Chapter 5: Modeling with Higher-Order Differential Equations - PowerPoint PPT Presentation

Linear Models: Initial-Value Problems Nonlinear Models Summary Chapter 5: Modeling with Higher-Order Differential Equations Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw October 29, 2013 1 / 25 DE Lecture 8 王奕翔 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models Summary 1 Linear Models: Initial-Value Problems 2 Nonlinear Models 3 Summary 2 / 25 DE Lecture 8 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models Summary Modeling with Second Order Linear Differential Equation We focus on two linear dynamical systems modeled by the following: The two systems are: Spring/Mass Systems LRC Series Circuits 3 / 25 DE Lecture 8 ay ′′ + by ′ + cy = g ( t ) , y (0) = y 0 , y ′ (0) = y 1 , where the initial conditions are at time t = 0 . 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models 4 / 25 = = Hence, by Newton’s Second Law, = Due to Hooke’s Law, net force Assume that the equilibrium position is DE Lecture 8 Hooke’s Law + Newton’s Second Law Summary x = 0 , and x 向下為正 = mg − k ( s + x ) . l l l + s Note that at equilibrium 淨力為零 ⇒ mg = ks . unstretched s m x equilibrium position mx ′′ = mg − k ( s + x ) = − kx m mg − ks = 0 motion ⇒ mx ′′ + kx = 0 (a) (b) (c) ⇒ x ′′ + k mx = x ′′ + ω 2 x = 0 √ where ω = k / m . 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models 5 / 25 where DE Lecture 8 Summary Free Undamped Motion Solution to x ′′ + ω 2 x = 0 : x ( t ) = c 1 cos ω t + c 2 sin ω t . Free : No external force ⇐ ⇒ Homogeneous Equation Undamped : Motion is periodic (period = 2 π ω ), no loss in energy. Alternative Representation of x ( t ) : x ( t ) = A sin ( ω t + φ ) √ c 2 1 + c 2 A := 2 denotes the amplitude of the motion φ := tan − 1 c 1 c 2 denotes the initial phase angle 王奕翔

Linear Models: Initial-Value Problems with a resisting force proportional to the velocity. 6 / 25 Nonlinear Models Hence, by Newton’s Second Law, = Assume that the mass is in a surrounding medium Free Damped Motion Summary DE Lecture 8 Net force = mg − k ( s + x ) − β x ′ . mx ′′ = mg − k ( s + x ) − β x ′ = − kx − β x ′ ⇒ x ′′ + β mx ′ + k mx = x ′′ + 2 λ x ′ + ω 2 x = 0 m √ where ω = k / m and λ = β /2 m . 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models 7 / 25 DE Lecture 8 Summary Solutions of Free Damped Motion √ D 2 + 2 λ D + ω 2 has two roots − λ ± λ 2 − ω 2 . Solution to x ′′ + 2 λ x ′ + ω 2 x = 0 : Overdamped λ 2 > ω 2 : x ( t ) = e − λ t ( √ √ λ 2 − ω 2 t ) λ 2 − ω 2 t + c 2 e − c 1 e Critically damped λ 2 = ω 2 : x ( t ) = e − λ t ( c 1 + c 2 t ) Underdamped λ 2 < ω 2 : x ( t ) = e − λ t ( ) √ √ ω 2 − λ 2 t + c 2 sin ω 2 − λ 2 t c 1 cos 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models 8 / 25 (c) Underdamped (b) Critically damped (a) Overdamped Summary DE Lecture 8 x x t t undamped x underdamped t 王奕翔

Linear Models: Initial-Value Problems vertically oscillating. 9 / 25 = Nonlinear Models Hence, by Newton’s Second Law, DE Lecture 8 applied to the system. For example, the support is Driven Motion Summary Assume that the certain external force f ( t ) is Net force = mg − k ( s + x ) − β x + f ( t ) . mx ′′ = mg − k ( s + x ) − β x ′ + f ( t ) = − kx − β x ′ + f ( t ) ⇒ x ′′ + β mx ′ + k mx = x ′′ + 2 λ x ′ + ω 2 x = F ( t ) √ where ω = k / m , λ = β /2 m , and F ( t ) = f ( t )/ m . m 王奕翔

