ENGG2013 Unit 24 Linear DE and Applications Apr, 2011.

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

kshum 2

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

kshum 3

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

kshum 4

SEPARABLE DE ANDMETHOD OF SEPARATING VARIABLES

kshum 5

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

kshum 6

Example

kshum 7

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

kshum 8

-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

kshum 9

Proportionality constant

LINEAR NON-HOMOGENEOUS DE METHOD OF INTEGRATING FACTOR

kshum 10

Example: RC in series

• Physical laws– Voltage drop across resistor = VR(t) = R I(t)

– Voltage drop across inductor = C VC(t) = Q(t)

kshum 11

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

kshum 12

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.

kshum 13

Illustrations

1. Solve the initial value problem

2. Find the general solution to

kshum 14

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.

kshum 15

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)

kshum 17

Integrating factor

• Is there any function u(t) such that u’(t) = u(t)/RC ?

• Choose u(t) = exp(t/RC)

kshum 18

Now we can integrate

kshum 19

Use a standard fact from calculus

Solution to RC in series

• General solution

• If it is known that Q(0) = 0, then

kshum 20

approaches zeroas t

Steady-state solution

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

kshum 21

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

kshum 22

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

kshum 23

General form of a system of linear differential equation

• System variables: x1(t), x2(t), …, xn(t).

• A system of DE

kshum 24

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:

kshum 25

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).

kshum 26

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

kshum 27

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

kshum 28

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.

kshum 30

C1’ = – 6C1 + C2+ 10C2’ = 6C1 – 6 C2

x1’ = – 6x1 + x2

x2’ = 6x1 – 6 x2

Phase plane of a system with stable node

kshum 31

-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

kshum 32

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]

kshum 33

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.

kshum 34

-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.

kshum 35

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.

kshum 36

Phase Plane of a system with unstable node

kshum 37

-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

kshum 38

Phase Plane of a system with saddle node

kshum 39

-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.

kshum 40