The Minimum Speed for a Blocking Problem on the Half Plane

The Minimum Speed for a Blocking Problem on the Half Plane Alberto Bressan and Tao Wang Department of Mathematics, Penn State University University Park, Pa 16802, USA e-mails: bressan@mathpsuedu, wang t@mathpsuedu March 8, 2009 Abstract We consider a blocking problem: fire propagates on a half plane with unit speed in all directions To block it, a barrier can be constructed in real time, at speed σ We prove that the fire can be entirely blocked by the wall, in finite time, if and only if σ > 1 The proof relies on a geometric lemma of independent interest Namely, let K IR 2 be a compact, simply connected set with smooth boundary We define d K x, y as the minimum length among all paths connecting x with y and remaining inside K Then d K attains its maximum at a pair of points x, ȳ both on the boundary of K 1 Introduction Aim of this note is to analyze the blocking problem introduced in [4], originally motivated by the control of wild fires or of the spatial spreading of a contaminating agent At each time t 0, we denote by Rt IR 2 the burned region In absence of control, we assume that the set Rt grows uniformly in all directions, namely Rt = { BR 0, t = x IR 2 ; dx, R 0 t Here R 0 IR 2 is a fixed nonempty bounded open set, describing the region invaded by the fire at the initial time t = 0 In our model, the spreading of the fire can be controlled by constructing barriers In mathematical terms, we thus assume that the controller can construct a one-dimensional rectifiable curve γ which blocks the spreading of the contamination Calling γt IR 2 the portion of the wall constructed within time t 0, we make the following assumptions: H1 For every t 2 > t 1 0 one has γt 1 γt 2 1

H2 For every t 0, the total length of the wall satisfies m 1 γt σt 11 Here m 1 denotes the one-dimensional Hausdorff measure, normalized so that m 1 Γ yields the usual length of a smooth curve Γ The constant σ > 0 is the speed at which the wall can be constructed A strategy γ satisfying H1-H2 will be called an admissible strategy In addition, we say that the strategy γ is complete if it satisfies H3 For every t 0 there holds m 1 γt = σt, γt = s>t γs 12 Moreover, if γt has positive upper density at a point x, ie if m 1 Bx, r γt lim sup > 0, r 0+ r then x γt Here Bx, r is the ball centered at x with radius r As proved in [6], for every admissible strategy t γt one can construct a second admissible strategy t γt γt, which is complete When a wall is being constructed, the burned set is reduced Indeed, we define R γ t = { xt ; x absolutely continuous, x0 R 0, ẋτ 1 for ae τ [0, t], xτ / γτ for all τ [0, t] 13 In the above setting, we consider the problem BP1 Blocking Problem 1 Find an admissible strategy t γt such that the corresponding reachable sets R γ t remain uniformly bounded, for all times t 0 In other words, calling B r = {x IR 2 ; x < r, we seek a strategy such that for some radius r sufficiently large R γ t B r for all t 0 We recall that the Hausdorff distance between two compact sets X, Y is defined as d H X, Y = { max max dx, Y, max dy, X, x X y Y where dx, Y = inf y Y dx, y and dx, y = x y is the Euclidean distance on IR 2 In its original formulation, a strategy is a mapping t γt describing the walls γt IR 2 constructed at any given time t 0 This blocking problem can be reformulated in a simpler 2

way, where a strategy is entirely determined by assigning one single rectifiable set Γ IR 2 Indeed, consider a rectifiable set Γ IR 2 We assume that Γ is complete, in the sense that it contains all of its points of positive upper density: lim sup r 0+ m 1 Bx, r Γ r > 0 = x Γ Define the set reached at time t > 0 by trajectories which do not cross Γ: R Γ t = { xt ; x absolutely continuous, x0 R 0, ẋτ 1 for ae τ [0, t], xτ / Γ for all τ [0, t] 14 Throughout the following, S will denote the closure of a set S We say that the rectifiable set Γ is admissible for the construction speed σ if, for every t 0, the set γt = Γ R Γ t 15 satisfies 11, ie it can be constructed within time t We observe that the set γt in 15 represents the relevant portion of the barrier Γ which needs to be put in place at time t The remaining part Γ \ R Γ t has not been yet reached by the fire, and can thus be constructed at a later time We now consider: BP2 Blocking Problem 2 Find an admissible rectifiable set Γ IR 2 such that the union of all connected components of IR 2 \ Γ which intersect R 0 is bounded As proved in [7], the two above formulations of the blocking problem are equivalent Namely, the problem BP1 has a solution if and only the same is true for BP2 The analysis in [4] shows that in the entire plane a blocking strategy exists if σ > 2, and cannot exist if σ < 1 This result is not sharp, leaving a gap between the existence and the non-existence case In the light of the computations in [5] one is led to Conjecture: On the entire plane, a blocking strategy exists if and only if σ > 2 In the present paper we study the case where the fire is constrained to the half plane IR + 2 = {x 1, x 2 IR 2 ; x 2 > 0 For this case, the critical speed which discriminates between the existence or non-existence of blocking strategies can be precisely determined Theorem 1 If the fire is restricted to a half plane, then, for any initial open bounded set R 0 IR 2 +, a blocking strategy exists if and only if σ > 1 The remainder of the paper is organized as follows In Section 2 we give further equivalent forms of the blocking properties BP1, BP2 In particular we show that, if any blocking strategy exists, then a solution of BP2 can be found, consisting of a rectifiable set Γ with finite length and finitely many compact connected components In Section 3 we prove a geometric lemma on the distance function on a simply connected domain Namely, call d K x, y the 3

