introducing elliptic functions via simulations of

-15 -10 -5 0 5 10 15

-1

-0.5

0

0.5

1

sn(u) cn(u) dn(u)

Introducing Elliptic Functions via Simulations

of Nonlinear Differential Equations

MAA-SE 2018, Clemson University

Dr. R. L. Herman

Mathematics & Statistics, UNC Wilmington

Table of Contents

1. Nonlinear Pendulum

2. Jacobi Elliptic Functions

3. Simulink and ODEs

4. ODE Examples

5. Trig-Elliptic Systems

Elliptic Functions R. L. Herman MAA-SE, March 24, 2018 1/23

Nonlinear Pendulum

The Nonlinear Pendulum

m

θL

Figure 1: A point mass m is attached to a string of length L and released from

rest at θ = θ0.

θ + ω2 sin θ = 0. (1)


Nonlinear Pendulum - Quadrature θ + ω2 sin θ = 0

Multiply Equation (1) by θ,

θθ + ω2 sin θθ = 0,

and noted

dt

[1

2θ2 − ω2 cos θ

]= 0.

Therefore,1

2θ2 − ω2 cos θ = c . (2)

Using the initial conditions, θ(0) = θ0, θ(0) = 0, we have

c = −ω2 cos θ0 and

θ2 = 2ω2(cos θ − cos θ0).


Second Order ODE from θ2 = 2ω2(cos θ − cos θ0)

θ2 = 4ω2

(sin2 θ

2− sin2 θ0

2

)Let kx = sin θ

2 , then 2kx =√

1− k2x2 θ. Then,

x2 = ω2(1− x2)(1− k2x2) (3)

Differentiating,

2x x = ω2x[−2x(1− k2x2) + (1− x2)(−2k2x)

],

yields

z = ω2[2k2z3 − (1 + k2)z

](4)

Solution: x(t) = sn (ωt, k). - a Jacobi elliptic function


Jacobi Elliptic Functions

Jacobi Elliptic Functions:

x(t) = sn (t, k), y(t) = cn (t, k) z(t) = dn (t, k).

Solutions of [κ =√

1− k2]

x2 = (1− x2)(1− k2x2), x(0) = 0, x(0) = 1,

y2 = (1− y2)(κ2 + k2y2), y(0) = 1, y(0) = 0,

z2 = (1− z2)(1− κ2), z(0) = 1, z(0) = 0.

E.g., integration gives

dx

dt=√

(1− x2)(1− k2x2),

or

t =

∫ x= sn (t,k)

0

dξ√(1− ξ2)(1− k2ξ2)

.


Elliptic Integrals - Arclength of Ellipse

Recall that ∫ x

0

dξ√1− ξ2

= sin−1 x ≡ u

or

u =

∫ sin u

0

dξ√1− ξ2

.

Similarly,

F (sinφ, k) =

∫ sinφ

0

dξ√(1− ξ2)(1− k2ξ2)

=

∫ φ

0

dθ√1− k2 sin2 θ

, 0 ≤ φ ≤ π

2. (5)

sinφ = sn (u, k), cosφ = cn (u, k),√

1− k2 sin2 θ = dn(u, k).


Jacobi Elliptic Functions vs Trigonometric Functions

Trigonometric Functions

d

dtsin x = cos x ,

d

dtcos x = − sin x .

Initial Value Problem

x = y , x(0) = 0,

y = −x . y(0) = 1.


d

dtsn (t) = cn (t) dn (t),

d

dtcn (t) = − sn (t) dn (t),

d

dtdn (t) = −k2 cn (t) sn (t).

Initial Value Problem

x = yz , x(0) = 0,

y = −xz , y(0) = 0,

z = −k2xy , z(0) = 0.