Linear Models: Initial-Value Problems 1 Find the complementary solution: 10 / 25 2 Find a particular solution: Nonlinear Models DE Lecture 8 Summary Solve When F ( t ) is Periodic x ′′ + 2 λ x ′ + ω 2 x = F 0 sin γ t .  e − λ t ( λ 2 − ω 2 t ) √ √ λ 2 − ω 2 t + c 2 e − λ 2 > ω 2  ,  c 1 e  e − λ t ( c 1 + c 2 t ) , λ 2 = ω 2 x c ( t ) = √ √  e − λ t ( )  λ 2 < ω 2 ω 2 − λ 2 t + c 2 sin ω 2 − λ 2 t  , c 1 cos  A sin γ t + B cos γ t , λ ̸ = 0   λ = 0 , ω 2 ̸ = γ 2 x p ( t ) = A sin γ t + B cos γ t ,  λ = 0 , ω 2 = γ 2  At sin γ t + Bt cos γ t , 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models 11 / 25 steady-state transient DE Lecture 8 Summary Driven Damped Motion: Steady-State vs.Transient When λ ̸ = 0 , it is a damped system, and the general solution is x ( t ) = x c ( t ) + A sin γ t + B cos γ t , where  ( λ 2 − ω 2 t ) √ √ λ 2 − ω 2 t + c 2 e − λ 2 > ω 2  ,  c 1 e  x c ( t ) = e − λ t λ 2 = ω 2 ( c 1 + c 2 t ) , √ √  ( )  λ 2 < ω 2 ω 2 − λ 2 t + c 2 sin ω 2 − λ 2 t  , c 1 cos Note that if λ > 0 , x c ( t ) → 0 as t → ∞ . ∴ x ( t ) → A sin γ t + B cos γ t as t → ∞ . Decompose x ( t ) into two parts: x ( t ) = x c ( t ) + A sin γ t + B cos γ t ���� � �� � 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models Summary Pure Resonance solution is 12 / 25 DE Lecture 8 When λ = 0 and ω 2 = γ 2 , it is a undamped system, and the general x ( t ) = c 1 cos ω t + c 2 sin ω t + At sin ω t + Bt cos ω t Note that x ( t ) → ∞ as t → ∞ , which is because of resonance . x t 王奕翔

Linear Models: Initial-Value Problems Recall from Chapter 1 that the voltage drop across 13 / 25 have Nonlinear Models DE Lecture 8 Series Circuit Summary the three elements are L dI dt , IR , and q C respectively. L R Using the fact that I = dq E ( t ) dt and Kirchhoff’s Law, we Lq ′′ + Rq ′ + q / C = E ( t ) . C Overdamped R 2 > 4 L / C Critically damped R 2 = 4 L / C Underdamped R 2 < 4 L / C 王奕翔

Linear Models: Initial-Value Problems steady-state current. 14 / 25 3 Superposition principle of nonhomogeneous linear DE: if 2 We just need to find the particular solution q p . . Nonlinear Models Observation : DE Lecture 8 Steady-State Current Summary Example L R E ( t ) For the external voltage E ( t ) = E 0 sin γ t , find the C E 0 e i γ t − E 0 e − i γ t ) { E 0 e i γ t } 1 ( 1 E 0 sin γ t = Im = 2 i q p , 1 is a particular solution of Lq ′′ + Rq ′ + q / C = E 0 e i γ t q p , 2 is a particular solution of Lq ′′ + Rq ′ + q / C = E 0 e − i γ t then q p := 2 i ( q p , 1 − q p , 2 ) is a particular solution of the original DE. 1 1 4 q ∗ p , 1 = q p , 2 and therefore q p := 2 i ( q p , 1 − q p , 2 ) = Im { q p , 1 } . 王奕翔

