Basic Mathematical Models-Mathematical Modeling and Simulation-Lecture Slides, Slides of Mathematical Modeling and Simulation

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First Order Linear Models

Basic Mathematical Models

• Differential equations are equations containing

derivatives.

• The following are examples of physical phenomena

involving rates of change:

• Motion of fluids
• Motion of mechanical systems
• Flow of current in electrical circuits
• Dissipation of heat in solid objects
• Seismic waves
• Population dynamics
• A differential equation that describes a physical process

is often called a mathematical model.

Example 1

Added information: the rate at which the temperature of the body

changes is proportional to the difference of current temperature

of the body and the temperature of the surrounding medium.

Let us then define variable name (T). It is instantaneous

temperature of the body in degree C. Let t be the time variable

in minutes.

Then mathematically, we can say

 T  ; dt

dT   20  k  T  20  dt

dT

• This is a mathematical model. It is linear, homogeneous and first order.
• What we need to know is the value of k.
• Let us use the observation and find a rough value of k:

ΔT = 100 – 90.0 = 10.0 oC

Δt = 1 minutes

Therefore putting into the model, we get

k = [ΔT/ Δt ] /(T – 20) = [ 10/1 ]/( 90 – 20) = 0.

Example 1

• Here A is the constant of integration and we can find its value.

T (t)  20  80 exp(kt)

A A

T(t) Aexp(kt)

When t = 0, T = 100;

• Hence the mathematical model has solution form as

We can go one step further to again find value of k from the data known to us for the system i.e., when t = 1min, T = 90 degree C.

k.

k ln( / )

/ exp(k )

exp(k )

T (t )  20  80 exp(  0_._ 1335 t )

Example 1

• Now let us simulate this first order linear mathematical model using MATLAB/SIMULINK.
• First we draw a patch diagram in an empty workspace.

Initial condition: t = 0, T = 100.

 k  T  20  dt

dT

simout To Workspace

Scope 1/s Integrator

-K- Gain

-K- Gain

1 Constant

Example 1

Example 2 : Free Fall

• Let us Formulate a differential equation describing

motion of an object falling in the atmosphere near sea

level.

m

g v

mg

Formulate a differential equation describing motion of an object falling in the atmosphere near sea level.

Variables: time t , velocity v

Newton’s 2nd^ Law: F = ma = m (d v /d t )  net force

Force of gravity: F = mg  downward force

Force of air resistance: F = g v  upward force

Then

mg v

dt

dv

m   g

v

dt

dv

m

g v

mg

Taking g = 9.8 m/sec^2 , m = 10 kg, g = 2 kg/sec,

we obtain model equation as

Example 2 : Free Fall

Example 2:

Direction Field & Equilibrium Solution (4 of 4)

• Arrows give tangent lines to solution curves, and indicate where solution is increasing & decreasing (and by how much).
• Horizontal solution curves are called equilibrium solutions.
• Use the graph below to solve for equilibrium solution, and then determine analytically by setting v' = 0.

49

1. 2

2. 8

3. 8 0. 2 0

Set 0 :

 

 

  

 

v

v

v

v

v  9. 8  0. 2 v

Equilibrium Solutions

• In general, for a differential equation of the form

find equilibrium solutions by setting y' = 0 and solving for y :

• On your own: Find equilibrium solutions of the following model equations.

y  ay  b ,

a

b

y ( t )

y ^  2  y y  5 y  3 y  y ( y  2 )

Free Fall: Graphs for Part (a)

The graph of the solution found in part (a), along with the direction field for the differential equation, is given below.

v  e t 

v

v v

491.^2

 ^ 

Free Fall Part (b): Time, Speed of Impact

• Next, given that the object is dropped from 300 m. above

ground, how long will it take to hit ground, and how fast

will it be moving at impact?

• Solution: Let s ( t ) = distance object has fallen at time t.

It follows from our solution v ( t ) that

• Let T be the time of impact. Then
• Using a solver, T  10.51 sec, hence

. 2 . 2. 2



  t

t t

s C s t t e

s t v t e s t t e C

s ( T ) 49 T  245 e .^2 T  245  300

v ( 10. 51 ) 49  1  e ^0.^2 (^10.^51 )  43. 01 ft/sec

Example : Skydiver

• This skydiver falls from rest towards earth and parachute

opens when skydiver’s speed is 10 m/sec. Let us call

this as initial value or v(0) = 10 m/sec.

• Assume weight is W = mg = 700 Newton.
• Assume air resistance acts upward and is proportional to

the square of the velocity.

• Now we try to develop a mathematical model and

simulate it.

Example : Skydiver

Physical Laws and assumptions:

• Mass means force of gravity; W = mg. This force will be

downward.

• Air resistance means force upward and it is proportional to v^2.

Then this force is Fair = bv2.^ The parameter b is proportionality

constant.

• We assume value of b = 30 N-sec^2 /m^2.
• Now we start setting up the mathematical model.