AN ABSTRACT OF THE THESIS OF
for the
DAVID WARREN AKERS
(Name)
in
MASTER OF SCIENCE
(Degree)
presented on
MATHEMATICS
(Major)
61Y(Dat,e )
Title: FIRST ORDER DIFFERENTIAL CORRECTIONS IN THE
EIGENVALUE PROBLEM
Abstract approved:
n
4
Redacted for Privacy
Ggheen
Let A be an
nXn
characteristic values Xl' X2'
real, symmetric matrix with distinct
, X.
gonal matrix P such that PAP T
metric change,
sulting changes,
AA,
AP,
= A = (Xi).
in the matrix
and
AA,
We next assume that the change in
Then there exists an ortho-
we can calculate the re-
A,
in
A
Given a small sym-
P
and A respectively.
is dependent on time
t.
particular, we assume that A(t) is a differentiable function of
Then, if
A
0
A(0)
t.
has distinct characteristic values, we show that
for sufficiently small t, P
t,
In
and A are differentiable functions of
and that A(t) also has distinct characteristic values.
First Order Differential Corrections in the
Eigenvalue Problem
by
David Warren Akers
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Master of Science
June 1969
APPROVED:
Redacted for Privacy
Professor o
athematic s
in charge of major
Redacted for Privacy
Chairman of Department of Math(tmatics
Redacted for Privacy
Dean o`f Graduate School
Date thesis is presented
Typed by Clover Redfern for
David Warren Akers
TABLE OF CONTENTS
Chapter
I. INTRODUCTION
II.
III.
Page
1
CHANGES IN THE CHARACTERISTIC VALUES AND
DIAGONALIZING MATRIX
3
PRINCIPAL THEOREM
8
IV. NUMERICAL EXAMPLE
BIBLIOGRAPHY
19
25
FIRST ORDER DIFFERENTIAL CORRECTIONS
IN THE EIGENVALUE PROBLEM
I.
INTRODUCTION
From an historical standpoint, the theory of matrices and their
characteristic values has been of significant interest and importance
to many areas in the physical sciences. These characteristic values,
n X n matrix
or eigenvalues as they are also called, of a square
A,
are determined by solving the characteristic equation
det (A-XI) = 0.
This equation is a polynomial of degree
the Fundamental Theorem of Algebra has exactly
n,
and by
(not necessar-
n
ily distinct) roots, each of which is called a characteristic value of
A,
If
X
is a characteristic value of
a characteristic vector belonging to
X
A,
if
then
X
is said to be
XA = XX.
As we have said, in many fields the characteristic values and
their corresponding vectors of a given matrix A may have a special physical significance. For example, consider the theory of
molecular vibrations (2). A certain symmetric matrix arises in connection with a particular kind of motion of a given molecule. By cal-
culating the characteristic values and vectors of that matrix, we determine the frequency, amplitude, and phase of the vibration of each
atom of the molecule about its equilibrium position. Now suppose
that an atom of that molecule is replaced by an isotopic atom of the
same element. The result is a change in the frequencies of vibration
2
because the original matrix, and consequently its characteristic values, have been changed due to the change in mass of the atom.
We can consider this problem entirely out of its physical context and in a purely mathematical one. The problem can be stated
as follows. Suppose we have a given symmetric matrix A with
distinct characteristic values, and suppose we change A by a small
symmetric amount
AA.
What will be the change in the character-
that is, the ortho-
istic values and in the diagonalizing matrix P,
gonal matrix
P
such that PAP T = A, where A is the diagonal
matrix of characteristic values of
A?
A reasonable extension of the problem is to next assume that
the change in A
is dependent on time t.
differentiable function of t,
and A 0
If,
= A(0)
in fact,
A(t)
is a
has distinct charac-
teristic values, we are able to show that for sufficiently small t,
both
P
and A are differentiable functions of t,
and
A(t)
also has distinct characteristic values.
