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Classification trace determinant Kaper & Engler, 213
Topological 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).
Topological Classification saddle source trace sink determinant
Topological Classification Saddles : det < 2 1 1 saddle source trace/2 2 2 2 1 1 2 trace determinant 1 sink 1 determinant 2 2 1 1 2
Topological Classification Saddles : det < source saddle trace sink determinant
Topological Classification Sources det >, trace > 2 1 1 saddle source trace 2 2 2 1 1 2 trace determinant 1 sink 1 determinant 2 2 1 1 2
Topological Classification Sources det >, trace > source saddle trace sink determinant
Topological Classification Sinks det >, trace < 2 1 1 saddle source trace 2 2 2 1 1 2 trace determinant 1 sink 1 determinant 2 2 1 1 2
Topological Classification Sinks det >, trace < source saddle trace sink determinant
Topological Classification All sources look the same. Everything leaves.
Topological Classification All sinks look the same. Everything approaches the origin asymptotically.
Topological Classification One Variable x u u Let x u u u u u u 1 du u x u du u 1 Continuous, but not smooth at. u x du 1 du u u x 1 x u u du u
Topological Classification Two Variables Saddles dy x x 1 x u u 1 y v v du dv u v Continuous, but not smooth at (,). topologically conjugate
Topological Classification Sinks dy x x 1 x u u 1 y v v du dv u v Continuous, but not smooth at (,). topologically conjugate
Topological Classification What about spirals? dz ( 1 ) ix i ilog w z w we w Continuous, but not smooth at. dw w
Topological Classification Computation i dz i i 21 ( 1 iz ) ( ) i i 2 i 2 i 21 i 2 1 1 i 2 i 2 z w w ww w w w dz z w ww w w w 2 2 i Multiply by w w i i i 1 i w 2 w w w w 2 w complex conjugate i i (1 iz ) (1 i) w w 1 ww i ww (1 i) ww ww ww 2ww i i 1 ww ww 2 ww (1 i) ww 2 2 2 2 i i 2 2 ww 1 ww (1 i) ww ww ww 2ww ww iww (1 i) ww ww ww dw w add
Topological Classification What about spirals? dz ( 1 ) ix i ilog w z w we w Continuous, but not smooth at. dw w
Topological Classification All sinks are topologically conjugate, as are all sources and all saddles.
Topological Classification saddle source trace sink determinant
Rest points: Dynamical Systems Nonlinear Systems One Variable f ( x) f( x) If f( p), then x( t) p (constant) is a solution. Example x x 3 Rest points: 3 x x x 1 x 1 x 1,, and 1
Nonlinear Systems One Variable Rest points Example 2 x x 3 1 f(x) What about the stability of the rest points? 1 2 2 1 1 2 x
Nonlinear Systems One Variable Rest points Example 2 x x 3 1 f(x) What about the stability of the rest points? 1 2 2 1 1 2 x
Nonlinear Systems f ( x) Rest point p : f( p) Linear approximation: f ( x) f ( p) f ( p)( x p) f ( p)( x p) Introduce x p. Then f( x) f( p ) f( p) d f( x) f( p ) f( p) Basic Idea If is small, i.e., if x is close to p, d then solutions of f( p) d are close to solutions of f ( p). In particular, the rest point p is asymptotically stable for f ( x) d if the origin is asymptotically stable for f( p).
Nonlinear Systems f ( x) Rest point p : f( p) Criteria The rest point p is asymptotically stable for f( x) if f( p). It is unstable if f( p).
Nonlinear Systems Example f ( x) x x 3 2 stable Rest points: 1,, and 1 f( x) 13x f( 1) f(1) 2, f() 1 Rest points 1 and 1 are stable, rest point is unstable. 2 f(x) 1 1 2 2 1 1 2 x unstable
Nonlinear Systems Two Variables 1 f1( x1, x2) 2 f2( x1, x2) f( x), x, f : 2 2 2 f( x) If f( p), then x( t) p (constant) is a solution (rest point) f( p) f ( p, p ) 1 1 2 f ( p, p ) 1 1 2 x () t p 1 1 x () t p 1 1
Nonlinear Systems dy Example 3 x x y y Rest points: x x 3 y x x 3 x 1 x 1 x y y y ( 1,), (,), and (1,)
If y(), then y( t), for all t. invariant line : x axis Dynamical Systems Nonlinear Systems Example dy 3 x x y y y rest points x How do we analyze the full system?
Nonlinear Systems 3 x x y dy y Preview
Nonlinear Systems Jacobian Matrix 1 2 f ( x, x ) f 1 1 2 ( x, x ) 2 1 2 f( x), x, f : 2 2 2 Df ( x) f1 f1 ( x1, x2) ( x1, x2) x1 x 2 f2 f ( x 2 1, x2) ( x1, x2) x1 x 2
Nonlinear Systems Example dy 3 x x y y Df ( x, y) 2 1 3x 1 1 2 1 1 1 2 1 Df ( 1,) Df (,) Df (1,) 1 1 1
Nonlinear Systems f ( x) Rest point p : f( p) Linear approximation: f ( x) f ( p) Df ( p)( x p) Df ( p)( x p) Introduce x p. Then f( x) f( p ) Df( p) d f( x) f( p ) Df( p) Basic Idea If is small, i.e., if x is close to p, d then solutions of f( p) d are close to solutions of Df ( p). In particular, the rest point p is asymptotically stable for f ( x) d if the origin is asymptotically stable for Df ( p).
Nonlinear Systems f ( x) Rest point p : f( p) Criteria The rest point p is saddle for f( x) if det D f( p). It is a source if det D ( ) and trace D ( ). f p f p f p f p It is a source if det D ( ) and trace D ( ).
dy Dynamical Systems 3 x x y y Nonlinear Systems Example Rest points: ( 1,), (,), and (1,) 2 1 1 1 2 1 Df ( 1,) Df (,) Df (1,) 1 1 1 det D f ( 1,) 2 trace D f ( 1,) 3 sink 2 discriminant D f ( 1,) ( 3) 4 2 1 det D f (,) 1 saddle stable node
If y(), then y( t), for all t. invariant line : x axis Dynamical Systems Nonlinear Systems Example dy 3 x x y y y rest points x stable node saddle