the estimate
Am
IAJ 
Ami
:=:; Pm for each
j
E
1\ , where
Pm
=
A m p1.
The set y~ ,
yq ,
... ,
Y'N
form an orthonormal basis for
Sh ,
h e nc e
N
Py
=
L( Py, yJ)yJ. j= l
Cons e qu e ntly ,
Py  PmPy
=
I)py, yJ)yJ. j Ei\
If
y
E
E Arn
' then
L (Py , yJ?
jEi\
We now us e Proposition 5.2.7
.
~
CAJ
~mAml)
2
(y Py , yj)'
<
p~
L(y
 Py,
yJ)2
(Inequality (5.2.3)) jEi\
N
<
p~
'L) y

Py, yJ)2 j=l
(5.2.3)
D
CH A PTER
5 .
CONVERGE N CE
L e mma 5.2
.9
85
Proof
Ily 
PmPy11
<
Ily 
Pyll
+
IIPy 
PmPy11
<
(1
+
AmpI)
Ily 
pYI!.
o
C
orollar
y 5.2.3
PmP is invertible for h sufficiently small.
Proof
Let
y
E
Em
n
E
Am'
Then for
h
sufficiently small. Since it follows that
IIPmPyl1
>
~. Consequently
IIPmPyll>
1
Li"YI
for each
y
E
E
Am .
o
C
orollary
5.2.
4
If h is s'ufficiently small , then for each
j,
there exists a unique
Xj
E
EAm
w~th
IIxj II
=
1
such that
.i
=
1, 2 , ... , r
Proof
There exists a unique
y
E
EAm
such that
y
=
(PmP)lYm_r+j.
Hence ,
Let
{3 be a real number such that
1{31
= lIyll and let
Xj
=
{31y.
'vVe can choose
Xj
such that
{3
>
O. As a consequence lIyll
=
{3.
C H APTER
5.
CO N VERGENCE
86
Hence
II
Xj

Y~

r+jll
<
IIXj  yll
+
Ily

y~r+jll
<
2(1+p
1
A
m
)
llyPyll·
o
It is important to realise that one compute the approximation
y':nr+ j .
The result above guarantees the existence of an exact eigenvector, with norm one, close to the approxi mate one.
The following result from [S F] shows that an error es timate in the energy norm depends on error estimates in the norm
II
·II
and eigenvalue errors. We modified it slightly to m ake it useful for th e case of repeated eigenvalues.
L emma 5 .2.10
Proof
b(Ym

yj, Ym

yj) b(Ym, Ym )
 2b(Ym, yj)
+
b(yj , yj)
2
AmllYmll
 2A
m(Ym, yj)
+
AJIlY~1I2
Am  2Am(Ym, y~)
+
A J
Am[2

2(Ym,
y~)]
+
AJ

A m
Am [IIYmIl
2

2 ( Ym, yj)]
+
lIyJ II
2
]
+
AJ  Am
AmilY m yj ll2 +
AJ

Am.
o
T heorem 5.2.2
1. Let
c
>
0
be arbitrary. If h is sufficiently small ) then for each j, j
=
1, 2 , ..
. ,
r there exists a unique
Xj
E
EArn wi th IIx j
II
= 1
such that
CHAPTER
5.
CO N VERGENCE
87
2.
Suppose Problem
C
satisfies the regularity assumptwn.
If
h
IS
suffi ciently small, then for each
j, j
=
I , 2, ... ,
r
there
e.'LZsts
a unique
Xj
E
EArn
with
IIxj
II
=
1
such
that
.
II
Xj

h
II
E
:s;
C~C
\ a:
h(k*2)
.
Proof
1. Use Corollary 5.2.4: There exists a unique
Xj
E
EArn
such that
But
(Lemma 5.2.10). Hence
Ilxj

Y~r+jll~
:s;
4Am(1
+
p1Am)21Ixj
 PXjl12
+
A~_r+j

Am
(5.2.4)
N ow use Proposition 5.2.5 and Theorem 5.2.1.
2. Consider the Inequality (5.2.4). We have the estimates
(5.2.5) from Proposition 5.2.6 and from Theorem 5.2.1. The result follows from (5.2.5) and (5.2.6).
(5.2.6) o
CH A
PTER 5. CO
N VERGENCE
5.3
Vibration probl
e m
88
Our concern is the difference between the solution u of Problem
A
and the solution
Uh
of Problem
AG.
It is possible to estimate this error in terms of the projection error (Section 4.6) and errors for the initial conditions. See
[SF , Section 7.3]. This is called a projection method and was first used for parabolic problems. For second order hyperbolic problems , it appear that credit is due to [DJ, [De] and [SF]. Research in this direction was also done by [Ba].
After this it appear that abstract methods became popular. See for example
[Sh, Section
6.4].
In Section 3 of an invited paper, [BIt], very general results are given. (Incidentally they use results in [Sh].)
A general approximation theory , using functional analysis, is obviously im portant. However, we found that the basic error inequality mentioned before
([SF, Section
7.3] and [D , Lemma 1]) is valid for an abstract problem as general as P roblem
A.
As a final remark we mention the paper [FXX] where the autho rs also use what they term a "partial projection " me t hod to obtain
£ 2 error estimates.
5.3.
1
Di s c
retizat
ion e rror
In this section we show that the convergence proof sketched by Strang and Fix
[SF , Section 7.3] can be applied to Problem A in Section 3.4 and Problem AG in Section 4.l. In this proof the projection operator P defined in Section 4.6 is used to find an estimate on the discretization error
Ilu(t)  uh(t)IIE for
t
E
[0 ,
(0 ) . vVe also use the symbol
P
to denote the " projection"
Pu
of the solution
u
of
Problem A , i.e.
(Pu)(t)
=
Pu(t)
for each
t
?:
O.
Let
e(t)
=
Pu(t)  Uh(t) and
ep(t)
=
u(t)  Pu(t).
Then
Ilu(t)  uh(t)IIE::;
Ilep(t)IIE
+
Ile(t)IIE'
(5.3.1)
The following result is required for the main result of this section. N ote that differentiability with respect to the energy norm is required to prove that the projection function
Pu
is differentiable. This regularity requirement is not stated by [SF].
CHAPTER 5. CO
NV ERGENCE
89
Lemma 5.3.1
ffu
E
C
2
([O,
00) ,
V), then Pu
E C2([O , OO), V)
wdh
(Pu)'(t)
=
Pu'(t) and (Pu)"(t)
=
Pu//(t).
P roof
As the projection operator
P
is a bounded linear operator with norm less than one, it follows that
II(ot)l
(Pu(t
+ c5t) Pu(t))  Pu'(t)IIE ::;
11(c5t)l(u(t
+ c5t) 
u(t))

u'(t)IIE'
This implies that
Pu
E
C
1
([0,00),
V) and
(Pu)'(t)
=
Pu'(t).
In exactly the same way we prove that
(Pu)//(t)
=
Pu//(t).
(Pu)'
E Cl
([0 , 00),
V) and o
Since we already have an estimate for the projection error
e p(t)
,
it is only necessary to estimate the other part of the error.
In the next proof the following " energy" expression will be c onvenient:
E(t)
=
~(e'(t),
e'(t) )
+
~b(e(t),
e(t))
1 1
 211
e' ( t )
112
+
211
e
( t )
II
~.
(5.3
.
2)
Lemma 5 .3.2
Assume that u
E
C
2
([O, 00) ,
V). Then, for any t
2 0 ,
Ile(t)
liE::; IIPcx

CXhllE
+
liP;]
 ;]h
II
+
it
Ile~(s )
II
+
~[ Ile ~( s )
110
ds.
Proof
From P roblem A and the Galerkin approximation (Problem AG) we deduce that for any v
E
Sh ,
(u//(t)
u~(t) ,
v)
+
a(u'(t) u~(t) ,
v)
+
b(u(t)  Uh(t), v
)
=
O. (5.3.3)
Since
P
is a projection, we have
b(u(t)  Pu(t)
, v)
=
b(u'(t) 
Pu'(t), v)
=
0 for all
v
E
Sh.
Using the fact that
Pu//(t)
=
(Pu)//(t) ,
(5.3.3) can be written as
(el/(t), v)
+
b(e(t), v)
= (e~(t),
v) k(e~(t),
v ) o

k(e'(t) , v)o
Jib(e'(t),v)
for all
v
E
Sh. (5.3.4)
CHAPTER 5. CO
N VERGENCE
( ote that
a(u, v)
=
fJb(u, v)
+
k(u, v)o
where
fJ
or
k
or both can be z ero.)
W e will use the fact that
E'(t)
=
(el/(t), e'(t))
+
b(e(t), e'(t)).
As
e(t)
E
Sh it follows that
e'(t)
E
Sh. Choose
v
=
e'(t)
in (5.3.4), then
E'(t)

(e~(t),
e'(t))

(ke~(t),
e'(t))o  k(e'(t),
e'(t))o
 fJb(e'(t)
, e'(t))
~ (1/e~(t)11
+
~llle~(t)llo)
Ile'(t)ll·
From (5.3.2),
1/e'(t)11 ~
J2E(t).
Thus
E'(t)
~
J2E(t)
(1Ie~(t)11
+
~I
II
~(t)llo) and consequently
90
This yields that
JE(i5
~
JE(O)
+
~
1t
(1Ie~(s)11
+
~Ille~(s)llo)
ds. (5.3 .
5)
As
E
(
0)
=
1
2
1
2
'2
IIP
P' 
P'hll
+
'2 IIPQ

