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The reference here is Pinchover & Rubinstein's An introduction to partial differential equations, pages 36-37. It's about the existence and uniqueness of a solution to the equation $a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)$ with initial condition (curve on the integral surface) $x(0,s)=x_0 (s), y(0,s)=y_0(s), u(0,s)=u_0(s)$.

It involves the construction of a surface, by solving a family of systems of ODEs (whose solutions $x(t,s), y(t,s), u(t,s)$) are curves on the desired surface and thus "knit it together"). They claim that "the transversality condition ($x_t(0,s)y_s(0,s)-y_t(0,s)x_s(0,s)\neq 0$) implies that the parametric representation provides a smooth surface". Why is it enough to check this along the initial curve?

Then they check that this surface satisfies the PDE. On they go: "to show that there are no further integral surfaces, we prove that the characteristic curves we constructed must lie on an integral surface". I don't know what this means, because I thought that was precisely what they had just done.

Most likely I'm just completely clueless and their writing is not to blame.

I appreciate any comments that may enlighten me on understanding this proof.

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I may be completely wrong, but I think that this condition (assuming everything is continuous) just locally implies that you will get a smooth surface. Since, if a continuous function is nonzero at a point, it must also be nonzero on a neighborhood of that point. So the transversality condition will hold on some neighborhood of the initial curve. So, I think the solution may break down outside this neighborhood. I hope this helps in any way ... –  Dejan Govc Jun 8 '12 at 9:49
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I think @DejanGovc is right: the statement about integral surface is local (has an unspecified $\epsilon$ in it). To understand the second part, read it as "... we prove that the characteristic curves we constructed must lie on any integral surface". Previously they checked that the union of curves is an integral surface. Now they show that if there is any other integral surface, it must also contain these curves. Hence, it's the same surface. –  user31373 Jun 9 '12 at 22:57

1 Answer 1

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+100

You can use the inverse function theorem as follows. Take an extra parameter, $r$ say, and map $(t,s,r)$ to $(x(t,s), y(t,s), z(t,s)+r)$. The Jacobian is ${\rm det}\begin{bmatrix}x_t & x_s & 0 \cr y_t & y_s & 0 \cr z_t & z_s & 1\end{bmatrix}$, which has been assumed nonzero by choice of a noncharacteristic initial curve, at $(t,s,r)=(0,s_0,0)$, for each $s_0$ within the domain of the initial curve. By the inverse function theorem, there is a diffeomorphism between a neighborhood of $(0,s_0,0)$ and a neighborhood of $(x(0,s_0),y(0,s_0),z(0,s_0))$. In particular this diffeomorphism carries a piece of the $(t,s,0)$ plane onto a piece of smooth surface. The theorem is local, so the piece might be small, and you have a solution $u$ in a neighborhood of $(x_0(s_0),y_0(s_0),u_0(s_0))$.

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