A Low Complexity Algorithm for Non-Monotonically Evolving Fronts

Abstract

A new algorithm is proposed to describe the propagation of fronts advected in the normal direction with prescribed speed function F. The assumptions on F are that it does not depend on the front itself, but can depend on space and time. Moreover, it can vanish and change sign. The novelty of our method is that its overall computational complexity is predicted to be comparable to that of the Fast Marching Method (Sethian in Proceedings of the National Academy Sciences 93:1591–1595, 1996); (Vladimirsky in Interfaces Free Bound 8(3):281–300, 2006) in most instances. This latter algorithm is $\mathcal {O}(N^{n}\log N^{n})$ if the computational domain comprises $N^{n}$ points. We use it in regions where the speed is bounded away from zero—and switch to a different formalism when $F \approx 0$. To this end, a collection of so-called sideways partial differential equations is introduced. Their solutions locally describe the evolving front and depend on both space and time. The well-posedness and geometric properties of those equations are addressed. We then propose a convergent and stable discretization of the PDEs. The resulting algorithm is presented together with a thorough discussion of its features. The accuracy of the scheme is tested when F depends on both space and time. Each example yields an $\mathcal {O}(1/N)$ global truncation error. We conclude with a discussion of the advantages and limitations of our method.

Appendices

Appendix 1: Direct Method to Compute $\psi _{\mathrm {II}}$ in the t-FMM, in 2D

We provide a direct method for solving the minimization problem appearing in Equation (17), in two dimensions. Introducing $\tau (y)=h/|F(\mathbf {x}_{ij},\psi (y))|$, we first use linear interpolation to simplify the quantity we wish to minimize:

$$\begin{aligned} \psi (\tilde{\mathbf {x}}) + \sqrt{\xi ^{2}+(1-\xi )^{2}}~ \frac{~h}{|F(\mathbf {x}_{ij},\psi (\tilde{\mathbf {x}}))|}= & {} \psi (\tilde{\mathbf {x}}) + \sqrt{\xi ^{2}+(1-\xi )^{2}}~ \tau (\tilde{\mathbf {x}}) \nonumber \\\approx & {} \xi \psi (\mathbf {x}_{i-1,j})+(1-\xi ) \psi (\mathbf {x}_{i,j+1})\nonumber \\&+ \sqrt{\xi ^{2}+(1-\xi )^{2}}\nonumber \\&\times \, \left( \xi \tau (\mathbf {x}_{i-1,j})+(1-\xi ) \tau (\mathbf {x}_{i,j+1}) \right) \nonumber \\=: & {} f(\xi ) \end{aligned}$$

(37)

Minimizing f over $\xi \in (0,1)$ amounts to finding the roots of $0 = c_{4}\lambda ^{4} + c_{3}\lambda ^{3} + c_{2}\lambda ^{2} + c_{1}\lambda + c_{0}$ where $\lambda \in (0,1)$ is such that $f'(\lambda )=0$. This quartic can be solved either directly with closed formulas, or with Newton’s method—we use the latter. For each root $r_{i} \in (0,1)$ the corresponding value of $\psi$ is computed as $\psi _{\mathrm {II},r_{i}}=f(r_{i})$. If $\psi _{\mathrm {II},r_{i}}< \psi (\mathbf {x}_{i-1,j})$ or $\psi _{\mathrm {II},r_{i}}<\psi (\mathbf {x}_{i,j+1})$, then $\psi _{\mathrm {II},r_{i}}$ is discarded. Values arising from minimization in one dimension are also computed as $\psi _{\mathrm {II},0} = \psi (\mathbf {x}_{i,j+1}) + \tau (\mathbf {x}_{i,j+1})$ and $\psi _{\mathrm {II},1} = \psi (\mathbf {x}_{i-1,j}) + \tau (\mathbf {x}_{i-1,j})$. The global minimum is found by comparing all those values.

Appendix 2: Implementation Details for the Examples

We give some details about the examples presented in Sect. 7. All tests were performed using Matlab $^\circledR$ [51]. In particular, finding the minimum value in $\mathcal {N}$  is done using min.

Computing the outward normal In regions where the level-set function $\phi$ is $C^{1}$, we have $\hat{\nu }= (\phi _{x},\phi _{y},\phi _{t})/|(\phi _{x},\phi _{y},\phi _{t})|$. We use the Implicit Function Theorem: e.g., If $\psi (y,t)=x$ satisfies $\phi (\psi (y,t),y,t)=0$, then $\phi _{y} = -\phi _{x}\psi _{y}$ and $\phi _{t} = -\phi _{x}\psi _{t}$. Since $\phi _{x} \ne 0$, we set $\mathbf {\nu }= (+\mathrm {sign}(\phi _{x}),\psi _{y},\psi _{t})$ and $\hat{\nu }= \mathbf {\nu }/|\mathbf {\nu }|$. In the (t-)FMM as well as in the sideways representations, the points used to approximate the derivatives using finite-differences are the ones involved in the computation of the new point. E.g., In the t-FMM, if two-dimensional optimization was used in Quadrant III to obtain $\psi _{ij}$, then $\hat{\nu }(p_{ij}) = \mathbf {\nu }/|\mathbf {\nu }|$ where

$$\begin{aligned} \mathbf {\nu } = \left( \, \left( \psi _{ij}-\psi _{i-1,j} \right) / h \,,\, \left( \psi _{ij}-\psi _{i,j-1} \right) /h \,,\, -\mathrm {sign}(\hat{\nu }_{3}(p_{ij-1})) \, \right) \end{aligned}$$

(38)

Choice of parameters In all examples, the number of points in each dimension is $N+1$, and the spatial grid spacings are even: $h=dx=dy$. The size of the local grid in Algorithm 3 is $s= \lfloor \frac{N}{3} \rfloor$. When F depends on time, we use adaptive time-stepping. Before the time where $F=0$, we set $\varDelta t = r_{1} h$. Passed that time, we let $\varDelta t = r_{2} h$. To assess the convergence of the sideways methods, a yt-grid with spacings h and $\varDelta t = h/2$ is built. The exact normal $\hat{\nu }$ is assigned to the points as they are accepted in all the examples, except Example 1 where it is computed as explained in the above paragraph.

