Given the pde $$xu_x+yu_y+uu_z=0$$ where $$u(x,y,0)=xy$$ for $x>0$ and $y>0$, the solution, gotten from the method of characteristics is $$u(x,y,z) = xye^{\frac{-2x}{u}}$$
My question is, how would one describe this solution? I have no idea how to even imagine it.
Thanks.
EDIT: Workings
Characteristics : $$\frac{dx}{x} = \frac{dy}{y} = \frac{dz}{u} = \frac{du}{0}$$
$\frac{du}{dx} = 0 \implies u = K_1 = $ constant
$\frac{dx}{dy} = \frac{x}{y} \implies \frac{y}{x} = K_2$
$\frac{dy}{dz} = \frac{y}{u} = \frac{y}{K_1} \implies \ln (y) - \frac{z}{u} = K_3$
So, $u = F(K_2, K_3) = F(\frac{y}{x},\ln (y) -\frac{z}{u}) = F(X,Y)$
Using the initial condition give above, $xy = F(\frac{y}{x},\ln(y))$. So, $X = \frac{y}{x}$ and $Y = \ln (y)$. Solving for $x$ and $y$, we get that $xy = \frac{e^{2Y}}{X}$
Then, $u(x,y,z) = \frac{e^{2Y}}{x} = \frac{xe^{2(\ln(y)-z/u)}}{y} = xye^{\frac{-2z}{u}}$
It's an inavoidably implicit function. Some nice way to visualize some of its characteristics is using a 3D plotter. For a three variables function the surfaces of $u$ constant are interesting to see. It's easy to isolate $z$ as function of $x$, $y$, and $u=k$. I've used GeoGebra to plot these surfaces. It has an slider to change the value of $k$
$$k=xy^{-2z/k}\implies z=-(k/2)\ln(k/xy)$$
Isolating $z$ has problems as function, so, you can isolate $x$ and change variables $\bar x=z$, $\bar z=x$ and $\bar y=y$ for the program could plot the function (entered always as $z=f=f(x,y)$