Separation of variables (integration)

Question

If $u=\frac{1}{2} E^2$ and $ v=\frac{1}{2}B^2$

and we have that $$\frac{\partial L}{\partial u} \frac{\partial L}{\partial v} = -1$$

The author says:

to obtain explicit solution of the above, one must resort to techniques such as separation of variables in particular coordinate systems. For example, if one supposes that the solution separates multiplicatively in (u,v) coordinates one obtains: $$L = ± \sqrt{\alpha - \beta E^2}\sqrt{\gamma - \delta B^2}$$ where $\beta \gamma =1$.

How was this obtained? I didn't get this method of integration?

Answer 1

The author assumed separation of variables, $$ L(u,v)=E(u)B(v) $$ which leads to $$ \frac{\partial L}{\partial u}\frac{\partial L}{\partial v}=B\frac{\partial E}{\partial u}E\frac{\partial B}{\partial v}=-1 $$ Rearranging, $$ E\frac{\partial E}{\partial u}=-\left(B\frac{\partial B}{\partial v}\right)^{-1}\tag{1} $$ Note that $$ E\frac{\partial E}{\partial u}=\frac{\partial}{\partial u}\left(\frac12E^2\right) $$ So if $u=\frac12E^2$ and $v=\frac12B^2$ then Equation (1) is identically -1. All that's left is finding inverting $E$ and $B$ to get $L$. This is where a slight difficulty comes because $E=\sqrt{2u}$ and $B=\sqrt{2v}$ doesn't quite do it. Thus, we toy with some constants, because the derivative of a constant is always zero, and find the relation the author gives: \begin{align} L&=\pm\sqrt{\alpha-\beta E^2}\sqrt{\gamma-\delta B^2}\\ &=\pm\sqrt{\alpha-2\beta u}\sqrt{\gamma-2\delta v} \end{align} Choosing this form, we do require that $\beta\delta=-1$ to satisfy Equation (1).