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23.1 Ordinary Differential Equations
The function lsode
can be used to solve ODEs of the form
using Hindmarsh's ODE solver LSODE.
 Loadable Function: [x, istate, msg] = lsode (fcn, x_0, t, t_crit)
Solve the set of differential equations
dx  = f(x, t) dt
with
x(t_0) = x_0
The solution is returned in the matrix x, with each row corresponding to an element of the vector t. The first element of t should be t_0 and should correspond to the initial state of the system x_0, so that the first row of the output is x_0.
The first argument, fcn, is a string, inline, or function handle that names the function f to call to compute the vector of right hand sides for the set of equations. The function must have the form
xdot = f (x, t)
in which xdot and x are vectors and t is a scalar.
If fcn is a twoelement string array or a twoelement cell array of strings, inline functions, or function handles, the first element names the function f described above, and the second element names a function to compute the Jacobian of f. The Jacobian function must have the form
jac = j (x, t)
in which jac is the matrix of partial derivatives
 df_1 df_1 df_1     ...    dx_1 dx_2 dx_N     df_2 df_2 df_2     ...   df_i  dx_1 dx_2 dx_N  jac =  =   dx_j  . . . .   . . . .   . . . .     df_N df_N df_N     ...    dx_1 dx_2 dx_N 
The second and third arguments specify the initial state of the system, x_0, and the initial value of the independent variable t_0.
The fourth argument is optional, and may be used to specify a set of times that the ODE solver should not integrate past. It is useful for avoiding difficulties with singularities and points where there is a discontinuity in the derivative.
After a successful computation, the value of istate will be 2 (consistent with the Fortran version of LSODE).
If the computation is not successful, istate will be something other than 2 and msg will contain additional information.
You can use the function
lsode_options
to set optional parameters forlsode
.
 Loadable Function: lsode_options (opt, val)
When called with two arguments, this function allows you set options parameters for the function
lsode
. Given one argument,lsode_options
returns the value of the corresponding option. If no arguments are supplied, the names of all the available options and their current values are displayed.Options include

"absolute tolerance"
Absolute tolerance. May be either vector or scalar. If a vector, it must match the dimension of the state vector.

"relative tolerance"
Relative tolerance parameter. Unlike the absolute tolerance, this parameter may only be a scalar.
The local error test applied at each integration step is
abs (local error in x(i)) <= ... rtol * abs (y(i)) + atol(i)

"integration method"
A string specifying the method of integration to use to solve the ODE system. Valid values are
 "adams"
 "nonstiff"
No Jacobian used (even if it is available).
 "bdf"
 "stiff"
Use stiff backward differentiation formula (BDF) method. If a function to compute the Jacobian is not supplied,
lsode
will compute a finite difference approximation of the Jacobian matrix.

"initial step size"
The step size to be attempted on the first step (default is determined automatically).

"maximum order"
Restrict the maximum order of the solution method. If using the Adams method, this option must be between 1 and 12. Otherwise, it must be between 1 and 5, inclusive.

"maximum step size"
Setting the maximum stepsize will avoid passing over very large regions (default is not specified).

"minimum step size"
The minimum absolute step size allowed (default is 0).

"step limit"
Maximum number of steps allowed (default is 100000).

Here is an example of solving a set of three differential equations using
lsode
. Given the function
function xdot = f (x, t) xdot = zeros (3,1); xdot(1) = 77.27 * (x(2)  x(1)*x(2) + x(1) \  8.375e06*x(1)^2); xdot(2) = (x(3)  x(1)*x(2)  x(2)) / 77.27; xdot(3) = 0.161*(x(1)  x(3)); endfunction 
and the initial condition x0 = [ 4; 1.1; 4 ]
, the set of
equations can be integrated using the command
t = linspace (0, 500, 1000); y = lsode ("f", x0, t); 
If you try this, you will see that the value of the result changes dramatically between t = 0 and 5, and again around t = 305. A more efficient set of output points might be
t = [0, logspace (1, log10(303), 150), \ logspace (log10(304), log10(500), 150)]; 
See Alan C. Hindmarsh, ODEPACK, A Systematized Collection of ODE
Solvers, in Scientific Computing, R. S. Stepleman, editor, (1983) for
more information about the inner workings of lsode
.
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