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There seems to be a major inconsistency (perhaps due to my lack of understanding) between what Folland calls a "characteristic" and what I had previously thought was a characteristic.

For example, Folland says that the characteristics of the equation $\partial_x u = 0$ are $\{ \xi : \xi_1 = 0\}$ (in $\mathbb{R}^2$). This confuses me to no end since I thought the characteristic hypersurface was the $x_1$ axis here? According to this definition it's the orthogonal compliment to this. The same issue arises with the heat operator $L=u_t-u_{xx}$ where he says the characteristics are $\{\xi \neq 0 : \xi_x = 0\}$. Aren't the characteristics $t=$ cosntant which are orthogonal to these?

Sorry if this question is elementary but it's given me a real headache and I'm not sure what it is I'm missing here.

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You're confusing the so-called "characteristic variety", which is the set of all $\xi$ where the symbol vanishes and lives in the cotangent fiber, and a "characteristic hypersurface", which is a hypersurface in the domain of the PDE such that its conormal cotangent vector (known classically as the normal vector) at each point lies inside the characteristic variety. In other words, a hypersurface is characteristic if the symbol of the PDE vanishes when evaluated on a conormal cotangent vector. –  Deane Yang Sep 26 '10 at 17:01
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Dorian, it's a lot easier to explain this stuff in person than on the internet. It says on your profile that you're a grad student at Courant. Couldn't you consult a classmate or professor there? Or, if you're willing to take the A train to Brooklyn, you're welcome to drop by my office some time and ask me. –  Deane Yang Sep 26 '10 at 17:13
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I assume you mean Folland's "Intro to PDE" book. Take a look at the end of Section 1.A (p.34 in the second edition). "A hypersurface is called characteristic for $L$ at $x \in S$ if the normal vector $\nu(x)$ to $S$ at $x$ is in $\mathrm{char}_x(L)$". –  Hans Lundmark Sep 26 '10 at 17:31
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I haven't looked at Folland or John in a long time, but I am pretty sure that if you parse what they say very carefully, you'll figure it out. It is confusing. I also recommend spending the time to learn the more modern language of cotangent and conormal vectors. It is also confusing at first, but makes things much clearer in the long run. –  Deane Yang Sep 26 '10 at 17:43
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Just to add one more remark to Deane's comment: one of the reasons to consider characteristic varieties rather than characteristic surfaces is that in more than two dimensions, for variable coefficient PDEs, characteristic varieties are not necessarily integrable. So that you may have a good description of the characteristic variety without having any hypersurface that realizes it. –  Willie Wong Sep 29 '10 at 17:48

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