Numerical Solutions for Linear 2nd Order Non-Homogeneous ODEs, Slides of Calculus for Engineers

Download Numerical Solutions for Linear 2nd Order Non-Homogeneous ODEs and more Slides Calculus for Engineers in PDF only on Docsity!

Engr/Math/Physics 25

Chp9: ODE’s

Numerical Solns

Learning Goals

• List Characteristics of Linear, MultiOrder,

NonHomgeneous Ordinary Differential

Equations (ODEs)

• Solve ANALYTICALLLY, Linear, 2nd^ Order,

NonHomogeneous, Constant Coefficient

ODEs

• Use MATLAB to determine Numerical

Solutions to Ordinary Differential Equations

(ODEs)

Numerical ODE Solution

• Today we’ll do some

MTH

• We’ll “look under the

hood” of NUMERICAL Solutions to ODE’s

• The BASIC Game-Plan for even Sophisticated Solvers: - Given a STARTING POINT, y(0) - Use ODE to find dy/dt at t= - ESTIMATE y 1 as

0

1 0

≈ +∆ ⋅ dt t

dy y y t

Numerical Solution - 1

• Notation (^)  Exact Numerical

Method (impossible to achieve) by Forward Steps t (^) n = n ×∆ t

y (^) n = y ( tn )

f (^) n = f ( tn , yn )

n ≡Step Number

∆ t ≡ TimeStep Length

f ( t , y ) dt

dy

 Now Consider

yn+

t (^) n

yn

t (^) n+

t ∆ t

Numerical Solution -

• However, we can calculate the TANGENT slope at any point FROM the differential equation itself

 The Basic Concept for all numerical methods for solving ODE’s is to use the TANGENT slope, available from the R.H.S. of the ODE, to approximate the chord slope

 Recognize dy/dt as the Tangent Slope

( (^) n , n ) t t

n f t y dt

dy m

n

=

tangent slope= f ( t , y ) (^ n n ) t

n n f t y dt

dy

t

y y

n

(^1) ≈ = , ∆

• −

Euler Method – 1

st

Order

• Solve 1 st^ Order ODE

with I.C.

 ReArranging

 Use: [Chord Slope]

≈ [Tangent Slope at start of time step]

f ( t , y ) dt

dy

y ( 0 ) = b yn^ + 1 ≈ yn + ∆ t ⋅ fn

 Then Start the “Forward March” with Initial Conditions

t (^) 0 = 0 y 0 = b

( (^) n n ) t

n n f t y dt

dy

t

y y

n

(^1) ≈ = , ∆

• −

1 n^ or t

n y dt

dy y t n

• ≈ ∆^ +

Euler Exmpl Calc

n tn yn fn= – yn+1 yn+1= yn+ ∆ t f (^) n

0 0 0.000 1.000 0.

1 0.1 0.100 0.900 0. 2 0.2 0.190 0.810 0. 3 0.3 0.271 0.729 0. 4 0.4 0.344 0.656 0. 5 0.5 0.410 0.590 0. 6 0.6 0.469 0.531 0. 7 0.7 0.522 0.478 0. 8 0.8 0.570 0.430 0. 9 0.9 0.613 0.387 0.

= − y + 1 ∆ t = 0. 1 dt

dy

Plot

Slope

Euler vs Analytical

0

0 0.25 0.5 0.

1

t

Exact

Numerical

y

t y e

− = 1 −

 The Analytical Solution

Analytical Soln

• And

= − y + 1 y^ ( t^ = 0 )^ = 0 dt

dy

 Thus Soln u(t) t u Ke

−

 Sub u = 1−y

 Now use IC

 The Analytical Soln

t C C t t

u

e e e Ke

e u

− + − − = =

=

ln

t y Ke

− 1 − =

1

1 0

0

⇒ =

− =

−

K

Ke

t y e

− 1 − = 1 ⋅

t y e

− = 1 −

Predictor-Corrector - 1

• Again Solve 1st^ Order

ODE with I.C.

 Mathematically

 This Time Let:

Chord slope ≈ average of tangent slopes at start and END of time step

f ( t , y ) dt

dy

y ( 0 ) = b

 BUT , we do NOT know y (^) n+1 and it appears on the RHS ...

