solve a complex second order differential equation

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嘉杰 程
嘉杰 程 el 20 de Oct. de 2021
Comentada: 嘉杰 程 el 22 de Oct. de 2021
the ode has a form:
and for, given , how could I use ode45 to solve it with plot? thx

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Bjorn Gustavsson
Bjorn Gustavsson el 20 de Oct. de 2021
Editada: Bjorn Gustavsson el 20 de Oct. de 2021
The first step is to convert your second-order ODE to two coupled first-order ODEs:
Then you should write that ODE-system as a matlab-function:
function [dphidt_domegadt] = yourODEs(t,phi_w)
phi = phi_w(1);
w = phi_w(2);
dphidt = w;
if t == 0 % Here I assume that domegadt/t goes to zero as t -> 0+, perhaps there are solutions for other finite values of that ratio...
domegadt = phi^3;
else
domegadt = -2/t*dphidt - phi^3;
end
dphidt_domegadt = [dphidt;
domegadt];
This should be possible to integrate with ode45:
phi0w0 = [1 0];
t_span = [0 exp(2)]; % some limits of yours
[t,phi_w] = ode45(@(t,phi_w) yourODEs(t,phi_w),t_span,phi0w0);
HTH
  9 comentarios
嘉杰 程
嘉杰 程 el 21 de Oct. de 2021
every t determine certain curvature of phi and eta, when eta=0, phi(eta)=1, phi'(eta)=0, that's one side, another side is determined by t
嘉杰 程
嘉杰 程 el 22 de Oct. de 2021
actually this is only a specific function when n=3

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Walter Roberson
Walter Roberson el 20 de Oct. de 2021
You cannot use any numeric solver for that. You have initial conditions at η = 0, but at 0 you have a division by 0 which gets you a numeric infinity. That numeric infinity is multiplied by the boundary condition of 0, but numeric infinity times numeric 0 gives you NaN, not 0.
If you work symbolically you might think that the infinity and the 0 cancel out, but that only works if the φ' approaches 0 faster than 1/η approaches infinity, which is something that we do not immediately know to be true.
  3 comentarios
嘉杰 程
嘉杰 程 el 20 de Oct. de 2021
If I let to be a quite small number?
Bjorn Gustavsson
Bjorn Gustavsson el 20 de Oct. de 2021
That is not enough. The ratio of 1/t*dphi/dt has to behave well for t = 0.

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