9
$\begingroup$

I see that the form of a (e.g.) parabolic equation is $$Au_{xx} + 2Bu_{xy} + Cu_{yy} + Du_x + Eu_y + F = 0$$ with $B^2-4AC=0$ whereas the equation of a parabola is $$Ax^2 + 2Bxy + Cuy^2 + Dx + Ey + F = 0$$ with $B^2-4AC=0$.

They are similar, but is there a deeper relationship between these two mathematical concepts than the mere analogy on their notations?

$\endgroup$
  • $\begingroup$ See math.stackexchange.com/questions/21525/…. $\endgroup$ – KCd Jan 4 '15 at 1:50
  • $\begingroup$ @KCd: I did not know! I suppose my answer is not as good. $\endgroup$ – Bombyx mori Jan 4 '15 at 1:58
  • $\begingroup$ @Bombyxmori: your answer has the virtue of being short. $\endgroup$ – KCd Jan 4 '15 at 2:21
  • $\begingroup$ @KCd: I had seen that, but that question is about their definitions, in particular if the PDE is nonlinear and above second-order. My question is about the existence of any relation between a parabolic PDE and a parabola beyond their notations. $\endgroup$ – toliveira Jan 4 '15 at 21:42
  • $\begingroup$ @toliveira: I think this is from the principal symbol map, as I wrote down below. But this would certainly not work for non-linear PDE, in which people then talk about semilinear, fully non-linear, etc. $\endgroup$ – Bombyx mori Jan 5 '15 at 15:39
4
$\begingroup$

It is because the principal symbol of the partial differential operator involved in the process. You can think a typical linear PDE as $$ Pu=0 $$ where $u$ is in some function space with possible boundary condition given, and $P$ is a linear differential operator like $\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial xy}$. The symbol of this operator is given by the polynomial $\xi_{2}^{2}+\xi_1^{2}+\xi_1\xi_2,\xi=(\xi_1,\xi_2)$. So on a purely formal level, (linear) PDEs are classified by their (principal) symbols. For example, an equation is called elliptic if its principal symbol is invertible.

But this leaves the question that why the symbol matters at all in the classification process. Intuitively, it seems "clear" that the highest derivative should matter most. However to show this rigorously is in fact quite deep; a proof of elliptic operators satisfies some basic nice properties takes at least half a page and usually require one knew pseudo-differential operators. I suppose others might be better at answering this. If you can follow, I also think Klainerman's article on PDE might be helpful, in which he discussed thoroughly the classification of PDEs.

$\endgroup$
  • $\begingroup$ Thank you, but it still seems as if we can notate a parabola and an parabolic PDE in similar ways, so we will call them similarly. Or are there properties studied on polynomial theory that apply to parabolas as well as to parabolic PDEs? $\endgroup$ – toliveira Jan 5 '15 at 20:04
  • $\begingroup$ @toliveira: I guess the name parabolic does come from that, but nowadays parabolic equations means a lot more than just second order PDEs of the type your described. So any attempt from this side is not exhaustive. $\endgroup$ – Bombyx mori Jan 5 '15 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.