minimum length among all continuous paths γ : [0, 1] K which connect the points x, y K If K IR 2 is a simply connected, compact domain whose boundary K is a smooth Jordan curve, we show that the maximum of the path-distance function d K is attained at a couple of points x, ȳ both lying on the boundary of K We remark that this conclusion is generally false if the domain K is not simply connected Using this geometric lemma, in Section 4 we give a proof of Theorem 1 For an introduction to geometric measure theory and rectifiable sets we refer to [1] The basic properties of Jordan curves and of simply connected sets in the plane, used in this paper, can be found in any textbook on algebraic topology, for example [11, 12, 13] 2 Equivalent properties Throughout this section, we assume that the initial set R 0 is contained in the half plane IR + 2 = { x 1, x 2 ; x 2 > 0, and that all trajectories of the fire are restricted to IR + 2 as well The following lemma collects various equivalent formulations of the blocking property In the following, we call B ρ + = B0, ρ IR + 2 the upper half of the open disc centered at the origin, with radius ρ The reachable set at time t, starting from R 0 and avoiding the wall Γ will be denoted as R Γ t, R 0 Lemma 1 Given a construction speed σ > 0, the following statements are equivalent: i A blocking strategy for BP1 exists when R 0 = B + ρ, for some ρ > 0 ii For every initial bounded open set R 0, a blocking strategy for BP1 exists iii For every initial bounded open set R 0 there exists a set Γ IR 2 with m 1 Γ <, which solves the blocking problem BP2 and consists of finitely many compact connected components Proof The implication ii= i is trivial, while iii = ii is an immediate consequence of Theorem 1 in [7] To prove that i = ii, let R 0 IR 2 + be any bounded set Choose a radius r > 0 such that R 0 B + r Let t γt be a complete, admissible strategy which blocks a fire initially starting from the half disc B + ρ As observed in [4], the rescaling γt = r ρ γ ρ r t defines another complete admissible strategy Indeed, for every t 0 one has m 1 γt = r ρ ρ m 1 γ r t σ t Moreover, the corresponding sets reached by the fire starting from R 0 B r + R γ t, R 0 R γ t, B r + = r ρ ρ Rγ r t, B+ ρ 4 satisfy 21

for every t 0 Since by assumption the sets on the right hand side of 21 remain uniformly bounded, we conclude that γ is a blocking strategy, when the fire starts from R 0 Finally, we prove the implication ii = iii Let ii hold In particular, there must exist some admissible strategy t γt which solves the blocking problem BP2 in connection with the initial set R 0 = B + 3 Say, R γ t, B + 3 B+ r, for some radius r > 3 and all t 0 It is clearly not restrictive to assume that all walls γt are contained in the closed half disc B + r Choose a time T large enough such that T > 2 σ m 2B + r = πr2 σ 22 Consider the complete rectifiable set Γ = γt + 1 By the analysis in [7], we can assume that the totally disconnected component of Γ has 1-dimensional measure zero The set Γ can thus be decomposed as a countable union of connected components with positive length, plus a totally disconnected set Γ 0 of zero length, say Γ = Γ 0 Γ 1 Γ N Γ N+1 Here we choose the integer N large enough so that, setting Γ = j>n γ j, there holds m 1 Γ < 1 2 23 Let S be the union of all bounded connected components of the complement IR + 2 \ Γ We observe here that the boundary of S satisfies S Γ Hence, every component Γ i with 1 i N is either entirely contained in S, or else it does not intersect S at all We now define Γ = Γ i \ S 1 i N as the union of those among the first N components which are outside S For 1 < t < T + 1, the analysis in [7] yields R Γ t, B + 1 RΓ t 1, B + 2 RΓ t 1+m 1 Γ, B + 2 S R Γ t, B + 2 S B+ r 24 By 22 and 24 there exists a time τ [1, T ] such that m 2 R Γ τ + 1, B 1 + m 2 R Γ τ, B 1 + < σ 25 Since Γ is the union of finitely many compact, connected, rectifiable sets, the minimum time function V x = inf {t 0 ; x R Γ t, B 1 + is locally Lipschitz continuous outside Γ Indeed see for example [2], it provides a viscosity solution to the eikonal equation V x 1 = 0 An application of the coarea formula yields m 2 R Γ τ + 1, B + 1 \ RΓ τ, B + 1 = 5 τ+1 τ m 1 {x / Γ ; V x = τ dt

