Neighbor-Gradient Single-Pass Method for Solving Anisotropic Eikonal Equation

Abstract

We develop a single pass method for approximating the solution to an anisotropic eikonal equation related to the anisotropic min-time optimal trajectory problem. Ordered Upwind Method (OUM) solves this equation, which is a single-pass method with an asymptotic complexity. OUM uses the search along the accepted front (SAAF) to update the value at considered nodes. Our technique, which we refer to as “Neighbor-Gradient Single-Pass Method”, uses the minimizer of the Hamiltonian, in which the gradient is substituted with neighbor gradient information, to avoid SAAF. Our technique is considered in the context of control-theoretic problem. We begin by discussing SAAF of OUM. We then prove some properties of the value function and its gradient, which provide the key motivation for constructing our method. Based on these discussions, we present a new single-pass method, which is fast since it does not require SAAF. We test this method with several anisotropic eikonal equations to observe that it works well while significantly reducing the computational cost.

Appendix

Lemma 1

The viscosity solution $u(x)$ of (6) is semi-concave.

Proof

Let us prove that $H(x,p)$ is Lipchitz continuous in $x$.

For arbitrary $x,y$, suppose that.

$$ H(x,p) = - \mathop {\min }\limits_{{a \in S_{1} }} \{ p \cdot af(x,a)\} - 1 = - p \cdot a_{0} f(x,a_{0} ) - 1, $$

$$ H(y,p) = - \mathop {\min }\limits_{{a \in S_{1} }} \{ p \cdot af(y,a)\} - 1 = - p \cdot a_{1} f(y,a_{1} ) - 1. $$

If $p \cdot a_{0} f(x,a_{0} ) \ge p \cdot a_{1} f(y,a_{1} )$, then.

$$ \begin{gathered} \left| {H(x,p) - H(y,p)} \right| = \left| {p \cdot a_{0} f(x,a_{0} ) - p \cdot a_{1} f(y,a_{1} )} \right| \hfill \\ \le \left| {p \cdot a_{0} f(y,a_{0} ) - p \cdot a_{0} f(x,a_{0} )} \right| \le \left\| p \right\|L\left\| {x - y} \right\| \hfill \\ \end{gathered} $$

If $p \cdot a_{0} f(x,a_{0} ) < p \cdot a_{1} f(y,a_{1} )$, then.

$$ \begin{gathered} \left| {H(x,p) - H(y,p)} \right| = \left| {p \cdot a_{0} f(x,a_{0} ) - p \cdot a_{1} f(y,a_{1} )} \right| \hfill \\ \le \left| {p \cdot a_{1} f(x,a_{1} ) - p \cdot a_{1} f(y,a_{1} )} \right| \le \left\| p \right\|L\left\| {x - y} \right\| \hfill \\ \end{gathered} $$

Therefore $H(x,p)$ is Lipchitz continuous in $x$. We also know that $H{(}x,p)$ is convex in the second argument and thus $u$ is semi-concave in $\Omega$. ([31]) □

Lemma 2

Assume that $u$ is differentiable at $x \in \Omega$ and suppose that $\alpha ( \cdot )$ is an optimal control for $x$ and $y( \cdot )$ is the optimal trajectory for $x$ . Then $\nabla u(y( \cdot ))$ is continuous at $[0,\; + \infty )$ .

Proof

For arbitrary $t > 0$, $u$ is differentiable at $y{(}t)$ because characteristics never emanate from the shocks-non differentiable point. ([16]).

If $\nabla u$ is defined at $x \in \Omega \backslash \partial \Omega$, there exists constant $L$ such that $\left\| {\nabla u} \right\| \le L$. [16].

We fix any $t \ge 0$. For arbitrary sequence $t_{n} \to t(t_{n} \ge 0)$, $y{(}t_{n} ) \to y(t)$. The $\nabla u(y(t_{n} ))$ is bounded. Therefore, it has convergent subsequence. Since $u$ is semi-concave, limits of all convergent subsequences are the same.i.e.,$\nabla u(y(t))$. [36] Therefore $\nabla u(y(t_{n} )) \to \nabla u(y(t))$ and thus $\nabla u(y( \cdot ))$ is continuous at $[0,\; + \infty )$. □

Lemma 3

Consider a grid node $x \in X$ and assume that u is differentiable at $x.$ Suppose that the characteristic for $x$ intersects the line segment $x_{1} x_{2}$ at $x_{0}$ , where $x_{1} ,x_{2} \in ND(x),\;x_{1} \in N(x_{2} )$ . Then.

$$ \left\| {\nabla u(x_{0} ) - \nabla u(x)} \right\| = O(h). $$

Proof

Suppose that $\alpha ( \cdot )$ is an optimal control for $x$ and $y( \cdot )$ is the optimal trajectory for $x$.

If $y(t_{{0}} ) = x_{0}$, since $\nabla u(y( \cdot ))$ is continuous at 0, we get.

$$ \left\| {\nabla u(y(t_{{0}} )) - \nabla u(y(0))} \right\| = O(t_{{0}} ). $$

By the Bellman’s optimality principle, we get $t_{{0}} \le \frac{{x_{0} - x_{1} }}{{F_{2} }}$ and $\frac{{x_{0} - x_{1} }}{{F_{1} }} \le t_{{0}}$.

Therefore $O(t_{0} ) = O(h)$. From the fact that $y(0) = x,y(t_{0} ) = x_{0}$, we conclude that.

$\left\| {\nabla u(x_{0} ) - \nabla u(x)} \right\| = O(h) \,$.□

Lemma 4

If $xx_{1} x_{2}$ is a sufficiently small simplex, which contains the characteristic for $x$ , then.

$$ \min (u(x_{1} ),\;u(x_{2} )) < u(x). $$

Proof

By the Bellman’s optimality principle, we get.

$$ \begin{gathered} u(x) = u(x_{s} ) + \frac{{\left\| {x_{s} - x} \right\|}}{{f\left( {x,\,\,{{{(}x_{s} - x{)}} \mathord{\left/ {\vphantom {{{(}x_{s} - x{)}} {\left\| {x_{s} - x} \right\|}}} \right. \kern-\nulldelimiterspace} {\left\| {x_{s} - x} \right\|}}} \right)}} + O(h^{2} ) = \hfill \\ \lambda u(x_{1} ) + \left( {1 - \lambda } \right)u(x_{2} ) + \frac{{\left\| {x_{s} - x} \right\|}}{{f\left( {x,\,\,{{{(}x_{s} - x{)}} \mathord{\left/ {\vphantom {{{(}x_{s} - x{)}} {\left\| {x_{s} - x} \right\|}}} \right. \kern-\nulldelimiterspace} {\left\| {x_{s} - x} \right\|}}} \right)}} + O(h^{2} ). \hfill \\ \end{gathered} $$

Since $xx_{1} x_{2}$ is sufficiently small simplex, we obtain that $u(x) > \lambda u(x_{1} ) + \left( {1 - \lambda } \right)u(x_{2} )$.

Therefore, $\min (u(x_{1} ),\;u(x_{2} )) < u(x)$. □