Code covered by the BSD License  

Highlights from
Chebfun V4

image thumbnail

Chebfun V4

by

 

30 Apr 2009 (Updated )

Numerical computation with functions instead of numbers.

Editor's Notes:

This file was selected as MATLAB Central Pick of the Week

Wikipedia ODE examples

Wikipedia ODE examples

Mark Richardson, 27 September 2010

Contents

(Chebfun example ode/WikiODE.m)

Here, we solve three simple problems considered in the Wikipedia article on ODEs:

http://en.wikipedia.org/wiki/Linear_differential_equation

The problems are solved in the order they appear in the article, with boundary conditions imposed to make the solutions unique.

Problem 1: Second-order problem

L(y)  =   y'' - 4y' + 5y = 0
                   y(-1) = exp(-2)*cos(-1)
                    y(1) = exp(2)*cos(1)

Begin by defining the domain d, chebfun variable x and operator N.

d = [-1 1];
x = chebfun('x',d);
N = chebop(d);

The problem has Dirichlet boundary conditions.

N.lbc = exp(-2)*cos(-1);
N.rbc = exp(2)*cos(1);

Define the linear operator.

N.op = @(y) diff(y,2) - 4*diff(y,1) + 5*y;

Define the RHS of the ODE.

rhs = 0*x;

Solve the ODE using backslash.

y = N\rhs;

Analytic solution.

y_exact = exp(2*x).*cos(x);

How close is the computed solution to the true solution?

norm(y-y_exact)
ans =
   6.4927e-12

Plot the computed solution.

plot(y,'linewidth',2), grid on

Problem 2: Simple Harmonic Oscillator

L(y)  =   y'' + pi^2*y =  0
                 y(-1) = -1
                 y'(1) = -pi
d = [-1 1];
x = chebfun('x',d);
N = chebop(d);

This problem has a Dirichlet BC on the left,

N.lbc = -1;

and a Neumann condition on the right.

N.rbc = @(u) diff(u) + pi;

Define the linear operator.

N.op = @(y) diff(y,2) + pi^2*y;

Define the RHS of the ODE.

rhs = 0*x;

Solve the ODE using backslash.

y = N\rhs;

Analytic solution.

y_exact = cos(pi*x)+sin(pi*x);

How close is the computed solution to the true solution?

norm(y-y_exact)
ans =
   4.5112e-13

Plot the computed solution.

plot(y,'linewidth',2), grid on

Problem 3: First-order problem

     L(y)  =  y' + 3*y  = 2
                   y(0) = 2
d = [0 1];
x = chebfun('x',d);
N = chebop(d);

First-order problems require only one boundary condition.

N.lbc = 2;

Define the linear operator.

N.op = @(y) diff(y) + 3*y - 2;

Define the RHS of the ODE.

rhs = 0*x;

Solve the ODE using backslash.

y = N\rhs;

Analytic solution, usually found with integrating factors.

y_exact = 2/3 + 4/3*exp(-3*x);

How close is the computed solution to the true solution?

norm(y-y_exact)
ans =
   1.8245e-15

Plot the computed solution

plot(y,'linewidth',2), grid on

Contact us