Find the solution to the nonlinear PDE: $U - XUx - (1/2)*(Uy)^2 + X^2 = 0$ with $U(X,0) = X^2 - (1/6)*(x^4), 0Step 1: Rewrite the PDE
$P= Ux$ and $Q= Uy$
$$F = U - XP - (1/2)Q^2 + x^2$$
Step 2: Charpits Equations
$dx/dτ = Fp = -x$
$dy/dτ = Fq = -2Q$
$dp/dτ = -Fx-P*Fu = -2x$
$dq/dτ = -Fy-Q*Fu = -Q$
$du/dτ = PFp+QFq = -xp - 2Q^2$
Step 3: Integrate
At this point I am confused as to where to start?
Next Steps: Parameterise Initial data
$Xo = S, Uo = S^2 - (1/6)*S^4, Yo = 0$
$dUo/dS = Po*dxo/ds + Qo*dyo/ds$
$Po = 2S -(2/3)*S^3$
 A: Solving with reduction of to a first order PDE and method of characteristics :

A: Hint:
Let $U=X^2-V$ ,
Then $U_X=2X-V_X$
$V_Y=-U_Y$
$\therefore 2X^2-V-X(2X-V_X)-\dfrac{(-V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$
$XV_X-V-\dfrac{(V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$
Let $V=XW$ ,
Then $V_X=XW_X+W$
$V_Y=XW_Y$
$\therefore X(XW_X+W)-XW-\dfrac{(XW_Y)^2}{2}=0$ with $W(X,0)=\dfrac{X^3}{6}$
$X^2W_X=\dfrac{X^2(W_Y)^2}{2}$ with $W(X,0)=\dfrac{X^3}{6}$
$(W_Y)^2=2W_X$ with $W(X,0)=\dfrac{X^3}{6}$
$W_Y=\pm\sqrt{2W_X}$ with $W(X,0)=\dfrac{X^3}{6}$
$W_{XY}=\pm\dfrac{W_{XX}}{\sqrt{2W_X}}$ with $W(X,0)=\dfrac{X^3}{6}$
Let $Z=W_X$ ,
Then $Z_Y=\pm\dfrac{Z_X}{\sqrt{2Z}}$ with $Z(X,0)=\dfrac{X^2}{2}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dY}{dT}=1$ , letting $Y(0)=0$ , we have $Y=T$
$\dfrac{dZ}{dT}=0$ , letting $Z(0)=Z_0$ , we have $Z=Z_0$
$\dfrac{dX}{dT}=\mp\dfrac{1}{\sqrt{2Z}}=\mp\dfrac{1}{\sqrt{2Z_0}}$ , letting $X(0)=f(Z_0)$ , we have $X=f(Z_0)\mp\dfrac{T}{\sqrt{2Z_0}}=f(Z)\mp\dfrac{Y}{\sqrt{2Z}}$ i.e. $Z=F\left(X\pm\dfrac{Y}{\sqrt{2Z}}\right)$
$Z(X,0)=\dfrac{X^2}{2}$ :
$F(X)=\dfrac{X^2}{2}$
$\therefore Z=\dfrac{\left(X\pm\dfrac{Y}{\sqrt{2Z}}\right)^2}{2}$
$2Z=X^2\pm\dfrac{\sqrt2XY}{\sqrt{Z}}+\dfrac{Y^2}{2Z}$
