Numerical Solution of Differential Equations

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Numerical Solution of

Differential Equations

3^rd year JMC group project ● Summer Term 2004

Supervisor: Prof. Jeff Cash

Saeed Amen Paul Bilokon

Adam Brinley Codd Minal Fofaria

Tejas Shah

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Agenda

z Adam: Differential equations and numerical methods

z Paul: Basic methods (FE, BE, TR). Problem:

“circle”

z Saeed: Advanced methods (4^th order).

Problem: Kepler’s equation

z Minal: Stiff equations

z Tejas: Derivation of 6^th order method

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Agenda

z Adam: Differential equations and numerical methods

z Paul: Basic methods (FE, BE, TR). Problem:

“circle”

z Saeed: Advanced methods (4^th order).

z Minal: Stiff equations

z Tejas: Derivation of 6^th order method

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Introduction to Differential

Equations

z Equations involving a function y of x, and its derivatives

z Model real world systems

z General Equation:

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Introduction to Differential

Equations

z Simple example

z Solution obtained by integrating both sides

z Initial values can determine c

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Introduction to Differential

Equations

z Special case when equations do not involve x

z E.g.

z Initial values y(0) = 1 and y'(0) = 0, solution is

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Introduction to Differential

Equations

z Kepler’s Equations of Planetary Motion

z Difficult to solve analytically – we use numerical methods instead

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Introduction to Differential

Equations

z Forward Euler and Backward Euler

z Trapezium Rule

z General method

– Use initial value of y at x=0

– Calculate next value (x=h for small h) using gradient

– Call this y₁ and repeat

z Need formula for y_n+1 in terms of y_n

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Agenda

z Adam: Differential equations and numerical methods

“circle”

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Agenda

z Adam: Differential equations and numerical methods 3

“circle”

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Euler’s Methods

z Simplest way of solving ODEs numerically

z Does not always

produce reasonable solutions

z Forward Euler (explicit) and Backward Euler

(implicit)

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Forward Euler

z y(x + h) = y(x) + h y’(x)

z Explicit

z E^loc = ½h²y⁽²⁾(ξ) ∝ h²

z First order

z Asymmetric

x₀ x₁ x₂ x

y

y₀ y₁ y₂

h h

(x₀, y₀)

(x₁, y₁)

(x₂, y₂) True solution

Approximated solution Slope f(x1, y₁) Slope f(x₀, y₀)

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Backward Euler

z y(x + h) = y(x) + h y’(x + h)

z Implicit

z E^loc = - ½h²y⁽²⁾(ξ⁾∝ ^h²

z First order

z Asymmetric

x₀ x₁ x₂ x

y

y₀ y₁ y₂

h h

(x₀, y₀)

(x₁, y₁)

(x₂, y₂)

True solution

Approximated solution Slope f(x₂, y₂) Slope f(x₁, y₁)

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Trapezium Rule

z y(x + h) = y(x) + ½ h [y’(x) + y’(x + h)]

z Average of FE and BE

z Implicit

z E^loc = - (h³/12) f⁽²⁾(ξ) ∝ h³

z Second order

z Symmetric

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Example Problem: “Circle”

z Consider y’’ = -y, with initial conditions

– y(0) = 1

– y’(0) = 0

z The analytical solution is y = cos x, that can be

used for comparison with numerical solutions

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Plots for the True Solution

Time series plot (xy) Phase plane plot (zy)

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Time Series

y

x

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Phase Plane Plots: FE & BE

Forward Euler Backward Euler

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Phase Plane Plots: TR

z FE & BE fail: non-periodic

z TR: OK, periodic solⁿ

W h y ?

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Symmetricity

z TR is symmetric, whereas FE & BE are not

y(x + h) = y(x) + ½ h [y’(x) + y’(x + h)]

h Æ -h

y(x - h) = y(x) - ½ h [y’(x) + y’(x - h)]

X := x - h

y(X + h) = y(X) + ½ h [y’(X) + y’(X + h)]

Still TR!

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Symmetricity

z TR is symmetric, whereas FE & BE are not

y(x + h) = y(x) + h y’(x) h Æ -h

y(x - h) = y(x) - h y’(x) X := x - h

y(X + h) = y(X) + h y’(X + h)

Still FE?

