Variational Problems for Tree Roots and Branches
aa r X i v : . [ m a t h . O C ] J a n Variational Problems for Tree Roots and Branches
Alberto Bressan, Michele Palladino, and Qing SunDepartment of Mathematics, Penn State University,University Park, Pa. 16802, USA.e-mails: [email protected], [email protected], [email protected] 17, 2020
Abstract
This paper studies two classes of variational problems introduced in [7], related to theoptimal shapes of tree roots and branches. Given a measure µ describing the distribution ofleaves, a sunlight functional S ( µ ) computes the total amount of light captured by the leaves.For a measure µ describing the distribution of root hair cells, a harvest functional H ( µ )computes the total amount of water and nutrients gathered by the roots. In both cases,we seek a measure µ that maximizes these functionals subject to a ramified transportationcost, for transporting nutrients from the roots to the trunk or from the trunk to the leaves.Compared with [7], here we do not impose any a priori bound on the total massof the optimal measure µ , and more careful a priori estimates are thus required. Inthe unconstrained optimization problem for branches, we prove that an optimal measureexists, with bounded support and bounded total mass. In the unconstrained problem fortree roots, we prove that an optimal measure exists, with bounded support but possiblyunbounded total mass. The last section of the paper analyzes how the size of the optimaltree depends on the parameters defining the various functionals. MSC:
Keywords: shape optimization, branched transport, sunlight functional, optimal harvesting.
In the recent paper [7], two of the authors introduced a family of variational problems, aimedat characterizing optimal shapes of tree roots and branches. All these optimization problemstake place in a space of positive measures on a d -dimensional space R d . In the case of roots,calling µ the distribution of root hair cells, one seeks to maximize the total amount of waterand nutrients harvested by the roots, minus a cost for transporting these nutrients to thebase of the trunk. In the case of branches, calling µ the distribution of leaves, one seeks tomaximize the total sunlight captured by the leaves, minus a cost for transporting water andnutrients from the base of the trunk to the tip of every branch.1he main results in [7] established the semicontinuity of the relevant functionals and theexistence of optimal solutions, under a constraint on the total mass of the measure µ . Inessence, by fixing the total mass µ ( R d ) one prescribes the size of the tree. In turn, themaximization problem determines an optimal shape .In the present paper we study the corresponding unconstrained optimization problems, withoutany a priori bound on the total mass of the measure µ . Roughly speaking, this aims atdetermining the optimal size of a tree, in addition to its optimal shape.Compared with [7], proving the existence of optimal solutions for the unconstrained prob-lems requires a much more careful analysis. Following the direct method of the Calculus ofVariations, we consider a maximizing sequence of measures ( µ k ) k ≥ . Two main issues arise.(i) First, one needs to establish an a priori bound on the support of the measures µ k . Atfirst sight this looks easy, because if a measure contains some mass far away from theorigin, its transportation cost will be very large. However, since we are here consideringa ramified transportation cost [1, 12, 19, 20], there is an economy of scale : as the totaltransported mass increases without bound, the unit cost decreases to zero. For thisreason, in order to achieve a uniform bound on Supp( µ k ), we first establish an a prioribound on the transportation cost. At a second stage, this yields a bound on the totalpayoff. Finally, we obtain an estimate of the support of the optimal measure.(ii) Next, we seek an a priori bound on the total mass µ k ( R d ). This does not follow from abound on the transportation cost, because as k → ∞ the measures µ k may concentratemore and more mass in a small neighborhood of the origin. Concerning the optimizationproblem for branches, our analysis yields the existence of an optimal measure µ suchthat µ ( R d ) < + ∞ . On the other hand, in the optimization problem for tree roots, weprove that an optimal measure µ exists, with bounded support but possibly unboundedtotal mass. Indeed, for any ρ > µ (cid:0) { x ∈ R d ; | x | > ρ } (cid:1) < + ∞ .However, we cannot rule out the possibility that µ ( R d \ { } ) = + ∞ .The remainder of the paper is organized as follows. Section 2 reviews the three main in-gredients of our variational problems: the sunlight functional , the harvest functional , and the ramified transportation cost . In Section 3 we prove the existence of a bounded measure µ whichsolves the unconstrained optimization problem for tree branches. The proof relies on the con-struction of a maximizing sequence of measures ( µ k ) k ≥ with uniformly bounded support anduniformly bounded total mass. In this direction, a key step is to prove a uniform bound onthe ramified transportation cost for all measures µ k . Section 4 deals with the unconstrainedoptimization problem for tree roots. The existence of an optimal measure µ is proved, withbounded support but possibly infinite total mass. Finally, in Section 5 we discuss how theoptimal size of tree roots and branches is affected by the various parameters appearing inthe equations. Here the key step is to analyze how the various functionals behave under arescaling of coordinates.The theory of ramified transport for general measures was developed independently in [12]and [19]. See also [1] for a comprehensive introduction, and [20] for a survey of the field.Further results on optimal ramified transport can be found in [2, 5, 13, 14, 17]. An interestingcomputational approach, based on Gamma-convergence, has been developed in [15, 18]. Ageometric optimization problem involving a ramified transportation cost was recently studied2n [16]. The “sunlight functional” was introduced in [7], in a slightly more general settingwhich also takes into account the presence of external vegetation. The “harvest functional”,in a space of Radon measures, was first studied in [6] in connection with a problem of optimalharvesting of marine resources. Given a positive, bounded Radon measure µ on R d , three functionals were considered in [7].The corresponding optimization problems determine the optimal configurations of roots andbranches of a tree. n n ⊥ Figure 1:
Sunlight arrives from the direction parallel to n . Part of it is absorbed by the measure µ ,supported on the grey regions. Let µ be a positive, bounded Radon measure on R d . Thinking of µ as the density of leaveson a tree, we seek a functional S ( µ ) describing the total amount of sunlight absorbed by theleaves. As shown in Fig. 1, fix a unit vector n ∈ S d − = { x ∈ R d ; | x | = 1 } and assume that all light rays come parallel to n . Call E ⊥ n the ( d − n and let π n : R d E ⊥ n be the perpendicular projection. Each point x ∈ R d can thus be expressed uniquely as x = y + s n (2.2)with y ∈ E ⊥ n and s ∈ R .On the perpendicular subspace E ⊥ n consider the projected measure µ n , defined by setting µ n ( A ) = µ (cid:16)(cid:8) x ∈ R d ; π n ( x ) ∈ A (cid:9)(cid:17) . (2.3)Call Φ n the density of the absolutely continuous part of µ n w.r.t. the ( d − E ⊥ n . 3 efinition 2.1 The total amount of sunshine from the direction n captured by a measure µ on R d is defined as S n ( µ ) . = Z E ⊥ n (cid:16) − exp (cid:8) − Φ n ( y ) (cid:9)(cid:17) dy . (2.4) Given an integrable function η ∈ L ( S d − ) , the total sunshine absorbed by µ from all directionsis defined as S η ( µ ) . = Z S d − Z E ⊥ n (cid:16) − exp (cid:8) − Φ n ( y ) (cid:9)(cid:17) dy ! η ( n ) d n . (2.5)We think of η ( n ) as the intensity of light coming from the direction n . We recall two estimatesproved in [7]. Lemma 2.2
