A mechanical model of the human lung Benjamin Mauroy∗ and Nicolas Meunier† November 13, 2006

Abstract In this paper we construct a model to describe some aspects of the deformation of the bronchial tree composing the human lung considered as a continuous elastically deformable medium. To achieve this purpose, we study the interaction between an elastic tree and the fluid that goes through it. The key feature of our approach is the use of a fixed point theorem in order to find the optimal flow associated to a deformed tree. We also give some numerical results with interesting consequences on human lung deformation during expiration, particularly concerning the localization and progression of the equal pressure point (EPP).

1 Introduction The goal of this paper is to construct a simple and global mechanical model of the central region of the human lung in the case of small deformations. The bronchial tree of the human lung can be viewed as a dyadic net of pipes composed of 23 generations. More precisely, according to [5, 6, 7, 13], we can distinguish three parts in this tree. In the first part, mainly from the first generation to the fifth generation, there is some cartilage and the pipes can be assumed to be rigid. Moreover, the effects of inertia in the flow are large and correspond to the nonlinear Navier-Stokes regime. In the middle part, mainly from the sixth generation to the sixteenth generation, cartilage does not exist and the effects of inertia are smaller. This validates the Poiseuille regime (see [6, 5, 7, 3, 13]), at least for simulations at rest respiratory regime. In this part of the tree, there are interactions between the fluid and the walls of the pipes. In the last part, the tree function becomes different (beginning of gas exchange with blood). The plan of the paper is as follows. Assuming that some viscous fluid flows through a single pipe, our first step, in Section 2, consists in modeling the deformation mechanism of the pipe. Then, in Section 3, we consider a finite dyadic tree and express the deformation for the whole tree when air flows through it according to Poiseuille’s law. In Section 4, we build the optimal air flow distribution at the leaves of the tree that minimizes an energy functional corresponding to the dissipated viscous energy for a given tree. Then, in Section 5, using a fixed point theorem, we prove the existence of a deformation and of an air flow distribution such that this air flow applied to the tree minimizes the energy functional associated to this tree after deformation. In Section 6, we build a numerical scheme to compute the fixed point defined in Section 5 and we give some estimates to determine the convergence condition and convergence speed of the scheme. Finally, in Section 7, we present a few numerical simulations and applications of this theory to the human lung. ∗

Laboratoire MSC, Université Denis Diderot (Paris 7), 2 place Jussieu, building 33/34, 2nd floor, 75251 Paris Cedex 05, France. ([email protected]) † Laboratoire MAP5, Université René Descartes (Paris 5), 45 Rue des Saints Pères, 75006 Paris, France. ([email protected])

1

2 Branch deformation mechanism We consider in this section a viscous fluid which flows through a single elastic pipe. The pipe is characterized by its resistance which is the ratio of the pressure jump between its ends over the flux. Assuming that the branch stays cylindrical after deformation and that the constitutive law of the wall is linear, we build an elastic model of the branch deformation mechanism (which is nonlinear because of geometrical considerations) depending on the pressure jump between the inside and the outside of the branch. More precisely, in this section B is a cylindrical elastic pipe of radius r and of length L, in which an incompressible, viscous, non-inertial fluid flows. The pressure is supposed to be uniform over each end section. Moreover, the inlet is referred to a (for above) and the outlet to b (for below). The flow q going through B is chosen positive when the fluid goes from exit to entry. We assume that the branch is submitted to a uniform external pressure Pext . Remark 1. In the case of the human lung, since there exists an external strain which acts on all pipes through smooth muscles (with spiral shapes along the pipe wall), in a first approximation, the hypothesis that the pipes stay cylindrical can be assumed for those pipes located between the 6th and the 16th generations. Cylindrical coordinates (r, θ, z) will be used.

2.1 The Poiseuille law We first assume that the pipe is rigid so that there is no interaction between the fluid and the pipe. In such a case, the external forces acting on the fluid can be characterized by both values Pa and Pb . The linearity of the Stokes equations ensure the existence of a coefficient R > 0 which relates the flux q (considered positive if the fluid goes from b to a) and the pressure jump Pb − Pa Pb − Pa = Rq.

(2.1)

By analogy with electric conductors (flux and pressure respectively play roles of intensity and potential), R is called the resistance of the pipe. It depends on the geometrical characteristics of the pipe and on the viscosity µ of the fluid: L 8µL = C 4. (2.2) R= 4 πr r

2.2 Flow through an elastic pipe Let us now consider interactions between the fluid and the pipe. Let r(z) denote the equilibrium radius of a section of the branch. It depends on the position z on the axis [0, L]. We will make the hypothesis that the fluid flow is stationary (i.e. flux is not time dependent). Let us now detail the interaction between fluid and branch. Consider a small portion of a branch δB (see figure 1). The superficial lineic tension is given by a function t which depends on the radius r of the branch. In order to force the branch to stay cylindrical, we introduce the following tangential force Ft . It is tangent to the branch surface and the resulting force on δB is: dFt = t(r)τ (θ)L − t(r)τ (θ + dθ)L, where τ is the tangential vector.

2

n(θ)

Pext −t τ(θ+dθ)L

t τ(θ)L Pint(z)

dθ x

0 y z

L

Figure 1: Tangential forces orientations on an element dS of the branch surface. The external pressure force on a small area dS = r dθ dz is given by the external pressure times the surface, ie Pext dS. Its direction is normal to the surface (along vector n(θ)) and inward the center of the branch. On δB : Z L Z θ+ dθ Z θ+ dθ dFPext = − n(η) dη. Pext n(η)r dη dz = −Pext rL 0

θ

θ

The internal pressure Pint (z) at position z ∈ [0, L] is due to the flow q inside the branch and is given by Poiseuille relation (2.1) with (2.2): Pint (z) = Pa + C

z q. r4

Hence the internal pressure forces acting on the portion of branch is directed along n(θ) and from the center of the branch toward its surface: Z L Z θ+ dθ z (Pa + C 4 q)n(η)r dη dz dFPint = r 0 θ Z θ+ dθ Cq L2 = (Pa rL + 3 n(η) dη. ) r 2 θ Moreover, we have: Z

θ+ dθ

n(η) dη = −τ (θ + dθ) + τ (θ) = n(θ) dθ.

θ

Hence, if δB is in a quasi-static state: h CqL2 i n(θ). 0 = dFt + dFPext + dFPint = t(r)L + (Pa − Pext )rL + 2r 3

The equilibrium state of the branch is such that its radius is a positive root (if it exists) of the following equation: CqL = 0. (2.3) −t(r)r 3 + (Pa − Pext )r 4 + 2 Three specific branch radii corresponding to different values of pressure and flux appear through this model. We will use the following convention: 3

Definition 2.1. Unconstrained radius: Let P0 be a fixed pressure. We will denote by r 0 the branch radius such that t(r 0 ) = 0.

(2.4)

It is a solution of (2.3) when Pext = Pa = P0 and when there is no flow going through the branch, i.e. q = 0. Remark 2. For human lung, this state will correspond to the case of a dead body. The lung is then almost collapsed. Definition 2.2. Initial radius: We denote by r e the radius which corresponds to the branch geometry (i.e. solution of (2.3)) when 1 , P and P are given and such that P 1 the pressures Pext , Pext 0 a ext = Pext 6= P0 , Pa = P0 and when there is no flow inside the branch. In this case, this geometry r e satisfies the following equilibrium: 1 −t(r e ) + (P0 − Pext )r e = 0

(2.5)

In the sequel, it will be referred as an initial state. Definition 2.3. Final radius: 1 , P and P We denote by r a solution of equation (2.3), if it exists, when the pressures Pext , Pext 0 a 2 are given and satisfy Pa , Pext = Pext and the flux q is given, assumed to be non-negative. This situation corresponds to a deformed branch with a flow q inside. Since we are interested in the modelling 2 − P 1 > 0. It will be referred as a final state. of expiration, we assume that Pext ext The different radii of a branch B are linked together through mechanical equilibrium equations. This branch is assumed to have an unconstrained radius r 0 under a pressure P0 (see (2.4)). Then, we 1 and consider that there is no flow inside the branch. The radius modify the exterior pressure to Pext e of the pipe is r solution of (2.5). r e corresponds to Weibel’s data [13] and r 0 is calculated from r e 2 and to equilibrium equation (2.5). The last step consists in modifying the exterior pressure to Pext apply a non-negative flux q through the pipe. Hence, we obtain the final radius r. Furthermore, since we consider expiration, we assume that 2 1 Pext ≥ Pext .

(2.6)

Remark 3. In the sequel, we will consider the following linearly elastic law: r ˜ − 1), t(r) = E(w)( r0 ˜ where E(w) depends on the Young modulus and the width w of the branch. More precisely, the term ˜ E(w) is a lineic force and corresponds to the resultant of elasticity forces on a unit section of bronchi wall, hence this corresponds to Ew where E is the Young modulus, see Figure 2. According to data from [9], we use a linear dependence between bronchi radius and bronchi wall thickness of the type : w = τ r e . In [9], estimated values of τ are between 2/5 and 1/2. In the following we will use τ = 2/5, hence r 2 t(r) = Er e ( 0 − 1), 5 r

(2.7)

Although such a law is not realistic in the sense that t(r) does not tend to −∞ when r goes to zero (which should be the case in order to describe the fact that the branch cannot collapse in vivo), it is a good approximation in a first study. 4

w F=ES=Ew S x y

1

z

˜ Figure 2: The lineic elastic force E(w) acting on the gray section S can be written F = ES = E × w × 1, where E is the material Young modulus and w its thickness, F units are N/m. Hence ˜ E(w) = Ew. 1 + 2E , which we assume from now Remark 4. The existence of r e depends on the sign of P0 − Pext 5 e 0 on to be positive. Moreover, we assume that r ≥ r , which corresponds to 1 P0 ≥ Pext .

