Math 211 Math 211 Review for the Final Exam December 8, 2002 2 - PDF document

1 Math 211 Math 211 Review for the Final Exam December 8, 2002 2 The Final Exam The Final Exam • The final will be comprehensive, covering material from the entire semester. • The final will emphasize the material covered since the last exam. • These slides will cover primarily the material covered since the last exam. They do not cover all of the material on the exam. • Questions about any of the material of the course will be answered. Return 3 The Themes of the Course The Themes of the Course • Modeling. � Population, finance, mixing, motion, vibrating spring, electrical circuits, . . . • Exact solutions. � Separable and linear equations in dimension 1. � Linear equations in higher dimension. ◮ Matrix algebra. � Second order equations. • Numerical solutions. • Geometric analysis. Return 1 John C. Polking

4 Solving x ′ = A x Solving x ′ = A x • A is an n × n matrix. • Solution strategy: Look for a fundamental set of solutions, i.e., n linearly independent solutions. • The function x ( t ) = e tA v solves the initial value problem x ′ = A x with x (0) = v . • Refined strategy: Compute e tA v for n linearly independent vectors v . � Computing e tA v is hard except for specially chosen vectors v . Return 5 Key to Computing e tA v Key to Computing e tA v Suppose that A an n × n matrix, and λ a number (an eigenvalue). Then e tA v = e λt · e t ( A − λI ) v = e λt · [ v + t ( A − λI ) v + t 2 2!( A − λI ) 2 v + · · · ] • If λ is an eigenvalue and v is an associated eigenvector, then ( A − λI ) v = 0 , so e tA v = e λt v . • If ( A − λI ) 2 v = 0 , then e tA v = e λt [ v + t ( A − λI ) v ] . Return 6 Generalized Eigenvectors Generalized Eigenvectors Definition: If λ is an eigenvalue of A and ( A − λI ) p v = 0 for some integer p ≥ 1 , then v is called a generalized eigenvector associated with λ. • Then v + + t ( A − λI ) v + t 2 � e tA v = e λt 2!( A − λI ) 2 v t p − 1 � ( p − 1)!( A − λI ) p − 1 v + · · · + • We can compute e tA v for any generalized eigenvector. Return 2 John C. Polking

7 Multiplicities Multiplicities A an n × n matrix with distinct eigenvalues λ 1 , . . . , λ k . • The characteristic polynomial has the form 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 . � q 1 + q 2 + . . . + q k = n. • The geometric multiplicity of λ j is d j , the dimension of the eigenspace of λ j . � 1 ≤ d j ≤ q j . • There is an integer k j ≤ q j for which null(( A − λ j I ) k j ) has dimension q j . Return Generalized eigenvectors Strategy 8 Procedure for Solving x ′ = A x Procedure for Solving x ′ = A x • Find the eigenvalues of A and their algebraic multiplicities. • For each eigenvalue λ with algebraic multiplicity q : � Find the smallest integer k for which null(( A − λI ) k ) has dimension q . � Find a basis for null(( A − λI ) k ) . � For each vector v in the basis compute the solution x ( t ) = e tA v . • The set of all of these solutions is a fundamental set of solutions. Return Key 9 Replacing Complex Solutions with Real Replacing Complex Solutions with Real Solutions Solutions • If A has complex eigenvalues, the fundamental set of solutions contains complex valued solutions. • Complex solutions occur in complex conjugate pairs z ( t ) = x ( t ) + i y ( t ) and z ( t ) = x ( t ) − i y ( t ) . • Replace z ( t ) and z ( t ) with the real solutions x ( t ) = Re( z ( t )) and y ( t ) = Im( z ( t )) . Return 3 John C. Polking

10 Solutions to Higher Order Equations Solutions to Higher Order Equations Homogenous linear equation with constant coefficients: y ′′ + py ′ + qy = 0 • Look for exponential solutions y ( t ) = e λt . • Characteristic polynomial: λ 2 + pλ + q . • If λ is a root of the characteristic polynomial then y ( t ) = e λt is a solution. Return 11 Fundamental sets of solutions Fundamental sets of solutions • Two distinct real roots λ 1 and λ 2 : y ( 1 t ) = e λ 1 t y 2 ( t ) = e λ 2 t . and • One real root λ of multiplicity 2: y 1 ( t ) = e λt y 2 ( t ) = te λt . and • Complex conjugate roots λ = α ± iβ : y 1 ( t ) = e αt cos βt y 2 ( t ) = e αt sin βt. and Return Solutions 12 Inhomogeneous Equations Inhomogeneous Equations y ′′ + Py ′ + Qy = f ( t ) • The method of undetermined coefficients finds a particular solution y p ( t ) . • The general solution is y ( t ) = y p ( t ) + C 1 y 1 ( t ) + C 2 y 2 ( t ) , where y 1 and y 2 are a fundamental set of solutins to the homogeneous equation. • If the forcing term f ( t ) has a form which is replicated under differentiation, look for a particular solution of the same general form as the forcing term. Return 4 John C. Polking

