Replacing in equation introduces more solutions

Let’s say I have an equation $y=2-x^2-y^2$.
now, since I know that $y$ is exactly the same as $2-x^2-y^2$ I can create the following, equation by replacing $y$ with $2-x^2-y^2$.

$y=2-x^2-(2-x^2-y^2)^2$

doing this replacement introduces new solutions such as $(-1, 0)$. Replacements in other various equations have similar results, although some do not change the equation at all!

What mechanic introduces these new solutions, and what are they?

Edit: one such example of an equation where no solutions are introduced via replacement is $y=x^2+y^2$. That will give $y=x^2+(x^2+y^2)^2$ which upon graphing is the same graph as the original, $y=x^2+y^2$.

Here is an image of the iteration of this replacement on the same function, just for fun.

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Suppose first that $f$ is a real-valued function of one variable. The equation
$$x = f(x) \tag{1}$$
acts as a condition, selecting values of $x$ for which (1) is true. Substituting (1) into itself gives a new condition,
$$x = f\bigl(f(x)\bigr) = f^{[2]}(x), \tag{2}$$
and so forth.

Certainly every solution of (1) is a solution of (2). If the function $f$ is not injective (one-to-one), however, (2) can have solutions that are not solutions of (1).

For example, if $f(x) = 4x(1 – x)$, then $f$ maps $[0, 1]$ onto $[0, 1]$. Each point except $x = \frac{1}{2}$ has two preimages, and $f$ maps each interval $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ bijectively to $[0, 1]$. It follows that $f \circ f$ maps each half-interval onto $[0, 1]$, and each point of $[0, 1]$ has two preimages in each half-interval, so $f^{[2]} = f \circ f$ has four fixed points, etc. (Diagram below.) In this example, the $n$-fold composition of $f$ with itself, $f^{[n]}$, has $2^{n}$ fixed points, i.e., the equation
$$x = f^{[n]}(x) = (\underbrace{f \circ \dots \circ f}_{n \text{times}})(x) \tag{n}$$
has $2^{n}$ solutions, even though (n) is obtained from (1) by successively substituting (1) into itself.

Your situation is analogous: Starting from
$$y = 2 – x^{2} – y^{2} = f(x, y), \tag{1a}$$
you substitute (1a) into itself, obtaining
$$y = 2 – x^{2} – f(x, y)^{2} = f\bigl(x, f(x, y)\bigr), \tag{2a}$$
and so forth.

In your example, $f$ has qualitatively similar behavior along the $y$-axis to the “logistic map” $f(x) = 4x(1 – x)$, and the solution sets
$$y = f\Bigl(\dots f\bigl(x, f(x, y)\bigr)\dots\Bigr)$$
become increasingly complicated with successive iteration.

This type of phenomenon (e.g., the precise location/shape of the solutions) is generally complicated (chaotic in the technical sense). Wikipedia pages of possible interest include:

• The logistic map
• The tent map
• The horseshoe map
• Dynamical systems

We know that every $(x,y)$ such that
$$y=2-x^2-y^2\quad\text{i.e.}\quad x^2+y^2+y-2=0$$
satisfies
$$y=2-x^2-(2-x^2-y^2)^2\quad \text{i.e.}\quad (2-x^2-y^2)^2+x^2+y-2=0\tag1$$
This implies that $(2-x^2-y^2)^2+x^2+y-2$ is divisible by $x^2+y^2+y-2$ :
$$(2-x^2-y^2)^2+x^2+y-2=(x^2+y^2+y-2)(x^2+y^2-y-1)$$
Hence,
$$(1)\iff x^2+y^2+y-2=0\quad\text{or}\quad x^2+y^2-y-1=0.$$

Similarly, for the second example, we know that $(x^2+y^2)^2+x^2-y$ is divisible by $x^2+y^2-y$ :
$$(x^2+y^2)^2+x^2-y=(x^2+y^2-y)(x^2+y^2+y+1)$$

Note here that there are no $(x,y)\in\mathbb R$ such that
$$x^2+y^2+y+1=0\iff x^2+\left(y+\frac 12\right)^2=-\frac 34$$

Point of warning: research didn’t yield any useful results for your question, so this is the result of a quick session in which I looked at the problem. I managed to come up with something, but it is not a very general or rigorous approach. Still, I hope you find it useful (or at least interesting). It only helps you find analytic expressions for some of the solutions I actually wonder if the other solutions have analytic expressions at all: they kind of look like they don’t.

