# Classification of this nonlinear PDE into elliptic, hyperbolic, etc.

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}

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

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.

• 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} Apr 10, 2015 at 23:08
• 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.
• 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.