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In this paper http://arxiv.org/abs/hep-th/9506035 equation (3.11) was written as: $$\frac{\partial L}{\partial u}\frac{\partial L}{\partial v} = -1$$

The author then said p.9 that "approximate solutions to equation (3.11) can be obtained by a power series expansion. It is convenient to define new variables $x$ and $y$ by:"

$x= u+v$ and $y=u-v$

Then he said equation (3.11) becomes $$(\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1 $$

How come? I didn't understand his argument.

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    $\begingroup$ Have you tried writing $\partial_u$ and $\partial_v$ in terms of $\partial_x$ and $\partial_y$? $\endgroup$
    – ACuriousMind ♦
    Commented Dec 4, 2014 at 15:10
  • $\begingroup$ Ah the chain rule!! Thank you a lot! @ACuriousMind $\endgroup$
    – Fluctuations
    Commented Dec 4, 2014 at 15:26

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This is just the chain rule: $\frac{\partial L}{\partial u}=\frac{\partial L}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial L}{\partial y}\frac{\partial y}{\partial u}$. And similar thing for $\frac{\partial L}{\partial v}$. We have the derivative of $x$ and $y$ with respect to $u$ and $v$.

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