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I need an existence and uniqueness check for the system of equations, and wondering that if following idea makes sense:

Suppose our system of equations, where functions are continuous and infinitely differentiable in both variables, as follows:

$$f(x,y) = 0 \qquad \text{and } \qquad g(x,y) = 0$$ and also assume that we have a unique solution $x (y)$ for a given $y (x).$ Under what conditions can we have the argument: start with any $y$ and find $x^*(\tilde{y})$ and then find $y^*(x^*(\tilde{y}))$, and this method will converge to solution and such solution exists and unique?

I guess the condition should be about contraction and fixed point theorems, but could not manage to come up with formal way to explain it. Do you have any idea?

Can we say that it is worthless to have unique solution for given value of the other variable to find existence and uniqueness of such system?

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I"ve worked on a similar problem in the past. Typically this is a hard problem.

One strategy that can work: Use the IVT to find a base solution $(x_0,y_0)$ such that, say, $f(x_0,y_0)=0$. By the implicit function theorem, establish a curve $f(x,y_1(x))=0$. You can compute $y_1'(x)=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$ and determine if $y_1'(x)$ is strictly positive (or negative).

Repeat for $g$ with $x_0$ fixed. Find a curve (with a solution for $x_0$) such that $g(x_0,y_2(x_0))=0.$ If $y_2'$ is the opposite of $y_1'$, you may be able to prove that they intersect. There needs to be things proven; for example, if the strictly increasing function is bigger than the strictly decreasing function, we cannot have $y_1(x)$ and $y_2(x)$ intersect.

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  • $\begingroup$ thanks very much @zugzug. $\endgroup$ – user229519 Jan 4 at 18:25

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