Geometric Interpretation of the Solutions of the Eikonal Equation (Arnold’s PDE)

Let $(x_1, x_2)$ be the coordinates in $\mathbb{R}^2$ , then the eikonal equation in $\mathbb{R}^2$ is given by:

$\frac{\partial u}{\partial x_1}^2 + \frac{\partial u}{\partial x_2}^2 = 1$

Goal: We will show that any solution of the equation is locally the sum of a constant and the distance to a curve.

Key outline of proof:

Step 1: We consider the gradient vector field $(\frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2})$ and its corresponding integral curves.

IMPORTANT: these integral curves happen to be straight lines.

Step 2: We consider how the solution $u$ changes along an integral curve. Turns out, in this example, $u$ is linear.

Step 3: Find $\Gamma$ the curve that satisfies $u(x) = C + \text{dist}(\Gamma, x)$ . Notice: once we have picked a curve $\Gamma$ that is orthogonal to each integral curve at the intersection point, then, we probably have what we are looking for. What sort of curve would this be?

Proof:

Let $x(t) = (x_1(t), x_2(t))$ be an integral curve of the vector field $(\frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2})$ .

We will show that $x(t)$ is a straight line, i.e. $\ddot{x} = 0$ .

By definition, $\dot{x} = (\frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2})$ .

Differentiating $\frac{\partial u}{\partial x_1}^2 + \frac{\partial u}{\partial x_2}^2 = 1$ by $\frac{\partial}{\partial x_i}$ for $i = 1, 2$ , we have:

$2[\frac{\partial u}{\partial x_1} \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial u}{\partial x_2} \frac{\partial^2 u}{\partial x_1 \partial x_2}] = 0$

$2[\frac{\partial u}{\partial x_1} \frac{\partial^2 u}{\partial x_2 \partial x_1} + \frac{\partial u}{\partial x_2} \frac{\partial^2 u}{\partial x_2^2}] = 0$ .

Thus, it follows that:

$\text{Hess}(u) \nabla(u) = \begin{bmatrix} \frac{\partial u}{\partial x_1} \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial u}{\partial x_2} \frac{\partial^2 u}{\partial x_1 \partial x_2} \\ \frac{\partial u}{\partial x_1} \frac{\partial^2 u}{\partial x_2 \partial x_1} + \frac{\partial u}{\partial x_2} \frac{\partial^2 u}{\partial x_2^2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Hence,

$\ddot{x} = \frac{d}{dt}(\frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2}) = \text{Hess}(u) \dot{x} = \text{Hess}(u) \nabla(u) = 0$ .

This shows that the integral curves are straight lines.

Now, we look at how $u$ changes along an integral curve.

$\frac{d}{dt} u(x(t)) = \nabla(u) \dot{x}(t) = \frac{\partial u}{\partial x_1}^2 + \frac{\partial u}{\partial x_2}^2 = 1$ .

Thus, along an integral curve,

$u(x(t)) = u_0 + t$ .

This gives us the idea that a level curve $\Gamma = \{u = c\}$ might work.

Let $p$ be a point on an integral curve and $c = u(p)$ . Set $\Gamma = \{u = c\}$ .

Key: since the integral curves are that of the gradient vector field (gradient vector is orthogonal the tangent vector of the level curve), it follows that the integral curve is orthogonal to $\Gamma$ at the point of intersection.

In a small neighbourhood of $p$ , every point $x$ lies on exactly one integral curve orthogonal to $\Gamma$ .

Let $q \in \Gamma$ and $t$ be the distance travelled along the integral curve passing through q.

Then, $|x(t) - q| = t$ . Hence, it follows that:

$u(x) = c + t = c + \text{dist}(x, \Gamma)$

(or, depending on which side of the curve the point lies on $u(x) = c - t = c - \text{dist}(x, \Gamma)$ $\square$

Part 2: Understanding the wave-particle duality of this equation.

Recall: reducing the study of wave propagation to the study of evolution of particles mathematically means reducing a first-order PDE to a system of ODEs.

We let $\frac{\partial u}{\partial x_1} = p, \frac{\partial u}{\partial x_2} = q$ . Then, the equation becomes: