I am interested in solutions to the Monge-Ampere equation for a smooth function $h(x,y)$ of two variables(though I suppose I could try to make do with $C^2$ solutions). The equation is:

$$\det[\nabla^2(h(x,y))]=0 $$

Here $\det$ denotes the determinant and $\nabla^2$ denotes the Hessian. Geometrically this says that we have a smooth developable surface $(x,y,h(x,y))$ of constant Gaussian curvature.

I want this solution to be defined a half-plane where $x > R $ for some $ R >> 0$. And I want it to satisfy the following properties.

$$h(x,y)=x \quad when \quad |y| < \epsilon $$ $$ h(x,y)=\sqrt{x^2 + y^2} \quad when \quad |y| > 2\epsilon $$

Do such solutions exist?


The answer in the smooth case is 'no', because the differential equations governing the second fundamental form of the graph (which has Gauss curvature $K\equiv0$) show that the entire part of the domain $y>0$ (and $x>R$) will have to belong to the non-planar locus, i.e., the open set where the second fundamental form of the graph $\bigl(x,y,h(x,y)\bigr)$ is nonzero. That rules out patching together the solutions in the way that you want.

Here is the basic argument: Suppose that one had a smooth solution $h$ of the given equation defined on the half-plane $x>R>>0$, agreeing with $\sqrt{x^2+y^2}$ when $|y|\ge2\epsilon$ and agreeing with $x$ when $|y|\le \epsilon$. The surface $$ S = \{ \bigl(x,y,h(x,y)\bigr)\ \|\ x>R, y\in\mathbb{R} \}\subset\mathbb{R}^3 $$ is then a surface with Gauss curvature $K\equiv 0$, i.e., its first fundamental form $\mathsf{I}$ is flat.

Now, the second fundamental form $\mathsf{I\!I}$ must have rank at most $1$ everywhere (by the Gauss Equation). Let $N\subset S$ denote the non-planar domain, i.e., the open subset where $\mathsf{I\!I}$ is nonvanishing, and let $P\subset S$ denote the planar subset, i.e, the closed subset of $S$ on which $\mathsf{I\!I}$ vanishes. Note that $N$, because it is open, must properly contain the part of $S$ that lies above the two 'quadrants' $x>R$ while $|y|\ge2\epsilon$. Meanwhile $P$, which is closed, at least contains the part of $S$ that lies above the 'plank' $x>R$ while $|y|\le \epsilon$.

Now, a classic fact from surface theory: Suppose that $U\subset\mathbb{R}^3$ is a connected and simply connected surface whose first fundamental form is flat while its second fundamental form is nowhere vanishing (and hence, after an orientation change if necessary, can be assumed to be nonnegative). Then there exist smooth functions $s$ and $t$ on $U$, unique up to additive constants, and smooth functions $a$ and $b$ on $U$ that satisfy $\mathrm{d}a\wedge\mathrm{d}t = \mathrm{d}b\wedge\mathrm{d}t = 0$ (i.e., locally, at least, $a$ and $b$ are functions of $t$) such that $a-bs>0$ on $U$ and $$ \mathsf{I} = \mathrm{d}s^2 + (a-bs)^2\,\mathrm{d}t^2 \qquad\text{and}\qquad \mathsf{I\!I} = (a-bs)\,\mathrm{d}t^2 $$ This is easily proved via the structure equations, so I won't go into that here, except to remark that the level sets of $t$ (which are the kernel of $\mathsf{I\!I}$) are the lines that rule the surface $U$. (Moreover, there is a converse, in the sense that, for any two smooth functions $s$ and $t$ on a simply-connected surface $U$ that satisfy $\mathrm{d}s\wedge\mathrm{d}t\not=0$ and any two functions $a$ and $b$ that satisfy $\mathrm{d}a\wedge\mathrm{d}t = \mathrm{d}b\wedge\mathrm{d}t = 0$ and $a-bs>0$, the above quadratic forms $\mathsf{I}$ and $\mathsf{I\!I}$ satisfy the Gauss and Codazzi equations and so define a flat, nonplanar surface in $\mathbb{R}^3$, unique up to rigid motion.)

For example, when $U$ is the graph $z = \sqrt{x^2+y^2}$, one finds that $$ (x,y,z) = \left(\frac{s\ \cos(t\sqrt2)}{\sqrt2},\frac{s\ \cos(t\sqrt2)}{\sqrt2},\frac{s}{\sqrt2}\right) $$ with $s>0$ and $t$ can be regarded as periodic of period $\pi\sqrt2$ (after all, $U$ is not simply-connected). One also finds, in this case that $a\equiv0$ and $b\equiv -1$. I.e., in this case, $\mathsf{I} = \mathrm{d}s^2 + s^2\,\mathrm{d}t^2$ and $\mathsf{I\!I} = s\,\mathrm{d}t^2$.

Now, go back to our putative example $S$, which, by hypothesis, agrees with the above cone above the quadrant where $x>R$ and $y\ge 2\epsilon$. It follows that in the connected component of the region $N\subset S$ that contains that part of the cone, we can extend the functions $s$ and $t$ to any $1$-connected open set $U$ that contains this conical part and that lies inside $N$. But this forces $U$ to also be part of the cone (since the first and second fundamental forms are the same). The point is that the ruling lines in $U$ will have to be extensions of the ruling lines over the conical part, and the issue is that the second fundamental form cannot go to zero along such a line until you get to $s=0$, which, because $x>R$, can never happen along such a line with $t$ in the range $0<t<\pi/(2\sqrt2)$.

It follows that $N$ must contain the entire locus lying above the quadrant $x>R$ while $y>0$, but this is impossible, since, by hypothesis, $S$ contains the planar locus above the 'plank'.

Thus, such a smooth solution joining the two partial solutions cannot exist. The argument would work just as well for $C^k$ when $k\ge 3$, but I'm not sure about $C^2$. Probably, it's OK, because the second fundamental form will still be continuous, but the argument for the above local coordinates would have to be examined carefully to see whether you could still establish it when you can't differentiate the coefficients of the second fundamental form.

Note that this kind of argument would not work if, instead of trying to join the conical 'quadrant' solutions to the planal 'plank' solution, you had specified the domains as wedges centered on the origin, for then the boundaries of the domains would be ruling lines, which are characteristics for the PDE. It just so happens that you are trying to modify the solution along a *non-*characteristic curve, and that doesn't work.

  • $\begingroup$ Thank you, Professor Bryant, for this really nice answer! $\endgroup$
    – user36931
    Jun 20 '14 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.