Linear Models: Initial-Value Problems steady-state current. 15 / 25 Hence the steady-state (complex) current Nonlinear Models q s We just need to solve the following: DE Lecture 8 Example Summary Steady-State Current L R E ( t ) For the external voltage E ( t ) = E 0 sin γ t , find the C Lq ′′ + Rq ′ + q / C = E 0 e i γ t . Note that the particular solution take the form q s e i γ t . Plug it in we get L ( i γ ) 2 + R ( i γ ) + 1/ C E 0 ( ) + i γ Re i γ t . = E 0 = ⇒ q p , 1 ( t ) = ( 1 C − L γ 2 ) E 0 I p , 1 ( t ) = ) e i γ t ( 1 R + i γ L − γ C 王奕翔

Linear Models: Initial-Value Problems steady-state current. 16 / 25 R . The steady-state (real) current is just the imaginary part of the above: Nonlinear Models Let’s further manipulate the steady-state (complex) current DE Lecture 8 Example Summary Steady-State Current L R E ( t ) For the external voltage E ( t ) = E 0 sin γ t , find the C ) e i γ t = E 0 E 0 I p , 1 ( t ) = R + iXe i γ t , ( 1 R + i γ L − γ C 1 where X := γ L − γ C is called the reactance of the circuit. R 2 + X 2 ( R sin γ t − X cos γ t ) = E 0 E 0 I p ( t ) = Im { I p , 1 ( t ) } = Z sin ( γ t − φ ) , √ R 2 + X 2 is called the impedance of the circuit, φ = tan − 1 X where Z := 王奕翔

Linear Models: Initial-Value Problems Nonlinear Models Summary 1 Linear Models: Initial-Value Problems 2 Nonlinear Models 3 Summary 17 / 25 DE Lecture 8 王奕翔

Linear Models: Initial-Value Problems Consider a suspended cable with the weight per unit 18 / 25 dx = dx dy dx is the total cable length dx Nonlinear Models DE Lecture 8 Summary Suspended Cable length = ρ . We would like to find the shape of the cable, that is, the function y ( x ) . At a point ( x , y ) of the suspended cable, we have y { T 1 = T 2 cos θ, Horizontal Net Force = 0 T 2 . T 2 sin θ W = ρ s = T 2 sin θ, Vertical Net Force = 0 P 2 θ T 2 cos θ wire P 1 T 1 √ (0, a ) ) 2 ∫ x ( dy W Here s = 1 + x ( x , 0) 0 between (0 , a ) and ( x , y ) . dx = tan θ = ρ s Since dy T 1 , we get 2 √ ∫ x T 1 = ρ y ′′ = ρ √ 1 + ( y ′ ) 2 . dx = W 1 + dy ⇒ T 1 T 1 0 王奕翔

Linear Models: Initial-Value Problems du 19 / 25 cosh = = Nonlinear Models DE Lecture 8 Summary Suspended Cable √ Solve y ′′ = 1 + ( y ′ ) 2 (dependent variable y ρ T 1 missing) by substituting u := y ′ : u ′ = ρ ∫ ∫ ρ √ 1 + u 2 = ⇒ √ 1 + u 2 = T 1 dx T 1 ⇒ sinh − 1 u = ρ y T 1 x + c 1 . T 2 T 2 sin θ P 2 θ T 2 cos θ wire Since u (0) = y ′ (0) = 0 , we have c 1 = 0 . Therefore, P 1 T 1 ( ρ ( ρ (0, a ) W ) ) u = y ′ = sinh ⇒ y = T 1 x + c 2 . ( x , 0) ρ cosh T 1 x T 1 x Since y (0) = a , we have c 2 = a − T 1 ρ . Hence, ( ρ { ) } y ( x ) = a + T 1 − 1 . ρ T 1 x 王奕翔