QhllE and
Ile(t)IIE
~
J2E(t) ,
again from (5.3.2) , the result follows from (5.3.5). 0
Theorem 5 .3
.1
Assume that u
E
C
2
([0 ,
(0),
V). Then ,
JOT any t
~
O .
Ilu(t) 
uh(t
)
IIE
~
Ilep(t)IIE
+
IIPQ 
QhllE
+
IIPp'  P'hll
+
1t
(1Ie~(s)11
+
~I Ile~(s)llo)
ds.
Proof
Use L emma 5.3.2 and Equation (5.3.1). o
To prove the convergence results, it is now necessary to consider the terms on the right side of the inequality in this theorem.
C HA PTER
5. CO
N VERGENCE
5.3.2
Convergence
The main factor that determines the rate of convergence of the solution
Uh
of Problem AG to the solution
U
of Problem A as
h
tends to zero, is the regularity of the weak solution u. The regularity of u depends on the regularity of the initial values
Q
and
f3,
as we pointed out in Section 3.4. [Raj gave an example to show that the regularity of the solution is necessary to obtain optimal order convergence.
Th e rate of convergence is also directly influenced by the choice of the initial values
Qh
and
f3h
for the solution
Uh
of Problem AG. V·..,re will c onsider two cases, i.e.
Qh
=
ITQ, f3h
=
ITf3
and
Qh
=
PQ , f3h
=
Pf3.
In the following result we show that the rate of convergence in the en e rgy norm is of order
h
2
if certain regularity c onditions ar e satisfied. The estimates are expressed in terms of the constants
C
J
and
C
E
d e fined in Section 3 .4 as well a s
C d e fin ed in Section 4.5.
Theorem 5.3.2
LetQh
=
ITQ andf3h
=
ITf3. A ssu me thatu
E C
2
([0 , 00 ),
V) and that u(t)! u'(t) and ul/(t) are in H4
n
V
for t
2:
O.
Then ,
Ilu(t)  uh(t)IIE::;
C
(IQI4
+
C e
1 1f314
+
lu(t)14
+
k(CECJ)lt
max
lu'(s)14 sE[a,t[
+Ce1t
max
lul/(s)14) h
2
SE[a,f)
for
t
E [0 , 00).
Proof
From Theor e m 5.3.1,
Ilu(t) 
uh(t)IIE
::;
Ilep(t)IIE
+
II
PQ  ITQll
E i
IIP
f3 
IT
) I
I
+ it
( 1Ie~ (s ) 1 1
+
~J Il e ~(s)lla)
ds.
All that remains to be done is to apply the a ppro x imation results from
Co
rollary 4.6.1 to ea c h of the terms in this expression:
91 and
CHAPTER
5.
CONVERGENCE
92
From Lemma 5.3
.1 ,
( Pu)'
=
Put
and hence e~(t)
=
u'(t) Pu'(t).
This yields that and
Similarly, o
Under less strict regularity conditions we can still show that the solution
Uh
of P roblem AG converges to the solution
u
of Problem A in the energy norm if
h
tends to zero .
Th
e
orem 5.3.3
Let O:h
=
and u
E
C
2
([0,
00),
V ) , th e n
ITo:
and Ph
=
ITp. Assum e that
0:
E
V ,
P
E
V lim
h+O
Ilu(t) 
uh (t) II E
=
0
jar
t
E
[0 ,
T].
Proof
From Theorem 5.3.1,
Il
u(t) 
uh(t )
IIE ::;
Il ep(t) II E
+
IIPo:
ITo:II
E
+
IIP
p 
TIp ll
+
fo·t
( 1I e ~( s ) 1 1
+
~J Ile ~(s)l l o
dS) .
From the approximation results we know that for any
E
>
0 , each term is less th an
E,
provided that
h
is suffi c i e ntly small. 0
5.3.3 In
e
rtia norm
es
timat
e
I n a final result we show that the AubinNits c he tri ck can also be applied to this probl e m to find inertia norm estimates for th e dis c retization e rror.
CH A
PTER 5. CONVERGENCE 93
Theorem 5.3.4
ul/ (t) are all in
V
Let O'.h
=
PO'. and !3h
=
P!3. Assume that u(t) , u'(t) and
n
H4 for all t
;::
0.
Then ,
Ilu(t) uh(t)11 :::;
C
(lu(t)14
+
kt(CJCE)l
m ax lu'(s)14
+
tCi/
max
lul/(S)14) h4
SE[O,tl
SE[O,tj
for t
E
[0,
(0).
Proof
From Theorem 5.3.2,
. I/u(t)
 Uh(t)/1
<
l/ep(t)/1
+
/Ie(t ) /1
<
/Ie p(t ) /1
+
Ci
< l/e p(t)
/1
+
Ci1
1
1I e(t)/IE l t
( lI e ~(s) 1I
+
~J ll e~(s ) ll o)
ds.
For a fixed
t;::
0, we consider
ep(t)
=
u(t)
 Pu(t
) .
We conclude from Corollary 4 .6
.2 that
II
e
p (
t )
1/ :::;
C
u (
t)14
h
4
.
Simil ar arguments yield that
lIe~(t)l/o
:::;
C!llle~(t)1I
:::;
Ci1Clu'(t)14h4
and lie~(t)1I
:::;
Clul/(t)14h4. (5.3.6)
o
A useful result is also obtained if the AubinNits c he trick is us e d only for the terms containing the integrals .
T heorem 5 .3.
5
Let O'.h and ul/ (t) are all in
V
=
ITO'.
and!3h
=
IT l)',
Assume that
0'..
.
;3 ,
u(t). u'(t)
n
H4 for all t. Then, lI u(t)

Uh(t)
liE :::;
C
(10'.14 + Ci
1
1!314
+
lu(t)14) h2
+
C
(ktCi2
max
l u'(s)14
+
t
max lul/(s)14)
h4 for t
E [0,
(0).
SE[O , t] sE [O,tj
Proof
T he proof is exactly the same as the proof of Theorem 5.3.2. The estimates in (5.3
.
6) are used for the terms containing the integr a l. D
Remark
We consider thi s result to be significant. It is advantageous to have an error estimate in the energy norm, while the terms containing tare
" suppressed" by
h4.
CHA PT
ER 5. CO
N VERGENCE
5.
4
Finite Differ
e
nces
94
In this section we consider the system of ordinary differential equations ,
Problem A D in Section 4.1, and the finite difference method for approxima ting the solution. The objective is to prove that the solution of the discre tized problem converges to the solution of the Galerkin approximation. This method has been extensively studied e ven in the context of finite difference methods for second order hyperbolic partial differential equations. However , one must be careful when matching the estimates. Although all norms are equivalent in the finite dimensional space
Sn,
the " c onstants" may depend on the dimension of Sh. Presenting error estimates for semidiscrete and fully discrete systems in the same presentation is a line also followed by others .
See for example [ D ], [Ea] and [FXX].
W e consider Problem AG in Section 4.1 and the finite difference scheme proposed in Section 4.4. In the first subsection we estimate the local error and then proceed to establish stability results.
5
.4
.1
L
ocal error
The first step is to derive finite difference formula.s similar to the Newma.rk schemes [Zi]. Since we need error estimates in terms of the unknown fun c tion or its derivatives, it is necessary to derive the formulas .
We will use Taylor's theorem in the following form:
g(t )
=
g(to)
(t t
) n l
+
(t  to)g'
( to)
+ ... +
(~ _ o
1)1
g(nl)(to)
+
R(t)
where
R ( t)
=
( n ~l)1
ft:(t

e)nlg(n)(e ) de .
It is also true for
t
<
to·
See [el, p 179] or [Ap, p 279].
The following notation is introduced for convenience.
1
N
otation
R~(t)=
( _ )1
n
1 .
t j tHt
(t+6tet
1 g( n )(e)de a nd
The first proposition contains wellknown results and the proofs are trivial.
CHAPTER
5. C
ON V ERGENCE
95
Proposition 5.4.1
l.
If the real valued function 9 is
m
C3 [t 
bt, t
+
bt], then g(t
+
bt)
g(t  bt)
=
2btg'(t)
+
Rj(t)
 R3(t). (5.4.1)
2. If the real valued function
9 is
in C4[t  bt, t
+
btl, then g(t
+
bt)

2g(t)
+
g(t

bt)
=
(bt)2 gl/(t)
+
R t(t)
+
Ri(t). (5.4.2)
Proof
1.
Use Taylor 's theorem to get:
g(t
(bt
) 2
+
bt)
=
g(t)
+
bt g'(t)
+
2 
gil ( t)
+
Rj (t)
and
g(t 
bt)
(bt)
2
=
g(t) 
btg'(t)
+
2 gl/(t )
+
R3 (t).
Clearly
9 (t
+
bt)

9(t

bt)
=
2bt g' (t) +
Rj
(t
)

R3
( t ) .
2. A pproximate
9
by a polynomial of degree three and co mpute
g ( t
+
bt)
+
g(t

bt).
o
W e gave the proof of part one in detail b eca u se we us e the re s ult. in the n e xt proposition.
Proposition
5.4.2
Let Po and
PI
be real numbers such that Po
+
2PI
= l. l.
If th e real valued function 9 is in C4[t

bt, t
+
bt], then g(t
+
bt)  g(t 
bt)
=
2bt (PIg'(t
+
bt)
+
Pog'(t)
+
PIg'(t  bt))
+
R
4(t),
(5.4.3)
CHAPTER
5.
CONVERGE N CE
96
2. Suppose the real valued function 9 is
m
C5[t  ot, t
+
M], then g(t
+
M)  2g(t)
+
g(t  M)
=
(M)2 (Plg//(t
+
M)
+
Pog"(t)
+
Plg//(t  M))
+
R5(t), (5.4.4) where R5(t)
=Wl
{Rt(t) +Ri(t)} +W2{ Rt(t) +R5(t)

(M)2jt+6t
24
t
(t
+ M  8)2g(5)(8) d8  
(M)2jtc5t
24
t
}
(t
 M  8)2g(5)(8) d8 .
Proof
1.
Use Taylor's theorem to get:
9
(t
(M)2 (M)3
+
M)
=
9
(t)
+
Mg' (t)
+
2
g"
(
t)
+
6
g'"
(
t)
+
R; (t
) and
9
(t
 M)
=
9
(
t)
 M g'
(t)
(ot)2 (ot) 3
+
2 
g// (t)

6
g'" (t)
+
Ri
(t ) .
This yields
g(t
(M)3
+
M)  g(t  M)
2M g'(t)
+ 
g"'(t)
+
Rt(t)  R
;
(t).
3
(5.4.5)
Applying Taylor ' s theorem once more on
g' we obtain
g' (t
(M)2
+
M)
= g' (
t)
+
M g" (t)
+
2
g//' (t) +
11
2
t t 6t

+ (t
+
M

8)2g(4)(e) de
and
g' (t  M)
(M)2
=
g'
( t) 
M g"
(
t)
+
2 
g//'
(
t) +

1
2 t
1 t c5t
(tM8)2g(4)(8)d8.
The two equations yield
CHAPTER
5.
CO N VERGENCE g'(t
+
5t)
+
g'(t  5t)
=
2g'(t)
+
(5t)2 gl/'(t)+
97
From this we get an expression for into (5 .4.5). The result is
(5t)2g//'(t)
which can substituted
g(t
+
bt)

g(t 
bt)
(5.4.6)
Finally we combine (5.4.1) and (5 .4.6
) with weights
WI
and
W2
to get the desired result.
2. This proof is similar to the proof in
(1) .
D
Remark
The results above will also be used in the case where the function
9
is not defined for
t
<
O. In this case we ma y extend
9
by using the polynomial approximation on
[t 
5t ,
0). This will only influence the result in so far as there will be fewer remainder terms.
The second step is to apply the difference formulas to
Problem
AG a nd to estimate the errors.
A ss umption
We assume that
f
E
C3[O ,
T]
so that the solution
Uh
of Pro blem
AG is in
C
5
[0,T].
N otation
Illuhlllf
max
=
,
L
5
k=O
max
tE[O,T ]
Ilu~k)(t)IIE'
Notation
In the rest of this section
C b
will denote a generic constant that depends on the bilinear forms, i.e.
C b
is a combination of
C
E
and
C
I .
Notation
Illflhmax
=
L
3
k=O
max
tE[O , T I
Ilf(k)(t)ll·
CHAPTER 5. CONVERGENCE
98
Propo
s
i
t
ion
5.4.3
Suppose Ui
E
C
5
[t

M, t
{¢1 , ¢2, ... ,
+
Ml
faT
i
= 1, 2, ... ,
nand
¢n} is the basis faT Sh and let Uh(t)
=
L ~ = l
Ui(t)¢i' Suppose also that Po and P1 aTe Teal numbeTs such that Po
+
2P1
=
1 .
If Uh(t
+
M)

