# When does the eikonal equation $\lvert Du \rvert^2 = f$ admit a local solution?

Let $$f$$ be a smooth function defined on the unit disc $$D \subset \mathbf{R}^2$$ with $$\begin{equation} f \geq 0 \text{ in D and } f(0) = 0. \end{equation}$$ This is allowed to have a degenerate minimum at the origin, namely it is allowed that $$D^2 f(0) = 0.$$

Question. When is there $$\rho \in (0,1)$$ and $$u \in C^1(D_\rho)$$ so that $$\lvert D u \rvert^2 = f$$? I would be more than happy with an answer specialised to the case where $$f$$ is the polynomial $$(xy)^{2N}$$—say with $$N \geq 2$$—if a general discussion is too onerous.

• As far as I understand, the equation was initially considered with a strictly positive right-hand side. One classical example is where $$\lvert \nabla u \rvert^2_g = 1$$, with respect to the some Riemannian metric $$g$$ on $$D$$. One may attempt to rescale the Euclidean metric to $$g = f^2 g_e$$, in order to get $$\lvert \nabla u \rvert_g^2 = f^{-2} \lvert \nabla u \rvert_{g_e} = 1$$. However $$g$$ is unfortunately degenerate where $$f = 0$$.

• When the zero of $$f$$ at the origin is non-degenerate, then one can construct a solution to the eikonal equation via a sort of dynamic argument, as is explained in this answer.

For $$n=1$$ or $$2$$, there is no $$u\in C^1(D_\rho)$$ for any $$\rho>0$$ that satisfies $$|\nabla u|^2 = (xy)^{2n}$$. (Note that $$f=(xy)^2$$ has a degenerate minimum at $$(0,0)$$, so $$n=1$$ should be allowed in this discussion.) Meanwhile, for $$n\ge 3$$, there do exist $$u\in C^1(\mathbb{R}^2)$$ that satisfy $$|\nabla u|^2 = (xy)^{2n}$$.

The above non-existence result is special to the case $$f = (xy)^{2n}$$. For example, in the case $$f=(xy)^{2n}(x^2{+}y^2)$$, which has an even more degenerate minimum at $$(0,0)$$, there is a real-analytic, global solution $$u(x,y) = (xy)^{n+1}/(n{+}1)$$.

A few preliminaries before sketching the main argument are in order.

First, note that the disk radius $$\rho>0$$ actually plays no role in the problem. If $$u\in C^1(D_\rho)$$ were to satisfy $$|\nabla u|^2 = (xy)^{2n}$$, then for any $$r>0$$ the scaled function $$\tilde u(x,y) = r^{-(2n+1)}\,u(rx,ry)$$ would satisfy $$\tilde u\in C^1(D_{\rho/r})$$ and $$|\nabla \tilde u|^2 = (xy)^{2n}$$. Thus, one can assume that $$\rho$$ be arbitrarily large.

Second, one can assume, by adding a constant to $$u$$, that $$u(0,0)=0$$, so I will assume this normalization made henceforth. Then, the obvious integral inequality arising from $$|\nabla u| = |xy|^n$$ and the Cauchy-Schwartz inequality would imply that $$|u(x,y)| \le \frac{|xy|^n\sqrt{x^2+y^2}}{(2n{+}1)}.$$ In particular, $$u$$ would vanish to order $$2n{+}1$$ at $$(0,0)$$ and would satisfy $$u(x,0) = u(0,y) = 0$$. It follows from this that $$u$$ could not be of differentiability class $$C^{2n+2}$$, since, if it were, the limit function $$p(x,y) = \lim_{r\to0} \frac{u(rx,ry)}{r^{2n+1}}$$ would exist and be a polynomial homogeneous of degree $$2n{+}1$$ that satisfied $$|\nabla p|^2 = (xy)^{2n}$$, and it is easy to show that there is no such polynomial. However, as will be seen, when $$n\ge3$$, there exists a $$u\in C^{n-1}(\mathbb{R}^2)$$ satisfying $$|\nabla u|^2 = (xy)^{2n}$$ and the homogeneity condition $$u(rx,ry) = |r|^{2n+1}\,u(x,y)$$ for all $$r$$. This $$u$$ is real-analytic away from the lines $$x\pm y = 0$$ but fails to be $$C^n$$ on these two lines.

From now on, I will assume that $$u\in C^1(D_\rho)$$ (with $$\rho>>0$$ as large as necessary for the argument) satisfies $$u(0,0)=0$$ and $$|\nabla u|^2 = (xy)^{2n}$$. In particular, as the OP points out, $$u$$ satisfies the eikonal equation $$|\nabla^g u|^2 = 1$$ for the singular 'metric' $$g = (xy)^{2n}(\mathrm{d}x^2+\mathrm{d}y^2)$$, so that the gradient flow lines of $$\nabla^g u$$ are $$g$$-geodesics in the four quadrants of the $$xy$$-plane where $$xy\not=0$$.

Now, the metric $$g$$ has some interesting properties: First, it is homogeneous of degree $$2n{+}2$$, so that its family of geodesics is preserved under the scaling homothety, and, moreover, it is invariant under the discrete symmetries $$(x,y)\to(-x,y)$$, $$(x,y)\to(x,-y)$$, and $$(x,y)\to(y,x)$$. Consequently, it suffices to study the behavior of its geodesics in the 'first' quadrant, where $$x>0$$ and $$y>0$$, and it is immediate that the lines $$y\pm x = 0$$ are geodesics in the quadrants. Let the ray $$\{(x,x)\ |\ x>0\}$$ in the first quadrant be known as the fundamental geodesic. Of course, $$g$$ is not complete in the first quadrant, as the two boundary rays can be reached from, say, $$(1,1)$$, by curves of finite $$g$$-length.

