Math 5490 11/10/2014 Dynamical Systems Math 5490 Nonlinear Systems - PDF document

Math 5490 11/10/2014 Dynamical Systems Math 5490 Nonlinear Systems November 10, 2014 dx dt  f p  Topics in Applied Mathematics: f x ( ) Rest point : ( ) p 0 Introduction to the Mathematics of Climate Linear approximation:      f x ( ) f p ( ) Df p ( )( x p ) Df p ( )( x p ) Mondays and Wednesdays 2:30 – 3:45 http://www.math.umn.edu/~mcgehee/teaching/Math5490-2014-2Fall/ Basic Idea    Streaming video is available at Introduce x p .  If is small, i.e., if is close to , x p      http://www.ima.umn.edu/videos/ Then ( ) f x f p ( ) Df p ( )  d    Click on the link: "Live Streaming from 305 Lind Hall".  then solutions of f ( p ) d dx        f x ( ) f p ( ) Df p ( ) dt Participation: dt dt  d   are close to solutions of Df p ( ) . https://umconnect.umn.edu/mathclimate dt dx  In particular, the rest point is asymptotically stable for p f x ( ) dt  d   if the origin is asymptotically stable for Df p ( ) . dt Math 5490 11/10/2014 Dynamical Systems Dynamical Systems Flows Flows “Vector Fields Determine Flows” “Vector Fields Determine Flows” Example     n   n A flow is a continuous map : satisfying        vector field: x ax , x , initial value: (0) x x ( ,0) x x , for all , x 0       at solution: x e x ( , x t s ) ( ( , ), ), for all , x t s x t , and . s 0   at flow: ( , ) x t e x 0 0  t   Alternate Notation ( ) x ( , ) x t   0 id n  Smooth : C , n 0     0 0   t s   t   s Group Property ( times continuously n differentiable) Big Theorem Check properties:   n  n If : f is smooth, then the initial value problem Group Property    a 0    ( x ,0) e x x  x f x ( ), x (0) x , 0 0 0 0     a t s ( )  at as  at     ( x t , s ) e x e e x e ( x , ) s ( ( x , ), ) s t   n     n defines a flow : satisfying 0 0 0 0 0        ( x t , ) f ( ( x t , )), ( x ,0) x . 0 0 0 0  Also, is a smooth as . f Math 5490 11/10/2014 Math 5490 11/10/2014 Dynamical Systems Dynamical Systems Flows Flows “Vector Fields Determine Flows” “Vector Fields Determine Flows” Example Example     n    2   vector field: x Ax , x initial value: (0) x x vector field: x x , x 0   tA initial value: (0) x x solution: x e x 0 0   tA flow: ( x t , ) e x 0 0 Calculus Check properties: dx x t x Group Property       2  2   2    1  x x dx dt x dx dt x t    0 A  dt x 0 x ( x ,0) e x x 0 0 0 0 0  1 1 1 1 1 x t x              ( t s A )  tA sA  tA     0 0 ( x t , s ) e x e e x e ( x , ) s ( ( x , ), ) s t t t x 0 0 0 0 0  x x x x x 1 x t 0 0 0 0 x  0 solution: x  1 x t 0 Experts Only x   0 flow: ( x t , ) 0  1 x t 0 Math 5490 11/10/2014 Math 5490 11/10/2014 Richard McGehee, University of Minnesota 1

Math 5490 11/10/2014 Dynamical Systems Dynamical Systems Calculus Flows Flows “Vector Fields Determine Flows” “Vector Fields Determine Flows” Example Example   2      2    vector field: x x , x initial value: (0) x x vector field: x x , x initial value: (0) x x 0 0 x x     0 0 flow: ( x t , ) flow: ( x t , ) 0  0  1 x t 1 x t 0 0 Issue: group property Check properties: Solutions do not exist for all time. x    0 ( x ,0) x 0   0 x 1 1 x 0      0 0 ( x t , ) as t 0  1 x t x x 0 0 0   ( x t , ) 1 x t x x          0 0 0 0 ( ( x t s , ), ) ( x t , s ) 0       0 local flow : Solutions exist for some time interval. x 1 ( x t s , ) 1 x t x s 1 x ( t s )   0 0 1 s 0 0 0 1 x t 0 Math 5490 11/10/2014 Math 5490 11/10/2014 Dynamical Systems Dynamical Systems Flows Flows “Vector Fields Determine Flows” “Vector Fields Determine Flows” Example Example    2   vector field: x 1 x , x    2    vector field: x 1 x , x initial value: (0) x x 0  initial value: x (0) x  x tanh t 0   0 flow: ( x t , ) 0  1 x tanh t 0 Calculus dx x t x        2   2 1    2 1   1  1 x (1 x ) dx dt (1 x ) dx dt tanh x t Group Property x 0 dt x 0  0 tanh t tanh s         x 1  1   1  1    1       x tanh( t s ) 0  x x tanh tanh t s tanh t tanh s tanh x tanh x t tanh x tanh x t x tanh tanh x t     1 tanh tanh t s  0 0 0 0 0 0 ( x t , s )  0      tanh t tanh s 1 x tanh( t s ) 1 tanh tanh t s x (tanh t tanh ) s   1   0 1 x 0 tanh(tanh x ) tanh t x tanh t   0  0 0 1 tanh tanh t s x   1  1 x tanh t equal 1 tanh(tanh x )tanh t 0 0  x tanh t  0 tanh s        ( x t , ) tanh t 1 x tanh t x tanh t tanh s x tanh tanh t s x tanh t    0  0  0 0   0 ( ( x t s , ), ) flow: ( x t , )  0      0  1 ( x t , )tanh t x tanh t 1 x tanh t x tanh s tanh tanh t s   1 x tanh t 0 0 1 tanh s 0 0 0 1 x tanh t 0 Math 5490 11/10/2014 Math 5490 11/10/2014 Dynamical Systems Dynamical Systems Flows Flows “Vector Fields Determine Flows” Backwards Time Example    n   vector field: x f x ( ), x (0) x , x 0    2   vector field: x 1 x , x       flow: ( x t , ) f ( x t , ) 0 0  x tanh t   0 local flow: ( x t , ) 0  1 x tanh t     Let ( x t , ) ( x , t ) 0 0 0   Issue:                     then ( x t , ) ( x , t ) ( x , t ) f ( x , t ) f ( x t , )  0  0 0 0 0 Solutions do not exist for all time. t t   So ( x ,t) satisfies x tanh t  0         1  0 x 1 ( x t , ) as t tanh ( 1 x )     0 0  0 x f x ( ), x (0) x 1 x tanh t 0 0 Going backward in time is the same as following the  negative of the vector field. x tanh t           0 1 x 1 ( x t , ) as t tanh ( 1 x ) 0 0 0  1 x tanh t 0 Math 5490 11/10/2014 Math 5490 11/10/2014 Richard McGehee, University of Minnesota 2