Technique of nondimensionalization

Aim: – To remove physical dimensions– To reduce the number of parameters– To balance or distinguish different terms in the equation– To choose proper scale for different variables

Method:– Set a scale for each variable – Plug into the equation & balance different terms– Determine the scales

* with : chosen scale;

*: dimensionless variables su u u u

u

Example 1

For Malthus model

Set By the chain rule

Plug into the equation

0 0 0 0

( )( ) ( ) : ( ), 0, (0) 0

d N tb d N t r N t t N N

dt

( )( ), ( ) : s

s s s

N tt N tu

t N N

( ( ))( ) ( ) ( ) 1 ( )ss s s

s

d N ud N t d u d u d d uN N N

dt dt dt d dt t d

0 0 0

( ) ( )( ) ( ), 0, (0) 0s

s ss

N du d N tr N t r N u N u N

t d dt

Example 1

Simplify the equation

Choose

Plug back, we get the dimensionless equation

No parameter !!

00

( )( ), 0, (0)s

s

Ndur t u u

d N

00 0

0

11, 1 ,s s s

s

Nr t t N N

N r

( )( ), 0, (0) 1

duu u

d

Example 2

For logistic model

Set By the chain rule

Plug into the equation

( )( ), ( ) : s

s s s

N tt N tu

t N N

( ( ))( ) ( ) ( ) 1 ( )ss s s

s

d N ud N t d u d u d d uN N N

dt dt dt d dt t d

0

( )( ) ( ) ( )( )(1 ) ( )(1 ), 0,

(0) 0

s ss

s

s

N N udu d N t N tr N t r N u

t d dt K K

N u N

0

( )(1 ), 0 with (0)

d N t Nr N t N N

dt K

Example 2

Simplify the equation

Choose

Plug back, we get the dimensionless equation

No parameter in the equation !!!

0( )( )( )(1 ), 0, (0)s

ss

N u Ndur t u u

d K N

11, 1 ,s

s s s

Nr t t N K

K r

0( )( )(1 ( )), 0, (0)

Nduu u u

d K

Example 3

For insect outbreak model

SetBy chain rule:

Plug into the equation

22

0 1 0 2 2

( )d N t B Nb N b N d N

dt A N

( )( )

, ( ) : s

s s s

N tt N tu

t N N

( ( ))( ) ( ) ( ) 1 ( )ss s s

s

d N ud N t d u d u d d uN N N

dt dt dt d dt t d

22

0 1 0 2 2

2 22 2

0 1 0 2 2 2

( ) ( )

s

s

ss s s

s

N d u d N t B Nb N b N d N

t d dt A N

B N ub N u b N u d N u

A N u

Example 3

Simplify the equation

Scaling 1: Choose

Plug back, we get

22

0 1 0 2 2 2

21

0 0 220 0

( )

( ) 1

s ss s s s

s

ss

s

s s s

B t N ud ub t u b t N u d t u

d A N u

b N ub d t u u

NAb du

B t N B t

2

1, 1 ,ss s

s s s

NA At N A

B t N B t B

0 0 0 0

1

( )( )( )

A b d b dB t N tu r q

A A B A b

2

2(1 )

1

d u u ur u

d q u

Example 3

Scaling 2: Choose

Plug back, we get

The scaling is NOT unique. All are correct.Different ones are good for different parameter regime.

1 0 00 0

0 0 0 0 1

1( ) 1, 1 ,s

s s s

b N b db d t t N

b d b d b

2220 01

21

(b -d )b A(1 ) with = =

B b B

d u uu u

d u

Analytical and numerical solutions

We obtain first ODE from different applications– Mixture problem– Population models

• Malthus model• Logistic model• Logistic model with harvest

– Point-mass motion– Maximum profit– Rocket, ……….

Analytic & Numerical Solutions of 1st Order ODEs

General form of first order ODE:

– t: independent variable (time, in dimensionless form) – y=y(t): state variable (population, displacement, in dimensionless form)– F=F(t,y): a function of two variables

Solution: y=y(t) satisfies the equation– Existence & uniqueness ???– Geometric view point: at various points (t,y) of the two-dimensional coordinate

plane, the value of F(t,y) determines a slope m=y’(t)=F(t,y)!!– A solution of this differential equation is a differential function with graph having

slope y’(t) at each point through which the graph passes.

( )y y t

0

( )( , ( )), ( . ., 0)

dy tF t y t t t e g

dt

Analytic & Numerical Solutions of 1st Order ODEs

– Solution curve: the graph of a solution of a differential equation – A solution curve of a differential equation is a curve in the (t,y)- plane whose

tangent at each point (t,y) has slope m=F(t,y).