Now, computation shows that the Gauss curvature of $$g$$ is $$K = n(x^2{+}y^2)/(xy)^{2n+2}>0$$, which suggests that nearly all of the geodesics of $$g$$ will avoid going to the singular boundary where $$xy=0$$, and, indeed, this turns out to be the case (see below). To parametrize the geodesics, it turns out to be convenient to use a parameter $$t$$ other than arc length. A curve $$\bigl(x(t),y(t)\bigr)$$ in the first quadrant parametrizes a $$g$$-geodesic when there is a function $$\phi(t)$$ satisfying the ODE system $$\dot x = xy\cos\phi, \quad \dot y = xy\sin\phi, \quad \dot\phi = n\,(x\cos\phi-y\sin\phi),\tag1$$ and every $$g$$-geodesic in the first quadrant has such a parametrization, unique up to replacing $$t$$ by $$t+t_0$$ for some constant $$t_0$$. In this case, arclength $$s(t)$$ along the geodesic satisfies $$\dot s = (xy)^{n+1}$$. [The advantage of writing the geodesic equations this way is that they extend smoothly across the singular locus $$xy=0$$.] Note that these equations are invariant under the homothetical scaling $$(t,x,y,\phi)\to(t/r,rx,ry,\phi)$$. Because of the scaling symmetry of the equations, one can extract a 2D phase portrait that makes clear the behavior of the geodesics as follows: Let $$x+iy = \mathrm{e}^{u+iv}$$. Then the above equations become (after a change of independent variable) $$u' = \cos(\phi{-}v)\,\cos v\sin v,\qquad v' = \sin(\phi{-}v)\,\cos v\sin v,\qquad \phi' = n\,\cos(\phi{+}v).\tag2$$

One can now draw the $$v\phi$$-phase portrait, concentrating on the strip $$0\le v\le \pi/2$$, which represents the first quadrant in the $$xy$$-plane, and bearing in mind that these equations are invariant under the involution $$(v,\phi)\to(\tfrac12\pi{-}v,\tfrac12\pi{-}\phi)$$ and reverse under $$(v,\phi)\to (v,\phi{+}\pi)$$.

There are a sink at $$S_- = (v,\phi)=(\pi/4,\pi/4)$$, a source at $$S_+ = (v,\phi)=(\pi/4,-3\pi/4)$$, and saddles at $$S_1 = (v,\phi)=(0,\pi/2)$$, $$S_2 = (v,\phi)=(\pi/2,0)$$, $$S_3=(v,\phi)=(0,-\pi/2)$$, and $$S_4=(\pi/2,-\pi)$$. In addition to the 'trivial' separatrices that make up the boundary lines $$v=0$$ and $$v=\tfrac12\pi$$, there are four 'non-trivial' separatrices: $$L_1$$ leaving $$S_1$$ and going to $$S_-$$, $$L_2$$ leaving $$S_2$$ and going to $$S_-$$, $$L_3$$ leaving $$S_+$$ and going to $$S_3$$, and $$L_4$$ leaving $$S_+$$ and going to $$S_4$$.

Here is where the difference between the cases $$n=1,2$$ and the cases $$n\ge 3$$ becomes evident. When $$n\le 2$$, the fixed points $$S_\pm$$ are spiral, i.e., the linearization of the flow at these two points has eigenvalues that are non-real (and complex conjuate), while, when $$n\ge 3$$, the linearizations of the flow at these two points have distinct real eigenvalues whose ratio is a real number strictly between $$n{-}2$$ and $$n{-}1$$.

One then finds that, when $$n\ge 3$$, the union of the two separatrices $$L_1$$ and $$L_2$$, together with their endpoints $$S_1$$, $$S_2$$, and $$S_-$$ is the graph $$\phi = f_n(v)$$ of a function $$f_n:[0,\tfrac12\pi]\to[0,\tfrac12\pi]$$ that is real-analytic except at the midpoint $$v=\tfrac14\pi$$, where it is $$C^{n-2}$$. One then shows that the corresponding $$g$$-geodesics in the first quadrant define a $$C^{n-2}$$ foliation by geodesics that meet the boundary rays $$x=0$$ and $$y=0$$ orthogonally. Moreover, by reflecting this foliation across the lines $$x=0$$ and $$y=0$$, one can construct a foliation $$\mathcal{F}$$ of $$\mathbb{R}^2$$ minus the origin by $$g$$-geodesics that is real-analytic except along the lines $$x\pm y=0$$, where it is $$C^{n-2}$$. It then follows easily that there is a unique $$C^{n-1}$$ function $$u$$ that vanishes on the axes $$x=0$$ and $$y=0$$, satisfies $$|\nabla u|^2=(xy)^{2n}$$ and, away from the axes, the gradient lines of $$u$$ are the leaves of the foliation $$\mathcal{F}$$.

Meanwhile, when $$n\le 2$$, the spiral nature of the two fixed points $$S_\pm$$ leads to an analysis that shows that there is no foliation of the 'first quadrant' quarter of a disk $$D_\rho$$ by $$g$$-geodesics, which leads to the conclusion that there is no $$C^1$$ solution $$u$$ on any $$D_\rho$$ to the equation $$|\nabla u|^2=(xy)^{2n}$$.

If there is interest, I can supply details of these arguments when I get the time.

• Thank you for the answer! I'll read through it more carefully and see whether I can figure out the details. Aug 4, 2021 at 17:26