<!–html_preserve–>RxODE<!–/html_preserve–>

Stiff ODEs with Jacobian Specification

Occasionally, you may come across a stiff differential equation, that is a differential equation that is numerically unstable and small variations in parameters cause different solutions to the ODEs. One way to tackle this is to choose a stiff-solver, or hybrid stiff solver (like the default LSODA). Typically this is enough. However exact Jacobian solutions may increase the stability of the ODE. (Note the Jacobian is the derivative of the ODE specification with respect to each variable). In RxODE you can specify the Jacobian with the df(state)/dy(variable)= statement. A classic ODE that has stiff properties under various conditions is the van dar Pol differential equations.

In RxODE these can be specified by the following:

library(RxODE)

Vtpol2 <- RxODE({
    d/dt(y)  = dy
    d/dt(dy) = mu*(1-y^2)*dy - y
    ## Jacobian
    df(y)/dy(dy)  = 1
    df(dy)/dy(y)  = -2*dy*mu*y - 1
    df(dy)/dy(dy) = mu*(1-y^2)
    ## Initial conditions
    y(0) = 2
    dy(0) = 0
    ## mu
    mu = 1 ## nonstiff; 10 moderately stiff; 1000 stiff
})

et <- eventTable();
et$add.sampling(seq(0, 10, length.out=200));
et$add.dosing(20, start.time=0);

s1 <- Vtpol2 %>%  solve(et, method="lsoda")
rxHtml(s1)

Solved RxODE object
Parameters ($params):
mu
1
Initial Conditions ( $inits):
y dy
2 0
First part of data (object):
time y dy
0.0000000 22.00000 0.0000000
0.0502513 21.99781 -0.0455530
0.1005025 21.99552 -0.0455578
0.1507538 21.99323 -0.0455625
0.2010050 21.99094 -0.0455673
0.2512563 21.98864 -0.0455721

While this is not stiff at mu=1, mu=1000 is a stiff system

s2 <- Vtpol2 %>%  solve(c(mu=1000), et)
rxHtml(s2)

Solved RxODE object
Parameters ($params):
mu
1000
Initial Conditions ( $inits):
y dy
2 0
First part of data (object):
time y dy
0.0000000 22.00000 0.00e+00
0.0502513 22.00000 -4.55e-05
0.1005025 22.00000 -4.55e-05
0.1507538 21.99999 -4.55e-05
0.2010050 21.99999 -4.55e-05
0.2512563 21.99999 -4.55e-05

While this is easy enough to do, it is a bit tedious. If you have RxODE setup appropriately, that is you have:

You can use the computer algebra system sympy to calculate the Jacobian automatically.

This is done by the RxODE option calcJac option:

Vtpol <- RxODE({
    d/dt(y)  = dy
    d/dt(dy) = mu*(1-y^2)*dy - y
    ## Initial conditions
    y(0) = 2
    dy(0) = 0
    ## mu
    mu = 1 ## nonstiff; 10 moderately stiff; 1000 stiff
}, calcJac=TRUE)

To see the generated model, you can use rxCat:

> rxCat(Vtpol)
d/dt(y)=dy;
d/dt(dy)=mu*(1-y^2)*dy-y;
y(0)=2;
dy(0)=0;
mu=1;
df(y)/dy(y)=0;
df(dy)/dy(y)=-2*dy*mu*y-1;
df(y)/dy(dy)=1;
df(dy)/dy(dy)=mu*(-Rx_pow_di(y,2)+1);