On the Optimal Strategy

for an Isotropic Blocking Problem

Alberto Bressan and Tao Wang

Department of Mathematics, Penn State UniversityUniversity Park, Pa. 16802, U.S.A.

e-mails: bressan@math.psu.edu , wang t@math.psu.edu

September 27, 2011

Abstract

The paper is concerned with a dynamic blocking problem, originally motivated by thecontrol of wild fires. It is assumed that the region R(t) ⊂ R2 burned by the fire is initiallya disc, and expands with unit speed in all directions. To block the fire, a barrier Γ can beconstructed in real time, so that the portion of the barrier constructed within time t haslength ≤ σt, for some constant σ > 2. We prove that, among all barriers consisting of asingle closed curve, the one which minimizes the total burned area is axisymmetric, andconsists of an arc of circumference and two arcs of logarithmic spirals.

1 Introduction

A class of dynamic blocking problems, motivated by the control of wild fires, was introducedin [4]. The region burned by the fire at time t ≥ 0 is denoted by R(t) ⊂ R2. In the casewhere no blocking strategy is implemented, the set R(t) is described as the reachable set fora differential inclusion

x ∈ F (x) , x(0) ∈ R0 , (1.1)

where the upper dot denotes a derivative w.r.t. time. Here x 7→ F (x) is a Lipschitz continuousmultifunction with compact, convex values, containing the origin. The initial set R(0) = R0

is supposed to be open and bounded.

To restrain the growth of the burned region, it is assumed that a barrier can be constructed,in real time. We denote by γ(t) ⊂ R2 the portion of the barrier constructed within timet ≥ 0. Given a construction speed σ > 0, we say that a strategy t 7→ γ(t) is admissible if thefollowing conditions hold.

(H1) For any t1 < t2, one has γ(t1) ⊆ γ(t2).

(H2) For every t ≥ 0, γ(t) is a rectifiable set with length

m1(γ(t)) ≤ σ t .

1

Here m1 denotes the one-dimensional Hausdorff measure, normalized so that m1(Γ) yields theusual length of a smooth curve Γ.

The reachable set (burned by the fire) at time t, determined by the blocking strategy γ, isdefined as

Rγ(t).=

x(t); x(·) is absolutely continuous, x(0) ∈ R0 ,

x(τ) ∈ F (x(τ)) for a.e. τ ∈ [0, t], x(τ) 6∈ γ(τ) for all τ ∈ [0, t].

(1.2)In other words, Rγ(t) is the set reached by trajectories of the differential inclusion (1.1) whichdo not cross the constructed portion of the barrier.

As proved in [7], one can equivalently describe a blocking strategy t 7→ γ(t) in terms of onesingle set

Γ.=

⋃t≥0

γ(t)

\⋃t≥0

Rγ(t)

. (1.3)

Given a rectifiable set Γ ⊂ R2, the corresponding reachable sets for the differential inclusion(1.1) restricted to R2 \ Γ are defined as

RΓ(t).=

x(t); x(·) is absolutely continuous, x(0) ∈ R0 ,

x(τ) ∈ F (x(τ)) for a.e. τ ∈ [0, t], x(τ) 6∈ Γ for all τ ∈ [0, t].

(1.4)We say that Γ is admissible if

m1

(Γ ∩RΓ(t)

)≤ σ t for all t ≥ 0. (1.5)

Here and in the sequel, an overline denotes the closure of a set. Clearly, the set Γ is admissibleif and only if

t 7→ γ(t).= Γ ∩RΓ(t) (1.6)

is an admissible strategy. Notice that the set γ(t) in (1.6) is the portion of the barrier Γtouched by the fire within time t. We shall denote by

RΓ∞

.=

⋃t≥0

RΓ(t) (1.7)

the whole region burned by the fire. In connection with this model, two issues naturally arise:

Blocking problem: Find an admissible set Γ such that RΓ∞ is bounded.

Optimization problem: Given two constants κ1 ≥ 0 and κ2 ≥ 0, find an admissible set Γwhich minimizes the cost

J(Γ).= κ1 ·m1(Γ) + κ2 ·m2(RΓ

∞). (1.8)

Here m2(·) denotes the two-dimensional Lebesgue measure of a set. The functional in (1.8)thus takes into account (i) the cost of constructing the barrier, and (ii) the value of the entireregion destroyed by the fire.

2

Various results on the existence or non-existence of an admissible strategy that confines thefire within a bounded set were proved in [4, 5, 8]. A general theorem on the existence ofoptimal blocking strategies was proved in [6, 10]. Moreover, various necessary conditions foroptimality were derived in [4, 9, 17], providing explicit descriptions of the optimal barriers, insome basic cases.

On the other hand, sufficient conditions for optimality are not yet known. In particular, not asingle case is known of a blocking strategy which is provably optimal. The difficulty is due tothe fact that all necessary conditions given in [4, 9, 17] require some regularity assumptions,which may not be satisfied by a general optimal strategy. The present paper represents a firststep in the derivation of sufficient conditions for optimality, developing techniques which allowa direct comparison between different strategies.

We shall consider the isotropic case, where the fire is initially burning on the open unit discand propagates with unit speed in all directions. This corresponds to (1.1), taking R0 = B1

.=

y ∈ IR2 ; |y| < 1 and letting F (x).= B1 = y ∈ IR2 ; |y| ≤ 1 be the closed unit disc, for

every x ∈ IR2. Let a construction speed σ > 2 and constants κ1 ≥ 0, κ2 > 0 be given. Assuggested by the necessary conditions derived in [4, 9], the optimal barrier Γ∗ which minimizesthe cost (1.8) should consist of an arc of circumference and two arcs of logarithmic spirals.

P+

1+ τ

P_

α

1x

−

τγ

γτ

+

~γτ

α

Figure 1: Construction of the barrier Γτ = γτ ∪ γ+τ ∪ γ−τ .