For any positive Radon measure µ on R d , one has S η ( µ ) ≤ k η k L · µ ( R d ) . (2.6) If µ is supported inside a closed ball with radius r , calling ω d − the surface of the unit spherein R d , one has S η ( µ ) ≤ k η k L · ω d − r d − . (2.7) We now consider a utility functional associated with roots. Here the main goal is to collectmoisture and nutrients from the ground. To model the efficiency of a root, in the followingwe let u ( x ) be the density of water+nutrients at the point x , and consider a positive Radonmeasure µ describing the distribution of root hair.Consider the half space Ω . = { x = ( x , . . . , x d ) ; x d < } . Let µ be a positive, bounded Radonmeasure supported on the closure Ω, such that µ ( V ) = 0 for every set V having zero capacity.Consider the elliptic problem with measure source∆ u + f ( u ) − u µ = 0 (2.8)and Neumann boundary conditions ∂ n ( x ) u = 0 on ∂ Ω . (2.9)By n ( x ) we denote the unit outer normal vector at the boundary point x ∈ ∂ Ω, while ∂ n u isthe derivative of u in the normal direction. Of course, in this case (2.9) simply means x d = 0 = ⇒ ∂∂x d u = 0 . If µ is a general measure and u is a discontinuous function, the integral (2.14) may not be welldefined. To resolve this issue, calling − Z V u dx = 1meas( V ) Z V u dx u on a set V , for each x ∈ Ω we consider the limit u ( x ) = lim r ↓ − Z Ω ∩ B ( x,r ) u ( y ) dy. (2.10)As proved in [10], if u ∈ H (Ω) then the above limit exists at all points x ∈ Ω with the possibleexception of a set whose capacity is zero. If the measure µ satisfies (A3), the integral (2.14)is thus well defined. Our present setting is actually even better, because in (2.8) u and µ arepositive while f is bounded. Therefore, if the constant C is chosen large enough, the function u + C | x | is subharmonic. As a consequence, the limit (2.10) is well defined at every point x ∈ Ω.Elliptic problems with measure data have been studied in several papers [3, 4, 9] and are nowfairly well understood. A key fact is that, roughly speaking, the Laplace operator “does notsee” sets with zero capacity. Following [3, 4] we thus call M b the set of all bounded Radonmeasures on Ω. Moreover, we denote by M ⊂ M b the family of measures which vanish onBorel sets with zero capacity, so thatcap ( V ) = 0 = ⇒ µ ( V ) = 0 . (2.11)For the definition and basic properties of capacity we refer to [11]. Every measure µ ∈ M b can be uniquely decomposed as a sum µ = µ + µ s , (2.12)where µ ∈ M while the measure µ s is supported on a set with zero capacity. In the definitionof solutions, the presence of the singular measure µ s is disregarded. Definition 2.3
Let µ be a measure in M b , decomposed as in (2.12). A function u ∈ L ∞ (Ω) ∩ H (Ω) , with pointwise values given by (2.10), is a solution to the elliptic problem (2.8)-(2.9)if − Z Ω ∇ u · ∇ ϕ dx + Z Ω f ( u ) ϕ dx − Z Ω uϕ dµ = 0 (2.13) for every test function ϕ ∈ C ∞ c ( R d ) . In connection with a solution u of (2.8), the total harvest is defined as H ( u, µ ) . = Z Ω u dµ . (2.14)Throughout the following we assume (A1) f : [0 , M ] R is a C function such that, for some constants M, K , f ( M ) = 0 , ≤ f ( u ) ≤ K, f ′′ ( u ) < u ∈ [0 , M ] . (2.15)5 .3 Optimal irrigation plans Given α ∈ [0 ,
1] and a positive measure µ on R d , the minimum cost for α -irrigating the measure µ from the origin will be denoted by I α ( µ ). Following Maddalena, Morel, and Solimini [12],this can be described as follows. Let M = µ ( R d ) be the total mass to be transported and letΘ = [0 , M ]. We think of each θ ∈ Θ as a “water particle”. A measurable map χ : Θ × R + R d (2.16)is called an admissible irrigation plan if(i) For every θ ∈ Θ, the map t χ ( θ, t ) is Lipschitz continuous. More precisely, for each θ there exists a stopping time T ( θ ) such that, calling˙ χ ( θ, t ) = ∂∂t χ ( θ, t )the partial derivative w.r.t. time, one has (cid:12)(cid:12) ˙ χ ( θ, t ) (cid:12)(cid:12) = ( t ∈ [0 , T ( θ )] , t ≥ T ( θ ) . (2.17)(ii) At time t = 0 all particles are at the origin: χ ( θ,
0) = for all θ ∈ Θ.(iii) The push-forward of the Lebesgue measure on [0 , M ] through the map θ χ ( θ, T ( θ ))coincides with the measure µ . In other words, for every open set A ⊂ R d there holds µ ( A ) = meas (cid:16) { θ ∈ Θ ; χ ( θ, T ( θ )) ∈ A (cid:9)(cid:17) . (2.18)Next, to define the corresponding transportation cost, one must take into account the factthat, if many paths go through the same pipe, their cost decreases. With this in mind, givena point x ∈ R d we first compute how many paths go through the point x . This is describedby | x | χ = meas (cid:16)(cid:8) θ ∈ Θ ; χ ( θ, t ) = x for some t ≥ (cid:9)(cid:17) . (2.19)We think of | x | χ as the total flux going through the point x . Definition 2.4 (irrigation cost).
For a given α ∈ [0 , , the total cost of the irrigation plan χ is E α ( χ ) . = Z Θ Z T ( θ )0 (cid:12)(cid:12) χ ( θ, t ) (cid:12)(cid:12) α − χ dt ! dθ. (2.20) The α -irrigation cost of a measure µ is defined as I α ( µ ) . = inf χ E α ( χ ) , (2.21) where the infimum is taken over all admissible irrigation plans. emark 2.5 In the case α = 1, the expression (2.20) reduces to E α ( χ ) = Z Θ (cid:18)Z R + | ˙ χ t ( θ, t ) | dt (cid:19) dθ = Z Θ [total length of the path χ ( θ, · )] dθ . Of course, this length is minimal if every path χ ( · , θ ) is a straight line, joining the origin with χ ( θ, T ( θ )). Hence I α ( µ ) = inf χ E α ( χ ) = Z Θ | χ ( θ, T ( θ )) | dθ = Z | x | dµ . On the other hand, when α <
1, moving along a path which is traveled by few other particlescomes at a high cost. Indeed, in this case the factor (cid:12)(cid:12) χ ( θ, t ) (cid:12)(cid:12) α − χ becomes large. To reduce thetotal cost, is thus convenient that particles travel along the same path as far as possible.For the basic theory of ramified transport we refer to [5, 12, 19], or to the monograph [1]. Thefollowing lemma provides a useful lower bound to the transportation cost. In particular, werecall that optimal irrigation plans satisfy Single Path Property: If χ ( θ, τ ) = χ ( θ ′ , τ ′ ) for some θ, θ ′ ∈ Θ and 0 < τ < τ ′ , then χ ( θ, t ) = χ ( θ ′ , t ) for all t ∈ [0 , τ ] . (2.22) Lemma 2.6
For any positive Radon measure µ on R d and any α ∈ [0 , , one has I α ( µ ) ≥ Z + ∞ (cid:16) µ (cid:0) { x ∈ R d ; | x | ≥ r } (cid:1)(cid:17) α dr . (2.23) In particular, for every r > one has I α ( µ ) ≥ r · µ (cid:0) { x ∈ R d ; | x | ≥ r } (cid:1) α . (2.24) Proof.
Let χ : Θ × R + R d be an optimal transportation plan for I α ( µ ). For any given t >
0, let Θ t . = n θ ∈ Θ ; T ( θ ) ≥ t o be the set of particles whose path has length ≥ t . By the Single Path Property (see Chapter 7in [1]), if χ ( θ, τ ) = χ (˜ θ, ˜ τ ) , for some θ, ˜ θ ∈ Θ and 0 ≤ τ ≤ ˜ τ , then χ ( θ, t ) = χ (˜ θ, t ) for all t ∈ [0 , τ ] . (2.25)As a consequence, if t ≤ T ( θ ), then (cid:12)(cid:12) χ ( θ, t ) (cid:12)(cid:12) χ ≤ meas(Θ t ) . (2.26)7n addition, since all particles travel with unit speed, we have the obvious implication x = χ ( θ, T ( θ )) = ⇒ T ( θ ) ≥ | x | , hence meas(Θ t ) ≥ µ (cid:16)(cid:8) x ; | x | ≥ t (cid:9)(cid:17) . (2.27)Always relying on the optimality of χ , by (2.26) and (2.27) we conclude I α ( µ ) = E α ( χ ) . = Z Θ (cid:18)Z R + (cid:12)(cid:12) χ ( θ, t ) (cid:12)(cid:12) α − χ · | ˙ χ ( θ, t ) | dt (cid:19) dθ = Z Θ Z T ( θ )0 (cid:12)(cid:12) χ ( θ, t ) (cid:12)(cid:12) α − χ dt ! dθ ≥ Z Θ Z T ( θ )0 (cid:2) meas(Θ t ) (cid:3) α − dt ! dθ = Z + ∞ Z { T ( θ ) ≥ t } (cid:2) meas(Θ t ) (cid:3) α − dθ ! dt = Z + ∞ (cid:2) meas(Θ t ) (cid:3) α dt ≥ Z + ∞ h µ (cid:0) { x ∈ R d ; | x | ≥ t } (cid:1)i α dt . This proves (2.23). The inequality (2.24) follows immediately.
In this section we study a problem related to the optimal shape of tree branches. (OPB) Optimization Problem for Branches.
Given a function η ∈ L ( S d − ) and constants α ∈ [0 , , c > , maximize: S η ( µ ) − c I α ( µ ) , (3.1) among all positive Radon measures µ , supported on closed the half space R d + . = (cid:8) ( x , . . . , x d ) ; x d ≥ (cid:9) , (3.2) without any constraint on the total mass. In [7] the existence of an optimal solution to the problem (3.1) was proved under a constrainton the total mass of the measure µ , namely µ ( R d ) ≤ κ . Our present goal is to prove the existence of an optimal solution of (3.1) without any constraint.Throughout the following, it will be natural to assume1 − d − . = α ∗ < α ≤ . (3.3)Indeed, if a measure µ is supported on a set whose ( d − S η ( µ ) = 0. On the other hand, if α < α ∗ , then any set with positive ( d − α < α ∗ the optimization problem (3.1) becomestrivial: the zero measure is already an optimal solution. We can now state the main result ofthis section. 8 heorem 3.1 Suppose that d ≥ and α > α ∗ , as in (3.3). Then the unconstrained opti-mization problem (3.1) admits an optimal solution µ , with bounded support and bounded totalmass. Proof.
