math 54902014/11/05 · but not smooth at 0. dw w dt 2221 21 22 22 1 1 (1 ) (1 ) iiiiii iiii i dz z...
TRANSCRIPT
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Topics in Applied Mathematics:Introduction to the Mathematics of Climate
Mondays and Wednesdays 2:30 – 3:45http://www.math.umn.edu/~mcgehee/teaching/Math5490-2014-2Fall/
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Click on the link: "Live Streaming from 305 Lind Hall".Participation:
https://umconnect.umn.edu/mathclimate
Math 5490November 5, 2014
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Dynamical Systems
Math 5490 11/5/2014
Classification
Kaper & Engler, 2013
determinant
trace
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Dynamical SystemsTopological Classification
If neither eigenvalue has zero real part, then the system is called hyperbolic, in which case, there are only three classes:
1. saddles: One positive eigenvalue and one negative. The determinant is negative.
2. sources: Both eigenvalues have positive real part. The determinant is positive, and the trace is positive.
3. sinks: Both eigenvalues have negative real part. The determinant is positive, and the trace is negative.
Every system in one of the three categories looks the same to a topologist (topological conjugacy).
Math 5490 11/5/2014
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Dynamical SystemsTopological Classification
trace
0
0determinant
saddle
source
sink
Math 5490 11/5/2014
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Dynamical SystemsTopological Classification
trace
0
0determinant
saddle
source
sink
Math 5490 11/5/2014
‐2
‐1
0
1
2
‐2 ‐1 0 1 2
‐2
‐1
0
1
2
‐2 ‐1 0 1 2
determinant
trace/2
Saddles : det
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Dynamical SystemsTopological Classification
Math 5490 11/5/2014
trace
0
0determinant
saddle
source
sink
Saddles : det
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Dynamical SystemsTopological Classification
trace
0
0determinant
saddle
source
sink
Math 5490 11/5/2014
determinant
trace
Sources det > 0, trace >0
‐2
‐1
0
1
2
‐2 ‐1 0 1 2
‐2
‐1
0
1
2
‐2 ‐1 0 1 2
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Dynamical SystemsTopological Classification
Math 5490 11/5/2014
trace
0
0determinant
saddle
source
sink
Sources det > 0, trace >0
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Dynamical SystemsTopological Classification
trace
0
0determinant
saddle
source
sink
Math 5490 11/5/2014
determinant
trace
Sinks det > 0, trace
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Dynamical SystemsTopological Classification
Math 5490 11/5/2014
trace
0
0determinant
saddle
source
sink
Sinks det > 0, trace
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Dynamical SystemsTopological Classification
All sources look the same.Everything leaves.
Math 5490 11/5/2014
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Dynamical SystemsTopological Classification
All sinks look the same.Everything approaches the
origin asymptotically.
Math 5490 11/5/2014
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Dynamical SystemsTopological Classification
One Variable
0dx xdt
1x u u
1dx duudt dt
x u
du udt
1dx duudt dt
x u
du udt
0u 0u
0dx xdt
1 0Let 0
u ux u u
u u
du udt
Continuous, but not smooth at 0.
Math 5490 11/5/2014
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Dynamical SystemsTopological Classification
Two Variables
1
1
x u u
y v v
0
0
dx xdtdy xdt
du udtdv vdt
Continuous, but not smooth at (0,0).
topologically conjugate
Math 5490 11/5/2014
Saddles
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Dynamical SystemsTopological Classification
Sinks
1
1
x u u
y v v
0
0
dx xdtdy xdt
du udtdv vdt
Continuous, but not smooth at (0,0).
topologically conjugate
Math 5490 11/5/2014
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Dynamical SystemsTopological Classification
What about spirals?
Math 5490 11/5/2014
( 1 )dz i xdt
logi i wz w w e w
Continuous, but not smooth at 0. dw w
dt
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2 1 2 12 22 2
2 2
1
1
(1 ) (1 )
i ii ii i
i ii i
i
dzz w ww w w wdt
ww w w ww
i z i w w
Dynamical SystemsTopological Classification
Computation
Math 5490 11/5/2014
( 1 )dz i zdt
2 12 2( )i ii iz w w ww w w w
dw wdt
2 2
2 2
1 (1 )
1 (1 )
2
i i
i i
ww ww i ww
ww ww i ww
ww ww ww
Multiply by iw w
complex conjugate
add
2ww ww ww
2 21 2 (1 )(1 )
i iww ww ww i ww
ww iww i ww
ww ww
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Dynamical SystemsTopological Classification
What about spirals?
Math 5490 11/5/2014
( 1 )dz i xdt
logi i wz w w e w
Continuous, but not smooth at 0. dw w
dt
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Dynamical SystemsTopological Classification
All sinks are topologically conjugate, as are all sources
and all saddles.
