Given the following functions ('fields') in one variable $h(\rho),\Phi(\rho),\chi(\rho)$ and a parameter $k$ with a system of three coupled second order differential equations:
$$h\chi''+h'\chi'-2h\chi'\Phi'=h\chi \left(h^2+\frac{1}{4h^2}-3 \right)\tag{1}$$ $$\Phi''-\frac{h'}{h}\Phi'=(\chi')^2+ \left( h^2-\frac{1}{4h^2} \right)\chi^2 \tag{2}$$ $$\frac{2}{k}+2\Phi''-2(\Phi')^2=(\chi')^2+ \left(3h^2-\frac{1}{4h^2}-3 \right)\chi^2 \tag{3}$$ such that for $\psi \in {h,\Phi,\chi}$: $$\psi'=\frac{\partial\psi}{\partial \rho},\psi''=\frac{\partial^2\psi}{\partial \rho^2}$$
I want to make a simplification for those equations, such that they will have the form: $$h'=f(\Phi',h)\tag{4}$$ $$\Phi'=g(h,\chi)\tag{5}$$ $$\chi'=m(h,\chi)\tag{6}$$ where $f,g,m$ are functions that have "fields" as variables, in general: $$f(\Phi',h)=f(\Phi',h,\chi)$$ $$m(h,\chi)=m(h,\chi,\Phi')$$ such that any solution to the equations $(4)-(6)$ will be the solution for $(1)-(3)$ (but the opposite side in general is wrong).
There is a similar example of such a system where such simplification is existing, I will show the 2nd-order equation then the simplification, and give proof that any solution to the 1st-order equations is also a solution for the 2nd order, The 2nd-order system is given below: $$h\chi''+h'\chi'-2h\chi'\Phi'=h\chi \left(h^2-2 \right)\tag{7}$$ $$\Phi''-\frac{h'}{h}\Phi'=(\chi')^2+ h^2\chi^2 \tag{8}$$ $$\frac{2}{k}+2\Phi''-2(\Phi')^2=(\chi')^2+ \left(3h^2-2 \right)\chi^2 \tag{9}$$ The simplified 1st-order equations are: $$h'=f(\Phi',h)=h\Phi'+1\tag{10}$$ $$\Phi'=g(h,\chi)=-h\left(\chi^2+\frac{1}{k}\right)\tag{11}$$ $$\chi'=m(h,\chi)=-h\chi\tag{12}$$
Now we will prove that using equations $(10)-(12)$ one can restore equations $(7)-(9)$:
We begin with $(7)$ we will prove that LHS is equal to RHS, using $(12)$ we will write all $\chi' $ in terms of $h,h',\chi$:
$$h\chi''+h'\chi'-2h\chi'\Phi'=h^3\chi-2h'h\chi+2h^2\chi\Phi'$$
Now we will express $\Phi'$ using $h'$ by the use of $(10)$
$$h\chi''+h'\chi'-2h\chi'\Phi'=h^3\chi-2h'h\chi+2h^2\chi\frac{(h'-1)}{h}=h\chi(h^2-2)$$
indeed what we wanted. now moving next to prove $(8)$ we will look at the LHS,using $(11)$ we will express $\Phi',\Phi''$ in terms of $h,\chi$ and then express $\chi'$ in terms of $h,\chi$ according to $(12)$:
$$\Phi''-\frac{h'}{h}\Phi'=-2h\chi\chi'=2h^2\chi^2$$
Using $(12)$ and raising the power of two, one can show:
$$h^2\chi^2=(\chi')^2\tag{13}$$
we use this and conclude:
$$\Phi''-\frac{h'}{h}\Phi'=h^2\chi^2+h^2\chi^2=(\chi')^2+h^2\chi^2$$
indeed the RHS of $(8)$.
Finally, we prove $(9)$ by looking again in the RHS of $(8)$ we have shown that:
$$\Phi''-\frac{h'}{h}\Phi'=2h^2\chi^2$$
Now I express $h'$ using $\Phi'$ by the use of $(10)$
yielding the following expression:
$$\Phi''-(\Phi')^2-\frac{\Phi'}{h}=2h^2\chi^2=\frac{1}{2}h^2\chi^2+\frac{3}{2}h^2\chi^2$$
Using $(11)$ to express $\Phi'$ by $h,\chi$ in the last element in the RHS yields:
$$\Phi''-(\Phi')^2+\frac{1}{k}+\chi^2=\frac{1}{2}h^2\chi^2+\frac{3}{2}h^2\chi^2$$
moving the last element in the RHS to the LHS and using $(13)$ for the last element in the LHS yields:
$$\frac{1}{k}+\Phi''-(\Phi')^2=\frac{1}{2}(\chi')^2+\frac{3}{2}h^2\chi^2-\chi^2$$
multiplying everything in 2 and we got exactly $(9)$
$$\frac{2}{k}+2\Phi''-2(\Phi')^2=(\chi')^2+ \left(3h^2-2 \right)\chi^2 $$
In my case of equation $(1)-(3)$ I tried many guesses based on the simplified equations $(10)-(12)$ corresponds to $(7)-(9)$ that has a similar form to my equations $(4)-(6)$, but unfortunately I couldn't yet find any consistent simplified set by going in the same lines of the previous proof. I would be glad if someone might find such a set, or alternatively, if it is impossible I'll be glad to get proof explaining why for the specific set $(1)-(3)$ it is impossible.