Therefore, by 25 there exists a time τ [τ, τ + 1] such that m 1 {x / Γ ; V x = τ < σ 26 We claim that the set Γ = Γ {x / Γ ; V x = τ is a closed, admissible set which solves the blocking problem BP2 when R 0 = B + 1 To check that Γ is closed, let x n x be a convergent sequence with x n Γ for every n Then, either x Γ Γ, or else x / Γ In this second case, V x n = τ for all n sufficiently large, hence V x = τ by continuity In both cases x Γ By construction, we have R Γ t, B + 1 = RΓ τ, B + 1 for all t τ 27 To prove that Γ is admissible, for t < τ we estimate m 1 Γ R Γ t, B 1 + m 1 Γ R Γ t, B 1 + m 1 Γ R Γ t, B + 2 m 1 Γ R Γ t 1, B + 3 = m 1 Γ R Γ t, B + 1 σt 1 28 Here we used 24 and the fact that Γ does not intersect S Moreover, for t τ a similar argument yields m 1 Γ R Γ t, B + 1 = m 1 Γ R Γ τ, B 1 + m 1 Γ R Γ τ 1, B 3 + + m 1 {x / Γ ; V x = τ < στ 1 + σ σt Therefore, Γ is admissible and provides a solution to the blocking problem In general, however, the level set in 26 may have infinitely many connected components In order to satisfy the last property in iii, we consider the compact connected set R Γ τ, B 1 + and call Ω the unbounded component of its complement Then the boundary of Ω is connected and satisfies Ω Γ Replacing Γ with its closed subset Γ = Γ Ω, it is clear that Γ satisfies all the properties in iii 29 We have thus established the implication ii = iii in the case where the initial set is R 0 = B 1 + The general case follows from a simple rescaling argument 3 The distance function on a simply connected domain Let K IR 2 be a compact, path-connected set For x, y K we define the distance d K x, y as the minimum length among all absolutely continuous paths γ : [0, 1] K with γ0 = x, γ1 = y As shown in fig 1, if the set K is not simply connected the function d K may attain its global maximum at a couple of interior points 6

y _ K _ x Figure 1: For the multiply connected domain K, max x,y K d K x, y is attained at the couple of interior points x, ȳ In this section we will prove that, if K is simply connected, then the maximum of the distance d K is always attained at boundary points We recall that a Jordan curve is a homeomorphic image of the circumference {x 1, x 2 ; x 2 1 + x2 2 = 1, see [11, 13] Lemma 2 Let K IR 2 be the compact region enclosed by a smooth Jordan curve Γ = K Then max d Kx, y = max d Kx, y 31 x,y K x,y Γ As a consequence, max d Kx, y 1 x,y K 2 m 1 K 32 Of course, the assumption implies that K is simply connected, with smooth boundary The lemma will be proved in various steps We begin by proving an intermediate result Lemma 3 Let K be a compact, simply connected polygon Then for any two points x, y K, the shortest path γ : [0, 1] K joining x with y is unique, up to reparametrizations Proof 1 Let p 1,, p n be the vertices of the polygon, with n 3 Then K can be covered with triangles 1,, n 2, whose vertices lie in the set {p 1,, p n This is a well known result, used in numerical analysis [3] For reader s convenience, we sketch her a proof The result is trivially true when n = 3 By induction, assume that the result holds for all polygons whose number n of vertices satisfies 3 n < m Let K be a polygon with m vertices p 1,, p m For notational convenience, set p 0 = pm Choose three consecutive vertices p j 1, p j, p j+1 such that the angle at p j formed by the consecutive edges p j 1, p j and p j, p j+1 has amplitude < π Such an index j certainly exists Two cases can occur see figure 2 CASE 1: The triangle j with vertices p j 1, p j, p j+1 does not contain any other vertex Then we can add the edge p j 1, p j+1 The original polygon K is thus decomposed as the union of a triangle and a sub-polygon K having m 1 vertices We can thus apply the inductive assumption to K 7