2Uh(t)
+
Uh(t

M)
=
(M?(P1U~(t
+
ot)
+
Pou~(t)
+
P1U~(t

M))
+ e~,
(5.4.7)
Uh(t
+
M)

Uh(t

M)
=
2M
(P 1U~(t
+
M)
+
Po ' u~(t)
+
P1U; 1(
t 
M))
+ e ~
(5.4.8)
and
(5.4.9)
then
lle~ll
S;
K(M)3 { max
eE[t,tHt j
I lu~3)(e)11
+ max
eE[t,tHtj
Ilu~4)(e ) 11
+ max
eE [ t,tHt)
Ilu~5)(e)ll }
and
P roof
Consider (5.4.7) as an example: Use (5.4.4) in P roposition 5.4.2 for denote the remainder by
U i
and
R
5i
(t),
N ow, each term in (5.4
.
7)can be written as a linear combination, for example,
Uh(t
+
M)
=
L n
i = l
Ui(t
+
M)¢i.
Consequently we have (5.4.7), if we set e~(t)
=
L~l
R
5i
(t)¢i'
CHAPTER 5.
CO N VERGENCE
It remains to estimate the error term e~(
t),
which is ac tually the sum of six error terms. Consider one of the terms: For any
(W I v
E
Sh t
[+01
(t
+
8t

e)3 u
;41 (e )t/>, de, v)
 1 [+01
W ,
(t
+
8t

e)3 (
u~41
(e), v) del
<
I t +bt
< t
IWII(t+Me) 31Iu~ 4)(e ) llll v llde
~
(5
t)4IwIl
llv
ll
max
Ilu~4)(
e
)
ll·
4
eE[t ,l + M)
H ence there exists a constant
K,
which depends only on the wei g hts
W 2
such that
( l e~ (t),v )1
:S
K (M)3
1Ivll { max
eE[t,t+6tj e El t , t +ot)
WI a nd
Ilu~3)(e)11
+ max
Ilu ~4)(e) 11
+ ma x
eElt,Hot)
lIu~5)(e)II}.
N ote that the worst of the errors are of order
(M)3.
Since
v
is a rbitr a ry, we have the desired result. The same procedure yields estimates in the energy norm. o
Lemma 5.4.
1
S up po se ' uh 'is the s olut ion of Problem
AG.
L e t Po and P I be real nu mb ers suc h that Po
+
2Pl
= l.
If u*(t ,
M )
i s defined by
(u*(t,
M) 
2Uh(t)
+
Uh
(t 
5t), v)
+
(M ) a(u *(t,
M) 
Uh(t

5t ) , v )
2 .
+ ( M )2 b(PIU *( t,
M)
+
POUh(t)
+
PIUh(t
 M
), v)
(M)2 (pI! ( t
+ M) +
Pof(t)
+
pI!(t
 M)
, v)o for ea x h v
E
S\
(5.4.10)
99
Proof
Using Proposition 5.4.3 we have
(Uh(t
+ M) 
2Uh(t)
+
Uh(t  M),
v)
+

(M)
2
a(uh(t
+ M) 
Uh(t

M ), v)
(M)2
( PIU~ ( t + M ) + Pou~(t) + PIU~(t

M), v)
+
(e 7 , v)
+(M )2
a(pIu~(t
+ M) +
Po ' u~(t)
+
PIU~(t

M), v )
+
(~t) a(e~,
v).
CHAPTER
5.
CO N VERGENCE
100 ow use the fact that
Uh
is the solution of Problem AG to prove that
Uh
satisfies (5.4
.
10) with
u(t
+
M) in stead of
u*(t,
M) provided that the error terms
(e~,
v)
and
(~t)
a(
e~,
v)
are included.
Consequently,
(u(t
+
M) 
u*(t, M), v)
=
(e~,
v)
+
(~t) a(e~,
v)
for each
v
E
Sil.
Re place
v
by
u*(t,
M) 
Uh(t
+
M) to obtain the estimate.
0
Reconsider the semi discrete system in Section 4.4.
Mul/(t)
+
Lu'(t)
+
Ku(t)
=
J(t)
(5.4.11)
u(O)
=
Ci ,
u'(O)
=
/1
To estimate the local errors for a finite difference scheme, we consider a onetoone correspondence between
Sh
and
lR n .
D
efini
t
ion
5
.4.1
For uh
E
Sh, the vector u n
=
Quh has components Ui wh ere u h
=
LUi(/Ji.
i = l
If we use the norm
1
IluliM
=
(Mu· up
for
lR n ,
then
IIQuhllM
=
Iluhll·
In our next result use the fact that
u
is a solution of (5.4.11) if and only if
Uh
i s a solution of Problem AG.
C
orollary 5.4.1
If u is a solution of the system of differential equatzons
(5.4.11) andu*(t,M) is defined by
M [u*(t ,
M) 
2u(t)
.
+
u(t

M) ]
+
(M) L [u* (t,
M) 
u(t

M)]
2
+(M)2 K [PIU*(t,
M)
+
Po u(t)
+
PIU(t

M ) ]
(M)2[PIJ(t
+
M)
+
PoJ(t )
+
PIJ(t

M)], then Ilu*(t,
M) 
u(t
+
M)IIM :::; C
b
(M)311Iuhlllf.max·
(5.4.12)
Proof
Consider the terms in (5.4. 10) .
If
fj
=
Qv h
)
then
(Uh( t
+
M), vh)
=
j '\1u(t
+
M) .
v.
In this way we can associate each term in (5.4
.12) with a corresponding term in (5.4.10). The resul t follows fom the fact that
u(t
+
M)
=
QUh(t
+
M). u*(t ,
M)
=
Qu*(t ,
M) and
0
C H
APTER 5. C
ONV ERGENCE
102
2. If we assume that
f
is merely conti nuous and hence
Uh
twice continu ous l y differentiable, we could still estimate the local errors but not obtain the same order. The results would be of the form:
Given
c
>
0 ,
there exists a real number
6.
>
0
such that the error will be less than
cOt
K for Ot
<
6..
(K
a constant depending on
Uh
and
f. )
5.4.2
Transformation
Due to symmetry considerations , it will be more conve nient to consider a transformed system for stability ana l ysis. Since
!vI
is symmetric and posit ive definite, there exists a symmetric positive definite matrix
N
such that
N 2
=
M.
Set
v(t)
=
Nu(t),
then
v
is a solution of the problem
v"
+
N  1LN1v'
+
N 1 K N 1v
=
N 1
J
or
v"
+
Lv'
+
Kv
=
.9. where
L
=
N  1 LN1
,
K
=
N l
K
N 
1 and
9
=
N l
].
(5.4.14)
The advantage of the transformation is that the matrix
K
is symmetric , and hence has orthogon al eigenve ctors.
Let
y
=
Ni ,
then
Ki
=
A j Vff
if and on l y if
The eigenvalues of K are the eigenvalues of the eigenvalue problem Pro blem CG (See Sections 4.1 and 5.2.)
W e use the norm
IIi l12
=
(i . i)
~ , and in the remaining part of this sectio n
II . I I
will refer to
I I
.
11 2 unless stated otherwise.
Corollary 5.4.3
If v is a solution of the system of differential equations
(5.4.14), and v*(t, Ot) is defined by
[v * ( t , Ot)
 2v(t)
(Ot)
~
+
v(t 
Ot)]
+
2
L [v * (t, c5t)

v(t

Ot)]
+(Ot f i( [ PIV*(t, Ot)
+
Pov(t)
+
PIV(t

Ot)]
(Ot)2 [ Plg(t
+
Ot )
+
POg(t)
+
Plg(t  Ot
) ],
CHAPTER
5.
CO NV
ERGENCE
103
Proof
Direct from Corollary 5.4.1, since
v*(t,M)
=
N u*(t,M).
0
C
orollar
y 5
.4
.4
If v is a solution of the system of dIfferential equations
(5.4.14), and v* *(t,
M)
is defined by
2
[v** (t,M)

v(t )]
(
M)2
+
2K
[v**(t,
M) +
v(t)]
(6t)2
2[g(t
(6t)3
+
M)
+
g(t)]
+
2Mv '(t)

(M)2 Lv'(t)
+
2Kv'(t)
(M)3 _'(
2
t
)
'
then
Il v* *(t,
6t)

v(t
+
M)II ::;
C b
(6t)3
(1IIuhl
l
l f,max
+
(M )311Ifl hmax) .
Pro
o
f
See Corollary 5.4.3.
Remark
The result remains true for
t = o. o
5.4
.
3 G l o bal e rror
v V e approximate the solution of (5.4
.14) on the interval
[0 ,
T ] .
Let M indicate the time step len gt h , i.e.
6t
=
T
/
N,
and let
Wk
denote the approxi mation for
v(t k )
.
W e use the difference sc heme (which co rrespond s to (5.4
.12))
(

l 
2
+ )
+
(M)L(

Wkl
+(M)2 K(PIWk+l
+
POWk
+
PIWkd
=
(6t)2(Plgk+l
+
POgk
+
Plgkd·
(5.4. 15)
The initi a l conditions for the system of differenti al equations are
v(O)
=
No:
and
v'(O)
=
N iJ,
and the initial conditions of the finite difference system are
C HA PTER
5.
C ON V E R GEN C E
To estimate local errors the following scheme will also be used:
104
To estimate the global error tions
w
N
V
(T)
we introduce artificial numerical sol u
wk')
For each i,
wk i
)
satisfies
(5.4.14) with
w?)
=
v(t i )
and
(i) 
W i
l
WHl 
2
s:t/(t)
U
V i .
N
0 t th t 

_ _
(0)
.
For the global error we have
Ilv(T)  WNII
~
Ilv(T)  wtl)11
+
I l wtl
) w~V2)11
+ ... +
Ilw~)

~wNII·
(5.4.17)
( N ote that the global error fo r the original system can be derived fro m this error. )
It is clearly necessary to estimate
Ilw~)

~w~l)ll.
The next t w o subsections will be devoted to the estimation of the differences between "neighbouring numerical solutions" .
5.4.
4 Consistency
In this subsection we consider the differences
II
+
2 
(i+l)
.
. with the "starting" error. l"t d t
Ilw~21

v(ti+d
II and
(t)
b Th fi t I d
1
L emma 5 .4.3
Proof
This is a direct consequence of Corollary 5.4.4.
1 ext we have the error at the second step. o
C HA P T ER
5.
CONVERGENCE
Lemma 5
.4.4
105
Proof
Combine the results of Corollary 5.4.3 and Lemma 5.4.3. 0
Lemma 5.4.3 provide an estimate for the difference
W~i)

W~21.
The following result provide an estimate for the difference at the second step.
C
orollary
5 .4.
5
Ilw;22 
w;~~l)11
::;
C
b(bt)3
(1IIud;'max
+
Illflhmax) .
P
roof
Use Lemmas 5.4.3 and 5.4.4.
II
WH2 
W
(Hl)
H2
II
<
II
_(i
i
+
)
2