(2.8)

In the next paragraph, we will study hypothesis under which equation (2.3) admits a unique solution.

2.3 Study of equation (2.3) and definition of q → r(q): 1 , P 2 , P > 0 are given and satisfy (2.6), (2.8). Moreover, let In this section, we assume that Pext 0 ext max Pa be given and suppose that the inlet pressure satisfies Pa ∈]Pamin ; Pamax [. Furthermore, let r 0 > 0 and r e > 0 be given by (2.4) and (2.5) respectively with t(r) given by (2.7) and L = 6r e . (2.9)

Pamin ,

Recalling (2.5) and (2.7), equation (2.3) becomes: −

2E 5

2 − P1 ) + (P0 − Pa ) + (Pext ext 2Er e 5

r+1+

15Cq = 0, 2Er 3

(2.10)

which, for simplicity, we rewrite as follows: gα (r, q) = α with α = −1 −

q r + 1 + η 3, e r r

  2 − P1 ) 5 (P0 − Pa ) + (Pext ext 2E

and η =

(2.11)

15C > 0. 2E

(2.12)

Proposition 2.4. Let q max , αmin and αmax be given such that q max > 0 and αmin ≤ αmax < 0. Assume that q ∈ [0; q max ] and that α ∈]αmin ; αmax [, then gα (r, q) = 0, with gα (r, q) given by (2.11), admits a unique solution that we denote by r α (q). Moreover, α → r α (q) and q → r α (q) are increasing functions.

5

Proof. Let q ≥ 0 and α < 0 be fixed. Using the strictly decreasing character of the function r → gα (r, q) together with limr→0+ gα (r, q) = +∞ if q > 0 or limr→0+ gα (r, q) = 1 if q = 0 and limr→+∞ gα (r, q) = −∞, we obtain the existence and uniqueness of the root r α > 0. Let α1 , α2 ∈]αmin ; αmax [, be such that α1 < α2 . By definition, gα1 (r α1 , q) = gα2 (r α2 , q) = 0 α and gα2 (r α1 , q) = (α2 − α1 ) rre1 > 0, then we deduce the increasing character α → rα (q) from the decreasing character of r → gα (r, q). The increasing character of q → r α (q) is obtained similarly. Remark 5. Similarly, it can be shown that re → r α (q) is a decreasing function and that limre →0 r α (q) = +∞. This means that if there is a strictly positive flow in a branch with quasi zero radius, then the strain is very large and induces a very large dilatation. Remark 6. The second assumption αmax < 0 in the previous Proposition is satisfied when 2 1 Pamax < P0 + (Pext − Pext )+

2E . 5

(2.13)

In the following, we will assume that this condition is always satisfied. This corresponds to a value 2 and/or to values for E that are large enough. for Pext Remark 7. Under the assumptions of Proposition 2.4, we have min

0 < rα

max

(q min ) ≤ r α (q) ≤ r α min

In the sequel, we will simply denote r α

(q max ). max

(q min ) by r min and r α

(2.14)

(q max ) by r max .

Remark 8. The bound properties on r α (q) together with the regularity of gα prove that q → r α (q) is C ∞. Remark 9. We easily deduce from (2.14) together with (2.1) and (2.2) that Pbmin ≤ Pb ≤ Pbmax , with Pbmin = Pa + 6C

3 Finite tree

re r max


4 q

min

and Pbmax = Pa + 6C

(2.15) re r min


4 q

max

.

(2.16)

In this section, we consider a full dyadic tree and we investigate the deformation mechanism described in the previous section for the elastic pipes in which an incompressible, viscous, non-inertial fluid flows. We start with some notations for the tree, then for a rigid tree we state the relations between the fluxes at the leaves and the pressure in the branches. In this case, there is no interaction between the fluid and the tree. Then, we consider the case of an elastic tree. Note that the study of a rigid tree with constant radii for each generation was done in [3]. A more theoretical approach was also studied in [11].

6

3.1 Conventions on a tree: indexing and paths In the sequel, we will consider a finite dyadic three dimensional tree with N + 1 generations (of height N ). It will be denoted by TN . In such a tree, there are the root, 2N leaves, 2N +1 nodes and 2N +1 − 1 branches. We will denote by XN the set of the nodes and by BN the set of the branches. Moreover, nodes will be indexed by 0 for the root node and i ∈ {1, ..., 2N +1 − 1} for the other nodes with the convention that the two nodes steaming from Xi are X2i and X2i+1 , see Figure 3. Branches are indexed similarly with the convention that branch i ends at node i.

Xi X2i+1 X2i Figure 3: A four generation tree scheme (N = 3): the nodes are represented by the disks (total number 2N +1 = 16), while the branches are represented by lines (number 2N +1 − 1 = 15). The root node is the gray filled disk, the 2N = 8 leaf exits are the blank disks. Remark 10. Sometimes, it will be convenient to use a different indexing: 0 for the root node and (k, l) for the other nodes. In such a case, k ∈ {0, ..., N } indicates the generation number and l ∈ {0, ..., 2k − 1} is the position on the k-th generation. The relation between the two different indexings is: i = l + 2k . (3.1) For simplicity, when i is given, we denote by g(i) = k its generation number. With the previous conventions, we have: XN

= {X0 , (Xk,l )0≤k≤N,0≤l≤2k −1 } = {X0 , (Xi )1≤i≤2N+1 −1 },

BN

= {(Bk,l )0≤k≤N,0≤l≤2k −1 } = {(Bi )1≤i≤2N+1 −1 }.

Let us now define the notions of path and sub-path on TN . Definition 3.1. Let i ∈ {1, ..., 2N +1 − 1} be given, the set of the branches (resp. indices of branches) corresponding to the g(i) + 1 branches that link the root node to the i-th node is denoted by path(0 → i) (resp. Π0→i ). More precisely, Π0→i is the set of strictly increasing integers: Π0→i = {[

i i ] = 1, ..., [ ], i}, k 2 2

(3.2)

where [.] denotes the integer part. Moreover, assume that m ∈ {0, .., g(i)}, we denote by path(0 → i)(m) (resp. Π0→i (m)), the sub-path of path(0 → i) (resp. subset of Π0→i ) which corresponds to the branches (resp. indices of branches) starting at root node and ending at node j which is such that j ∈ path(0 → i) and g(j) = m: i i (3.3) Π0→i (m) = {[ g(i) ] = 1, ..., [ g(i)−m ]} 2 2 7

3.2 Flow through a rigid dyadic tree We consider an incompressible, viscous and non-inertial fluid which flows through a tree TN of connected pipes. Each pipe is characterized by its resistance see (2.1) and (2.2). Our first step, as in [3], consists in establishing the relation between pressures and fluxes at leaf exits. 3.2.1

Pressure, flux, resistance and radius associated to TN

In the sequel, we will denote by Proot the pressure at the root node which we assume to be nonnegative and we still denote by P0 a reference pressure that we also assume to be non-negative. Remark 11. In the case of the human lung, the pressure P0 is the atmospheric pressure and assumed to be zero. We denote by pTN (resp qTN , rTN and RTN ) the pressure vector (resp. flux, radius and resistance vectors) which components are the pressure (resp. flux, radius and resistance) on the nodes (resp. branches) of TN . Since the pressure at the root node is given by Proot , this can be written as: pTN = t (Proot , p0,0 , p1,0 , p1,1 , ..., pN,2N −2 , pN,2N −1 ) = t (Proot , p˜1 , p˜2 , ..., p˜2N+1 −1 ), qTN = t (q0,0 , q1,0 , q1,1 , ..., qN,2N −2 , qN,2N −1 ) = t (˜ q1 , q˜2 , ..., q˜2N+1 −1 ). The vectors rTN and RTN are defined similarly to qTN . Furthermore, the total resistance associated with path(0 → i) is: X ˜j . R Rpath(0→i) = RΠ = 0→i

j∈Π0→i

Remark 12. When no confusion arises, we will omit the subscript, i.e. r and R will denote the radius and resistance vectors on the tree respectively. 3.2.2

Relation between pressure and flux at leaf exits of TN

From now on, leaf exits (resp. leaves) of the tree will be indexed by 0, ..., 2N − 1. Moreover, the pressure (resp. flux) vector at leaf exits will simply be denoted by p (resp q) with: p = t (p0 , ..., p2N −1 ) q = t (q0 , ..., q2N −1 ). Remark 13. Recalling notations introduced in the previous section we have p = t (pN,0 , ..., pN,2N −1 ). Let us first give a definition, see [3], which will be useful in the sequel. Definition 3.2. Given two positive integers i and j and their binary expansions i=

∞ X k=0

αk 2k , j =

∞ X

βk 2k , with αk , βk ∈ {0, 1}, ∀k,

k=0

we define νi,j as νi,j = inf{k ≥ 0, αl = βl ∀l ≥ k}. We now state the relation between pressure and flux at leaf exits.

8

(3.4)

Proposition 3.3. We consider a full dyadic tree TN characterized by its radius r and its resistance R. Supposing that the root node is at pressure 0, then pressures and fluxes at leaf exits are related by p = B N (r)q, B N (r) = (B N (r)i,j )0≤i,j≤2N −1 ∈ M2N (R), with B N (r)i,j = RΠ0→i+2N (N −νi,j ) .

