# Solving 1st order PDE with three variables using method of characteristics

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 $$x=t, y=s, z = e^{e^{4ts}}$$

• where is the boundary data? – qbert Nov 30 '17 at 5:54
• I got no boundary data, I am supposed to write them as an autonomous system and express my answer as a generic function of the first integrals. – NCAD Nov 30 '17 at 6:48

$$u_x + x \,y \,u_y + 2 x^2 \,z\,\ln z\,u_z = 0$$ $$\frac{dx}{1}=\frac{dy}{xy}=\frac{dz}{2x^2z\ln|z|}=\frac{du}{0}$$

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.

• I don't agree with the doraemonpaul's result. I think that $z^{-e^\frac{2x^3}{3}}$ is not correct and should be $z^{ e^{-\frac{2}{3}x^3}}$. May be a typo ? In fact, after correction : $$u=f\left(ye^{-\frac{x^2}{2}},z^{ e^{-\frac{2}{3}x^3} }\right)=\text{F}\left( \frac{x^2}{2}-\ln|y| \:\:,\:\: \frac{2}{3}x^3-\ln|\ln|z|| \right)$$ where $f(\alpha,\beta)=F(X,Y)$ with $\alpha=e^{-X}$ and $\beta=e^{e^{-Y}}$. Except the discrepancy, both results are consistent and are two equivalent form of function. – JJacquelin Nov 30 '17 at 15:09

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=xy=ty$ , letting $y(0)=y_0$ , we have $y=y_0e^\frac{t^2}{2}=y_0e^\frac{x^2}{2}$
$\dfrac{dz}{dt}=2x^2z\ln z=2t^2z\ln z$ , letting $z(0)=z_0$ , we have $z=z_0^{e^\frac{2t^3}{3}}=z_0^{e^\frac{2x^3}{3}}$
$\dfrac{du}{dt}=0$ , letting $u(0)=f(y_0,z_0)$ , we have $u(x,y,z)=f(y_0,z_0)=f(ye^{-\frac{x^2}{2}},z^{e^{-\frac{2x^3}{3}}})$
• So I'm mistaken in my understanding of the theory of autonomous systems in that $\frac{dy}{dt}$ and $\frac{dz}{dt}$ do not have to be independent of t? – NCAD Nov 30 '17 at 13:58