Two-Point Boundary Value Problems - Numerical Methods in Engineering with MATLAB - page 299

3. Having determined the valueof u , solve the differentialequationsonce more and

record the results.

If the differentialequationislinear, any root-finding methodwill need only one
interpolation to determine u .Butsince Brent's methoduses quadratic interpolation,
it needsthree points: u 1 , u 2 and u 3 , the latterbeing providedbyabisection step. This
is wasteful, since linear interpolationwith u 1 and u 2 wouldalso result in the correct
valueof u . Therefore, we replace Brent's methodwith linear interpolationwhenever
the differentialequationislinear.

linInterp

Here is the algorithm for linear interpolation:

function root = linInterp(func,x1,x2)
% Finds the zero of the linear function f(x) by straight
% line interpolation between x1 and x2.
% func = handle of function that returns f(x).

f1 = feval(func,x1); f2 = feval(func,x2);
root=x2-f2*(x2-x1)/(f2-f1);

EXAMPLE 8.1
Solve the nonlinear boundary value problem

y + 3 yy = 0 y (0) = 0 y (2) = 1

Solution The equivalent first-order equations are

y 1
y 2

y 2
−

y = =

3 y 1 y 2

with the boundary conditions

y 1 (0) = 0 y 1 (2) = 1

y (0), the
unspecified initialcondition. Wecouldalways pick two numbers at randomand hope
for the best. However, it is possible to reduce the elementofchance with a little
detective work.Westart by making the reasonable assumption that y issmooth (does
not wiggle) in the interval0

Now comes the daunting task of estimating the trial values of y 2 (0) =

≤ ≤

x 2. Next we note that y hastoincrease from 0to1,

which requires y > 0.Since both y and y are positive, weconcludethat y must be