(3.5)

Remark 14. B N (r) is a symmetric matrix. Remark 15. When the pressure at the root node is Proot > 0, the relation between pressures and fluxes at leaf exits is obtained by adding Proot to the pressure in the previous Proposition. Definition 3.4. We denote by B N the set of matrices B N (r) of M2N (R) which are given by (3.5). The proof of Proposition 3.3 is similar to the proof of Proposition 1.2 of [3]. Remark 16. From now on we will use the convention that a vector (a0 , ..., a2N −1 ) is such that ai is at position i. Proof. Using the linearity of the relation between q and p, which is obvious, it is sufficient to compute the pressure vectors associated to the elements of the canonical basis of R2N . Let us first consider a flux vector q = (1, 0, 0, ..., 0). This corresponds to the situation where some fluid flows through leaf 0 (node XN,0 ), and only through the latter. By conservation, the flux which exits the domain through the root is exactly 1, so that the pressure at node X0,0 is R0,0 . As there is no flow in the right-hand subtree stemming from X0,0 , the pressure at its leaves (leaves with indices between 2N −1 and 2N − 1) is exactly R0,0 . Similarly, the pressure at node X1,0 is R0,0 + R1,0 = RΠ0→2N (1) and so is the pressure at the leaf of the right-hand subtree stemming from X1,0 . Following this approach recursively, we find pressure at nodes X2,0 , X3,0 ,..., XN,0 to be RΠ0→2N (2) , RΠ0→2N (3) ,..., RΠ0→2N (N ) respectively. Consequently, the pressure at leaf j ∈ {0, ..., 2N − 1} is RΠ0→2N (N −ν0,j ) . The same argument can be applied to any vector q = (0, ..., 0, 1, 0, ..., 0) (with the 1 at position i), to which corresponds the pressure field (RΠ0→i+2N (N −νi,0 ) , RΠ0→i+2N (N −νi,1 ) , ..., RΠ0→i+2N (N −νi,2N −1 ) ). Consequently, we obtain pi =

N −1 2X

qj RΠ0→i+2N (N −νi,j ) .

j=0

Remark 17. We can also express the previous relation as follows: p = B N q,

9

(3.6)

with B

N

=

R0,0 I0N

+



R1,0 I1N 0

0 R1,1 I1N



 0 0 0 R2,0 I2N   0 0 0 R2,1 I2N  + N   0 0 0 R2,2 I2 N 0 0 0 R2,3 I2   RN,0 0 ... 0  0  RN,1 0 0 , +... +   ...  ... ... ... 0 0 0 RN,2N −1 

where 0 is used for 0IkN with IkN ∈ M2N−k (R) is a matrix of ones. Remark 18. We also have X

pi = p˜i+2N =

˜ j q˜j . R

(3.7)

j∈Π0→i+2N ,

Remark 19. When the tree TN is homogeneous, that is to say characterized by its generation-wise resistances R0 , R1 , ..., RN , the relation between pressures and fluxes at leaf exits is given by, see [3]: N p = AN q, AN = (AN i,j )0≤i,j≤2N −1 ∈ M2N (R), Ai,j = SN −νi,j ,

(3.8)

where Sn is the cumulative resistance R0 + R1 + ... + Rn . The matrix AN was studied in [3]. Up to a multiplicative constant, it is a doubly stochastic matrix which admits the Haar basis as eigenvector basis. The next result easily follows from the semi-definite positive character of IkN , 0 ≤ k ≤ N − 1, together with the positive definite character of   RN,0 0 ... 0  0  RN,1 0 0    ...  ... ... ... 0 0 0 RN,2N −1 Proposition 3.5. The matrix B N (r) is positive definite and its eigenvalues λ0 , ..., λ2N −1 satisfy: min

λi ≥ min(RN,0 , RN,1 , ..., RN,2N −1 ),

max

λi ≤ 2N R0,0 + 2N −1 max(R1,0 , R1,1 ) + ...

i∈{0,...,2N −1} i∈{0,...,2N −1}

+ max(RN,0 , RN,1 , ..., RN,2N −1 ). The following result will be useful in the sequel. Let J = t (1, ..., 1). Lemma 3.6. The matrix B N (r) belongs to GL2N (R) and   B N (r)−1 J > 0, ∀i ∈ {0, ..., 2N − 1}. i

10

Proof. Since B N (r) is a positive definite matrix, it belongs to GL2N (R). Let (B N (r))−1 J = t (β0 , ..., β2N −1 ). From B N (r)(B N (r))−1 J = J together with the definition 3.3 of B N (r), we easily deduce that     B N (r)(B N (r))−1 J − B N (r)(B N (r))−1 J = R2,0 β0 − R2,1 β1 = 0, 0

1

hence β0 β1 ≥ 0 and β0 β1 = 0 if and only if β0 = β1 = 0. Similarly, we obtain that β2i β2i+1 ≥ 0 and β2i β2i+1 = 0 if and only if β2i = β2i+1 = 0 for all i ∈ {0, ..., 2N −1 − 1}. Moreover, we also have     B N (r)(B N (r))−1 J − B N (r)(B N (r))−1 J = 0

2

R2,0 β0 − R3,1 β2 + R1,0 (β0 + β1 ) − R1,1 (β2 + β3 ) = 0.

Using next that β1 = 

R2,0 R2,1 β0

and β3 =

R1,0 + R2,0 +

R2,2 R2,3 β2 ,

we obtain

   R2,2 R1,0 R2,0 β0 = R1,1 + R2,2 + R1,1 β2 , R2,1 R3,3

hence β0 β2 ≥ 0 and β0 β2 = 0 if and only if β0 = β2 = 0. Similarly, we obtain that β2i β2i+2 ≥ 0 and β2i β2i+2 = 0 if and only if β2i = β2i+2 = 0 for all i ∈ {0, ..., 2N −1 − 2}. Hence, following this approach recursively, from     B N (r)(B N (r))−1 J − B N (r)(B N (r))−1 J = 0, i

j

we deduce that all the β have the same sign and if one of them vanishes, so do all of the others. The latter case is not possible since B N (r)t (β0 , ..., β2N −1 ) = J. Moreover from B N (r)t (β0 , ..., β2N −1 ) = J again, we deduce that βi > 0. 3.2.3

Relation between pressure of TN and flux at leaf exits

In this section, we state the relations between the flux (the pressure) that crosses the branch (at node) i and the fluxes at the leaves. Proposition 3.7. We consider a full dyadic tree of height N , TN , with given fluxes q at the leaves. The flux in branch (k, l) (resp i = l + 2k ) of the tree and fluxes at the leaves are related by: qk,l = q˜i = t Jl,2N−k q,

(3.9) N

where (Jl,2N−k )0≤k≤N,0≤l≤2k −1 is the family of vectors of R2 defined by: Jl,2N−k = t (0, ..., 0, 1, ..., 1, 0, ..., 0), where the first 1 is at position l2N −k and which has 2N −k non zero terms (ones indeed). Note that J = t (1, ..., 1) = J0,2N . Moreover, ∂pk,l = RΠ (k−νl,[ i ] ) , i [ ] ∂qi 2N−k 0→l+2 2N−k for i ∈ {0, ..., 2N − 1}, 0 < k ≤ N and l ∈ {0, ..., 2k − 1}. 11

Proof. Let q = (q0 , ..., q2N −1 ), and let us consider the branch (k, l) with k ∈ {0, ..., N } and l ∈ {0, ..., 2k − 1}. The indices of the leaves of the subtree stemming from Xk,l are between l2N −k and (l + 1)2N −k − 1. Hence by conservation, the flux that exits branch (k, l) is exactly qk,l = P(l+1)2N−k −1 qi = t Jl,2N−k q. Let us consider the subtree Tk of TN , with 0 < k ≤ N , starting at the i=l2N−k root node in which fluxes at the leaves are qk = {qk,0 , ..., qk,2N−k −1 }. Applying next Proposition 3.3 on this subtree, we obtain k pk = B k qk , B k = (Bi,j )0≤i,j≤2k −1 ∈ M2k (R), with k Bi,j = RΠ0→i+2k (k−νi,j ) ,

where pk = t (pk,0 , ..., pk,2k −1 ). The result then follows from the expression of qk . 3.2.4

Path matrix subset M2N+1 −1 of GL2N+1 −1 (R)

In this section, we state some properties of matrices belonging to a particular set M2N+1 −1 . This will be useful in the sequel to study the convergence speed of an iterative process, see below Section 6. Definition 3.8. We denote by M2N+1 −1 the subset of lower triangular matrices M of GL2N+1 −1 (R) that satisfy Mij = 0 if j ∈ / Π0→i , for i, j ∈ {1, ..., 2N +1 − 1}. For example, the gradient matrix ∇qTN pTN (qTN ) of the application qTN → pTN (qTN ) belongs to M2N . We have then the following properties: Proposition 3.9. If A ∈ M2N+1 −1 then A−1 ∈ M2N+1 −1 . Actually, (M2N+1 −1 , ×) is a subgroup of GL2N+1 −1 (R). Moreover the inverse of an element in M2N+1 −1 can be obtained through an iterative process: Proposition 3.10. Let A = (aij ) ∈ M2N+1 −1 then its inverse B = (bij ) ∈ M2N+1 −1 is such that:     X −1  aik bkj  for j ∈ Π0→i , j 6= i bij = aii k∈Π0→i \Π0→j ,k6=i

and

bii = 1/aii for i ∈ {1, ..., 2N +1 − 1}. Remark 20. With the previous formula, the inverse B can be built column-wise, from the beginning of the column to the end (i.e., we first calculate b1i , then b2i , etc). Moreover, this leads to an easy way to numerically inverse matrix A. Proof. We easily see that B belongs to M2N+1 −1 and that AB = I2N+1 −1 .

