Mathe III Lecture 5
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Transcript of Mathe III Lecture 5
Mathe IIILecture 5Mathe IIILecture 5Mathe IIILecture 5Mathe IIILecture 5
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Stability:
In the long run, the solution should be independent of the initial conditions.
The general solution of
t+2 t+1 t tx + ax + bx = c
is: *1 2t t t tx = Au + Bu + u
if :
it
tlim u = 0
The system is stable.
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i tt i
t tlim u = lim tm = 0
where
solves
i tt i i
2
u = m m
m + am + b = 0.
i tt i
t tlim u = lim m = 0
if im < 1
1-1
2f m = m + am + b
m
2a - 4b 0, f -1 = f 1 > 0, f' -1 < 0, f' 1 > 0
The root (s) are in (-1, 1) iff:
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1-1
2f m = m + am + b = 0
m
2a - 4b 0, f -1 = f 1 > 0, f' -1 < 0, f' 1 > 0
2a 4b,
1 - a + b > 0, 1+ a + b > 0
-2 + a < 0, 2 + a > 0 a < 2
a < 1+ b
b < 1
a < 1+ b, b < 1The system is stable iff:
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Differential Equations
First Order Differential Equations
wheredx
= f x,t x = x tdt
first order, ordinary equation (single variable)
Differential EquationsDifferential Equations
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x
dx= f x,t
dt
t
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dx= x = f t
dt
x t = f t dt + C
The simplest possible equation:
x
t
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wherex = f x,t x = x t
x t + τ - x t= f x,t
τ
An approximation:
For a given τ let: t = 0,τ,2τ,3τ, ....nτ
we obtain a difference equation , solve it and let τ 0
or graphically:
x t + τ = x t + τf x t ,t
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x
wherex = f x,t x = x t
For t = 0, assume x(0) = x0
x0
t
0x 0 = f x ,0
τ
x1
1x τ = f x ,τ
2τ
x2
etc.
x t + τ = x t + τf x t ,t
0 0x τ = x + τf x ,0 x 2τ = x τ + τf x τ ,τ
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x
wherex = f x,t x = x t
For t = 0, assume x(0) = x0
x0
tτ
x1
2τ
x2
Now choose a smaller τ
2ττ
As we approach a curve which solves
τ 0
x = f x,t
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,
There exists a unique satisf
ying
0
x = x t
x = f x,t x 0 = x
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Separable Differential Equations
x = f t g x
dx= f t g x
dt
A formal ‘trick’:
dx= f t dt
g x dxf= t dt + C
g x
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Is this ‘trick’ valid ???
If solves the equa tio n (t) x = f t g x
then : (t) = f t g t
when then : (t)
g t 0 = f tg t
hence : (t)dt f t dt C
g t
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hence : (t)dt f t dt C
g t
change the varia : ble : x = t dx = t dt
dxf t dt C
g x
G x = F t + C
1 G' x = , F' t = f t
g xThis defines x as an implicit function of t
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3
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tx =
x + 1EXAMPLE:
6 3x + 1 dx = t dt
6 3x + 1 dx = t dt C
7 41 1x + x = t + C
7 4
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(compound interest)
r t EXAMPLE:
1
d= r t dt C
ln 1= R t C R t = r t dt
1C + R tt = e
1 R tCt = e e R t= Ce 0R0 = Ce R 0
0C =
e
R t
R 0
0t = e
e
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r t EXAMPLE:
1 R tCt = e e R t= Ce 0R0 = Ce R 0
0C =
e
R t
R 0
0t = e
e
R t -R 0= 0 e
t
0
R t - R 0 r s ds
R t -R 0= 0 e
t
0
r s ds
t 0 e
exp t
0
0 r s ds
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Separable Differential Equations
0 0
x t = xx = f t g x
(again)
We have seen that if solves the equation,
with then : 0 0
(t)
(t ) = x
(t)= f t
g t
0
and
0
t
t t
t(s)
ds f s dsg s
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Separable Differential Equations (again)
0
0
t
t t
t(s)
ds f s dsg s
change the variable :
0 0= s , d = s ds, = t
0
0
t
x t
xd
f s dsg
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0
2 x = x 0dx
= -2x tdt
EXAMPLE:
2
dx- = 2tdt
x
x t
0
2
x 0
d- = 2sds
1
0
xt2
0x
s
2= t0
1 1- =
x x
2
0
1 1= t
x x 2
0
1t +
x
1x =
Graphic description of the solution
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0
2 x = x 0dx
= -2x tdt
EXAMPLE:
2
0
1t +
x
1x =Graphic description of the solution
denote0
1 C =
x
2t + C
1x =
0assume first 00
1 x > 0 : C =
x
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0 0
2 x = x 0dx
= -2x tdt
EXAMPLE:
2t + C
1x =
t
x
Graphic description of the solution
x 0
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0
2 x = x 0dx
= -2x tdt
EXAMPLE:
2t C
1x =
-
t
x
Graphic description of the solution
< 0
, 02
0
01
t +x
1 1x = C = -
x
C C
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0
2 x = x 0dx
= -2x tdt
EXAMPLE:
2t C
1x =
-
t
x
Graphic description of the solution
< 0
C C
0 < t < Ct > C
0 > x 0
C2t C
1x =
+
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Example
1-α α
λt0
X = AK L
K = sX
L = L e
national product
capital stock
number of workers
X t =
K t =
L t
1-α αK = sAK L α1-α λt0= sAK L e
λt0L e
1-α αAK L
sX
α 1-α λαt0
dKK = = sAL K e
dtα-1 α λαt
0K dK = sAL e dt
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Example
α-1 α λαt0K dK = sAL e dt
0
α-1 α λατ0
K t
K 0
d = sAL e dτ
0
K tα α λατ
0K 0
1 1sAL e
α αλ
α α α λαt0 0
1 1Κ - K sAL e - 1
α αλ
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1-α α
λt0
X = AK L
K = sX
L = L e
Example
α α α λαt0 0
1 1Κ - K sAL e - 1
α αλ
1/αα α λαt0 0
sK = K + A L e - 1
λ
This enables us to study how the evolution of capital changes with the parameters s,α, λ, A
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1-α α
λt0
X = AK L
K = sX
L = L e
Example
1/αα α λαt0 0
sK = K + A L e - 1
λ
How does K/L behave in the long run?
α λαtα α00
α α λαt0
L e - 1KK s= + A
L L λ L e
α α α λαt0 0
sK = K + A L e - 1
λ
λt0L e
αL/
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1-α α
λt0
X = AK L
K = sX
L = L e
Example
1/αα α λαt0 0
sK = K + A L e - 1
λ
How does K/L behave in the long run?
α λαtα α00
α α λαt0
L e - 1KK s= + A
L L λ L e 1α
-λαt0α
K s= + A e
L λ
t
sA
λ