Math 5490 11/5/2014
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Dynamical SystemsTopological Classification
trace
0
0determinant
saddle
source
sink
Math 5490 11/5/2014
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear SystemsOne Variable
0 ( ) 0dx f xdt
Rest points:
3dx x xdt
If ( ) 0, then ( ) (constant) is a solution.f p x t p
( )dx f xdt
Example
3 0 1 1 01, 0, and 1
x x x x x
Rest points:
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear SystemsOne Variable
3dx x xdt
Example
Rest points
‐2
‐1
0
1
2
‐2 ‐1 0 1 2
f(x)
x
What about the stability of the rest points?
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear SystemsOne Variable
3dx x xdt
Example
Rest points
‐2
‐1
0
1
2
‐2 ‐1 0 1 2
f(x)
x
What about the stability of the rest points?
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
If is small, i.e., if is close to ,
then solutions of ( )
are close to solutions of ( ) .
x pd f pdt
d f pdt
Rest point : ( ) 0p f p ( )dx f xdt
Introduce .Then ( ) ( ) ( )
( ) ( ) ( )
x pf x f p f p
d dx f x f p f pdt dt
Basic Idea
Linear approximation:( ) ( ) ( )( ) ( )( )f x f p f p x p f p x p
In particular, the rest point is asymptotically stable for ( )
if the origin is asymptotically stable for ( ) .
dxp f xdt
d f pdt
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
Rest point : ( ) 0p f p ( )dx f xdt
Criteria
The rest point is asymptotically stable for ( ) if ( ) 0.
It is unstable if ( ) 0.
dxp f x f pdt
f p
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear SystemsExample
3( )dx f x x xdt
stable
‐2
‐1
0
1
2
‐2 ‐1 0 1 2
f(x)
x
Rest points: 1, 0, and 12( ) 1 3
( 1) (1) 2, (0) 1f x x
f f f
Rest points 1 and 1 are stable,rest point 0 is unstable.
unstable
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear SystemsTwo Variables
0 ( ) 0dx f xdt
If ( ) 0, then ( ) (constant) is a solution (rest point)f p x t p
2 2 2( ), , :dx f x x fdt
11 1 2
22 1 2
( , )
( , )
dx f x xdtdx f x xdt
1 1 2
1 1 2
( , ) 0( ) 0
( , ) 0f p p
f pf p p
1 1
1 1
( )( )
x t px t p
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
3dx x x ydtdy ydt
Example
3 3 1 1 00 000 0
( 1,0), (0,0), and (1,0)
x x xx x y x xyy y
Rest points:
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
3dx x x ydtdy ydt
Example
How do we analyze the full system?
x
yrest points
“invariant line”:x‐axis
If (0) 0, then ( ) 0, for all .y y t t
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems3dx x x y
dtdy ydt
Preview
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
Jacobian Matrix
2 2 2( ), , :dx f x x fdt
1 11 2 1 2
1 2
2 21 2 1 2
1 2
( , ) ( , )( )
( , ) ( , )
f fx x x xx x
Df xf fx x x xx x
11 1 2
22 1 2
( , )
( , )
dx f x xdtdx f x xdt
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
Example
21 3 1( , )0 1
xDf x y
3dx x x ydtdy ydt
2 1 1 1 2 1( 1,0) (0,0) (1,0)
0 1 0 1 0 1Df Df Df
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
If is small, i.e., if is close to ,
then solutions of ( )
are close to solutions of ( ) .
x pd f pdt
d Df pdt
Rest point : ( ) 0p f p ( )dx f xdt
Introduce .Then ( ) ( ) ( )
( ) ( ) ( )
x pf x f p Df p
d dx f x f p Df pdt dt
Basic Idea
Linear approximation:( ) ( ) ( )( ) ( )( )f x f p Df p x p Df p x p
In particular, the rest point is asymptotically stable for ( )
if the origin is asymptotically stable for ( ) .
dxp f xdt
d Df pdt
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
Rest point : ( ) 0p f p ( )dx f xdt
Criteria
The rest point is saddle for ( ) if det D ( ) 0.
It is a source if det D ( ) 0 and trace D ( ) 0.
It is a source if det D ( ) 0 and trace D ( ) 0.
dxp f x f pdt
f p f p
f p f p
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Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
3dx x x ydtdy ydt
Example
Rest points: ( 1,0), (0,0), and (1,0)
2 1 1 1 2 1( 1,0) (0,0) (1,0)
0 1 0 1 0 1Df Df Df
det D (0,0) 1 0f
det D ( 1,0) 2 0
trace D ( 1,0) 3 0
f
f
2discriminant D ( 1,0) ( 3) 4 2 1 0f
saddlesink
stable node
-
Dynamical Systems
Math 5490 11/5/2014
Nonlinear Systems
3dx x x ydtdy ydt
Example
x
yrest points
“invariant line”:x‐axis
If (0) 0, then ( ) 0, for all .y y t t
saddlestable node