CASE 2: The triangle j with vertices p j 1, p j, p j+1 contains some other vertices, say p α for α I {1,, m Call n the outer normal to the triangle at the point p j, perpendicular to the opposite edge p j 1, p j+1 Choose an index β I such that n, pβ = max n, pα 33 α I Here, denotes an inner product We claim that the segment p j p β does not intersect any other edge of the polygon Indeed, assume on the contrary that this segment intersected the edge p l 1, p l Since p l 1, p l does not intersect the two edges p j 1, p j and p j, p j+1, at least one of the vertices p l 1, p l must be contained in the interior of the smaller triangle j = j { p IR 2 ; n, p n, p β But this contradicts the maximality condition 33 Adding the edge p j, p β, the original polygon K is decomposed in two sub-polygons, each with a number of edges < m By the inductive hypothesis, each of these can be triangulated p j 1 p j p j 1 j p β p j+1 p j+1 j n p j Figure 2: The two cases of the inductive step 2 Let now 1,, n 2 be a triangulation of the n-polygon K We say that two triangles are connected if they have an edge in common With this relation, we claim that the set of triangles is a simply connected graph To establish the simple connectedness, we argue by contradiction Assume that the graph contains a nontrivial cycle, say i0, i1,, iν, with iν = i0, and such that each two consecutive triangles ik 1, ik have an edge in common, say Γ ik For every k = 1,, ν, fix a point q k Γ ik, different from the endpoints see fig 3 Consider the closed polygonal γ having vertices q 1,, q ν This is a simple, closed curve, contained in the interior of K The set IR 2 \ γ has two connected components, one bounded and the other unbounded Since K is simply connected, all points in the bounded component of IR 2 \ γ belong to the interior of K Hence this bounded component does not contain any of the vertices p 1,, p n This yields a contradiction, because every edge Γ ik has two endpoints, say p ik and p+ ik, located in distinct components of IR2 \ γ 3 Let any two points x, y K be given Choose indices α, β {1,, n 2 such that x α, y β We observe that these indices may not be unique However, any choice will suffice for our purposes 8

ik 1 ik q k? γ Figure 3: Proving the simple connectedness of the set of triangles By the previous step, the set of triangles { 1,, n 2 forms a simply connected graph Therefore, there exists a unique chain of triangles α = i0, i1,, iν = β connecting α with β Given a path γ : [0, 1] K connecting x with y, for l = 0, 1,, ν we define the times t l = min {t [0, 1] ; γt il, t + l = max {t [0, 1] ; γt il Since γ0 = x i0 and γ1 iν, we clearly have t 0 = 0, t+ ν = 1 Assume that, among all paths joining x with y and remaining inside K, the path γ has minimum length Since the points γt l and γt+ l are both contained in the convex triangle il, the restriction of γ to each sub-interval [t l, t+ l ] must be a segment Up to a reparameterization, the curve γ must therefore be a polygonal line, say with vertices x = z 0, z 1,, z ν = y Moreover, for l = 1,, ν 1, the vertex z l lies on the edge Γ l = il 1 il 4 Now let γ, γ : [0, 1] K be two distinct paths joining x with y, both with minimum length Call z 0,, z ν and z 0,, z ν the corresponding vertices, as in the previous step A contradiction will be obtained by showing that the polygonal γ with vertices z l = z l + z l 2, l = 0,, ν has strictly smaller length Clearly z 0 = x and z ν = y Moreover, each segment [z l 1, z l ] is entirely contained in the triangle jl We observe that, for each l = 0,, ν, the convexity of the euclidean norm yields z l z l 1 = z l + z l 2 z l 1 + z 2 l 1 z l z l 1 2 + z l z l 1 2 Moreover, equality holds if and only if the two vectors z l z l 1 and z l z l 1 are parallel If we assume that the polygonals γ and γ are distinct, the above vectors cannot be parallel for every l = 0,, ν Hence the length of γ is strictly less than the length of γ and of γ This contradiction completes the proof of the lemma 9

