ENGG2013 Unit 22

Modeling byDifferential Equations

Apr, 2011.

FREE FALLING BODY

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Height, velocity and acceleration

• Parabola

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0 0.5 1 1.5 2-10

0

10

20

t

y

0 0.5 1 1.5 2-20

-10

0

t

v

0 0.5 1 1.5 2-11

-10

-9

t

a

y = –5t2+15

v = –10t

a = –10

Newton’s law of motion

• F = ma – Force = mass acceleration– a = y’’(t)

• F = mg – Gravitational Force is proportional to the mass,

the proportionality constant g –10 ms-2.

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y’’(t) = g

Assume no air friction

Differential equation

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• A differential equation is an equation which involves derivatives.

• Examples:

The variable x is a function of time t.

The variable y is a function of t.

dx/dt = x + 2t

d2y/dt2 = t2 + y2

Initial conditions

• y(0)=0• y’(0)=10

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0 0.5 1 1.5 220

22

24

26

t

y0 0.5 1 1.5 2

-10

0

10

t

v

0 0.5 1 1.5 2-11

-10

-9

t

a

y(t) = –5t2+10t+20

y(t) = –10t+10

y’’(t) = –10

Initial conditions

• y(0)= –5• y’(0)= –10

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y(t) = –5(t+1)2

y(t) = –10t –10

y’’(t) = –10

0 0.5 1 1.5 2-60

-40

-20

0

t

y

0 0.5 1 1.5 2-30

-20

-10

t

v

0 0.5 1 1.5 2-11

-10

-9

t

a

Variables and parameters

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• The dependent variable is called the system state, or the phase of the system. The independent variable is usually time.

• A constant which does not change with time is called a parameter.

• In the example Newton’s law of motion y’’(t) = g– Phase = system state = height of the mass– Independent variable = time– g is parameter.

Solutions to a differential equation• A solution is a function which satisfies the given

differential equation.• In solving differential equation, the solutions are

function of time.• In general, there are many solutions to a given

differential equation. We have different solutions for different initial condition.

• Deriving a solution is difficult, but checking whether a given function is a solution is easy.

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y(t) = –5t2+15 is a solution to y’’(t) = –10because after differentiating –5t2+15 twice, we get –10.

y(t) = 4t2 is not a solution to y’’(t) = –10because after differentiating 4t2 twice, we get 8, not –10.

General solution

• If every solution to a differential equation can be obtained from a family of solutions

f(t,c1,c2,c3,…,cn)

by choosing the constants c1, c2, c3,…, cn

appropriately, then we say that f(t,c1,c2,c3,…,cn) is a general solution.

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General solution to y’’(t) = -10

• For this simple example, just integrate two times.• Integrate both sides of y’’(t) = – 10 y’(t) = –10t+c1

• Integrate both sides of y’(t) = – 10+c1

y(t) = – 5t2+c1t+c2 (general solution)

• The constants c1 and c2 can be obtained from the initial conditions.

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FIRST-ORDER DIFFERENTIAL EQUATION

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Brief review of derivatives

• Derivative is the slope of tangent line.– Tangent line is a line touching a curve at a point

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-2 -1.5 -1 -0.5 0 0.5 1 1.5 20

0.5

1

1.5

2

2.5

3

3.5

4

y=x2.

Slope of the tangent lineat (x,x2) equals 2x.

Derivative of x2.

Slope of tangent line

• Derivative is the instantaneous rate of change.

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-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-10

-8

-6

-4

-2

0

2

4

6

8

10y=x3+x

Slope = 4 at (-1,-2).

y’ =3x2+1y’ evaluated at x=-1 is 3(-1) 2+1=4.

First-order differential equation• No second or higher derivative, for example

• First-order derivative defines slope.• Example

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dx/dt = a function of x and t

General solution

constant

An illustration• If an initial condition is given, then we can solve for the

constant C.• Suppose that x(0) = 2. C=3.

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-3 -2 -1 0 1 2 30

10

20

30

40

50

60

t

x

-3 -2 -1 0 1 2 30

10

20

30

40

50

60

t

x

0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4

4

5

6

7

8

9

t

x

Zoom in at (1, 6.1548)

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Line segment with slope -1+e1=7.1548.

Zoom in at (2, 19.1672)

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-3 -2 -1 0 1 2 30

10

20

30

40

50

60

t

x

1.8 1.85 1.9 1.95 2 2.05 2.1 2.15 2.2 2.25

16

17

18

19

20

21

22

t

x

Line segment with slope –1 + 3e2=21.1672

Direction field or slope field

• A graphical method for solving differential equation.

• Systematically evaluate f(x,t) on a grid on points.

• On a grid point (t,x), draw a short line segment with slope f(x,t).

• A solution must follow the flow pattern.

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Direction Field for x’=x+t

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Sample solution for x(0)=2

-3 -2 -1 0 1 2 3

-10

-8

-6

-4

-2

0

2

4

6

8

10

t

xx ' = x + t

Each grid point (t,x) is associated with a line segment with slope x+t.

Newton’s law of cooling

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• Imagine a can of coffee in an air-conditioned room.

• The rate of change of the temperature T(t) is directly proportional to the difference between T and the temperature T0 of the environment.

• Rate of change in temperature is directly proportional to (T – T0).– k is a positive constant.– T > T0, T decreases with rate k (T – T0).– T < T0, T increases with rate k (T0 – T).

dT/dt = – k (T – T0) (k>0)

Rate of change in temperature

10 12 14 16 18 20 22 24 26 28 30-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

T

rate

of c

ha

ng

e in

tem

p.

T T

dT/dt = – 0.2 (T – 23)

T0

0 5 10 15 20 25 30

0

5

10

15

20

25

30

35

40

t

T

T ' = - 0.2 (T - 23)

Direction fielddT/dt = – 0.2 (T – 23)

Sample solutions

Some typical solution paths, corresponding initial temperature 0, 5, 10, 15, 20, 25, 30, 35, 40, are shown in the graph.

0 5 10 15 20 25 30

0

5

10

15

20

25

30

35

40

t

TT ' = - 0.2 (T - 23) dT/dt = – 0.2 (T – 23)

Autonomous DE and Phase line

• Autonomous DE: x’(t) = a function of x only.– no independent variable on the R. H. S.

• For autonomous DE, we can understand the system via the phase line.

T0T T

dT/dt = – k (T– T0)Phase line

Stable equilibrium

Critical point at T0

(k>0)

Direction field for x’=2x(1-x)

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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-1

-0.5

0

0.5

1

1.5

2

t

xx ' = 2 x (1 - x)

The pattern is the same on every vertical line.

Slopesare zeroon thesetwo criticallines.

Phase line for x’=2x(1-x)

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1x x

Phase line

Critical points at x=0 and x=1

0 x

Stable equilibriumUnstable equilibrium

Without solving the differential equation explicitly, we know that the solution x(t) converges to 1 if it starts at positive x(0), butdiverges to negative infinity if it starts at negative x(0).

Main concepts

• Independent variable, dependent variable and parameters

• Initial conditions• General solution• Direction field• Autonomous differential equations.

– Phase line• Equilibrium

– Stable and unstable