Math 211 Math 211 Lecture #36 The Use of the Linearization - PDF document

1 Math 211 Math 211 Lecture #36 The Use of the Linearization November 19, 2003 2 Linearization of a Planar System Linearization of a Planar System x ′ = f ( x, y ) y ′ = g ( x, y ) • Assume ( x 0 , y 0 ) is an equilibrium point, so f ( x 0 , y 0 ) = g ( x 0 , y 0 ) = 0 • We have by Taylor’s theorem f ( x 0 + u, y 0 + v ) = ∂f ∂x ( x 0 , y 0 ) u + ∂f ∂y ( x 0 , y 0 ) v + R f ( u, v ) g ( x 0 + u, y 0 + v ) = ∂g ∂x ( x 0 , y 0 ) u + ∂g ∂y ( x 0 , y 0 ) v + R g ( u, v ) R f ( u,v ) R g ( u,v ) where u 2 + v 2 → 0 and u 2 + v 2 → 0 . √ √ Return 3 Linearization at ( x 0 , y 0 ) Linearization at ( x 0 , y 0 ) • Set x = x 0 + u and y = y 0 + v . The system becomes u ′ = ∂f ∂x ( x 0 , y 0 ) u + ∂f ∂y ( x 0 , y 0 ) v + R f ( u, v ) v ′ = ∂g ∂x ( x 0 , y 0 ) u + ∂g ∂y ( x 0 , y 0 ) v + R g ( u, v ) • Approximate by the linear system u ′ = ∂f u + ∂f ˜ ∂x ( x 0 , y 0 )˜ ∂y ( x 0 , y 0 )˜ v v ′ = ∂g u + ∂g ˜ ∂x ( x 0 , y 0 )˜ ∂y ( x 0 , y 0 )˜ v Return 1 John C. Polking

4 Matrix Form of the Linearization Matrix Form of the Linearization v ) T and introduce the Jacobian matrix • Set u = (˜ u, ˜ ∂f ∂f   ∂x ( x 0 , y 0 ) ∂y ( x 0 , y 0 ) J =  ∂g ∂g  ∂x ( x 0 , y 0 ) ∂y ( x 0 , y 0 ) The linearization becomes u ′ = J u . • We can solve the linear system explicitly. • The linearization gives us information about the original system. Return Original system 5 Theorem 1 Theorem 1 Consider the planar system Theorem: x ′ = f ( x, y ) y ′ = g ( x, y ) where f and g are continuously differentiable. Suppose that ( x 0 , y 0 ) is an equilibrium point. If the linearization at ( x 0 , y 0 ) has a generic equilibrium point at the origin, then the equilibrium point at ( x 0 , y 0 ) is of the same type. • Generic types: Saddle, nodal source, nodal sink, spiral source, and spiral sink. — All occupy large open subsets of the trace-determinant plane. • Nongeneric types: Center and others. — Occupy pieces of the boundaries between the generic points. Return Matrix form 6 Examples Examples • Center. x ′ = y + αx ( x 2 + y 2 ) y ′ = − x + αy ( x 2 + y 2 ) � α > 0 ⇒ (0 , 0) T is unstable. � α < 0 ⇒ (0 , 0) T is a sink. • Competing species. x ′ = (5 − 2 x − y ) x y ′ = (7 − 2 x − 3 y ) y • Default system in pplane. Return Theorem 2 John C. Polking

7 Linear Analysis of Equilibrium Points Linear Analysis of Equilibrium Points • Provides a good qualitative picture of how solutions behave near generic equilibrium points. • Provides limited qualitative information about the solutions near nongeneric equilibrium points. � A linear center could be a spiral source or a spiral sink. • Provides no information about the global behavior of solutions to nonlinear systems. Return 8 Higher Dimensional Systems Higher Dimensional Systems Autonomous equation y ′ = f ( y ) . • y = ( y 1 , y 2 , · · · , y n ) T , y 0 is an equilibrium point. • f ( y ) = ( f 1 ( y ) , f 2 ( y ) , · · · , f n ( y )) T • J is the Jacobian matrix • f ( y 0 + u ) = J ( y 0 ) u + R ( u ) where lim u → 0 R ( u ) = 0 . | u | • Set y = y 0 + u . The system becomes u ′ = J ( y 0 ) u + R ( u ) . • The linearization is u ′ = J ( y 0 ) u . Return Planar system Planar linearization 9 Theorem 2 Theorem 2 Theorem: Suppose that y 0 is an equilibrium point for y ′ = f ( y ) . Let J be the Jacobian of f at y 0 . 1. Suppose that the real part of every eigenvalue of J is negative. Then y 0 is an asymptotically stable equilibrium point. 2. Suppose that J has at least one eigenvalue with positive real part. Then y 0 is an unstable equilibrium point. Return Linearization Theorem 1 3 John C. Polking

10 Example Example x ′ = − 2 x − 4 y + 2 xy y ′ = x − 6 y + x 2 − y 2 • The origin (0 , 0) is an equilibrium point. • The Jacobian has one eigenvalue, λ = − 4 , of algebraic multiplicity 2. • First theorem does not apply. • Second theorem ⇒ the origin is a sink. Return Linear analysis 11 The Lorenz System The Lorenz System x ′ = − ax + ay y ′ = rx − y − xz z ′ = − bz + xy • Use a = 10 , b = 8 / 3 , and r = 5 , 20 , 28 , 100 . 4 John C. Polking