It is very important that we assume that
A
0
has distinct
characteristic values, which, in the physical context, means that we
do not have any degenerate frequencies. If we do not make this as-
sumption, the procedures and calculations used here will clearly not
hold. For a treatment of the degeneracy problem in the setting of
quantum mechanics, see Kemble (1), who refers to a method first
used by Van Vleck.
3
U. CHANGES IN THE CHARACTERISTIC VALUES AND
DIAGONALIZING MATRIX
Let A be a real, symmetric,
tinct characteristic values
Xi, X2,
n X n matrix with
,
n
dis-
Then by the Principal
Xn.
Axes Theorem for real, symmetric matrices, we know that A is
simultaneously similar to and congruent to the diagonal matrix
A = (X.).
Thus there exists an orthogonal matrix P such that
PAP-1 = PAP T = A.
Since
PP T
is orthogonal,
P
=
PTP = I.
Suppose now that we change the matrix A by some small
amount AA such that A + LA is still symmetric. Then A
will be changed by some small amount,
AA,
and A + LA will
still be diagonal. The diagonalizing matrix P will then be changed
and
will also be orthogonal. Thus we have the following
P + AP
two equations.
(P+AP)(A+AA)(P+AP)T = A + LA
(2. 1)
(2. 2)
(P+AP)(P+AP)T
LP
We wish to determine
AA,
in the matrix
A.
=
(P+AP)T(P+AP) = I.
and
LA given a certain small change,
For notational convenience in the following
development, we will denote
(LP)
T
by
APT.
Carrying out the multiplication on the left hand side of Equation
(2. 1) we obtain
4
(P+LP)(A+LA)(P+LP)T
(2. 3)
= PAP
+
T
+ A PAP
PALPT
T
+
(P+LP)(A+LA)(PT+LPT)
PLAPT + A PAAPT
LPALPT
+
=
+
PLALPT
+
LPLALPT
= A + LA.
is sufficiently small, then so is
LP.
Thus any term
in the above expression containing at least two
"L"
terms can be
Now, if
neglected.
LA
(This fact is readily seen when considering A, P,
and
A as functions of time and the proof will be left until then). Since
PAPT = A we now have
(2. 4)
A PAP
T
+
PLAPT
PALPT = AA.
+
Similarly, from Equation (2. 2) we obtain
LPPT + PAP T = 0
(2. 5)
and this implies
LP = -PLPTP.
(2. 6)
Now PAP T = A implies that PA = AP.
Substituting this
expression and Equation (2. 6) into Equation (2. 4) we have
(-PLPTP)APT
+
PLAPT
+
APLPT = DA.
Thus
(-PLP T )(PAP T ) + PLAPT
+
APLPT = AA.
5
and so
(-PAP T )A +A (PAP T) = AA - PAAP T.
(2.7)
Recalling that we wish to solve for AP and
consider the
be equal. First, the
T
(i, j)th
is the
(PAP )..
13
elements must
(i, j)th
element of A(PAPT) is X.(PAP T ).
element of PAP T.
(i, j)th
(-PAP T )A is
j)th element of
AA,
element of each side of Equation (2.7). Since
(i, j)th
(2. 7) is an equation of two matrices, these
where
AA given
Also, the
(-PAP T )..X. = -X.( APT )ij.
P(i,
J
Hence the
(X.-X.)(PAP
1
3
T)...
Next, the
13
(i, j)th
element of
is the Kronecker delta; that is,
5..
5..13 = 1
if
i = j.
of (2.7) is
A X. 6 .
1
is
AA
A VS..,
1
where
13
element of the left hand side of Equation (2. 7) is
(i, j)th
Thus the
if
i
j
and
element of the right hand side
Since these
(PAAPT)...
-
.
13
(i, j)th
5..
=0
13
13
(i, j)th
elements are
equal, it follows that
(2.8)
(X.1 -X.)(PAP
Suppose now that
zero,
5.. = 1,
11
(2. 9)
Since
T
= A X.6.. - (PAAPT)..
1
3
i = j.
Then the left side of Equation (2. 8) is
and we have
E X. =
AA = (AX.)
13
PAAPT)ii
for
i = 1, 2, ...
n.
is a diagonal matrix, it is completely determined
6
by Equation (2. 9).
