ColoState
Notes on Finite Difference Method
MATH451
for Linear 2nd Order ODE Boundary Value Problems
Jiangguo (James) Liu
∗
March 2, 2016
We consider a typical two-point boundary value problem for a linear 2nd order ODE
( 00
x (t) = u(t) + v(t)x(t) + w(t)x0 (t), t ∈ [a, b]
x(a) = α,
x(b) = β
(1)
After applying respectively the 2nd order and 1st order central finite differences for discretization x00 (ti ), x0 (ti ), we obtain
1
1
− 1 − hwi yi−1 + 2 + h2 vi yi + − 1 + hwi yi+1 = −h2 fi
(2)
2
2
Here is an example with [a, b] = [0, 1], α = 0, β = 1 and
u(t) = −π 2 sin(πt) − 2t(1 + π cos(πt)) − (t2 + 1)(t + sin(πt));
v(t) = t2 + 1,
w(t) = 2t
along with a known exact solution x(t) = t + sin(πt).
FDM for 2nd order ODE BVP: Numerical vs exact solutions
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 1: Time step h = 0.1; ODE: x00 = u(t) + v(t)x + w(t)x0 , BVP: x(0) = 0, x(1) = 1; Exact
solution x(t) = t + sin(πt)
∗
Department of Mathematics,
(liu@math.colostate.edu)
Colorado
State
University,
1
Fort
Collins,
CO 80523-1874,
USA
Shown below is Matlab code.
close all; clear all;
format compact; format long e;
a = 0;
b = 1;
alpha = 0;
beta = 1;
h = 0.1; n = (b-a)/h - 1;
t = a:h:b; % All nodes including boundary nodes
wval
vval
uval
xval
=
=
=
=
2*t;
t.*t + 1; % Componentwise multiplication
-(pi^2)*sin(pi*t) - 2*t.*(1+pi*cos(pi*t)) - (t.*t+1).*(t+sin(pi*t));
t + sin(pi*t);
dtmp
atmp
ctmp
btmp
= 2 + (h*h)*vval;
= -1 - 0.5*h*wval;
= -1 + 0.5*h*wval;
= -(h*h)*uval;
d = dtmp(2:n+1);
a = atmp(3:n+1);
c = ctmp(2:n);
b = btmp(2:n+1);
b(n) = b(n) - ctmp(n+1)*beta;
% Applying bndry cond. at the right endpoint
% Solving the linear system
A = diag(d,0) + diag(a,-1) + diag(c,1);
NumerSln = A\b’;
%
ExactSln = xval(2:n+1)’;
norm(ExactSln-NumerSln,inf)
% Padding with boundary values
x = [alpha; ExactSln; beta];
y = [alpha; NumerSln; beta];
figure(1);
plot(t,x,’k-’,t,y,’r*’);
title(’FDM 2nd order ODE BVP: Numerical vs exact solutions’);
return;
2