1 Math 211 Math 211 Lecture #30 Solutions of Systems and Stability November 5, 2003 2 Multiplicities Multiplicities A an n × n matrix • Distinct eigenvalues λ 1 , . . . , λ k . • The characteristic polynomial is p ( λ ) = ( λ − λ 1 ) q 1 ( λ − λ 2 ) q 2 · . . . · ( λ − λ k ) q k . • The algebraic multiplicity of λ j is q j . • The geometric multiplicity of λ j is d j , the dimension of the eigenspace of λ j . Return 3 Properties of Multiplicities Properties of Multiplicities • q 1 + q 2 + · · · + q k = n . • 1 ≤ d j ≤ q j . • There are d j linearly independent exponential solutions corresponding to λ j . • If d j = q j for all j we have n linearly independent exponential solutions. • If d j < q j we can use our Proposition. Return Multiplicities 1 John C. Polking
4 Generalized Eigenvectors Generalized Eigenvectors If λ is an eigenvalue of A and [ A − λI ] p v = 0 Definition: for some integer p ≥ 1 , then v is called a generalized eigenvector associated with λ. Theorem: If λ is an eigenvalue of A with algebraic multiplicity q , then there is an integer p ≤ q such that null([ A − λI ] p ) has dimension q . • For each generalized eigenvector v we can compute e tA v . • We can find q linearly independent solutions associated with the eigenvalue λ . Return 5 Procedure for Solving x ′ = A x Procedure for Solving x ′ = A x • Find the eigenvalues and their algebraic multiplicities. • For each eigenvalue λ with algebraic multiplicity q find q linearly independent solutions associated with λ : � Find the smallest integer p such that null([ A − λI ] p ) has dimension q. � Find a basis v 1 , v 2 , . . . , v q of null([ A − λI ] p ) . � For j = 1 , 2 , . . . , q compute x j ( t ) = e tA v j . • This results in n linearly independent solutions. Return 6 Procedure for a Complex Eigenvalue Procedure for a Complex Eigenvalue • If λ is complex of algebraic multiplicity q . Then λ also has multiplicity q . � Find the smallest integer p such that null([ A − λI ] p ) has dimension q. � Find a basis w 1 , w 2 , . . . , w q of null([ A − λI ] p ) . � For j = 1 , 2 , . . . , q compute z j ( t ) = e tA w . � Compute x j ( t ) = Re( z j ( t )) and y j ( t ) = Im( z j ( t )) . • This results in 2 q linearly independent real solutions corresponding to the eigenvalues λ and λ . Return 2 John C. Polking
7 Stability Stability Autonomous system x ′ = f ( x ) with an equilibrium point at x 0 . The basic question is: What happens to all solutions that start near x 0 as t → ∞ ? • x 0 is stable if for every ǫ > 0 there is a δ > 0 such that a solution x ( t ) with | x (0) − x 0 | < δ ⇒ | x ( t ) − x 0 | < ǫ for all t ≥ 0 . � Every solution that starts close to x 0 stays close to x 0 . � In dimension 2 centers and sinks are stable. Return 8 • x 0 is asymptotically stable if it is stable and there is an η > 0 such that if x ( t ) is a solution with | x (0) − x 0 | < η , then x ( t ) → x 0 as t → ∞ . � Every solution that starts close to x 0 approaches x 0 . � In d = 2 sinks are asymptotically stable, centers are not. � x 0 is called a sink. • x 0 is unstable if there is an ǫ > 0 such that for any δ > 0 there is a solution x ( t ) with | x (0) − x 0 | < δ with the property that there are values of t > 0 such that | x ( t ) − x 0 | > ǫ . � There are solutions starting arbitrarily close to x 0 that move away from x 0 . � In d = 2 sources and saddles are unstable. Return 9 Dimension 2 Dimension 2 • Sinks are asymptotically stable. � The eigenvalues have negative real part. • Sources are unstable. � The eigenvalues have positive real part. • Saddles are unstable. � One eigenvalue has positive real part. • Centers are stable but not asymptotically stable. � The eigenvalues have real part = 0 . Return 3 John C. Polking
10 Stability Theorem Stability Theorem Theorem: Let A be an n × n real matrix. • Suppose the real part of every eigenvalue of A is negative. Then 0 is an asymptotically stable equilibrium point for the system x ′ = A x . • Suppose A has at least one eigenvalue with positive real part. Then 0 is an unstable equilibrium point for the system x ′ = A x . Notice that if there are eigenvalues with real part equal to 0, no conclusion is made. Return Dimension 2 Procedure 11 Examples Examples • Suppose the dimension is 2 and T 2 − 4 D = 0 . � T < 0 ⇒ sink. T > 0 ⇒ source. • y ′ = A y , ⎛ − 2 − 18 − 7 − 14 ⎞ 1 6 2 5 ⎜ ⎟ A = ⎠ . ⎜ ⎟ 2 2 − 3 0 ⎝ − 2 − 8 − 1 − 6 � A has eigenvalues − 1 , − 2 , & − 1 ± i . � 0 is asymptotically stable. Theorem 12 Propostion Propostion Proposition: Suppose that A is an n × n matrix, λ is a number, and v is a vector. 1. If [ A − λI ] v = 0 , then e tA v = e λt v . 2. If [ A − λI ] 2 v = 0 , then e tA v = e λt ( v + t [ A − λI ] v ) . 3. If [ A − λI ] k v = 0 , then � v + t [ A − λI ] v + t 2 e tA v = e λt 2! [ A − λI ] 2 v + · · · t k − 1 � ( k − 1)![ A − λI ] k − 1 v + . Return 4 John C. Polking
Recommend
More recommend