We now describe more precisely this barrier (Fig. 1). For every τ > 0 small enough, thereexists a unique arc of circumference γτ with the following properties:

(i) γτ is symmetric w.r.t. the x1-axis and has length m1(γτ ) = στ

(ii) the endpoints P−, P+ lie on the circumference |x| = 1 + τ

(iii) the angle α between the two circumferences at P− and at P+ satisfies

sinα =2

σ. (1.9)

In addition, consider the two arcs of logarithmic spirals γ+τ , γ

−τ , defined as

γ±τ =

(r cos θ , ± r sin θ) ; r = r0 eλ θ, r ≥ 1 + τ , θ ≤ π

. (1.10)

3

Here

λ =

√4

σ2 − 4, (1.11)

while the constant r0 is chosen so that the two arcs start from the points P+, P− respec-tively. Notice that the conditions (1.9)–(1.11) imply that the arcs γ±τ meet the circular arc γτtangentially at P±.

For every fixed τ , the union of these three arcs

Γτ.= γτ ∪ γ+

τ ∪ γ−τ (1.12)

is a simple closed curve. By minimizing the cost J(Γτ ) over the scalar parameter τ , we singleout the curve

Γ∗.= Γτ∗ , τ∗

.= argmin

τ>0J(Γτ ) . (1.13)

As shown in [9], the optimal value τ∗ in (1.13) could also be singled out by the requirementthat the Lagrange multiplier t 7→ W (t), describing the “instantaneous value of time”, becontinuous at t = τ .

The necessary conditions proved in [4, 9] strongly suggest that Γ∗ should be optimal for theoptimization problem described at (1.8). In this direction, our main result is

Theorem 1. The barrier Γ∗ is optimal within the family of all admissible Jordan curves withfinite length.

In other words, if Γ is any simple closed curve with m1(Γ) <∞, which is admissible accordingto (1.5), then

J(Γ∗) ≤ J(Γ). (1.14)

Indeed, our analysis shows that the inequality in (1.14) is strict, except when Γ is the imageof Γ∗ by a rotation around the origin. Observe that, if κ1 > 0, then any curve providing theminimum to (1.8) must have finite length.

Ω

Γ

Γ

Ω∼

Ω

Γ’

’’

Figure 2: The three mains steps in the proof of Theorem 1.

The proof of Theorem 1 is accomplished in three steps, which are outlined below (Fig. 2).

4

STEP 1. Consider any admissible, simple closed curve Γ. Thinking of Γ as a wall which blocksthe light, we let Ω ⊂ IR2 be the set of points illuminated by a light source located at the origin0 ∈ IR2. As it will be proved in Section 3, the boundary Γ′

.= ∂Ω is also an admissible, simple

closed curve. Moreover, J(Γ′) ≤ J(Γ).

STEP 2. By the previous construction, the domain Ω is star-shaped. Indeed, it can berepresented in polar coordinates as

Ω =

(r cos θ , r sin θ) ; r ≤ r(θ) θ ∈ [−π, π]

(1.15)

for some (possibly discontinuous) function θ 7→ r(θ) having bounded variation. Letting θ 7→r(θ) be the symmetric, non-decreasing rearrangement of the map θ 7→ r(θ), we obtain asymmetric domain

Ω =

(r cos θ , r sin θ) ; r ≤ r(θ) θ ∈ [−π, π]

(1.16)

According to the analysis in Section 4, the boundary Γ′′.= ∂Ω is an admissible, simple closed

curve. Moreover, J(Γ′′) ≤ J(Γ′).

STEP 3. The curve Γ′′ has sufficient regularity, so that the necessary conditions for optimalityderived in [9] can finally be applied. Relying on these necessary conditions, in Section 4 weshow that Γ′′ can be optimal only if Γ′′ = Γ∗, concluding the proof.

In the remainder of the paper, Section 2 introduces some further definitions and notations.The three main steps in the proof of Theorem 1 are then worked out in Sections 3, 4, and 5.

For an introduction to geometric measure theory and rectifiable sets we refer to [1]. The basicproperties of Jordan curves and of simply connected sets in the plane, used in this paper, canbe found in most books on algebraic topology, see for example [13, 16]. Regarding set-valuedfunctions, and the Hausdorff distance in the family of compact sets, our basic reference is [3].

2 Preliminaries

Given an admissible barrier Γ, we define the corresponding minimum time function as

TΓ(x).= inf

t ≥ 0; x ∈ RΓ(t)

. (2.1)

The set of times where the constraint (1.5) is saturated, i.e. it is satisfied as an equality, willbe denoted as

S .=t ≥ 0; m1

(Γ ∩RΓ(t)

)= σt

. (2.2)

The saturated and the free portions of the barrier Γ are defined respectively as

ΓS.=x ∈ Γ; TΓ(x) ∈ S

, ΓF

.=x ∈ Γ; TΓ(x) 6∈ S

. (2.3)

For future use, we now prove

Lemma 2.1. If Γ is a closed, admissible barrier, then the set S in (2.2) is closed.

5

Proof. Consider a convergent sequence of times tn → t∗, with tn ∈ S for every n ≥ 1. Weneed to show that t∗ ∈ S. Observe that from this sequence we can always extract a monotonesubsequence. We thus need to consider two cases.

CASE 1: The sequence is monotone increasing, so that tn ≤ tn+1 ≤ t∗ for all n ≥ 1.

In this case RΓ(tn) ⊆ RΓ(tn+1) ⊆ RΓ(t∗). Hence

σt∗ ≥ m1

(RΓ(t∗) ∩ Γ

)≥ lim sup

n→∞m1

(RΓ(tn) ∩ Γ

)= lim sup

n→∞σtn = σt∗. (2.4)

Of course, the first inequality holds because Γ is admissible. Therefore all terms in (2.4) areequal, and t∗ ∈ S.

CASE 2: The sequence is monotone decreasing, so that tn ≥ tn+1 ≥ t∗ for all n ≥ 1.

In this case RΓ(tn) ⊇ RΓ(tn+1) ⊇ RΓ(t∗) for every n. Hence RΓ(t∗) ⊆⋂n≥1R

Γ(tn). Weclaim that the opposite inequality also holds:

RΓ(t∗) ⊇⋂n≥1

RΓ(tn) . (2.5)

Indeed, assume y /∈ RΓ(t∗). Since this set is closed, the distance must be strictly positive:

δ.= d

(y ; RΓ(t∗)

)> 0

Recalling that the fire propagates with unit speed, we conclude

y /∈ RΓ(t) for all t < t∗ + δ .

Hence the point y is not contained in the right hand side of (2.5), and our claim is proved.

From(2.5) it follows

σt∗ ≥ m1

(RΓ(t∗) ∩ Γ

)= lim inf

n→∞m1

(RΓ(tn) ∩ Γ

)= lim inf

n→∞σtn = σt∗,

hence t∗ ∈ S.

3 A barrier bounding a star-shaped domain

Let Γ be an admissible barrier, consisting of a simple closed curve of length `.= m1(Γ). Notice

that the open unit disc B1 must be contained in the bounded component of R2 \Γ. Otherwisethe set RΓ

∞ is unbounded and the cost J(Γ) is infinite.

Let s 7→ γ(s) be a Lipschitz parameterization of Γ, with s ∈ [0, 1], γ(0) = γ(1), and |γ(s)| = `for a.e. s ∈ [0, 1]. It is convenient to extend γ by periodicity to the entire real axis, settingγ(s) = γ(s+N) for every integer N . We shall denote by (ρ(s), θ(s)) the polar coordinates ofthe point γ(s) ∈ R2. Thinking of a light source located at the origin, we define the illuminatedregion (see Fig. 2) as

Ω.=x ∈ IR2 ; λx /∈ Γ for all 0 ≤ λ < 1

. (3.6)

6

The boundary of this region will be denoted by Γ′.= ∂Ω. In this section we work toward a

proof of the following properties:

(P1) Γ′ is an admissible, simple closed curve.

(P2) J(Γ′) ≤ J(Γ), with equality holding only if Γ′ = Γ.

3.1 The shaded set

We define the shaded set as

Z.=s ∈ R ; γ(s) /∈ Ω

=s ∈ R ; λγ(s) = γ(s′) for some s′ ∈ IR , 0 < λ < 1

. (3.7)

This is the set of parameter values s such that the point γ(s) is in the shade.

Lemma 3.1. The set Z is a countable union of disjoint nontrivial intervals.

Proof. Since Z ⊂ R, every connected component of Z is an interval, possibly reduced to asingle point.

Suppose that Z contains a connected component consisting of a single point, say s∗. In thiscase, we can find two monotone sequences sk → s∗, s′k → s∗, with

sk < s∗ < s′k , sk /∈ Z, s′k /∈ Z for all k ≥ 1 . (3.8)

The following steps show that this assumption leads to a contradiction.

γ

γ

γ

γ

γ

(s )

(s )

(s )

(t )j

*

k

k

0

’

(t )*

θ*

Figure 3: If γ(s∗) is in the shade, then it must be part of a nontrivial arc of points γ(s) all in the

shade.

1. Call (ρ(s), θ(s)) the polar coordinates of the point γ(s) along the curve. Since γ(s∗) is inthe shade, as shown in Fig. 3 we can find t∗ ∈ [0, 1] \ Z such that θ∗

.= θ(t∗) = θ(s∗) and

ρ(t∗) < ρ(s∗). Observe that t∗ cannot be in the interior of the set s ; θ(s) = θ∗. Otherwise,since the map s 7→ γ(s) is one-to-one, the point γ(t∗) would also be in the shade, contrary tothe choice of t∗. We can thus assume that there exists a sequence tj → t∗ such that θ(tj) < θ∗

7

for all j ≥ 1 (the case θ(tj) > θ∗ is entirely similar). Set δ.= ρ(s∗)−ρ(t∗)

2 > 0. Choosing tjsufficiently close to t∗, by continuity we have ρ(s) ≤ ρ(t∗) + δ for all s ∈ [t∗, tj ]. Hence all thepoints with polar coordinates (r, θ) such that

r > ρ(s∗)− δ , θ ∈ [θ(tj), θ(t∗)]

lie in the shade. Hence there exists a neighborhood N of P ∗ = γ(s∗) such that every pointγ(s) ∈ N with θ(s) ≤ θ∗ lies in the shade.

Since all points γ(sk), γ(s′k) are illuminated, we must have

θ(sk) > θ∗ , θ(s′k) > θ∗ (3.9)

for all k sufficiently large.

2. For notational convenience we write P ∗.= γ(s∗). For ε > 0 sufficiently small, consider the

two arcs γ1, γ2 ⊂ Γ, defined as

γ1.=γ(s); s ∈ ]s∗ − ε, s∗[

, γ2

.=γ(s); s ∈ ]s∗, s∗ + ε[

. (3.10)

0

α

θ*

B

A

C

D P*

1P

P2

γ1

γ2

Figure 4: The construction used in the proof of Lemma 3.1.

Choose an angle α > θ∗ close to θ∗. In view of (3.9), the arcs γ1 and γ2 must both containpoints along the ray

~ζα.=

(r cosα, r sinα) ; r ≥ 0,

provided that α− θ∗ is chosen sufficiently small.

For i=1,2, let Pi.= γ(sPi) be the first point where γi intersects the ray ~ζα, as shown in Fig. 4.

In other words, chooses∗ − ε < sP1 < s∗ < sP2 < s∗ + ε

such that

θ(sP1) = θ(sP2) = α , and θ(s) < α for all s ∈ ]sP1 , sP2 [ .

8

To fix the ideas, assume that ρ(sP1) < ρ(sP2), the other case being entirely similar.

By (3.8) and (3.9), we can choose a point A on γ2 with A = γ(sA) = γ(s′j) for some j ≥ 1,such that θ∗ < θ(sA) < α. We then consider the point B = γ(sB) on γ1 which satisfies

ρ(sB) = minρ(s); θ(s) = θ∗, s ∈ [sP1 , s

∗].

This choice of B implies that the union

Σ.= OP1 ∪ OB ∪

γ(s); s ∈ [sP1 , sB]

. (3.11)

is a Jordan curve. Let U be the open, bounded connected component of IR2 \ Σ, so thatΣ = ∂U . We observe that A ∈ U while P2 6∈ U. Therefore, the connected arc

ωAP2

.= γ2(s) ; s ∈ [sA , sP2 ] ⊆ γ2

must intersect the boundary ∂U = Σ at some point C. Since γ2 ∩ OP1 = γ2 ∩ γ1 = ∅, thisintersection must lie on the segment OB. Choose the intersection point C

.= γ(sC) such that

sC.= min

s ∈ [sA, sP2 ]; θ(s) = θ∗ and ρ(s) < ρ(sB)

. (3.12)

By (3.12) we can now choose a new point D = γ(sD) sufficiently close to C, such that

sD < sC , θ(sD) > θ∗, ρ(s) < ρ(s∗) for all s ∈ [sD, sC ].

