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**performs numerical integration of vector
differential equations of the form**

**CALL ODE(***, "dername", , , ,***J=**"jacobian"<,**EPS=**eps<, "data">**);**

The ODE subroutine returns the following values:

- is a numeric matrix that contains the results of the integration over connected subintervals. The number of columns in is equal to the number of subintervals of integration as defined by the argument . In case of any error in the integration on any subinterval, partial results will not be reported in .

The inputs to the ODE subroutine are as follows:

*"dername"*- specifies the name of an IML module
used to evaluate the integrand.
- specifies an initial value vector for the variable .
- specifies a sorted vector that describes the
limits of integration over connected subintervals.
The simplest form of the vector contains only
the limits of the integration on one interval.
The first component of should contain the
initial value, and the second component should
be the final value of the independent variable.
For more advanced usage of the ODE subroutine, the
vector can contain more than two components.
The components of the vector must be sorted in ascending order.
Two consecutive components of the vector
are interpreted as a subinterval.
The ODE call reports the final result of integration
at the right endpoint of each subinterval.
This information is vital if
has internal points of discontinuity.
To produce accurate solutions, it is essential that
you provide the location of these points in the
variable , since the continuity of the forcing
function is vital to the internal control of error.
- specifies a numeric vector that contains three components:
the minimum allowable step size,
*hmin*; the maximum allowable step size,*hmax*; and the initial step size to start the integration process,*hinit*. *"jacobian"*- optionally specifies the name of an IML module
that is used to evaluate the Jacobian analytically.
The Jacobian is the matrix , with

If the*"jacobian"*module is not specified, the ODE call uses a finite-difference method to approximate the Jacobian. The keyword for this option is J. *eps*- specifies a scalar indicating the required accuracy.
It has a default value of 1E-4.
The keyword for this option is EPS.
*data*- (scalar, optional, character, input) a valid predefined SAS data set name that is used to save the successful independent and dependent variables of the integration at each step.

The ODE subroutine is an adaptive, variable order, variable step-size, stiff integrator based on implicit backward-difference methods. Refer to Aiken (1985), Bickart and Picel (1973), Donelson and Hansen (1971), Gaffney (1984), and Shampine (1978). The integrator is an implicit predictor-corrector method that locally attempts to maintain the prescribed precision

Consider the differential equation

/* Define the integrand */ start fun(t,y); v = -t*y; return(v); finish; /* Call ODE */ c = 0.5; t = { 0 1}; h = { 1E-12 1 1E-5}; call ode(r,"FUN",c,t,h); print r[format=E21.14];

In this case, the integration is carried out over to give the value of at . The optional parameter

/* Define the integrand */ start fun(t,y); v = -t*y; return(v); finish; /* Call ODE */ c = 0.5; t = {0 1}; h = {1E-12 1. 1E-5}; call ode(r,"FUN",c,t,h) eps=1E-7; print r[format =E21.14];

Compare this value to E-01 and observe that the result is correct through the sixth decimal digit and has an error relative to 1 that is E.

If the solution was desired at 1 and 2 with an accuracy of 1E-7, you would use the following statements:

/* Define the integrand */ start fun(t,y); v = -t*y; return(v); finish; /* Call ODE */ c = 0.5; t = { 0 1 2}; h = { 1E-12 1. 1E-5}; call ode(r,"FUN",c,t,h) eps=1E-7; print r[format=E21.14];

Note that R contains the solution at in the first column and at in the second column.

Now consider the smoothness of the forcing function . For the purpose of estimating errors, adaptive methods require some degree of smoothness in the function . If this smoothness is not present in over the interior and including the left endpoint of the subinterval, the reported result will not have the desired accuracy. The function must be at least continuous. If the function does not meet this requirement, you should specify the discontinuity as an intermediate point. For example, consider the differential equation

To find the solution at , use the following statements:

/* Define the integrand */ start fun(t,y); if t < 1 then v = t; else v = .5*t*t; return(v); finish; /* Call ODE */ c = 0; t = { 0 2}; h = { 1E-12 1. 1E-5}; call ode(r,"FUN",c,t,h) eps = 1E-12; print r[format =E21.14];

In the preceding case, the integration is carried out over a single interval, . The optional parameter

Note that the value of R does not have the required accuracy (it should contain a 12 decimal-place representation of 5/3), although no error message is produced. The reason is that the function is not continuous at the point . Even the lowest-order method cannot produce a local reliable error estimate near the point of discontinuity. To avoid this problem, you can create subintervals so that the integration is carried out first over and then over . The following code implements this method:

/* Define the integrand */ start fun(t,y); if t < 1 then v = t; else v = .5*t*t; return(v); finish; /* Call ODE */ c = 0; t = { 0 1 2}; h = { 1E-12 1. 1E-5}; call ode(r,"FUN",c,t,h) eps=1E-12; print r[format=E21.14];

The variable R contains the solutions at both and , and the errors are of the specified order. Although there is no interest in the solution at the point , the advantage of specifying subintervals with no discontinuities is that the function is infinitely differentiable in each subinterval.

When is continuous, the ODE subroutine can compute the integration to the specified precision, even if the function is defined piecewise. Consider the differential equation

/* Define the integrand */ start fun(t,y); if t < 1 then v = t; else v = t*t; return(v); finish; /* Call ODE */ c = 0.5; t = { 0 2}; h = { 1E-12 1. 1E-5}; call ode(r,"FUN",c,t,h) eps=1E-12; print r[format=E21.14];

This example compares the ODE call to an eigenvalue decomposition for stiff-linear systems. In the problem

The following statements produce two solutions, one by using the ODE call and the other by using the eigenvalue decomposition:

/* Define the integrand */ start fun(t,x) global(a,count); count = count+1; v = a*x; return(v); finish; /* Define the Jacobian */ start jac(t,x) global(a); v = a; return(v); finish; a = {-1000 -1 -2 -3, -1 -2 3 -1, -2 3 -4 -3, -3 -1 -3 -5 }; /* Call ODE */ count = 0; t = { 0 1 2}; h = {1E-12 1 1E-5}; eps = 1E-9; c = {1, 0, 0, 0 }; call ode(z,"FUN",c,t,h) eps=eps j="JAC"; print z[format=E21.14];

print count;

/* Do the eigenvalue decomposition */ start eval(t) global(d,u,c); v = u*diag(exp(d*t))*u`*c; return(v); finish; call eigen(d,u,a); free z1; do i = 1 to nrow(t)*ncol(t)-1; z1 = z1 || (eval(t[i+1])); end; print z1[format=E21.14];

The question now is whether or not this is an E result. Note that for the problem

The ODE subroutine can fail for problems with unusual qualitative properties, such as finite escape time in the interval of integration (that is, the solution goes towards infinity at some finite time). In such cases, try testing with different subintervals and different levels of accuracy to gain some qualitative information about the behavior of the solution of the differential equation.

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