Following the direct method in the Calculus of Variations, we consider a maximizingsequence of measures ( µ k ) k ≥ . While each µ k is a bounded positive measure, at this stagewe cannot exclude the possibility that µ k ( R d ) → + ∞ . By showing that all measures µ k are uniformly bounded and have uniformly bounded support, we shall be able to select asubsequence, weakly converging to an optimal solution. The proof is given in several steps. As a first step, we claim that the irrigation costs I α ( µ k ) are uniformly bounded.Indeed, given a radius r >
0, we can decompose any measure µ as a sum µ = µ − r + µ + r . = χ { x ≤ r } · µ + χ { x>r } · µ . (3.4)Here χ A denotes the characteristic function of a set A ⊂ R d . Calling ω d − the volume ofthe unit ball in R d − , using (2.6)-(2.7) and then (2.24), the sunlight functional can now bebounded as S η ( µ ) ≤ S η ( µ − r ) + S η ( µ + r ) ≤ k η k L · ω d − r d − + k η k L · µ (cid:16) { x ; | x | > r } (cid:17) ≤ k η k L · " ω d − r d − + (cid:18) I α ( µ ) r (cid:19) /α . (3.5)In the above inequality, the radius r > r such that ω d − r d − = (cid:16) I α ( µ ) r (cid:17) /α . This choice yields ω αd − r α ( d − = I α ( µ ) , r = I α ( µ ) ω αd − ! α ( d − . (3.6)Inserting (3.6) in (3.5) one obtains the a priori bound S η ( µ ) ≤ C (cid:16) I α ( µ ) (cid:17) d − α ( d − , (3.7)for some constant C depending only on α, d , and k η k L .In connection with the original problem (3.1), this implies S η ( µ ) − c I α ( µ ) ≤ C (cid:16) I α ( µ ) (cid:17) d − α ( d − − c I α ( µ ) . (3.8)We now observe that the assumption (3.3) is equivalent to d −
11 + α ( d − < . κ large enough so that I α ( µ ) ≥ κ = ⇒ S η ( µ ) − c I α ( µ ) ≤ . (3.9)In the remainder of the proof, without loss of generality we shall seek a global maximum forthe functional in (3.1) under the additional constraint I α ( µ ) ≤ κ . (3.10)In turn, by (3.7) one has a uniform bound S η ( µ ) ≤ κ (3.11)for all µ satisfying (3.10). Let ( µ k ) k ≥ be a maximizing sequence. In this step we construct a second maximizingsequence (˜ µ k ) k ≥ such that all measures ˜ µ k are supported inside a fixed ball B ρ centered atthe origin with radius ρ .Toward this goal, let χ be an optimal irrigation plan for a measure µ , as in (2.16). By (2.24)and (3.10), for any radius r > µ (cid:16) { x ∈ R d ; | x | ≥ r } (cid:17) ≤ (cid:18) I α ( µ ) r (cid:19) /α ≤ (cid:16) κ r (cid:17) /α . (3.12)Consider two radii 0 < r < r . As in (3.4), we can decompose the measure µ as a sum: µ = µ ♭ + µ ♯ . = χ { x ≤ r } · µ + χ { x>r } · µ . (3.13)By possibly relabeling the set Θ = Θ ♭ ∪ Θ ♯ , we can assume that • χ ♭ : Θ ♭ × R + R d is an irrigation plan for the measure µ ♭ • χ ♯ : Θ ♯ × R + R d is an irrigation plan for the measure µ ♯ .Note that χ ♭ and χ ♯ are not necessarily optimal. If µ ♯ is removed, by (3.12) the difference inthe gathered sunlight is S η ( µ ♭ + µ ♯ ) − S η ( µ ♭ ) ≤ S η ( µ ♯ ) ≤ k η k L · µ ♯ ( R d ) ≤ k η k L · (cid:18) κ r (cid:19) /α . (3.14)On the other hand, by the Single Path Property (2.22), for any x ∈ R d with | x | ≥ r one has I α ( µ ) ≥ | x | αχ · r . Therefore | x | χ ≤ (cid:18) I α ( µ ) r (cid:19) /α ≤ (cid:18) κ r (cid:19) /α . (3.15)We now estimate the difference of the irrigation cost, if part of the measure is removed. Twocases will be considered. 10ASE 1: 0 < α <
1. By (3.15) we can then choose r large enough so that | x | ≥ r = ⇒ α | x | α − χ ≥ . (3.16)According to Proposition 4.8 in [1], the cost of an irrigation plan χ can be equivalently de-scribed as E α ( χ ) = Z R d | x | αχ d H ( x ) . (3.17)where H denotes the 1-dimensional Hausdorff measure. If χ is an optimal irrigation plan for µ = µ ♭ + µ ♯ , then I α ( µ ♭ + µ ♯ ) = Z R d | x | αχ d H ( x ) = Z R d (cid:16) | x | χ ♯ + | x | χ ♭ (cid:17) α d H ( x ) ≥ Z R d (cid:16) | x | αχ ♭ + α (cid:0) | x | χ ♭ + | x | χ ♯ (cid:1) α − | x | χ ♯ (cid:17) d H ( x ) ≥ Z R d | x | αχ ♭ d H ( x ) + Z {| x |≥ r } α | x | α − χ · | x | χ ♯ d H ( x ) ≥ I α ( µ ♭ ) + Z {| x |≥ r } | x | χ ♯ d H ( x ) ≥ I α ( µ ♭ ) + ( r − r ) µ ♯ ( R d ) . (3.18)We now choose r large enough so that c ( r − r ) ≥ k η k L . By the second inequality in (3.14)and (3.18) it follows S η ( µ ♭ + µ ♯ ) − S η ( µ ♭ ) ≤ c (cid:16) I α ( µ ♭ + µ ♯ ) − I α ( µ ♭ ) (cid:17) . (3.19)Let now ( µ k ) k ≥ be a maximizing sequence. We decompose each measure as µ k = µ ♭k + µ ♯k . = χ { x ≤ r } · µ k + χ { x>r } · µ k . (3.20)By (3.19), the sequence ( µ ♭k ) k ≥ is still a maximizing sequence, where all measures are sup-ported inside the fixed ball B r .CASE 2: α = 1. In this case we simply choose r = 1 c k η k L . (3.21)In connection with the decomposition (3.20), we have S η ( µ ♭ + µ ♯ ) − S η ( µ ♭ ) ≤ S η ( µ ♯ ) ≤ k η k L · µ ♯ ( R d )= r c µ ♯ ( R d ) ≤ c I ( µ ♯ ) = c (cid:16) I ( µ ♯ ) − I ( µ ♭ + µ ♯ ) (cid:17) . Again, this shows that ( µ ♭k ) k ≥ is a maximizing sequence, where all measures are supportedinside the ball B r . In this step, relying on the assumption that the space dimension is d ≥
3, we prove theexistence of a maximizing sequence (˜ µ k ) k ≥ with uniformly bounded total mass.11ndeed, let µ be any measure with I α ( µ ) ≤ κ . For any integer j , consider the radius r j . = 2 − j and the spherical shell V j . = (cid:8) x ∈ R d ; r j +1 < | x | ≤ r j (cid:9) . (3.22)Moreover, call µ j . = χ V j · µ the restriction of the measure µ to the set V j . For every j ≥ S η ( µ ) − S η ( µ − µ j ) ≤ S η ( µ j ) ≤ k η k L · ω d − r d − j . (3.23)We now estimate the difference in the irrigation costs. By (3.15), for every x ∈ R d one has | x | χ ≤ (cid:18) κ | x | (cid:19) /α , (3.24)hence min (cid:8) | z | α − χ ; | z | ≥ r j +2 (cid:9) ≥ κ α − α · r α − j +2 . = κ r α − j , for a suitable constant κ . This implies I α ( µ ) − I α ( µ − µ j ) ≥ r j · Z V j κ r α − j dµ = κ r /αj · µ ( V j ) . (3.25)Comparing (3.23) with (3.25) we see that, if κ r /αj · µ ( V j ) ≥ k η k L · ω d − r d − j , (3.26)then the difference S η ( µ ) − c I α ( µ ) will increase if we remove from µ all the mass located inside V j .We can repeat the above procedure, removing from µ the mass contained in all regions V j such that (3.26) holds. More precisely, let J be the set of all integers j ≥ µ . = µ − X j ∈ J µ j . (3.27)By the previous analysis, S η (˜ µ ) − c I α (˜ µ ) ≥ S η ( µ ) − c I α ( µ ) . (3.28)Moreover, by (3.26) we have the implication j / ∈ J = ⇒ µ ( V j ) ≤ k η k L · ω d − κ r d − − α j . = κ r d − − α j . The total mass of ˜ µ can thus be estimated by˜ µ ( R d ) = µ (cid:16) { x ; | x | > } (cid:17) + X j / ∈ J µ ( V j ) ≤ µ (cid:16) R d \ B (cid:17) + X j ≥ κ r d − − α j = µ (cid:16) R d \ B (cid:17) + κ X j ≥ − j (cid:16) d − − α (cid:17) < + ∞ , (3.29)12rovided that d − − α >