12

Now we obtain an upper bound for the coefficients of A−1 depending on boundary properties on the coefficients of the matrix A in M2N+1 −1 . Let A = (aij ) be given in M2N+1 −1 . We assume uniform boundedness conditions: ∀i 6= j, |aij | ≤ α and − aii ≥ β > 0.

(3.10)

Recalling Propositions 3.9 and 3.10, we want to obtain an upper bound for A−1 = (bij ). First let us introduce the following real sequence:  u1 = 1/β P un+1 = αβ np=1 up

Lemma 3.11. This sequence can be rewritten for n ≥ 2:   α n−2 α un = 1 + β β2 Proof. We have un+1 − un = (α/β)un for n ≥ 2, hence using u2 = α/β 2 gives the result. Proposition 3.12. Let i, j be in {1, ..., 2N +1 − 1} such that i ≥ j. According to the branch numbering, this implies that the corresponding generations of the branch i and j verify g(i) ≥ g(j). With this hypothesis, we have: |bij | ≤ u(g(i)−g(j)+1) Proof. If path(j → i) = ∅ then bij = 0 and the inequality is true. Now assume path(j → i) 6= ∅. According to the definition of bij and the bound hypothesis on αij , we can write if i 6= j: X α |bkj | |bij | ≤ β k∈path(j→i),k6=i

and of course bii ≤ 1/β. The important point is that the set of bkj corresponds to the branches linking branch j of generation g(j) branch i of generation g(i). Hence there is exactly one k in the sum for each generation in {g(j), g(j) + 1, ..., g(i) − 1} (recall that we suppose k 6= i). Hence, there are exactly g(i) − g(j) terms in the sum. This also shows that the upper bound of bij only depends of bkj which have smaller generations and consequently such that their indexes k verify k < i. Hence, we will use a recursive proof indexed by the difference between generations g(i) − g(j). Recall that we assume path(j → i) 6= ∅. The n ranked induction hypothesis is : "if g(i) − g(j) ≤ n then bij ≤ ug(i)−g(j)+1 ". Assume first g(i) − g(j) = 0, this means i = j and |bii | =

1 1 ≤ = u1 |aii | β

This is true at rank 0. Now assume g(i) − g(j) = n + 1 and assume true the n ranked induction hypothesis, then: X α |bij | ≤ |bkj | β k∈path(j→i),k6=i

13

But we recall that if k is different from i and belongs to the set path(j → i), then its associated generation is smaller than i, hence g(k) − g(j) < g(i) − g(j) and g(k) − g(j) ≤ n. Moreover such k, (i.e. different from i and belonging to the set path(j → i)), cover each generations between g(j) and g(i) − 1. Consequently for each p in {g(j), ..., g(i) − 1} there exists a unique kp in the sum such that g(kp ) = p. According to the n ranked induction hypothesis we have: |bkp j | ≤ ug(kp )−g(j)+1 Now putting this in bij : g(i)−1 g(i)−1 α X α X |bkp j | ≤ ug(kp )−g(j)+1 |bij | ≤ β β p=g(j)

which leads to: α |bij | ≤ β

p=g(j)

g(i)−g(j)

X p=1

n+1

αX up = up = un+2 β p=1

This shows the n + 1 ranked induction hypothesis and the result is true for every n ∈ N∗ . Now if the tree has N generations and because n → un is increasing, we have:  N −2 |bij | ≤ 1 + αβ |bii | ≤

α β2

for i 6= j

1 β

and denoting by |||.|||2 the matrix norm subordinate to the euclidean norm (i.e. if M = (mij ) then X||2 |||M |||2 = supX6=0 ||M ||X||2 ) we have: Proposition 3.13. Let A belong to M2N+1 −1 , then −1

|||A

N +1

|||2 ≤ (2

"

α − 1) max 2 β



α 1+ β

N −2

1 , β

#

3.3 Tree deformation mechanism We are interested in the modeling of air flow in the bronchial tree and we focus on the expiration phase. In such a phase, since the pressure at the root node Proot is assumed to be non-negative, the pressures on the whole tree p˜i are also non-negative. More precisely, going from the root of the tree to the leaves, the pressures on a path are increasing. This is a consequence of the assumption that the leaf’s fluxes are non-negative. Hence, the fluxes on the whole tree are also non-negative. N

Definition 3.14. We say that a leaf flux vector q ∈ R2 is ǫ-admissible when it satisfies t Jq = Φ and qi > ǫ, ∀i ∈ {0, ..., 2N − 1}.

(3.11)

We denote by Ωǫ the set of ǫ-admissible leaf flux vector. Moreover, we denote Ω = Ω0 and q ∈ Ω is simply called admissible. The following lemma easily follows from the Kirchoff’s law for the fluxes on the tree. 14

N

Lemma 3.15. Let Φ > 0 be fixed and q ∈ R2 be an ǫ-admissible leaf flux, then for all i ∈ {1, ..., 2N +1 − 1}, 2N −j ǫ < q˜i < Φ − (2j − 1) 2N −j ǫ, (3.12) where j = g(i) ∈ {1, ..., N } is the generation number of i. Remark 21. In the sequel, we will use the following notations: q˜jmin = 2N −j ǫ and q˜jmax = Φ − (2j − 1) 2N −j ǫ, with ǫ depending on Φ well chosen, see Proposition 5.3 below. Let us now study the case of an elastic dyadic tree TN in which flows an incompressible, viscous and non-inertial fluid with a given total flux Φ. For every pipe of the net, we assume that the branch mechanism is similar to the one described in Section 2, that the fluid follows Poiseuille’s law and that it flows in a stationary way. Similarly to Section 2, we consider three different states of the tree: unconstrained tree, initial tree and final tree, when every pipe of the tree has a radius which is respectively unconstrained, initial and final state. We neglect the gravity and the exterior pressure is assumed to be uniform all around 1 and P 2 the exterior pressures associated with the initial and final state the tree. We denote by Pext ext 1 . 2 respectively with Pext ≥ Pext More precisely, the unconstrained tree, which is denoted by TN0 is such that its radius vector, N+1 denoted by r0 ∈ R2 −1 , satisfies: (3.13) ti (˜ ri0 ) = 0, on every branch Bi , i ∈ {1, ..., 2N +1 − 1}, with 2 e r r ( − 1), ti (r) = E˜ 5 i r˜i0

(3.14)

The initial tree, which is denoted by TNe is such that its radius, denoted by re satisfies on every branch: 1 −ti (˜ rie ) + (P0 − Pext )˜ rie = 0. (3.15) Finally, let q ∈ Ω be given, the final tree, which is denoted by TN is such its radius vector, denoted by r, is a solution (if it exists), for all i ≥ 1, of the following equation: gi,αi (r, q) = 0, with gi,αi (r, q) = αi and αi = −1 −

(3.16)

q˜i r˜i + 1 + η 3, e r˜i r˜i

  2 − P1 ) 5 (P0 − p˜[ i ] ) + (Pext ext 2

2E

(3.17)

and η =

15C , 2

(3.18)

with the convention that p˜0 = Proot . Next, we define the tree deformation function G by: N+1 −1

G : R2

N

× [0, +∞[2

N+1 −1

→ R2

(r, q) → (Gi (r, q) = gi,αi (r, q))1≤i≤2N+1 −1 , where gi,αi (r, q) is given by (3.17). This is obviously a C ∞ -function.

15

(3.19)

P2N+1 −1 Remark 22. The matrix ∇r G, is a sparse matrix which has at most k=0 (k + 1)2k non vanishing k terms (it has at most (k + 1)2 non vanishing terms on line k). Moreover ∇r G belongs to M. N

2 − P 1 ) + 2E and that q ∈ [0, +∞[2 is such that Proposition 3.16. Assume that Proot < P0 + (Pext ext 5 t Jq = Φ. There exists a ∈ RN such that the following holds : if T is a tree verifying for all i ∈ N +,∗ < 0 such that < αmax {1, ..., 2N +1 −1−2N }, r˜ie < aj if j = g(i), then there exists real numbers αmin 1 1 αi α max min ri (˜ qi ))i∈{1,...,2N+1 −1} , for all α1 ∈]α1 ; α1 [, there exists a unique solution, denoted by r (q) = (˜ of G(r, q) = 0 with G given by (3.19). Furthermore, there exists strictly positive real numbers (˜ rjmin , r˜jmax )1≤j≤N such that for all i in {1, ..., 2N +1 − 1}:

r˜jmin ≤ r˜iαi (˜ qi ) ≤ r˜jmax ,

(3.20)

with j = g(i). Proof. We will proceed recursively on the generations. Let TN be a tree of height N . Recalling 2 − Proposition 2.4, condition (2.13) with Pa = Proot is exactly the hypothesis Proot < P0 + (Pext 2E max min 1 < 0 such that for all < α1 Pext ) + 5 , hence we know that there exists real numbers α1 α1 min max α1 ∈]α1 ; α1 [, there exists a unique solution, denoted by r1 (q), of g1,α1 (r, q) = 0. Moreover, recalling Remark 9, we deduce that p˜min ≤ p˜1 ≤ p˜max (3.21) 1 1 where p˜min = Proot + 6C 1 with

r˜1e

min


4 q˜1 r˜1max αmin

r˜1α1 ,min (q) = r˜1 1

and p˜max = Proot + 6C 1 αmax

(Φ) and r˜1α1 ,max (q) = r˜1 1

r˜1e

(3.22)

(Φ).