Graphical method for constructing approximate solution: – Direction field (or slope field): through each of a representative collection of

points (t,y), we draw a short line segment having slope m=F(t,y).– Sketch a solution curve that threads its way through the direction field in such a

way that the curve is tangent to each of the short line segments that it intersects. – Isoclines: An isoclines of the differential equation y’(t)=F(t,y) is a curve of the

form F(t,y)=c (c is a constant) on which the slope y’(t) is constant.

Direction fields and solution curves

Direction fields and solution curves

Direction fields and solution curves

Direction fields and solution curves

Direction fields and solution curves

Direction fields and solution curves

Direction fields and solution curves

Direction fields by software

In Mathematica:– plotvectorfield

In Maple:– DEplot

In Matlab: – dfield: drawing the direction field

Solutions of ODE

Infinitely many different solutions!!!Solution structure of simple equation

– If y1(t) is a solution, then y1(t)+c is also a solution for any constant c !

– If y1(t) and y2(t) are two solutions, then there exists a constant c such that y2(t)=y1(t)+c.

– If y1(t) is a specific solution, then the general solution is (or any solution can be expressed as) y1(t)+c

0

( )( ),

dy tq t t t

dt

Solutions of ODE

Infinitely many different solutions!!!Solution structure of linear homogeneous problem

– If y1(t) is a solution, then c y1(t) is also a solution for any constant c !

– If y1(t) and y2(t) are two nonzero solutions, then c1 y1(t) + c2

y2(t) is also a solution (superposition) and there exists a constant c such that y2(t)=c y1(t).

– If y1(t) is a specific solution, then the general solution is c y1(t)

0

( )( ) ( ),

dy tp t y t t t

dt

Solutions of ODE

Solution structure of linear problem

– If y1(t) is a solution of the homogeneous equation, y2(t) is also a specific solution. Then y2(t)+c y1(t) is also a solution for any constant c !

– If y1(t) and y2(t) are two solutions, y1(t)-y2(t) is a solution of the homogeneous equation.

– If y1(t) and y2(t) are specific solutions of the homogeneous equation and itself respectively, then any solution can be expressed as y2(t) +c y1(t)

0

( )( ) ( ) ( ),

dy tp t y t q t t t

dt

Solutions of IVP

Initial value problem (IVP):

Solution: – Existence – Uniqueness

Examples– Example 1:

• Solution: • There exists a unique solution

0 0

( )( , ( )), ; ( )

dy tF t y t t t y t b

dt

0 0

( ): ( , ), ; ( )

dy ty F t y t t y t b

dt

00( ) ,t ty t b e t t

Solutions of IVP

– Example 2

• Solution: separable form

• General solution

• Different cases:– b=0: these is at least one solution y=y(t)=0– b>0 (e.g. b=1): these is no solution!! (F(t,y) is not continuous near (0,

b>0)!!!

22 2

20, (0) : ( , ), (0)

dy dy yy t y b F t y y b

dt dt t

2 2 2 2

1 1dy dt dy dtc

y t y t y t

( )1

ty y t

ct

Solutions of IVP

Theorem: Suppose that the real-valued function F(t,y) is continuous on some rectangle in the (t,y)-plane containing the point (t0,b) in its interior. Then the above initial value problem has at least one solution defined on some open interval J containing the point t0. If, in addition, the partial derivative is continuous on that rectangle, then the solution is unique on some (perhaps smaller) open interval J0 containing the point t=t0.

Proof: Omitted

/F y

Solutions of IVP

Example 1:– Condition: F and are continuous

– Conclusion: There exists a unique solution for any initial data (t0,b)

Example 2: – Condition:F is continuous for y>=0 and is continuous for y>0

– Conclusion: • There exists a unique solution for any initial data (t0,b>0)• For b=0, e.g. y(0)=0, there are two solutions

: ( , ), 1dy F

y F t ydt y

12 : ( , ),

dy Fy F t y

dt y y

/F y

/F y

21 2( ) , ( ) 0y t t y t

Classification

Classification– Linear ODE, i.e. the function F is linear in y

– Otherwise, it is nonlinear – Autonomous ODE, i.e. F is independent of t

Some cases which can be solved analytically

( , ) ( ), . . (1 )dy dy

F t y f y e g y ydt dt

( , ) ( ) ( ), . . (1 ) sindy dy

F t y p t y q t e g t y tdt dt