0. This is indeed the case if α satisfies (3.3) and d ≥ By the previous steps, we can choose a maximizing sequence ( µ k ) k ≥ such that all measures µ k have uniformly bounded total mass and are supported on a fixed ball. By possibly takinga subsequence, we achieve the weak convergence µ k ⇀ µ for some bounded measure µ . By theupper semicontinuity of sunlight functional S η proved in [5] and by the lower semicontinuity ofthe irrigation cost I α , see [1, 12], this limit measure µ provides a solution to the optimizationproblem (3.1). d = 2 . In dimension d = 2 we have d − − α ≤ α ≤
1, hence the estimate (3.29) on the totalmass breaks down. We develop here a different approach, which is valid for √ − < α ≤ . (3.30) j r r r j jj +1+2 V Γ j j P Figure 2:
In dimension d ≥
3, if µ ( V j ) is large, then we can increase the payoff (3.1) by simplyremoving all the mass contained in the spherical shell V j . This idea is used in step of the proof ofTheorem 3.1. In dimension d = 2, if µ ( V j ) is large, to increase the payoff (3.1) we replace the measure µ j = χ V j · µ with a new measure ˜ µ j uniformly distributed over the half circumference Γ j . Notice that˜ µ j can be irrigated by moving the water particles from the origin to P j , and then along Γ j . Theorem 3.2 If d = 2 and α satisfies (3.30), then the unconstrained optimization problem(3.1) admits an optimal solution µ , with bounded support and bounded total mass. Indeed, repeating the steps in the proof of the Theorem 3.1, we obtain a maximiz-ing sequence ( µ k ) k ≥ of positive measures with uniformly bounded support. Moreover, theirrigation costs I α ( µ k ) remain uniformly bounded.In order to achieve a uniform bound on the total mass µ k ( R d ), an auxiliary result is needed. Lemma 3.3
Let α satisfy (3.30) and let κ > be given. Then there exists an integer j ∗ and an exponent ε > such that the following holds. Given any bounded measure µ with I α ( µ ) ≤ κ , there exists a second measure ˜ µ satisfying (3.28) and such that, setting r j . = 2 − j , ˜ µ (cid:16)(cid:8) x ∈ R ; r j +1 < | x | ≤ r j (cid:9)(cid:17) ≤ − εj for all j ≥ j ∗ . (3.31)13 roof. 1. If (3.30) holds, we can find 0 < ε < β < αβ + 1 > ε + 1 α . (3.32)Let µ j be the restriction of the measure µ to the spherical shell V j defined at (3.22). Moreover,let ˜ µ j be the positive measure with total mass˜ µ j ( R d ) = πr βj , uniformly distributed on the half circumferenceΓ j . = { x = ( x , x ) ∈ R ; | x | = r j , x > } . As shown in Fig. 2, there is a simple irrigation plan χ for ˜ µ j . Namely, we can first move allwater particles on a straight line from the origin to the point P j = ( − r j , P j toall points along the half circumference Γ j . The total cost of this irrigation plan satisfies E α ( χ ) ≤ [total mass] α × [maximum length traveled] ≤ ( πr βj ) α · ( π + 1) r j . (3.33)Therefore, the minimum irrigation cost for ˜ µ j satisfies I α (˜ µ j ) ≤ π α +1 r αβ +1 j . (3.34)On the other hand, assuming µ ( V j ) ≥ r εj , by (3.25) we have I α ( µ ) − I α ( µ − µ j ) ≥ κ r ε + α j . (3.35)By (3.32), for all r j small enough it follows h I α ( µ ) − I α ( µ − µ j ) i − I α (˜ µ j ) ≥ κ r ε + α j . (3.36) Next, we estimate how the sunlight functional changes if we replace µ j by ˜ µ j . We claimthat S η ( µ ) − S η ( µ − µ j + ˜ µ j ) ≤ [total amount of light hitting V j ] − [light captured by ˜ µ j ] ≤ k η k L · exp (cid:0) − r β − j (cid:1) . (3.37)Indeed, consider any unit vector n ∈ S . As shown in Fig. 3, let [ a j , b j ] = π n (Γ j ) be theperpendicular projection of Γ j on the orthogonal subspace E ⊥ n . By construction, the projectedmeasure π n ˜ µ j is absolutely continuous w.r.t. 1-dimensional Lebesgue measure on E ⊥ n . Itsdensity Φ n j satisfies ( Φ n j ( y ) ≥ r β − j if y ∈ [ a j , b j ] , Φ n j ( y ) = 0 if y / ∈ [ a j , b j ] . y b j a E n ⊥ Γ j n r j Figure 3:
Let ˜ µ j be the measure supported on the half circumference Γ j , with constant density r β − j w.r.t. 1-dimensional measure. Then, for any unit vector n , the projected measure π n ˜ µ j has density ≥ r β − j on [ a j , b j ] = π n (Γ j ). For j ≥ V j in the direction n ] − [light parallel to n captured by ˜ µ j ] ≤ ( b j − a j ) − Z b j a j (cid:16) − e − Φ n j ( y ) (cid:17) dy ≤ Z b j a j exp (cid:0) − r β − j (cid:1) dy ≤ exp (cid:0) − r β − j (cid:1) , (3.38)because b j − a j ≤ We now observe that, since β <
1, when j ≥ j ∗ is sufficiently large the right hand side of(3.38) is smaller than the right hand side of (3.36). By possibly choosing a larger j ∗ , we canalso assume that πr βj < r εj for all j ≥ j ∗ . (3.39)Defining the set of indices J . = (cid:8) j ≥ j ∗ ; µ ( V j ) > r εj (cid:9) , we claim that the modified measure˜ µ . = µ + X j ∈ J (˜ µ j − µ j ) . (3.40)satisfies all conclusions of the lemma. Indeed, from (3.36) and (3.37)-(3.38) it follows that ˜ µ achieves a better payoff than µ , i.e. (3.28) holds. In addition, the bounds (3.31) on the totalmass follow from (3.39).We observe that (3.31) implies an a priori bound on the total mass ˜ µ (cid:16)(cid:8) x ∈ R ; | x | ≤ r j ∗ (cid:9)(cid:17) .On the other hand, a bound on ˜ µ (cid:16)(cid:8) x ∈ R ; | x | ≥ r j ∗ (cid:9)(cid:17) is already provided by (3.12).15hanks to the above lemma, the proof of Theorem 3.2 is now straightforward. Indeed, byLemma 3.3 we can replace each µ k by a new measure ˜ µ k we can one can construct a maximizingsequence of measures with uniformly bounded support and uniformly bounded total mass.Taking a weak limit, the existence of an optimal solution can thus be proved using the uppersemicontinuity of S η and the lower semicontinuity of I α , as in [7]. In this section we study the optimal shape of tree roots. (OPR) Optimization Problem for Roots. maximize: H ( u, µ ) − c I α ( µ ) , (4.1) subject to ( ∆ u + f ( u ) − uµ = 0 , x ∈ R d − . = (cid:8) ( x , . . . , x d ) ; x d < (cid:9) ,u x d = 0 , x d = 0 . (4.2) Here µ is a positive measure concentrated on the set Ω . = (cid:8) ( x , . . . , x d ) = (0 , . . . ,
0) ; x d ≤ (cid:9) , (4.3) without any constraint on its total mass. We recall that H ( u, µ ) = Z { x d ≤ } u dµ = Z R d − f ( u ) dx (4.4)is the harvest functional introduced at (2.14), while I α ( µ ) is the minimum irrigation costdefined at (2.21).As in Section 2, we assume that the function f satisfies all conditions in (2.15). In order toconstruct an optimal solution, we consider a maximizing sequence ( u k , µ k ) k ≥ . By suitablyadapting the arguments used in the previous section, we will prove a priori bounds on thetotal irrigation costs I α ( µ k ) and on the total harvesting payoffs H ( u k , µ k ). Our first lemmashows that the total harvest achieved by a measure supported on a closed ball B ρ , centeredat the origin with a large radius ρ , grows at most like ρ d . Lemma 4.1
Let f satisfy the assumptions (A1). Then there exists a constant C f such thatthe following holds. For any ρ ≥ , if µ is a positive measure supported inside the closed ball B ρ , then for any solution u of (4.2) one has H ( u, µ ) ≤ C f ρ d . (4.5) Proof. 1.