(3.23)

max
4 q˜1 , min r˜1

< 0 and such that αmax and αmax Next, we want to obtain αmin 2 2 2 αmin ≤ α2 ≤ αmax along with αmin ≤ α3 ≤ αmax 2 2 2 2

(3.24)

where αmin = −1 − 2 αmax = −1 − 2

  2 − P1 ) 5 (P0 − p˜min ) + (P ext ext 1

2E   2 − P1 ) max 5 (P0 − p˜1 ) + (Pext ext

,

2E

2 − P1 ) + < P0 + (Pext < 0 is equivalent to p˜max Inequality αmax ext 1 2 possible :

. 2E 5 .

Two situations are

• The inequality is verified, we choose a1 = r1e and we can go on to the next step. • The inequality is false, then working with a smaller r˜1e leads to a larger radius r1α1 (q) (because re → r α (q) is a decreasing function in Proposition 2.4, see Remark 5). Smaller r˜1e and larger . Because r1α (q) goes to infinity with r1e going r1α1 (q) corresponds to a smaller pressure p˜max 1 e max to zero, p˜1 goes to Proot when r˜1 goes to 0, hence there exists a1 > 0 such that if r˜1e < a1 2 − P 1 ) + 2E . then p˜max < P0 + (Pext ext 1 5

16

Note that the second situation can be reproduced in the downward parts of the tree because the pressure in one branch only depends on what happens between this branch and the root of the tree. Hence, the next steps, modifying only downward branches in the tree, will not modify the properties (pressure, flow or radius) of the current branch. Hence, reproducing this scheme in the next generations of the tree leads to a = (ai )i=0,...,N−1 such that if, for all i verifying g(i) = j, r˜ie < aj , then 2 − P 1 ) + 2E and αmax < 0. Finally, note that no similar condition need to be p˜max < P0 + (Pext ext j j 5 imposed to the last 2N generation branches. Hence, following this approach recursively, we obtain the result. max ) Remark 23. In Proposition 3.16, the real numbers (αmin 1≤k≤N are constructed in such a way k , αk N +1 that for all i in {1, ..., 2 − 1}:

αmin ≤ αi ≤ αmax ≤ 0, j j

(3.25)

with j = g(i). N

Remark 24. Since the roots are bounded, it is easy to see that q → rα (q) is C ∞ for q ∈ [0, Φ]2 , see remark 8. The previous Proposition leads to the definition of the function q → rα (q). Definition 3.17. Under the same assumptions as Proposition 3.16, we define the C ∞ mapping rα as follows: α

r

2N

: [0; Φ]

→

2N+1 Y−1

j

]˜ rjmin ; r˜jmax [2

j=1

q → rα (q) such that G(rα (q), q) = 0.

< 0 are fixed and we will < αmax From now on, we will assume that the real numbers αmin 1 1 α simply note r = r .

4 Viscous energy minimization Let TN be given and B N (r) be the resistance matrix associated to it by Proposition 3.3 and Definition 3.4. We recall that B N (r) is a symmetric matrix of M2N (R). Let us denote by ED the viscous dissipated energy of the tree. It is a function of the flux vector q at leaves, and it is the sum of the viscous dissipated energy in each branch of the tree (for the branch ˜ i q˜2 ). It is easy to prove that the total viscous dissipated energy in the tree is i, this loss of energy is R i given by ED (q) = t qB N (r)q Assuming that the flow Φ going through the first generation branch (the root node or the “trachea” depending on which part of the human lung we consider) is given, we want to minimize ED over all N fluxes q ∈ [0; Φ]2 such that F (q) = t Jq = Φ, with J = t (1, 1, ..., 1). 17

Now using Lagrange multipliers, at an extremum q0 , we have ∇ED (q0 ) = λ∇F (q0 ), N

hence, for all h in R2 , 2t q0 B N (r)h = λt Jh. Therefore, we have: 2B N (r)q0 = λJ t Jq0 = Φ, whence q0 = λ2 (B N (r))−1 J and Φ =

λ t N −1 2 J(B (r)) J,

q0 =

this gives λ = 2Φ/t J(B N (r))−1 J and

(B N (r))−1 J ×Φ t J(B N (r))−1 J

Remark 25. In particular, the optimal flow q0 is the image of an homogeneous distribution of pressures at exits equal to Φ/(t JB N (r)−1 J) (note that the term (t JB N (r)−1 J)−1 represents the equivalent hydrodynamic resistance of the whole tree). Remark 26. Note that if TN is homogeneous, see Remark 19, J is an eigenvector of A and q0 = Φ/2N J. Let us now define a flux optimization mapping as follows: Definition 4.1. We define the mapping Q by: N

→ R2 Q : S2+∗ N (R)

A → Q(A) =

A−1 J Φ, t JA−1 J

Definition 4.2. Let TN be given and B N (r) be its resistance matrix by Proposition 3.3 and Definition 3.4, the mapping f is defined as follows: N+1 −1

f : R∗+ 2

N

→ R2


(B N (r))−1 J Φ, r → f (r) = q0 = Q o B N (r) = t J(B N (r))−1 J N

We recall that Ω = {q ∈]0; Φ[2 such that t Jq = Φ}. Proposition 4.3. The mapping f satisfies Im(f ) ⊂ Ω.

Proof. Since t Jf (r) = Φ, it is enough to prove that f (q)i > 0, for all i ∈ {0, ..., 2N − 1}. This easily follows from Lemma 3.6.

5 Optimization flux for a deformable tree In this section, we state our main result. We consider an elastic dyadic tree TN with given radii r0 and re , we prove that under some assumptions on α1 (r, q), there exists an optimal flux q ∈ Ω for the deformed tree of radius r(q). Let us first introduce a new function.

18

Definition 5.1. We denote by F the application defined by F : Ω → Ω
−1 J B N (r(q)) q → F (q) =
−1 Φ = f ◦ r(q), t J B N (r(q)) J

where we recall that r(q) is given by Definition 3.17.

Proposition 5.2. Under the same assumptions as Proposition 3.16, there exists q in Ω such that F (q) = q. Proof. Since Im(F ) ⊂ Ω ⊂ Ω and Ω is a compact and convex set and because F is continuous, we deduce the existence of qF ∈ Ω such that F (qF ) = qF , from the Brouwer fixed point theorem. Because Im(F ) ⊂ Ω, it follows that qF ∈ Ω. It is however possible to obtain a better localization of the fixed point qf using the fact that for each tree TN there exists ǫTN > 0 such that Im(F ) ⊂ ΩǫTN . This property is a consequence of the limitation of deformation range of radii. Actually, and because the flow are bounded by Φ and positive, the tree branches cannot collapse (zero radius) or infinitely dilate. Proposition 5.3. Under the same assumptions as Proposition 3.16, there exists ǫTN > 0 such that Im(F ) ⊂ ΩǫTN . Proof. Let us define the application m : Ω →]0; Φ[ such that m(q) = mini qi . The application m is continuous on Ω along with F on Ω. Hence the applications m ◦ F is continuous on the compact Ω. Then m ◦ F reaches its minimum η in Ω and because F (q)i > 0 for each q ∈ Ω, η > 0. Taking ǫTN = η/2 leads to the result by definition, because F (q)i > ǫTN for each q ∈ Ω. Finally, the following result holds: Proposition 5.4. Under the same assumptions as Proposition 3.16, the restriction F|ΩǫT ΩǫTN admits a flow qf ∈ ΩǫTN such that F|ΩǫT (qf ) = qf .

of F on

N

N

Proof. Using ΩǫTN in the same way than proposition 5.2 leads to the result. Remark 27. According to Remark 21, we will use ǫTN as ǫ to define the different qjmin and qjmax .

6 Iterative process In the previous section, we found a fixed point of F . However, we did not prove uniqueness nor supplied a constructive method. In this part, we will prove that under more restrictive hypothesis, the Picard fixed point theorem can be applied. In the following, we will assume that Proposition 3.16 is verified. Let us begin with the convergence and convergence speed of the iteration scheme defined by: q0 ∈ Ω, q1 = F (q0 ), q2 = F (q1 ), ... To do so, we look for a constant 0 < C < 1 such that |F (q2 ) − F (q1 )| ≤ C|q2 − q1 | 19

(6.1)

More precisely, we will prove that ∇q F is bounded and that its bounds can be adjusted thanks to models parameters in order to apply Picard Theorem. First, recalling that F = f ◦ r, from the chain rule together with G(r, q) = 0, it follows that : h i ∇q F (q).dq = ∇r f (r(q)).∇q r(q).dq = −∇r f (r(q)). [∇r G(r(q), q)] −1 .∇q G(r(q), q).dq (6.2) Remark 28. The gradient of r is ∇q r(q) = [∇r G(r(q), q)] −1 × ∇q G(r(q), q). The expression of ∇q F leads us to study the three gradients of the right-hand side of equation 6.2.

6.1 Gradients In this part, we calculate the three gradients of the right-hand side of equation 6.2. 6.1.1

Calculation of ∇r G

Let us start with the matrix ∇r G. First, we recall that for all i ∈ {1, ..., 2N +1 − 1}: p˜i = Proot +

X 3C r˜je

j∈Π0→i

r˜j4

q˜j ,

hence (r, q) → αi (r, q), given by (3.18), has zero derivatives Furthermore, we have:

(6.3) ∂αi ∂rj (r, q)

if j ∈ / Π0→i or j = i.

