I was given an initial value problem and I'm really struggling with how to solve it. My instructor hasn't taught any other method for solving yet apart from method of characteristics, so I guess one needs to solve using that.
$$u_y=xuu_x,\quad u(x,0)=x$$ I referred to multiple sources and each of them had some variation in the method. My instructor himself hasn't explained how one gets a particular characteristic equation. Please help :)
The PDE $$ u_y=xuu_x, \quad u(x,0)=x, \label{1}\tag{1} $$ resembles the inviscid Burger's equation $$ u_t+uu_x=0, \quad u(x,0)=f(x),\label{2}\tag{2} $$ the solution of which is given in implicit form by $$ u(x,t)=f(x-ut). \label{3}\tag{3} $$ Inspired by this solution, let's try to solve \eqref{1} with the ansatz $$ u(x,y)=F(g(x)+v(x,y)y)=:F(\xi), \label{4}\tag{4} $$ where $g$ and $v$ are functions to be determined. Substituting \eqref{4} in \eqref{1}, we obtain \begin{align} u_y-xuu_x&=(v_yy+v)F'(\xi)-xF(\xi)(g_x+v_xy)F'(\xi) \\ &=(v_y-xF(\xi)v_x)yF'(\xi)+(v-xF(\xi)g_x)F'(\xi). \label{5}\tag{5} \end{align} The RHS of \eqref{5} vanishes if we choose $v(x,y)=u(x,y)=F(\xi)$ and $xg_x=1$, or $g(x)=\ln(Cx)$. Substituting these results in \eqref{4} we obtain $$ u(x,y)=F(\ln(Cx)+uy). \label{6}\tag{6} $$ To determine $F$, we apply the condition $u(x,0)=x$ to \eqref{6}: $$ F(\ln(Cx)+x\cdot0)=x\implies F(\xi)=C^{-1}e^{\xi}, \label{7}\tag{7} $$ so that $$ u(x,y)=C^{-1}e^{\ln(Cx)+uy}=xe^{uy}. \label{8}\tag{8} $$ Equation \eqref{8} gives the solution to \eqref{1} in implicit form. We can also write $u$ as an explicit function of $x$ and $y$ using the Lambert W function: \begin{align} u=xe^{uy}&\implies -uye^{-uy}=-xy \\ &\implies-uy=W_0(-xy) \\ &\implies u(x,y)=-y^{-1}W_0(-xy). \label{9}\tag{9} \end{align} Finally, a remark is in order here: the solution \eqref{9} is real iff $xy\leq\frac{1}{e}$.