Lemma 4 Let K IR 2 be the compact region enclosed by a smooth Jordan curve Γ Fix any interior point q int K Then the function x V q K x = d K x, q is C 1 on int K \ {q Proof We first establish the result assuming that K is a polygon, then in the general setting considered in the lemma We observe that the distance function V q K provides a viscosity solution of the eikonal equation V 1 = 0, 34 on int K \ {q In the interior of K, the function V q K is locally Lipschitz continuous with constant 1 By Rademacher s theorem, the gradient V q K x is well defined for ae x 1 If K is a polygon and q int K, we argue by contradiction If the conclusion of the lemma fails, there exists a point x int K \ {q where the gradient V q K is not continuous By a translation of coordinates we can assume that x = 0 In this setting, there exist sequences of points x n 0, x n 0 and unit vectors v n = V q K x n v, v n = V q K x n v, 35 with v v For each n, let γ n be the unique shortest path connecting x n to q, parameterized by arc-length Similarly, let γ n be the shortest path connecting x n to q Since 0 lies in the interior of K, we can choose r > 0 such that the open disc centered at the origin with radius r satisfies B0, r K The necessary conditions for the optimality of the paths γ n, γ n imply d ds γ ns = v n, d ds γ ns = v n 36 as long as γ n s, γ ns B0, r By a compactness argument, after extracting a subsequence we can assume γ n γ, γ n γ, for some paths γ, γ By 36 and 35 it follows γs = s v γ s = s v 0 < s < r Hence the two paths are distinct Moreover, both γ and γ connect 0 with q, and their lengths satisfy γ lim inf γ n = lim inf V q n n K x n = V q K 0, γ lim inf n γ n = lim inf n V q K x n = V q K 0 Therefore both paths are optimal Since this would be in contradiction with Lemma 3, we conclude that the distance function V q K must be continuously differentiable in the interior of the polygon K 2 Next, assume that K is a compact set whose boundary is a smooth Jordan curve, say K = Γ We can then construct a sequence of simply connected polygons K n which invade K More precisely, given any compact subset K contained in the interior of K, there exists an integer N such that K K n for all n N Let any point q int K be given Then, for all n large enough, the functions V q K n : K n IR + are well defined We claim that, for each x int K, the sequence V q K n x decreases monotonically to V q K x Indeed, the fact that the sequence is non-negative and monotone decreasing follows immediately from the definition To compute its limit, let γ : [0, 1] K 10

be a path of minimum length connecting x with q For each ε > 0 we can approximate γ with a second path γ, with the same initial and terminal point, taking values strictly in the interior of K Choose an integer N such that γ is entirely contained in the polygon K N For all n N we thus have This proves the pointwise convergence V q K n x γ γ + ε = V q K x + ε lim V q n K n x = V q K x x int K 3 Still assuming that K = Γ is a smooth Jordan curve, we now argue by contradiction If V q K is not C1 in the interior of K, the same construction as in step 1 shows that, after a translation of coordinates, there exists a radius r > 0 and two unit vectors v v such that V q K rv = V q K rv = V q K 0 r 37 On the other hand, all functions V q K n are C 1 on a neighborhood of the closed disc B0, r By possibly taking a subsequence we can assume the convergence v n = V q K n 0 v, for some unit vector v From the identities taking the limit as n we deduce V q K n sv n = V q K n 0 + s s [ r, r], V q K rv = V q K 0 + r 38 However, from 37 it follows V q K {V rv min q K rv + rv rv, V q K rv + rv rv = V q K { v 0 r + r min v, v v < V q K 0 + r This yields a contradiction with 38, completing the proof of Lemma 4 Proof of Lemma 2 Using the regularity result stated in Lemma 4, we now provide a proof of Lemma 2 Let K be a compact, simply connected domain, whose boundary K is a smooth Jordan curve Choose points x, ȳ K such that d K x, ȳ = max x,y K d Kx, y If x, ȳ K, we are done Otherwise, assume x int K Observe that this implies ȳ K Indeed, by Lemma 4 the function x V x K x = d K x, y provides a C 1 solution to the eikonal equation 34 on the open set int K \ { x Therefore, it cannot have any local maximum in the interior of K Granted that ȳ K, we now consider a sequence of points y n ȳ with y n int K for every n 1 For each n, by Lemma 4 the function x V yn K x = d K x, y n 11