Hence there exists a neighborhood N of P ∗ = γ(s∗) such that every point γ(s) ∈ N withθ(s) ≥ θ∗ lies in the shade produced by the arc γCD = γ(s) ; s ∈ [sD, sC ].

Together with the previous step, this implies that an entire neighborhood of P ∗ lies in theshade. Hence s∗ is an interior point of Z, contrary to our initial assumption. This contradictionproves the lemma.

The next lemma shows that each connected component of Z is precisely a half-open interval.

Lemma 3.2. Let I be a maximal interval contained in Z, with closure I = [sa, sb]. Then

θ(sa) = θ(sb). (3.13)

Furthermore, I is half-open. Indeed,

ρ(sa) < ρ(sb) =⇒ I = ]sa , sb] ,ρ(sa) > ρ(sb) =⇒ I = [sa , sb[ .

(3.14)

Proof. 1. Assume that, on the contrary, θ(sa) 6= θ(sb). Choose s′a ≤ sa < sb ≤ s′b such thats′a, s

′b /∈ Z, with |s′a − sa| and |s′b − sb| sufficiently small. Consider the points

Pa = γ(sa), Pb = γ(sb), P ′a = γ(s′a), P ′b = γ(s′b).

As in Fig. 5, call Qa, Qb respectively the intersections of the unit circumference with thesegments OP ′a, OP

′b. These points divide the unit circumference into two sub-arcs, which we

call ω+, ω−. We shall also split the original curve Γ as

Γ.= Γ′ ∪ Γ′′, Γ′

.=γ(s) ; s′a ≤ s ≤ s′b

, Γ′′ = Γ \ Γ′.

9

’

Γ

Γ

P

PP

PQa

a

a

b

b

P

P ""

’

’

ω

bQ ω+

_

θ* ’

Figure 5: The construction used in the proof of (3.13) in Lemma3.2. In this case, one would choose

ω = ω+.

Consider the curveΓ]

.= P ′aQa ∪ ω ∪ P ′bQb ∪ Γ′

where we choose either ω = ω+ or ω = ω−. Observe that, regardless of the choice of the arcω, the curve Γ] is a simple, closed curve. We now choose the arc ω ∈ ω+, ω− so that theorigin remains in the unbounded component of IR2 \ Γ].

Let θ∗ be the angle corresponding to the mid-point of the arc ω. We claim that the arc Γ′

must contain a point P ′ = γ(s′) with θ(s′) = θ∗. Indeed, in the opposite case, by our choiceof ω all points along the ray

(r cos θ∗, r sin θ∗) ; r ≥ 0

(3.15)

(except for r = 1) would belong to the unbounded connected component of R2 \ Γ]. But thisis impossible, because Γ] is a Jordan curve, and intersects the ray (3.15) exactly once. Thisproves the claim.

If |s′a − sa| and |s′b − sb| were chosen sufficiently small, then this intersection point P ′ = γ(s′)satisfies sa < s′ < sb and hence s′ ∈ Z. Since P ′ is in the shade, there exists a point in thelight, say P ′′ = γ(s′′) ∈ Γ′′, with

s′′ /∈ Z, θ(s′′) = θ(s′) = θ∗, ρ(s′′) < ρ(s′).

Next, observe that Γ′′ contains the point P ′′ and does not intersect Γ]. Therefore, it is entirelycontained in the bounded component of R2 \ Γ]. We conclude that the bounded componentof R2 \Γ = R2 \ (Γ′ ∪Γ′′) is contained in the bounded component of R2 \Γ], hence it does notcontain the origin. This is a contradiction, proving (3.13).

2. To prove (3.14), assume that ρ(sa) < ρ(sb), the other case being similar. Then by (3.13)the point Pb is in the shade of the point Pa. Hence sb ∈ Z.

10

θ

cP

Pb

aPk

P

Figure 6: The construction used in the proof of (3.14) in Lemma 3.4.

Finally, assume that sa ∈ Z. A contradiction is then obtained as follows.

Choose sc 6∈ Z, such that θ(sc) = θ(sa).= θa and

ρ(sc) = minρ(s) ; θ(s) = θ(sa)

.

As in Fig. 6, we set Pc.= γ(sc). Then, as s ranges in a small neighborhood of sc, we can

assume that the angle θ(s) covers a right neighborhood of the angle θa (the case of a leftneighborhood is entirely similar). As shown in (Fig. 6), all points γ(s) in a neighborhood ofγ(sa) with θ(s) ≤ θa thus lie in the shade.

On the other hand, by assumption, there exists a sequence sk → sa with sk /∈ Z. This ispossible only if θ(sk) > θa. By continuity, there exists k large enough and δ > 0 such that

ρ(s) < ρ(sb)− δ for all s ∈ [sk, sa] .

Therefore, all points γ(s) in a neighborhood N of Pb with θ(s) > θa lie in the shade. But wealready know that points with θ(s) ≤ θa also lie in the shade. Hence γ(s) ∈ Z for all s in aneighborhood of sb, against the assumptions. This completes the proof of the lemma.

3.2 Cutting out a shaded portion

Let Γ be an admissible, simple closed curve, parameterized as s 7→ γ(s). Recalling (3.6)-(3.7),assume that

γ].=γ(s); s ∈ ]s0, s1]

⊂ Γ (3.16)

is a maximal arc contained in the shaded region. One can then replace γ] with the segmentγ[

.= P0P1 connecting the two endpoints P0 = γ(s0), P1 = γ(s1). Aim of this section is to

prove

Lemma 3.3. In the above setting, the curve

Γ[.= γ[ ∪ (Γ \ γ]) (3.17)

is an admissible, simple closed curve. Moreover

J(Γ[) ≤ J(Γ) (3.18)

with equality holding only if Γ[ = Γ.

11

Proof. 1. If γ] is itself a segment, then Γ[ = Γ and the conclusion is trivial. Otherwise, it sclear that

m1(γ[) < m1(γ]).

Moreover, the global burned areas satisfy RΓ∞ ⊂ RΓ[

∞ . The difference in the areas is computedas

m2(RΓ∞) = m2(RΓ[

∞) +m2(W ) ,

where W is the union of all bounded components of R2 \ (γ] ∪ γ[). This proves (3.18).