+ 1 ≈ 0. 5 [ ( , )+ ( + 1 , + 1 )]

∆

− n n n n

n n f t y f t y t

y y

Avg of the Tangent Slopes at (t (^) n,y (^) n) & (t (^) n+1,y (^) n+1)

Predictor-Corrector Example

• Solve ODE with IC

 The Corrector step

= − y + 1 dt

dy (^) y ( 0 ) = 0

[ ]

∗ yn + 1 = yn + 0. 5 ∆ t fn + fn + 1

 The next Step Eqn for dy/dt = f(t,y)= –y+

1. 5 [ ( 1 ) ( 1 )]

yn + 1 = yn + ∆ t − yn + + − yn + 1 +

 Numerical Results on Next Slide

Predictor-Corrector Example

n

0 0 0.000 1.000 0.100 0.900 0.

1 0.1 0.095 0.905 0.186 0.815 0.

2 0.2 0.181 0.819 0.263 0.737 0.

3 0.3 0.259 0.741 0.333 0.667 0.

4 0.4 0.329 0.671 0.396 0.604 0.

t (^) n y (^) n f (^) n

y (^) n + 1

f (^) n + 1 yn + 1

f = dy dt = − y + 1 [ ]

∗ yn + 1 = yn + 0. 5 ∆ t fn + fn + 1

Slope Slope

ODE Example:

• Euler Solution with ∆t =
  1. • The Solution Table

= 3. 9 cos ( 4. 2 y ) − 3. 9 ln( 5. 1 t + 6 ) dt

dy

(^220 1 2 3 4 5 6 7 8 9 )

24

26

28

30

32

34

36

38

t

y(t) by Euler

Euler Solution to dy/dt = 3.9cos(4.2y)-ln(5.1t+6)

n 0 (^) t (^) 0 37.0000 -1.7457 -0.4364 36.5636 y dy/dt dely y (^) n+ 12 0.25 36.56360.5 36.9143 -1.3492 -0.3373 36.5769 1.4027 0.3507 36. 34 0.75 36.57691 36.8872 -1.2264 -0.3066 36.5806 1.2410 0.3103 36. 56 1.25 36.58061.5 36.8418 -0.7108 -0.1777 36.6641 1.0448 0.2612 36. 78 1.75 36.66412 36.9608 -2.5004 -0.6251 36.3357 1.1868 0.2967 36. 109 2.25 36.3357 -2.6357 -0.6589 35.67682.5 35.6768 -1.6265 -0.4066 35. 1112 2.75 35.27013 35.2882 -0.2436 -0.0609 35.2273 0.0722 0.0181 35. 1314 3.25 35.22733.5 35.3380 -1.1420 -0.2855 35.0526 0.4430 0.1107 35. 1516 3.75 35.0526 -0.0139 -0.0035 35.04914 35.0491 -0.1072 -0.0268 35. 1718 4.25 35.0223 -0.5255 -0.1314 34.89094.5 34.8909 -2.6041 -0.6510 34. 1920 4.75 34.2399 -1.1497 -0.2874 33.95245 33.9524 -3.0108 -0.7527 33. 2122 5.25 33.1997 -3.0006 -0.7502 32.44965.5 32.4496 -3.0151 -0.7538 31. 2324 5.75 31.6958 -2.9862 -0.7466 30.94926 30.9492 -3.0384 -0.7596 30. 2526 6.25 30.1897 -2.9328 -0.7332 29.45646.5 29.4564 -3.1419 -0.7855 28. 2728 6.75 28.6710 -2.6916 -0.6729 27.99817 27.9981 -3.5484 -0.8871 27. 2930 7.25 27.1110 -1.7458 -0.4365 26.67457.5 26.6745 -2.8722 -0.7180 25. 3132 7.75 25.9565 -2.4562 -0.6141 25.34248 25.3424 -0.4717 -0.1179 25. 3334 8.25 25.2245 -2.2562 -0.5641 24.66048.5 24.6604 -0.0369 -0.0092 24. 3536 8.75 24.6512 -0.0977 -0.0244 24.62689 24.6268 -0.2699 -0.0675 24. 3738 9.25 24.5593 -1.0481 -0.2620 24.29739.5 24.2973 -3.9863 -0.9966 23. 3940 9.75 23.3007 -0.9318 -0.2329 23.067810 23.0678 -1.0551 -0.2638 22.

Compare Euler vs. ODE

Euler Solution

• Euler is Much LESS accurate

ODE45 Solution

(^220 1 2 3 4 5 6 7 8 9 )

24

26

28

30

32

34

36

38

t

y(t) by Euler

Euler Solution to dy/dt = 3.9cos(4.2y)-ln(5.1t+6)

34.5 0 1 2 3 4 5 6 7 8 9 10

35

36

37

T by ODE

Y by ODE