provides a C 1 solution to the eikonal equation 34 on the open set int K \ {y n Hence, it cannot have local maximum on this set Choose x n K such that By the previous argument, x n K d K x n, y n = max x K d Kx, y n By taking a subsequence, we can assume x n x for some x K Clearly x K, because x n K for every n By the uniform continuity of the distance function d K on K K, we now have max d Kx, y d K x, ȳ = lim d Kx n, y n x,y K n lim d K x, y n = d K x, ȳ = max d Kx, y, n x,y K proving 31 The statement 32 is now clear 4 Proof of the main theorem Relying on the lemmas proved in the previous sections, we give here a proof of Theorem 1, in several steps 1 Assume σ > 1 By Lemma 1, it suffices to prove that a blocking strategy exists in the case where R 0 = B 1 + is the upper half of the unit disc centered at the origin Set λ = σ 2 1 1/2, so that σ = 1 + λ 2 /λ, and consider the strategy { γt = e λθ cos θ, e λθ sin θ ; 0 θ λ 1 ln1 + t This strategy is admissible because m 1 γt = λ 1 ln1+t 0 e λθ 1 + λ 2 dθ = σt Next, consider the time T such that λ 1 ln1 + T = π We claim that, for t [0, T ], the reachable set is R γ t {r = cos θ, r sin θ ; 0 < θ < π, 0 < r < min{1 + t, e λθ 41 Indeed, call St the set on right hand side of 41 Since for each t 0 the wall γt is contained inside the arc of spiral Σ = { e λθ cos θ, e λθ sin θ ; 0 θ π, the inclusion St R γ t is clear To prove the converse inclusion, consider any point p = r cos θ, r sin θ with r e λθ If p R γ t, then there exists an absolutely continuous path s xt with x0 < 1, xt = p, ẋs 1 for ae s [0, t], 12

xs / γs for all s [0, t] Since Σ splits the upper half plane in two connected components, there must be some τ ]0, t] such that xτ Σ We now observe that xτ < 1 + τ, and hence xτ γτ = {x Σ ; x 1 + τ This contradiction shows that R γ t St, completing the proof of 41 In particular, when t = T we have R γ T {r = cos θ, r sin θ ; 0 < θ < π, 0 < r < e λθ, γt = Σ We observe that the boundary of R γ T is entirely contained in the union of the arc γt = Σ and the x 1 -axis Hence the reachable set cannot become any larger: R γ = R γ t = Rγ T for every t T This shows that the strategy γ blocks the fire within a bounded set 2 Next, assume σ 1 We argue by contradiction If some blocking strategy exists, then by Lemma 1 we can again assume R 0 = B 1 + and we can find a blocking strategy consisting of finitely many compact connected components Γ = Γ 1 Γ N, with m 1 Γ < By assumption, the reachable set R Γ = t 0 RΓ t is bounded It is not restrictive to assume that Γ R Γ, 42 otherwise we can simply replace Γ by the intersection Γ R Γ We can also assume that Γ B 1 + = Notice that the reachable set RΓ is precisely the union of all connected components of the open set IR + 2 \ Γ which intersect R 0 = B 1 + For each x RΓ, let V x = inf { t 0 ; x R Γ t be the minimum time needed for the fire to reach x We observe that admissibility condition 11, 15 implies t = sup V x m 1 Γ 43 x R Γ Otherwise, by 42 we would have m 1 Γ R Γ t, B + 1 against the assumption that Γ is admissible = m 1 Γ R Γ = m 1 Γ > t, To prove Theorem 1, a contradiction will be achieved by showing that every point x R Γ can be reached from B 1 + in time V x m 1 Γ 1 4 44 More precisely, consider the point y = 0, 1/2 B + 1 If we show that every point x RΓ can be connected to y by a path γ of length m 1 Γ + 1/4, without crossing the wall Γ, we are done Indeed, if x / B + 1, the path γ must contain a portion of length 1/2 inside B+ 1 Hence 44 must hold 13

