Reducing order of a Non Linear Second Order ODE

Question

Let be $u=u(x)$ so: $$u''=f(u) \left[ (u')^2+g(x)u'\right]$$ Where the apex denotes x-derivative. In my case: $f(u)=-1/u$ and $g(x)=x/2$. Is there a change of variable in order to reduce the order of ODE?

Answer 1

The differential equation does permit a scaling symmetry of $x$ and $u$. Note, that the differential equation is invariant under the transformation $$\tilde{x}=\exp(\varepsilon)x$$ $$\tilde{u}=\exp(2\varepsilon)u.$$

You can then try the method of canonical coordinates to obtain the substitutions: $$s(x,u)=\ln x \implies x(r,s) = \exp(s)$$ $$r(x,u)=u/x^2 \implies u(r,s) = r\exp(2s).$$

The resulting ODE is given by (calculated with Maple):

$$-{\frac {-2\, \left( {\frac {\rm d}{{\rm d}r}}s \left( r \right) \right) ^{3}r-3\, \left( {\frac {\rm d}{{\rm d}r}}s \left( r \right) \right) ^{2}+{\frac {{\rm d}^{2}}{{\rm d}{r}^{2}}}s \left( r \right) }{ \left( {\frac {\rm d}{{\rm d}r}}s \left( r \right) \right) ^{3}}}= -1/2\,{\frac { \left( 2\,r{\frac {\rm d}{{\rm d}r}}s \left( r \right) +1 \right) \left( 4\,r{\frac {\rm d}{{\rm d}r}}s \left( r \right) +{ \frac {\rm d}{{\rm d}r}}s \left( r \right) +2 \right) }{r \left( { \frac {\rm d}{{\rm d}r}}s \left( r \right) \right) ^{2}}} $$

Note that the differential can be expressed as a nonlinear first order ODE by the substitution $s'(r)=k(r)$ and you can further simplify this.