Numerical Analysis of Differential Equations Initial Value Problems (I)

Fakult¨ at f¨ ur Mathematik und Informatik 17. April 2012 TU Bergakademie Freiberg

W. Queck/M. Helm

Numerical Analysis of Differential Equations Initial Value Problems (I)

==== Part A – to be prepared before the date of problem session =====

Exercise 1

Reactivate your prior knowledge about systems of linear differential equations of first order with constant coefficients. Solve the following initial value problems.

a)

y

⁰

= y

₁⁰

(t)

y

₂⁰

(t)

= Ay =

3 0 0 −2

y

1

(t) y

2

(t)

,

y

1

(0) y

2

(0)

= 4

7

b)

y

⁰

= y

₁⁰

(t)

y

₂⁰

(t)

= Ay = 1 2

4 3

y

1

(t) y

₂

(t)

,

y

1

(0) y

₂

(0)

= 1

0

Hints:

– Determine the eigenvalues and corresponding eigenvectors of the matrix A first.

– Diagonalize A to D = V

⁻¹

AV , where V is generated by the eigenvectors of A.

– The substitution z = V

⁻¹

y leads to the system z

⁰

= Dz of differential equations. Solve this system and determine y using the backward substitution y = V z.

============ Part B – Main Part ========================

Exercise 2

Determine the exact solution of the following linear initial value problems of first order.

a)

y

⁰₁

(t) y

⁰₂

(t)

= 1 2

4 3

y

1

(t) y

2

(t)

+ 2

0

,

y

1

(0) y

2

(0)

= 1

0

b)

y

₁⁰

(t) y

₂⁰

(t)

=

0 1

−1 2

y

₁

(t) y

₂

(t)

+ e

^2t

0

,

y

₁

(0) y

₂

(0)

= 3

−1

c)



 y

₁⁰

(t) y

₂⁰

(t) y

₃⁰

(t)



 =





−1 1 1

0 −2 0

0 0 −2







 y

₁

(t) y

2

(t) y

3

(t)



 ,



 y

₁

(0) y

2

(0) y

3

(0)



 =



 y

⁰₁

y

⁰₂

y

⁰₃





Initial Value Problems 2 Exercise 3

Rewrite the following higher order initial value problems as first order systems.

(a) y

⁰⁰

− 2y

⁰

+ y = te

^t

− t , 0 ≤ t ≤ 1 y(0) = 0 , y

⁰

(0) = 0 .

(b) y

⁰⁰⁰

= 1

2 y

²

, t ≥ 1

y(1) = 1, y

⁰

(1) = −1, y

⁰⁰

(1) = 2 . (c) t

²

y

⁰⁰

(t) − 2ty

⁰

(t) + 2y(t) = t

³

ln t , t ≥ 1

y(1) = 1, y

⁰

(1) = 0 . Exercise 4

Consider a cone-shaped tank, standing upside down and filled with water flowing out through a circular aperture. The outflow velocity is given by

dx

dt = 0.6πr

²

p

−2g

√ x A(x) ,

where r is the aperture radius, x is the height of fluid surface, and A(x) is the cross sectional area of the tank x units above the aperture.

Assume that r = 3 cm, g = −9.81 m/s

²

, and that the tank has a water level of 2.4 m and a volume of 14.55 m

³

in the beginning.

a) Determine the water level after 10 minutes using a fourth–order Runge–Kutta method with a stepsize of h = 20 s.

Alternatively, you can use another single-step method or an ode... function in Matlab .

b) Determine the time when the tank will be empty up to one minute precision.