# How to determine where a non-linear PDE is elliptic, hyperbolic, or parabolic?

I'm trying to understand the classification of PDEs into the categories elliptic, hyperbolic, and parabolic. Frustratingly, most of the discussions I've found are "definition by examples.''

I think I more or less understand this classification in the case of quasi-linear second-order PDE, which is what's described on Wikipedia. I want to understand what makes a non-linear second-order PDE elliptic, parabolic, hyperbolic. MathWorld says something vague about this: "A non-linear PDE is elliptic at a solution u if its linearization is elliptic at u"   (http://mathworld.wolfram.com/EllipticPartialDifferentialEquation.html). What is meant by "linearization" here?

To be concrete, I want to be able to answer a question like the following: where (in the $x-y$ plane) is the non-linear second-order PDE

$$a \partial_x^2 u + b \partial_x \partial_y u + c \partial_y^2 u + d (\partial_x u)^2 + e \partial_y u = 0$$

elliptic, hyperbolic, or parabolic (here all coefficients depend on $x,y,$ and $u$)? This doesn't really fit the form of any discussions I've found, so I'm not sure how to make sense of it.

EDIT: I am aware of the question at Mathematical precise definition of a PDE being elliptic, parabolic or hyperbolic, but I believe that the answer there really only addresses linear PDE.

Several people have marked this question as already being answered at that link, but I don't see how. I would be much obliged if you could sketch a procedure for classifying the qualitative behavior for a non-linear example operator of the form I describe.

• I saw that question and I do not think that the answer there addresses the generality I'm interested in. It seems to really only address the quasi-linear case, which the example PDE above is not. – Tony Apr 14 '14 at 16:14

## 1 Answer

At your link to MathWorld, "linearization" is to be understood as a Fréchet derivative of the appropriate nonlinear mapping, which has become quite habitual nowadays. Your second-order PDE is quasilinear, i.e., linear w.r.t. the highest-order derivatives. A strict formal definition of a quasilinear nonlinear PDE is generally being omitted mostly due to its rather awkward styling that can hardly be avoided. Indeed, in your case, the strict formal definition could look like this. First, we describe $a,b,c,d,e\,$ as functions of $x,y,u$, with $u$ being some unknown function of independent variables $x,y$. Second, we introduce a function $F=F(x,y,p,q_1,q_2,r_{11},r_{12},r_{22})\,$ of eight independent variables being a linear polynomial in its last three variables $\,r_{11},r_{12},r_{22}\,$ reserved for the highest-order deivatives. Namely, in your case, it is $$F=a(x,y,p)r_{11}+b(x,y,p)r_{12}+c(x,y,p)r_{22}+d(x,y,p)(q_{1})^2+e(x,y,p)q_{2}\,.$$ And finally third, we proclaim your equation to be of the form $$F\bigl(x,y,u(x,y),u_x(x,y),u_y(x,y),u_{xx}(x,y),u_{xy}(x,y),u_{yy}(x,y)\bigr)=0\quad \forall\, x,y.$$
At any point $(x,y)$, classification of your equation is determined by the value of the expression $$D(x,y)\overset{\rm def}{=}\Bigl(b\bigl(x,y,u(x,y)\bigr)\Bigr)^2 -4a\bigl(x,y,u(x,y)\bigr)\!\cdot\! c\bigl(x,y,u(x,y)\bigr),$$ with the equation being called elliptic whenever $D(x,y)<0$, hyperbolic whenever $D(x,y)>0$, and parabolic whenever $D(x,y)=0$. There can be absolutely nothing else to it, though one can't help making a remark that identifying the parabolic type with just $D(x,y)=0$ now amounts to a thorough anachronism still tolerated in PDE as something like Historical Landmark.

Fully nonlinear PDE. In case a nonlinear PDE is not quasilinear, classification is made judging by the linear part of the nonlinear mapping, i.e., by its Fréchet derivative that dominates questions of local solvability for the nonlinear mpapping. Just to illustrate how it works, consider some simple example of the second-order nonlinear partial differential operator, say, $$L(u)\overset{\rm def}{=}F(u_{xx},u_{xy},u_{yy}), \quad F\in C^1(\mathbb{R}^3),$$ defined on differentiable functions $u=u(x,y)$ with some suitable choice of function spaces. The Fréchet derivative of nonlinear mapping $L$ at the solution $u$ is a linear partial differential operator with variable coefficients $$L_u(v)\overset{\rm def}{=}F_p(u_{xx},u_{xy},u_{yy})v_{xx}+ F_q(u_{xx},u_{xy},u_{yy})v_{xy}+F_r(u_{xx},u_{xy},u_{yy})v_{yy}$$ where notations $F_p\,,F_q\,,F_r$ are meant to signify the partial derivatives $$F_p={\partial_p}F(p,q,r),\quad F_q={\partial_q}F(p,q,r), \quad F_r={\partial_r}F(p,q,r).$$ At any point $(x,y)$, classification of the linear partial differential operator $L_u(v)$ is of course determined by the value of the expression $$D_u(x,y)\overset{\rm def}{=}\bigl(F_q(u_{xx},u_{xy},u_{yy})\bigr)^2 -4F_p(u_{xx},u_{xy},u_{yy})\!\cdot\!F_r(u_{xx},u_{xy},u_{yy}).$$ Hence, the nonlinear partial differential operator $L(u)$ at the solution $u$ is called elliptic at a point $(x,y)$ whenever $D_u(x,y)<0$, and likewise so on.