θ

0

γ*

P0 P

0n

γ #

P1

Γ

A

B

’A’

B

n

εE*

Figure 7: The construction used to show that Γ[ is admissible.

2. It remains to prove that Γ[ is admissible. Relying on the fact that the original curve Γ isadmissible, it suffices to show that

m1

(Γ[ ∩RΓ[(t)

)≤ m1

(Γ ∩RΓ(t)

)≤ σ t for all t ≥ 0 . (3.19)

As described in Lemma 3.2, in the present situation we have ρ(s0) = |P0| < |P1| = ρ(s1).Define the times

t0.= ρ(s0)− 1 t1

.= ρ(s1)− 1 .

For t < t0, the reachable sets RΓ(t) and RΓ[(t) coincide. Indeed, their closure does not

intersect γ], nor γ[.

For t ≥ t0, the portion of the segment γ[ touched by the fire has length

m1

(γ[ ∩RΓ[(t)

)= min t− t0 , t1 − t0 . (3.20)

To prove (3.19) it thus suffices to show that

m1

(γ] ∩RΓ(t)

)≥ t− t0 for all t ∈ [t0, t1] . (3.21)

Let ζ 7→ γ](ζ) be an arc-length parameterization of the curve γ], starting at the point γ](0) =P0. We claim that

γ]τ.=γ](ζ) ; ζ ∈ [0, τ ]

⊂ RΓ(t0 + τ) whenever 0 < τ < t1 − t0 . (3.22)

12

In other words, given any point P = γ](τ), there exists a curve Σ of length ≤ t0 + τ (atrajectory for the fire) which starts inside the unit disc R0, does not cross Γ, and reaches apoint arbitrarily close to P . If the curve γ] is smooth, this fact would be obvious. Indeed,one could define Σ as the union of a segment from R0 to a point close to P0, together with acurve of points having constant distance ε > 0 from γ] (Fig. 7). In general, however, we onlyknow that γ] is a Lipschitz continuous curve. In this case, the set of points at a small distanceε > 0 from γ] may not be a connected curve. A more careful analysis is thus needed.

Since the map ζ 7→ γ](ζ) is Lipschitz continuous, it is differentiable for a.e. ζ. By slightlymoving the point P along the curve, we can thus assume that the tangent and the normalvectors to the curve γ] are well defined at the point P . As in Fig. 7, we denote by n the unitnormal to γ] at P , oriented toward the interior of the set Ω enclosed by Γ. We recall that Γis a Jordan curve, and Ω is the unique bounded connected component of R2 \ Γ. Moreover,we denote by n0 the unit vector perpendicular to the segment P0P1, oriented so that, at P0,it points toward the interior of Ω. By choosing ε0 > 0 sufficiently small, the two points

A.= P0 + ε0n0 , B

.= P + ε0n

are both contained in the interior of Ω. Since the open set Ω is connected, there exists asmooth path γ∗, entirely contained inside Ω, with endpoints A, B. Consider the curve

Σ.= P0A ∪ γ∗ ∪ PB ∪ γ](ζ) ; 0 ≤ ζ ≤ τ . (3.23)

Notice that the last portion of Σ is the portion of the arc γ] from P0 to P . By construction,Σ is a Jordan curve. Call E the bounded connected component of R2 \ Σ. By Theorem 5 in[2] on the one-sided Minkowski content, one has

limε→0+

m1

(x ∈ E ; d(x,Σ) = ε

)= m1(∂E) = m1(Σ) = ε0 +m1(γ∗) + ε0 + τ . (3.24)

In general, the open setEε

.= x ∈ E ; d(x,Σ) > ε

has many connected components. Since γ∗ is smooth, there will be one connected component,which we call E∗ε , whose boundary contains a smooth curve γ∗ε of points remaining at a distanceε from γ∗. Let A′, B′ be the endpoints of γ∗ε . Since both E∗ε and its complement are connected,the boundary ∂E∗ε is a Jordan curve. Therefore, one can start from A′ and reach B′, movingalong the curve ∂E∗ε in two different ways: either along the curve γ∗ε , or along the complement∂E∗ε \ γ∗ε . In the second case the total length traveled is

m1(∂E∗ε )−m1(γ∗ε ) ≤ m1(∂Eε)−m1(γ∗ε ) = m1(Σ)−m1(γ∗)+o(1) = 2ε0 +τ+o(1) , (3.25)

where the Landau symbol o(1) denotes a quantity that tends to zero as ε→ 0,

To complete the proof of (3.22), it only remains to observe that the point A can be connectedto the unit disc R0 by a segment of length < |P0| + ε0, without crossing Γ. Moreover, thesegments AA′ and BB′ have lengths which approach zero as ε0, ε→ 0.

In turn, (3.22) imples (3.21), completing the proof of the lemma.

13

3.3 A star-shaped domain

As proved in Lemma 3.2, the set of points P ∈ Γ lying in the shade can be expressed as theunion of its connected components:

γ(s) ; s ∈⋃j≥1

Ij

,

where the Ij ⊂ [0, 1] form a countable family of disjoint half-open intervals: either Ij = [aj , bj [ ,or Ij =]aj , bj ].

For k ≥ 0, let Γk be the Jordan curve obtained by replacing the first k arcsγ(s); s ∈ Ij

,

1 ≤ j ≤ k, by the segments

Sj.=θγ(aj) + (1− θ)γ(bj) ; θ ∈ [0, 1]

(3.26)

having the same endpoints. By induction on k, from Lemma 3.3 it follows that curve Γk isadmissible. Moreover

m1(Γk+1) ≤ m1(Γk) ≤ m1(Γ) , Ωk+1 ⊆ Ωk ⊆ Ω (3.27)

for every k ≥ 1, with equalities holding if and only if the sets are equal. The Hausdorff distancebetween the above curves satisfies∑

k≥1

dH(Γk, Γk−1) ≤∑k≥1

m1

(γ(s) ; s ∈ [ak, bk]

)≤ m1(Γ) .