Proposition 6.1. The matrix ∇r G is triangular. It belongs to GL2N+1 −1 (R) and is given by:  αi (r,q)  − 3ηr˜4q˜i if j = i  r˜ie  i ∂Gi 30C r˜je r˜i (r, q) = q ˜ if j ∈ Π0→i , j 6= i − E r˜ie r˜j5 j  ∂rj   0 elsewhere.

Proof. The fact that

∂Gi ∂ri (r, q)

6= 0 comes from the assumption αi (r, q) < 0 together with q˜i ≥ 0.

Remark 29. ∇r G is triangular and belongs to the set M2N+1 −1 studied in Section 3.2.4. 6.1.2

Calculation of ∇q G

To evaluate the gradient of G at q, it is important to note that for all j in {1, ..., 2N +1 − 1}: X q˜j = qk , k s.t. j∈Π0→2N +k

hence:

∂ q˜j = ∂qk



1 0

if j ∈ Π0→2N +k else

where we recall that (qk )0≤k≤2N −1 are the leaf fluxes, while (˜ qj )1≤j≤2N+1 −1 are the fluxes on the tree with the convention that q˜2N +k = qk . 20

Obviously, from (6.3), we deduce that: X 6C r˜je ∂ q˜j ∂ p˜i (r, q) = = ∂qk r˜j4 ∂qk j∈Π0→i

6C r˜je

X

j∈Π0→i ∩Π0→k

r˜j4

.

(6.4)

Moreover, recalling the expression of G, we have ∂Gi η ∂ q˜i ∂αi r˜i + 3 = , e ∂qk ∂qk r˜i r˜i ∂qk hence, using the expression of αi , we obtain  ∂Gi 15C r˜i  =  ∂qk E˜ rei

6.1.3

X

j∈Π0→[ i ] ∩Π0→k 2



r˜je   r˜j4

+

η ∂ q˜i . r˜i3 ∂qk

(6.5)

Calculation of ∇r f

Note that f can be decomposed as f (r) = Q ◦ B N (r) with Q(A) =

A−1 J Φ for A ∈ S2+∗ N t JA−1 J

and B N (r) being the matrix associated to a tree TN with an r distribution of radii. A simple calculation gives the differential of Q:  t −1  JA HA−1 J −1 A−1 HA−1 A − DA Q(A).H = JΦ. t JA−1 J (t JA−1 J)2

(6.6)

The differential of the application r → B N (r) is easy to calculate, because every coefficient (i, j) ˜ k = 6Cr e /r 4 with k in a subset Ni,j of {1, ..., 2N +1 − 1}. of B N (r) is a sum of resistance terms R k k Ni,j has the property that if k, l ∈ Ni,j then g(k) 6= g(l) (hence there is a maximum of N terms, reached on diagonal), see Proposition 3.3 and Definition 3.4). Then, we can write for all (i, j) in {1, ..., 2N +1 − 1}: X 6C r˜e k B N (r)i,j = r˜k4 k∈Ni,j

and consequently : X −24Cr k
e hk . ∇r B N (r).h i,j = 5 rk

(6.7)

k∈Ni,j

Finally, the chain rule yields:

6.2 Estimates

  ∇r f (r).h = DA Q(B N (r)). ∇r B N (r).h .

In this part we give the estimates of the three gradients of the right-hand side of equation 6.2.

21

6.2.1

∇r G : evaluation of α and β

Recalling the definition of ∇r G, we know that it belongs to M2N+1 −1 , hence we can apply Proposition 3.13. For simplicity, we will now denote A = (aij ) the matrix ∇r G. Estimate of α : We assume in this paragraph that i 6= j. We recall that: aij = −

30C r˜je r˜i

if j ∈ path0→i

q˜j E˜ rie r˜j5 aij = 0

Hence: |aij | ≤

else.

max 30C r˜je r˜g(i) min )5 E˜ rie (˜ rg(j)

max q˜g(j) ,

and α=

max 60C r˜je r˜g(i)

max

i∈{1,...,2N −1},

min )5 ˜ie (˜ rg(j) j∈path(0→i),j6=i r

max q˜g(j) .

(6.8)

Estimate of β: According to the previous definition: aii =

αi (r, q) 3η q˜(i) − r˜ie r˜i4

where we recall that αi = −1 −

  2 − P1 ) 5 (P0 − p˜[ i ] ) + (Pext ext 2

2E

and η =

15C . 2

Within our hypothesis (which are the same as Proposition 3.16) and recalling Proposition 3.16 together with Remark 23, for all i in {1, ...2N +1 − 1}, we have −aii ≥ −

αmax g(i) r˜ie

min 3η q˜g(i)
. max 4 r˜g(i)

+

According to the data of our model, the right-hand side of the previous inequality is always strictly positive, hence we can conclude: β=

min

i∈{1,..,2N −1}

−

αmax g(i) r˜ie

Then, along with Proposition 3.13, this yields −1

|||(∇r G(r, q))

min 3η q˜g(i) +
> 0. max 4 r˜g(i)

"

α |||2 ≤ 2 max 2 β N

with α and β given by (6.8) and (6.9). 22



α 1+ β

N −2

(6.9)

# 1 , , β

6.2.2

Estimate of ∇q G

From equation (6.5), for all i in {1, ..., 2N +1 − 1} and k ∈ {0, ..., 2N −1 }, we have:   max X 15C r˜g(i)  r˜je  ∂Gi η |≤ | 
4  + max
3 = Mi,k . i min ∂qk E˜ re r˜ r˜ i g(j)

j∈Π(0→[ 2 ])∩Π(0→k)

g(i)

Next, we know that:

|||∇q G(q)|||2 ≤ 2N (2N +1 − 1)

∂Gi |, ∂qk

max

|

max

Mi,k .

(i,k)∈{1,...,2N+1 −1}×{0,...,2N−1 }

hence, |||∇q G(q)|||2 ≤ 2N (2N +1 − 1) 6.2.3

(i,k)∈{1,...,2N+1 −1}×{0,...,2N−1 }

Estimate of ∇r f

Let us note A = B N (r). Here, we will use the spectral properties of the matrix A, from Proposition 3.5 together with the trace and |||.|||2 properties we know that: ˜i . if λ ∈ sp(A) then λ ≥ min R

(6.10)

i|g(i)=N

λmin = min(sp(A)) verifies λmin ≤ tr(A)/N.

(6.11)

λmax = max(sp(A)) verifies tr(A)/N ≤ λmax ≤ tr(A).

(6.12)

−1

(6.13)

|||A

||2 = 1/λmin .

||J|22

N

= 2 .

(6.14)

According to the formula of ∇r f given in Section 6.1.3, we need to estimate the following terms DA Q(A).H and A−1 with correct norms.

t JA−1 J,

Estimate of t JA−1 J: Since A−1 is a symmetric positive definite matrix (because A is, see Proposition 3.5), we know that it is coercive and that its coercivity constant is its smallest eigenvalue, i.e.:   1 ||J||22 t −1 JA J ≥ min . × ||J||22 = λmax λ∈sp(A) λ Moreover, N

tr(A) =

2 X

(N − g(j)) Rj ,

j=1

and recalling that Rj ≤

6Crje min )4 (rg(j)

, we have:

tr(A) ≤ 6C

N X k=1



 N −k (˜ rkmin )4 23

X

j st g(j)=k



rje  .

Let us call m1 (TN ) the right-hand side term of the previous inequality divided by 2N N , then tr(A) ≤ 2N N m1 (TN ), where m1 (TN ) only depends on the parameters of the model (which are the unconstrained radius, the pressures Pext , the root pressure, P0 , E and Φ). Then, using (6.12), we obtain: N 2N N t JA−1 J ≥ ||J||22 ≥ N ≥ m1 (TN )−1 . (6.15) tr(A) 2 N m1 (TN ) Estimate of DA Q(A).H: Thanks to the previous inequality, and from Section 6.1.3 we have:   |||DA Q(A).H|||2 ≤ 2N m1 (TN )|||A−1 |||32 + |||A−1 |||22 m1 (TN )Φ2N/2 ||H||2 .

(6.16)

Next, taking H = ∇r A.h in (6.16), and recalling equation (6.7) we obtain an upper bound: 5 N  X 1 ||∇r A.h||2 ≤ 2 max |aij | ≤ 2 24c||re ||∞ ||h||2 . r˜gmin i,j∈{1,...,2N } N

N

g=1

P Let us set m2 (TN ) = 2N 24c||re ||∞ N rgmin )5 , which only depends on the parameters of g=1 1/(˜ the model. We have: ||∇r A.h||2 ≤ m2 (TN )||h||2 . Estimate of A−1 : Using (6.10) and (6.13), we deduce that ||A−1 ||2 ≤

1 mini|g(i)=N Ri

≤

max )4 (˜ rN = m3 (TN ), 6c mini|g(i)=N (rei )

(6.17)

and m3 (TN ) only depends on the parameters of the model. Estimate of ∇r f : Using previous estimates (6.15– 6.17), we deduce that: |||∇r f (r).h|||2 = |||DA Q(∇r A.h)|||2 (6.18)   N N 2 2 ≤ 2 2 2 m1 (TN )m3 (TN ) + 1 m1 (TN ) m2 (TN )m3 (TN ) Φ||h||2 . 6.2.4

Final estimate

Combining all results (6.15– 6.18) from this section allows us to get an upper bound of |||∇q F (q)|||2 : h i ∇q F (q).h = −∇r f (r(q)). [∇r G(r(q), q)] −1 .∇q G(r(q), q).h . N

Hence there exists a constant C(TN , Φ) such that for all h ∈ R2 , ||∇q F (q).h||2 ≤ C(TN , Φ)Φ||h||2 , therefore: |||∇q F (q)|||2 ≤ C(TN , Φ)Φ.

