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.Iterating this replacement

Solutions Collecting From Web of "Replacing in equation introduces more solutions"

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.

Successive iterations of a quadratic polynomial

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.