3 Consider the interval [a, b] where a = { inf x 1 ; x 1, 0 R Γ, b = { sup x 1 ; x 1, 0 R Γ For later purposes, we define the segment S 0 = {x 1, 0 ; x 1 [a, b] Next, let Ω be the unique unbounded connected component of the open set IR 2 \R Γ Since R Γ is a compact connected set, the boundary Ω is a connected set, contained in Γ S 0 Let Γ 0 denote the union of Ω and of all connected components of Γ which intersect Ω Moreover, let Γ 1,, Γ m be a list of all the connected components of Γ which do not intersect Γ 0 We observe that m m 1 Γ 0 + m 1 Γ j m 1 Γ + b a 45 j=1 Let any point x R Γ be given In the remainder of the proof we will show that there exists a path γ joining the point y = 0, 1/2 to the point x, without crossing Γ 0 Γ 1 Γ m, with length γ m 1 Γ + 1/4 4 Since x / Γ S 0 and Γ 0,, Γ m are disjoint compact sets, we can choose 0 < ε < 1/4 such that dx, Γ S 0 > ε, dγ i, Γ j > 2ε 46 for all i, j = 0,, m with i j For each i = 2,, m, we construct a smooth Jordan curve γ i surrounding Γ i such that γ i BΓ i, ε, m 1 γ i < 2m 1 Γ i + 1 5m 47 Since each Γ i is a compact, rectifiable set, this can be done as in [7] Namely, we first choose a radius 0 < r 2 < ε/2 such that the neighborhood of radius r 2 around the set Γ i has measure m 2 BΓ i, r 2 < 2r 2 m 1 Γ i + 1 20m Choosing r 1 sufficiently small, with 0 < r 1 << r 2, we achieve m 2 BΓ i, r 2 \ BΓ i, r 1 < 2r 2 r 1 m 1 Γ i + 1 48 10m Define the distance function with cutoff V i x = dx, Γ i if r 1 < dx, Γ i < r 2, r 1 if dx, Γ i r 1, r 2 if dx, Γ i r 2 Let V i,ε = ϕ ε V i be a mollification of V i, where the smooth kernel ϕ ε is supported inside the disc B0, ε, for some ε with 0 < ε < r 1 The functions V i, V i,ε are both Lipschitz continuous with constant one Using the co-area formula and then 48, we obtain r2 m 1 {x ; V i,ε x = s ds = V i,ε dx V i dx r 1 IR 2 IR 2 = m 2 BΓ i, r 2 \ BΓ i, r 1 14 < 2r 2 r 1 m 1 K + 1 10m 49 410

Since V i,ε is smooth, by Sard s theorem almost every level set Σ r = {x ; Vi,ε x = r is the union of finitely many smooth curves By 410, there exists some ρ, with r 1 < ρ < r 2 such that the level set Σ ρ is the finite union of smooth curves an moreover The construction of V i,ε Consider the sub-level set clearly implies m 1 Σ ρ < 2m 1 Γ i + 1 5m 411 Σ ρ BΓ i, r 2 + ε \ BΓ i, r 1 ε 412 Σ ρ = {x ; V i,ε x < ρ Observe that this open set need not be connected However, since Σ ρ contains the connected neighborhood BΓ i, r 1 ε, we can uniquely define the set Σ ρ as the connected component of Σ ρ which contains BΓ i, r 1 ε Clearly, its boundary satisfies Σ ρ Σ ρ 413 The curve γ i = Σ ρ is then a Jordan curve, satisfying all our requirements 5 We now perform a similar construction for the set Γ 0, but in a more careful way Let a continuous path be given, say γ xy : [0, 1] IR 2 \ Γ 0, joining x with y without crossing Γ 0 By possibly shrinking the value of ε in 46, we can assume that min dγs, Γ 0 > 2ε 414 s [0,1] The same procedure used in the previous step now yields radii 0 < r 1 < r 2 < ε and a finite family of smooth closed curves γ 0,0,, γ 0,k BΓ 0, ε such that the following holds: k m 1 γ 0,l 2m 1 Γ 0 + 1 4 l=1 415 and moreover, if p 1, p 2 IR 2 and dp 1, Γ 0 < r 1 while dp 2, Γ 0 > r 2, then p 1 and p 2 lie inside distinct connected components of the complement IR 2 \ k l=1 γ 0,l We claim that, among the closed curves γ 0,l, there exists at least one curve, say γ 0,µ such that Γ 0 is contained in the bounded connected component of IR 2 \ γ 0,µ Indeed, let Ω 0 be the connected component of the complement IR 2 \ k l=1 γ 0,l which contains the compact connected set Γ 0 The boundary of the unbounded connected component of IR 2 \ Ω 0 is the desired curve Next, we claim that there exists a second curve, say γ 0,ν, such that the path γ xy is contained in the bounded component of IR 2 \ γ 0,ν while Γ 1 is contained in the unbounded connected component of IR 2 \ γ 0,ν Indeed, let Ω 1 be the connected component of the complement IR 2 \ k l=1 γ 0,l which contains the path γ xy Notice that the open set Ω 1 must be disjoint from the previous component Ω 0 The boundary of the unbounded connected component of IR 2 \ Ω 1 now provides the desired curve By a simple relabeling, we can assume that µ = 1 and ν = 2 We thus have m 1 γ 0,1 + m 1 γ 0,2 k m 1 γ 0,l 2m 1 Γ 0 + 1 4 l=1 15 416

