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Let's start with a fully non-linear equation of first order

$$ F(x,y,z,z_x,z_y) = a(x,y,z,z_x,z_y) \frac{dz}{dx} + b(x,y,z,z_x,z_y) \frac{dz}{dy} - c(x,y,z,z_x,z_y) = 0 $$

To solve such an equation, I employ the methods of characteristics.

$$ \frac{dx}{ds} = \frac{d}{dp}F \\ \frac{dy}{ds} = \frac{d}{dq}F \\ \frac{dp}{ds} = - \frac{d}{dx}F - p\frac{d}{dz}F \\ \frac{dq}{ds} = - \frac{d}{dy}F - q \frac{d}{dz}F \\ \frac{dz}{ds} = p \frac{d}{dp}F + q \frac{d}{dq}F $$ where $ p = z_x $ and $ q = z_y $.

I want to provide more information about the type of such an equation based on the characteristics of solutions.

  1. Can you do the following to classify the PDE? Assume that u is infinitely differentiable. You could differentiate each side and use the classification of second-order PDE to determine if the system is hyperbolic, parabolic, or elliptic. Is it a valid approach? I think that this approach is wrong, but I can't provide a rigorous argument for it.
  2. The methods of characteristics involve derivating F based on p,q,x,y, and z. In the characteristics forms, is the PDE a first or a second-order PDE?

NB: If you could provide me with resources on the subject, I would be pleased. I read Strauss but didn't find something insightful on the subject.

NBB: I have seen multiple resources saying that first-order PDEs are hyperbolics, but without providing insight.

Lödrik
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  • Indeed, a scalar first order PDE is always hypnotic. The fact that it can be solved using the method of characteristics is an indication of this. – Deane Nov 07 '23 at 20:18
  • hypnotic? Do you mean hyperbolic?

    Can you elaborate on "The fact that it can be solved using the method of characteristics is an indication of this"?

     – Lödrik Nov 07 '23 at 20:23
  • Yes, hyperbolic. Hyperbolic PDE can be solved using an initial value problem. Elliptic PDE cannot. Parabolic PDE can be but the details are quite different from hyperbolic. – Deane Nov 07 '23 at 20:34
  • To elaborate, if $F$ is assumed to be a real analytic function of its inputs and has at least one noncharacteristic direction, then a suitable initial value problem can be solved using the Cauchy-Kovalevski theorem. If $F$ is assumed only to be smooth, then it is known that a necessary condition for the initial value problem be solvable for any smooth initial data is that the PDE be at least weakly hyperbolic. If the PDE is strongly hyperbolic, then it is known that the initial value problem always has solutions. There's a gray area in between. – Deane Nov 08 '23 at 14:44

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