24

Proposition 6.2. There exists η > 0 such that if Φ belongs to [0, η[ then the Picard fixed-point theorem applies for F on ΩǫTN . This leads to uniqueness of the fixed point qf of F in ΩǫTN and to convergence of scheme 6.1 toward qf . Proof. We just need to prove that if Ψ is a strictly positive real number, then for all Φ in [0, Ψ], there exists K(TN ) > 0 such that C(TN , Φ) ≤ K(TN ). This property is a consequence of the fact that qjmin ≥ 0 and qjmax ≤ Ψ for all Φ in [0, Ψ] and that the estimates from this chapter can be obtain in a similar way but independently of Φ, with qjmin = 0 and qjmax = Ψ. Remark 30. In the same way, it can be shown there exists Emin > 0 such that if E > Emin then the Picard Theorem applies on F . Remark 31. The Proposition 6.2 and the Remark 30 prove that the Picard Theorem better applies for small deformations, which is consistent with the linear elasticity model used for bronchial wall deformation.

7 Numerical simulations 7.1 Methodology The simulations were performed with Matlab 7. The evaluations of the function q → r(q), defined by G(r(q), q) = 0 (Proposition 3.16), were obtained through a Newton method. The numerical process uses the characteristic geometrical structure of this problem and calculates most of the different variables (pressures, radii) from the top of the tree down to the lower part. 7.1.1

Physiological hypotheses

We study the deformation of an idealized subtree of the lung during stationary expiration. Because of the linear deformation model, the simulation will be focused on conditions leading to small deformations. In the previous parts of this paper, we have proved that, for a deformable tree, there exists an unique flow q that minimizes the energy dissipation after deformation. This particular flow is interesting in the sense that living systems tend to minimize their energy cost [7, 15, 1]. Hence, in the case of non negligible deformations, it is reasonable to assume that ventilation should take into account possible deformations of the tree in order to minimize viscous dissipation in the deformed tree. In a time-dependent world, this minimization could be achieve through a feedback system (which is known to exist in the lung [12] ). However in a stationary model, minimization is not trivially reached because the response is not linear. Hence, the minimizing flow is calculated through a non biological iterative process corresponding to Proposition 6.2. This model only applies for lung generations without cartilage, mainly from the sixth generation to the sixteenth after which the tree function becomes different (beginning of gas exchange with blood, see [2, 13]). Working on these generations significantly reduces the effects of inertia and validate the Poiseuille regime, at least for simulations at rest respiratory regime (but not for exercise and forced expiration regimes, for which we will however neglect inertia). The Young modulus has been estimated for rat lungs in [8, 10], we choose to apply this value for human lung since mammals lungs are very similar to each other [12]. This value is however relative to a macroscopic view of the lung. To obtain the Young modulus value for tissue, and for bronchi 25

wall, it is important to note that one fifth of lung volume consists in tissue and four fifth consist in air. Hence the Young modulus of tissue should be evaluated at around five times the macroscopic value measured in [8, 10]. The value used in our application is consequently E = 5 × 1250 = 6250 P a. To model the geometry, Weibel’s data [13] have been used to build a simplified model as in [7]. The lung is assumed to be fractal with an homothetic ratio h = 0.82 linking the dimensions between two successive generations of the tree. This value reflects an inflated lung at the end of inspiration and numerical data are reasonably in accordance with [13]. At the end of inspiration, air is assumed to be motionless and the whole tree to be at equilibrium. This equilibrium defines the initial radii re . The unconstrained radii r0 are then calculated from forces balance. The parenchyma and alveolar pressures are estimated from data obtained from [14, 4], moreover parenchyma pressure will be assumed to be linearly linked with velocity in trachea on the range 0.5 m.s−1 to 45 m.s−1 , for sake of simplicity. The subtree root pressure Proot is calculated from the lung global resistance, which itself is calculated from pressure drop and flow at rest in an idealized fractal tree (this assumes that resistance does note depend on ventilation regime, which is an approximation). Zero pressure corresponds to atmospheric pressure. 7.1.2

Indicators

The relative contraction of branch i, Ci (in percents), is calculated through Ci = 100 ×

ri − rie rie

Note that if Ci is negative then the bronchia is constricted while if Ci is positive the bronchia is dilated. Let τi denote the proportion of deformation due to the flow in the branch i (percents). It is calculated as the difference in deformation between a tree which is only subject to a pressure change in parenchyma (with zero air flow in bronchia) - deformation called dri0 = ri (0) − rie - and the same tree with the same pressure drop in parenchyma and with optimal flow qΦ inside bronchia - deformation called driφ = ri (qΦ ) − rie . More precisely, we have: τi = 100 ×

driΦ − dri0 ri (qΦ ) − ri (0) = 100 × Φ ri (qΦ ) − rie dri

Again, the sign of τi gives the consequence of air flow circulation on bronchial radii. During expiration, the effect of the flow is always to increase radii and hence τi is always positive, it is negative during inspiration. The mean l2 -deformation of the branches or tree deformation is given by: v u2N+1 −1 u X 1 t d = N +1 Ci2 2 −1 i=1

Finally, to study the convergence speed of scheme 6.1, a local estimate of the Lipschitz constant k of the application F has been calculated. Convergence velocity is given by the following inequality, which holds true for all n ∈ N∗ : ||qn − q|| ≤

kn ||q1 − q0 || 1−k 26

To locally estimate k, the value Err = ln ((qn+1 − qn )/(q1 − q0 )) has been stored for each sequence index n. According to the inequality ||qn+1 − qn || ≤ kn ||q1 − q0 || going along with Picard theorem, if convergence occurs, Err should be smaller than a line with negative slope ln(k) (k ∈ ]0, 1[) and hence should be decreasing to −∞ with n.

7.2 Results 7.2.1

Convergence

To illustrate the scheme convergence, we exhibit an example corresponding to a fractal tree (h = 0.82) of eleven generations with one of the third generation branches partly collapsed (the radius has been reduced to one third of its original value). The flow and root pressure have been adjusted such that the velocity in trachea corresponds to forced expiration and reaches 15 m.s−1 (remember that the first generation of our tree corresponds to the sixth generation of the lung). The Young modulus has been chosen to be E = 1250 P a (five times smaller than in the following sections). The results are presented on Figure 4. On the left part Err has been represented and is decreasing very fast. From the numerical results, the local Lipschitz constant is smaller than 0.071. Hence for the sixth iteration, the inequality ||q6 − q||2 ≤ 1.3 × 10−7 ||q1 − q0 ||2 holds. The right part of Figure 4 shows the leaf flow profile in the tree along the iterative process. Because the convergence is fast, the difference between the initial profile (dashed line) and the first iteration profile (dashed dotted line) is large. The sixth iteration (continuous line) is very close to optimal flow. The mean l2 deformation of the branches in this example is of 1.8%, with a larger value reached on root branch (5.8%) and smaller values on leaf branches (particularly on the daughters of the collapsed branch, 1.4%). Convergence.

−7

0

1.5

Optimal flow evolution during iterations.

1.25

−4 1 −6

Flow

ln(||qn+1−qn||2/||q1−q0||2)

−2

x 10

−8

q0 q1

0.75

q6 0.5

−10 0.25

−12 −14 1

2

3

4

5

0 0

6

Iteration

200

400

600

800

1000

Leaf index

Figure 4: Convergence of iterative scheme qn+1 = F (qn ). Left : “convergence curve” (log of relative error from one step to the next), this curve helps us to bound the Lipschitz constant of F , which is smaller than 0.071 here, hence the convergence is fast. Right : flow during iterative scheme, initial flow q0 is represented by the dashed line. Note that as stated in Proposition 6.2, reducing too much the parameter E (lower than 219 P a in this particular case, to compare to the 6250 P a for the lung) or increasing too much the parameter Φ (larger than 84 m.s−1 ) leads to non convergent schemes. Moreover, these two thresholds depend on each other, for instance a 80 m.s−1 velocity in the trachea leads to a threshold on E of 1182 P a, while E = 6250 P a leads to a threshold on trachea velocity of 417 m.s−1 ).

