First order partial differential equation How to find L if the form is:
$$\left(\frac{\partial L}{\partial x}\right)^2-\left(\frac{\partial L}{\partial y}\right)^2=-1$$
The author wrote, $$L=y+ax^2+...$$
but I didn't get how?
Edit: where L is Maxwell Lagrangian and $x=1/2(E^2+B^2)$ and $y=1/2(E^2-B^2)$
It was then said: Assuming that L coincides with the Maxwell Lagrangian for small values of E and B, i.e,$ L = y+ O(x^2,y^2)$ then it is found that the quadratic terms in x and y must be proportional to $x^2$ and then it was written that $$L = y + ax^2 + O(x^3,y^3)$$
 A: I would say that he considered the L to be quadratic in nature with 2 variables. Why quadratic ? because the second derivative w.r.t x and the second derivative w.r.t y (summed up) give a constant (-1). Therefore you have 2 conditions to start up with, 


*

*second derivative w.r.t x is equal to a constant

*second derivative w.r.t y is also a constant
meaning:

*first derivative w.r.t x is a line equation

*first derivative w.r.t y is a line equation as well


and so on. 
I would say that without any more information, L can have multiple solutions. Maybe the author just wanted to portray something, main point is
L has this form:
L(x,y) = ax^2 + by^2 + cx + my + k
hope it helps a bit
A: $\left(\dfrac{\partial L}{\partial x}\right)^2-\left(\dfrac{\partial L}{\partial y}\right)^2=-1$
$\left(\dfrac{\partial L}{\partial x}\right)^2=\left(\dfrac{\partial L}{\partial y}\right)^2-1$
$\dfrac{\partial L}{\partial x}=\pm\sqrt{\left(\dfrac{\partial L}{\partial y}\right)^2-1}$
$\dfrac{\partial^2L}{\partial x\partial y}=\pm\dfrac{\dfrac{\partial L}{\partial y}\dfrac{\partial^2L}{\partial y^2}}{\sqrt{\left(\dfrac{\partial L}{\partial y}\right)^2-1}}$
Let $U=\dfrac{\partial L}{\partial y}$ ,
Then $\dfrac{\partial U}{\partial x}=\pm\dfrac{U}{\sqrt{U^2-1}}\dfrac{\partial U}{\partial y}$
$\dfrac{\partial U}{\partial x}\mp\dfrac{U}{\sqrt{U^2-1}}\dfrac{\partial U}{\partial y}=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dU}{dt}=0$ , letting $U(0)=U_0$ , we have $U=U_0$
$\dfrac{dy}{dt}=\mp\dfrac{U}{\sqrt{U^2-1}}=\mp\dfrac{U_0}{\sqrt{U_0^2-1}}$ , letting $y(0)=f(U_0)$ , we have $y=\mp\dfrac{U_0t}{\sqrt{U_0^2-1}}+f(U_0)=\mp\dfrac{Ux}{\sqrt{U^2-1}}+f(U)$ , i.e. $U=F\left(y\pm\dfrac{Ux}{\sqrt{U^2-1}}\right)$