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Time Step

h = 10 ^-1 h = 10 ^-2

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Agenda

“circle”

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Agenda

“circle” 3

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After Euler and TR

z Can create higher order methods, which have far smaller global errors

z Methods are more complex and require more computation on each step

z But for same step size more accurate

z Introduce concept of half step

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Two Fourth Order Methods

z y_n+1 – y_n = y_n + h/2(y’_n) + (h²/12)(4y’’_n+½+ 2y’’_n)

z y’_n+1– y’_n = h/6(y’’_n+1+ 4y’’_n+½+ y’’_n) (*)

z Then to calculate the half-step we use either A or B

– A) y_{n+ ½}= ½(y_n+1+ y_n) – h²/48(y’’_n+1 + 4y’’ _n+½ + y’’_n)

– B) y_{n+ ½}= y_n + ½y’_n– h²/192(-2y’’_n+1 + 12y’’ _n+½ + 14y’’_n) z Which one is more accurate? (next slide)

z Solve iteratively and then apply solution to find derivative (*)

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Creating Method A

z Start with

– y_n+1 – y_n = h/6[y’_n+1 + 4y’_n+½ + y’_n] (1)

– diff. y’_n+1 – y’_n = h/6[y’’_n+1 + 4y’’_n+½ + y’’_n] (2)

– y_n+½= ½(y_n+1 + y’_n) – h/8(y’_n+1 – y’_n) (3)

– diff. y’_n+½= ½(y’_n+1 + y’’_n) – h/8(y’’_n+1 – y’’_n) (4)

z subs (4) into (1) (to eliminate y’_n+½ ) and eliminate y’_n+1 using (2),

z then use (2) in (3) to get half-step

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Comparing Fourth Order Methods

z Comparing errors when solving circle problem

z Both A and B produce a much smaller order of error than Euler’s

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Finding Earth’s Orbit Around Sun

z Kepler’s Equation

z Can use it to find the orbit of planets

z Use to find orbit of earth around the sun

z Work in two dimensions z and y

z z’’ = -(GM z) / (y² + z²)^3/2 y’’ = -(GM y) / (y² + z²)^3/2

z Constants and initial conditions in report

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Results Plot

z Solution uses small step size h = 0.01

z Becomes difficult to tell difference between

methods visually

z Using h = 0.1 difference is more marked

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Agenda

“circle” 3

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Agenda

“circle” 3

Problem: Kepler’s equation 3

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Stiff Equations

z Certain systems of ODEs are classified as stiff

z A system of ODEs is stiff if there are two or more very different scales of the independent variable on which the dependent variables are changing

z Some of the methods used to find numerical solutions fail to obtain the required solution

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Example

z Consider the equation:

z With initial conditions:

z There are two solutions to this problem

0 )

1

2 (

2 + + + y =

dx dy dx

y

d λ λ

1 )

0 ( = y

1 )

0 (

' = − y

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Example continued

z Analytical solution: y=e^-x

z Unwanted solution: y=e^-λx

z We will now show how forward Euler is not

stable when solving this problem under certain circumstances

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Example continued

z Let λ = 10³ and let (the step size) h = 0.1

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Increasing the range of the x-axis

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When h = 0.001

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Agenda

“circle” 3

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Agenda

“circle” 3

z Minal: Stiff equations 3

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A Sixth Order Numerical Method

z Has an error term with smallest h degree as h⁷

z Idea is to find values for α, A, B, C, D, E, F, C, D, E, F such that:

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Derivation

z Compare Taylor’s Expansion of LHS and RHS

z So for:

– First expand the LHS

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Derivation

• Now expand individual terms on RHS of our expression:

• This gives:

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Derivation

z Equate both sides and solve for constants:

z Similarly we can find C, D, E, F and C, D, E, F

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Applying 6

^th

Order Method

z How to obtain results using derived method.

z Produce a set of simultaneous equations and solve.

z Find y’_n+1 from old values and those just obtained.

z Set x, y_n , y’_n etc. to new values.

z Repeat above procedure with updated variables.

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Circle Problem Example

z Circle is produced for phase plane plot

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Kepler’s Equations

• Solution obtained from z-y plot reinforces fourth order method results.

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The End

But perhaps you want more?

z Read our report

z Visit our website:

http://www.doc.ic.ac.uk/~pb401/DE