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Imagine you are given $f(x,y) := y^2-\sin(x)^2$ and you want to answer the question, if there is a neighbourhood of $x=0$ such that $f(x,y(x))=0$ with $y(0)=0$.

One idea that comes to mind is the implicit function theorem: We readily verify that $f(0,0)=0$. Also $f$ is smooth in both its elements.

However, $\partial_2 f(0,0)=0,$ which means that the assumptions of the implicit function theorem are not met.

This is clear since $y = \sin(x)$ and $y=-\sin(x)$ are both solutions. Therefore, there is no unique solution in the neighbourhood of zero.

For our particular $f$ there is of course no need to invoke any heavy tools, since everything is explicit. However, I wonder if for functions of this type there is a different abstract method to argue that there exists at least one solution $y(x)$ in a neighbourhood of $x=0$ (disregarding the uniqueness).

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    $\begingroup$ Another relevant example is $y^2 + \sin(x)^2 = 0$, where $(0,0)$ again is a solution but there is no solution in a neighbourhood of $x=0$. So you need something more than just $f(0,0) = 0$ to get existence. $\endgroup$ Commented Dec 10, 2023 at 18:14
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    $\begingroup$ What comes to mind is that if at $(0,0)$ the gradient of $f(x,y)$ is $(0,0)$ and the Hessian matrix is indefinite (one positive and one negative eigenvalue), then you have a saddle point and there should be two solution curves passing through $(0,,0)$. $\endgroup$ Commented Dec 10, 2023 at 18:36
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    $\begingroup$ The function $y^2-x^2$ exhibits locally the same behaviour (and is of course locally diffeomorphic to your example). It is dealt with by the abstract theory of algebraic geometry (resolution of singularities).. $\endgroup$
    – terceira
    Commented Dec 10, 2023 at 19:51
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    $\begingroup$ Parentheses can disambiguate when you use them to distinguish between $(\sin x)^2$ and $\sin(x^2),$ but I wonder what their purpose is when one writes $\sin(x)^2. \qquad$ $\endgroup$ Commented Dec 10, 2023 at 20:43
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    $\begingroup$ @MichaelHardy: parentheses can also be used to indicate functional evaluation. $\endgroup$ Commented Dec 13, 2023 at 13:58

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It is hard to tell exactly what level of generalization you are aiming at, as the implicit function theorem is extremely general (holding, for example, for mappings between Banach spaces) and it is unclear to me whether you are seeking a result as strong as that.

But if you are not, here are some thoughts:

  1. The specific example you gave can be considered as a special case of the situation where you have a real-valued function on a manifold $f:M\to\mathbb{R}$. Implicit function theorem tells you that when $df(x_0) \neq 0$, then in a neighborhood $U$ of $x_0$ you have $\{x\in U: f(x) = f(x_0)\}$ is a codimension-1 submanifold.

    The case where $df(x_0) = 0$ is dealt with by Morse Theory, when the Hessian (which is well-defined at a critical point) is non-singular. In this case the Morse Lemma gives you the canonical form of $f$ near $x_0$. Note however that in the coordinates provided by the Morse Lemma, the level set $\{x\in U: f(x) = f(x_0)\}$ takes the form of a cone, and so with the exception of $\dim(M) = 2$ you cannot generally find a "implicit function" that is smooth.

  2. If you are interested in the case with higher dimensions in the codomain, I think a lot of our current understanding traces to Whitney's paper that invented modern singularity theory. Whitney's paper treated singularities of mappings from the plane to itself, and led to a lot of modern developments. I am not an expert in it, maybe someone else will comment, but at the very least you can just look at papers that cite Whitney to get an idea.

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Both your functions are analytic, and in this situation there is a general approach called Newton-Puiseux series (the results in Wiki are formulated for polynomials, but the theory applies more generally to analytic functions of several complex variables, see e.g. Bieskorr-Knörrer, Plane algebraic curves, Theorem 1 on page 386).

Of course, everything is over complex numbers, so you need to additionally check if there are any real solutions.

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  • $\begingroup$ sorry, maybe from just your answer and the wikipedia article, I don't quite see how this method works....do you have a simple example at hand to illustrate ideas? $\endgroup$
    – Daisy_Duck
    Commented Dec 15, 2023 at 1:51
  • $\begingroup$ @Daisy_Duck, check the Bieskorr-Knörrer book, it is quite friendly and introduces the method by working out examples. $\endgroup$
    – Kostya_I
    Commented Dec 15, 2023 at 8:34

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