Vi+2
II
+
11
Vi+l 
(Hl) II
.
o
5.4.5
Stabilit
y
For the stability analysis we introduce the following matrices:
A
B
C
(bt)
1+L+Pl(bt)2K ,
2
=
21
+
PO(bt) 2
K,
1 
(bt)
2
Z
+
Pl(bt)2
K.
The system (5.4.15 ) is now
(5.4.18)
As mentioned at the end of Subsection 5.4.3, we need to estimate the differ ence w~) _W~+l) for each i. Since both
W;i)
and
wy+l)
satisfy the system
(5.4.18) it follows that the error
ej
=
wJ i)
w;Hl) must satisfy
(5.4.19)
CHAPTER
5.
C ONV ERGE N CE
106 with the starting values, the local errors
ei+l
and
ei+2,
already estimated.
For the case
L
=
J.1K ,
we derive the eigenvalues of
A,
Band
C. If
Ky
=
).,y,
then
By
Cy
+
J.1Ky
+
Pl(cSt)2Ky
=
(
1
+
2(J.1).,)
+
Pl(cSt)2/\
) y,
2
2y
+
Po
(cSt)2
Ky
=
(2 +
Po (cSt)2
).,)
y,
(1
~t(J.1).,)
+
Pl(<5t)2).,)
y.
It is now possible to solve (5.4.19). Let Yl, Y2, ... , Yn denote the normalized eigenvectors of
K
and suppose
e i +l
=
.L~1
TJ(Y i
and
ei+2
=
.L~1 ~iYi'
Since the eigenvectors are orthogonal , it is sufficient to solve difference equa tions of the form where
ai,
{3i and
Ii
denote the eigenvalues of the matrices
A,
Band
C respecti vel y.
The following result can be obtained by elementary calculations. Note that we do not use the subscripts for the coefficients
a,
{3 and i. We take
Tl =
~ and
TO
=
TJ·
Solution of t he differenc e equation
C ase 1
(32
<
4ai.
N ote that in this case
I
>
O. The solution is of the form
Tk
=
pk(A
coswk
+
B
sinwk),
wherep
=
Vi/Ct.,
cosw
=
(3/(2...flYY), A
=
TJ
and
B
=
(~
PTJCosw)/(sinw).
Cas e
2
{32
=
4Ct./, .
The solution is of the form
CHAPTER 5. CONVERGENCE
The solution is of the form
107 with
7'1 and
r2
the real roots of the equation
o:r2
+
;3r
+
r
=
O. The constants are
A
= (~
1]T2)/(rl  r2) and
B
= (~
1]Tr)/(r2  rr).
We now prove the stability result. Bear in mind that
)..k
+
(X) as
k
+ 00.
Lemma 5.
4 .5
Stabzlzty
If Po ::; 2pl) then there exists a constant K independent of the dimenszon
of Sh 
s
uch that
II
W N  W N
II
K
(II
::; 'Ui+l 
W i
+
1
II
+
II
(i) (i+l)
II)
.
Pr
o
of
For the eigenvalues
1
M
+
Pl(M)
2
)..
+
2)..jJ,
2
+
Po(M? )..,
l+Pl(M)
2
M
)..2)..jJ
of
A,
Band
C, we get
4Po(M)2)..
+
P6(c5t)4)..2  8Pl(M)2)"4pi(M)4)..2
+
(M)2 jJ2)..2
)"(M)2 [jJ2).. 
4
+
(P6  4Pi)(M)2)..] .
Consider the different cases:
C a se
1
If
;32
<
40: r
then
rk
is bounded if "/ ::;
0:.
This case,
;32 
40: r
<
0, is possible only for a finite number of small eigen values and only if
Po
~
2pl. Since
r/O:
<
I, the corresponding modes will not cause error growth.
C as e 2
If
;32
=
40: r
then
rk
is bounded if
1;3/0:1
<
2.
If
PO(M)2).. < 2, we have
1;31/10:1
<
f·
If
Po(M)2)..
>
2, we have
Po ( c5
t )
2 )..
Po .
1/31/10:1
<
(c5
)2)..
= ::;
2 If
Po::;
2Pl'
PI
t
PI
CH AP TER
5.
C ONV ERGE N CE
108
C ase 3
If
13
2
>
4CYi
then
Tk
is bounded if both roots of
CYT
2
+
j3T
+ /'
=
0 are less than one in absolute value. Let and let
T
max denote the absolute value of the root largest in absolute value.
Tmax
<
13
+
J75,
2cy
PO(bt)2)...
+
)"'6t
VM2
+
(P6  4pi)(6t)2
2pl(6t)2)...
+
6t)...M
Po6t
+
VM2
+
(P6  4pi)(6t)2
2P1 6t
+
M
Po6t
+
M
<
2P1 6t
+
M
<
1 ( if
Po
S;
2pl)·
o
Remark
If damping is excluded, the difference system is considered to be unconditionally stable for
PI
=
~ and
Po
=
~, see [RM] or [Zi]. However, the bound may depend on the eigenvalues. cosw
= '
2y1CFi
 Po ( 6
t )
2 )...
+
2
2(1
+
PI (6t)2)...)
po(bt?
+
2)",
1
Po
2Pl(bt)2
7 
+
2)...1
2Pl
as )...
7
00.
Consequently, sin
w
7
0 as )...
7
00 and sin
w
is present in the numerator of a constant.
R e marks
l. Exactly the same results hold if we assume that rotary inertia can be ignored and we have only viscous damping. In this case
L
=
kMo and
M=Mo·
2. The eigenvalues of
K i
=
)...j'vloi
are much larger than the eigenvalues of
Ki
=
)...j'vli
(with rotary inertia).
3. Rotary inertia and KelvinVoigt damping both enhance stability.
CHAPTER
5.
CO N VERGENCE
5.4.6
Convergenc
e
L
emma
5.4.6
Global error
109
(w here
N
is the number of steps).
Proof
IleN11
<
I
lv(T)
w ~ ~ lll
+ ... +
Ilw~ )
 wN11
<
K N
max{
2
Ilvi+l  W;2111
+
Ilv
i+
l w;2211}.
N ow use Lemmas 5.4.3 and 5.4.4.
To the se quence of finite differ ence vectors Vk, correspond a sequence of ap proximations for
uh:
u~k )
=
(Q J)lVk
E
Sh.
o
T
heorem
5.4.1
If
Uh is the solution of Probl em AG. then
Proof
Since
eN
=
QN(U n(
T) 
u;:),
we have
N ow use Lemma 5.4.6. o
Remark
Error estimates for the fully discrete system is obtained by com binin g Theorem 5.4.1 with the results of Section 5.3. Note that the error estim ates are with respect to the inertia norm.
Chapter
6
Application. Damaged beam
6.1 Introdu
c
tion
We consider Problem 1 (from Section 2.3). This model for a damaged beam was proposed in [VV] . See also [JVRV].
The detection of damage in structures or materials is clearly of great im portance. Ideally it should be possible to infer the location and extent of damage from indirect measurements or signals. To facilitate such deduction , a mathematical model of the object or structure is necessary. See [ VV ] for details and numerous other references.
Viljoen
et al.
[VV] use changes in the natural frequencies of the beam to loca te and quantify the damage. The natural angular frequencies for the damaged beam are calculated from the characteristic equation obtained from the as s ociated eigenvalue problem. As is wellknown, only the first few n atu ral angular frequencies and modes are usually calculated with this method, because of computational difficulties with the hyperbolic functions. Due to this limitation, the need arises for a numerical method to simulate the dynamical behaviour of the beam.
In a joint paper, [ZVV], we developed a finite element method (FEM) to ap proximate the solution of the model problem for arbitrary initial conditions.
(Ironically we also found it possible to calc ulate eigenvalue s a nd eigenfunc tions more accurately with the FEM .
)
It wa s nec es sary to adapt sta nd ar d procedures to deal with the discontinuity in the derivative that arises a s a
110
CHAPTER
6.
APPLICATION. DAMAGED BEAM
111 result of the elastic joint. W e made the assumption that damping would not influence t he so lution significantly on a small time scale. We now inves t igate the validity of this assumption, and als o deem it prudent to include the effect of rotary inertia.
In the paper [ZVV], only He rmite piecewise cubics were used as b as i s func tions.
In
this investigation we a lso demonstrate the effectiveness of Hermite piecewi se quintics.
From Section 3.1 we have the variational formulation. The Galerkin appro x i mations for the eigenvalue and initial value problems a re given in Section 4.l. eW e do not cons ider the equilibrium problem.) In Section 4.2 we showed ho"v the stand ard basis functions are adapt ed to deal with the discontinuity
111 the deriv ative.
In
Section 6.2 we compute the n atura l angular frequencies and modes of v ibr ation from the characteristic equation for comparison purpos es. The computation of the matrices is discussed in Section 6.3.
I n
Sections 6.4 and 6.5 numeric al results are presented that demonstrate not only the effect of dama ge on the motion of a beam but also the effect of dampin g and rotary inertia. We a lso investi gate t he use of Hermite piecewise quintics as basis functions instead of Hermite piecewise cubics.
CHAPTER
6 .
APPLICATION. DAMAGED BEAM
112
6.2
N
atural fr
e
qu
e
nci
e
s and
m
odes of vibra
t
i
o
n
One way to calculate the natural angular frequencies and modes of vibration for the damaged beam is to apply the method of separation of variables directly to Problem 1 (from Section 2.3).
For the case
T
=
0 (without rotary inertia), we have the following eigenvalue problem:
W(4) 
.Aw w(O) w(a+) w"(a
+)
w'"
(a +)
w" (a)
0,
0 < x
<
1,
x
I:a,
w'(O)

