engg2013 unit 25 second-order linear de
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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 - PowerPoint PPT PresentationTRANSCRIPT
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ENGG2013 Unit 25
Second-order Linear DE
Apr, 2011.
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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
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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
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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
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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.
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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
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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
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x
v
Sample solutions
Convergenceto the origin
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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.
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Solution to vibrating spring with damping
Characteristic equation
Eigenvalues
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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.
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Solution to vibrating spring with damping
Substitute by the diagonalized matrix
Change of variables
An uncoupled system
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Solution to vibrating spring with damping
Solve the uncoupled system
K1 and K2 are constants
Substitute back
(C1, C1, a and b are constants)
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Solution to vibrating spring with damping
0 2 4 6 8 10-1
-0.5
0
0.5
1
time
x
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A general strategy for linear system
Linear systemDecouple(Diagonalization)
Solve each subsystem separately
Piece themtogether
Solved
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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)
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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
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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
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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
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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
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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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