13 Cases Cases • If f ( t ) = Ce bt , try y p ( t ) = ae bt . • If f ( t ) = A cos ωt + B sin ωt , try y p ( t ) = a cos ωt + b sin ωt. � Or try the complex method. • If f ( t ) is a polynomial of degree n , let y p be a polynomial of degree n . • Exceptional cases: Multiply expected form of y p by t . • Combination cases: Solve the equation in pieces. Return Undetermined coefficients 14 Harmonic Motion Harmonic Motion • Spring: y ′′ + µ m y ′ + k m y = 1 m F ( t ) . • Circuit: I ′′ + R L I ′ + LC I = 1 1 L E ′ ( t ) . • Essentially the same equation. Use x ′′ + 2 cx ′ + ω 2 0 x = f ( t ) . � We call this the equation for harmonic motion. • ω 0 is the natural frequency. c is the damping constant. f ( t ) is the forcing term. Return 15 Unforced Harmonic Motion Unforced Harmonic Motion x ′′ + 2 cx ′ + ω 2 0 x = 0 • Undamped: c = 0 . • Underdamped: 0 < c < ω 0 . • Critically damped: c = ω 0 . • Over damped: c > ω 0 . Return Harmonic motion 5 John C. Polking

16 Forced Harmonic Motion Forced Harmonic Motion x ′′ + 2 cx ′ + ω 2 0 x = A cos ωt • A is the forcing amplitude and ω is the forcing frequency. • The general solution is x ( t ) = x p ( t ) + x h ( t ) . � x p is a particular solution. x h is the general solution of the homogenous equation. • Undamped: c = 0 . � ω � = ω 0 : Beats. � ω = ω 0 : Resonance. Return 17 Forced, Damped Harmonic Motion Forced, Damped Harmonic Motion x ′′ + 2 cx ′ + ω 2 0 x = A cos ωt • c > 0 implies that x h ( t ) → 0 as t increases, so x h is called the transient term . • x p ( t ) is called the steady-state solution. It has the form x p ( t ) = G ( ω ) A cos( ωt − φ ) � x p is oscillatory at the driving frequency. � The amplitude of x p is the product of the gain , G ( ω ) , and the amplitude of the forcing function. � x p has a phase shift of φ with respect to the forcing function. Return Forced harmonic motion 18 Qualitative Analysis Qualitative Analysis • Existence and uniqueness. • For an autonomous system x ′ = f ( x ) , the basic question is, What happens to all solutions as t → ∞ ? • The easy cases: equilibrium points f ( x 0 ) = 0 and equilibrium solutions x ( t ) = x 0 . • Local qualitative analysis: What happens as t → ∞ to all solutions that start near an equilibrium point x 0 ? � This is the question of stability. • Global qualitative analysis: What happens to all solutions as t → ∞ ? Return 6 John C. Polking

19 Stability Stability Suppose the autonomous system x ′ = f ( x ) has an equilibrium point at x 0 . • x 0 is stable if every solution that starts close to x 0 stays close to x 0 . • x 0 is asymptotically stable if every solution that starts close to x 0 stays near x 0 and approaches x 0 as t → ∞ . � x 0 is called a sink. • x 0 is unstable if there are solutions starting arbitrarily close to x 0 that move away from x 0 . Return Qualitative analysis 20 Stability for x ′ = A x Stability for x ′ = A x • D = 2 : Trace-determinant plane. • 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 . Return 21 Stability for x ′ = f ( x ) Stability for x ′ = f ( x ) • Suppose that x 0 is an equilibrium point. • The linearization at x 0 is the system u ′ = J u , where J is the Jacobian matrix of f at x 0 . x ′ = f ( x, y ) � � • For the planar system , the Jacobian is y ′ = g ( x, y ) ∂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 ) Return 7 John C. Polking