w"
(1)
w(a), w
"( a
)
,
w
'"
( a
)
,
= w"'(l)
0,
~(w'(a+)
 w'(a)).
For this eigenvalue problem it is possible to find so called exact solutions .
It is convenient to introduce the positive real number
u,
with
.A sequently
u
2
= vf).. is a natural angular
=
u
4
.
Con frequency. Analogous to the case of the undamaged beam, the corresponding mod e is of the form
_ ( Asin(ux)  Asinh(ux)
+
Bcos(ux)  Bcosh(ux) for 0
<
x
< a ,
w(x) (C
+
A)
sin(ux)
+
(D  A)
sinh(ux)
+(E
+
B) cos(ux)
+
(F
 B)
cosh(ux)
for a
<
x
<
1.
N ote that the boundary conditions at
x
=
0 have already been taken into account.
From the continuity conditions and the jump condition at
x
= a, the con stants
C, D, E, and
F
can be expressed
in
terms of
A
and
B.
Finally, from the two boundary conditions at x
=
1, the characteristic equation for u can be constructed from
(6.2.1)
CHAPTER
6.
APPLICATION. DAMAGED BEAM
113 where
(sin
v
+ sinh
v)
+
ov (sin
va
+ sinh
va)
x
2
(sin
v cos va sinh
v cosh va cos
v sin va
+ cosh
v
sinh
va),
( cos
v ov
+ cosh
v)
+ ( cos
va
+ cosh va) x
2
(sin v cos va sinh v cosh va cos v sin va
+ cosh v sinh va)
(cos
v ov
+ cosh
v)
+
(sin
va
+ sinh
va)
x
2
(cos
v
cos
va
cosh v cosh
va
+ sin v sin va
+ sinh
v
sinh
va),
(sin
v
sinh
v ) ov
+
(cos
va
+ cosh //a) x
2
(cos
v cos
va
cosh
v cosh va
+ sin
v sin va
+ sinh
v sinh va).
Solving equation (6.2.1) numerically using the NewtonRaphson method, yields the natural angular frequencies for the damaged beam. For each na tural angular frequency a corresponding mode can then be obtained. As is expected, only the first few natural angular frequencies and modes could be calculated, as it is difficult to handle the hyperbolic functions numerically for large values of
v.
N umerical results obtained using the finite element method using cubics as well as quintics as basis functions are given in Section6.4.
114
CHAPTER
6.
AP PLICATI O N. DAMAGED BEAM
6.3 Computation of Matrices
The matrices
K ,
Land
M
are defined in Section 4.1 in terms of the bilinear forms defined in Section 3.1. The computation of the matrices is complicated by the interface conditions which results in nonstandard basis elements.
In this section we give an indication of how we went about in computing th ese matric es. The first step is to reorder the basis elements constructed in
Subsection 4.2.2.
Consider the matrix
j \1 o:
[ J\!IoJi
j
= (¢i'
¢j)
=
l Q cPjl
+
1 1 o te that
¢i
(0,
cPi)
or
(cPi ,O)
except when we are dealing with a basis element associated with the node
xp
=
0: , the location of the damage. In general then, the entries will be those of the standard mass matrix for an undamaged beam. N ow suppose one of the basis elements are associated with
xp :
Again the result will be the same as in the standard case. The same for
(2)
cP P .
On the other hand, suppose
¢i
=
¢;
2 ) then
[j\1
0 l ij
=
let cPj1 cPi l
+
0 , which is not the same as for an undamaged beam. Similarly for
¢i
=
¢;~.
Thus the standard matrix h as to be modified for the damaged beam.
W e have the same situation for the matrix
M r
where we define
There is an additional complication for the
K
matrix:
CHAPTER
6.
APPLICATION. DAMAGED BEAM
115
Only four entries in the standard Kmatrix will change due to the additional term,
Cu~(ex)
 'U'l(ex))
(v~(ex)
v~(ex))
/6, in the bilinear form
b.
For greater clarity we will explain the procedure in another way. In the discussion that follows, we refer to
¢~k) as a Type k basis function.
In modifying the matrices for an undamaged beam to the matrices for a dam aged beam, we have to keep in mind that the Type 1 basis function associated with
xp
= ex, has changed. By replacing the row and column associated with the Type 1 basis function at
x
P'
in the matrix of the undamaged beam, by two rows and columns respectively, provision is made for the modified basis function. The values in the matrix in these two rows and columns have to be modified accordingly. For the
K matrix one must also keep the additional term in mind.
Having computed M
o,
Mr and K we are done since L
p,K
M
=
j\!fo
+
Mr.
+
kMo
and
CHAP T ER
6.
APPLICATION. DAMAGED BEAM
116
6.4
N
um
e
rical result
s
. Eig
e
nvalu
e
problem
Cubics as basis functions, are usually sufficiently accurate in solving one dimensional vibration problems with the finite element method. In a joint paper
[ ZV V] we discussed the use of cubics as basis functions for the damaged · beam.
In this section of the thesis we also consider numeri ca l converg e nce of th e eigenvalues. The order of convergence that is sugg e sted b y the numeri ca l results is also compared to the order obtained from the th eory. Additionally, quintics are considered as basis fun c tions. Th e main reason for this is that cubics are not compatible with reduced quintics in plate beam models . We also investigate the effe c t of rotary inertia.
6.4.1
C
ubics
N atural angular frequencies and modes for the vibration problem are calcu lated by solving the eigenvalue problem with the FEM. 'vVe developed the code to construct the relevant matrices in Matlab and use standard Matlab subroutines to calculate the eigenvalues and ei ge nvectors of the generalised eigenvalue problem.
It i s possible to compare only the first few FEM eigenvalues to the so called exact eigenvalues calculated from the characteristic equation. Thereafter the exact values can not be computed accurately a nd the F E M is used to calculate the eigenvalues.
In Table 6.1 we list values for the eigenvalues obtained from the characteristic equation (see Section 6.2) and values obtained by the FE M using cubic s a s b as is functions with 20, 40, 80 and 160 subintervals respectively. This give approximations for respectivel y the first 40, 80 , 160 and 320 eigenvalues.
CHAPTER
6.
APPLICATION.
DAMAGED BEAM
117
). (20) t
).
(80) t
1 11.81469 1l.81469
11.81469 11.81469 1l.81469
2 406.01614 406.01757 406.01623
406.01615 406.01615
3 3806.05283 3806.17742 3806.06067 3806.05332 3806.05287
4 12544.12940
12545.47137 12544.21439 12544.13473 12544.12972
5 39943.82322 39957.35763 39944.68387
39943.87724 39943.82661
Table 6.1:
Eigenvalv
,es
from the characteristzc equation as well as FEM ei genvalues using cub
ics
as basis functions with
c5
=
0.1
and
Ct
=
0.5.
Throughout. t.his sed .in
n
n
clellntp. th>:> nu m be r of subintervals. (All of e quoJ length. )
To investigate the convergence of the FEM eigenvalues, we calculate the relative difference between
FEM approximations, that is ( ).
(2n)
).
(n))
/ ).
(2n).
These differences are calculated and listed in Table 6.2 for n
=
20, 40, 80 and 160 subintervals respectively. n
=
20 n
=
40
I
n
=
80 n
=
160
6
6.1
X
10
4
3.9
X
10
5
2.5
X
10
6 l.6
X
10
7
12 l.1 x 10
'4
7.6 x 10
4
4.9 x 10
5
3.1 x 10
b
24 2.2 x
10
1 l.2
X
10
2
8 .6
X
10
4
5.5
X
10
5
48

2.2 x 10
1
1.3 x 10
2
9.2
X
10
4
Table 6.2:
functions.
Relat
ive
differences fOT FEM eigenvalues using cubics as basis
The tendency of the relative difFerence to decrease (by roughly a factor 10) each time that the number of subintervals is doubled , is empirical verifica tion that there is convergence of the FEM eigenvalues. Vve found th a t the eigenvalues computed from the characteristic equation were less dependable.
It is necessary to determine a relationship between the number of
FEM eigen values that is suffi c iently accurate (c riterion to be specified) and the number of subintervals used.
A relative difference strictly lCtls than 10
3 is considered sufficiently accurate for our purpose. Using this as criterion, we find that approximately a sev enth of the 2n eigenvalues calculated using n subintervals, yields a relative difference,
(
).(2n) ).(n))/).(2n) , strictly less than 10
3 , see Table 6.2.
CHAPTER
6.
APPLICATIO
N.
DAMAGED BEAM
118
The relative difference between the FEM eige nvalues with 160 an d 320 subin tervals is an indi cation of the relative error between the exact eigenvalue an d the FEM e igenv a lu e using 320 subinterval s.
Since we u se
(>,(320) >(160))/ >(320) as measure of the relative error,
(>>(3 20)) / >, we conclude that the first 90 eigenvalues obta ined usmg
320 subintervals y ield a relati ve e rror that is sufficiently accurate.
A n indic at ion of the order of conve rgence of the FE M eige nvalue s can be obtained from the ratio of two success ive differences
>(2n) _ >(n)I/I>(4n) _
1 t t t
>
(2n)
I t '
Typical results are listed in Table 6 .3
. z
1
3
6
12
24
l>i:3n
)
_
>~n) I /I >;4n) _
>?n)
I
n
=
20
n
=
40
n
=
80
10 .25
15.88
15.53
14.23
18.29
0.03
16.41
15.88
15.57
14 .47
0.14
6.57
15.72
15.90
15.62
Table 6.3
:
Relationship between
su ccessive
relative
differences wz th cubzcs
as basis functions.
These relative differen ces decrease b y roughly a factor 16 if the number of subintervals is doubled. F rom this it would appear tha t the co nver gence i s of order
h4
whi ch matches the theory, Section 5.2.
It is observed that those differences not yieldin g a factor 16 typically occur in the ri ght top pa rt as well as the lef t bottom part o f Table 6.3. These deviations are illus trate d by the first , third and 24th eig enva lue s:
Fir st l y, the accuracy of an a pproximation can decr ease if the number of s ubin te rvals is increased. This i s due to an increase in the roundoff error and has significant effe cts in situations where the errors ar e alr eady small. For example, FEM approximations for the first eigenvalue y i eld
(>i40) _ >i20))
=
1.1 x
10
13 while
(>i80 ) >i40 ))
=
3.8 x
10
12 .
From the theory, Se ct ion 5.2, we know that the FE M approxirnations of a n eigenvalue will decrease i f the number of subintervals is in crease d. This can
C H APTER
6.
A PPLICATION. DAMAGED BEAM
119 be used to detect cases where the effect of the roundoff error is greater than the advantageous effect of an increase in the number of subintervals used.
R ounding error also explain the decrease in the ratios for the third eigenvalue from roughly a factor 16 to 6.5. This situation differ from the first eigenvalue in that the decrease (improvement) in the relative difference was just partially cancelled by the increase in the roundoff error.
Secondly, as we have showed previously, there is a relationship between the number of FEM eigenvalues that can be calculated sufficiently accurately and the number of subintervals used. The 24th eigenvalue is such an example.
The effect of the poor approximation of
.\~~O ), is seen in Table 6.3 in that
18.29
>
14.47. This is expected as only the first six eigenvalues obtained , using 20 subintervals, yield relative errors less than
10
3
6.4.2
Quintic
s
W e now consider quintics as basis functions, and compare the results to the case where we used cubics.
In Table 6.4 we list values for the eigenvalues obtained from the characteristic equation and values obtained by the FEM using quintics as basis functions with 2, 4, 8 and 16 subintervals respectiv e l y . This gives approxim a tions for respectively the first 6 , 12, 24 and 48 eigenvalu e s .
1 11.81469 11.81469 11.81469 11.81469 11.81469
2 406.01614 406.01954 406.01618 406.01614 406.01614
3
3806.05283 3822.51900 3806 .
09344 3806.05297 3806 .
052 8 3
4
12544.12940 12844.88875 12544.53719 12544.13441
12544.12941
5 39943.82322
41569.18041 40042.72518 39944.02589 39943.82387
Table 6.4:
Eigenvalues
from the characteristic equation as well as
FEM
ei genvalues
using
quintics as
basis
functions with
<5
=
0.1
and
ex
=
0.5.
If these values are compared to those in Table 6.1, it seems as if the same accuracy can be obtained, using quintics as basis functions, with less subin tervals, than in the case where cubics were us e d as basis functions. For example, the fifth FEM eigenvalue using quintics as b a sis functions with
CH AP TER
6.
APPLICATION. DAlvIAGED BEAM
120
16 subintervals, already yields a better approximation than using cubics with
80 subintervals.
As in the case with cubics as basis functions, we investigate the convergence of the
F E M
eigenvalues by considering relative differences, PI
(2n) _)., (n)) /)., (2n).
These values are are list e d in Table 6.5 for 2, 4, 8 and 16 subintervals re spectively.
'/,
1
2.1
n =2 n=4 n = 8 n
=
16 x
10
8
7.9
X
10
11
1.2
X
10 
11
1.7
X
10
11
2 8
.
3
X
10
6
4 2.4
X
10
2
9.5
X
10
8
3.2
X
10
5
3.4
X
10
10
4.0
X
10
7
1.3
X
10
11
1.4
X
10
9
8 
3.9
X
10
2
7.2
X
10
5
8.2
X
10
7
Table 6.5:
functions .
Relative differences for FEM eigenvalues using quintics as basis
The numerical results suggest convergence of the FEM eigenvalues s ince the relativ e error decreases (by roughly a factor 100) each time that the number of subintervals is doubled, see Table 6.5.
For approximately a third of the
3n
eigenvalues computed , u s ing
n
s ubinter vals , the relative differ e nce
().,(2n)