Next suppose that
Then we have
i A j.
(Xi-Xj)(PAPT)ii = - (PAAPT)ij
5.. = 0.
since
13
X. - X.
1
0.
The characteristic values of A are distinct and so
Thus
3
T
(PAP )ij -
(2. 10)
1
X
.
-X.
(PAAPT).
.
3
and let
For convenience of notation, let B = PLAPT = (b..)
13
c.. = 0 and
where
C = (c..)
13
c
1
i3
bij.
3
Then,
PAP T = D + C
(2.11)
where D is some diagonal matrix. Clearly
since
AA
is, and hence
Consider now (PAP T ) T.
We see that
PAP
PAP T
+
is symmetric,
is skew-symmetric, that is,
C
T)T
APPT = (PAP
By Equation (2. 5),
B
T
+
=
(D+C)T
APPT = 0
=
DT + CT = D - C .
and so
APPT = (D+C) + (D-C) = 0.
CT = -C.
7
This implies that 2D = 0 and hence D = 0.
tion (2. 11) it follows that
PAP T = C.
(2. 12)
Thus
(2. 13)
APT
=
PTC
and so
AP = (PTC) T = CTP = -CP.
From this and Equa-
8
III. PRINCIPAL THEOREM
Equation (2. 9) of Chapter II shows how to determine a small
change in the characteristic values of a symmetric matrix
given
A,
a small symmetric change in that matrix. Similarly, Equation (2.13)
Suppose now that
gives the change in the diagonalizing matrix P.
the change in A
is dependent on time
that A is a differentiable function of
that A(t) = A0 + t(A1 -A0) for
are real, symmetric,
characteristic values
nXn
X
0
1'
X
0
2
Let us further suppose
t.
In particular, suppose
t.
where A 0
0 < t .51,
and
matrices, and A 0 has
...
X
0
n
.
Let
orthogonal diagonalizing matrix for A0.
PO
A(t)
(p0
ij)
distinct
be the
Then
)
and P PT
P 0A 0 PT0 =A 0= (Xi)
i
0 0
It is clear that at time t = 0,
=
n
Al
T
=PP
PO PO= I
.
takes on the value
Note also that A is symmetric for any t.
A0.
0
Thus, for any t,
our two basic equations hold; that is, there exists an orthogonal ma-
trix P such that
(3. 1)
(3. 2)
PAP T = A
PP T = PTP = I.
Since A is a function of time, it follows that P and A are also.
As in Chapter II, we obtain the following equation from (3. 1).
9
APAP T
(3. 3)
+
PLAPT
+
LPLAPT
LPAAPT+PAAAPT
Let us next assume that
t.
+
P
and
+
PALPT
+
APAAAPT = AA.
A are differentiable functions of
We will show that this assumption is a valid one when we prove
the existence of a set of continuous solutions to a certain system of
differential equations.
This gives
Now divide both sides of Equation (3. 3) by Lt.
(3.4)
,IN_PAPT
LA T
At
of P
+
AP
At
Since
A, P,
ADP
T
MDT
AP
Tt °AP + PA
+ PAt AP +APLALP
T
AA
T
At
_
AA
Lst
At
and A are differentiable functions of t,
lim
lim LA =
At-- 0
0
Also
lim
At--'
Alt
Lt -4- 0
AP
T
At
0
dA
LA
At
dP T
dt
AP = lim APT
At--' 0
dt '
'
and
0.
dP
dt
AP
lim
0
'
dA
dt
Lt
Lt--. 0
lim
Thus, taking the limit of Equation (3.4) as
the following equation.
=
At
LA
tends to zero gives
10
,oT
dP
T
dA T
d7 AP + P
P + PA dt
(3. 5)
dA
dt
We see now that the terms with two or more
terms as factors
"A"
become zero in the limit. This is why we were able, in Chapter II,
to neglect those terms, providing we chose
AA
small enough.
Similarly, from Equation (3. 1) we obtain
dP
(3. 6)
dPT
PT + P dt
Now suppose that we can choose
°.
