PDE Method of Characteristics with 3 independent variables. Any Idea?

$$u_x + u_y + zu_z = u^3$$

where $$u(x, y, 1) = h(x, y)$$

using characteristic curves.

As far as I have studied, the characteristic lines are as follow: (am I right?)

$$\frac{dx}{1} = \frac{dy}{1} = \frac{dz}{z} = \frac{du}{u^3}$$

I am trying to figure out how to write formula here. please excuse me for not being expert on this website yet. thank you

• Welcome to StackExchange! Could you please be so kind and edit your question so that it easier to understand? We like to help, make it easy to help. – Michael Paris Jan 27 at 22:43
• Can you write down the ordinary differential equations which give you the characteristic curves? – Christoph Jan 28 at 0:50

1. $$dx = \frac{dz}{z}$$: $$x = x_0 + \ln(z)$$,
2. $$dy = \frac{dz}{z}$$: $$y = y_0 + \ln(z)$$,
3. $$\frac{du}{u^3} = \frac{dz}{z}$$: $$u = \left(u_0^{-2} -2 \ln(z)\right)^{-1/2}$$,
where the initial values $$(x_0,y_0,u_0)$$ are all given at $$z=1$$ ($$\ln(z) = 0$$). From the initial condition we now obtain $$$$u_0 = u(x_0,y_0,1) = h(x_0,y_0) \stackrel{1., 2.}{=} h(x-\ln(z),y-\ln(z)),$$$$ and therefore $$$$u(x,y,z) \stackrel{3.}{=} \left(h\left(x-\ln(z),y-\ln(z)\right)^{-2} -2 \ln(z)\right)^{-1/2}.$$$$ You can now verify that this function $$u$$ satisfies indeed both the PDE (if $$h$$ is differentiable) and the initial condition.