How to add two loops for two variables in a nonlinear ode

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Long Cheng
Long Cheng 2016-6-23
评论： Long Cheng 2016-6-23
For a nonlinear ode:
dy/dt=a*(1-10^(b*y))
How to solve this ode (i.e y) for different conditions, for example:
a=0:1:10 and b=0:1:10 at t=10?

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回答（2 个）

Marc
Marc 2016-6-23
See nested functions...
Basically you need to define a function that takes a and b as inputs, solves the ode with one of the ode solvers with the function nested. Anything outside the nested function, a and b are 'known' to the nested function.
function [yout, tout]= odeIn(a, b)
t = your time interval
Define your initial cond
yout = ode15s(yournestedfunc, t, initial cond)
function dydt = yournestedfunc(t)
your function with a and b
end
end
Something like that... Or you can turn your function into an anonymous function that calls a and b as variables. There is a good section in the help documentation on this...
Roger Stafford
Roger Stafford 2016-6-23
This particular equation can be solved by ordinary calculus methods. It has the three possible analytic forms:
1) For y > 0, y = 1/(b*log(10))*log(1/(1-exp(-c*(t+K))))
2) y = 0, a constant in time
3) For y < 0, y = 1/(b*log(10))*log(1/(1+exp(-c*(t+K))))
where c = a*b*log(10) and where K is selected to satisfy whatever initial condition you apply to y.
You can derive these solutions by making the substitution
z = 10^(b*y)
and the resulting differential equation in z can readily be solved:
1/(z*(1-z))*dz/dt = a*b*log(10)
Evaluating the above expressions directly is presumably easier than making use of the numerical ‘ode’ functions in matlab (once you are convinced they are correct.)

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