Transcript Second-order Linear differential equations

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.)
kx
m x’’
• Newton’s law: F = m x’’
– m is the mass, m > 0
m x’’ + k x = 0
Assumptions:
• Spring has negligible weight
• No friction
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Second-order, autonomous
linear, constant-coefficient
and homogeneous
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•
Solutions to undamped springmass
model
Normalize by m
• Direct substitution verifies that
are 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.
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GRAPHICAL METHOD
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•
Graphical illustration of springmass
model
Define the displacement-velocity vector
• Reduction to system of two first-order
differential equations.
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Phase plane for the vibrating spring
x'=v
v ' = - k x/m
k=3
m=1
Sample solution
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3
2
v
1
0
-1
-2
-3
-4
-5
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
The trajectory is a ellipse
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Order reduction technique
equivalent
Second-order DE with constant coefficients is basically
the same as a system of two first-order DE.
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Vibrating spring with damping
• Vibrating spring in honey
m x’’ = – k x – d x’
honey
Force exerted by the spring
Damping due to viscosity
Equivalent form
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Assumption:
Magnitude of damping force
is directly proportional to x’.
d> 0
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Phase plane for damped spring
x'=v
v ' = - (k x + d v)/m
m=1
d=1
k=3
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Sample solutions
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3
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Convergence
to the origin
v
1
0
-1
-2
-3
-4
-5
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
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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.
Solution to vibrating spring with
damping
Characteristic equation
Eigenvalues
Solution to vibrating spring with
damping
Eigenvectors have complex components
Two eigenvectors are not
scalar multiple of each other,
because the two eigenvalues are
distinct . Hence inverse exists.
Concatenate
Diagonalize
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
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x
0.5
0
-0.5
-1
0
2
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4
time
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A general strategy for linear system
Decouple
Linear system
(Diagonalization)
Solve each
subsystem separately
Solved
Piece them
together
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.
General solution: x(t) = c1 e–2t+ c2e–t
2. Solve x’’–4x’+4x=0.
General solution: x(t) = c1 e2t+ c2 t e2t
3. Solve x’’+9x= 0.
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
Differential equation
Characteristic equation
Case
Roots
Basis of solutions
General Solution
1
Distinct real  and 
et, et
c1et+c2et
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Repeated root 
et, tet
c1et+c2tet
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Complex roots
 =r+i, =r–i
e(r+i)t, e(r–i)t
er(c1cos t+c2sin t)
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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)
A solution to the
homogeneous DE
x’’+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
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 = 1/45
Let xp(t) = e5t/45 as a particular solution.
General solution is c1e–2t+ c2e–t + e5t/45
Homogeneous solution
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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
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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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