27

7.2.2

Deformation during rest and forced expiration (N = 4)

In this section, more physiological parameters will be used and we will compare deformation in lung subtrees of five generations corresponding to generation six to ten (for sake of an easier study of deformations). There will be assumed that a branch of the third generation is partly collapsed (radius being one third of its original value). Young’s modulus has been fixed at E = 6250 P a. At rest, the flow is very low and the deformation of the subtree is mainly due to pressure jump between the parenchyma and the air inside the bronchia. In this regime, most part of the pressure drops inside the tree is a consequence of the passage in the nasal cavity and in the first generations of the lung. Hence measured deformation is quasi-homogeneous in the whole tree after reaching optimal flow distribution. For a trachea velocity of 0.5 m.s−1 , the mean l2 deformation of the branches is d = 4.09%. The deformations due to the flow τi are very small (less than 1%). During forced expiration, flows are much larger (between ten times and a hundred times than at rest) and they counterbalance constriction effects due to pleural pressure. Furthermore, flows are mostly different in every branch, leading to different pressures inside branches, hence an inhomogeneity of deformations will appear depending on the localization of the branch (and only depending on the generation in a homogeneous tree). This is illustrated in the following example, where a trachea velocity of 15 m.s−1 has been chosen to simulate forced expiration. Optimal flow

−6

8

x 10

6 5 Flow

x 10

gen. 1 gen. 2 gen. 3 gen. 4 gen. 5

7 Flow through branches

7

4 3 2 1 0

Flows

−5

8

6 5 4 3 2 1

2

4

6

8 10 Leaf index

12

14

0 0

16

10

20 Branches

30

Figure 5: Flow distributions. On the left flow distribution at leaves is represented while on the right flow repartition for all branches in the tree are shown. The collapse of the third generation branch (in the right part of the tree) induces a flow reorganization with larger distribution to the left part, where resistance is smaller. On Figure 5 are plotted the flow distributions. On the left flow distribution at leaves is represented, the collapsed third generation branch induces a reorganization of the optimal flow. The daughter branches of the collapsed one (13 to 16) receives practically zero flow, while other branches receive more, particularly the sister branch of the collapsed one. On the right, flow values in each branches are shown. The left part of the tree receives more flow than the right part where the collapse has occurred (at branch number 28). Branches deformations are represented on Figure 6. On the left part, the relative deformations of branches are plotted. They are dependent on the generation because of the relatively large values of the flow, unlike rest regime, where flow influence is negligible. The constricted branch (number 28) has a lower deformation than its sisters, thanks to its smaller radius with identical elastic properties. In the right part we represent which is the proportion of deformation that is due to the flow. For the daughters of the collapsed branch, there is quasi zero flow, hence it plays no role in deformation. However, the 28

Branches deformation

Flow influence on deformation Proportion of deformation due to flow (%)

Relative contraction of branches [100 × (r−ri)/ri] (%)

0 gen. 1 gen. 2 gen. 3 gen. 4 gen. 5

−0.2 −0.4 −0.6 −0.8 −1 −1.2 −1.4 0

10

20 Branches

30

40 gen. 1 gen. 2 gen. 3 gen. 4 gen. 5

35 30 25 20 15 10 5 0 0

10

20 Branches

30

Figure 6: Deformations of branches. Generation dependence is notable. On the left, global deformation is plotted. The constriction (branch number 28) induces a smaller global deformations for itself and for its daughters than their corresponding sisters. However, the collapsed branch has a higher resistance because of its small radius and hence is relatively far more deformed by the forces induced by the flow than its daughters as shown on the right plot. collapsed branch flow influence is large relatively to its daughters (approximately 20% to compare to less than 2% for its daughters) because of its smaller radius which implies a higher resistance and hence a higher pressure drop between its extremities. The pressures at the branches exit are shown on Figure 7, hence pressures variations are negligible from the daughters of the collapsed branch to the exit of the collapsed branch. As stated above, we see that there is a large pressure drop between the collapsed branch and its mother (rightest blue bar). The presence of higher flow in the tree, and consequently of higher pressures, prevents a collapse of the branches. Moreover, this region of our lung model is less deformed during moderate forced expiration than during rest respiration. Pressures 14 gen. 1 gen. 2 gen. 3 gen. 4 gen. 5

Pressure at leafs

12 10 8 6 4 2 0 0

10

20 Branches

30

Figure 7: Pressure drops from the end of branches to the tree root. Note the very small differences between pressures inside the daughter branches of the collapsed one (number 28), due to small flow circulation. Note the large pressure drop between the collapsed branch and its mother (number 24), this is due to the high resistance of the collapsed branch.

29

7.2.3

Study of Equal Pressure Point (EPP)

The behavior of bronchial wall (constricted or dilated) is defined by the difference between pressure increases in the bronchia and in the parenchyma (pleural pressure). There are two scenarios: • This difference is negative, this leads to bronchial dilatation. • This difference is positive, this leads to bronchial constriction. From the leaves of the tree to its root, the bronchial pressure pi decreases with generations up to the trachea, where it reaches atmospheric pressure (chosen to be P0 = 0 in our model). In the case when pleural pressure increase during expiration is lower than alveolar pressure increase, then both scenarios can happen in different bronchi of the tree. This creates a dilated region in the lower part of the tree and a constricted region in the higher part, as shown on Figure 8. The transition region (which is more precisely a set of generations in our model) is called the Equal Pressure Point, shortly named EPP. To track EPP, we have simulated a range of velocities in trachea and checked when both scenarios are present. The tree deformation is directly linked to the presence of EPP: deformation reaches its minima when EPP reached the lower generations. This is a natural consequence of EPP definition, because the pressures differences between parenchyma and bronchi are close to those from equilibrium. Hence we have used this criterion to detect EPP. Tree deformation

Tree deformation 1.5

L2 Deformation (%)

L2 Deformation (%)

1.5

1

0.5

0 16

17

18 19 Velocity in trachea

20

1

0.5

0 16

21

17

18 19 Velocity in trachea

20

21

Figure 8: Plots of tree deformations. The minimum point is used to localize the range of velocities where EPP occurs. On the left: a fractal tree of eleven generations (h = 0.82), on the right: the same fractal tree with a branch from the third generation being partly collapsed (radius divided by three). Note the important effect of this collapse on EPP localization. Two tree deformations plots for a range of trachea velocity have been drawn on Figure 8. The left plot corresponds to a fractal tree of eleven generation (h = 0.82) and shows a minima around 17.4 m.s−1 . The right plot shows the consequence on the minima on a tree with a third generation branch partly collapsed (radius divided by three). The minima is then shifted to the lower velocity 16.8 m.s−1 . Hence the global deformation can be directly linked to tree structure or defects, this criterion could be used to check tree pathologies due to geometrical changes. To understand how geometry defects can affect EPP localization, we have defined a number z measuring the average generation on which branch dilatation stops and branch constriction starts (which actually is the EPP) : 30

Position of EPP (mean generation)

Position of EPP (mean generation) 12 Position of EPP (mean generation)

Position of EPP (mean generation)

12 10 8 6 4 2 0 16

17

18 19 Velocity in trachea

20

10 8 6 4 2 0 16

21

17

18 19 Velocity in trachea

20

21

Figure 9: Mean generation z of EPP versus trachea velocities in an eleven generation fractal tree (h = 0.82) on the left and in the same tree with a third generation branch constricted (radius divided by three). EPP progression in the tree is highly dependent on tree structure.

z=

X

i∈{1,...,2N+1 −1} s.t. ri ≤r

1 2g(i)−1

(7.1)

In a homogeneous fractal tree, pressures and flows are the same for each generation : deformations are identical and hence z is an integer between 0 and N + 1 (as shown in the left plot on figure 9). More generally, it can be a real number belonging to [0, N + 1] depending on the tree structure (as shown in the right plot on figure 9). Mean EPP position z is dependent of tree structure, and the effect of the constriction of a branch of the third generation in a fractal tree of eleven generation is shown on figure 9 (right plot) in comparison with the same tree without constriction (left plot). The minima of tree deformation, plotted on figure 8, corresponds to the velocities for which EPP enters the tree. On figure 9, it can be seen that when velocity increases, EPP goes up into the tree. The exit velocity is the same for both trees (20.9 m.s−1 ), however entry velocity and EPP progress profile are different. Actually, the constriction of the third generation branch creates inhomogeneous paths from leaves to tree root, inducing differences in EPP progression. As shown on figure 10, the EPP enters sooner in the constricted tree (right) than in the homogeneous tree (left) and spreads very fast into the daughters of the constricted branch, because they have very similar inside pressure. However, EPP is stopped by the constricted branch and a jump of velocity is needed to cross it and reach the next generations. The constriction has also effects on the subtree down the sister of the collapsed branch, slowing down EPP progression inside it. However it slightly increases EPP progression in the other parts of the tree. 7.2.4

Conclusion about numerical results

The algorithm developed in this paper has been useful to obtain stationary optimal flows in a deformable tree, to mimic an active regulation in the lung that minimizes dissipated energy due to flow viscosity (which could be one of the reason of the existence of smooth muscles on bronchi wall, although their real functions are not really known [12]). This model has been applied in the case of small deformations, and more particularly when equal pressure point (EPP) occurs. We have established how a constricted branch (that is a local geometrical change) can modify the behaviors of the whole structure. We also have studied its influence on EPP progression. These results could be interesting for the understanding of lung structure and of its pathology. Our model should however be 31

Figure 10: Velocities inducing EPP in the branches of a fractal tree (h = 0.82, on the left) and of a constricted tree (on the right), identical to the fractal tree but with a partly collapsed branch at the third generation (radius reduced to one third of the original one). Note that these two-dimensional geometries are only two-dimensional representations of our three-dimensional trees, the only geometrical respected values here are the radii. improved with more realistic laws for elasticity, alveolar and pleural pressures determination. Hence this study could be extended to larger deformations with a point of view closer to medical applications.

Acknowledgments The authors wish to thank Professor Le Dret for very helpful discussions and the CMLA (ENS de Cachan, France) for access to Matlab 7.

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[10] F.G.S ALERNO and M.S.L UDWIG, Elastic moduli of excised constricted rat lungs, J. of Applied Physiology, 1999, vol 86, pp. 66-70. [11] C.VANIER, Memoire de DEA, 2005. [12] E.R.W EIBEL, The pathway for oxygen, Harvard University Press, December 1984. [13] E.R.W EIBEL, Morphometry of the human lung, Springer Verlag, 1963. [14] J.B.W EST, Physiologie respiratoire, Pradel, 1997. [15] G.B.W EST, J.H.B ROWN and B.J.E NQUIST, A general model for the origin of allometric scaling laws in biology, Science, 1997, vol 276, pp. 122-126.

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