If you write $f_x(y) = 2 – x^2 – y^2$, you are looking for points such that
$$y = f_x^k(y), k\geq 1$$
Points that satisfy this equation for $k = 1$ are called fixed points of $f_x$, points that satisfy this equation for any $k$ are called periodic points with period $k$. The sequence $(f_x^k(y))_{k \geq 0}$ is called the orbit of $y$, so periodic points have periodic orbits. I’m not gonna do anything fancy with these definitions (in fact, I hope I recited them correctly…), I just wrote them down to give the interested reader some terms to google. I don’t know any theorems or strategies for finding periodic points in general.

So, let’s look at this specific function. If we have a solution to $f(y) = 2 – x^2 – y^2$ we also have a solution to $f(y) = 2 – x^2 – (2 – x^2 – y^2)^2$ by substitution. The converse is not true: not every solution of the second equation is also a solution to the first one, as you noticed. This is equivalent to noting that every $y = f_x(y)$ is also a solution to $y = f(f(y))$, but, in general, the converse of the statement is not true: not every solution to the second equation is also a solution the first equation.

I am drifting away: we are seeking solutions to $y = 2 – x^2 – (2 – x^2 – y^2)^2$. If we assume $y = 2 – x^2 – y^2$ we can simplify to $y = 2 – x^2 – y^2$. Now I’m going to make some wild assumptions here, but bear with me. I’m interested if anyone can think of or knows a more straightforward way to do this.

I assume that any other solutions can be found in the same way: assume $y = g(y)$ for some function $g$, then replace $g(y)$ by $y$ to simplify the equation, and end up with the same equation $y = g(y)$. Of course, we choose $g \not = f$ here. But what $g$ should we choose? That is not easy to answer, but if we want get some analytic solution, it sure helps if we can easily substitute in $y = 2 – x^2 – (2 – x^2 – y^2)^2$ and doing so would simplify the expression. Since we substitute it in $(x^2 – y^2)^2$, an educated guess would be that after the substitution we have the square of a linear form in $y$ left, which would mean that $g$ is quadractic in $y$. This is both consistent with the solution $y = 2 – x^2 – y^2$.

Now, the point of this gibberish: we assume that replacing $g(y)$ by $y$ in $2 – x^2 – y^2$ yields some linear expression. Let’s say that we have $2 – (x^2 + y^2)$ and get $ay + b$ after substituting $y$ for $g(y)$. Then $y = g(y)$ is equivalent to $2 – x^2 – y^2 = ay + b$. Solving for $y$ yields $y = \frac{2 – b – (x^2 + y^2)}{a} = g(y)$. After substitution we get $y = 2 – x^2 – (ay + b)^2 = \frac{2 – b^2 – x^2 – a^2y^2}{2ab + 1}$. Since we only know (since we assumed it) that $y = \frac{2 – b – x^2 – y^2}{a} = g(y)$, we can make the whole thing true by picking $a, b$ that make $$\frac{2 – b – x^2 – y^2}{a} = \frac{2 – b^2 – x^2 – a^2y^2}{2ab + 1}$$

Now $a = 1, b = 0$ gives us $y = 2 – x^2 – y^2$, $a = -1, b = 1$ gives us $-y = 1 – x^2 – y^2$. I wanted to do the general case as well, but this has cost me enough time for now. I might add it in a few days.