I got stuck on the following problem:

Determine the subsets of $\mathbb{R}^2$ where the pde $$u_{xx}+2xu_xu_{xy}+yu_{yy}+yu_x=1$$ is elliptic, hyperbolic and parabolic respectively.

Now, at first I thought this to be an exceedingly simple task, all we need to do is look at the determinant of $$\begin{pmatrix}1 & xu_x \\xu_x & y\end{pmatrix}.$$

If it is positive, we are elliptic, if it's negative, we are hyperbolic, if it's $0$, we are parabolic. So far, so good.

However, what kind of ruined the notion that this was going to be oh so easy was the fact that suddenly I was left with the differential inequalitiy $$y> (xu_x)^2$$ with symmetric versions for $=,<$. Duh.

Now, think positively, at least one does get the trivial case that for $y<0$ we are hyperbolic. And for $(x,y)=(0,0)$ we are parabolic. Also, for $x=0,y>0$ we are elliptic. Only about half of the real numbers left, yay. So in the following we could assume that $y\geq0$, you never know, it might help.

In order to try and solve this, I figured I could at least try to do something for the parabolic case: $$\frac{y}{x^2}=u_x^2 \quad\Leftrightarrow \\ \frac{\sqrt{y}}{|x|}=|u_x|$$ and now I cannot even say anything definitive about the integral because, well, we have $|u_x|$ instead of $u_x$. Also, even if I could integrate, if I had $\sqrt{y}\ln|x|+c=|u(x,y)|$, I still would have no idea for which sets $(x,y)\in\mathbb{R}^2$ this is true. Same goes for the inequalities, if not worse, which goes to say I've basically not made any progress with this.

One other thing I've tried is to at least look at the case $x\neq 0,y=0$, where if $u_x\neq0$ we are also trivially hyperbolic, but inserting that into the original pde doesn't seem to give me any kind of contradiction ($u_{xx}$ does, unfortunately, not need to be $0$, too - also, we do not (well, at least, I don't) know anything about it's sign, since it only appears in our inequality quadratically).

Also note that it is explicitly stated that it is not necessary to solve the original pde. In case this is even possible; I'm fairly new to this subject.

Hence I assume that I have simply been unable to find anything resembling a "correct" approach, and would be quite thankful for anyone shedding light on what to do here.

  • $\begingroup$ This is a nonlinear equation, so its classification depends on what function you linearize around. If you linearize about a particular function $u$, then it will be elliptic wherever $y>(xu_x)^2$, hyperbolic where $y<(xu_x)^2$, and parabolic on the boundary between the two. But the location of the boundary depends on $u$. $\endgroup$
    – Jack Lee
    Oct 15, 2014 at 22:07
  • $\begingroup$ Pardon me, I am not really familiar with the concept of linearization. The definition given, however, applied on this particular pde, leads to the cases I've stated - and what I am actually interested in is the exact boundary you speak of, just in a more explicit form. $\endgroup$ Oct 15, 2014 at 22:10
  • $\begingroup$ There's not really any good way to make sense of ellipticity, etc., of a nonlinear equation other than looking at the linearization. But even if you don't know what that means, the main point is that the classification of such an equation depends on a choice of a function $u$, so you can't really say any more about where it's elliptic until you pick a particular $u$. For example, choosing $u(x,y) = y$ leads to the conclusion that it's elliptic exactly where $y>0$. Choosing $u(x,y)=x$ leads to $y>x^2$. As you can see, many different regions are possible. $\endgroup$
    – Jack Lee
    Oct 15, 2014 at 22:17
  • $\begingroup$ The task I was given, however, was to characterize the regions of $\mathbb{R}^2$, explicitly. If it helps, there was a smiliar, albeit trivial one, where the pde had the form $4u_{xx}+2xu_{xy}+(y^2+1)u_{yy}=r(x,y,u)$, leading to the border $4y^2-x^2=4$. Not depending on the choice of my $u$. This is basically what I should somehow find for above pde as well, just that I have no idea how to get there. $\endgroup$ Oct 15, 2014 at 22:23
  • $\begingroup$ Then the problem is not well posed. For the nonlinear pde you wrote down, there is no set in the plane with the property that the equation is elliptic exactly on that set regardless of what $u$ is. $\endgroup$
    – Jack Lee
    Oct 16, 2014 at 4:54

1 Answer 1


This is a nonlinear equation, so its classification as elliptic, parabolic, or hyperbolic depends on what function you linearize around. For example, if you linearize around $u(x,y)=y$, the equation is elliptic exactly where $y>0$. If you linearize around $u(x,y)=x$, then it's elliptic where $y>x^2$. Many different regions are possible.

Since you were given the task of explicitly characterizing the regions of $\mathbb R^2$ where the equation is elliptic, parabolic, or hyperbolic, the problem is not well posed. There is no set in the plane with the property that the equation is elliptic exactly on that set regardless of what $u$ is.


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