Math 5490 11/17/2014 Dynamical Systems Math 5490 November 17, - PDF document

Math 5490 11/17/2014 Dynamical Systems Math 5490 November 17, 2014 Topics in Applied Mathematics: Introduction to the Mathematics of Climate Mondays and Wednesdays 2:30 – 3:45 Bifurcation Theory http://www.math.umn.edu/~mcgehee/teaching/Math5490-2014-2Fall/ Streaming video is available at http://www.ima.umn.edu/videos/ Click on the link: "Live Streaming from 305 Lind Hall". Participation: https://umconnect.umn.edu/mathclimate Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Saddle ‐ Node Bifurcation Saddle ‐ Node Bifurcation stable manifold stable manifold     1 6 unstable manifold 1 6 c c  b  R R 2 2 b     0.3 0.33 a stable spiral stable spiral a saddle saddle Increase the Increase the flow resistance. flow resistance. stable node The saddle and stable node the stable node Not much different, but it is start to merge. easier to get to c . Math 5490 11/17/2014 Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Saddle ‐ Node Bifurcation Bifurcation Theory stable spiral Setup      n    m   x f x x ( , ), , 1 6 c  R 2 state variables parameters   0.4     x f rest point at 0 when 0: (0,0) 0 Increase the flow resistance. What happens when we change the parameters? The saddle and the stable node have disappeared. The Gulf Stream will eventually reverse. Math 5490 11/17/2014 Math 5490 11/17/2014 Richard McGehee, University of Minnesota 1

Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Bifurcation Theory Bifurcation Theory     n   m x f x x   ( , ), ,     n   m x f x x   ( , ), ,     x f rest point at 0 when 0 : (0,0) 0 x    f  rest point at 0 when 0 : (0,0) 0 Poincaré Continuation No Bifurcation (Poincaré Continuation)  D f If Jacobian matrix (0,0) is nonsingular, then, for small values of , 1  D f p The Jacobian matrix (0,0) is nonsingular, i.e., has no zero eigenvalues. there is a rest point ( ) satisfying 1     p f p (0) 0, ( ( ), ) 0. Conclusion Idea of Proof  p  For small values of , there is a rest point ( ) satisfying p  f p    (0) 0, ( ( ), ) 0. We can write        f x Ax B 2 x ( , ) O ( , ) 0, The rest point “continues” for small parameter values.   A D f B n m x where (0,0) and is an matrix and solve for : 1  x  p   A 1 B   2  ( ) O ( ). Math 5490 11/17/2014 Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Bifurcation Theory Classification x   f x  x   n    m ( , ), ,     x f rest point at 0 when 0: (0,0) 0 Poincaré Continuation Poincare  D f continuation If Jacobian matrix (0,0) is nonsingular, then, for small values of , 1  p there is a rest point ( ) satisfying trace     p f p (0) 0, ( ( ), ) 0. There’s more! f C 1 If is continuously differentiable ( ), then the Jacobian matrix D f p    ( ( ), ) varies continuously with , as do the eigenvalues and eigenvectors. 1 If the rest point at μ = 0 is hyperbolic (or a saddle, or a stable node, determinant or an unstable node, or a stable spiral, or an unstable spiral), then the Poincare continuation fails when determinant = 0. rest point p ( μ ) inherits the property for small values of μ . Kaper & Engler, 2013 Math 5490 11/17/2014 Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Bifurcation Theory Bifurcation Theory Example Example x   f x     x  x 2 ( , ) 2       2 x  f x x x ( , ) 2 D f x     x D f    ( , ) 2 2 , so (0,0) 2 0,       1 1 D f x x D f ( , ) 2 2 , so (0,0) 2, 1 1    x p p so there is a rest point ( ) satisfying (0) 0.     x p so the rest point ( ) has an eigenvalue near 2 for small       x p 2 ( ) O ( ). and hence is asymptotically stable. 2 For each value of μ close to 0 , there is a unique rest     x Note that there is another rest point at 2 for 0. point near x = 0 .       D f Its eigenvalue is ( 2,0) 2 2( 2) 2, so it is unstable. 1   p  Futhermore, for small values of , there is a unique rest point ( ) In this example, we can solve explicitly:   x near 2, and that rest point is unstable. 2     x x 2 0     2 4 4 x       1 1 . 2  p  2 0 Since (0) 0 , we take the "+" sign: small  x  p       ( ) 1 1 . Math 5490 11/17/2014 Math 5490 11/17/2014 Richard McGehee, University of Minnesota 2

Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Classification Bifurcation Theory Example   x  x 2  x             2 Two rest points: 2 0, 1 1 x f x x x ( , ) 2        p rest point: 1 1 eigenvalue: 2 1 1 Poincare     D f x x         1 ( , ) 2 2 p rest point: 1 1 eigenvalue: 2 1 continuation 2 trace  When μ = -1 , the rest points 4 merge, and the eigenvalue 3 becomes 0 . The rest point becomes a 2 “saddle ‐ node”. 1 x unstable 0 stable determinant ‐ 1 Poincare continuation fails when determinant = 0. ‐ 2 saddle ‐ node What happens when it fails? ‐ 3 ‐ 2 ‐ 1 0 1 Kaper & Engler, 2013 Math 5490 11/17/2014 Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Bifurcation Theory Bifurcation Theory Example Example Rest Points:          2 x f x x x ( , ) 2 x     x 2   x y     0 : two rest points: ( , ) ,0 vector field y    y      2 x y 0 : one rest point: ( , ) 0,0   0 : no rest point   x  2 0   D f x y    Jacobian ( , ),  4 1   0 2  unstable 3 μ < 0 2 stable       2 0       saddle 1 D f  ( ,0), 1   0 2  x   0 0        stable 2 0          D f  1 ( ,0), ‐ 1 1 node   0 2  saddle ‐ node    2 ‐ 2 ‐ 3 ‐ 2 ‐ 1 0 1 Math 5490 11/17/2014 Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Bifurcation Theory Bifurcation Theory Example Example Rest Points: Rest Points:         2           2       x x x y x x x y 0 : two rest points: ( , ) ,0 0 : two rest points: ( , ) ,0 vector field vector field           y y y y 2   x y  2   x y  0 : one rest point: ( , ) 0,0 0 : one rest point: ( , ) 0,0     0 : no rest point 0 : no rest point  x  2 0   x y    x D f Jacobian ( , ),  1    2 0 2 saddle μ = 0 1 saddle ‐ node    0 0 determinant = 0   D f    0 (0,0),  1    trace < 0 0 2 stable node ‐ 1 The local structure is not determined by the linearized equations. ‐ 2 ‐ 3 ‐ 2 ‐ 1 0 1 Math 5490 11/17/2014 Math 5490 11/17/2014 Richard McGehee, University of Minnesota 3

Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Bifurcation Theory Stommel Model Example stable manifold x     x 2 y    y 2   1 6 unstable manifold c stable node saddle saddle node R  b 2   0.3 stable spiral a saddle Increase the flow resistance. stable node Not much different, but it is easier to get to c .       0 0 0 Math 5490 11/17/2014 Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Stommel Model Stommel Model stable spiral stable manifold     1 6 1 6 c c   R R 2 2 b     0.33 0.4 a stable spiral saddle Increase the Increase the flow flow resistance. resistance. stable node The saddle and The saddle and the stable the stable node node have disappeared. start to merge. The Gulf Stream will eventually reverse. Math 5490 11/17/2014 Math 5490 11/17/2014 Dynamical Systems Dynamical Systems Bifurcation Theory Bifurcation Theory Example Example    3      3   x x x x x x 3 3 Rest Points Rest Points  3       3  3       3 x x x x x x x x 3 0, 3 3 0, 3 D f x    x 2 1 ( , ) 3 3 3 3 2 2    D f x x stable: ( , ) 0, if 1 1 1 1    D f x x unstable: ( , ) 0, if 1 1 μ 0 μ 0     D f x x ‐ 1 ( , ) 0, if 1 ‐ 1 1 ‐ 2 ‐ 2 ‐ 3 ‐ 3 ‐ 3 ‐ 2 ‐ 1 0 1 2 3 ‐ 3 ‐ 2 ‐ 1 0 1 2 3 x x Math 5490 11/17/2014 Math 5490 11/17/2014 Richard McGehee, University of Minnesota 4