).,(n))/).,(2n)
is strictly les s than 10
3
.
As with the cubics, we now consider the ratio of two successive differences to get an idea of the order of convergence. Typical results are listed in
Table 6.6.
As was the case in T able 6.3, the values in the top right of T able 6.6 exhibit effect of roundoff error and the values in the bottom left the result of eigen values not calculated sufficiently accurately. From this it would appear that the order of convergence is
h
8
which matches the theory, Section 5.2.
To compare the accuracy of the FEM eigenvalues using quintics as basis function s to the case using cubics as basis functions , we choose the number of subintervals in each of the cases such that the sizes of the matrices in the two cases are equal. For example, using 30 subintervals for cubics yie ld
61 x 61 matrices and 20 subintervals for quintics 62 x 62 matrices. We then
CHAPTER
6.
AP P LICATIO N. DAMAGED BEAM
121
~
IAr~n)
Ain)I/IAi4n)
Ai~n)1
n=2 n=4 n=8
1
265.60 6.75
0.69
2 87.21 278.22 25.45
3 405.90 278.67 319.70
4 745.69 80 .
59 291.89
5 15.43 488.57 311.65
6 16.24
330.71
307.27
7 304.98
166.95
286.96
8
302.37 540.94 87.70
Table 6.6:
Relationship
between successive relative
differen
ces with quintics basis functions.
compare the eigenvalues calculated in the two cases with the eigenvalues computed using cubics with 320 subintervals. (We use the first 90 FEM eigenvalues using cubics as basis functions with 320 subintervals as the FEM approximation to the first 90 exact eigen val ues.)
Our numerical experiments indicate that using quintics with
n
subintervals , yield at least double the number of eigenvalues to the prescribed accuracy
(relative error strictly le ss than 10
3
) than when cubics a re used with
3n/2
subintervals. In Table 6.7 we give an example of results obtained.
In Table 6.7 we use the following notation:
• Let Ai denote the ith FEM eigenvalue that we use as approximation for the exact eigenvalue. (In this case those FEM eigenvalues obtained using cubics as basis fun ct ions with 320 subintervals.)
• To distinguish between the FEM eigenvalues computed using quintics and cubics as basis functions, we denote the ith FEM eigenvalue using cubics with 30 subintervals by
A~c) and using quintics with 20 subinter vals by
A~q ).
N ote that the FEM approximations for the first eigenvalue are identical in both cases .
From Table 6.7 we see that using cubics, the first 9 eigenvalues (app roxim ately a seventh of the number of eigenvalues calculated, 61/7
~
8.7) have
CHAPTER
6.
APPLICATION. DAMAGED BEAM
122
1
2.6
X
10
6
5
6.8 x
10
5
9
8.5
X
10
4
10
1.2
X
10
3
15
6.8 x 10
3
20
1.9
X
10
2
2.6
X lO b
9.6
X
10
9
4.3
X
10
7
8.6
X
10
7
4.6
X
10
5
1.4
X
10
4
Table 6.7:
Comparzng FEM ezg envalues us zng quintics wit h
20
subzntervals to FEM ei genvalu es using cubics wzth
30
subintervals.
relative difference less than 10
3
.
Using quintics , the first 20 eigenvalues, that is approximately a third of the number of eigenvalues calculated , have relative difference less than 10
3
.
In conclusion , for the same computational effort (same size of the matrices), quintics yield twice as many eigenvalues sufficiently accurate than when cu bics are used i.e. to obtain the first
k
FEM eigenvalues with relative differ e nce less than 10 
3
)
7k/2
subintervals must be used with cubics as basis functions and
k
subintervals with quintics .
6.4
.3 T h e
effec
t
of
rot a
ry in
ert
ia
W e now consider the effect of rotary inertia on th e eigenvalues and use quintics as basis functions.
N ote that this eigenvalue problem differs from the one excluding rotary in e r tia, Section 3.5. The parameter
r
is a measure of the effect of rotary inertia,
Section 2.2.
W e start by establishing convergence of the FEM eigenvalues for the case where rotary inertia is included, thereafter, we investigate the effect of rotary inertia on the eigenvalues.
As for th e case without rotary inertia, the numerical re s ults indicate conver ge nce of the FE M eigenvalues. In Table 6.8 typical results for the relative differences,
(>(2n)
 >
(n))/ >(2n),
including rot ar y inertia , are listed for 2 , 4 , 8 and 16 subintervals respectively.
CHA P TER
6.
APPLICATI ON. DAMAGED BEAM
123
n=2 n=4
I
n=8
n
=
16
1
2.1
X
10
8
2 7.8
X
10
6
7.8
8.9
X
X
1O
10
11
8
2.9
3.2
X
X
10
10
11
10
7.6
1.4
X
X
10
10
10
11
4 1.6 x
10
3
4.0 x
10
6
2.0 x 10
tl
7 .
9
X
10
11
8

1.6
X
10
2
9.5
X
10
5
3 .
3
X
10
7
Table 6.8:
Relatwe differences for FEM eigenvalues including rotary inertza with 1 1 r
=
4800 .
The numeric al results again suggests convergence of the FEM eigenvalues .
The same pattern with respect to the order of convergence is observed as for the case without rotary inerti a.
The presen ce of rotary inertia decreases the values of corresponding eigenval ues in comparison to the case without rotary inertia. Furthermore, the bigger the parameter
r,
the greater the change in the eigenvalues in comparison to the case without rotary inertia. In Table 6.9 we list eigenvalues for different values of
r
as well as for the case without rotary inertia
(r
=
0). We use 32 subintervals for these approximations.
In Table 6.9
Ai
denotes the ith FEM eigenvalue.
I
i
I
A i with
r
=
0
I
11r
=
19200
I
11r
=
4800
I
11r
=
1200
I
1
2
11.81469
406.01614
11 .
81138
403.97925
11.80145
397.71100
11.76186
370.06960
4 12544.12940 11057.05461 5401.71657 3576.19925
8 273293.79309
169832.12061 158744.64019 125970.79578
Table 6.9:
FEM eigenvalues for different effects of rotar y inertia using 32 subintervals.
These results are for the dimensionless case. W here rotary inertia is included , two dimensionless constants ,
T
a nd
r,
must be calculated if the results is to be connected to a spec ific beam, Section 2.2.
M
ode
s
In [ Z VV] we showed that only up to the seventh so called exact mode can be computed before computational difficulties are encountered. Therefore we
CHA P TER
6.
APPLICATION. DAMAGED BEAM
124 consider the convergence of the FEM modes using quintics as basis functions and include rotary inertia.
Let w~n) denote the F E M approximation for the ith mode using
n
elements, normalised with respect to the infinity norm ,
II .
1100'
The way in which we ord ere d our ba5i5 elements implies that the firsL
n
components of w~n) are associated with the function values at the
n+
1
+
1 nodes.
The next
n
+
2 values represent the values of the first order derivatives at the nodes. Two values are associated with the point where the damage occurs.
Quintics as basis functions also yield approximations for the values of the second order derivatives, and the last
n
+
1 values of w~n) represent the values of the second order derivatives at the nodes.
In
Table 6.10 the numerical convergence of the FEM modes are illustrated.
W e list the differences
Ilw ~ 2n) w ~ n)lloo, II(w?n
)y ( w~n)Ylloo and
II ( w?n))" 
(w~n))"l l oo for different values of
n.
CHAPTER
6. APPLICATION. DAMAGED BEAM 125 z
1
2
4
8
14
16
Ilw;LnJ  wtJ
1100
II(wi2n)), (win))'lloo
II
(w?n))"  (win))"
II
00
n=4 n=8 n
=
16
6.83140
X
10
6
4.56592
X
10
7
2.578292
X
10
8
9.65478
X
10
0
6.45299
X
10
1 3.64876
X
10
I)
2.48714
X
10
0
6.33818
X
101) 8.43408
X
10»
2.49236
X
10
5
2.09579
X
10
6
1.47615
X
10
7
1.18092
X
10
4
9.93009
X
10
0
6.99423
X
10
1
1.96527
X
10
'I
6.46058
X
10
0
2.06282
X
10
I
1.46719
X
10
5
6.47026
X
10
6
1.61154
X
10
4
7.01558
X
10
5
5.44719
5.90054
X
X
10
10
7
0
8.16804
X
10
3
2.53189
X
10
4
8.2
8 3 69
X
10
0
3.70776
X
10
4
6.16539
X
10
6
1. 75076
X
10
0
7.90080
X
10
3
1.30886
X
10
4
3.71925
X lOb
6.26608
X
10
L
1.00243
X
10
 j
2.90161
X
10
b
3.20162
X
10
4
9.48592
X
10
5
9.65450
X
10
7
1.68312
X
10
£
4.23303
X
10
· 3
4.29
8 24
X lOb
1.17602 1.65933
X
10
1 1.00174
X lO
L



