symDynJac

Return the symbolically generated partial derivatives (required for sensitivity) from an ODE function handle

Syntax

[dfdz,dfdp] = symDynJac(ode)

[dfdz,dfdp] = symDynJac(ode,nstates)

[dfdz,dfdp] = symDynJac(ode,nstates,nparam)

[dfdz,dfdp,sym_dfdz,sym_dfdp] = symDynJac(ode)

Description

[dfdz,dfdp] = symDynJac(ode) converts the function handle ode into a symbolic expression, calculates the partial derivatives using the Symbolic Toolbox command Jacobian, then converts each derivative expression back to a function handle as dfdz (derivative with respect to states, z) and dfdp (derivative with respect to parameters, p).

[dfdz,dfdp] = symDynJac(ode,nstates) ensures the correct number (nstates) of states is used in the resulting Jacobian. This is useful if your function does not include all indices of z.

[dfdz,dfdp] = symDynJac(ode,nstates,nparam) ensures the correct number (nparam) of parameters is used in the resulting Jacobian. This is useful if your function does not include all indices of p.

[dfdz,dfdp,sym_dfdz,sym_dfdp] = symDynJac(ode) also returns the intermediate symbolic expressions of the partial derivatives.

Important Notes

This function uses the Matlab Symbolic Toolbox^® thus you must have this installed to use this function. This is also a very basic routine thus you may wish to confirm the solution via one of the other differentiation routines provided.

The form of your function handle must be as follows:

ode = @(t,z,p) z(1) + p(1)*z(2); where the function is only a function of t,z, and p, and all variables are indexed into the expression (no vectorized calls).

Example

>> ode = @(t,z,p) [z(2);
p(1)*(1-z(1)^2)*z(2) - z(1)];

>> [dfdz,dfdp] = symDynJac(ode)

dfdz =

@(t,z,p)[[0,1];[-2*p(1)*z(1)*z(2)-1,-p(1)*(z(1)^2-1)]]

dfdp =

@(t,z,p)[[0];[-z(2)*(z(1)^2-1)]]