At = dt
small enough so that
we do not go through a degenerate case. That is, suppose that we can
choose
t
()
small enough such that A(t) will have distinct char-
acteristic values for every t such that
0 < t < t0.
(Recall that
we have already assumed that A(0) = A 0 has distinct characteristic
values. This is essential. ) Once again, by showing the existence of
a unique set of continuous solutions to a certain system of differential
equations, we will show that this assumption is correct. This is, in
fact, the main result we are seeking.
We can now proceed with Equations (3. 5) and (3. 6) in an analo-
gous way to the method used in Chapter II. Doing so we obtain the
following equations.
dX.
(3. 7 )
dt
= (P dA P T ).
dt
for
i = 1, 2, .
n
11
dP
(3. 8)
-CP
dt
C = (c..)
13
where
is defined by c.. = 0,
c..
13
X.-1 X.
b..13 = (P dA
PT )
dt
ij
with
b.1 3
.
J
We can also write Equation (3. 8) in a more convenient fashion.
dp..
(3. 9)
dt
= -(CP).j
for
i, j = 1, 2, ... , n.
it is clear that
ddt
A
and let P.1 denote the
Al - A0 = M = (m..)
13
ith
Since
A(t) = A + t(A 1-A0),
0
= Al - A0.
row of
Let
P. Then
Equation (3.7) becomes
dX.
(3. 10)
dt
= P MP
T
for
i
i = 1, 2, ...
n.
We can write out Equations (3. 9) and (3. 10) explicitly by carrying out
the indicated matrix multiplication.
(3. 11)
dX.
1
dt
= P.MP.
= m 11 p.2i l + m 22 p.22 +...+m nnp.2 + 2
1
1
/
k< j
for
i = 1, 2, ...
n.
m kj.p. p..
12
dp..
dt
(3. 12)
c. p kj
=k=1
i- 1
1
=-
(PMP
T
).
p
k-Xi
(k=1
kj +
K
k=i+1
for
k
-X
i
(PMPT )ik pkj
i, j = 1, 2, ... , n.
We notice now that Equations (3. 11) and (3. 12) define a system
of
This is the system of differen-
differential equations.
n2 + n
tial equations that we will show has a unique set of continuous solutions. Furthermore, these solutions will assume the values
Xl' i', i'fp ,, P
0
0
2
0
11
0
'
n
0
nn
at time t= 0.
f
Before we can continue, we need several definitions and a gen-
eralized form of Taylor's Theorem.
Definition.
open interval
be a real-valued function defined on an
Let f(x)
(a, b),
and let xo be an interior point of
Suppose there exists a neighborhood
L
f
such that x E N'(x 0 )
implies
N(x0)
If(x)
(a, b).
and a positive number
f(x0)I < Llx
x01.
Then
is said to satisfy a Lipschitz condition at the point x0.
0
This definition can be extended easily to a function of several
variables as follows.
Definition. Let
(x
0
x
0
2
"
x0) be a point in
Rn.
Let
13
xn)
f(x1, x2,
be a real-valued function defined on the domain
where D is defined by
- xi 1
1x1
< al,
01
x1
< a2,
1x2
D,
,
01
< a n . Then f is said to satisfy a Lipschitz condition at
x
n
n.
0
0
L
x2,
... , x0) if there exists constants L1, L
the point (xi,
1x
2"n
such that, for all interior points
(x
1,
x2,
,
of the domain D,
xn)
we have that
lf(x l' x 2'
< LI 1x - x 1
1
0
'
0
xnn - f(x l' x 2'
n
+ L x - x201 + ...+ Lnixn -
0
In order to state a generalized Taylor's Theorem for a function
of several variables in a form analogous to the theorem for a function
of one variable, we first need to define the concept of higher order
differentials.
Definition.
set of Rn.
function of
f
Let f be a real-valued function defined on a sub-
The (first order) differential of
2n
f,
denoted
variables, defined for those points
xE
has all its partial derivatives, and for every t E Rn,
equation
n
df(x ; t) =
D.1 f (x)t.
1
i=1
df,
Rn
is a
where
by the
14
x = (x1, x2, ... , xn), t = (t1, tz,
where
D.1 f(x) -
,
tn)
and
(x).
ax.
1
The second order differential of
f,
denoted
d
2f,
is defined
by the equation
n
d
2f
n
x; t
D. .f(x)t.t.
31
1, 3
i =1 j =1
where
Di,
3.f
(x)
-
of
(x).
ax.ax.
1
3
The third order differential of
denoted
f,
d
3f,
is defined
by the equation
n
d
3
n
n
f(x; t) =
D.
i=1 j=1 k=1
The
3,f(x)t
kkt.tj i
.
mth order differential, denoted
d
f,
.
is defined similarly.
We can now state the following
Theorem (Taylor). Let f have continuous partial derivatives
of order m at each point of an open set
b E S, a
b,
and if the line segment,
S
L(a, b),
in
Rn.
joining
a c S,
If
a
and
b
15
lies entirely in
then there exists a point
S,
on
z
L(a, b)
such
that
m-1
f(b) = f(a)
kli d
k
f(a;b-a)+ m! dmf(z ;b-a).
1
k=1
The proof of this theorem can be found in many analysis texts and
will be omitted here.
We can again consider our system of
differential
n2 + n
equations. Instead of writing the actual expressions, we will use the
following function notation for convenience.
dX
(3.13)
1
dt
'1"1'
' Xn' P11'
21'
Xn' P11'
fn(X1'
' Xn' P11'
' Pnn)
dX2
dt
'
Pnn)
dXn
dt
' Pnn)
dpil
dt
dPnn
dt
fn+1 (X 1'
= fn2+n (X l'
'
X
n'
'
X
1311'
n' P11'
' P nn)
'
Pnn).
It is clear from Equations (3. 11) and (3. 12) that each of the above
16
functions
f
1"
is continuous and satisfies the conditions
f 2 ,
n +n
of Taylor's Theorem for the open set
by to.
Now let
b = (X1,
... '
expand
f.
o
a=
'
o
p
' p11,
2
Rn+n that is defined
in
0
. .
n " nn
S
,
).
X0 , P0 1'
' Pnn'
and let
Then for
i = 1,
..,n 2+n,
we can
into the following Taylor's series.
m-1
1
f.(b) = f.(a) +
d
k!
k
f.(a;b-a) + R.
k=1
where R.1
1
=
m!
m any fixed integer,
dmf (z ; b-a),
i
z. E L(a, b).
i = 1, ...,n 2+n,
Hence, for
f. (b) = fi .(a) + dfi(a;b-a) +
2
1
d f
+ ...+ (m-1)! drn-1 f (ab-a) + R.
1
Now if we choose
t
.
small enough, then b is not too far from a.
Thus the higher order differentials can be neglected since they can be
made as small as we wish. Hence f.(b) - f.(a) will be approximately df.(a; b-a).
We now have the following system.
(3. 14)
dt
dXn
dt
8f1
8f1
dX1
- f 1 (b) = f 1 (a) +
1
ax
of
= f (b) = f (a) +
n
n
ax
) + ...+
(a)(X -X
(a)(pnn-pnn)
uPnn
0
(a)(X. -X, )
i
8f af
+ ...+
n
uPnn
0
(a)(p-p)
17
cIP11
- f n+1 (b) - f n+1 (a) +
dt
dpnn
afn+1
ax 1
afn+1
0
(a )(X -X ) +
1
.
. .
+
(a)(p-p0
)
Pnn
- f n2+n (b) - fn2+n (a) + a nf 2+n(am 1 ..x0)+...+afn2+n
apnn (a)(pnn-Pnn)
1
dt
Now let
afn2+n
L11
L12
'
,
,
Ln
(a)
n2+n
Then, by the triangle inequality, we see that, for
aPnn
i = 1, ...,n 2+n,
satisfies the following Lipschitz condition.
f.
f. (b) - f. (a)I < L.3.1
Since each
-X
IX
01
+...+ L.1, n2+n Pnn
1
Pnn
is continuous and also satisfies the above
f.