As shown in Fig. 4, right, let ψ = ψ ( r ) be the solution to the ODE ψ ′′ ( r ) + f ( ψ ( r )) = 0 , r > , (4.6)16 (0) = 0 , lim r → + ∞ ψ ( r ) = M. (4.7)We claim that ψ is a monotonically increasing function such that ψ ( r ) → M as r → + ∞ , with an exponential rate of convergence.Indeed, let F ( s ) = R s f ( ξ ) dξ . Then, for any solution of (4.6), the energy E ( r ) . = (cid:0) ψ ′ ( r ) (cid:1) F (cid:0) ψ ( r ) (cid:1) (4.8)is constant. The second limit in (4.7) implies that E = F ( M ). We thus obtain the ODE ψ ′ ( r ) = p F ( M ) − F ( ψ ( r )) . (4.9)Since f ( M ) = 0 and f ′ ( M ) <
0, for ψ ∈ [0 , M ] one has F ( M ) − F ( ψ ) ≥ γ ( M − ψ ) , (4.10)for some constant γ > f itself. Therefore, ψ ′ ( r ) > p γ ( M − ψ )We thus conclude that the solution ψ of (4.6)-(4.7) satisfies ψ ( r ) ≥ M (1 − e −√ γ r ) . (4.11)Therefore f ( ψ ( r )) ≤ f ′ ( M )( ψ ( r ) − M ) ≤ C e −√ γ r , C . = − f ′ ( M ) M > . (4.12) Let u be a solution to (2.8)-(2.9), where the measure µ is supported on the ball B ρ . Weclaim that u ≥ v , where v is the radially symmetric function defined by v ( x ) = ( ψ ( | x | − ρ ) if | x | ≥ ρ, | x | < ρ. (4.13)Indeed, for | x | > ρ , by (4.13) and (4.6) one has∆ v ( x ) + f ( v ) = ψ ′′ ( | x | − ρ ) + d − | x | ψ ′ ( | x | − ρ ) + f (cid:0) ψ ( | x | − ρ ) (cid:1) = d − | x | ψ ′ ( | x | − ρ ) ≥ , (4.14)showing that v is a lower solution on the region where | x | > ρ . Hence u ( x ) ≥ v ( x ) for all x ∈ R d . Since u ≥ v , an upper bound on the total harvest is now provided by: H ( u, µ ) = Z Ω f ( u ( x )) dx ≤ ω d − Z + ∞ r d − b f ( ψ ( r − ρ )) dr, (4.15)17here (see Fig. 4, left) b f ( s ) = max (cid:8) f ( ξ ) ; ξ ∈ [ s, M ] (cid:9) = ( f ( s ) if s ≥ u max ,K if s ≤ u max . (4.16)Here u max ∈ ]0 , M [ is the unique point at which the function f attains its maximum. By the previous steps, the solution ψ of (4.6)-(4.7) is a monotonically increasing functionconverging to M as s → + ∞ . We can thus find a radius r ∗ > ψ ( r ) ≥ u max for all r ≥ r ∗ . (4.17)Indeed, one can choose r ∗ . = − ln(1 − u max M ) √ γ . (4.18)Using (4.16),(4.12) and performing the variable change r = ρ + s , one finds Z ∞ r d − b f ( ψ ( r − ρ )) dr = Z ρ + r ∗ r d − f ( u max ) dr + Z + ∞ ρ + r ∗ r d − f ( ψ ( r − ρ )) dr ≤ Kd ( ρ + r ∗ ) d + Z + ∞ ρ + r ∗ r d − C e −√ γ ( r − ρ ) dr ≤ Kd ( ρ + r ∗ ) d + ρ d − Z + ∞ r ∗ (1 + s ) d − C e −√ γ s ds ≤ C ρ d + C ρ d − . (4.19)Combining (4.19) with (4.15) one obtains the desired inequality (4.5).The next lemma provides an estimate on the total harvest achieved by a measure supportedin a small ball B ρ , as ρ → Lemma 4.2
Let f satisfy (A1). Then there exists a positive continuous function η , with lim s → η ( s ) = 0 , such that the following holds. Let ( u, µ ) be any solution of (4.2). If µ issupported on the closed ball B ρ , then the total harvest satisfies H ( u, µ ) ≤ η ( ρ ) . (4.20) Proof. 1.
Let U = U ( r ) be a solution to U ′′ ( r ) + d − r U ′ ( r ) + f ( U ) = 0 , ρ < r < + ∞ , (4.21) U ( ρ ) = 0 , lim r → + ∞ U ( r ) = M. (4.22)A lower bound on U will be achieved by constructing a suitable subsolution.Observing that f ( U ) ≥ U ′ >
0, such a subsolution can be obtained by patching togethera solution to U ′′ ( r ) + d − r U ′ ( r ) = 0 , ρ < r < r ∗ , (4.23)18ith a solution of U ′′ ( r ) + f ( U ) = 0 , r ∗ < r < + ∞ . (4.24)As in the proof of the previous lemma, let ψ be the solution to (4.6) and (4.7). In addition, asolution of (4.23) with boundary condition U ( ρ ) = 0 (4.25)is found in the form U ( r ) = ( κ (ln r − ln ρ ) if d = 2 ,κ ( ρ − d − r − d ) if d ≥ . (4.26)By linearity, here κ can be any constant. f u max M u0K f ε rR ε r ε M ψ U Figure 4:
Left: a function f satisfying the assumptions (2.15) and the corresponding function b f in(4.16). Right: a subsolution obtained by patching together the functions U ε and ψ . To patch together the two solutions U and ψ , we proceed as follows. Recalling (4.16), forany ε >
0, choose R ε large enough so that R ε (cid:0) ψ ( R ε ) − M (cid:1) < ε, Z + ∞ R ε b f ( ψ ( s )) ds ≤ ε. (4.27)This is certainly possible because f ( M ) = 0 and ψ ( s ) → M exponentially fast as s → + ∞ .Next, we claim that there exists r ε > κ ε > U ε ( r ) . = ( κ ε (ln r − ln r ε ) if d = 2 ,κ ε ( r − dε − r − d ) if d ≥ , (4.28)satisfies (see Fig. 4, right) U ε ( R ε ) = ψ ( R ε ) , U ′ ε ( R ε ) ≤ ψ ′ ( R ε ) , (4.29) Kr ε + Z R ε r ε b f ( U ε ( s )) ds < ε. (4.30)Here K is the maximum value of f , as in (2.15).To prove our claim, having fixed R ε , for any ρ > κ ε so that the function U ρ ( r ) . = ( κ ε (ln r − ln ρ ) if d = 2 ,κ ε ( ρ − d − r − d ) if d ≥ , (4.31)19atisfies U ρ ( R ε ) = ψ ( R ε ). As ρ →
0, one now haslim ρ → U ′ ρ ( R ε ) = 0 , lim ρ → U ρ ( r ) = ψ ( R ε ) for any 0 < r ≤ R ε . (4.32)This is proved by a direct computation. When d = 2 we have κ ε = ψ ( R ε )ln R ε − ln ρ , U ρ ( r ) = ψ ( R ε ) ln r − ln ρ ln R ε − ln ρ ,U ′ ρ ( r ) = 1 r ψ ( R ε )ln R ε − ln ρ . Hence the limits in (4.32) hold. On the other hand, when d ≥ κ ε = ψ ( R ε ) R − dε − ρ − d , U ρ ( r ) = ψ ( R ε ) ρ − d − r − d ρ − d − R − dε ,U ′ ρ ( r ) = ψ ( R ε ) ( d − r − d ρ − d − R − dε , and the limits in (4.32) again hold.Having determined R ε according to (4.27), if we now choose r ε = ρ > U ′ ε ( R ε ) < ψ ′ ( R ε ).Moreover, thanks to the second limit in (4.32) and the first inequality in (4.27), by choosing r ε > Kr ε + Z R ε r ε b f ( U ε ( s )) ds < ε + R ε f ( ψ ( R ε )) ≤ ε + R ε L f ( ψ ( R ε ) − M ) ≤ ( L f + 1) ε , where L f denotes the Lipschitz constant of f . Since ε > Let u be a solution to (4.2), where the measure µ is supported on the closed ball B ρ . Bya comparison argument, we conclude that u ≥ v , where v is the function defined by v ( x ) = | x | ≤ r ε ,U ε ( r ) if r ε < | x | < R ε ,ψ ( r ) if R ε ≤ | x | . By (4.27)-(4.30), an upper bound on the total harvest is now provided by H ( u, µ ) = Z R d − f ( u ( x )) dx ≤ ω d − Z + ∞ r d − b f ( v ( r )) dr = ω d − (cid:18)Z r ε b f (0) dr + Z R ε r ε U ε ( r ) dr + Z + ∞ R ε ψ ( r ) dr (cid:19) ≤ ω d − ε . Since ε > orollary 4.3
In the same setting as Lemma 4.2, let µ be a positive measure on R d and let ( u, µ ) be a solution to (4.2). Then, for every ρ > , one has Z | x |≤ ρ u dµ ≤ η ( ρ ) , (4.33) where η is the function in (4.20). Proof.
Call µ ρ . = χ {| x |≤ ρ } · µ the restriction of the measure µ to the closed ball of radius ρ .Let u ρ ≥ u be a corresponding solution of (4.2). Using Lemma 4.2 we now obtain Z | x |≤ ρ u dµ = Z u dµ ρ ≤ Z u ρ dµ ρ ≤ η ( ρ ) . Using Lemma 4.1, we now prove that an analogue of (3.9) holds also for the harvesting problem.
Lemma 4.4
Let α > − d . Under the assumptions (A1), there exists a constant κ suchthat, for any positive bounded measure µ on Ω . = { x = ( x , . . . , x d ) ∈ R d ; x d ≤ } , I α ( µ ) ≥ κ = ⇒ H ( u, µ ) − c I α ( µ ) ≤ . (4.34) Proof.
