Eikonal equation

The eikonal equation (from German Eikonal, which is from Greek εἰκών, image[1][2]) is a non-linear partial differential equation encountered in problems of wave propagation, when the wave equation is approximated using the WKB theory. It is derivable from Maxwell's equations of electromagnetics, and provides a link between physical (wave) optics and geometric (ray) optics.

The eikonal equation is of the form

$| \nabla u(x)|=F(x), \ x\in \Omega$

subject to $u|_{\partial \Omega}=0$, where $\Omega$ is an open set in $\mathbb{R}^n$ with well-behaved boundary, $F(x)$ is a function with positive values, $\nabla$ denotes the gradient and |·| is the Euclidean norm. Here, the right-hand side $F(x)$ is typically supplied as known input. Physically, the solution $u(x)$ is the shortest time needed to travel from the boundary $\partial \Omega$ to $x$ inside $\Omega,$ with $F(x)$ being the time cost (not speed) at $x$.

In the special case when $F=1$, the solution gives the signed distance from $\partial \Omega$.

One fast computational algorithm to approximate the solution to the eikonal equation is the fast marching method.

Physical interpretation

The physical meaning of the eikonal equation is related to the formula

$E = -\nabla V$

where $E$ is the electric field strength and $V$ is the electric potential. There is a similar equation for velocity potential in fluid flow and temperature in heat transfer. The physical meaning of this equation in the electromagnetic example is that any charge in the region is pushed to move at right angles to the lines[clarification needed] of constant potential, and along lines of force determined by the field of the E vector and the sign of the charge.

Ray optics and electromagnetism are related by the fact that the eikonal equation gives a second electromagnetic formula of the same form as the potential equation above where the line of constant potential has been replaced by a line of constant phase and the force lines have been replaced by normal vectors coming out of the constant phase line at right angles. The magnitude of these normal vectors is given by the square root of the relative permittivity. The line of constant phase can be considered the edge of one of the advancing light waves. The normal vectors are the rays the light is traveling down in ray optics.

Mathematical description

An eikonal equation is one of the form

$H(x,\nabla u(x)) = 0$
$u(0,x') = u_0(x'),\text{ for } x = (x_1,x')$

The plane $x = (0,x')$ can be thought of as the initial condition, by thinking of $x_1$ as $t.$ We could also solve the equation on a subset of this plane, or on a curved surface, with obvious modifications.

The eikonal equation shows up in geometrical optics, which is a way of studying solutions of the wave equation $c(x)^2 |\nabla_x u(x,t)|^2 = |\partial_t u(x,t)|^2$. In geometric optics, the eikonal equation describes the phase fronts of waves. Under reasonable hypothesis on the "initial" data, the eikonal equation admits a local solution, but a global smooth solution (e.g. a solution for all time in the geometrical optics case) is not possible. The reason is that caustics may develop. In the geometrical optics case, this means that wavefronts cross.

We can solve the eikonal equation using the method of characteristics. One must impose the "non-characteristic" hypothesis $\partial_{p_1} H(x,p) \neq 0$ along the initial hypersurface $x = (0,x')$, where H=H(x,p) and p=(p1,...,pn) is the variable that gets replaced by ∇u. Here x=(x1,...,xn) = (t,x').

First, solve the problem $H(x,\xi(x)) = 0$, $\xi(x) = \nabla u(x), x\in H$. This is done by defining curves (and values of $\xi$ on those curves) as

$\dot x(s) = \nabla_\xi H(x(s),\xi(s)), \;\;\;\; \dot \xi(s) = -\nabla_x H(x(s),\xi(s)).$
$x(0) = x_0, \;\;\;\; \xi(x(0)) = \nabla u(x(0)).$ Note that even before we have a solution $u$, we know $\nabla u(x)$ for $x = (0,x')$ due to our equation for $H$.

That these equations have a solution for some interval $0 \leq s < s_1$ follows from standard ODE theorems (using the non-characteristic hypothesis). These curves fill out an open set around the plane $x = (0,x')$. Thus the curves define the value of $\xi$ in an open set about our initial plane. Once defined as such it is easy to see using the chain rule that $\partial_s H(x(s), \xi(s)) = 0$, and therefore $H = 0$ along these curves.

We want our solution $u$ to satisfy $\nabla u = \xi$, or more specifically, for every $s$, $(\nabla u)(x(s)) = \xi(x(s)).$ Assuming for a minute that this is possible, for any solution $u(x)$ we must have

$\frac{d}{d s} u(x(s)) = \nabla u(x(s)) \cdot \dot x(s) = \xi \cdot \frac{\partial H}{\partial \xi}$,

and therefore

$u(x(t)) = u(x(0)) + \int_0^t \xi(x(s))\cdot \dot x(s)\, ds.$

In other words, the solution $u$ will be given in a neighborhood of the initial plane by an explicit equation. However, since the different paths $x(t)$, starting from different initial points may cross, the solution may become multi-valued, at which point we have developed caustics. We also have (even before showing that $u$ is a solution)

$\xi(x(t)) = \xi(x(0)) - \int_0^s \nabla_x H(x(s),\xi(x(s))) \, ds.$

It remains to show that $\xi$, which we have defined in a neighborhood of our initial plane, is the gradient of some function $u$. This will follow if we show that the vector field $\xi$ is curl free. Consider the first term in the definition of $\xi$. This term, $\xi(x(0)) = \nabla u(x(0))$ is curl free as it is the gradient of a function. As for the other term, we note

$\frac{\partial^2}{\partial x_k \, \partial x_j} H = \frac{\partial^2}{\partial x_j \, \partial x_k} H.$

The result follows.