0
Lipschitz condition at
lA
0
;
'
P 13Pnnh
0
11
0
we know from the
theory of differential equations that for small enough t,
""
(3. 13) has a unique set of continuous solutions
p
X
11
0
(t)
p
...
X
0
'
p
nn
(t).
1 1
X
(t),
the system
Xn(t),
Furthermore, these solutions assume the values
p
0
nn
respectively at time
t = 0.
We now state this result formally as a theorem.
Theorem.
Let A 0
and
Al
matrices, and suppose that A 0 has
values. Let A(t)
be real, symmetric, n X n
n
distinct characteristic
be a differentiable function of t.
In particular,
18
let A (t) = A 0 + t(A 1-AO) for
chosen small enough,
values. Furthermore,
is
will also have distinct characteristic
A(t)
P(t)
functions with respect to t,
nalizing matrix for A(t),
Then, providing t
0 < t < 1.
and
A(t)
where
and
A(t)
will also be differentiable
P(t)
is the orthogonal diago-
is the diagonal matrix of
characteristic values of A(t).
It should be noted here that we know that the system (3. 13) has
a set of solutions even if we only know that each function
f.
is con-
tinuous, regardless of whether or not we know that each satisfies a
Lipschitz condition. However, we want to be sure that the solutions
are unique, since we want a uniquely determined diagonalizing matrix
and unique characteristic values for a given matrix A(t).
To
guarantee this, it is sufficient (but not necessary) to show, as we
have done, that each of the functions satisfies a Lipschitz condition.
19
IV. NUMERICAL EXAMPLE
Let us now consider an example to illustrate some of the meth-
ods and results of the previous chapters.
Let A 0
2).
1
= (2
Then A 0 is clearly symmetric, and to
1
determine the characteristic values, we solve the characteristic
equation
det (A0- XI) = 0.
Thus we solve
det (A0 -XI) = det
(12X
12 X) =
0.
Then
(1402 - 4 = 0
- 2X + 1
X2 -
- 4 =0
2X - 3 = 0
(X-3)(X+1) = 0 .
Hence the characteristic values of
A
0
are
X1 = 3
and
0
X2 = -1.
We next find the diagonalizing matrix P0.
It is easy to prove
that the characteristic vectors belonging to distinct characteristic
values of a real, symmetric matrix are orthogonal. Thus, to determine
X2
P0,
we need only find the characteristic vectors
belonging to
X
0
=3
and
0
X2 = -1
X1
and
respectively, and normalize
1
them. Then Po will be the matrix whose rows are
Let X1 = (x1, y1)
and
X2 = (x2, y2).
X1
and
X2.
Then X1A0 = XiXi
20
and
X2 A0 =
XoX
We first solve for
2
(x
1
2
1, y 1)(2
1
) = 3(x
1'
X
1
y 1) = (3x 1' 3y 1)
xi + 2y1 = 3)(1
2x
+ y 1 = 3y 1
1
Solving this system of linear equations, we see that
y1.
So let
x1 -
1
- yi.
1
Then
1
X1 = (-- ,-N-r_z)
x1
and
must equal
X1
is
already normalized, that is, has length one. Similarly, we can show
that
X
1
2
=
-1
,
and so
/
1
P
1
Nrz0
1
1
\\72T
0)=
(1
I.
A check shows that P 0A 0P 0 = ( 0 -10) = A 0 and that P0 P=
0
1
Then by Equation (2. 9) we have
Now let A A = (
. 02
. 04)'
0
= P 0A 0 PT0 )..ii for i = 1, 2.
A
(X.
Now
T
P 0AA0P 0
=
1
1
NFT
NI-Z-
1
1
.03 .02
.02 .04
.055 -. 005
-.005
.
015
21
Hence
AX
and
. 055
1
AX
2
We see then that the charac-
= . 015.
03 2. 02
should be
teristic values of the matrix AO + AA 0 = ( 1.