Given any radius r ≥
1, we can decompose the measure µ as µ = µ − r + µ + r . = χ { x ≤ r } · µ + χ { x>r } · µ (4.35)Let u be a corresponding solution of (4.2). 0 ≤ u ( x ) ≤ M . Then there exists a solution u − tothe same elliptic problem with µ replaced by µ − r , such that0 ≤ u ( x ) ≤ u − ( x ) for all x ∈ R d − . (4.36)Using (4.36), and recalling that µ is concentrated on the domain Ω at (4.3), the harvestfunctional can be estimated by H ( u, µ ) . = Z Ω u dµ = Z Ω u dµ − r + Z Ω u dµ + r ≤ Z Ω u − dµ − r + M µ + r (Ω ) ≤ C f r d + M (cid:16) I α ( µ ) r (cid:17) α . (4.37)Here the last inequality was obtained by applying (4.5) to the measure µ − r and (2.24) to themeasure µ + r .Next, assuming I α ( µ ) sufficiently large, we can find a radius ρ ≥ C f ρ d = M (cid:16) I α ( µ ) ρ (cid:17) α . (4.38)21hoosing r = ρ in (4.37), we obtain H ( u, µ ) ≤ C (cid:16) I α ( µ ) (cid:17) d αd . (4.39)for some constant C depending only on α, d, and f .In connection with the original problem (4.1), this implies H ( u, µ ) − c I α ( µ ) ≤ C (cid:16) I α ( µ ) (cid:17) d αd − c I α ( µ ) . (4.40)Assuming that α > − d , it follows that d αd <
1. Hence by (4.40) there exists a constant κ large enough so that (4.34) holds. Lemma 4.5
Let α > − d and let the assumptions (A1) hold. Consider a maximizing se-quence ( u k , µ k ) k ≥ for the functional in (4.1). Then there exists another maximizing sequence (˜ u k , ˜ µ k ) k ≥ such that I α (˜ µ k ) ≤ κ , H (˜ u k , ˜ µ k ) ≤ κ , (4.41) for some constants κ , κ and all k ≥ . Moreover, all measures ˜ µ k are supported in a commonball B ρ . Proof. 1.
By (4.34), any maximizing sequence must satisfy the first inequality in (4.41). Thesecond inequality then follows from (4.39). It remains to prove the existence of a maximizing sequence with uniformly bounded support.Toward this goal, let χ be an optimal irrigation plan for a measure µ . By (2.24) and (3.10),for any radius r > µ (cid:16) { x ∈ R d ; | x | ≥ r } (cid:17) ≤ (cid:16) I α ( µ ) r (cid:17) /α ≤ (cid:16) κ r (cid:17) /α . (4.42)Consider two radii 0 < r < r , where r large enough such that (3.16) holds. As in (4.43),we can decompose the measure µ as a sum: µ = µ − + µ + . = χ { x ≤ r } · µ + χ { x>r } · µ . (4.43)By the same argument used in (3.18), for 0 < α ≤
1, one has I α ( µ − + µ + ) − I α ( µ − ) ≥ ( r − r ) µ + (Ω ) , (4.44)where Ω is the domain in (4.3). We now estimate the decrease in the harvest functional, when µ is replaced by µ − . Let u be a solution of (4.2), corresponding to the measure µ . Then there exists a solution u − to thesame problem, with µ replaced by µ − , such that0 ≤ u ( x ) ≤ u − ( x ) for all x ∈ R d − . (4.45)22sing (4.45), the harvest functional can be estimated by H ( u, µ ) . = Z Ω u dµ = Z Ω u dµ − + Z Ω u dµ + ≤ Z Ω u − dµ − + Z Ω M dµ + = H ( u − , µ − ) + M µ + (Ω ) . (4.46)Taking r large enough so that c ( r − r ) ≥ M , by (4.44) and (4.46) it follows H ( u, µ ) − H ( u − , µ − ) ≤ c (cid:16) I α ( µ ) − I α ( µ − ) (cid:17) . (4.47) Let now ( u k , µ k ) k ≥ be a maximizing sequence. We decompose each measure as µ k = µ − k + µ + k . = χ { x ≤ r } · µ k + χ { x>r } · µ k . (4.48)Choose u − k the corresponding solution to the same elliptic problem with µ k replaced by µ − k ,such that 0 ≤ u k ( x ) ≤ u − k ( x ) for all x ∈ R d − . (4.49)By (4.47), ( u − k , µ − k ) k ≥ is still a maximizing sequence, where all measures are supported insidethe closed ball B r .The next lemma yields a more detailed estimate on the support of the optimal measure. Lemma 4.6
Suppose ( u k , µ k ) k ≥ is a maximizing sequence for the optimization problem (4.1),with irrigation costs I α ( µ k ) ≤ κ for all k ≥ . Then there exists a second maximizing sequence (˜ u k , ˜ µ k ) k ≥ such that ˜ µ k (cid:16)(cid:8) x ∈ R d ; ˜ u k ( x ) < C | x | /α (cid:9)(cid:17) = 0 (4.50) for all k ≥ . Here C . = c − α κ − α . Proof. 1.
Given a positive measure µ and a corresponding solution u of (4.2), consider theset A . = (cid:8) x ∈ Ω ; u ( x ) ≥ C | x | /α (cid:9) . Moreover, let ˜ µ . = χ A · µ (4.51)be the measure obtained from µ by removing all the mass that lies outside A .By (3.25) it follows c I α ( µ ) − c I α (˜ µ ) ≥ C Z Ω \ A | x | /α dµ, (4.52)23 . To estimate the difference in the harvest functional, let ˜ u be a solution to the same ellipticproblem (4.2) with µ replaced by ˜ µ , such that u ( x ) ≤ ˜ u ( x ) ≤ M for all x ∈ Ω . (4.53)We compute H ( u, µ ) − H (˜ u, ˜ µ ) = Z Ω u dµ − Z Ω ˜ u d ˜ µ ≤ Z Ω \ A u dµ ≤ C Z Ω \ A | x | /α dµ . (4.54)Comparing (4.52) with (4.54) we thus obtain H ( u, µ ) − c I α ( µ ) ≤ H (˜ u, ˜ µ ) − c I α (˜ µ ) . (4.55)Recalling that u ≤ ˜ u , by (4.51) it follows˜ µ (cid:16)(cid:8) x ∈ Ω ; ˜ u ( x ) < C | x | /α (cid:9)(cid:17) ≤ ˜ µ (cid:16)(cid:8) x ∈ Ω ; u ( x ) < C | x | /α (cid:9)(cid:17) = 0 . If now ( u k , µ k ) k ≥ is any maximizing sequence, for every k ≥ A k . = (cid:8) x ∈ Ω ; u k ( x ) ≥ C | x | /α (cid:9) , ˜ µ k . = χ A k · µ k . Moreover, we let ˜ u k ≥ u k be a solution to (4.2) corresponding to the measure ˜ µ k . By theprevious analysis, (˜ u k , ˜ µ k ) k ≥ is another maximizing sequence, satisfying (4.50).When f satisfies the assumptions (2.15), any solution u of (2.8) will take values inside theinterval [0 , M ]. By the previous arguments, it thus follows the existence of a maximizingsequence ( u k , µ k ) k ≥ , where the measures µ k satisfy(i) Supp( µ k ) ⊂ B r , where C r /α = M .(ii) I α ( µ k ) ≤ C .In particular, for every r > k ≥ µ k (cid:0) { x ∈ R d ; | x | > r } (cid:1) < + ∞ . (4.56)This does not necessarily imply that the total mass of the measures µ k is uniformly bounded.Indeed, they may concentrate more and more mass close to the origin.To achieve the existence of an optimal measure, we thus need to work in the wider classof positive measures µ on the domain Ω in (4.3), possibly with infinite total mass. As apreliminary, the definition of irrigation plan and irrigation cost must be extended to thesemore general measures.If µ ( R d ) = + ∞ , an irrigation plan for µ is a map χ : R + × R + R d with the properties(i)–(iii) introduced in Section 1. For every m ≥
1, call χ m : [0 , m ] × R + R d the restrictionof χ to [0 , m ]. Then the cost of χ is defined as E α ( χ ) . = lim m → + ∞ E α ( χ m ) = sup m ≥ Z m Z τ ( θ )0 (cid:12)(cid:12) χ m ( θ, t ) (cid:12)(cid:12) α − χ m dtdθ. (4.57)24n the other hand, the harvest functional is defined as H ( u, µ ) . = sup ε → Z | x | >ε u dµ . (4.58)It is clear that the right hand sides in (4.57) and (4.58) are well defined, possibly taking thevalue + ∞ .We can now state our main result on the existence of an optimal measure. Theorem 4.7
Let the function f satisfy the assumptions (A1). Then the maximization prob-lem for roots (OPR) has an optimal solution ( u, µ ) , where µ is a positive measure on thedomain Ω defined at (4.3). The optimal measure µ has bounded support, but possibly un-bounded total mass. Proof. 1.