Therefore, the sequence of compact sets (Γk)k≥0 is Cauchy w.r.t. the Hausdorff distance.Hence it admits a unique limit:

Γ∞.= lim

k→∞Γk . (3.28)

By Golab’s theorem (see Theorem 3.18 in [14]), it follows

m1(Γ∞) = limk→∞

m1(Γk) ≤ m1(Γ) . (3.29)

We will show that Γ∞ coincides with the set

Γ′.= (Γ ∩ Ω) ∪

( ⋃j≥1

Sj

). (3.30)

In other words, Γ′ is defined as the illuminated portion of Γ, together with all the segmentsSj . A further characterization of Γ∞ will be given in terms of the set

Ωo .=x ∈ IR2 ; λx /∈ Γ for all 0 ≤ λ < 1

. (3.31)

From the definition, it is clear that Ωo is star shaped, and its closure contains the illuminatedset Ω in (3.6). We observe that Ωo is open. Indeed, assume that x ∈ Ω and the segment

Sx.=λx ; λ ∈ [0, 1]

(3.32)

does not intersect Γ. Then, by the compactness of Sx and Γ, there exists δ > 0 such thatevery point y ∈ Sx has distance ≥ δ > 0 from Γ. This implies B(x, δ) ⊂ Ωo. We concludethat Ωo is open, and coincides with the interior of Ω. The next lemma relates the above sets,defined at (3.28), (3.30), and (3.31).

14

Lemma 3.4. One has the identities Γ′ = Γ∞ = ∂Ωo. Moreover, Γ′ is an admissible, simpleclosed curve.

Proof. 1. As a first step, we prove Γ′ ⊆ Γ∞. Indeed, if x is a point in the illuminated portionof Γ, then x ∈ Γk for every k, hence x ∈ Γ∞. On the other hand, if x ∈ Sj , then x ∈ Γk forevery k ≥ j. This again implies x ∈ Γ∞.

2. In this step we show that Γ∞ ⊆ ∂Ωo. Indeed, if x ∈ Γ∞, there exists a sequence of pontsxk ∈ Γk such that xk → x as k →∞. We consider two cases.

CASE 1: There exists an index j such that xk ∈ Sj for infinitely many k. In this case, x ∈ Sj ,and hence x /∈ Ωo. To fix the ideas, assume that the segment Sj has endpoints γ(aj), γ(bj),with Euclidean norm 1 ≤ |γ(aj)| < |γ(bj)|. Let x = λγ(bj) for some λ ∈ [0, 1]. Since theinterval ]aj , bj ] is a maximal interval in the shaded set, there exists a sequence of illuminatedpoints γ(sn) with sn → bj . For all n ≥ 1, we now have

yn.=

(λ− 1

n

)γ(sn) ∈ Ωo

Moreover, yn → x. Hence x ∈ ∂Ωo.

CASE 2: There exists a subsequence such that xk /∈ S1 ∪ · · · ∪ Sn(k), with n(k) → ∞ ask → ∞. In this case, we claim that we can find a second sequence of illuminated pointsyk = γ(sk) ∈ Γ ∩ Ω such that |yk − xk| → 0. Indeed:

(i) If xk ∈ Γ ∩ Ω, we simply take yk = xk.

(ii) If xk ∈ Sj for some Sn(k) < j ≤ k, we take yk = γ(aj) or yk = γ(bj). By Lemma 3.2,exactly one of these two points is illuminated. Then |yk − xk| ≤ m1(Sj). Since j → ∞as k →∞, the length of the segment Sj approaches zero.

(iii) If xk = γ(s) for some s ∈ [aj , bj ] with j > k, we again take yk = γ(aj) or yk = γ(bj).Then |yk − xk| ≤ m1

(γ(s) ; s ∈ [aj , bj ]

). Since j → ∞ as k → ∞, this length

approaches zero.

Defining

zk.=

k − 1

kyk ∈ Ωo,

we now havex = lim

k→∞xk = lim

k→∞yk = lim

k→∞zk .

Hence x ∈ Γ ∩ ∂Ω0.

3. We now show that ∂Ωo ⊆ Γ′. Let x ∈ ∂Ωo. Two cases will be considered.

CASE 1: x is an illuminated point, i.e., x ∈ Ω. If x /∈ Γ, then x ∈ Ωo, against the assumption.Therefore x ∈ Γ ∩ Ω ⊆ Γ′.

15

CASE 2: x /∈ Ω. In this case, taking λ.= minλ′ ; λ′x ∈ Γ, we determine a point y = γ(s) ∈

Γ ∩ Ω such that y = λx, for some 0 < λ < 1.

As in Section 3, we denote by (ρ(s), θ(s)) be the polar coordinates of the point γ(s). Ifthe map s 7→ θ(s) does not have a local extremum at s = s, then for every ε > 0 the setθ(s) ; s ∈ [s − ε, s + ε] covers a whole neighborhood of θ(s), hence there exists an entireneighborhood of x consisting of points in the shade. This contradicts the assumption x ∈ ∂Ωo.Therefore, θ

.= θ(s) is either a local minimum or a local maximum of the angular function

θ(·). Hence, s is in the boundary of the shaded set Z in (3.7). Using Lemma 3.1, we canassume that s = aj , where ]aj , bj ] is a maximal interval contained in the shaded set Z, and1 ≤ |γ(aj)| < |γ(sj)|, the other case being entirely similar.

Since we already know that γ(aj) = λx for some λ < 1, to prove that x ∈ Sj it remains toshow that |x| ≤ |γ(bj)|. To fix the ideas, let aj be a point where the map s 7→ θ(s) attainsa local maximum; the case of a local minimum is entirely similar. Given ε > 0, there existsδ > 0 such that all points z = r(cos θ, sin θ) with r > |γ(aj)|+ ε and θ ∈ [θ − δ, θ] are in theshade. On the other hand, by the maximality of the interval ]aj , bj ], there exist a sequenceof illuminated points γ(sn) ∈ Γ ∩ Ω, with sn ↓ bj . By the previous analysis, we must haveθ(sn) > θ. By continuity of the maps s 7→ (ρ(s), θ(s)), for every ε > 0 there exists δ > 0such that all points z = r(cos θ, sin θ) with r > |γ(bj)|+ ε and θ ∈ [θ, θ + δ] are in the shade.Hence, if |x| > |γ(bj)|, we conclude that x has a whole neighborhood consisting of points inthe shade, against the assumption x ∈ ∂Ωo.

4. Together, the three previous steps prove the identities Γ′ = Γ∞ = ∂Ωo. Being the boundaryof the simply connected, star shaped set Ωo, it is clear that Γ′ = Γ∞ is a simple closed curve.It remains to prove that Γ′ is admissible.

Call Ωk the open region enclosed by the simple closed curve Γk, i.e. the bounded connectedcomponent of IR2 \ Γk. Observe that RΓ′(t) is the set reached at time t by trajectories of thedifferential inclusion

|x| ≤ 1 , |x(0)| < 1 , (3.33)

which remain inside Ωo at all times. On the other hand, RΓk(t) is the set reached at time tby trajectories of (3.33) which remain inside Ωk. Since Ωo ⊆ Ωk for every k, we clearly have

RΓ′(t) ⊆ RΓk(t) for all t ≥ 0.

For every t ≥ 0 and k ≥ 1, we can write

m1(Γ′ ∩RΓ′(t)) ≤ m1(Γ′ ∩RΓk(t)) ≤ m1(Γk ∩RΓk(t)) +m1(Γ′ \ Γk)

≤ σt+∑j>k

m1(Sj) .

Letting k → ∞ the sum on the right hand side approaches zero, proving the admissibility ofthe closed curve Γ′.

Being the boundary of a star-shaped set, the simple closed curve Γ′ can be represented inpolar coordinates as

Γ′ =

(r cos θ , r sin θ) ; r−(θ) ≤ r ≤ r+(θ) , θ ∈ [−π, π], (3.34)

16

for some suitable functions r−, r+. Here r− is lower semicontinuous, while r+ is upper semicon-tinuous. Moreover r−(θ) = r+(θ) for all but countably many values of θ. The total variationof these functions on [−π, π] satisfies

Tot.Var.r−(·) = Tot.Var.r+(·) ≤ m1(Γ′). (3.35)

4 Symmetric rearrangement

Based on the analysis of the previous section, we can now assume that the curve Γ = Γ′ isalready the boundary of a star-shaped, open domain

Ωo =

(r cos θ, r sin θ) ; 0 ≤ r < r(θ) , θ ∈ [−π, π],

for some lower semicontinuous BV function θ 7→ r(θ), with r(−π) = r(π). To simplify thenotation, we are here taking r(θ)

.= r−(θ), dropping the superscript.

In this section, we consider the new domain

Ω =

(r cos θ, r sin θ) ; 0 ≤ r < r(θ),

where r(·) is the symmetric, nondecreasing rearrangement of the function r−(·). In otherwords, r : [−π, π] 7→ IR+ is the unique lower semicontinuous function with the properties

meas(θ ; r(θ) < ρ

)= meas

(θ ; r−(θ) < ρ

)for all ρ > 0 , (4.1)

r(−θ) = r(θ) , r(θ) ≤ r(θ′) for all 0 ≤ θ ≤ θ′ ≤ π. (4.2)

From (4.1) one immediately obtains

m2(Ω) =

∫ ∞0

ρ·meas(θ ; r(θ) > ρ

)dρ =

∫ ∞0

ρ·meas(θ ; r−(θ) > ρ

)dρ = m2(Ωo) .

(4.3)

To prove that Γ.= ∂Ω is an admissible curve, we need to show that, for every t > 0 one has

m1

(Γ ∩B(0, 1 + t)

)≤ m1

(Γ ∩B(0, 1 + t)

)≤ σt . (4.4)

Notice that the second inequality in (4.4) follows from the assumption that Γ is admissible. Thefirst inequality apparently cannot be deduced directly from general theorems on symmetricrearrangements in [11, 15]. However, the ideas involved in the proof are quite similar. Westart with an elementary inequality.

Lemma 4.1. Given any numbers b1, . . . , bn, with n ≥ 2, one has

n∑i=1

√1 + b2i ≥ 2

√1 +

(∑ni=1 bi2

)2

. (4.5)

Proof. Applying Jensen’s inequality to the convex function ϕ(s) =√

1 + s2 we obtain

n∑i=1

√1 + b2i ≥ n

√1 +

(∑ni=1 bin

)2

≥ m

√1 +

(∑ni=1 bim

)2

(4.6)

whenever n ≥ m. Taking m = 2, one obtains (4.5).

17

Using the lemma, the first inequality in (4.4) will be proved first in the case where r(·) is apiecewise affine function, then in the general case, using an approximation argument.

1. Assume that the function θ 7→ r(θ) is continuous, piecewise affine, and satisfies

r(−π) = r(π) ,dr

dθ(θ) 6= 0 for a.e. θ . (4.7)

For each ρ > 0, consider the sets

Θ(ρ).=θ ∈ [−π, π] ; r(θ) = ρ

, Θ(ρ)

.=θ ∈ [−π, π] ; r(θ) = ρ

.

By the assumption (4.7), every set Θ(ρ) contains at most two elements, say θ(ρ), −θ(ρ), whileΘ(ρ) contains at most finitely many elements, say θ1(ρ), . . . , θN (ρ), with N also depending onρ. Writing

dθ

dr(θ) =

(dr

dθ(θ)

)−1

,dθ

dr(θ) =

(dr

dθ(θ)

)−1

,

the assumption that r(·) is a rearrangement of r(·) implies∣∣∣∣dθdr (θ)

∣∣∣∣ =

∣∣∣∣dθdr (−θ)∣∣∣∣ =

1

2

N∑i=1

∣∣∣∣dθdr (θi)

∣∣∣∣ . (4.8)

Using (4.5) with bi.= ρ

∣∣∣dθdr (θi)∣∣∣, the two expressions in (4.4) can now be estimated as

m1

(Γ ∩B(0, 1 + t)

)=

∫ 1+t

0

∑θ∈Θ(ρ)

√1 + ρ2

∣∣∣∣dθdr (θ)

∣∣∣∣2dρ =

∫ 1+t

02

√1 + ρ2

∣∣∣∣dθdr (θ(ρ))

∣∣∣∣2 dρ

≤∫ 1+t

0

N(ρ)∑i=1

√1 + ρ2

∣∣∣∣dθdr (θi(ρ))

∣∣∣∣2 dρ ≤∫ 1+t

0

∑θ∈Θ(ρ)

√1 + ρ2

∣∣∣∣drdθ (θ(ρ))

∣∣∣∣2dρ= m1

(Γ ∩B(0, 1 + t)

)≤ σt .

