Page 298 - Applied Numerical Methods Using MATLAB

BOUNDARY VALUE PROBLEM (BVP) 287

%nm654.m
% to solve a stiff differential eqn called Van der Pol equation
global mu
mu=25, t0=0; tf = 100; tspan = [t0 tf]; xo = [2 0];
[tH1,xH1] = ode_Ham(’df_van’,tspan,x0,8700);
subplot(221), plot(tH1,xH1)
tic,[tH2,xH2] = ode_Ham(’df_van’,tspan,x0,9000); time_Ham = toc
tic,[t45,x45] = ode45(’df_van’,tspan,x0); time_o45 = toc
subplot(222), plot(tH2,xH2), subplot(223), plot(t45,x45)
mu = 200; tf = 200; tspan = [t0 tf];
tic,[t45,x45] = ode45(’df_van’,tspan,x0); time(1) = toc;
tic,[t23,x23] = ode23(’df_van’,tspan,x0); time(2) = toc;
tic,[t15s,x15s] = ode15s(’df_van’,tspan,x0); time(3) = toc;
tic,[t23s,x23s] = ode23s(’df_van’,tspan,x0); time(4) = toc;
tic,[t23t,x23t] = ode23t(’df_van’,tspan,x0); time(5) = toc;
tic,[t23tb,x23tb] = ode23tb(’df_van’,tspan,x0); time(6) = toc;
plot(t45,x45, t23,x23, t15s,x15s, t23s,x23s, t23t,x23t, t23tb,x23tb)
disp(’ ode23 ode15s ode23s ode23t ode23tb’)
time
function dx = df_van(t,x)
%Van der Pol differential equation (6.5.27)
global mu
dx=zeros(size(x));
dx(1) = x(2); dx(2) = mu*(1-x(1).^2).*x(2) - x(1);

6.6 BOUNDARY VALUE PROBLEM (BVP)

A boundary value problem (BVP) is an Nth-order differential equation with some
of the values of dependent variable x(t) and its derivative specified at the initial
time t 0 and others specified at the final time t f .

(2)

[BVP] N : x (N) (t) = f(t, x(t),x (t), x (t), ...,x (N−1) (t))

with the boundary values x(t 1 ) = x 10 ,x (t 2 ) = x 21 ,..., x (N−1) (t N ) = x N,N−1
(6.6.1)
In some cases, some relations between the initial values and the final values may
be given as a mixed-boundary condition instead of the initial/final values spec-
ified. This section covers the shooting method and the finite difference method
that can be used to solve a second-order BVP as

[BVP] 2 : x (t) = f(t, x(t), x (t)) with x(t 0 ) = x 0 ,x(t f ) = x f (6.6.2)
6.6.1 Shooting Method

The idea of this method is to assume the value of x (t 0 ), then solve the differential
equation (IVP) with the initial condition [x(t 0 )x (t 0 )] and keep adjusting the value