Modeling with High-Order Differential Equations National Chiao Tung University Chun-Jen Tsai 10/16/2019

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Modeling with High-Order Differential Equations

National Chiao Tung University Chun-Jen Tsai 10/16/2019

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Linear Models: IVP

 Many linear dynamic systems can be represented using a 2^nd order DE with constant coefficients:

In this formulation, g(t) is the input or forcing function of the system, the output of the system is a solution y(t) of the DE that satisfies the initial conditions

y(t₀) = y₀, y(t₀) = y₁ on an interval containing t₀. )

0 (

2 1 2

2 a y g t

dt a dy dt

y

a d   

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Free Undamped Motion

 Hooke’s law describes the restoring force:

F = ks

 Newton’s 2^nd law (F = ma) describes the motion:

m(d²x/dt²) = –k(s + x) + mg

= –kx + (mg – ks) = –kx

 DE of free undamped motion:

 Solution of the motion:

x(t) = c₁ cos



t + c₂ sin



t.

1 0

2 2

2

) 0 ( ' , )

0 ( ,

0 x x x x

dt x x

d    

equilibrium position

s l

m

mg - ks = 0 m

motion x l + s

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Alternative Form of Solution

 By applying trigonometric formula, we have:

2 2 1

2 2

1 , tan

), sin(

)

( c

c c c

A t

x 







 





x negative

x positive x = 0

5

1 x

0 x

20

 17

 x

0 x

20

 17 x

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Aging Spring

 In real world, the spring constant k usually varies as the spring gets old. Replace k with k(t) = ke^–^^t, k > 0,



> 0, we have a more realistic system model:

mx + ke^–^^tx = 0

 Non-constant coefficient 2^nd-order linear DE!

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Free Damped Motion

 DE of free damped motion:



 The roots of the auxiliary eq.:

dt kx dx

dt x

m d²₂    

0

2 ²

2

2   x 

dt dx dt

x

d  

2

2 



  





m ^m

m

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Three Cases of Damped Motion

 Case I: Over-damped

 Case II: Critically damped

 Case III: Under-damped

) (

)

(t e ^t c₁e ² ²^t c₂e ² ²^t x  ^^ ^ ^^  ^ ^ ^^

) (

)

(t e c₁ c₂t x  ^^^t 

) sin

cos (

) (

2 2

2

2 2

1

t c

e t

x ^t

















 ^

t x

x

t

x

t undamped

underdamped

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Driven Motion

 Now, consider the effect of external

force f(t) on the damped motion system:



)

2 (

2

t dt f

kx dx dt

x

m d     

) (

2 ²

2 2

t F dt x

dx dt

x

d     

m

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Transient and Steady-State Terms

 When F(t) is a periodic function and



> 0, the solution is the sum of a non-periodic function x_c(t) and a

periodic function x_p(t). Moreover lim_t x_c(t) = 0.

1

-1

transient

steady-state x

t

1

-1

x( t) = transient + steady-state x

t

/2 /2

xp(t)

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Example: Transient/Steady State

 The solution of

is x(t) = (x₁ – 2)e^–t sin t + 2 sin t,

2 1 2

) 0 ( ,

0 )

0 ( ,

sin 2 cos

4 2

2 x t t x x x

dt dx dt

x

d       

transient steady-state ₄

- 4

x₁= 7 x

t - 2

2

x₁= 3 x₁= 0

x₁= -3

2



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Undamped Forced Motion

 The solution of

is

where c₁ = 0, c₂ = –γF₀ /ω(ω² – γ²).



 There is no transient term.

0 )

0 ( ,

0 )

0 ( ,

0 sin

2 2

2  x  F t x  x 

dt x

d  

F t t

c t

x



 



sin sin

cos )

( ₁ ₂ ₂ ⁰ ₂

 











 



 ^ ^ ^

 ( sin sin ) ,

) ) (

( ₂F⁰ ₂ t t

t x

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Pure Resonance

 In the previous example, when







^{, the}

displacement of the system become large as t  .

