Transcript ENGG2013 Lecture 24 - Chinese University of Hong Kong

ENGG2013 Unit 24
Linear DE and Applications
Apr, 2011.
Outline
• Method of separating variable
• Method of integrating factor
• System of linear and first-order differential
equations
– Graphical method using phase plane
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Nomenclatures
• “First-order”: only the first derivative is involved.
• “Autonomous”: the independent variable does
not appear in the DE
• “Linear”:
– “Homogeneous”
– “Non-homogeneous”
c(t) not identically zero
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Separable DE
• “Separable”: A first-order DE is called separable if
it can be written in the following form
• Examples
–
–
–
–
–
x’ = cos(t)
x’ = x+1
x’ = t2sin(x)
t x’ = x2–1
All linear homogeneous DE
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SEPARABLE DE AND
METHOD OF SEPARATING
VARIABLES
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How to solve separable DE
• Write x’= f(x) g(t) as
.
• Separate variable x and t (move all “x” to the
LHS and all “t” to the RHS)
• Integrate both sides
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Example
Solve
(1) Write the DE as
(2) Separate the variables
(3) Integrate both sides
General solution to x’=t/x
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Solution curves
• The solutions are hyperbolae
Sample solutions
x ' = t/x
Some constant
4
3
2
x
1
0
-1
-2
-3
-4
-4
-3
-2
-1
0
1
2
3
4
t
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Example: Newton’s law of cooling
• Suppose that the room temperature is Tr = 24
degree Celsius. The temperature of a can of
coffee is 15 oC at T=0 and rises to 16 oC after
one minute.
– T(0) = 15, T(1) = 16.
• Find the temperature after 10 minutes
Proportionality constant
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LINEAR NON-HOMOGENEOUS DE
METHOD OF INTEGRATING FACTOR
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Example: RC in series
• Physical laws
– Voltage drop across resistor = VR(t) = R I(t)
– Voltage drop across inductor = C VC(t) = Q(t)
Charge
Vc
C
From Kirchoff voltage law
VC(t) + VR(t) = sin(wt)
sin(wt)
R
Linear non-homogeneous
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Linear DE in standard form
• Linear equation has the following form
• By dividing both sides by p(t), we can write
the differential equation in standard form
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Product rule of differentiation
• Idea: Given a DE in standard form
Multiply both sides by some function u(t)
so that the product rule can be applied.
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Illustrations
1. Solve the initial value problem
2. Find the general solution to
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Example: Mixing problem
Water tank
1000 L
Initial Caesium concentration
= 1 Bq/L
• In-flow of water: 10 L per minute
• Out-flow of water: 10 L per minute
• In-flowing water contains Caesium with concentration
5 Bq/L
• Describe the concentration of Ce in the water tank as a
function of time.
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http://en.wikipedia.org/wiki/Henri_Becquerel
Henri Becquerel
• French physicist
• Dec 1852 ~ Aug 1908
• Nobel prize laureate of Physics
in 1903 (together with Marie
Curie and Pierre Curie) for the
discovery of radioactivity.
• Bq is the SI unit for radioactivity
– Defined as the number of nucleus
decays per second.
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Back to the RC example
• Write it in standard form
• Multiply by an unknown function u(t)
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Integrating factor
• Is there any function u(t) such that
u’(t) = u(t)/RC ?
•  Choose u(t) = exp(t/RC)
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Now we can integrate
Use a standard fact from calculus
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Solution to RC in series
• General solution
• If it is known that Q(0) = 0, then
Steady-state solution
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approaches zero
as t  
20
Sample solution curves
• Take R=C = 1, w=10 for example.
1
0.8
0.6
0.4
Q
0.2
0
-0.2
-0.4
-0.6
Different solutions
correspond to different initial
values.
-0.8
-1
0
1
2
3
4
5
6
7
8
t
Transient state
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Steady state
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SYSTEM OF DIFFERENTIAL
EQUATIONS
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Interaction between components
• If we have two or more objects, each and they
interact with each other, we need a system of
differential equations.
• Metronomes synchronization
– http://www.youtube.com/watch?v=yysnkY4WHyM
• Double pendulum
– http://www.youtube.com/watch?v=pYPRnxS6uAw
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General form of a system of linear
differential equation
• System variables: x1(t), x2(t), …, xn(t).
