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I am dealing with the following problem of multivariate calculus but I haven't been able to solve it yet: "If $u$ and $v$ are class 2 functions in $\mathbb{R}$ and $z=z(x,y)$ is class 2 in $\mathbb{R^2}$,verifying that $(u(x)+v(y))^2e^{z(x,y)}=2u'(x)v'(y)$ with $u(x)+v(y) \neq 0 \forall x,y \in \mathbb{R}$, prove that $z_{xy} \neq 0$."

I have tried to compute the derivatives but it's long and fruitless, so I thought that some property needs to be used, but I don't know which one. Thanks for your help.

CharlesJA
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1 Answers1

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The given equality, first of all, implies that $u'(x) \neq 0$ and $v'(y) \neq 0$ for every $x,y \in \mathbb{R}$. Moreover, the left-hand side is positive so the right-hand side is positive. Now, take logarithms on both sides so that you get: $$2\log(u(x)+v(y)) + z(x,y) = \log(2) + \log(u'(x)) + \log(v'(y))$$ Since $z$ is of class $\mathcal{C}^2$, it doesn't matter if we calculate $z_{xy}$ or $z_{yx}$, by the Schwarz Symmetric Theorem. I will denote these by $z_{12}$ or $z_{21}$, where $1$ and $2$ indicate the variable that I am differentiating with respect to. So: $$z_1(x,y) = \frac{u''(x)}{u'(x)} -\frac{2u'(x)}{u(x)+v(y)}$$ $$z_{12}(x,y) = \frac{2u'(x)v'(y)}{(u(x)+v(y))^2} = e^{z(x,y)}$$ This implies that $z_{12} \neq 0$ and we are done.

Abhijeet Vats
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