engg2013 unit 22 modeling by differential equations apr, 2011
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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