3 Systems of second order differential equations

The decoupling method discussed above can be readily extended to this situation which could arise, for example, in a mechanical system consisting of coupled springs.

A typical example of such a system with two unknowns has the form

ẍ = a x + b y ÿ = c x + d y

or, in matrix form,

Ẍ = A X where X = x y A = a b c d , ẍ = d 2 x d t 2 , ÿ = d 2 y d t 2

Task!

Make the substitution X = P Y where Y = r ( t ) s ( t ) and P is the modal matrix of A , A being assumed here to have distinct eigenvalues λ 1 and λ 2 . Solve the resulting pair of decoupled equations for the case, which arises in practice, where λ 1 and λ 2 are both negative.

Exactly as with a first order system, putting X = P Y into the second order system   Ẍ = A X gives

Ÿ = P − 1 A P Y that is Ÿ = D Y where D = λ 1 0 0 λ 2  and Ÿ = r ̈ s ̈  so

r ̈ s ̈ = λ 1 0 0 λ 2 r s

That is, r ̈ = λ 1 r = − ω 1 2 r and s ̈ = λ 2 s = − ω 2 2 s (where λ 1 and λ 2 are both negative.)

The two decoupled equations are of the form of the differential equation governing simple harmonic motion. Hence the general solution is

r = K cos ω 1 t + L sin ω 1 t and s = M cos ω 2 t + N sin ω 2 t

The solutions for x and y are then obtained by use of  X = P Y .

Note that in this second order case four initial conditions, two each for both x and y , are required because four constants K , L , M , N arise in the solution.

Exercises
  1. Solve by decoupling each of the following first order systems:
    1. d X d t = A X where A = 3 4 4 − 3 , X ( 0 ) = 1 3

    2. ẋ 1 = x 2   ẋ 2 = x 1 + 3 x 3 ẋ 3 = x 2 with x 1 ( 0 ) = 2 , x 2 ( 0 ) = 0 , x 3 ( 0 ) = 2
    3. d X d t = 2 2 1 1 3 1 1 2 2 X , with X ( 0 ) = 1 0 0
    4. ẋ 1 = x 1 ẋ 2 = − 2 x 2 + x 3 ẋ 3 = 4 x 2 + x 3 with x 1 ( 0 ) = x 2 ( 0 ) = x 3 ( 0 ) = 1
  2. Matrix methods can be used to solve systems of second order differential equations such as might arise with coupled electrical or mechanical systems. For example the motion of two masses m 1 and m 2 vibrating on coupled springs, neglecting damping and spring masses, is governed by m 1 ÿ 1 = − k 1 y 1 + k 2 ( y 2 − y 1 ) m 2 ÿ 2 = − k 2 ( y 2 − y 1 )

    where dots denote derivatives with respect to time.

    Write this system as a matrix equation Ÿ = A Y and use the decoupling method to find Y if

    1. m 1 = m 2 = 1 , k 1 = 3 , k 2 = 2

      and the initial conditions are y 1 ( 0 ) = 1 , y 2 ( 0 ) = 2 , ẏ ( 0 ) = − 2 6 , ẏ 2 ( 0 ) = 6

    2. m 1 = m 2 = 1 , k 1 = 6 , k 2 = 4

      and the initial conditions are y 1 ( 0 ) = y 2 ( 0 ) = 0 , ẏ 1 ( 0 ) = 2 , ẏ 2 ( 0 ) = 2 2

    Verify your solutions by substitution in each case.

    1. X = 2 e 5 t − e − 5 t e 5 t + 2 e − 5 t
    2. X = 2 cosh 2 t 4 sinh 2 t 2 cosh 2 t
    3. X = 1 4 e 5 t + 3 e t e 5 t − e t e 5 t − e t
    4. X = 1 5 5 e t 2 e 2 t + 3 e − 3 t 8 e 2 t − 3 e − 3 t
    1. Y = cos t − 2 sin 6 t 2 cos t + sin 6 t
    2. Y = sin 2 t 2 sin 2 t