would you please help me with solving the 1st order PDE below?

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

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

  • $\begingroup$ 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. $\endgroup$ – Michael Paris Jan 27 at 22:43
  • $\begingroup$ Can you write down the ordinary differential equations which give you the characteristic curves? $\endgroup$ – Christoph Jan 28 at 0:50

Your characteristic equations are correct. You may solve them as

  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 \begin{equation} u_0 = u(x_0,y_0,1) = h(x_0,y_0) \stackrel{1., 2.}{=} h(x-\ln(z),y-\ln(z)), \end{equation} and therefore \begin{equation} u(x,y,z) \stackrel{3.}{=} \left(h\left(x-\ln(z),y-\ln(z)\right)^{-2} -2 \ln(z)\right)^{-1/2}. \end{equation} You can now verify that this function $u$ satisfies indeed both the PDE (if $h$ is differentiable) and the initial condition.

  • $\begingroup$ Thank you very much. It is always reassuring to have other opinions for me. $\endgroup$ – Shay Feb 4 at 13:58

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