) (

) sin

sin (

lim

) (

sin lim sin

) (

2 0 3

2 0 2



 





 































 



d d

t d t

d F

t F t

t x

t F t

F t 

 

 sin 2 ^cos 2

0 2

0 



t x

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Tacoma Narrow Bridge, WA, USA

 Opened in July 1, 1940, collapsed in Nov. 7, 1940.

 The wind-blow frequency matched the natural frequency of the bridge, which caused a pure resonance effect that destroyed the bridge.

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Damping System of Taipei 101

 Taipei 101 uses a 730-ton damping ball^† to stabilize the building under wind-blow effect

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Linear Models: BVP

 The deflection of a flexible beam can be modelled by a 4^th-order differential equation:

load per unit length flexural rigidity

A straight flexible beam The deflection curve of the beam

)

4 (

4

x dx w

y EI d 

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Flexible Beam Applications

 For precise robot arm control, we must take into account the bending effect of the robot links:

Base position (x₀, y₀)

True end-point position (x₄, y₄)

Estimated end-point position (𝑥 , 𝑦 ) without considering bending effect

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Boundary Conditions

 Boundary conditions of a flexible beam:

End of beam Boundary conditions

embedded y = 0 y = 0

free y = 0 y = 0

supported y = 0 y = 0

x = 0 x = L x = 0 x = L x = 0 x = L

Embedded at both ends Free at the right end Supported at both ends

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Eigenvalue Problems

 An eigenvalue problem in DE is a homogeneous BVP such that the boundary conditions evaluate to 0 and there is a parameter



at the coefficient of y:

y" + p(x)y' +



q(x)y = 0, y(a) = 0, y(b) = 0.

The eigenvalue problem tries to find a



(eigenvalue) such that the BVP has a nontrivial solution.

 The non-trivial solution that corresponding to an eigenvalue



is then called an eigenfunction.

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Example: ^{y" +}  y = 0, y(0) = y(L) = 0 (1/2)

 The problem can be solved by enumerating different cases when



= 0,



< 0, and



> 0.

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

= 0, we have y = 0,

 the general solution is y(x) = Ax + B.

 y = 0 is the only solution for the BVP





= 0 is not an eigenvalue of the BVP

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

< 0, let



= –



²,



> 0, we have y –



²y = 0,

 the general solution is y(x) = c₁e^^x + c₂e^–^^x.

 y = 0 is the only solution for the BVP





< 0 do not have eigenvalues of the BVP

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Example: ^{y" +}  y = 0, y(0) = y(L) = 0 (2/2)

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

> 0, let



=



²,



> 0, we have y +



²y = 0,

 the solution is y(x) = c₁cos(



x) + c₂sin(



x).

 y(0) = 0 implies c₁ = 0

 y(L) = 0 implies sin(



L) = 0, or



L = n



, n Z

 The BVP has infinitely many eigenvalues:

and the corresponding eigenfunctions are:

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Nonlinear Spring Models (1/2)

 The general mathematical model of an undamped spring has the form:

for a linear spring model, F(x) = kx. However, spring are quite often nonlinear, e.g. F(x) = kx + k₁x³.

x F

soft spring hard spring

linear spring

0 )

2 (

2  F x 

dt x m d

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Nonlinear Spring Models (2/2)

 Damping force of a spring system can be nonlinear as well:

 Restoring force F(x) is usually an odd function such as kx + k₁x³. The reason is that we want F(–x) = –F(x).

0 )

2 (

2   F x 

dt dx dt

dx dt

x

m d 

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Nonlinear Pendulum

 The pendulum system can be modeled as

Using Maclaurin series of sin



, we have

O

l

mg sin



mg cos

P  W = mg

. 0

2 sin

2   

l g dt

d

! , 5

! sin 3

5 3







  



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Linearization of Nonlinear Systems

 Assuming that sin







–



³/6, we have:

System can be linearized by assuming sin







:

 Impact of initial values:

^{(0) = 1}₂^,

. 6 0

3 2

2  



 







 



   



l g l

g dt

d

.

2 0

2  



 



  



l g dt

d

A nonlinear model

 similar to the spring systems!



𝜃 0 = , 𝜃 0 = 2

𝜃 0 = , 𝜃 0 =

Modeling with High-Order Differential Equations National Chiao Tung University Chun-Jen Tsai 10/16/2019