Jacobian determinant question
Hint Let $p=u^3v^44,\ q=2u^23v^48,$ so that your system is $$x^2y^2=p, \\ 2xy+y^2=q. \tag{1}$$ If the second equation of $(1)$ is solved for $x$ one has $$x=\frac{qy^2}{2y}. \tag{2}$$ 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, $$3y^4+2(2p+q)y^2q^2=0. \tag{3}$$ 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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user108896
Updated on August 01, 2022Comments

user108896 over 1 year
Show that the pair of equations
$$x^2y^2u^3+v^2+4=0\;,\;\; 2xy+y^22u^2+3v^4+8=0$$
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 $(2,1)$

DonAntonio almost 10 yearsWhat've you done? Tried to apply the implicit function theorem or something?

user108896 almost 10 yearsI've never come across that theorem in the syllabus I'm doing so I guess there must be a way to do this without that theorem. I haven't really got anywhere with it

DonAntonio almost 10 yearsPerhaps the Open map theorem...? One stems from the other, and I really can't see how to attack this kind of problems without it...

user108896 almost 10 yearsI'm not familiar with that one either

DonAntonio almost 10 yearsThen I can't help, sorry. Perhaps someone else.

coffeemath almost 10 years@user108896 Please have a look at my answer below, and respond as to whether it fits your needs.

user108896 almost 10 years@coffeemath Yep. I followed your hint and solved for y^2 yielding a discriminant when plugging in p=3 and q=3 of 48. i sure i've gone wrong with that answer. i do see how this method gives local functions x and y. and then the partial derivative du/dx can be evaluated

coffeemath almost 10 yearsWhen I plugged in $p=3,q=3$ to get the discriminant for equation $(3)$ of my answer, I got $144$. With $p,q$ left in it, the discriminant is $16(p^2+pq+q^2).$ Of course to go further one would have to plug in for $p,q$ here the two fourth degree terms at the start of the answer, maybe a mess.

user108896 almost 10 years@coffeemath oh right i follow. i got the same discriminant as you and a equation for y in terms and p and q, and consequently another for x. The plugginig does seem incredibly messy. Especially since one is then expected to find the partial derivatives of x and y in terms of u and v and then take the inverse of the Jacobi Matrix to have a value for du/dx at (2,1). Unless anyone knows a simpler way once x and y are established?
