The Computable Differential Equation Lecture ... - Bruce E. Shapiro

154 CHAPTER 8. BOUNDARY VALUE PROBLEMS
We will modify the Euler program slightly to take the number of mesh points and
the two boundary points – rather than the step size – as parameter. This is because
we want to force the problem to stop at the second boundary point. The following
code will return a single Euler solution for the variable u (we don’t really care about
v since it was not in the original problem).
EulerIVP[f , ya , a , b , n ] :=
Module[{ t, y, h, solution},
h = (b - a)/n;
solution = {{a, ya[[1]]}};
y = ya;
For[t = a, t < b, t += h,
y = y + h*f[t, y];
AppendTo[solution, {t + h, y[[1]]}];
];
Return[solution];
]
Suppose we want to estimate the solution using a mesh with 100 intervals. Then
our first guess would be obtained with
f[t , u , v ] := {v, u/(t*t) - v/t}
EulerIVP[f, 1, 0, 1, 2., 100]
which will return
{{1, 1}, {1.01,1}, {1.02, 1.0001},...,{2.,1.25041}}
where the intervening values have been omitted. The very last value is what we are
interested in: our first guess gives u(2) (0) = 1.25041. Our goal is to iterate on this
value: if we have two guesses at c and the corresponding two guess at u(2), we can
use linear interpolation to make a third guess:
m (i) = u(b)(i) − u(b) (i−1)
c (i) − c (i−1) (8.9)
c (i+1) = c (i) +
u(b) − u(b)(i)
m
(8.10)
The problem is that we don’t have a good second estimate for c, so we pick one
arbitrarily, namely
c (1) = c (0) + 1 = 1 (8.11)
Then
returns
EulerIVP[f, 1, 1, 1, 2., 100]
{{1, 1}, {1.01,1.01}, {1.02, 1.02},...,{2.,2}}
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007