Let ( u k , µ k ) be a maximizing sequence. By the previous analysis we can assumethat all measures µ k are supported inside a fixed ball B ρ , and the quantities I α ( µ k ), H ( u k , µ k )remain uniformly bounded.By possibly selecting a subsequence, we can assume the existence of a positive measure µ suchthat the weak convergence µ k ⇀ µ holds on R d \ { } . In other words, Z ϕ dµ k → Z ϕ dµ for every continuous function ϕ ∈ C c ( R d \ { } ).Let µ ε be the restriction of µ to the subset {| x | > ε } . Then I α ( µ ) = sup ε> I α ( µ ε ) . (4.59)On the other hand, calling µ εk the restriction of µ k to the set {| x | ≥ ε } , the lower semicontinuityof the irrigation cost for bounded measures implies I α ( µ ε ) ≤ lim inf m →∞ I α ( µ εk ) ≤ lim inf m →∞ I α ( µ k ) . (4.60)Combining (4.59) with (4.60) we obtain I α ( µ ) ≤ lim inf m →∞ I α ( µ k ) . (4.61) To complete the existence proof, we need to find a solution u of (4.2) and show that H ( u, µ ) ≥ lim sup k →∞ H ( u k , µ k ) . (4.62)Toward this goal, choose radii ρ n → µ (cid:16) { x ∈ Ω ; | x | = ρ n } (cid:17) = 0 (4.63)25or all n ≥
1. Let µ ρ n k , µ ρ n be the restrictions of the measures µ k , µ to the closed sets V n . = { x ∈ Ω ; | x | ≥ ρ n } . Thanks to (4.63), for each n ≥ µ ρ n k ⇀ µ ρ n . Let u ρ n k , u ρ n be the corresponding solutions of (4.2). By the analysis in [7], since all measures µ ρ n k haveuniformly bounded mass, for each n ≥ Z u ρ n dµ ρ n ≥ lim sup k →∞ Z u ρ n k dµ ρ n k . (4.64)By Corollary 4.3 it follows Z u k dµ k ≤ Z u ρ n k dµ ρ n k + η ( ρ n ) . (4.65) We now observe that, as n → ∞ , the sequence of measures µ ρ n is increasing while thesequence of solutions u ρ n is decreasing. Setting u ( x ) . = inf n ≥ u ρ n ( x ) , one checks that ( u, µ ) is a solution to (4.2). Moreover, for any δ > Z | x |≥ δ u dµ = inf n ≥ Z | x |≥ δ u ρ n dµ. (4.66)Given ε >
0, we can find δ > b n large enough so that ρ b n ≤ δ, η ( ρ b n ) ≤ η ( δ ) < ε, Z | x |≥ δ u ρ b n dµ < Z | x |≥ δ u dµ + ε . (4.67)Using (4.65), then (4.64), and finally (4.67), we concludelim sup k →∞ Z u k dµ k ≤ lim sup k →∞ Z u ρ b n k dµ ρ b n k + η ( ρ b n ) ≤ Z u ρ b n dµ ρ b n + η ( ρ b n ) < Z | x |≥ δ u dµ + ε ! + ε ≤ Z u dµ + 2 ε. Since ε >
Let α ∈ [0 ,
1] be given. In Sections 3 and 4 we have proved the existence of an optimalconfiguration of tree roots and tree branches, where the optimal measure µ has boundedsupport. Here we are interested in how this support depends on parameters. More precisely,given a measure µ on R d , let R ( µ ) . = inf n r > µ (cid:0) {| x | > r } (cid:1) = 0 o (5.1)26e the radius of the smallest ball centered at the origin which contains the support of µ .We first consider the optimization problem (OPB) for tree branches. We seek an upper boundon R ( µ ), depending on the dimension d , the constants α, c , and the L norm of the function η in (3.1), measuring the intensity of light from various directions.As a preliminary, we recall how the irrigation cost behaves under rescalings. Given a measure µ and a constant λ >
0, we define the measures λµ and µ λ respectively by setting( λµ )( A ) . = λ µ ( A ) , µ λ ( A ) . = µ ( λ − A ) , (5.2)for every Borel set A ⊆ R d . Lemma 5.1
For any positive Radon measure µ on R d and any λ > , ≤ α ≤ , thefollowing holds: I α ( λµ ) = λ α I α ( µ ) , I α ( µ λ ) = λ I α ( µ ) . (5.3) Proof. 1.
To prove the first identity in (5.3), let Θ = [0 , M ] and let χ : [0 , M ] × R + R d bean admissible irrigation plan for µ . Then the map χ λ : [0 , λM ] × R + R d , defined by χ λ ( θ, t ) = χ ( λ − θ, t ) (5.4)is an admissible irrigation plan for λµ . Its cost is computed by E α ( χ λ ) = Z [0 ,λM ] (cid:16) Z R + | χ λ ( θ, t ) | α − χ λ · | ˙ χ λ ( θ, t ) | dt (cid:17) dθ = Z [0 ,M ] (cid:16) Z R + (cid:16) λ | χ ( θ, t ) | χ (cid:17) α − · | ˙ χ ( θ, t ) | dt (cid:17) λ dθ = λ α E α ( χ ) . (5.5)Taking the infimum over all irrigation plans we obtain I α ( λµ ) ≤ λ α I α ( µ ). Replacing λ by λ − we obtain the opposite inequality. To prove the second identity, consider any λ > χ : Θ × R + R d be an irrigationplan for µ . Then χ † : Θ × R + R d defined by χ † ( θ, t ) = λ · χ ( λ − θ, t )is an admissible irrigation plan for µ λ . Performing the change of variables ˜ θ = λ − θ , its costis computed as E α ( χ † ) = Z Θ (cid:16) Z R + | χ † ( θ, t ) | α − χ † · | ˙ χ † ( θ, t ) | dt (cid:17) dθ = Z Θ (cid:16) Z R + | χ (˜ θ, t ) | α − χ · | ˙ χ (˜ θ, t ) | dt (cid:17) λ d ˜ θ = λ E α ( χ ) . (5.6)Taking the infimum over all irrigation plans we obtain I α ( µ λ ) ≤ λ I α ( µ ). Replacing λ by λ − we obtain the opposite inequality. 27imilar formulas relate the sunlight captured by a rescaled measure. Namely, as proved in [7],one has S bη ( µ ) = b S η ( µ ) , S η ( λ d − µ λ ) = λ d − S η ( µ ) . (5.7)Thanks to the rescaling properties (5.3) and (5.7), the solution to the problemmaximize: S η ( µ ) − I α ( µ ) (5.8)can be related to the solutions to the family of problemsmaximize: S bη ( µ ) − c I α ( µ ) , (5.9)for any constants b, c > Lemma 5.2
Assume − d − . = α ∗ < α and assume that the measure µ is optimal for theproblem (5.8). Then, for any given constants b, c > , the measure ˜ µ = λ d − µ λ , λ . = (cid:16) bc (cid:17) α − d − (5.10) provides an optimal solution to the problem (5.9). Proof.
Given any measure µ , define ˜ µ by setting λ d − µ λ = ˜ µ . (5.11)By the rescaling formulas (5.3) and (5.7), one has S bη (˜ µ ) − c I α (˜ µ ) = S bη ( λ d − µ λ ) − c I α ( λ d − µ λ )= bλ d − S η ( µ ) − cλ α ( d − I α ( µ ) . (5.12)By the definition of λ in (5.10), it follows S bη (˜ µ ) − c I α (˜ µ ) = b α ( d − α − d − c − d α − d − (cid:16) S η ( µ ) − I α ( µ ) (cid:17) . (5.13)Therefore, S bη (˜ µ ) − c I α (˜ µ ) attains the maximum possible value if and only if S η ( µ ) − I α ( µ )attains the maximum possible value. This completes our proof.Our next result provides an estimate on the size of the support of the optimal measure µ . Proposition 5.3
In the same setting as Theorem 3.1, for any d ≥ and − d − < α < ,there is a constant C α,d such that any optimal measure µ for the problem (3.1) is supportedinside a ball of radius R ( µ ) ≤ C d,α (cid:16) k η k L c (cid:17) α − d − . (5.14) When α = 1 one simply has R ( µ ) ≤ k η k L c . (5.15)28 roof. 1. Consider first the special case where k η k L = c = 1. By (3.5)-(3.6) and (3.8), wethen have e C d,α (cid:16) I α ( µ ) (cid:17) d − α ( d − − I α ( µ ) ≥ , (5.16)where e C d,α is a constant which only depends on d and α . Therefore, in (3.9) one can take theconstant κ . = e C α ( d − α − d − d,α . (5.17) In the case 1 − d − < α <
1, we choose the radius r . = α αα − κ . (5.18)By (3.15)-(3.16), this yields | x | ≥ r = ⇒ α | x | α − χ ≥ α (cid:16) κ r (cid:17) − α ≥ . (5.19)By the argument following (3.18), the optimal measure is supported on the ball B r , wherethe radius r satisfies r − r = 1 . (5.20)Recalling (5.18) and (5.17), we obtain an upper bound on R ( µ ), namely R ( µ ) ≤ r = r + 1 = α αα − κ + 1 = α αα − e C α ( d − α − d − d,α + 1 . = C d,α . (5.21) To cover the general case, let b . = k η k L . Then we can write η = b ˜ η , where k ˜ η k L = 1. ByProposition 5.2, a measure ˜ µ is optimal for the problemmaximize: S ˜ η ( µ ) − I α ( µ )if and only if the measure µ . = λ d − ˜ µ λ , λ = (cid:18) bc (cid:19) α − d − (5.22)is optimal for the problem (3.1).Since k ˜ η k L = 1, by the previous step the measure ˜ µ is supported on a ball of radius R (˜ µ ) ≤ C d,α . In turn, by (5.22), the measure µ is supported on a ball of radius R ( µ ) = λR (˜ µ ) ≤ λ C d,α .This proves (5.14). When α = 1, the estimate (5.15) is an immediate consequence of (3.21).29 emark 5.4 The radius of the smallest ball containing the support of µ can be regarded asthe “size” of the tree. As expected, the above analysis indicates that the optimal size increaseswith the amount of sunlight k η k L , and decreases with the factor c multiplying the irrigationcost.Similar questions can be asked in connection with the optimization problem (OPR) for treeroots. More precisely, assume that the diffusion depends on a parameter σ >
0, and let afunction f : R + R be given, as in (2.15). Consider the optimization problemmaximize: H ( u, µ ) − c I α ( µ ) , (5.23)subject to: ( σ ∆ u + a f ( bu ) − uµ = 0 , x ∈ R d − ,u x d = 0 , x d = 0 . (5.24)Let µ be an optimal measure and call R ( µ ) the radius of the smallest closed ball, centered atthe origin, which contains the support of µ . We wish to understand how this radius dependson the parameters a, b, c , and σ .Throughout the following, we assume that d ≥ − d < α ≤
1, while f satisfies (A1).As a first step, we consider the problemmaximize: H (˜ u, ˜ µ ) − ˜ c I α (˜ µ ) , (5.25)subject to: ( ∆˜ u + f (˜ u ) − ˜ u ˜ µ = 0 , x ∈ R d − , ˜ u x d = 0 , x d = 0 , (5.26)and prove a rescaling result, similar to Proposition 5.2. Lemma 5.5
A couple (˜ u, ˜ µ ) is an optimal solution to (5.25)-(5.26) if and only if ( u, µ ) is anoptimal solution to (5.23)-(5.24), where u ( x ) . = 1 b ˜ u ( λ − x ) , µ . = σλ d − ˜ µ λ , λ . = r σab , ˜ c . = cσ α a λ − (1 − α ) d − α . (5.27) Proof. 1.
