Find a complete integral of $4uu_{x}-u_{y}^{3}=0$, and then that solution which satisfies $u=4at$ on $x=0$, $y=t$
Solution:
The PDE $f\equiv 4up - q^{3}=0$
Thus the last pair of terms integrate to give $\dfrac{p}{q}=a$, which is a compatible equation. Thus the PDE becomes $q^{2}=4u$, so $f_{x}=0\qquad f_{y}=0\qquad f_{u}=4p\qquad f_{p}=4u\qquad f_{q}=-3q^{2}$
It follows that
$\dfrac{dx}{f_{p}}=\dfrac{dy}{f_{q}}=\dfrac{du}{pf_{p}+qf_{q}}=-\dfrac{dp}{f_{x}+pf_{u}}=-\dfrac{dq}{f_{y}+qf_{u}}$
$\dfrac{dx}{4u}=\dfrac{dy}{-3q^{2}}=\dfrac{du}{4pu-3q^{3}}=-\dfrac{dp}{4p^{2}}=-\dfrac{dq}{4pq}$
Using the last pair of the equations will give
$\dfrac{dp}{dq}=\dfrac{4p^{2}}{4pq}=\dfrac{p}{q}$
$\dfrac{dp}{p}=\dfrac{dq}{q}\implies \ln p=\ln q+\ln a$ \begin{equation} \dfrac{p}{q}=a \end{equation}
$q=\sqrt{4u}$ which simplifies to give $q=2\sqrt{u}$ and $p=2a\sqrt{u}$
Now, $du=pdx+qdy$
$4u\left( \dfrac{p}{q}\right)-q^{2}=0$
$4ua=q^{2}\Rightarrow q^{2}=4ua$
$q=\sqrt{4ua}=2\sqrt{ua}$
therefore $p=2a\sqrt{ua}$
$\begin{equation} du=2a\sqrt{u}dx + 2\sqrt{u}dy \end{equation} $
Thus, $du=pdx+qdy$ $du=2a\sqrt{au}dx+2\sqrt{au}dy$
$\dfrac{du}{\sqrt{u}}=2a\sqrt{a}dx+2\sqrt{a}dy$
$2u^{1/2}=2ax\sqrt{a}+2y\sqrt{a}+b$
$2u^{1/2}=2\sqrt{a}(ax+y)+b$
Dividing both sides by 2 will give
$u^{1/2}=\sqrt{a}(ax+y)+b$
$\begin{equation} u=\left[ \sqrt{a}(ax+y)+b\right]^{2} \end{equation}$
Using the Cauchy conditions $4t=\left[ \sqrt{a}(t)+b\right]^{2}$
Take the square root of both sides
$\begin{equation} 2t^{1/2}=t\sqrt{a}+b \end{equation}$
Find the $t$-derivative of eqn (30), we have
$t^{-1/2}=\sqrt{a}$
Substitute eqn (31) into eqn (30), we have
$2\left(\dfrac{1}{\sqrt{a}}\right)=\dfrac{1}{\sqrt{a}}(\sqrt{a})+b$
Clearly,
$\begin{equation} b=\dfrac{2}{\sqrt{a}} \end{equation}$
(a) the initial data, applied directly to this complete integral, yields $4t=[\sqrt{a}(a\times 0)+t)+b]^{2}$
which is satisfied with $b=0$ and $a=4$
Hint:
Let $u=v^2$ ,
Then $u_x=2vv_x$
$u_y=2vv_y$
$\therefore4v^22vv_x-(2vv_y)^3=0$ with $v(0,t)=\pm2\sqrt{at}$
$8v^3v_x-8v^3v_y^3=0$ with $v(0,t)=\pm2\sqrt{at}$
$v_x-v_y^3=0$ with $v(0,t)=\pm2\sqrt{at}$
$v_{xy}-3v_y^2v_{yy}=0$ with $v(0,t)=\pm2\sqrt{at}$
Let $w=v_y$ ,
Then $w_x-3w^2w_y=0$ with $w(0,t)=\pm\sqrt{\dfrac{a}{t}}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{ds}=1$ , letting $x(0)=0$ , we have $x=s$
$\dfrac{dw}{ds}=0$ , letting $w(0)=w_0$ , we have $w=w_0$
$\dfrac{dy}{ds}=-3w^2=-3w_0^2$ , letting $y(0)=f(w_0)$ , we have $y=-3w_0^2s+f(w_0)=-3xw^2+f(w)$ , i.e. $w=F(3xw^2+y)$
$w(0,t)=\pm\sqrt{\dfrac{a}{t}}$ :
$F(t)=\pm\sqrt{\dfrac{a}{t}}$
$\therefore w=\pm\sqrt{\dfrac{a}{3xw^2+y}}$
$w^2=\dfrac{a}{3xw^2+y}$
$3xw^4+yw^2-a=0$
$w^2=\dfrac{-y\pm\sqrt{12ax+y^2}}{6x}$
$w=\sqrt{\dfrac{-y\pm\sqrt{12ax+y^2}}{6x}}$ or $-\sqrt{\dfrac{-y\pm\sqrt{12ax+y^2}}{6x}}$