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Analytically solving this system of 2nd order PDEs

Posted 2 days ago

Hello,

I try to solve this system of 2nd order differential equations in variables $x(t,k),y(t,k)$ and $z(t,k)$

$k^2y + 3 H \dot{y} = \frac{1}{2z_0} ( H \dot{z} - \dot{z} ) + \frac{\dot{z}_0}{2z_0} ( 2 H z + \dot{z} ), ~~~~~~~~ (1)$

$H x + \dot{y} =- \frac{a^2}{2 z_0} ( \dot{z}- H z ) + \frac{a^2 \dot{z}_0}{2 z_0} x, ~~~~~~~~ (2)$

$\left[\ddot{y} + H (2 \dot{y} + \dot{x}) + ( 2 \dot{H} + H^2 )x \right] \delta_{ij} + 1/2 ( \Delta \delta_{ij} - \partial_i \partial_j ) ( x-y) = \frac{a^2}{2z_0} ( \partial_i \partial_j z - H \dot{z} \delta_{ij}) ~~~~~~~~~ (3)$

Taking the trace of (3) yields:

$ \ddot{y} + H ( 2 \dot{y} + \dot{z} ) + ( 2 \dot{H} + H^2 ) x = - \frac{a^2}{2z_0} H \dot{z}, ~~~~~~~~ (3a) $

While the off-diagonal terms are:

$x= y- \frac{a^2 z}{2z_0} ~~~~~~~ (3b)$

And the last equation

$\ddot{z} - H \dot{z} + 3k^2 z + 2 (x-3y) \ddot{z}_0 - 2 H (x-3y) z_0 + 2 (\dot{x} - 3 \dot{y} ) \dot{z}_0 =0 ~~~~~~~~~~(4)$

where ( $\dot{z}$) means the derivative with respect to time. $H(t), a(t), z0((t)$ are given functions as in the attached MA file

Now since I try to get three unknown functions, one should to solve only three equations to prevent redundant . So I wonder which equation should be used with (1) and (2), is it Equ. (4)? cause Equ. (3) has two parts?

Anyway here is my procedure in solving so if any help to know whether it's correct or no and what is the correct procedure.

I used the redundancy of the equations to substitute by the higher order derivatives as following and as in the MA code:

  • Substitute by $x$ from (3b) in (2) and (3a)
  • Substitute by $\dot{y}$ from (2) in (1) and (3)
  • Substitute by $\ddot{z}$ from (4) in (1) and (3)
  • Now I get a couple of equations in $(\ddot(y), y, z, \dot{z})$, which could be solved by AsymptoticDSolveValue

Any help is appreciated!

POSTED BY: Safinaz Salem
Posted 2 days ago

Your problem has lots of powers of and multiples of (2+3t)^(1/3) in numerators and denominators and products and sums.

If you think very carefully, would it be possible for you to transform all your expressions into z==(2+3t)^(1/3) for powers and multiples of z and do that in a way that greatly simplifies your problem instead of making it more complicated? You are going to have to handle t very carefully to do this.

If you can correctly do that then is there any chance that Mathematica might be able to find a solution for you?

POSTED BY: Bill Nelson
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