Let (˜ u, ˜ µ ) be an optimal solution to (5.26). For any test function e ϕ ∈ C c ( R d ), set ϕ ( x ) . = e ϕ ( λ − x ). By the definition of the rescaled measure ˜ µ λ in (5.2), one has Z ϕ d ˜ µ λ = Z e ϕ d ˜ µ . Therefore Z R d − σ ∇ u ( x ) ∇ ϕ ( x ) dx = σbλ Z R d − ∇ ˜ u ( λ − x ) ∇ e ϕ ( λ − x ) dx = σλ d − b Z R d − ∇ ˜ u ( y ) · ∇ e ϕ ( y ) dy , Z R d − a · f ( bu ( x )) ϕ ( x ) dx = Z R d − a f ( u ( λ − x )) e ϕ ( λ − x ) dx = aλ d Z R d − f ( u ( y )) e ϕ ( y ) dy , Z R d − u ( x ) ϕ ( x ) dµ = Z R d − σλ d − b ˜ u ( λ − x ) e ϕ ( λ − x ) d ˜ µ λ = σλ d − b Z R d − ˜ u ( y ) e ϕ ( y ) d ˜ µ . (5.28)30ince (˜ u, ˜ µ ) is a solution to (5.26), we have − Z R d − ∇ ˜ u ( y ) · ∇ e ϕ ( y ) dx + Z R d − f (˜ u ( y )) e ϕ ( y ) dx − Z R d − ˜ u ( y ) e ϕ ( y ) dµ = 0 . (5.29)Taking λ . = p σab , from (5.28)-(5.29) it follows − σ Z R d − ∇ u ( x ) · ∇ ϕ ( x ) dx + Z R d − a · f ( bu ( x )) ϕ ( x ) dx − Z R d − u ( x ) ϕ ( x ) dµ = 0 . (5.30)Since the test function ϕ is arbitrary, we conclude that ( u, µ ) is a solution to (5.24). We now claim that ( u, µ ) is actually a solution to the optimization problem (5.23)-(5.24).Indeed, given any measure ν , there is a unique measure ˜ ν such that ν = σλ d − ˜ ν λ , λ . = r σab . (5.31)By the preceding argument, if ˜ v is a solution to (5.26) corresponding to the measure ˜ ν , then v ( x ) . = b − ˜ v ( λ − x ) is a solution to (5.24) corresponding to the measure ν . Moreover, H ( v, ν ) = Z R d − a · f ( bv ( x )) dx = a Z R d − f (˜ v ( λ − x )) dx = aλ d H (˜ v, ˜ ν ) . (5.32)On the other hand, by (5.3) it follows c I α ( ν ) = c I α ( σλ d − ˜ ν λ ) = cσ α λ α ( d − I α (˜ ν ) . (5.33)By the choice of ˜ c in (5.27), from (5.32)-(5.33) we conclude H ( v, ν ) − c I α ( ν ) = aλ d (cid:16) H (˜ v, ˜ ν ) − ˜ c I α (˜ ν ) (cid:17) . (5.34)Therefore, H ( v, ν ) − c I α ( ν ) attains its maximum if and only if H (˜ v, ˜ ν ) − ˜ c I α (˜ ν ) attains itsmaximum. This completes the proof. Lemma 5.6
Let ( u, µ ) be an optimal solution for (4.1)-(4.2). Then there is a constant C ,depending on d , α and f , such that µ is supported inside the ball of radius R ( µ ) ≤ C · c − − (1 − α ) d . (5.35) In the special case where α = 1 , one has the simpler estimate R ( µ ) ≤ Mc . (5.36)
Proof. 1.
Assume that 1 − d < α <
1. By (4.40), there is a constant C , depending on d , α , and f , such that H ( u, µ ) − c I α ( µ ) ≤ C (cid:16) I α ( µ ) (cid:17) d αd − c I α ( µ ) . C (cid:16) I α ( µ ) (cid:17) d αd − c I α ( µ ) ≥
0, one obtains an a priori bound κ for the irrigationcost I α , namely I α ( µ ) ≤ κ . = C · c − αd α − d , (5.37)for some constant C depending on d , α and f .For any γ >
0, by same argument as in (3.15) we can find a radius r such that | x | ≥ r , = ⇒ α | x | α − χ ≥ γ . (5.38)Indeed, by (3.15) one can choose r . = α αα − γ α − α κ . (5.39)We now split µ = µ ♭ + µ ♯ as in (3.13), choosing r so that cγ ( r − r ) = M , r = r + Mcγ . (5.40)Using (3.18) and (5.38), we now obtain I α ( µ ♭ + µ ♯ ) ≥ I α ( µ ♭ ) + γ µ ♯ ( R d ) . By (5.37), (5.39), and (5.40) it follows r = C · γ α − α c αd α − d , r . = r + Mcγ = C · γ α − α c αd α − d + Mcγ . (5.41)To achieve the best estimate on the radius r , we minimize the right hand side of (5.41) overall possible choices of γ >
0. Taking γ . = (cid:16) MC (cid:17) − α c (1 − α ) d α − d , one obtains r = 2 M α · C − α c α − d = C · c − − (1 − α ) d , (5.42)where the constant C only depends on d , α and f . Next, if α = 1, by (3.20) one has I ( µ ♭ + µ ♯ ) − c (cid:16) I ( µ ♭ ) (cid:17) = c I ( µ ♯ ) ≥ crµ ♯ ( R ) . (5.43)On the other hand, by the assumptions in (A1), any solution u of (2.8) takes values inside[0 , M ]. Therefore, any measure containing some mass outside the ball B ρ , centered at theorigin with radius ρ = M/c , cannot be optimal.32ombining Lemmas 5.5 and 5.6, we now obtain an estimate on the support of the optimalmeasure for the general optimization problem for tree roots.
Proposition 5.7
Assume that d ≥ and − d < α ≤ , and let f satisfy the assumption ( A . Then there is a constant C only depending on the dimension d , α and f such that anyoptimal measure µ for the problem (5.23)-(5.24) is supported inside a ball of radius R ( µ ) ≤ C a − α − (1 − α ) d b − α − (1 − α ) d c − − (1 − α ) d . (5.44) When α = 1 one simply has R ( µ ) ≤ Mbc . (5.45)
Proof. 1.
Consider first the case α <
1. Let ( u, µ ) be an optimal solution to (5.23)-(5.24).Then by Lemma 5.5 the couple (˜ u, ˜ µ ) in (5.27) provides an optimal solution to (5.25)-(5.26).Using Lemma 5.6 and performing the variable transformations in (5.27), this yields R ( µ ) = λR (˜ µ ) ≤ r σab · C ˜ c − α − d = C r σab · " cσ α a (cid:16) σab (cid:17) − (1 − a ) d − α − − (1 − α ) d . (5.46)After some simplifications, from (5.46) one obtains precisely (5.44). Since every solution u of (5.24) satisfies 0 ≤ u ( x ) ≤ Mb , the estimate (5.45) is clear. Remark 5.8
By assumption, in (5.44) all denominators are positive: 1 + ( α − d >
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