(Particular) Solution to a nonlinear PDE Consider a function of type $u=u(x,y)$ that satisfies
$$
u_x^2 - u_y^2 = a y^2,
$$
where $a \in \mathbb{R}$. What is the solution to this PDE? A particular solution may be also interesting.
 A: We can make some progress in finding a complete solution to this equation by attempting to find the characteristics.
Consider the general boundary condition $f(x,y)|_\mathcal{C}=F(s)$ where the curve $\mathcal{C}(x_c(s),y_c(s))$ is parametrized by $s$. We would like to use the Lagrange-Charpit equations. We quickly notice that the system of equations does not close in the variables $x(t), y(t), f(x(t),y(t))$ so we extend the system by defining the quantities
$$p_1(t)=f_x(x(t),y(t)), p_2(t)=f_y(x(t),y(t))$$
We note that
$$\dot{p_1}=f_xf_{xx}(x(t), y(t))-f_yf_{xy}(x(t), y(t))\\
\dot{p_2}=f_xf_{xy}(x(t), y(t))-f_yf_{yy}(x(t), y(t))$$
However applying $\partial_x, \partial_y$ to our equation respectively we find
$$\dot{p_1}=0~~, ~~ \dot{p_2}=ay\\
\dot{x}=p_1~~,~~ \dot{y}=-p_2\\\dot{f}=\dot{x}f_x+\dot{y}f_y=ay^2\Rightarrow p_1^2-p_2^2=ay^2$$
which closes the extended system of LC equations. Additionally, the last equation imposes a constraint between the quantities above. Solving this linear system of ODE's is a simple matter (below we denote $x(0)=x_0, y(0)=y_0, p_1(x_0,y_0)=p_1(0), p_2(x_0,y_0)=p_2(0)$):
$$p_1(x(t),y(t))=p_1(0)\\x=x_0+p_1(0)t\\y(t)=y_0\cos(t\sqrt{a})-\frac{p_2(0)}{\sqrt{a}}\sin(t\sqrt{a})\\p_2(t)=p_2(0)\cos(t\sqrt{a})-y_0\sqrt{a}\sin(t\sqrt{a})$$
Here $a\in\mathbb{R}$ but one needs to define $\sqrt{a}\equiv i\sqrt{|a|}~,
~ a<0$. To find an explicit formula for the characteristics we need to eliminate $p_1(0), p_2(0), t$. To do this, we also need to differentiate the boundary condition with respect to the parameter $s$ and evaluate it at the parameter value $s_0$, such that
$(x_c(s_0), y_c(s_0))=(x_0,y_0)$. We obtain the linear constraint
$$\dot{x}_c(s_0)p_1(0)+\dot{y}_c(s_0)p_2(0)=\dot{F}(s_0)$$
which can be solved along with the quadratic constraint
$$p_1^2(0)-p_2^2(0)=ay_0^2$$
Rewriting $p_1(0), p_2(0)$ explicitly in terms of $x_0, y_0$ or equivalently $s_0$ is slightly tedious, so I won't go into it here, but bear in mind it also produces constraints on the form of the boundary curve $\mathcal{C}$. Defining an angle $\theta$ such that
$$p_2(0)=|p_1(0)|\cos\theta~,~y(0)=\frac{|p_1(0)|}{\sqrt{a}}\sin\theta$$
we can write the characteristic that passes through $(x_0,y_0)\in \mathcal{C}$ in the simple form, with all of the quantities other than $x,y$ can be expressed in terms of $s_0$:
$$y=\frac{|p_1(0)|}{\sqrt{a}}\sin\left(\sqrt{a}\frac{x-x_0}{p_1(0)}-\theta\right)$$
For $a<0, p_2(0)>0$ the characteristics can be written in the manifestly real form
$$\Lambda_{s_0}: ~~y=\frac{|p_1(0)|}{\sqrt{-a}}\sinh\left(\sqrt{-a}\frac{x-x_0}{p_1(0)}-\varphi\right)~~~~~ p_2(0)=|p_1(0)|\cosh\varphi, y_0=\frac{|p_1(0)|}{\sqrt{-a}}\sinh\varphi$$
We note that, for $a>0$, if the characteristics are a continuous family of curves, for almost all points of the plane, they must cross at finite time, so we expect to observe shockwave behavior.
Finally, to obtain the most general solution to this equation, find $\tau(X,Y)$ s.t the characteristic $\Lambda_\tau$ passes through the point $(X,Y)$ and integrate the equation $\dot{f}(x(t;\tau), y(t;\tau))=ay^2(t;\tau)$ along the characteristic; it is convenient to integrate using the variable $x$ here.
A: Particular Solution:
$$u(x,y) = -\frac{i}{2}\sqrt{a}y^2$$
if $a \geq 0$ and
$$u(x,y) = -\frac{1}{2}\sqrt{-a}y^2$$
if $a \leq 0$.
