Integrating two ODE'S

2 次查看（过去 30 天）
Ian DSouza
Ian DSouza 2017-12-28
评论： Ian DSouza 2017-12-30
I'm trying to integrate these two equations: d/da(rho*a^3) = -3*w*rho*a^2; ((da/dt)/a)^2 + k/a^2 = 8*pi*G/3*rho
where a==a(t), rho==rho(a),w=w(rho).
For now, I am taking 'w' to b a constant. I found out d(rho)/dt = d(rho)/da * da/dt to get 'ode1' and expressed rhoas a function of time 't'. 'ode2' is just the second equation.
Code:
---------------------
syms a(t) rho(t) w(rho)
G=1;
w=1/3;
k=0;
ode1 = diff(rho) == -3*rho*(w+1)*sqrt(8*pi*G/3*rho);
ode2 = (diff(a)/a)^2 + k/a^2 == 8*pi*G/3*rho;
odes = [ode1;ode2];
Sol = dsolve(odes)
aSol(t) = Sol.a;
rhoSol(t) = Sol.rho
---------------------------- Output:
Sol =
struct with fields:
a: [1×1 sym]
rho: [1×1 sym]
aSol(t) =
C5
rhoSol(t) =
0
K>>
This is of course not the solution. Can someone please tell me where I'm going wrong. Thanks.
I am open to numerically integrating these as well
  2 个评论
David Goodmanson
David Goodmanson 2017-12-30
Hello Ian, For w = constant, since rho = rho(a) only depends on time implicitly through a(t), it appears that the solution to the first equation is simply
rho = C*a^(-3(w+1))
and then a(t) can be calculated from the second equation.
Ian DSouza
Ian DSouza 2017-12-30
Thanks David. Yes,that is the solution for w=constant. I decided to keep the ODE because I wanted to eventually vary 'w' as a function of rho. But I got it to work using ODE45 now.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Ordinary Differential Equations 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by