I am trying to use the method of characteristics to solve the following first order PDE in three variables:
$$u_x + x \,y \,u_y + 2 x^2 \,z\,\ln z\,u_z = 0 $$
I have begun with the following: $ \frac{dx}{dt} = 1, \frac{dy}{dt}=x\,y, \frac{dz}{dt}=2x^2\,z\,\ln z$
I note that $x = t \,$ (if $\, x(0)=x_0$), and thus y and z are not independent of t, so I think I need to introduce another variable. Here is where I am confused.
$\frac{dy}{ds}=\frac{dy^2}{d^2t}=y, \frac{dz}{ds}=\frac{dz^2}{d^2t}=4\,t\,z\ln z$
This leads to $\ln y = s, y = e^s$ and $\frac{dz}{z\ln z}=4t \,ds, \ln (\ln z)=4ts$ and $z = e^{e^{4ts}}$
so $$x=t, y=s, z = e^{e^{4ts}}$$
Best Answer
$$u_x + x \,y \,u_y + 2 x^2 \,z\,\ln z\,u_z = 0 $$ $$\frac{dx}{1}=\frac{dy}{xy}=\frac{dz}{2x^2z\ln|z|}=\frac{du}{0}$$
First family of characteristic curves from $\quad du=0 \quad\to\quad u=c_1$
Second family of characteristic curves from $\quad \frac{dx}{1}=\frac{dy}{xy} \quad\to\quad \frac{x^2}{2}-\ln|y|=c_2$
Third family of characteristic curves from $\quad\frac{dx}{1}=\frac{dz}{2x^2z\ln|z|} \quad\to\quad \frac{2}{3}x^3-\ln|\ln|z||=c_3 $
General solution of the PDE expressed on the form of implicit equation : $$\Phi\left(u\:\:,\:\: \frac{x^2}{2}-\ln|y| \:\:,\:\: \frac{2}{3}x^3-\ln|\ln|z|| \right)=0$$ $\Phi$ is any differentiable function of three variables.
Or equivalently, general solution of the PDE on explicit form :
$$u(x,y,z)=\text{F}\left( \frac{x^2}{2}-\ln|y| \:\:,\:\: \frac{2}{3}x^3-\ln|\ln|z|| \right)$$ F is any differentiable function of two variables.