Krdytkyu

I would like to characterise how the solution of a nonlinear system of equations change if a perturbation term is added.

Namely, I have the system
begin{array}{lcl} 2y [F(x) - varepsilon frac{1}{y(xy-1)} ] & = & 0 \ y^2 [F^prime (x) -varepsilon frac{y}{xy-1}] & = & 0end{array}

In the unperturbed case $varepsilon = 0$ the solution is handy $$ (x_0,0)$$ where $x_0$ is such that $F^prime (x_0) = 0$.

How to describe the solution for small $varepsilon$??

I would have tried to expand the perturbed terms around the solution of the unperturbed system, but the term $frac{1}{y(xy-1)}$ is not even defined there.

Alternatively, following the perturbation theory one could assume that the perturbed solution can be expressed as
begin{array}{lcl} x & = & x_0 + varepsilon f_1 +varepsilon^2 f_2 + dots \ y = & = & varepsilon g_1 +varepsilon^2 g_2 + dots end{array}

and let me substitute in the first equation of the system.
I get
$$ 2(varepsilon g_11 + varepsilon^2 g_2 + dots ) Big[F(x_0+varepsilon f_1 + dots) - varepsilon frac{1}{(varepsilon g_11 + varepsilon^2 g_2 + dots)((x_0+varepsilon f_1 + dots)(varepsilon g_1 + varepsilon^2 g_2 + dots))}Big]$$ which could develop into

$$2(varepsilon g_1 + varepsilon^2 g_2 + dots )Big[[F(x_0) + F ^{prime prime}(x_0)frac{1}{2}varepsilon^2 f_1^2] - varepsilon frac{1}{x_0varepsilon^2g_1^2 + x_0varepsilon^3g_1g_2 + varepsilon^3 f_1g_1^2 + varepsilon^3x_0 g_1g_2 + dots}Big] $$

Now I need to collect terms that are first order in $epsilon$.

But how to do that systematically?
I am struggling to handle the fraction $$- varepsilon frac{1}{x_0varepsilon^2g_1^2 + x_0varepsilon^3g_1g_2 + varepsilon^3 f_1g_1^2 + varepsilon^3x_0 g_1g_2 + dots}$$

I would be most grateful for any hint.

Thanks and Happiest New 2019

EDIT:
I would like to describe one more attempt of mine.
I thought of replacing the original perturbed system
begin{array}{lcl} 2y [F(x) - varepsilon frac{1}{y(xy-1)} ] & = & 0 \ y^2 [F^prime (x) -varepsilon frac{y}{xy-1}] & = & 0end{array}
with the system

begin{array}{lcl} 2y [F(x) - varepsilon (-frac{1}{y}-x) ] & = & 0 \ y^2 [F^prime (x) -varepsilon (-y)] & = & 0end{array}
using the approximations

$$frac{1}{y(xy-1)} approx -frac{1}{y} -x$$ and $$ frac{y}{xy-1} approx -y$$ first-order valid around $y=0$.
Then I get something tractable, would this be a workaround?

EDIT
Following the comment by User121049, I would like to add, should it be of any interest, that the problem I have is equivalent to finding the stationary point of the function

$$ Z(x,y) = y^2 Big[ F(x) - epsilon [log(frac{1}{y}-x) +1)] Big]$$

the system I originally described is obtained by setting the partial derivatives to zero.

Write $Z(x,y,epsilon) = Z_0(x,y) + epsilon Z_1(x,y)$. For $epsilon=0$, stationary points of $Z = Z_0$ satisfy
$$
y^2 F'(x) = 0 quad text{and} quad 2 y F(x) = 0.
$$
Notice that any point $(x_0,0)$ obeys these equations, regardless of the value of $x_0$, as long as $x_0$ is in the domain of $F$. Furthermore, double roots of $F$ (for which $F(x_0) = 0$ and $F'(x_0) = 0$) are automatically stationary points of $Z_0$; therefore, the total set of stationary points of $Z_0$ consists of the union of lines
$$
{(x,y);|;y=0} ,cup,{(x,y) ;|; F(x)=0;text{and};F'(x)=0}.
$$

So far, so good. Now, however, we take a closer look at the perturbation
$$
Z_1(x,y) = -y^2left[1 + text{log}left(frac{1}{y}-xright)right].
$$
Here, we encounter the problem that the logarithm is only defined when $frac{1}{y}-x > 0$. This region in the plane is bounded by the hyperbolas $x y = 1$ and by the line $y=0$, it looks like this:

In particular, the (orange) line $left{ (x,y),|,y=0right}$ is on the boundary of the logarithm domain. However, the limit $lim_{y downarrow 0} Z_1(x,y)$ exists (check this!), so the horizontal axis is included in the domain of the logarithm. Furthermore, both limits $lim_{ydownarrow 0} frac{partial Z_1}{partial x}$ and $lim_{ydownarrow 0} frac{partial Z_1}{partial y}$ exist and are equal to zero (check this!), so the entire horizontal axis consists of stationary points of $Z_1$. We have just seen that the horizontal axis consists of stationary points of $Z_0$, so we conclude that the entire horizontal axis consists of stationary points of $Z = Z_0 + epsilon Z_1$. This is true for all $epsilon$, not necessarily small. No need for perturbation theory here.

What about the other stationary points of $Z_0$, that are characterised by double roots of $F$? Here, we do need perturbation theory. So, let's take a point $(x_0,y_0)$ such that $x_0$ is a double zero of $F$, i.e. $F(x_0) = 0$ and $F'(x_0)=0$. The value of $y_0$ is as yet unspecified. However, if we want to have any chance at all for $(x_0,y_0)$ to be a stationary point of $Z = Z_0 + epsilon Z_1$, we must at the very least have that $(x_0,y_0)$ is in the domain of $Z_1$ -- otherwise it doesn't make any sense to talk about the value of $Z_1(x_0,y_0)$, because it doesn't exist. So, we must assume that $y_0$ is such that $(x_0,y_0)$ lies somewhere inside the blue region in the image above. Since we've covered the case $y_0 = 0$ already above, we have the following conditions on $(x_0,y_0)$:
$$
F(x_0) = 0,quad F'(x_0) = 0,quad frac{1}{y_0} - x_0 > 0quadtext{and}quad y_0 neq 0.
$$

Only for these points you can start to apply perturbation theory. As the perturbation is regular, you just have to substitute $x = x_0 + epsilon x_1 + mathcal{O}(epsilon^2)$ and $y = y_0 + epsilon y_1 + mathcal{O}(epsilon^2)$, and expand the resulting expressions up to second order in $epsilon$. I leave this up to you, but I can tell you that I get, at order $epsilon$, the equations
$$
x_1 F''(x_0) + frac{y_0}{1 - x_0 y_0} = 0
$$
and
$$
frac{1}{1-x_0 y_0}-2-2,text{log}left(frac{1}{y_0}-x_0right) = 0.
$$