Topics in Applied Mathematics:Introduction to the Mathematics of Climate

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Math 5490November 5, 2014

Dynamical Systems

Math 5490 11/5/2014

Classification

Kaper & Engler, 2013

determinant

trace

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

Dynamical SystemsTopological Classification

trace

0

0determinant

saddle

source

sink

Math 5490 11/5/2014

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

Dynamical SystemsTopological Classification

Math 5490 11/5/2014

trace

0

0determinant

saddle

source

sink

Saddles : det

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

Dynamical SystemsTopological Classification

Math 5490 11/5/2014

trace

0

0determinant

saddle

source

sink

Sources det > 0,  trace >0

Dynamical SystemsTopological Classification

trace

0

0determinant

saddle

source

sink

Math 5490 11/5/2014

determinant

trace

Sinks det > 0,  trace 

Dynamical SystemsTopological Classification

Math 5490 11/5/2014

trace

0

0determinant

saddle

source

sink

Sinks det > 0,  trace 

Dynamical SystemsTopological Classification

All sources look the same.Everything leaves.

Math 5490 11/5/2014

Dynamical SystemsTopological Classification

All sinks look the same.Everything approaches the 

origin asymptotically.

Math 5490 11/5/2014

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

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

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

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

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

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

Dynamical SystemsTopological Classification

All sinks are topologically conjugate, as are all sources 

and all saddles.

Math 5490 11/5/2014

Dynamical SystemsTopological Classification

trace

0

0determinant

saddle

source

sink

Math 5490 11/5/2014

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:

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?

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?

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

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

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

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

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:

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

Dynamical Systems

Math 5490 11/5/2014

Nonlinear Systems3dx x x y

dtdy ydt

Preview

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

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

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

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

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