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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): | |||||||||||||||||||||||||
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Initial Conditions ( $inits): | |||||||||||||||||||||||||
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While this is not stiff at mu=1, mu=1000 is a stiff system
s2 <- Vtpol2 %>% solve(c(mu=1000), et)
rxHtml(s2)
Solved RxODE object | |||||||||||||||||||||||||
Parameters ($params): | |||||||||||||||||||||||||
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Initial Conditions ( $inits): | |||||||||||||||||||||||||
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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);