Transcript Math III - uni

Mathe III
Lecture 4
Linear Equations
X t+2 + at X t+1 + bt X t = ct
The associated homogeneous equation:
X t+2 + at X t+1 + bt X t = 0
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We have shown that:
 1
if ut , ut
 Au
t
 1
 2
are linearly independent solutions then
+ But
2

A, B is the set of ALL solutions.
The set of solutions to the homogeneous linear
difference equation xt+2 = at xt+1 + bt xt ,
is a linear space of dimension 2.
Hence: If we have two independent solutions,
then we have ALL.
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The nonhomogeneous equation
xt+2 = at xt+1 + bt xt + ct
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Lemma:
 1
if wt , wt
 2
are solutions to the
nonhomogeneous equation xt+2 = at xt+1 + bt xt + ct
then wt
 1
- wt
2
is a solution to the
homogeneous equation xt+2 = at xt+1 + bt xt .
Proof:
wt+ 2
-w

 1
 2
t+ 2
 1
 1
 2
 2
= at wt+1 + bt wt
= at wt+1


+ bt wt
+ ct
+ ct
 
wt+2  1 - wt+2  2  = at wt+1 1 - wt+1 2  + bt wt  1 - wt  2 

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We have shown that:
 1
if wt , wt
 2
are solutions to the
nonhomogeneous equation xt+2 = at xt+1 + bt xt + ct
 1
and ut ,ut
 2
are two independent solution to the
homogeneous equation xt+2 = at xt+1 + bt xt .
then wt
1
- wt
 2
= Aut
1
+ But
 2
for some A,B.
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Or equivalently:
 1
If ut ,ut
 2
are two independent solution to the
homogeneous equation xt+ 2 = at xt+1 + bt xt ,
and wt
 1
a solution to the nonhomogeneous
equation xt+ 2 = at xt+1 + bt xt + ct
then all other solutions of the nonhomogeneous
equation can be represented as
Aut
 1
+ But
 2
+ wt
 1
for some A, B.
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A simple geometric analogue
Consider the equation:
y = 5x + 15
and the associated homogeneous equation:
y = 5x
The set of solutions to the homogeneous equation
 x, y  y = 5x
is a linear space
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A simple geometric analogue
y = 5x
y
20
=
+
1
 1,20  +  x, y  y = 5x

 x, y  y = 5x + 15
x
y = 5x +15
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econd Order equations with
Constant coefficients.
Homogeneous equation :
xt+2 + axt+1 + bxt = 0
Assume that b  0 ,
(else the equation is of the first order).
search for a solution of the form xt = mt :
m
t+2
+ am
t+1
t
+ bm = 0
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econd Order equations with
Constant coefficients.


m t m 2 + am + b = 0
2
m + am + b = 0
2
m1,2
m
-a ± a - 4b
=
2
t+2
+ am
t+1
t
+ bm = 0
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cond Order equations with Constant coefficie
-a ± a 2 - 4b
2
if
a
- 4b > 0
m1,2 =
2
-a + a 2 - 4b
-a - a 2 - 4b
m1 =
, m2 =
2
2
m1  m2
t
t
the two solutions m1 ,m2 are
linearly independent
since for t = 0,1 the vectors  1,m1  , 1,m2 
are linearly independent.
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cond Order equations with Constant coefficie
2
-a ± a - 4b
2
if a - 4b = 0
m1,2 =
2 -a
m = m1 = m 2 
m t is a solution
2
t
tm is also a solution since :
t + 2 m
t+ 2
+ a  t + 1 m
t+1
+ btm 
t
2

= m  t + 2  m + a  t + 1  m + bt 
t


 


