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 ANDMETHOD 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

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Solve

(1) Write the DE as

(2) Separate the variables

(3) Integrate both sides

General solution to x’=t/x

Solution curves

• The solutions are hyperbolae

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

-4

-3

-2

-1

0

1

2

3

4

t

x

x ' = t/x

Some constant

Sample solutions

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

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Proportionality constant

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)

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Charge

R

C

sin(wt)

Vc

From Kirchoff voltage lawVC(t) + VR(t) = sin(wt)

Linear non-homogeneous

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

• 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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Water tank1000 L

Initial Caesium concentration = 1 Bq/L

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.kshum 16

http://en.wikipedia.org/w

iki/Henri_B

ecquerel

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

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Use a standard fact from calculus

Solution to RC in series

• General solution

• If it is known that Q(0) = 0, then

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approaches zeroas t

Steady-state solution

Sample solution curves• Take R=C = 1, w=10 for example.

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0 1 2 3 4 5 6 7 8

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t

Q

Steady stateTransient state

Different solutions correspond to different initial values.

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

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Some functions

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

• 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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Water tank 1

Volume = V1 m3

Concentration = C1(t)

Water tank 2

Volume = V2 m3

Concentration = C2(t)

f1

f12

f21

f2

Modeling

• Consider a short time interval [t, t+t]• C1 = C1(t+t)–C1(t) = cf1t + f21C2t – f12C1t

• C2 = C2(t+t)–C2(t) = f12C1t – f21C2t – f2C2t

• Take t 0, we haveC1’ = – 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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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

C1

C2

Phase Plane

• Suppose– f1 = 5

– f2 = 5

– f12 = 6

– f21 = 1

– c = 2– Initial concentrations

are zerokshum 29

Converges to (2,2)

C1’ = – 6C1 + C2+ 10C2’ = 6C1 – 6 C2

Convergence

• (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.

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C1’ = – 6C1 + C2+ 10C2’ = 6C1 – 6 C2

x1’ = – 6x1 + x2

x2’ = 6x1 – 6 x2

Phase plane of a system with stable node

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-4 -2 0 2 4

-4

-3

-2

-1

0

1

2

3

4

x1

x 2

All arrows points towardsthe origin

Sample solution curves

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The origin is a stable node

-4 -2 0 2 4

-4

-3

-2

-1

0

1

2

3

4

x1

x 2

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 point in a straight line.

• This is another geometric interpretation of the eigenvectors.

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-4 -2 0 2 4

-4

-3

-2

-1

0

1

2

3

4

x1

x 2

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.

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i1(t) i2(t)

Physical Laws

• Resistor: V=R i• Inductor: V=L i’• Capacitor: V=Q/C• KVL, KCL

Homework exercise

An expanding system

• Both eigenvalues are positive.

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Phase Plane of a system with unstable node

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

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

yThe origin is an unstable node.The red arrows indicate the eigenvectors

A system with saddle point

• One eigenvalue is positive, and another eigenvalue is negative

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Phase Plane of a system with saddle node

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

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

yThe origin is a saddle point.The thick red arrows indicatethe eigenvectors

Conclusion

The convergence and stability of a system of linear equations is intimately related to the

signs of eigenvalues.

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