Prove that for $(x, y) \in \Bbb R^2$

\begin{cases} x+y+\sin(xy)=2c \\ \sin(x^2 + y) = c^2 \\ \end{cases}

can give a solution for all $c ∈ \Bbb R$ close enough to the origin, by the inverse function theorem.

I am familiar with the IFT, its conditions, and the result. However, I am having a hard time seeing connection between proving the problem and IFT, let alone solving it.

  • 2
    $\begingroup$ For $c\in\mathbb R$, define $f_c : \mathbb R^2 \to \mathbb R^2$ by $f_c(x,y) = (x+y+\sin(xy)-2c,\sin(x^2+y)-c^2)$. Now verify the conditions of the implicit function theorem with this function. $\endgroup$ – azif00 Feb 12 at 4:12
  • $\begingroup$ got it, can you plz further explain how this function proves the system of equations has a solution once I prove that it satisfies the conditions? $\endgroup$ – james black Feb 12 at 6:49
  • $\begingroup$ i need a little in depth explanation as i cant make the connection between the thm and how the system of equations have a solution $\endgroup$ – james black Feb 12 at 6:50

Consider the function $$f(x,y,c):=\bigl(x+y+\sin(xy)-2c, \ \sin(x^2+y^2)-c^2\bigr)\ .$$ We obviously have $f(0,0,0)=(0,0)$. This means that when $c=0$ then the point $(x_0,y_0):=(0,0)$ solves your two equations. The implicit function theorem says that under a certain "technical assumption" we have $$f(x_c,y_c,c)=(0,0)$$ for all $c$ near $0$ and uniquely defined $x_c=\phi(c)$, $\>y_c=\psi(c)$ with $\phi$, $\psi\in C^1$ and $\phi(0)=\psi(0)=0$. Therefore your equations will have a single solution $(x_c,y_c)$ also for $c$ near $0$, whereby $x_c$, $y_c$ depend differentiably on $c$, and $x_0=y_0=0$.

The "technical assumption" is that $f\in C^1$ near $(0,0,0)$ (obviously satisfied), and that $$\det\left[\matrix{{\partial f_1\over\partial x}&{\partial f_1\over\partial y} \cr {\partial f_2\over\partial x}&{\partial f_2\over\partial y}\cr}\right]_{(0,0,0)}\ne0\ .$$ Therefore you have to compute this determinant and to check whether it is $\ne0$.

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