**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!



Solutions to the eikonal equation
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. 


