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(This page & the page before it are a source for my question if required)

When solving a nonlinear first order pde $F(x,y,z,p,q) = 0$ one can use the implicit function theorem to solve this for, say, $p$. At a fixed point $(x_0,y_0,z_0)$ we then have the relation $p = G(q)$, or $H(p,q) = 0$, which is the reason why one can generate the monge cone to a surface $z = z(x,y)$ at $(x_0,y_0)$. My question is about the quasilinear situation, this may be basic but I just don't see why one can't follow the exact same process & generate not a tangent line (monge axis) but a monge cone!?

For example, given the nonlinear $F(x,y,z,p,q) = pq - 1 = 0$ one solves for $p$ to get $p = 1/q$ thus we generate a one-parameter family of monge cones, the normal to each being $(1/q,q,-1)$. However for the quasilinear $F(x,y,zp,q) = ap + bq - 1 = 0$ if we follow the same process we find that $((-b/a)q+(1/a),q,-1)$ also generates out a one-parameter family of planes generating a monge cone - no? My guess is that this is incorrect based on something like the Cauchy-Kovalevsky uniqueness theorem, though from my reading of the theorem I don't see how it applies to either - any ideas?

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I struggled with this issue too...this is my explanation from my notes:

In the quasilinear case $F$ can be written $F =(p,q,-1)\cdot(a,b,c)$ for some $a,b,c$ (which are functions of the base point $(x_0,y_0,z_0)$. In this case, there is a specific direction that is in included in EVERY possible tangent plane. Namely, the direction is $(a,b,c)$; no matter what the value of $p,q$, the vector $(a,b,c)$ is always perpendicular to the plane's normal $(p,q,-1)$ and so is always in the tangent plane. Since every possible plane includes this direction, the envelope of all the possible tangent planes is degenerate; it is not a cone as in the general case. Instead it is just an “axis” in the direction $(a,b,c)$.

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  • $\begingroup$ Thanks for the help, I think I see it now, will work on it explicitly asap to make sure but yeah it makes sense intuitively, really appreciate the answer :) If you have any similar intuition for my question about Characteristics I'd really appreciate it cool smiley face $\endgroup$
    – bolbteppa
    Sep 5, 2013 at 15:48

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