The function lsode
can be used to solve ODEs of the form
dx -- = f (x, t) dt
using Hindmarsh's ODE solver Lsode.
Solve the set of differential equations
dx -- = f(x, t) dtwith
x(t_0) = x_0The 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 that names the function 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 two-element string array, 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 intial 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
.See also: daspk, dassl, dasrt, odessa.
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 specifing the method of integration to use to solve the ODE system. Valid values are
- "adams"
- "non-stiff"
- 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) \ - 8.375e-06*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
.