9.03027
X
10
4.78628
X
10
5
3
4.81429
X
10
6
2.60210
X
10
4
1.91004
X
10
1
2.44992
X
10
L
Table 6.10:
Convergence of FEM modes with 6
1/r
=
4800.
0.1,
Go
0.5 and
The rate at which convergence of the function values and the first order derivatives occur, differ from the convergence rate of the second order de rivatives, which is much slower. Those modes that are associated with the first eigenvalues, starting with the smallest, converges faster than the modes associated with later eigenvalues. (Convergence in the energy norm implies that the second order derivative converges in the mean.)
CHAPTER
6.
APPLICATION. DAMAGED BEAM
126
6.5
Numerical
r
es
ults. Initial
value
probl
e
m
Consider the initial value problem. From Section 4.1 we have the following system of differential equations
Mul/(t)
=
Lu'(t)  Ku(t).
For the numerical experimentations ) we choose the following initial condi tions: u~(O)
=
0 and
Uh(O)
a quintic "solitary wave)).
To approximate the solution of this problem we use the difference scheme in
Section 5.4 with
Po
=
2Pl
=
1 / 2.
(
M L
M2 2M 4
1)
Uk+l+
(!VI
bt
2
2
1 )
!VI L
(
M2 2M
1)
+
4
K
Ukl
=
O .
Since t he initi a l velocity is ze ro , we have
U
1
=
Ul'
The results obtained for the eigenvalue problem motivated us to use quintics as basis functions.
C onvergence
The verify convergence) we choose a fixed spacial discretization and a fixed final time
I.
Then) starting with 10 time intervals) we increased the num ber of intervals until the relative difference is strictly less than
10
3
.
This approximation is then considered as sufficiently accurate for the system of differential equations.
Decreasing the time step size , we found the first order derivative s needed approximately double the number of time steps to yield the same relative difference in
II
.
1100 than the function values do.
It seems as if the the second order derivatives do not converge point wise, if they do) the convergence is very slow. This is not altogether surprising (see Section 5.4).
To establish the number of elements needed for our approximation , we choose a fixed final time,
I)
and time step size, M. Then the number of elements, starting with
10, is doubled until the relative difference satisfy our criterion.
CHAPTER
6.
APPLICATION. DAMAGED BEAM
127
Simulation of th
e
motion of
bea
m
W e are primarily concerned with the detection of damage. In this section we give an indication of the effect of respectively damage, damping and rotary inertia on the motion of a beam.
Our experiments indicate that measurable differences between the undam aged and damaged beams occur in displacements as well as gradients.
(Table 6.11.) Viscous damping has no significant effect on the motion. Look ing at the modal analysis this was expected, since it only effects the first few modes. Adding KelvinVoigt damping, the differences between the damaged and undamaged cases decrease , but is still clearly detectable. (Table 6.12.)
The presence of rotary inertia can have a more significant effect on the dif ference between the motion of the damaged and undamaged beams. (Tables
6.13 and 6.14).
To illustrate the above effects we compare the motion of an undamaged beam to that of a damaged beam where the initial velocity is zero and the initial position a 'solitary wave". For this simulation we choose
C\'
=
0.4, 6
=
0.1 which is ra t her excessive , 80 elements ,
T
=
0.02 and 400 time subintervals.
Almost immediately after the first wave front pass through the damaged point, measurable differences in displacements as well as gradients between the two cases occur. (See Figure 6.l.) In Table 6.11 we compare the dis placement of the damaged and undamaged beams on
T
=
0.02 at
x
=
0.3 and
x
=
0.7.
I
x
I
Undamaged beam
I
Damaged beam
I
% difference
I
0.3
0.7
2.325 x 10
1
4 .
733 x 10
1 l.505
X
10
1
5.508
X
10
1
8.2
7.8
Table 6.11:
Eff·ect
oj damage
during motion where
6
=
0.1
,
C\'
=
0.4,
T
= 0.02.
CHAPTER
6 .
APPLICATION DAMAGED BEAM
128
1
Ini tial position
~
Q
Q)
E
Q) u
CIl
0
.~
Q 0
Damaged"",
.....
..:.
Undam a ed
0.4
Position
1
Figure 6.1:
Companng the moNon of an undamaged beam to that of a dam aged beam where
6
=
0.1,
Ct'
=
0.4
and
T
=
0.02. vVe now add KelvinVoigt damping to the same situation as in the previous case. We use
fL
=
3.469
X
10

5
.
This value for
fL
was obtained from [JVRVj.
I
x
I
Undamaged beam
I
Damaged beam
I
% difference
I
0.3
0.7
2.099 x
10
1
4.716 x
10
1
1.378
X
10
1
5.436
X
10
1
7.2
7.2
Table 6.12:
Effect of KelvinVoigt damping on the damage during motion where
6
=
0.1,
Ct'
=
0.4
,
T
=
0.02
and fL
=
3.469
X
10
5
.
The presence of rotary inertia can make the differences more difficult to detect. An example is given in Tables 6.13 and 6.14.
CHAPTER
6.
APPLICATION. DAMAGED BEAM
129
x
0 .
2
0.6
I
Und ama ged beam
I
Damaged beam
I
% difference
I
2.525 x 10
1
2.065 x 10
1
4 .
762 x 10
1
5.248
X
10
1
4 .
6
4.9
Table 6.13 :
Effect of
Rota r y
inertia on
the
damage during
motion
where
o
=
0 .1, c¥
=
0.4
,
T
=
0.02
and 1 1r
=
19200.
I
x
I
Undamaged b ea m
I
Dama g ed beam
I
% diff er enc e
I
0.2
2.236 x 10
1
0.6
4.140 x 10
1
2.092 x 10
1
4.287
X
10
1 l.4 l.5
T a bl e 6.14:
Effect
of
Rotary inertia
on
the
damage
during
motion where
o
=
0.1
, c¥
=
0.4
,
T
=
0.02
and
11r
=
4800 .
Chapter 7
Appli
c
ation. Pla
te
beam model
7 .
1
Introduc
t
ion
W e consider Problem 3 (from Section 2.6).
It is a mathematical model for a plate connected to two beams. Problems of this type are clearly of great practical importance. The plate can be rigidly connected to the beams or simply supported by the beams. The same model can be used for an Ishaped structural member (depending on the type of vibration) . For simplicity we restrict our investigation to the case of a plate supported by beams.
If the plate is rigidly connected to the beams, it may result in a problem with six unknown functions (excluding shear) due to dynamical effects. Even in our restricted case, one may easily encounter very large matrices.
In collaboration with others, [ZVGV1], we considered the equilibrium and eigenvalue problems of a rectangular plate supported by two beams at the boundary. In this thesis we extend the investigation and include the effect of rotary inertia.
The computation of the matrices is explained in Section 7.2. We use reduced quintics for the plate, which necessitates the use of quintics for the beams.
We treat the equilibrium problem in Section 7.3 and the eigenvalue problem
Section 7.4.
130
CHAPTER
7.
A
P
PLICATION. PLATE BEAM MODEL
7.2
Computation
of matrices
131
F or the numerical experimentation we consider a square plate, 0, rigidly supported at two opposing sides and supported by identical beams at the remaining sides. The plate has thickness h and the beams are of square profile with thickness
d.
Furthermore, we assume the plate and beams are of the same material. (These restrictions are evidently not necessary.)
The reference configuration 0 is the rectangle with 0
<
Xl
<
1 and o
<
X2
< l.
Thos e parts of the boundary where
Xl
=
0 and
Xl
=
1 are denoted by
~o and
~l respectively and correspond to the rigidly supported parts of the boundary . Those parts where
X2
=
0 and
X2
=
1 are denoted by
r
0 and
r
1 respectively and correspond to the sections of the boundary supported by beams.
7
.2.
1 B
asis elem
en
t
s
For the plate we use only reduced quintics as basis functions. These functions are in
H2(0)
or fully conforming , in finite element language . They are defined on a triangular mesh. The mesh for the rectangle 0 is generated in the following way: The interval
[0 , 1] is divided into nl subintervals and the interval
[0 , 1] into
n2
subintervals. This partition of the intervals yields n l x n2 rectangles. The final triangular mesh is then obtained by dividing each of these rectangles into two triangles by connecting the lower left corner with the upper right corner. The rectangle 0 is divided into
2nl x
n2
triangles.
Consequently we have 2nl x n2 elements
Oi.
Reduced quintics are defined in Section 4.3.l. The computation of the coeffi cients is not trivial and we describe it. in
ApppnniY
C The choice of reduced quintics "force" one to use quintics for the beams. Hermit.e piecewise quintics are defined in Section 4.2.l.
CHAPTER
7.
APPLICATIO N. PLATE BEAM MODEL
132
7
.2.2
St
andard Matr
ic es
First we compute standard matrices for the two beams with quintics. The procedure is the same as with cubics.
[Mci°Lj
=
JO\
Y
OcPi)
C'Y
O j
)
,
[MfO
L j
=
Jo\ ,
0cP
d C'Yo
cP
j)'
,
[Mci1Lj
=
JO\
Y
l
cPi
)C'YlcP
j) ,
[Mil
L j
=
.Jo
1
C'Yl
e,Di)
'
C'YI
e,D
j
) ' , as well as
N e xt we compute stand a rd m a trices for the plate. These computations are quite involved and we provide some detail in Appendix
C.
The bilinear forms are given in Section 3 .
3. Each basis element is of the form
¢
i
=
(cP
i, 1
0cPi,
I lcP i )'
(The forced boundary conditions are satisfied by eliminating certain basis elements .) Now , consider for example
C
O(
e,D
i
,
cPj
)
involves the restriction of basis functions e,D i and
e,D
j
to the bound ar y roo
These restrictions are nonzero only for some of the basis functions associ a ted with nodes on roo
( The restriction of a reduced quintic on ro is an onedimension a l quintic.
) The result is
Consequently
M
=
Mil.
+
{3
M
r o
+
{3
M
r l ,
Mij
=
lvtg
+
{3
MDo
+
{3Mrl.
where
The computation of the Kmatrix is the similar ,
CH AP TER
7.
APPLICATION. PLATE BEAM MODEL
7.3
Equilibrium probl
e m
133
To find the Galerkin approximation for the solutions of the equilibrium prob lem, we solve a system of linear equations.
Problem
BD
K
ii
=
F,
where
Fi
=
(f,
cPi)'
The parameter
a
gives an indication of the stiffness of the beams in c ompa rison to that of the plate . In c reasing the value of ex implies an increase in the stiffness of the beams and
a
=
0 corresponds to the case where two sides are free.
For different values of
a
we compare in Table 7.1 the FEN! approximations for the maximum displacement, to the so called exact solution. See [TW].
(Interesting historical remarks are found in [TW] .
)
Note that the maximum displacement occurs at the centre of the plate as a result of s ymmetry.
W e consider a square plate with the same number of equal intervals per side.
W e denote this number by
n ,
and use it to distinguish between different meshes.
Denote the maximum displacement obtained from th e so called exact solution by
U max
and the FEN! approximation of the maximum displacement where n subintervals are used, by u~2x.
Choose Poisson 's ratio
l/
= 0.3.
CHAPTER
7.
APPLICATION. PLATE BEAM MODEL
134
I
Q
I Exact
n=2
(U max

