Converting the first Painlevé equation into a different form.

Question

I am reading a paper which starts by saying that the first Painlevé equation can be converted into the equation

$u''+\frac{u'}{z}-\frac{3}{2}(u^2-1)-\frac{4u}{25z^2}=0$

Now I know that the first Painlevé equation can be written as

$y''=6y^2+x$

or equivalently

$P_I = \begin{cases} \frac{dx}{dz}=6y^2+z \\ \frac{dy}{dz}=x\ \end{cases}$

but I can't seem to find the transformation to convert it. The $\frac{3}{2}$ makes me think that I'll need a function with a constant added. I have found a function $y=\sqrt{x}u+\frac{1}{2}$ that gets me to something with a few correct terms:

$y'=\sqrt{x}u'+\frac{u}{2\sqrt{x}}$

so

$y''=\sqrt{x}u''+\frac{u'}{\sqrt{x}}-\frac{u}{4x\sqrt{x}}$

substituting gives

$\sqrt{x}u''+\frac{u'}{\sqrt{x}}-\frac{u}{4x\sqrt{x}}=6(\sqrt{x}u+\frac{1}{2})^2+x$

multiplying out gives

$\sqrt{x}u''+\frac{u'}{\sqrt{x}}-\frac{u}{4x\sqrt{x}}=6xu^2+6\sqrt{x}u+\frac{3}{2}+x$

dividing by $\sqrt{x}$ gets the first two terms as I want them but not much more

$u''+\frac{u'}{x}-\frac{u}{4x^2}=6\sqrt{x}u^2+6u+\frac{3}{2\sqrt{x}}+\sqrt{x}$

Any comments, advice etc most welcome ... however general.

Answer 1

The transformation is% $$ \frac{1}{x^{1/2}}u(x)=\alpha v(z),\quad z=\beta x^{5/4}, $$ where $\alpha$ and $\beta$ are two complex numbers different from zero. You can find it on this book book on page 17 formula (2.6).