





To this point, we have considered IVPs with 1st-order
Consider an nth order IVP which consists of an nth order ODE and n initial conditions:
y(t0) = y0
y(1)(t0) = y0(1)
y(2)(t0) = y0(2)
⋅
⋅
y(n − 1)(t0) = y0(n − 1)
For example,
y(0) = 1
y(1)(0) = 2
or
y(2) = 1
y(1)(2) = 2
y(2)(2) = 3
Suppose we want to find the estimate y1 of y(t1). To do this we will convert the nth-order IVP into a system of n 1st-order IVPs by defining:
y1(t) = y(1)(t)
y2(t) = y(2)(t)
⋅
⋅
yn − 1(t) = y(n − 1)(t)
We note that for i = 0, 1, ..., n − 2, the derivative yi(1)(t) = yi + 1(1)(t) and that yn - 1(1)(t) = f( t, y0(t), y1(t), y2(t), ..., yn − 1(t) ).
The initial conditions may now be converted to
y1(t0) = y0(1)
y2(t0) = y0(2)
⋅
⋅
yn − 1(t0) = y0(n − 1)
If we now define the vectors:
and
and the vector valued function:
then our initial value problem becomes the following vector-valued initial value problem:
y(t0) = y0
where the derivative of the vector y(t) is the vector of element-wise derivatives.
Any of the techniques we have seen, Euler's method, Heun's method, 4th-order Runge Kutta, or the backward-Euler's method may be applied to approximate y(t1).
Given our approximation y1 = (y1, i) of y(t1), we find that y(t1) ≈ y1, 1. We also get, for free, approximations of y(1)(t1) ≈ y1, 2, y(2)(t1) ≈ y1, 3, ..., and y(n − 1)(t1) ≈ y1, n.
Additionally, we may apply any of the multiple-step methods, including Runge Kutta Fehlberg.
Example
If we look at the two problems shown above, we could rewrite the first:
y(0) = 1
y(1)(0) = 2
by defining
where the 2nd-order ODE may now be written as the following system of ODEs
with the initial condition
The second example,
y(2) = 1
y(1)(2) = 2
y(2)(2) = 3
may be rewritten as a system of three functions:
where the 3rd-order ODE may now be written as the following system of ODEs
with the initial condition
We may now solve either of these systems of IVPs using the techniques explained in the previous topic on systems of IVPs.
Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.