I u~L)
/
U max
n=4
I
n=8
100 4.09
X
10
3
2.0421
X
10
3
2.9308
X
10
4
2.3653
X
10
4
30
4.16
X
10
3
1.0507
X
10
3
6.5975
X
10
4
7.1510
X
10
4
10
4.34 x
10
. j
1. 7896 x lOj
1. 7460
X
10 
4
1.2220
X
10
4
6
4.54
X
10
3
3.5724
X
10
3
5.0933
X
10
3
5.1428
X
10
3
4
4.72 x
10
j
2 .
9835 x
10
3
1.547 x lOj 1.5006
X
10
3
2
5.29
X
10
3
3.2719
X
10
3
2.0580
X
10
3
2.0181
X
10
3
1
6.24 x lOj 1.0228 x lOj
9.3231
X
10
5
6.2112
X
10
5
0.5
7 .5
6 x
10 
3
2 .
2617
X
10
3
1.6195
X
10 
3
1.5973
X
10 
3
0
1.309
X
10
2
3.2828
X
10
4
2.8584
X
10
4
2.8129
X
10
4
Ta .
ble 7.1:
Comparison oj exact v alues with FEM approximations oj ma :;; imum displacement.
The fact that the relative error originally improves if we double the number of intervals from 2 to 4 and then r emains almost the same, suggests that the so called exact solution is not very accurate, as could be expected since only a f e w significant digits are given.
The relative difference between consecutive FEM approximations strenghtens this observation as can be seen in Table 7.2.
Q
(u~L
u~~ x)/u~~ (u~~x
u~~x)
/ u~~x
100 1.748505
X
10 
3
5.653982
X
10
5
30
1.711612 x
10
3
10
1.614713
X
10
3
6 1.528720 x
10
3
4
2
1.433896
X
10
3
1.211367
X
10
3
9.294900
X
10
4
1
0.5
6.412267 x
10
4
0
4.242264
X
10
5
5 .
539486
5.238781
X
10
5
4.971930 x
10
5 x
10
5
4.677682
X
10
5
3.987090 x
10
5
3.111738
X
10
5
2.214781
X
10
5
4 .5
50660
X
10 
6
Table 7.2:
Compariso n oj FEM approximations Jar the max i mum displace ment.
CHAPTER
7.
APPLICATION. PLATE BEAM MODEL
7.4
E
ig
e
nvalu
e
probl
e
m
135
As mentioned before, Section 4.4, the occurence of eigenvalues has a highly irregular pattern in the twodimensional case. We have an elementary ex ample to illustrate this, and also to show how difficult it can be to identify eigenvalues with multiplicity.
7.4.
1
Multiplicity of eigenv
a
lue
s
Consider the following eigenvalue problem ,
 \]2U
=
AU
on the unit square with
' U
=
0 on the boundary.
Clearly,
u( x, y)
= sin(mrx) sin(m7fY) is an eigenfunction, for nand m integers. The corresponding eigenvalue is
A
=
n
2
+
m
2
.
The popular difference scheme is or where h is the length of a subinter va l.
Let
U i ,j
= sin(iwk) sin(Jwe), then
Hence
Ui ,j
We
satisfies the boundary conditions if
Wk
(br)/(n + 1) and
=
(t'7f)/(n+ 1). It satisfies the difference equations if
Ae
=
h2(2 2
cos
w e) .
Hence
Ui,j
is an eigenvector and every eigenvalue is of the form
Ak
+
Ae.
In
Table 7.3 we list the exact eigenvalues for this problem as well as the nu merical approximations obtained for different subinter va l lengths. We give
CHAPTER
7.
APPLICATION. PLATE BEAM MODEL
I i I Exact I
h
=
0.2 I
h
=
0 .
1 I
h
=
0.05 I
h
=
0.005 I
1
19 .
7 19.3
2 49.3 45.6
3
49.3 45.6
4 79.0 72
5 98.7 81.6
6 98.7 81.6
7 128
8 128
108
9
168
10
168
11 178
12
197
13 197
1 4 247
15
247
16
257
17 257
18 286
108
118
118
144
1
1
4 4
44
144
144
170
170
180
19 286
20 316
180
206
192
192
240
240
245
245
275
275
307
19.7
49 .
0
49.0
78.4
97.2
97.2
127
127
163
163
175
19.6
48 .
2
48.2
76.8
93.3
93.3
122
122
151
151
167
180
180
217
217
225
225
246
246
283
19.7
49 .3
196
196
245
245
254
254
283
283
313
49.3
78.8
98.3
98.3
128
128
167
167
177
136
T a ble 7.3:
lues
.
Comparison of finite
d~fference
eigenvalues to the exac
t
eigenva
only three significant digits as it is sufficient to illustrate difficulties of match ing exact eigenvalues and approximate eigenvalues.
Let i denote the number of the eigenvalue and
h
the length of a subinterval.
Interpreting numerical results with respect to multiplicity of eigenvalues is difficult . Great care should be taken to establish whether approximate ei genvalues that are close together are an indication of multiplicity of exact eigenvalues, or not.
For example, for
h
=
0.2 , the eleventh to fifteenth eigenvalues seem to be one eigenvalue with multiplicity five, while it actually approximates three diffe rent eigenvalues. Another example is the fifteenth and sixteenth eigenvalues for
h
=
0.05. These eigenvalues seem close and one might expect them to approximate the same eigenvalue with multiplicity more th a n one.
CHAPTER
7.
APPLICATION. PLATE BEAM MODEL
137
7
.4
.
2
P
la
t
e beam
Th e ge ner a lized eigenvalue problem associated with the plate beam model is given by
Probl e m C D
KiD
=
AM ' w.
In a joint report [ZVGV1] we consider this eigenvalue problem for th e plate be am model excluding rotary inertia . In this subsection we investigate th e effect of rotary inertia if included in the model.
The ratio
CY./{3
of the dimensionle ss co nstants,
CY.
{3
=
(Ebh)/(aD)
an d
=
(Pb A)(pah ),
defined in Section 2.6 , with the plate of thickness hand the beams of square profile with thickness
d,
is a measure of the stiffness of the beams in comparison to that of the plate.
In
the special case whe re both the beam s and the plate are of the same material, we have
CY.
= 
d
{3
(
h
)
2
2
(1 
l/
)
.
As the values of
d/ h
increase, i.e. the stiffness of the beams is increased, the s ituation approaches the plate problem where a ll four sides of the plate are rigidly supported. For this problem the eigenvalues an d eigenfunctions are known. The eigenvalues are of the form
((mr)2
+
(m7f)2)2 with corresponding eigenfunctions si n(n7f x) sin(m7fY).
Since the exact eigenvalue s for the plate beam problem are not available, the FEM approxim ati ons for the eigenvalues for l arg e values of
d/ h
can be compared to the eigenvalues of this limiting case, see Table 7.4.
Denote the ith eigenvalue for the case where a ll four sides are rigidly s up po rte d by
Ai.
The eigenvalues ar e ord ered ac cording to size. The FEM approximation of the ith eigenvalue i s denoted by
A~n) where
n
s ubintervals are used.
Throughout this subsection we use Poisson's ratio as
l/
=
0.3.
CHAPTER
7. APPLICATION.
PLATE
BEAM MODEL
'/,
Ai
8
)
for different values of
d/ h d/h
=
1
d/h
=
10
d/h
=
100
d/h
=
200
Ai
1 92.6654 386.3556 389.6361 389.6364 389.6364
2 250.7783 2359.4575 2435.2366 2435.2398 2435.2273
3
1264.1968 2433.5697 2435.2500 2435.2525 2435.2273
4 1514.0745 6221.0210 6234.2338 6234.2346 6234.1818
5 2142.1461 7345.9342 9741.4914 974l.5319 9740.9091
8 7725.6133 11308.1573 16463 .7086 16463.7127 16462.1364
10 11599.1799 16455 .3398 28158.9627 28159.0563 28151.2273
138
Table 7.4:
Comparison of FEM ezgenvalues for different values of d/ h with eigenvalues of a rigidly supported plate. Rotary inertza zs e:rcluded.
As
d/ h
increas es, the FEM approximation of the eigenvalues approaches the eigenvalues of the plate rigidly supported on all four sides.
For values of
d/ h
that do not correspond to the limit case, the numerical convergence of he FEM eigenvalues are illustrated in Table 7.5.
d/h
=
10
(
A~2n)
_
A~n))
/
A?n) d/h
=
100
i
A~2n)
A~n))
/
A?n)
i
n=2 n=4 n=2 n=4
1
6.94630 x
10
4
8.2136 x
10
b
7.0601 x
10
4
8.3506
X
10
6
2 2.1888
X
10
2
4.0239
X
10
4 l.2764
X
10
2
2.9014
X
10
4
3
4.1506
4
0.6013 x
10
2
4.2474 x
10
4
5.3144 x
10
2
5.6098
X
10
4
x
10
:J
7.:32b6
x
lU
' l
0.0438 x
10
'L
7.
3 537
X
10 
4
5
2.3179
X
10
2
9 .
5076
X
10
4
10
2.3327 x
10
1
5.0927
X
10
3
4.8512
X
10
2
5.7506
X
10
1
3.1069
X
10
4
1.3180
X
10
2
Table 7.5:
Numerical convergence of of d/ h. Rotary inertia is e:rcluded.
FEM eigenvalues for different values
Remark
Choosing
n
=
8, yields (486 x 486) matrices which are already very time consuming to handle with our availab le computer hardware and software. Therefore we do not consider more than 8 subintervals.
Including rotar
y
inertia in th
e
mod
e l
In addition to the joint report [ZVGV 1] we now establish the effect of rotary
CHAPTER
7.
APPLICATION.
PLATE BEAM
MODEL
139 inertia on the eigenvalues of the plate beam problem.
From Sections 2.5 and 2.6 we have the dimensionless constants
rb
and
=
hi
(a
2
d
2
) r
=
I
I (a
2 h).
In the experimentation we work with a fixed plate , i .e.
a
and
h
are fixed, and modify the beams by changing
d.
Consequently
r
depend on the relationship
rb
and
dl
h and indicate the effect
of rotary inertia.
I n
Table 7.6 we illustrate numerical convergence of the FEM eigenvalues if rotary inertia is included. For
d l
h
=
50 we have
rb
=
2.083 x 10
2 and
r
= 8.3~:3 x 10
6
.
~
(
A~4)
_
A~2))
I
A~4)
1 7.0599
X
10
4
5
4.8512
X
10
2
10
5.7479
X
10
1
(
A~8)
_
A;4))
I
A~8)
8.3271
X
10
0
3.1350
X
10 
3 l.312 x 10
2
Table 7.6:
Numerical
convergence of
FEM
e~genvalues
fOT
d l h
=
50.
Includ ing
rotary inertia.
CHAPTER
7.
APPLICATIO
N.
PLATE BEAM
MODEL
140
For the plate supported on all four sides, repeated eigenvalues are expected and indeed observed. For the plate beam problem the symmetry is partially lost, and the question arises if repeated eigenvalues will occur, and whether those F E M eigenvalues will be observed as repeated eigenvalues?
As with the case excluding rotary inertia, the exact solution is not avail able. Again, the exact eigenvalues for the plate rigidly supported on all four sides are used to give an indication of what can be expected of the FEM eigenvalues.
As is expected, the presence of rotary inertia decreases the eigenvalues in comparison to the case without rotary inertia. The effect of rotary inertia is illustrated in Table 7.7. For
d/ h
=
50 we have
rb r
=
8.333 x 10
6
.
=
2 .
083 x 10
2 and
6
7
8
9
3
4
5
10
11
Excluding rotary inertia
II
A
(8) t
1
389.6313
2 2435.1545
2435.2439
6234 .
2146
9740.7884
974l.5344
16463.2389
16463.6608
28154.6617
28158.7179
31564.1696
Including rotary inertia
A
( 8) t
389.5673
2434.1536
2434.2429
6230 .
1154
9732.7844
9733.5289
16445.6552
16446.0765
28115.3573
28119.4013
31517.5097
Exact value
Ai
389.6364
2435.2273
2435.2273
6234 .
1818
9740.9091
9740.9091
16462.1364
16462.1364
2815l.2273
2815l.2273
31560.5455
T able 7.7:
Effect
of rotary inertia on the e7gen v alues for d / h
=
10.
In Table 7.7 the multiplicity of eigenvalues of the plate rigidly supported on all four sides are observed. These repetitions give reason to expect that the corresponding FEM eigenvalues for the plate beam problem may also be repeated eigenvalues.
The question arises whether the FEM approximation will yield repeated ei genvalues or will the eigenvalues only be close?
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