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I am trying to show that the following equation system:

$$e^u+e^v+2-4e^{x^2-y}=0$$ $$\sin u+xu-v^4-y=0$$

Does not define $(y,x)$ as a function of $(u,v)$ implicitly around $(0,0,0,0)$.

Attempt What I tried to do was find $u,v$ such that in a small radius of (0,0,0,0) the equation would have no solution. My best guess was $u=0, v=R$ (where $R =$ radius), but I had difficulty showing that the system then has no solution.

Help would be appreciated, and thanks in advance!

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Differentiate implicitely with respect to u and v and evaluate at 0. Then see if the Jacobian matrix is singular. – PAD Jul 5 '12 at 7:32
We hadn't learned this technique yet (this is a question from a former exam I'm prepping with)... is there another solution? – ro44 Jul 5 '12 at 7:41
@ro44 You need to use the implicit function theorem. – user38268 Jul 5 '12 at 7:49
@BenjaLim: I know a version of the implicit function theorem that would let me prove this is an implicit function, but not one that would let me prove it isn't. Is there something I might be missing here? – ro44 Jul 5 '12 at 7:52
up vote 1 down vote accepted

You have the right idea. Setting $u=0$ simplifies the system to $e^v+3=4e^{x^2-y}$, $y=-v^4$. The key observation is that the exponent $x^2-y$ cannot be negative, and this is a problem for the first equation when $v<0$.

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Oh! I missed the fact that $y=-v^4$, thought it was $y=v^4$. – ro44 Jul 5 '12 at 16:40

The implicit function theorem guarantees you a nice solution $x=x(u,v)$, $\ y=y(u,v)$ with $x(0,0)=0$, $\ y(0,0)=0$ if a certain "technical condition" is satisfied, but has nothing to say if this condition is not satisfied (which is the case here).

On the other hand we can go and solve the system for $x$ and $y$ explicitly. From the second equation we deduce $y=x u+\sin u-v^4$, and plugging this into the first equation gives $$e^u+e^v+2=4\exp(x^2-x u-\sin u+v^4)$$ or $$x^2 - xu -\sin u +v^4-\log\Bigl(1+{e^u-1\over4}+{e^v-1\over4}\Bigr)=0\ .$$ This is a quadratic equation for $x$ which you can solve explicitly, but it might have complex solutions for some $(u,v)$ near $(0,0)$. Maybe you don't allow this.

To see what happens we retain only the terms of order $\leq2$ in $u$ and $v$. We then have the equation $$x^2-x u-{5\over4}u-{1\over4}v-{1\over32}(3u^2-2u v+3v^2)=0$$ with the formal solutions $$x={1\over2}\Bigl(u\pm\sqrt{5u+v+{\rm quadratic\ terms}}\Bigr)\ .$$ Here the radicand is negative on one side of a quadric going through the origin of the $(u,v)$-plane. This implies that in the immediate neighborhood of $(0,0)$ there are points $(u,v)$ to which correspond no real $(x,y)$ solving the given system of equations.

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