Asymptotic behaviour of solution of a nonlinear ODE near infinity

Question

Consider the nonlinear ODE \begin{align} y':=\frac{dy}{dx}=\sqrt{1-\frac{1}{y}} \end{align} where $y$ is (as usual) a function of $x$. I am interested to obtain a Laurent series solution $y$ of this ODE when $x$ is big. My guess is that since $y'>0$, $y$ should be increasing, and so when $x$ is big, $y$ is big, but then $y'$ would be close to $1$, which makes $y$ almost like a linear function when $x$ is big. Thus I expect \begin{align} y=x+a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots,\qquad x>>1 \end{align} but then I am not sure if this is correct and I would be glad to know any techniques to obtain such Laurent series solution.

Any comment, hint and answer are greatly appreciated. In particular, if there is any reference for problems alike, I would also be glad to know.

Answer 1

The d.e. is separable, and its general solution can be written as $$ x + c = \sqrt{y^2-y} + \frac{1}{2} \ln \left(-1 + 2 y + 2 \sqrt{y^2-y}\right)$$ As $y \to +\infty$ the right side is $y + \frac{1}{2} \ln(y) + O(1)$. Thus we should have $y \sim x - \frac{1}{2} \ln(x) + O(1)$. In particular, it is not a Laurent series because of the logarithmic term: the solution has a branch point, not a pole, at $\infty$.

Answer 2

We could solve this differential equation exactly using separation of variables.

$$\int \frac{dy}{\sqrt{1-\frac{1}{y}}} = x + C$$

Substitute $ y = \cosh^2 t$

$$ \implies \int \frac{2\cosh t \sinh t}{\sqrt{\tanh^2 t}} dt = \int 2 \cosh^2 t dt = \int 1 + \cosh 2t dt$$

$$ \implies \cosh^{-1}(y) + y\sqrt{y^2-1} = x + C$$

In other words, for large $x$ we get

$$ x \approx y^2 + \log(y) \implies y \approx \sqrt{\frac{\mathrm{W}(2e^{2x})}{2}} \approx \sqrt{x} + \sqrt{\frac{\log x}{2}}$$

from Lambert-W's known expansion for large x, but it would end up having $\log\log x$ terms and so on.

Answer 3

Almost repeating what Robert Israel answered, the general solution is $$x+c=\sqrt{(y-1) y}+\sinh ^{-1}\left(\sqrt{y-1}\right)$$ Expanded as a series for large values of $y$ $$x+c=\left(\log(2)-\frac 12\right)+y+\frac{1}{2}\log (y)-\frac{3}{8 y}+O\left(\frac{1}{y^2}\right)$$ So, for large values of $y$, solving $$x+d=y+\frac 12 \log(y)\implies y=\frac{1}{2} W\left(2 e^{2 (x+d)}\right)$$ where appears Lambert function (as already introduced in Ninad Munshi's answer).

For large values, using the first term of the expansion of Lambert function, this is $$y\sim \frac 12 \left(\log (t)-\log (\log (t))\right)\qquad \text{where} \qquad t=2 e^{2 (x+d)}$$