# Chapter 6: Eigenvalue Problems

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29 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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PHY 301: MATH AND NUM TECH

Chapter 6: Eigenvalue Problems

1.
Motivation

2.
Differential Equations

3.
Other Equations

PHY 301: MATH AND NUM TECH

6.1 Motivation

6.1 Motivation

Physical Systems are commonly described by ( usually differential) equations involving sets of
coupled variables.

If the equations are
linear and homogeneous (no constant term),
they can be

looked at as linear
transformations, and thus we can express them in matrix form.

This, in turn allows us, using
diagonalization, to decouple the equations in order to obtain simple
solutions. In addition, the solutions to the diagonalized problem exhibit the fundamental
characteristics of a physical system.

In this chapter,
we’ll
first work out the solution to the generic problem and

then apply
the method to
2 concrete examples: an vibrating string and an electrical circuit.
.

PHY 301: MATH AND NUM TECH

6.2. Differential Equations

6.2
-
A Linear Differential Equations: General Approach

Consider, for instance,
the
set of n differential equations at right:

All dependent variables,
x
1,
…,
x
n

, are all function of the
independent variable t; in addition note that the equations are
coupled together

variables are interrelated
-

but they are quite
special because only one order of derivative appears

here noted
“s”
-

Thus the approach, although quite general, works only for this
type of diff. eq.

1
11 1 12 2
2
21 1 22 2
...
(1-1)
d
d
...
......
d
d
s
s
s
s
x
a x a x
t
x
a x a x
t

 

11 12
21 22
.....
.....
..
(1-3)
....

a
a a
a
a
 
 

 
 
 
(1-2)
s
s
d x
ax
dt

1 2
,
,....
x x x

(1-4)
x PX

d
d
s
s
PX
a PX
t

d d
d d
s s
s s
PX X
P
t t

Inserting into (1
-
2) we get:

and since P is constant the left hand side can be re
-
written as:

By calling P the matrix of the eigenvectors of a, we can define the new variable x such that:

In general this is not so simple. What would make it easy is if the two equations in (1
-
1) were decoupled, i.e. the
first would be an equation involving only x
1

and the second equation an equation involving only x
2
. This can be
clearly achieved by diagonalizing the matrix a. Furthermore, in dimensions higher than 2, the solution would be
really horrendous, so that diagonalizing the matrix is even more important in that case.

And finally, the properties of the system are obscured by the complication of the solution. Diagonalizing the matrix
will exhibit the underlying characteristic of a system.

And where the matrix a is
:

These equations
can be re
-
written in matrix /vector form as:

where
:

PHY 301: MATH AND NUM TECH

6.2. Differential Equations

And thus finally the equations are decoupled. The solution is very easily found:

Example for s=2
:
then the solution in an exponential:

𝑋
𝑗
=
𝑋
0
𝑗
𝑒
+
/

𝜆
𝑗

𝑡

where

𝑋
0

is the initial value of
𝑋

and
𝜆
𝑗

is the
jth

eigenvalue of a. The sign is determined by the
physics of the problem (decaying or increasing solution)

If we instead have a first order diff
eq

then
s=1 and
𝑋
𝑗
=
𝑋
0
𝑗
𝑒
𝜆
𝑗

𝑡

6.2
-
A Linear Differential Equations: General Approach

Now, multiplying both sides by on
the left
we
get:

Since we know that is the diagonal
matrix of
the eigenvalues, we have replaced our
linear system of equations (1
-
1) by a decoupled linear system:

1
P

1
d
d
s
s
X
P aP X
t

1 1
1 1
1 1
1 1
2 2
2 2 2 2 2 2
0...
(1-2) 0....
............
.......
...
s s
s s
s s
s
s s s
d X d X
X
X
dt dt
X
d X d X
d X
DX X X X
dt dt dt

  
  
 

  
 
  
   
 
  
   
 
     

 
   
 

 
   
   
 

 
 

 
 
  
1
P aP

D
PHY 301: MATH AND NUM TECH

6.2. Differential Equations

6.2
-
A Linear Differential Equations: General Approach

This method can also handle derivatives of mixed order

Consider for example:

2
1 1
1
2

d d

0
d d
x x
x
t t
 
  
Now define:

then the above equation becomes:

1
2
d
d

x
x
t

2 1
1
d d
0
d d

x x
x
t t
 
  
2 1
2 1
2 1
2 1
d d
0 1 0
d d
d d
1 0
d d
x x
x x
t t
x x
x x
t t
 

  

  

The system can now be written in matrix
-

form as:

d

𝑑𝑡
=


𝑑

d
𝑡
=

1


So that now the system can be solved as previously.

Which together
with the definition of x
2

becomes a system of two
equations:

PHY 301: MATH AND NUM TECH

6.2. Differential Equations

6.2
-
A Linear Differential Equations: General Approach

E6.2
-
4

Consider the equations
𝑑
2

(
𝑡
)
𝑑
𝑡
2
=
2

(
𝑡
)


(
𝑡
)

and
𝑑
2

(
𝑡
)
𝑑
𝑡
2
=

(
𝑡
)
. Solve as explained above. To

1a. Write equation in matrix form. What is matrix a?

