I'm trying to solve $$xu_x+yu_y+uu_z = 0$$ subject to the condition $$u(x,y,0) = xy$$ where both $x>0$ and $y>0$, using the method of characteristics.
Setting up the characteristic equations, I have $$\frac{dx}{x} = \frac{dy}{y}=\frac{dz}{u}=\frac{du}{0}$$
And, solving the odes $$\frac{du}{dx} = 0 \implies u = K_1$$ where $K_1$ is constant. Then, working out $$\frac{dz}{dx} = \frac{u}{x} \implies z-u\ln(x)=K_2$$ where $K_2$ is a constant. Then, using the fact that $K_1 = F(K_2)$ for some arbitrary $F$, I get that $$u = F(z-u\ln(x)).$$ Applying the condition given, I arrive that $$xy = F(-xy\ln(x))$$ and, doing a substitution $w = -xy\ln x$ gives me $$F(w) = -\frac{w}{\ln x}$$
And hence $$u(x,y,z) = \frac{-(z-u\ln(y))}{\ln(y)}$$
But, this is wrong.The u's on both sides of the equation cancel out. Can anyone tell me what I'm doing wrong here? Thanks.
The Sorey's calculus is correct, but a characteristic equation $\ln(x)-\ln(y)=c $ is missing. The solution of the PDE requires an arbitrary function of two variables.
The difficulty encountered with the specified condition comes from the lost of one characteristic equation which restrict the generality of the solution of the PDE. With the full solution, there is no longer a contradiction.
In the calculus below, the form of the general solution on implicit form is apparently different, but is equivalent in fact.