Hamilton-Jacobi equation and Riemannian distance

Consider the cotangent bundle $T^\ast X$ as a symplectic manifold with canonical symplectic form $\omega$. Consider the Hamilton-Jacobi equation
$$\frac{\partial S}{\partial t} + H(x, \nabla S) = 0,$$
for the classical Hamilton function $S(x,t)$. Setting $x=x(t), p(t) = (\nabla S)(x(t), t)$ one sees immediately from the method of characteristics that this PDE is solved by the classical action
$$S(x,t) = \int_0^t (p \dot{x} - H) ds,$$
where the integral is taken over the solution $(x(s),p(s))$ of Hamilton's equations with $x(0)=x_0$ and $x(t) = x$. The choice of basepoint $x_0$ involves an overall additive constant of $S$, and really this solution is only valid in some neighbourhood $U$ of $x_0$. (Reason: $S$ is in general multivalued, as the differential "$dS$" is closed but not necessarily exact.)

Now consider the case where $X$ is Riemannian, with Hamiltonian $H(x,p) = \frac{1}{2} |p|^2$. The solutions to Hamilton's equations are affinely parametrized geodesics, and by a simple Legendre transform we have
$$S(x, t) = \frac{1}{2} \int_0^t |\dot x|^2 ds$$
where the integral is along the affine geodesic with $x(0) = x_0$ and $x(t) = x$. Since $x(s)$ is a geodesic, $|\dot x(s)|$ is a constant (in $s$) and therefore
$$S(x, t) = \frac{t}{2} |\dot x(0)|^2.$$
Now consider the path $\gamma(s) = x($|\dot x(0)|^{-1}$s)$. This is an affine geodesic with $\gamma(0) = x_0$, $\gamma(|\dot x(0)|t) = x$ and $|\dot \gamma| = 1$. Therefore, the Riemannian distance between $x_0$ and $x$ (provided $x$ is sufficiently close to $x_0$) is
$$d(x_0, x) = |\dot x(0)| t.$$
Combining this with the previous calculation, we see that
$$S(x, t) = \frac{1}{2t} d(x_0, x)^2.$$
Now insert this back into the Hamilton-Jacobi equation above. With a bit of rearranging, we have the following.

Theorem. Let $x_0$ denote a fixed basepoint of $X$. Then for all $x$ in a sufficiently small neighborhood $U$ of $x_0$, the Riemannian distance function satisfies the Eikonal equation
$$|\nabla_x d(x_0, x)|^2 = 1.$$

Now, for convenience set $r(x) = d(x_0, x)$. Then $|\nabla r|^2 = 1$, from which we obtain (by differentiating twice and contracting)
$$g^{ij} g^{kl}\left(\nabla_{lki} r \nabla_j r + \nabla_{ki}r \nabla_{lj} r\right) = 0.$$
Quick calculation shows that
$$\nabla_{lki} r = \nabla_{ilk} r - \left.R_{li}\right.^{b}_k \nabla_b r$$
Therefore, tracing over $l$ and $k$ we obtain
$$g^{lk} \nabla_{lki} r = \nabla_i ( \Delta r) + Rc(\nabla r, -)$$
Plugging this back into the equation derived above, we have
$$\nabla r \cdot \nabla(\Delta r) + Rc(\nabla r, \nabla r) + |Hr|^2 = 0,$$
where $Hr$ denotes the Hessian of $r$ regarded as a 2-tensor. Now, using $r$ as a local coordinate, it is easy to see that $\partial_r = \nabla r$ (as vector fields). So we can rewrite this identity as
$$\partial_r (\Delta r) + Rc(\partial_r, \partial_r) + |Hr|^2 = 0.$$

Now, we can get a nice result out of this. First, note that the Hessian $Hr$ always has at least one eigenvalue equal to zero, because the Eikonal equation implies that $Hr(\partial_r, -)=0$. Let $\lambda_2, \dots, \lambda_n$ denote the non-zero eigenvalues of $Hr$. We have
$$|Hr|^2 = \lambda_2^2 + \dots + \lambda_n^2,$$
while on the other hand
$$|\Delta r|^2 = (\lambda_2 + \dots + \lambda_n)^2$$
By Cauchy-Schwarz, we have
$$|\Delta r|^2 \leq (n-1)|Hr|^2$$

Proposition. Suppose that the Ricci curvature of $X$ satisfies $Rc \geq (n-1)\kappa$, and let $u = (n-1)(\Delta r)^{-1}$. Then
$$u' \geq 1 + \kappa u^2.$$

Proof. From preceding formulas, $|Hr|^2$ can be expressed in terms of the Ricci curvature and the radial derivative of $\Delta r$. On the other hand, $|\Delta|^2$ is bounded above by $(n-1) |Hr|^2$. The claimed inequality then follows from simple rearrangement.

Now, the amazing thing is that this deceptively simple inequality is the main ingredient of the Bishop-Gromov comparison theorem. The Bishop-Gromov comparison theorem, in turn, is the main ingredient of the proof of Gromov(-Cheeger) precompactness. I hope to discuss these topics in a future post.