ENGG2013 Unit 25

Second-order Linear DE

Apr, 2011.

Yesterday• First-order DE

– Method of separating variable– Method of integrating factor

• System of first-order DE– Eigenvalues determine the convergence behaviour near a

critical point.• Objectives:

– Solve the initial value problem• Given the initial value, find the trajectory• Transient-state analysis of electronic circuits

– Understand the system behaviour• Does the system converge?• Stable equilibrium point, unstable equilibrium point.

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Linear Second-order DE

• Homogeneous

• Non-homogeneous

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Linear Second-order constant-coeff. DE

• Homogeneous

• Non-homogeneous

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Vibrating spring without damping• x(t): vertical displacement• Hooke’s law: Force = k x

– k is the spring constant, k > 0(the constant k is sometime called the spring modulus.)

• Newton’s law: F = m x’’– m is the mass, m > 0

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m x’’

k x

m x’’ + k x = 0Assumptions:• Spring has negligible weight• No friction

Second-order, autonomouslinear, constant-coefficientand homogeneous

x

Solutions to undamped spring-mass model

• Normalize by m

• Direct substitution verifies that andare solutions, e.g.

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Natural frequency• Let

• We know that cos( t) and sin( t) are both solutions.

• is called the natural frequency.• Dimension check: The unit of is Hz = s-1.

– The spring constant k has unit kg s-2.– k/m has unit s-2.– Square root of k/m has unit s-1.

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Principle of superposition(aka linearity principle)

• For linear and homogenous differential equation, the linear combination of two solutions is also a solution.

• For any real numbers a and b, a cos( t) + b cos( t)

is a solution to x’’+ 2 x=0.kshum 8

GRAPHICAL METHOD

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Graphical illustration of spring-mass model

• Define the displacement-velocity vector

• Reduction to system of two first-order differential equations.

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x ' = v v ' = - k x/m

k = 3m = 1

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

v

Phase plane for the vibrating spring

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Sample solution

The trajectory is a ellipse

Order reduction technique

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equivalent

Second-order DE with constant coefficients is basicallythe same as a system of two first-order DE.

Vibrating spring with damping• Vibrating spring in honey

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m x’’ = – k x – d x’

Force exerted by the spring

Damping due to viscosity

Equivalent form

honey

Assumption:Magnitude of damping forceis directly proportional to x’.d> 0

Phase plane for damped spring

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x ' = v v ' = - (k x + d v)/m

m = 1d = 1k = 3

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

v

Sample solutions

Convergenceto the origin

METHOD OF DIAGONALIZATION

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Recall: Diagonalization

• Definition: an n n matrix M is called diagonalizable if we can find an invertible matrix S, such that

is a diagonal matrix, or equivalently,

• Example:

• Diagonalization is useful in decoupling a linear system for instance.

Solution to vibrating spring with damping

Characteristic equation

Eigenvalues

Solution to vibrating spring with damping

Eigenvectors have complex components

Diagonalize

Concatenate

Two eigenvectors are notscalar multiple of each other,because the two eigenvalues aredistinct . Hence inverse exists.

Solution to vibrating spring with damping

Substitute by the diagonalized matrix

Change of variables

An uncoupled system

Solution to vibrating spring with damping

Solve the uncoupled system

K1 and K2 are constants

Substitute back

(C1, C1, a and b are constants)

Solution to vibrating spring with damping

0 2 4 6 8 10-1

-0.5

0

0.5

1

time

x

A general strategy for linear system

Linear systemDecouple(Diagonalization)

Solve each subsystem separately

Piece themtogether

Solved

METHOD OF GUESS-AND-VERIFY

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2nd-order constant-coeff. DE• Homogeneous

– b and c are constants.

• Non-homogeneous– b and c are constants.– f(t) is a function of independent variable t.

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The homogeneous case

• Idea: try a function in the form ekt as a solution.– k is some constant.

• Substitute ekt into x’’+bx’+cx=0 and try to solve for the constant k.

• Apply the superposition principle: any linear combination of solutions is also a solution.

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Examples

1. Solve x’’+3x’+2x=0.

2. Solve x’’–4x’+4x=0.

3. Solve x’’+9x= 0.

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General solution: x(t) = c1 e–2t+ c2e–t

General solution: x(t) = c1 e2t+ c2 t e2t

General solution: x(t) = c1 e3i t+ c2 e-3i t

= d1 sin(t)+ d2 cos(t)

Summary of the three cases

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Case Roots Basis of solutions General Solution

1 Distinct real and et, et c1et+c2et

2 Repeated root et, tet c1et+c2tet

3 Complex roots =r+i, =r–i

e(r+i)t, e(r–i)t er(c1cos t+c2sin t)

Characteristic equation

Differential equation

The non-homogeneous case

• Property: If x1(t) and x2(t) are two solutions to

then their difference x1(t) – x2(t) is a solution to the homogeneous counterpart

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Consequence

• Suppose that xp(t) is some solution to x’’+bx’+cx=f(t) (given to you by a genie for example)

• Any solution of x’’+bx’+cx=f(t) can be written as xh(t) +xp(t)

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A solution to thehomogeneous DEx’’+bx’+c=0

A particular solution

Method of trial and error(aka as the method of

undetermined coefficient)To solve the non-homogeneous DE x’’+bx’+cx=f(t) 1. Find a particular solution by trial and error

(and experience)– Let the particular solution be xp(t).

2. Solve the homogeneous version x’’+bx’+cx=0.– Let the homogeneous solution be xh(t).

3. The general solution is xh(t)+ xp(t).

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How to guess a particular solution

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f(t) Choice for xp(t)

K xn cnxn+cn-1xn-1+…+c1x1+c0

K eat Ceat

K sin(t) c1 sin(t)+c2 cos(t)

K cos(t) c1 sin(t)+c2 cos(t)

K eat sin(t) eat (c1 sin(t)+c2 cos(t))

K eat cos(t) eat (c1 sin(t)+c2 cos(t))

K, C, a, , c0, c1, c2,… are constants

Example• Solve x’’+3x’+2x= e5t.• Try c e5t as a particular solution.

(c e5t )’’+3(c e5t )’+(c e5t )= e5t

25c e5t+15c e5t+5c e5t= e5t

25 c + 15c + 5c=1 c = 1/45Let xp(t) = e5t/45 as a particular solution.

General solution is c1e–2t+ c2e–t + e5t/45

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Homogeneous solution Particular solution

Summary

• Graphical method using phase plane.– Reduction to two 1st-order linear DE.

• Method of diagonalization– Need to reduce the second-order DE to a system of

first-order DE.– Time-consuming but theoretically sound.

• Method of undetermined coefficients– Find a solution quickly, but not systematic.– Good for calculation in examination.

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Final Exam

• Date: 6th May (Friday)• Venue: NA Gym• Time: 9:30~11:30• Coverage: Everything in Lecture Notes and

Homeworks• Close-book exam• You may bring a calculator, and a handwritten

A4-size and double-sided crib sheet.

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