System of equations and perturbation methods

Question

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}] & = & 0\end{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

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

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 $$- \varepsilon \frac{1}{x_0\varepsilon^2g_1^2 + x_0\varepsilon^3g_1g_2 + \varepsilon^3 f_1g_1^2 + \varepsilon^3x_0 g_1g_2 + \dots}$$

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}] & = & 0\end{array} with the system

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

Answer 1

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 $$ Z_1(x,y) = -y^2\left[1 + \text{log}\left(\frac{1}{y}-x\right)\right]. $$ 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 $\left\{ (x,y)\,|\,y=0\right\}$ 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 $\lim_{y\downarrow 0} \frac{\partial Z_1}{\partial x}$ and $\lim_{y\downarrow 0} \frac{\partial Z_1}{\partial y}$ 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 $(x_0,y_0)$: $$ F(x_0) = 0,\quad F'(x_0) = 0,\quad \frac{1}{y_0} - x_0 > 0\quad\text{and}\quad y_0 \neq 0. $$

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 $$ \frac{1}{1-x_0 y_0}-2-2\,\text{log}\left(\frac{1}{y_0}-x_0\right) = 0. $$