Simulink and ODEs

Simulink

• What is Simulink

• Graphical environment

for designing simulations

• Product of Mathworks

• Select and connect blocks

• Use in Differential Equations

• Project component of class

• Modeling applications

y' y

1-y

y

Logistic Equation

y' = r y (1-y)

1s

Integrator

1

Gain Scope

Product

1

Constant


Solving a Differential Equation

Consider initial value problem:

dx

dt= f (x), x(0) = x0.

Solution

x(t) = x0 +

∫ t

0

f (x(t)) dt.

Think of the solution as

x(t) =

∫x ′(t) dt.

input∫

outputxx′

Figure 2: Schematic for a general system.


Model of a Differential Equation

Modeling

x(t) =

∫x ′(t) dt =

∫f (x(t)) dt.

Schematic

input∫

outputxx′

Simulink Model

f (x)1

s

xx′

OutputIntegrator

Figure 3: Model for solving x ′ = f (x).


Simple First Order Differential Equation

Solve x ′ = −4x , x(0) = 1.

1

s

−4

xx′

ScopeIntegrator

Gain


Simulink Workspace


Scilab’s Xcos Workspace


ODE Examples

First Order Differential Equation - Example 1

Solve x ′ = 2 sin 3t − 4x , x(0) = 0.

1s

Integrator

4

Gain

ScopeSine Wave

Function


First Order Differential Equation - Example 2

Solve y ′ = 2t y + t2, y(1) = 1.

2/t y

dy/dt y

t

1/t 2/ty' = 2/t y+t , y(1)=1

Exact solution: y(t) = t

t2

2

3

1s

Integrator1 Scope1

2

Gain1

1

uMath

Function1

Clock

Product1

u2

Math

Function2


Second Order ODEs - Example 3

Solve ay ′′ + by ′ + cy = 0, y(0) = y0, y′(0) = v0.

y =

∫y ′ dx , y ′ =

∫y ′′ dx ,

y ′′ = −b

ay ′ − c

ay .

y' yy''

b/a y'

c/a y

Second Order Constant Coefficient ODE

1s

Integrator

1s

Integrator1

5

b/a

6

c/a

Scope


Linear Systems of Differential Equations - Example 4

x ′ = ax + by

y ′ = cx + dy .

x

Linear System of Differential Equations

y

y

x

y

x

x'=ax+byy'=cx+dy

1sxo

Integrator

1sxo

Integrator1

Scope

1

x(0)

2

y(0)

0

a

-1

c

0

d

1

b

XY Graph


Trig-Elliptic Systems

Trigonometric Simulink Model

x'=y

y'=-x

x

y

x(0)=0

y(0)=1x' = y

y' = - x

1s

Integrator

1s

Integrator1

Scope

Scope1

-1

Gain


Trigonometric Simulink Model - Results

Time offset: 0 Time offset: 0


Elliptic Function Simulink Model

x'=zy

y'=-zx

x

y

x(0)=0

y(0)=1

x' = yz

y' = - xz

z

z'= -k xy2

-x

z'= -k xy

z

2

y

z(0)=1

1s

Integrator

1s

Integrator1

x Scope

y Scope

-1

Gain

1s

Integrator2 z Scope

0.95

k

u2

Math

Function

Product

Product1

Product2


Elliptic Function Simulink Model - Results

Time offset: 0 Time offset: 0 Time offset: 0

-15 -10 -5 0 5 10 15

-1

-0.5

0

0.5

1

sn(u) cn(u) dn(u)


Conclusion

Nonlinear Pendulum


Simulink and ODEs

ODE Examples

Trig-Elliptic Systems


The End!

Thank you! Dr. R.L. Herman, [email protected]

Herman, R. L., Solving Differential Equations Using Simulink ,

http://people.uncw.edu/hermanr/MAT361/Simulink/index.htm

Meyer, K., M., Jacobi Elliptic Functions from a Dynamical Systems

Point of View, Amer. Math. Monthly 108 (2001) p. 729.


http://people.uncw.edu/hermanr/MAT361/Simulink/index.htm

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