[Math] Jacobian determinant question


Show that the pair of equations

$$x^2-y^2-u^3+v^2+4=0\;,\;\; 2xy+y^2-2u^2+3v^4+8=0$$

Determine local functions $x(u,v)$ and $y(u,v)$ defined for $(u,v)$ near $u = 2$ and $v = 1$ such that $x(2,1) = 2$ and $y(2,1) = -1$.

Compute $\dfrac{\partial u}{\partial x}$ at $(2,1)$

Best Answer

Hint Let $p=u^3-v^4-4,\ q=2u^2-3v^4-8,$ so that your system is $$x^2-y^2=p, \\ 2xy+y^2=q. \tag{1}$$ If the second equation of $(1)$ is solved for $x$ one has $$x=\frac{q-y^2}{2y}. \tag{2}$$ Now substitute this into the first equation of $(1)$ and move $p$ to the left side, obtaining an expression which must be zero. This expression has a denominator of $4y^2$ which is irrelevant for solutions, and the negative of its numerator is quadratic in $y^2$, i.e. only contains $y^4,y^2$ [and $p,q$ ] so it determines $y^2$ from $p,q.$ Specifically, $$3y^4+2(2p+q)y^2-q^2=0. \tag{3}$$ Now here's why this answer is only a hint: at this point you may be able to use the initial conditions that at $(u,v)=(2,1)$ you have $(x,y)=(2,-1).$ I did calculate that at these values of $u,v$ one has $p=3,\ q=-3$ and when plugged in this agrees with $(x,y)=(2,-1).$ When $(3)$ is solved for $y^2$ there will be a sign choice on the $y^2$, and after that on passing to $y$ itself there is another sign choice, for a total of four sign choices in all. To make the solution local satisfying the conditions, the right sign choices need to be made, which determines $y,$ and then $x$ is forced by $(2).$

Given the forms of $p,q$ this approach may give rather complicated formulas for $x,y$ in terms of $u,v$. There may be a simpler way but I don't see one.

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