while m 1 γ 0,1 > 2 diam Γ 0 2b a 417 6 Since γ 0,2 is a Jordan curve, the bounded connected component of IR 2 \ γ 0,2 is simply connected Moreover it contains the path γ xy By Lemma 2, we conclude that there exists a path γ : [0, 1] IR 2 \ Γ 0, connecting x with y, whose length satisfies γ 1 2 m 1γ 0,2 418 Combining 418 with 416 and 417, and recalling 45, we conclude γ 1 2 2m 1 Γ 0 m 1 γ 0,1 + 1 4 m 1 Γ 0 b a + 1 8 m 1Γ m m 1 Γ j + 1 8 j=1 419 Γ 1 γ 1 a x y b Figure 4: The shortest path γ joining y with x is not longer than the maximum path-distance between any two points in γ 0,2 7 The path γ constructed in the previous step does not intersect Γ 0, but it may cross the other components Γ 1,, Γ m To avoid these crossings, we need to go around the walls Γ j, using the paths γ 1,, γ m constructed at 47 For each index i {1,, m such that the closed curve γ i intersects the path γ, we define = inf {t [0, 1] ; γ t γ i, t + i = sup {t [0, 1] ; γ t γ i, t i An increasing sequence of times t i1 < t i2 < < t iν inductive procedure: Begin by setting t i1 = min 1 j m t j is now defined by the following Let t il be given If t j < t + il for every j = 1,, m, then we set ν = l and the induction terminates Otherwise we define t il+1 = min {t j ; t j > t+ il and continue It is clear that the indices i1, i2,, iν must be all distinct, and that the intervals [t il, t+ il ], l = 1,, ν are mutually disjoint In particular, ν m Since each γ i is a smooth, closed curve, for each l there is an arc of γ il joining γ t il with γ t + il having length 1 2 m 1γ il 16

A new path γ joining x with y is now obtained as follows Given the path γ, we replace each arc {γ t ; t [t il, t+il ] with an arc of the Jordan curve γ il having the same initial and terminal points It is clear that the path γ does not cross any of the sets Γ 0, Γ 1,, Γ m By the previous estimates 419 and 47, its length satisfies m 1 γ m 1 γ + 1 2 ν m 1 γ il m 1 Γ l=1 m m 1 Γ j + 1 m 8 + 1 2m 1 Γ j + 1 2 5m j=1 j=1 < m 1 Γ + 1 4 This yields the desired contradiction, proving the theorem Γ 1 a x γ 2 γ 1 y γ 3 b Figure 5: If the path γ as in fig 4 crosses some of the walls Γ 1,, Γ m, one has to go around these walls following portions of the smooth paths γ 1,, γ m Acknowledgements This work was supported by NSF through grant DMS-0807420, New problems in nonlinear control The authors also wish to thank Camillo De Lellis for useful suggestions References [1] L Ambrosio, N Fusco, and D Pallara, Functions of Bounded Variation and Free Discontinuity Problems Oxford University Press, 2000 [2] M Bardi and I Capuzzo-Dolcetta, Optimal Control and Viscosity Solutions of Hamilton- Jacobi-Bellman Equations Birkhäuser, Boston, 1997 [3] M de Berg, M van Kreveld, M Overmars, and O Schwarzkopf, Computational Geometry Algorithms and Applications Third edition Springer-Verlag, Berlin, 2008 [4] A Bressan, Differential inclusions and the control of forest fires, J Differential Equations special volume in honor of A Cellina and J Yorke, 243 2007, 179-207 [5] A Bressan, M Burago, A Friend, and J Jou, Blocking strategies for a fire control problem, Analysis and Applications 6 2008, 229-246 17

[6] A Bressan and C De Lellis, Existence of optimal strategies for a fire confinement problem, Comm Pure Appl Math, to appear [7] A Bressan and T Wang, Equivalent formulation and numerical analysis of a fire confinement problem, preprint 2008 [8] P Cannarsa and C Sinestrari, Semiconcave Functions, Hamilton-Jacobi Equations, and Optimal Control Birkhäuser, Boston, 2004 [9] C De Lellis, Rectifiable sets, densities and tangent measures Zürich lectures in advanced mathematics EMS Publishing House, 2008 [10] H Federer, Geometric measure Theory, Springer-Verlag, New York, 1969 [11] M Henle, A Combinatorial Introduction to Topology W H Freeman, San Francisco, 1979 [12] K Kuratovski Topology, VolII Academic Press, New York, 1968 [13] W S Massey, A Basic Course in Algebraic Topology, Springer-Verlag, New York 1991 18