Solve $$\ p^2-3q^2-u=0 $$ given $\ u(x,0)=x^2 $ ,using Charpit's method to find complete solution.
I have tried to do the following, since the given equation is of the form $\ f(p, q, u)=0$ as per Dover Books on Mathematics] Ian N. Sneddon - Elements of partial differential equations (2006, Dover Publications)
used the substitution $$\ p=qa $$ $$\ (qa)^2-3q^2-u=0 $$ $$\ q^2(a^2-3)=u $$ $$\ q^2 = u/ (a^2-3)$$ $$\ q = \sqrt(u/ (a^2-3)) $$ using this in given equation to find p in terms of q, $$\ p^2-3[u/(a^2-3)]-u=0 $$ $$\ p^2=u+3[u/(a^2-3)] $$ $$\ p= \sqrt(u+3(u/(a^2-3)) $$ put the values in $$\ du=pdx+qdy $$ $$\ du=\sqrt(u(1+3/(a^2-3)))dx+ \sqrt(u/ (a^2-3)) dy $$ $$\ du\sqrt u= \sqrt(1+3/(a^2-3))dx+ \sqrt(1/ (a^2-3)) dy $$ on integrating we get $$\ \sqrt u=\sqrt(1+3/(a^2-3)) x + \sqrt(1/ (a^2-3)) y $$
Am I going right? If yes how to use the given condition $$\ u(x,0)=x^2 $$ to find a. Kindly help out.
There is a typo in $$ du\sqrt u= \sqrt(1+3/(a^2-3))dx+ \sqrt(1/ (a^2-3)) dy $$ It is not $du\sqrt u$. It is $\frac{du}{\sqrt{u}}$.
Do not forget $\pm$ before the square roots. The next equation should be $$ \pm 2\sqrt u=\pm\sqrt{1+\frac{3}{a^2-3}}\: x \pm \sqrt{\frac{1}{a^2-3}}\: y +C$$ You forget the $2$ before $\sqrt{u}$.
The condition is $ u(x,0)=x^2$ thus with $y=0$ :
$\pm 2\sqrt{u(x,0)}=2(\pm x)=\pm\sqrt{1+\frac{3}{a^2-3}}\: x+C$
Thus $\sqrt{1+\frac{3}{a^2-3}}=2$ and $C=0$. This gives $a=\pm 2$.
$\sqrt{1+\frac{3}{a^2-3}}=2$ and $\sqrt{\frac{1}{a^2-3}}=1$.
$$\pm 2\sqrt{u}=\pm 2x\pm y$$ $$u=\left(\pm x\pm\frac{y}{2}\right)^2$$ Or $$u=x^2+\frac{y^2}{4}\pm xy$$