1. © Imperial College London 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

2. © Imperial College London Agenda Adam: Differential equations and numerical methods Paul: Basic methods (FE, BE, TR). Problem: “ circle ” Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation Minal: Stiff equations Tejas: Derivation of 6 th order method

3. © Imperial College London Agenda Adam: Differential equations and numerical methods Paul: Basic methods (FE, BE, TR). Problem: “ circle ” Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation Minal: Stiff equations Tejas: Derivation of 6 th order method

4. © Imperial College London Introduction to Differential Equations Equations involving a function y of x, and its derivatives Model real world systems General Equation:

5. © Imperial College London Introduction to Differential Equations Simple example Solution obtained by integrating both sides Initial values can determine c

6. © Imperial College London Introduction to Differential Equations Special case when equations do not involve x E.g. Initial values y(0) = 1 and y'(0) = 0, solution is

7. © Imperial College London Introduction to Differential Equations Kepler ’ s Equations of Planetary Motion Difficult to solve analytically – we use numerical methods instead

8. © Imperial College London Introduction to Differential Equations 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 y 1 and repeat Need formula for y n+1 in terms of y n

9. © Imperial College London Agenda Adam: Differential equations and numerical methods Paul: Basic methods (FE, BE, TR). Problem: “ circle ” Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation Minal: Stiff equations Tejas: Derivation of 6 th order method

10. © Imperial College London Agenda Adam: Differential equations and numerical methods  Paul: Basic methods (FE, BE, TR). Problem: “ circle ” Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation Minal: Stiff equations Tejas: Derivation of 6 th order method

11. © Imperial College London Euler ’ s Methods Simplest way of solving ODEs numerically Does not always produce reasonable solutions Forward Euler (explicit) and Backward Euler (implicit)

12. © Imperial College London Forward Euler y(x + h) = y(x) + h y’(x) Explicit E loc = ½h 2 y (2) (  )  h 2 First order Asymmetric x0x0 x1x1 x2x2 x y y0y0 y1y1 y2y2 hh (x 0, y 0 ) (x 1, y 1 ) (x 2, y 2 ) True solution Approximated solution Slope f(x 1, y 1 ) Slope f(x 0, y 0 )

13. © Imperial College London Backward Euler y(x + h) = y(x) + h y’(x + h) Implicit E loc = - ½h 2 y (2) (  )  h 2 First order Asymmetric x0x0 x1x1 x2x2 x y y0y0 y1y1 y2y2 hh (x 0, y 0 ) (x 1, y 1 ) (x 2, y 2 ) True solution Approximated solution Slope f(x 2, y 2 ) Slope f(x 1, y 1 )

14. © Imperial College London Trapezium Rule Average of FE and BE Implicit E loc = - (h 3 /12) f (2) (  )  h 3 Second order Symmetric y(x + h) = y(x) + ½ h [y’(x) + y’(x + h)]

15. © Imperial College London 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

16. © Imperial College London Plots for the True Solution Time series plot (xy)Phase plane plot (zy)

17. © Imperial College London Time Series y x

18. © Imperial College London Phase Plane Plots: FE & BE Forward EulerBackward Euler

19. © Imperial College London Phase Plane Plots: TR FE & BE fail: non-periodic TR: OK, periodic sol n W h y ?

20. © Imperial College London Symmetricity TR is symmetric, whereas FE & BE are not y(x + h) = y(x) + ½ h [y’(x) + y’(x + h)] h  -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!

21. © Imperial College London 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?

22. © Imperial College London Time Step h = 10 -1 h = 10 -2

23. © Imperial College London Agenda Adam: Differential equations and numerical methods  Paul: Basic methods (FE, BE, TR). Problem: “ circle ” Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation Minal: Stiff equations Tejas: Derivation of 6 th order method

24. © Imperial College London Agenda Adam: Differential equations and numerical methods  Paul: Basic methods (FE, BE, TR). Problem: “ circle ”  Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation Minal: Stiff equations Tejas: Derivation of 6 th order method

25. © Imperial College London 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

26. © Imperial College London Two Fourth Order Methods y n+1 – y n = y n + h/2(y’ n ) + (h 2 /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) y n+ ½ = ½(y n+1 + y n ) – h 2 /48(y’’ n+1 + 4y’’ n+½ + y’’ n ) – B) y n+ ½ = y n + ½y’ n – h 2 /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 (*)

27. © Imperial College London Creating Method A 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) 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

28. © Imperial College London Comparing Fourth Order Methods Comparing errors when solving circle problem Both A and B produce a much smaller order of error than Euler ’ s

29. © Imperial College London 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) / (y 2 + z 2 ) 3/2 y’’ = -(GM y) / (y 2 + z 2 ) 3/2 Constants and initial conditions in report

30. © Imperial College London 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

31. © Imperial College London Agenda Adam: Differential equations and numerical methods  Paul: Basic methods (FE, BE, TR). Problem: “ circle ”  Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation Minal: Stiff equations Tejas: Derivation of 6 th order method

32. © Imperial College London Agenda Adam: Differential equations and numerical methods  Paul: Basic methods (FE, BE, TR). Problem: “ circle ”  Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation  Minal: Stiff equations Tejas: Derivation of 6 th order method

33. © Imperial College London 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

34. © Imperial College London Example Consider the equation: With initial conditions: There are two solutions to this problem

35. © Imperial College London 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

36. © Imperial College London Example continued Let λ = 10 3 and let (the step size) h = 0.1

37. © Imperial College London Increasing the range of the x-axis

38. © Imperial College London When h = 0.001

39. © Imperial College London Agenda Adam: Differential equations and numerical methods  Paul: Basic methods (FE, BE, TR). Problem: “ circle ”  Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation  Minal: Stiff equations Tejas: Derivation of 6 th order method

40. © Imperial College London Agenda Adam: Differential equations and numerical methods  Paul: Basic methods (FE, BE, TR). Problem: “ circle ”  Saeed: Advanced methods (4 th order). Problem: Kepler ’ s equation  Minal: Stiff equations  Tejas: Derivation of 6 th order method

41. © Imperial College London A Sixth Order Numerical Method Has an error term with smallest h degree as h 7 Idea is to find values for α, A, B, C, D, E, F, C, D, E, F such that:

42. © Imperial College London Derivation Compare Taylor ’ s Expansion of LHS and RHS So for: – First expand the LHS

43. © Imperial College London Derivation Now expand individual terms on RHS of our expression: This gives:

44. © Imperial College London Derivation Equate both sides and solve for constants: Similarly we can find C, D, E, F and C, D, E, F

45. © Imperial College London Applying 6 th 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, y n, y ’ n etc. to new values. Repeat above procedure with updated variables.

46. © Imperial College London Circle Problem Example Circle is produced for phase plane plot

47. © Imperial College London Kepler ’ s Equations Solution obtained from z-y plot reinforces fourth order method results.

48. 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