Solving a PDE first order in three variables

Solve the following PDE

$$u_{x_1} + u_{x_2} + x_3 u_{x_3} = u^3$$

with cauchy data $$u(x_1,x_2,1) = \phi(x_1,x_2)$$. Also, determine the values of $$x_1,x_2$$ and $$x_3$$ for which the IVP exists.

Try:

The characteristics are given by

\begin{align*} \frac{dx_1}{ds} &= 1, \; \; \; x(r_1,r_2,0) = r_1 \\ \frac{dx_2}{ds} &= 1, \; \; \; x(r_1,r_2,0) = r_2 \\ \frac{dx_3}{ds} &= x_3, \; \; \; x(r_1,r_2,0) = 1 \\ \frac{d z}{ds} &= z^3, \; \; \; x(r_1,r_2,0) = h(r_1,r_2) \\ \end{align*}

We see easily that $$x_1(r_1,r_2,s) = s+r_1$$ and similarly $$x_2 = s + r_2$$ and $$x_3 = e^s$$. Next, we have

$$\frac{dz}{z^3} = ds \implies - \frac{1}{2 z^2} = s + C \implies -\frac{1}{2h^2} = C$$

Therefore,

$$z(r_1,r_2,s) = \frac{ 2 |h(r_1,r_2)| }{\sqrt{1 - 2h(r_1,r_2)} }$$

also, notice that $$s = \log x_3$$ and so $$r_1 = x_1 - \log x_3$$ and $$r_2 = x_2 - \log x_3$$. Therefore, our solution is

$$u(x_1,x_2,x_3) = \frac{ 2 | h( x_1 - \log x_3, x_2 - \log x_3) |}{\sqrt{1-2h( x_1 - \log x_3, x_2 - \log x_3)}}$$

and the solution exists for all $$(x_1,x_2,x_3)$$ since

$$Jac(x_1,x_2,x_3) = 1 \neq 0$$

Is this a correct solution?

You have the right idea; you've just wrong somewhere along the way in forgetting a square and dropping an $$s$$ it looks like. You can always check if your result is a solution or not by directly applying the PDE to it and checking the condition(s).

$$\begin{cases} u_x + u_y + z u_z = u^3 \\ u(x, y, 1) = \phi(x, y). \end{cases}$$

We have the characteristic

\begin{cases} \begin{align} \displaystyle &\frac{dx}{dt} = 1 &x(0) &= s_1 &\implies x(t) &= s_1 + t \\ \displaystyle &\frac{dy}{dt} = 1 &y(0) &= s_2 &\implies y(t) &= s_2 + t \\ \displaystyle &\frac{dz}{dt} = z &z(0) &= 1 &\implies z(t) &= e^t \\ \displaystyle &\frac{d\tilde{u}}{dt} = \tilde{u}^3 &\tilde{u}(0) &= \phi\big( s_1, s_2 \big) &\implies \tilde{u}(t) &= \sqrt{\frac{\phi^2(s_1, s_2)}{1-2\phi^2(s_1, s_2) t}}. \end{align} \end{cases}

The system is readily solvable for $$\{t, s_1, s_2\}.$$

$$u(x,y,z) = \sqrt{\frac{\phi^2(x - \log z, y - \log z)}{1 - 2\phi^2(x - \log z, y - \log z)\log z}}.$$

For the Jacobian, I get

$$J = z.$$

However, unless using the complex-valued logarithm, I would restrict $$z$$ to be positive.