© Imperial College London Numerical Solution of Differential Equations 3 rd year JMC group project ● Summer Term 2004 Supervisor: Prof. Jeff Cash Saeed

© Imperial College London

Numerical Solution ofDifferential Equations

3rd year JMC group project ● Summer Term 2004

Supervisor: Prof. Jeff Cash

Saeed Amen

Paul Bilokon

Adam Brinley Codd

Minal Fofaria

Tejas Shah


Agenda

Adam: Differential equations and numerical methods

Paul: Basic methods (FE, BE, TR). Problem: “circle”

Saeed: Advanced methods (4th order). Problem: Kepler’s equation

Minal: Stiff equations Tejas: Derivation of 6th order method


Agenda






Introduction to Differential Equations

Equations involving a function y of x, and its derivatives

Model real world systems General Equation:



Simple example

Solution obtained by integrating both sides

Initial values can determine c



Special case when equations do not involve x E.g.

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



Kepler’s Equations of Planetary Motion

Difficult to solve analytically – we use numerical methods instead



Forward Euler and Backward Euler Trapezium Rule

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

Need formula for yn+1 in terms of yn


Agenda






Agenda






Euler’s Methods

Simplest way of solving ODEs numerically

Does not always produce reasonable solutions

Forward Euler (explicit) and Backward Euler (implicit)


Forward Euler

y(x + h) = y(x) + h y’(x)

Explicit Eloc = ½h2y(2)() h2

First order 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)


Backward Euler

y(x + h) = y(x) + h y’(x + h)

Implicit Eloc = - ½h2y(2)() h2

First order 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)


Trapezium Rule

Average of FE and BE Implicit Eloc = - (h3/12) f(2)()

h3

Second order Symmetric

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


Example Problem: “Circle”

Consider y’’ = -y, with initial conditions– y(0) = 1– y’(0) = 0

The analytical solution is y = cos x, that can be used for comparison with numerical solutions


Plots for the True Solution

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


Time Series

y

x


Phase Plane Plots: FE & BE

Forward Euler Backward Euler


Phase Plane Plots: TR

FE & BE fail: non-periodic TR: OK, periodic soln

W h y ?


Symmetricity

TR is symmetric, whereas FE & BE are not

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

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

X := x - hy(X + h) = y(X) + ½ h [y’(X) + y’(X +

h)]

Still

TR!


Symmetricity

TR is symmetric, whereas FE & BE are not

y(x + h) = y(x) + h y’(x)h -h

y(x - h) = y(x) - h y’(x)X := x - h

y(X + h) = y(X) + h y’(X + h)

Still

FE?


Time Step

h = 10 -1

h = 10 -2


Agenda






Agenda






After Euler and TR

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

Methods are more complex and require more computation on each step

But for same step size more accurate Introduce concept of half step


Two Fourth Order Methods

yn+1 – yn = yn + h/2(y’n) + (h2/12)(4y’’n+½+ 2y’’n)

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

Then to calculate the half-step we use either A or B– A) yn+ ½ = ½(yn+1 + yn) – h2/48(y’’n+1 + 4y’’ n+½

+ y’’n)

– B) yn+ ½ = yn + ½y’n – h2/192(-2y’’n+1 + 12y’’ n+½ + 14y’’n)

Which one is more accurate? (next slide) Solve iteratively and then apply solution to find derivative

(*)


Creating Method A

Start with– yn+1 – yn = 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)

– yn+½= ½(yn+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)

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

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


Comparing Fourth Order Methods

Comparing errors when solving circle problem

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


Finding Earth’s Orbit Around Sun

Kepler’s Equation Can use it to find the orbit of planets Use to find orbit of earth around the sun Work in two dimensions z and y z’’ = -(GM z) / (y2 + z2)3/2

y’’ = -(GM y) / (y2 + z2)3/2

Constants and initial conditions in report


Results Plot

Solution uses small step size h = 0.01

Becomes difficult to tell difference between methods visually

Using h = 0.1 difference is more marked


Agenda






Agenda






Stiff Equations

Certain systems of ODEs are classified as stiff 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

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


Example

Consider the equation:

With initial conditions:

There are two solutions to this problem

0)1(2

2

ydx

dy

dx

yd

1)0( y

1)0(' y


Example continued

Analytical solution: y=e-x

Unwanted solution: y=e-λx

We will now show how forward Euler is not stable when solving this problem under certain circumstances


Example continued

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


Increasing the range of the x-axis


When h = 0.001


Agenda






Agenda






A Sixth Order Numerical Method

Has an error term with smallest h degree as h7

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


Derivation

Compare Taylor’s Expansion of LHS and RHS So for:

– First expand the LHS


Derivation

• Now expand individual terms on RHS of our expression:

• This gives:


Derivation

Equate both sides and solve for constants:

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


Applying 6th Order Method

How to obtain results using derived method.

Produce a set of simultaneous equations and solve. Find y’n+1 from old values and those just obtained. Set x, yn , y’n etc. to new values. Repeat above procedure with updated variables.


Circle Problem Example

Circle is produced for phase plane plot


Kepler’s Equations

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

The End

But perhaps you want more? Read our report Visit our website:

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

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

Documents

© Imperial College London Numerical Solution of Differential Equations 3 rd year JMC group project ● Summer Term 2004 Supervisor: Prof. Jeff Cash Saeed