• A system of DE
Some functions
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System of linear constant-coeff.
differential equations
• System variables: x1(t), x2(t), x3(t).
• Constant-coefficient linear DE
– aij are constants,
– g1(t), g2(t) and g3(t) are some function of t.
• Matrix form:
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Application 1: Mixing
Water tank 1
f1
f12
Water tank 2
f2
Volume = V2 m3
Volume = V1 m3
Concentration = C1(t)
Concentration = C2(t)
f21
• C1(t) and C2(t) are concentrations of a substance, e.g. salt, in tank 1 and 2.
• Given
–
–
–
–
–
Initial concentrations C1(0) = a, C1(0) = b.
In-low to tank 1 = f1 m3/s, with concentration c.
Flow from tank 1 to tank 2 = f12 m3/s
Flow from tank 2 to tank 1 = f21 m3/s
Out-flow from tank 2 = f2 m3/s
• Objective: Find C1(t) and C2(t).
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Modeling
•
•
•
•
Consider a short time interval [t, t+t]
C1 = C1(t+t)–C1(t) = cf1t + f21C2t – f12C1t
C2 = C2(t+t)–C2(t) = f12C1t – f21C2t – f2C2t
Take t  0, we have
C1’ = – f12C1 + f21C2+ cf1
C2’ = f12C1 – (f21+ f2) C2
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Graphical method
• For autonomous system,
• we can plot the phase plane (aka phase portrait) to
understand the system qualitatively.
• Select a grid of points, and draw an arrow for each
point. The direction of each arrow is
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Phase Plane
C1’ = – 6C1 + C2+ 10
C2’ = 6C1 – 6 C2
5
4.5
4
• Suppose
3.5
3
C2
– f1 = 5
– f2 = 5
– f12 = 6
– f21 = 1
–c=2
– Initial concentrations
are zero
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
C1
Converges to (2,2)
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Convergence
C1’ = – 6C1 + C2+ 10
C2’ = 6C1 – 6 C2
• (C1,C2)=(2,2) is a critical point.
– C1’ and C2’ are both zero when C1= C2=2.
• The analyze the stability of critical point, we
usually make a change of coordinates and
move the critical point to the origin.
• Let x1 = C1–2, x2 = C2–2.
x1’ = – 6x1 + x2
x2’ = 6x1 – 6 x2
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Phase plane of a system with
stable node
All arrows points towards
the origin
4
3
2
x2
1
0
-1
-2
-3
-4
-4
-2
0
x1
2
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Sample solution curves
The origin is a stable node
4
3
2
x2
1
0
-1
-2
-3
-4
-4
-2
0
2
4
x
1
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Theoretical explanation for
convergence
• The eigenvalues of the coefficient matrix
are negative. Indeed, they are equal to –3.5505
and –8.4495.
• The corresponding eigenvectors are
[0.3780 0.9258] and [–0.3780 0.9258]
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Eigen-direction
•
If we start on any point in
the direction of the
eigenvectors, the system
converges to the critical
4
point in a straight line.
This is another geometric 3
interpretation of the
2
eigenvectors.
1
x2
•
0
-1
-2
-3
-4
-4
-2
0
2
4
x1
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Application 2: RLC mesh circuit
• Suppose that the initial charge
at the capacity is Q0.
• Describe the currents in the
two loops after the switch is
closed.
i1(t)
i2(t)
Physical Laws
• Resistor: V=R i
• Inductor: V=L i’
• Capacitor: V=Q/C
• KVL, KCL
Homework exercise
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An expanding system
• Both eigenvalues are positive.
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Phase Plane of a system with
unstable node
The origin is an
unstable node.
The red arrows
indicate the
eigenvectors
5
4
3
2
y
1
0
-1
-2
-3
-4
-5
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
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A system with saddle point
• One eigenvalue is positive, and another
eigenvalue is negative
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Phase Plane of a system with
origin is a saddle point.
saddle node The
The thick red arrows indicate
the eigenvectors
5
4
3
2
y
1
0
-1
-2
-3
-4
-5
-5
-4
-3
-2
-1
0
x
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2
3
4
5
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Conclusion
The convergence and stability of a system of
linear equations is intimately related to the
signs of eigenvalues.
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