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Theorem. Let V be a $C^\infty$ function on a riemannian manifold $M$ and $p$ be a nondegenerate local minimum with $V(p)=0$. Then there is a unique positive function $\varphi \in C^\infty(U)$ such that $\varphi$ solves the eikonal equation $$\|\mathrm{grad} \varphi \|^2 = V.$$ Here, $U$ is some open neighborhood of $p$.

I found this statement (at least a quite similar one) in a physics paper without a real proof, just some motivation. It seems highly nontrivial and somehow I am struggling to find a proof anywhere. Can someone give me a good reference?

\Edit: I forgot to write down the (for uniqueness obviously essential) condition that $\varphi$ is positive. Sorry everyone!

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I don't know how helpful this will be, because it is outside my field of expertise. However I know someone who was a physicist and who did his PhD thesis on the eikonal equation. His thesis is available at . – Alan Haynes Nov 29 '11 at 23:25
Um, not true as stated: in the plane with Euclidean metric, $x^2+y^2$ and $x^2-y^2$ have the same $||grad||^2$, which is $4(x^2+y^2)$. – Tom Goodwillie Nov 30 '11 at 0:34
More generally, two functions $\phi$ and $\psi$ solve the equation if and only if $0 = \|\nabla\phi\|^2 - \|\nabla\psi\|^2 = \nabla(\phi - \psi)\cdot\nabla(\phi + \psi)$. This equation has many solutions, including $\phi = f(x) + g(y)$ and $\psi = f(x) - g(y)$, for any functions $f, g: R \rightarrow R$. – Deane Yang Nov 30 '11 at 5:03
Replacing $\varphi \mapsto \varphi + c$ gets you a new solution. By continuity of smooth functions, by possibly shrinking $U$ you can always translate any solution into one that is positive. So your "Edit" does not help at all with regards to the counterexamples given by Tom and Deane. – Willie Wong Nov 30 '11 at 10:19
I think what you want to require is that $\phi$ be nonnegative near $p$ while $\phi(p)=0$. – Robert Bryant Nov 30 '11 at 13:13
up vote 8 down vote accepted

Note: I have realized that, using the Stable Manifold Theorem, one can prove the smoothness of the solution $\phi$ that I describe below. Thus, I am modifying my answer to incorporate that.

Local existence and uniqueness of a smooth solution near $p$ satisfying $\phi(p)=0$ and $\phi\ge0$ near $p$ follows from an application of the Stable Manifold Theorem. Here is the argument.

This is a local question, so we might as well assume that $M=\mathbb{R}^n$, that $p=0$, that $g = g_{ij}dx^idx^j$ satisfies $g_{ij}(0)=\delta_{ij}$, and that the function $V$ has a Taylor expansion $V = h_{ij}x^ix^j + O(|x|^3)$, where $(h_{ij})$ is a symmetric, positive definite matrix. We are looking for a closed $1$-form $d\phi = f_i\ dx^i$ near $x=0$ so that $\phi$ satisfies the equation $g^{ij}f_if_j = V$ and, at the same time, satisfies $\phi(0)=0$ and $\phi\ge0$ near $x=0$.

Let $p_i$ be the coordinates on $T^*\mathbb{R}^n$ such that the canonical $1$-form has the expression $p_i\ dx^i$. Then the graph of $d\phi$, described by equations $p_i = f_i(x)$, will be a Lagrangian submanifold for the $2$-form $dp_i\wedge dx^i$ and will lie in the zero locus of the Hamiltonian $H(x,p) = g^{ij}(x)p_ip_j - V(x)$. Therefore, it will be a union of integral curves of the Hamiltonian vector field $$ X_H = 2g^{ij}p_i\frac{\partial\ \ }{\partial x^j} + \left(\frac{\partial V}{\partial x^k} - \frac{\partial g^{ij}}{\partial x^k}p_ip_j\right)\frac{\partial\ \ }{\partial p_k}. $$

This graph will have to pass through the unique singular point of $X_H$, i.e., $x = p = 0$ (since $\phi$ clearly must have a critical point at $x=0$ because it vanishes there and is nonnegative nearby), and the linear part of $X_H$ at $x=p=0$ is $$ Y_H = 2\delta^{ij}p_i\frac{\partial\ \ }{\partial x^j} + 2h_{ij}x^i\frac{\partial\ \ }{\partial p_j}. $$

The unstable manifold of $Y_H$ is the $n$-dimensional submanifold defined by $p_i = L_{ij} x^j$, where $L$ is the (unique) symmetric positive definite square root of $(h_{ij})$. The stable manifold of $Y_H$ is defined by $p_i = -L_{ij}x^j$. It follows from the Stable Manifold Theorem that $X_H$ has a smooth $n$-dimensional unstable submanifold $N_+$ given by $p_i = f_i(x) = L_{ij}x^j + O(|x|^2)$ and a smooth $n$-dimensional stable submanifold $N_-$ given by $p_i = -L_{ij}x^j + O(|x|^2)$.

From the dynamics of $X_H$, it is clear that the only possibility for $d\phi$, when $\phi$ satisfies the above conditions and is at least $C^2$, is to have its graph be $N_+$. Conversely, taking $d\phi = f_i(x)\ dx^i$ where $p_i = f_i(x)$ is the (necessarily Lagrangian) unstable manifold of $X_H$ and fixing the additive constant by requiring that $\phi(0)=0$ does give a smooth solution to the original equation.

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@Robert Bryant. Thanks for this answer. I was wondering if the solution you constructed, uniquely characterized among the smooth solutions by being $0$ in $p$ and nonnegative, could also be uniquely characterized among smooth solutions as the maximal solution. I'm pretty convinced this is true, but I don't see how to prove it. Does it follow somehow directly from your construction or are some other arguments needed? I would be very grateful if you or somebody else can comment on this! – Hans Apr 22 '12 at 13:39
@Hans: I'm not sure that I know what you mean by 'maximal'. Do you mean that this solution is greater than or equal to any other solution that vanishes at $p$, at least in a neighborhood of $p$, or do you just mean that there is no other solution that vanishes at $p$ that is greater than it near $p$? – Robert Bryant Apr 22 '12 at 14:05
@Robert Bryant. Sorry, I was imprecise: I mean the first option you mentioned. – Hans Apr 22 '12 at 15:02
@Hans: Well, any smooth solution $\phi$ that vanishes at $p$ will have a critical point at $p$ and hence have a $p$-centered Taylor expansion $\phi = \tfrac12 L_{ij} x^ix^j + O(3)$ for a symmetric matrix $L$ that satisfies $g^{ij}(p):L_{ik}L_{jl}=h_{kl}$. One can now show that if $L^+ = (L^+_{ij})$ is the (unique) positive definite symmetric solution to this equation, then the corresponding $\phi^+$ (as constructed above) is unique near $p$, and, since $L^+\ge L$ for $L$ any other solution, the corresponding $\phi$ will (at least near $p$) be dominated by $\phi^+$. – Robert Bryant Apr 23 '12 at 17:03
@Robert Bryant: hmm, this is what I thought at the beginning, but I'm not convinced. It seems to me that the quadratic approximation is not enough to decide. Why should $L^+\geq L$ imply $\phi^+\geq \phi$ near $p$? (We really have $L^+\geq L$ in general and not $L^+ > L$, otherwise it would be clear to me...). Do I miss something? – Hans Apr 24 '12 at 0:05

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