robousy
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Problem solving 2nd order ODE...not for the faint of heart!
Hey folks,
I'm having problems solving the following set of ODE's:
[tex]3H_a^2+H_b^2+6H_aH_b=k_1\rho[/tex] eq.1
[tex]\dot{H_a}+3H_a^2+2H_aH_b=k_2\rho[/tex] eq.2
[tex]\dot{H_b}+2H_b^2+3H_aH_b=k_3\rho[/tex] eq.3
These are cosmological equations. Note, [itex]\rho=\frac{1}{b^6}(1-b^2+b^4)[/itex], also the H's are Hubbles constant in a and b, eg
[tex]H_a=\frac{\dot{a}}{a}[/tex]
[tex]H_b=\frac{\dot{b}}{b}[/tex]
The a's and b's are functions of t (time) and the k's on the RHS are just constants. I want to put this all together and ultimately plot a as a function of t and b as a function of t.
The equations originate from the paper: http://arxiv.org/abs/0707.1062 , equations 9,10,11 and I am trying to duplicate the plots in fig1.
What I'm thinking:
Solve eqtn 1 for [itex]H_a[H_a] using the quadratic eqtn then plug that into 3 and use DSOLVE in mathematica.<br /> <br /> Can anyone let me know if this is the correct approach. <br /> <br /> Thanks in advance!<br /> <br /> <br /> Richard[/itex]
Hey folks,
I'm having problems solving the following set of ODE's:
[tex]3H_a^2+H_b^2+6H_aH_b=k_1\rho[/tex] eq.1
[tex]\dot{H_a}+3H_a^2+2H_aH_b=k_2\rho[/tex] eq.2
[tex]\dot{H_b}+2H_b^2+3H_aH_b=k_3\rho[/tex] eq.3
These are cosmological equations. Note, [itex]\rho=\frac{1}{b^6}(1-b^2+b^4)[/itex], also the H's are Hubbles constant in a and b, eg
[tex]H_a=\frac{\dot{a}}{a}[/tex]
[tex]H_b=\frac{\dot{b}}{b}[/tex]
The a's and b's are functions of t (time) and the k's on the RHS are just constants. I want to put this all together and ultimately plot a as a function of t and b as a function of t.
The equations originate from the paper: http://arxiv.org/abs/0707.1062 , equations 9,10,11 and I am trying to duplicate the plots in fig1.
What I'm thinking:
Solve eqtn 1 for [itex]H_a[H_a] using the quadratic eqtn then plug that into 3 and use DSOLVE in mathematica.<br /> <br /> Can anyone let me know if this is the correct approach. <br /> <br /> Thanks in advance!<br /> <br /> <br /> Richard[/itex]
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