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Numerical Analysis of Differential Equations Initial Value Problems (II)

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Fakultät für Mathematik und Informatik 29. April 2013 TU Bergakademie Freiberg

Prof. Dr. O. Rheinbach/Dr. M. Helm

Numerical Analysis of Differential Equations Initial Value Problems (II)

Exercise 1

Determine an approximate solution for the initial value problem y0=−(y+ 1)(y+ 3), 0≤t≤2 y(0) =−2

applying

a) the explicitEulerscheme, b) the modifiedEulermethod, c) the improvedEuler method

with step sizeh= 0.5 andh= 0.2. Compare your approximations with the corresponding values of the exact solutiony(t) =−3 + 2(1 +e−2t)−1.

Hint: For the improved Euler method take yn+1 =yn+h2[f(tn, yn) +f(tn+h, yn+hf(tn, yn))]

withy0=y(0) given.

Exercise 2

Rewrite each of the followingButcher/Runge-Kuttatableaus as a set of explicit formulae for the calculation ofyj+1. Do they describe explicit or implicit methods?

a) four-stageEnglandformula 0

1/2 1/2 1/2 1/4 1/4

1 0 -1 2

1/6 0 2/3 1/6

b) Butcher’s formula 0

1/8 1/8

1/4 0 1/4

1/2 1/2 -1 1

3/4 3/16 0 0 9/16

1 -5/7 4/7 12/7 -12/7 8/7

7/90 0 32/90 12/90 32/90 7/90 c) (one)RadauIIA method

1/3 5/12 -1/12

1 3/4 1/4

3/4 1/4

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Initial Value Problems II 2 Exercise 3

For the solutiony(x)of the initial value problem

y0 =−(y+ 1)(y+ 3), 0≤x≤1 y(0) =−2

find an approximation ofy(1) using the classicalRunge–Kuttamethod with step size h= 0.5.

Compare with the solutions in Exercise 1.

Exercise 4

Determine an approximate solution to the following system of ODEs

˙ y(t) =

6 −3 2 1

y(t), y(0) = 5

3

, 0≤t≤1.

Therefor apply

a) the explicitEulermethod with step sizeh= 0.25, b) Heun’s method with step size h= 0.5.

Compare the approximations with the exact solution y(t) =

exp(3t) 3 exp(4t) exp(3t) 2 exp(4t)

−1 2

.

Hint: You have to use vector versions of both methods, for instanceyn+1 =yn+hf(tn,yn) and y0=y(0)in the Euler case.

Exercise 5

Check the consistency order of Heun’s method by verification of the order conditions (for explicit Runge-Kuttamethods up to order three) known from the lecture.

0 1/3 1/3

2/3 0 2/3 0

1/4 0 3/4

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