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I wanted to know how one would classify a nonlinear PDE into elliptic, hyperbolic or parabolic forms. The particular PDE I would like to know about would be

\begin{align} \partial_t u &= D(\partial^2_{x} +\partial^2_y) u + AS((\partial_x u)^2+(\partial_y u)^2) +AA (\partial_xu)(\partial_yu) + c(1-c) \end{align}

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  • $\begingroup$ Why do you think this equation is nonlinear? $\endgroup$ Apr 7, 2015 at 3:03
  • $\begingroup$ I guess my notation was misleading. Now the (grad) squared terms are clearer. $\endgroup$
    – legolagon
    Apr 10, 2015 at 23:06
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    $\begingroup$ Thanks. What are $D$, $A$, $c$, etc.? $\endgroup$ Apr 11, 2015 at 1:55

1 Answer 1

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The particular PDE I would like to know about would be $$\partial_t u = D(\partial^2_{x} +\partial^2_y) u + AS((\partial_x)^2+(\partial_y)^2)u +AA \partial_x\partial_yu + c(1-c)$$


The classification of Partial Differential Equations into elliptic, parabolic and hyperbolic is formally defined for linear second order PDEs only.

Then, assuming $u=u(x,y)$, you can determine its type by bringing it to its canonical form

$$ \mathsf{A}\,u_{xx} + 2\,\mathsf{B}\,u_{xy} + \mathsf{C}\,u_{yy} + \mathsf{D}\, u_{x} + \mathsf{E} \, u_{y} + \mathsf{F} = 0 $$

In that case, if

  1. $\mathsf{B}^2 - 4\mathsf{A}\mathsf{C} <0$, then the equation is elliptic,
  2. $\mathsf{B}^2 - 4\mathsf{A}\mathsf{C} =0$, then the equation is parabolic,
  3. $\mathsf{B}^2 - 4\mathsf{A}\mathsf{C} >0$, then the equation is hyperbolic.

One way to apply this classification to a general (e.g. quasilinear, semilinear, nonlinear) second order PDE is to linearize it. It is actually unclear whether your original PDE is linear or not:

  • If $A$, $D$, and $S$ are constants, or functions which depend on $(x,y,t)$, then the equation is linear, so you can write out the canonical form and determine the type of PDE.

  • If, however, $A$, $D$, and $S$ are (nonlinear) functions, which depend on $u$ or its derivatives, then you might have to compute a linearization of PDE first, and then determine its type.

Note that for the classification purposes we do not make a distinction between spatial ($x,y$) and temporal ($t$) variables.

In your case function $ u = u(x,y,t) $ depends, apparently, on three variables. This means that you will have to use generalized classification for PDEs defined on $\mathbb{R}^{3}$, which can be found in the Wikipedia article about PDEs.

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  • $\begingroup$ How would I compute it's linearization? I am not sure how to linearize these terms: \begin{align} \partial_t u &= AS((\partial_x u)^2+(\partial_y u)^2) \end{align} $\endgroup$
    – legolagon
    Apr 10, 2015 at 23:08
  • $\begingroup$ If by $ AS\Big( \big(\partial_{x} u \big)^{2} + \big(\partial_{x} u \big)^{2} \Big) $ you mean $ A\cdot S\cdot \Big( \big(\partial_{x} u \big)^{2} + \big(\partial_{x} u \big)^{2} \Big) $, where $\cdot$ is regular multiplication, and if $A$ and $S$ do not depend on $u$ or its derivatives, then you already have it in the linear form. You might want to rewrite as $$ A\cdot S\cdot (\partial_x u)^2+A\cdot S\cdot (\partial_y u)^2 - \partial_t u = 0 $$ though, thus obtaining the canonical form of your PDE. $\endgroup$
    – Vlad
    Apr 13, 2015 at 1:32
  • $\begingroup$ I don't understand your comment. The terms I emphasized are gradient squared terms. They are clearly nonlinear as they are square terms. How can they be linearized . ( there is all the u(1-u) bit at the end which is nonlinear.) $\endgroup$
    – legolagon
    Apr 20, 2015 at 22:43
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    $\begingroup$ If you are talking about treating $$\begin{align} \partial_t u &= AS((\partial_x u)^2+(\partial_y u)^2) \end{align}$$ as an equation by itself, then it is of the first order, and the elliptic/parabolic/hyperbolic classification is not applicable. If, on the other hand, you are viewing these terms as a part of the original equation, then they play no role in 2-nd order PDE classification. The original equation is semilinear, i.e. it it linear w.r.t second order terms. The coefficients in front of $u_{xx},u_{yy}$, and $u_{xy}$ are the only ones used in determining the type of equation. $\endgroup$
    – Vlad
    May 27, 2015 at 0:42

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