1b. find eigenvalues and eigenvectors of a

1c. Find P

1d. Find P
-
1
and find
x

and
y

in terms of
X

and
Y

1e. Find the solutions for X and Y that
stricly

decreasing with time (decays)

1f. Then transform back to
x

and
y

to find the original variables as a function of time assuming they
decay from an original value

(
0
)
=
3
,
2

E6.2
-
2

Consider the equation
d

1

𝑑

=
2

1


2

and
𝑑

2

𝑑

=


1

+
2

2

Solve following the same
steps as E6.2
-
1

E6.2
-
1

Consider the equation
d

1

𝑑

=


1

+

2

and
𝑑

2

𝑑

=

1


2

1a. Solve following the same steps as the formal solution above. Find eigenvalues and
e.vectors

and then P.

When finding the matrix P do not get stopped by zero eigenvalue because the eigenvectors of a zero eigenvalue
don’t necessarily vanish!

1b. Then get the transformed solution for the Y’s

1c. Then transform back to get the solution in terms of the original y’s and plot in a 2dim graph (y1,y2) the
evolution of the solution starting from the point y=(1,
-
1).

E6.2
-
3

Consider the equation
d

1

𝑑

=
2

1

and
𝑑

2

𝑑

=

3

2

Write in matrix form and solve. Comment
on the solution process!

PHY 301: MATH AND NUM TECH

6.2. Differential Equations

6.2
-
B Linear Differential Equations: Example 1 Electrical Circuit

Writing that the voltage around each loop of current i
1

and i
2
, is
zero we get:

In the electrical
circuit
below the capacitance C
1

2

=2microF are initially charged at V
AB

=5Volts (say V
A
>V
B
). At t=0 the switch S is closed. Find the charge on the capacitors as a function of time.
Take R
1
=R
2
=100

Problem set up
:

1
1 1
1
2 1
2 2
2 1
0 ( ) ( )
0 ( ) ( ) ( )
A A A C C
A A A B B A
B B A
Q
V V V V V V Ri
C
Q Q
V V V V V V V V R i
C C
       
  

      
In addition we can write a relationship between q’s and i’s: and

So that the equations become:

1 2
1 2 2

dQ dQ
i i i
dt dt
  
1 1
1
1
2 2 1
2
2 1
2
0
0
(
)
d
dQ dQ Q
R
dt dt C
Q Q
R
C C
Q
dt

  

 

We massage
the
equations
a
little to put them in our generic form:

Which finally yields:

1 2 1 1
1
2 2 2 1 1
2 2 1
2 2 2 1
0
dQ Q Q Q
R
dt R C R C C
Q
dQ
dt
Q
R C R C
 
 

 
 
 
   

 

 
 

 
  
 

1 2 1 1 1 1 2
1 1 2
2 2 2 1 1 1 1 2 1 2 2 1 2 1 2 2
2 2 1 2 1 2
1 2
2 2 2 1 2 1
2
2 2 2 1 2 2
1 1 1 1 1

1
1 1

dQ Q Q Q dQ Q Q
Q Q Q
dt R C R C RC dt C R R R C C R R R C
Q Q Q Q
Q Q Q
R C R C R
dQ dQ
dt dt
C R C R C R C

 
     
           
 
     
 
     
   
 
 
 
  

PHY 301: MATH AND NUM TECH

5.2. Differential Equations

6.2
-
C Linear Differential Equations: Example 1 cont’d

PHY 301: MATH AND NUM TECH

5.2. Differential Equations

6.2
-
C Linear Differential Equations:
Example 1 cont’d

PHY 301: MATH AND NUM TECH

5.2. Differential Equations

6.2
-
C Linear Differential Equations: Example 2. Vibrating String & Normal Modes

Consider the following 3 masses m attached to a string under tension T. The string is fixed at
the ends and massless. The distances are indicated on the drawing. The masses are taken to
oscillate in a plane (2dim) and the tension constitutes the net force on the masses (i.e. no
gravity is present

outer space
-

or mg and normal force cancel by letting the masses oscillate
in a horizontal plane like a table top)

The tension in the rope is assumed to be
constant, say T.

PHY 301: MATH AND NUM TECH

5.2. Differential Equations

6.2
-
C Linear Differential Equations: Example 2. Vibrating String & Normal Modes

PHY 301: MATH AND NUM TECH

5.2. Differential Equations

6.2
-
C Linear Differential Equations: Example 2. Vibrating String & Normal Modes

PHY 301: MATH AND NUM TECH

5.2. Differential Equations

6.2
-
C Linear Differential Equations: Example 2. Vibrating String & Normal Modes

E6.2
-
5

Solve the vibrating string problem for 2 masses on a string under tension T. The masses are located at a
distance d from the fixed ends of the string, and are separated by a distance 2d. Find the general solution from the
position of the masses as a function of time and find the normal modes of the system.