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

Question

Long Cheng 2016-6-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/291763-how-to-add-two-loops-for-two-variables-in-a-nonlinear-ode

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?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Marc 2016-6-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/291763-how-to-add-two-loops-for-two-variables-in-a-nonlinear-ode#answer_226411

在 MATLAB Online 中打开

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...

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Long Cheng 2016-6-23

Thank you! Let me give a try :)

请先登录，再进行评论。

Answer 2

Roger Stafford 2016-6-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/291763-how-to-add-two-loops-for-two-variables-in-a-nonlinear-ode#answer_226429

在 MATLAB Online 中打开

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.)