Example 1

The exact solution to the Level-Set Equation is $\phi (x,y,t) = \sqrt{x^{2}+y^{2}} -R(t)$ where $R(t) = \left( r_{0} - \frac{e^{10t}-1}{10e}+t \right)$. Domain: $[-.321,.319]^{2}$. $T_{F}=0.3$. xt- and yt-rep.: $r_{1} = 1/3$, $r_{2} = 2$. Skewed rep.: $r_{1} = r_{2} = 1$. Domain for convergence of Algorithm 4: $(y,t) \in [-0.25,0.25]\times [0,0.3]$.

Example 2

The signed distance function to $\mathcal {C}_{t}$ is given as $\phi (x,y,t)=\sqrt{(x-x_{c}(t))^{2}+y^{2}}-r(t)$ where $x_{c}(t) = r_{0}\sinh t$ and $r(t) = r_{0}\cosh t$. Note that $\phi$ does not solve the Level-Set Equation. Domain: $[-1.01,0.99]^2$. $T_{F}=1$. xt- and yt-rep.: $r_{1} = 1/3$, $r_{2} = 2$. Skewed rep.: $r_{1} = 1/3$, $r_{2} = 5$. Domain for convergence of Algorithm 4: $(y,t) \in [-0.25,0.25]\times [0,1]$ and $(x,t) \in [-0.25,0.25]\times [0,1]$.

Example 3

The exact solution to the Level-Set Equation is $\phi (x,y,t) = \sqrt{(x-g t)^2+y^2}-\left( r_{0}+c t \right)$ where $b=10$, $c=1/2$ and $g(t) = \arctan \left( b(t-0.5)\right) + \frac{\pi }{2}$. The speed is

$$\begin{aligned} F = \frac{(x-gt)(g't+g)}{\sqrt{(x-gt)^{2}+y^{2}}} + c ~\Longrightarrow ~F \approx \left\{ \begin{array}{ll} c &{} \mathrm {for~} t \mathrm {~small} \\ \frac{(x-\pi t)\pi }{\sqrt{(x-\pi t)^{2}+y^{2}}} + c &{} \mathrm {for~} t \mathrm {~large} \\ \end{array} \right. \end{aligned}$$

(39)

We expect the circle to first expand (when t is small), and then expand while moving to the right with speed $\pi$ (when t is large). Domain: $[-1.51,+1.49]^{2}$. $T_{F}=0.5$. xt- and yt-rep.: $r_{1} = 1/3$, $r_{2} = 2$. Skewed rep.: $r_{1} = 1/3$, $r_{2} = 5$. Domain for convergence of Algorithm 4: $(y,t) \in [-0.25,0.25]\times [0,0.5]$.

Example 4

The set $\mathcal {C}_{0}$ consists of two disjoint circles of radius $r_{0}=0.25$, with centres at $(-0.3,0)$ and (0.3, 0). The speed is $F = 1-e^{2t-1}$. The circles touch along the y-axis when $t\approx 0.08$. When $t<0.5$ the exact solution to the Level-Set Equation is $\phi (x,y,t) = \min \left\{ \sqrt{(x+0.3)^{2}+y^{2}} \right.$ $\left. - R(t), \sqrt{(x-0.3)^{2}+y^{2}} - R(t) \right\}$ where $R(t)= r_{0} - \frac{e^{2t}-1}{2e}+t$. Domain: $[-1.5+ 0.01e,+1.5+ 0.01e]^{2}$. $T_{F}=1.2$. xt- and yt-rep.: $r_{1} = 1/3$, $r_{2} = 2$. Skewed rep.: $r_{1} = 1/3$, $r_{2} = 5$. Domain for convergence of Algorithm 4: $(y,t) \in [-0.5,0.5]\times [0.2,0.5]$ and $(y,t) \in [-0.5,0.5]\times [0.5,.52]$.

The Almond example The exact solution to the Level-Set Equation is

$$\begin{aligned} \phi (x,y,t)= & {} \left( \sqrt{x^{2}+y^{2}}-r_{0}+\frac{e^{ct}-1}{ce}-t(1+C) \right) + \frac{t|xt-y|}{\sqrt{1+t^{2}}} \end{aligned}$$

(40)

The constants are set to be: $r_0 = 1/4$, $c = 1$, and $C = .65$. The function $\phi$ is made up of two parts: The first one in brackets is qualitatively the same as in Example 1. Domain: $[-0.5,0.5]^2$. $T_{F}=1.9$. xt- and yt-rep.: $r_{1} = 1/3$, $r_{2} =2$. Skewed rep.: $r_{1} = 1/2$, $r_{2} = 6$.