I want to classify a PDE into (hyperbolic, elliptic, parabolic) which I know how to do when I am confronted with a linear PDE.

Can one classify nonlinear PDEs? If yes, how?

I have the following PDE which I want to classify: $$(u_{xx})^2+\exp{(y)}+(u_{yy})=\sin{(x)}$$

Another example is the following:$$(x^2-1)u_{xx}+2xyu_{xx}+(y^2-1)u_{yy}=xu_{x}+yu_{y}$$

with $u:(x,y)\rightarrow u(x,y)$.

The second example is linear.

  • 1
    $\begingroup$ What classification are you looking for? As you said, it is nonlinear. $\endgroup$
    – user9464
    Commented Jan 4, 2017 at 22:45
  • $\begingroup$ @Jack: I know. I am asking for the method to classify non linear PDE's $\endgroup$
    – XPenguen
    Commented Jan 4, 2017 at 22:46
  • $\begingroup$ @Jack:The problem I am given just says classify the non linear PDE. I thought it was about elliptic,hyperbolic,parabolic. $\endgroup$
    – XPenguen
    Commented Jan 4, 2017 at 22:48
  • $\begingroup$ @paulgarrett: The problem I am given just says 'classify the PDE'. I guess it could mean, either classify into linear/nonLinear PDE or what you said. I would like to know both though. I am gonna edit my post to make that clear and not to confuse everyone... $\endgroup$
    – XPenguen
    Commented Jan 4, 2017 at 22:57
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    $\begingroup$ Revising an earlier comment of mine: it is unclear whether one might be expected to approximate (in an ambiguous sense) a non-linear equation by a linear one, and invoke the linear classification. Or, perhaps, depending on context, since the equation is (by some standard) "fully non-linear" (I think), maybe the "right answer" is just "non-linear"? $\endgroup$ Commented Jan 4, 2017 at 23:01

1 Answer 1


Nonlinear PDEs do not always admit a nice classification. For your PDE

$$u_{xx}^2 + u_{yy} = f(x,y)$$

the classification depends on the sign of $u_{xx}$. If $u_{xx}>0$ the equation is elliptic. If $u_{xx}<0$ it is hyperbolic. If $u_{xx}$ switches sign, then the equation can be elliptic in some regions of the domain, and hyperbolic in other regions. This is one reason why nonlinear PDE are more interesting.

A classic example is the Monge-Ampere equation

$$u_{xx}u_{yy} - u_{xy}^2 = f(x,y),$$

which shares some similarities with your PDE. When $u$ is convex, the Monge-Ampere equation is elliptic. This is why people sometimes refer to the "elliptic Monge-Ampere" equation to mean they are studying convex solutions.

There is a nice definition of elliptic for nonlinear PDEs. A PDE

$$F(\nabla^2u,\nabla u, u,x) = 0$$

is called elliptic if

$$F(X,p,z,x) \geq F(Y,p,z,x)$$

whenever $X \leq Y$. Here, $X$ and $Y$ are symmetric $n\times n$ matrices and $X\leq Y$ means that $Y-X$ is non-negative definite.


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