{
{
= m t  t m 2 + am + b  2m 2 + am 
t
2

= m  t m + am + b  m  2m + a  = 0
0
0
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cond Order equations with Constant coefficie
2
-a ± a - 4b
m1,2 =
2 -a
m = m1 = m 2 
2
t
tm is also a solution
t
2
if a - 4b = 0
m t is a solution
t
The two solutions tm ,m are linearly independent
Since for t = 0,1 the two vectors  0,m  , 1,m 
are linearly independent.
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cond Order equations with Constant coefficie
2
-a ± a - 4b
2
if a - 4b < 0
m1,2 =
2
the quadratic equation has no real roots
Two independent solutions are :
r cos  θt  ,r sin  θt 
t
t
a
Where : r = b , cosθ = 2 b
This can be worked out directly by looking at the complex
t,
t
solutions m1 m2
or by using trigonometry to show that the two expressions above
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are solutions and are independent
cond Order equations with Constant coefficie
Summary:
The general solution of a second order equation is:
(for arbitrary A,B)
if
if
if
2
a - 4b > 0
2
a - 4b = 0
2
a - 4b < 0
t
Am1 + Bm2
t
Atm + Bm
t
m1,2 =
t
-a ±
2
a - 4b
2
m=-
a
2
Ar t cos θt  + Br t sin θt 
r=
b , cosθ = -
a
216 b
cond Order equations with Constant coefficie
Nonhomogeneous equations
xt+2 + axt+1 + bxt = ct
We need to find one particular solution
The general solution will be:
u*t
 1
 2
Aut + But + u
Consider the case ct is a constant
ct  c
Search, first, for a constant solution:
xt  C
C + aC + bC = c
1+ a + b  0
*
t
C  1+ a + b = c
17 
C = c 1 + a + b
-1
cond Order equations with Constant coefficie
Nonhomogeneous equations
Consider the case ct is a constant
What if
ct  c
1 + a + b = 0 ???
Search for a linear solution:
xt = Dt
D  t + 2  + aD  t + 1 + bDt = c
Dt  1+ a + b + D  2 + a  = c
-1
a  2
D = c 2 + a
a  2
c  2 + a  t is a solution.
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-1
cond Order equations with Constant coefficie
Nonhomogeneous equations
Consider the case ct is a constant
What if
ct  c
1+ a + b = 0 , and a = -2 ??? a = -2, b = 1
Search for a quadratic solution:
D t + 2
xt = Dt
2
- 2D  t + 1  + Dt = c
2
Dt  1 - 2 + 1 + Dt  4 - 4  + D  4 - 2  = c
a = -2, b = 1
2
2
2
c 2
t is a solution.
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xt+2 + axt+1 + bxt = ct
Consider the case ct is a linear combination of
at , tm
terms of the form
Undetermined Coefficients
Example:
t
2
xt+2 - 5xt+1 + 6xt = 4 + t + 3
Look for a solution of the form:
t
2
C4 + Dt + Et + F
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t
2
xt+2 - 5xt+1 + 6xt = 4 + t + 3
t
2
C4 + Dt + Et + F
C4 t+ 2 + D  t + 2 2 + E  t + 2  + F  

2
t+1

-5 C4 + D  t + 1  + E  t + 1  + F 




+6 C4 + Dt + Et + F 
t
2
= 2C4 + 2Dt +  -6D+ 2E  t +  -D - 3E + 2F 
t
2
4 +t +3
t
2
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= 2C4 + 2Dt +  -6D+ 2E  t +  -D - 3E + 2F 
t
2
4 +t +3
t
2
2C = 1, 2D = 1,
-6D + 2E = 0
-D - 3E + 2F = 3
1
3
C = D= , E= , F =4
2
2
1 t 1 2 3
4 + t + t+4
2
2
2
is a particular solution
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A general solution to the homogeneous equation
xt+2 - 5xt+1 + 6xt = 0
Solve:
2
m - 5m + 6 = 0
m1 = 2,m2 = 3
A general solution of the equation
t
2
xt+2 - 5xt+1 + 6xt = 4 + t + 3
is:
1 t 1 2 3
A2 + B3 + 4 + t + t + 4
2
2
2
t
t
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Stability:
In the long run, the solution should be
independent of the initial conditions.
The general solution of
xt+2 + axt+1 + bxt = ct
is:
 1
 2
*
t
xt = Aut + But + u
if :
i 
lim ut = 0
t 
The system is stable.
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i 
t
i
ut = m where mi
solves
2
m + am + b = 0.
i 
t
i
lim ut = lim m = 0
t 
t 
mi < 1
if
f  m  = m2 + am + b
-1
1
m
a2 - 4b > 0, f  -1 , f 1 > 0, f'  -1 < 0, f' 1 > 0
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