(4.9)

2. To cover the general case, let θ 7→ r(θ) be a lower semicontinuous function with boundedvariation. Then there exists a sequence of piecewise affine functions rn(·) converging to r(·)at a.e. θ ∈ [−π, π] and such that, calling

Γn.=rn(θ)(cos θ, sin θ) ; θ ∈ [−π, π]

, (4.10)

there holds

m1

(Γn ∩B(0, 1 + t)

)≤ σt+

1

n, (4.11)

for every t ≥ 0 and n ≥ 1. Call rn(·) the non-decreasing symmetric rearrangement of rn(·),and consider the curve Γn, defined as in (4.10) with rn replaced by rn. By the previous stepwe have

m1

(Γn ∩B(0, 1 + t)

)≤ σt+

1

n, (4.12)

18

for every t ≥ 0 and n ≥ 1.

To pass to the limit as n→∞, we proceed as follows. Let f : [−π, π] 7→ IR+ be a lower semi-continuous function, with f(−π) = f(π). Let Df = µac + µs be its distributional derivative,decomposed as the sum of an absolutely continuous measure with density f ′ w.r.t. Lebesguemeasure, plus a singular part. Consider the functional

Φ(f).=

∫ π

−π

√f2(θ) + (f ′)2(θ) dθ + |µs|

(]− π, π]

). (4.13)

The second term on the right hand side of (4.13) denotes the total mass of the singularmeasure µs. Observe that, if r = f(θ) is the polar coordinate representation of a smoothcurve Γ ⊂ IR2, then Φ(f) = m1(Γ) yields the total length of the curve. More generally, if f isa lower semicontinuous BV function and we choose r−(θ) = f(θ), r+(θ) = lim supθ′→θ f(θ′),then Φ(f) yields the length of the curve Γ′ in (3.34).

For a given radius ρ > 0, the portion of Γ′ contained inside the open ball B(0, ρ) can beexpressed as

m1(Γ′ ∩B(0, ρ)) = Φ(f ∧ ρ)− ρ ·m1

(θ ; f(θ) ≥ ρ

).

We use here the truncated function (f ∧ ρ)(θ).= minf(θ), ρ. In particular, for any t ≥ 0

and ε > 0 one has

m1

(Γ ∩B(0, 1 + t)

)≤ m1

(Γ ∩B(0, 1 + t+ ε)

)= Φ

(r ∧ (1 + t+ ε)

)− (1 + t+ ε) ·m1

(θ ; r(θ) ≥ 1 + t+ ε

),

(4.14)

By the lower semicontinuity of the integral functional Φ (see [12] or section 5.5 in [1]), itfollows

Φ(r ∧ (1 + t+ ε)

)≤ lim inf

n→∞Φ(rn ∧ (1 + t+ ε)

)≤ lim inf

n→∞Φ(rn ∧ (1 + t+ 2ε)

). (4.15)

By construction, for every n, t, ε we have

Φ(rn ∧ (1 + t+ 2ε)

)− (1 + t+ 2ε) ·m1

(θ ; rn(θ) ≥ 1 + t+ 2ε

)= m1

(Γn ∩B(0, 1 + t+ 2ε)

)≤ m1

(Γn ∩B(0, 1 + t+ 2ε)

)≤ σ(t+ 2ε) +

1

n.

(4.16)

Using (4.14)–(4.16) and the pointwise a.e. convergence rn(θ)→ r(θ), we conclude

m1

(Γ ∩B(0, 1 + t)

)≤ lim inf

n→∞

(σ(t+ 2ε) +

1

n

)

+(1 + t+ 2ε) · lim supn→∞

m1

(θ ; rn(θ) ≥ 1 + t+ 2ε

)−(1 + t+ ε) ·m1

(θ ; r(θ) ≥ 1 + t+ ε

)≤ σ(t+ 2ε) + ε · 2π .

(4.17)

Since ε > 0 was arbitrary, this proves the admissibility of the curve Γ.

19

5 The optimal barrier

In this final section we consider an admissible curve Γ = ∂Ω, enclosing a set of the form

Ω.=

(ρ cos θ, ρ sin θ) ; 0 ≤ ρ < r(θ) , θ ∈ [−π, π].

We assume that the map θ 7→ r(θ) satisfies

r(−θ) = r(θ) , r(θ) ≤ r(θ′) for all 0 ≤ θ ≤ θ′ ≤ π. (5.1)

We claim that, if Γ is optimal, i.e. if it minimizes the cost functional (1.8) among all admissiblecurves, then Γ must consist of the concatenation of an arc of circumference and two arcs oflogarithmic spirals.

Indeed, consider the set of times where the admissibility condition (1.5) is satisfied as anequality:

S .=t ≥ 0 ; m1(Γ ∩B(0, 1 + t)) = σt

. (5.2)

By Lemma 2.1, this set is closed.

Assume that the open interval ]τa, τb[ is a maximal connected set contained in the complementof S, with ta > 0. This means that

m1(Γ ∩B(0, 1 + τa)) = στa , (5.3)

m1(Γ ∩B(0, 1 + τb)) = στb, (5.4)

m1(Γ ∩B(0, 1 + t)) < σt for all t ∈ ]τa, τb[ . (5.5)

By the necessary conditions for optimality proved in [4, 9], the restriction of Γ to the set ofpoints

x ; TΓ(x) ∈ ]τa, τb[ = x ; 1 + τa < |x| < 1 + τb

consists of two arcs of circumferences, say with endpoints A1, B1 and A2, B2, as shown inFig. 8.

However, this is impossible. Indeed, call Q the point where the circumference is tangent to aray from the origin. From (5.3) and (5.5) it follows that, at the point A1, the angle α between

the arc A1B1 and the circumference x ∈ IR2 ; |x| = |A1| must satisfy

σ

2sinα ≥ 1. (5.6)

However, if (5.6) holds, there exists no point B1 along the arc A1Q where (5.4) is satisfied.

The previous analysis shows that, for the optimal barrier Γ, the set S in (5.2) is connected.Therefore, Γ must be obtained as the concatenation of an arc of circumference, and twologarithmic spirals.

Acknowledgment. This research was supported in part by NSF through grant DMS-1108702, “Problems of Nonlinear Control”.

20

x1

B2

A2

B1

A1

0

α

Q

Figure 8: The shaded area corresponds to points P ∈ IR2 with |P | − 1 ∈ S. If Γ is an optimal curve,then the set S ⊂ [0,∞[ of times where the constraint is saturated must be an interval.

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