2. 02 1. 04)
= 3. 055 and X 2 = -. 985. Calculating the characteristic values
X
1
of A 0
1
+ AA
= 3. 04
directly from the characteristic equation we obtain
o
and
X
(If we had chosen
= -. 98.
2
AA()
even smaller,
and had carried out the arithmetic to more decimal places, we could
have made the results even closer. )
Ap.. = -(CP0)ii.
From Equation (2. 13) we see that
c
IO
1
12
1
c
\
12
c1 2
Nr2-
Nrr
Nr2.-
Now
CPO =
c
0
21
c21
\ c21
Nrz-
where
1
c12
X
0
0 1-
2-X
1
0AOPO )12
c
12
and
c
Nr2-
= -c 12'
21
(-. 005) =
=- 1
4
.
Thus
00125
and
c
21
= -. 00125.
Hence
-1
Ap11 = NT,2-X (. 00125),
Ap
1
21
=
NI-2
X (. 00125),
Ai)
and
1
12
=
Ap
NfT
(. 00125),
X
1
22
=
NI-2-
X (. 00125)
It is obvious from our example so far that the calculations for even a
22
2X2
case are rather tedious, and for larger cases, calculation by
hand becomes quite impractical.
2
1
Again let AO = (2
1)
and let Al =
by the equation A(t) = AO + t(A1 -A0) for
Al
-31).
(21
0 < t < 1.
Define
A(t)
Let
Then
AO = M.
dA
M=
dt
By Equation (3. 11) we have that
dX.
dt1
2
= m 11 p.i l + m
2
22
p.
2
+2
/
m kj p. p..
for
i = 1, 2
k< j
By Equation (3. 12) we have
i- 1
dp..
13
dt
-
2
1
k=1
k
-X.
1
(PMPT ). pkj
k=i+1
1
for
k
(PMPT) ikpkj
i, j = 1, 2
We thus obtain the following system of six differential equations.
dX
1
dt
dX
- p 211 + 2p 212 - 6p 1 1 p 12
2
p
+ 2p
22
dt
21
2
- 6p
21
p
22
23
dp 11
dt
dpi2
1
2
(P 11 P 21 P 22 -3P 11 P 22 -3P 12 P 21 P 22 +2P 12 P22)
2
1
X -X 1
2
(P11P21-3P11P22-3P11P12P21+2P11P12 P22)
2
dp22
dt
-X
2
2
1
dt
dp2i
dt
2
2
1
X1 -X 2 (P 11 P 21 -3P 11 P 21 P 22 -3P 12 P 21 +2P 12 P 21 P22)
2
1
X
2
-X
2
(P11 P 12 P 21 -3P 11 P 12 P 22 -3P 12 P 21 +2P 12 P22).
1
As we have shown in Chapter III, each of these functions satisfies a
Lipschitz condition at the point
0
0
0
0
0
0
1
1
1
(X1' X2' P11' P12' P21' P22) = (3, -1, Nr2' Nr2' Nr7
providing we choose t
small enough.
1
Nrz-
)
To demonstrate the
Lipschitz conditions explicitly for the system, it is necessary to calculate 36 Lipschitz constants. These are determined by taking the
partial derivative of each function with respect to each of X1' X2'
successively, and evaluating that partial
P11' P12' P21 and p22
1
1
derivative at the point (3' -1' 11'71'1 77'
Since each function is continuous and satisfies a Lipschitz con-
dition, for sufficiently small
continuous solutions
t,
the system has a unique set of
X 1(t), X2(t), p11(t), pip), p21(t) and p22(t)
24
that assume the values
tively, at time
t = 0.
3, -1
1
1
1
' NI2' Nr2" Nr2"
and
re spec-
25
BIBLIOGRAPHY
1.
Kemble, Edwin C. The fundamental principles of quantum
mechanics. New York, Dover, 1958. 611 p.
2.
Wilson, E. Bright, Jr., J. C. Decius and Paul C. Cross.
Molecular vibrations: the theory of infrared and Raman vibrational spectra. New York, McGraw-Hill, 1955. 388 p.