Numerical analysis of a finite volume scheme for the
simulation of a nonlinear degenerate breast cancer model
Françoise Foucher, Moustafa Ibrahim, Mazen Saad
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Françoise Foucher, Moustafa Ibrahim, Mazen Saad. Numerical analysis of a finite volume
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Numerical analysis of a finite volume scheme for the simulation
of a nonlinear degenerate breast cancer model
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
Abstract In this paper, a finite volume method for the simulation of a degenerate breast cancer model is carried
on. This model consists in modeling the stepwise mutations from a normal breast stem cell to a tumor cell.
Furthermore, the model presents a degenerate parabolic equation modeling the invasion and interaction of the
solid tumor growth with its surrounding by means of releasing degradative enzymes which their dynamics
are modeled with a partial differential equation. We show that the proposed finite volume scheme ensures the
discrete maximum principle and we prove the convergence of existed discrete solutions toward a weak solution.
Finally, a numerical experiment is carried out to show the breast cancer development.
1 Introduction
Breast cancer is the second most common type of cancer on the list after skin cancer. It can occur in both men
and women, but it is very rare in men. The main cause of breast cancer remains unknown. However, recent
research in breast biology has provided support for the cancer stem-cell hypothesis, meaning that breast cancers
grow from breast stem cells in the way that healthy organs do [15, 33].
Breast cancer is a disease that occurs when the genetic information carried by the DNA is corrupted, leading to
abnormal patterns of gene expression. The genetic code exists in the form of deoxyribonucleic acid (DNA) that
is packaged in chromosomes in the cellular nucleus, which are made of two different chromatids that are joined
at the centromere. On each chromatid are so-called alleles with different information from the parent individuals, and the combination of these alleles encodes a gene. Cancer is driven by two classes of genes: oncogenes
and tumor suppressor genes (TSGs).
Oncogenes are derived from mutated versions of normal cellular genes (called proto-oncogenes) that control
cell proliferation, survival and invasion/motility. In cancer, activating mutations of proto-oncogenes cause uncontrolled cell division and enhanced survival. Oncogenes are described as phenotypically dominant, meaning
that a single mutated copy of a proto-oncogene is sufficient to promote cancer–a process known as tumorigenesis.
Tumor suppressor genes (TSGs) are normal cellular genes that function to inhibit cell proliferation and survival.
They are frequently involved in controlling cell cycle progression and programmed cell death/apoptosis. TSGs
are phenotypically recessive, meaning that both copies (alleles) must be functionally altered to promote cancer.
Françoise Foucher
École Centrale de Nantes. UMR 6629 CNRS, laboratoire de mathématiques Jean Leray. F-44321, Nantes, France.
e-mail: [email protected]
Moustafa Ibrahim
American College of the Middle East. Math and Science Division. 220 Dasman, 15453, Kuwait.
e-mail: [email protected]
Mazen Saad
École Centrale de Nantes. UMR 6629 CNRS, laboratoire de mathématiques Jean Leray. F-44321, Nantes, France.
e-mail: [email protected]
1
2
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
When both the alleles are either mutated or non-functional, then the suppressing tumorigenesis effect is lost
which changes the cell’s phenotype causing tumors [19]. Therefore, allelic loss of one allele of specific chromosomal regions in tumors implies TSGs at those loci and is called “loss of heterozygosity”(LOH). LOH on
TSGs is important for the initiation or early progression of breast cancer development [6, 24].
According to the breast cancer stem-cell hypothesis, one makes evidence that mutations of an oncogene or a
tumor suppressor gene are necessary to model the transformation of normal breast stem cells to tumor cells. In
this paper, we adopt the mathematical model proposed by Enderling and al. in [8] which consists of an extension
to the original model of the same authors [7]. In [8] and for simplicity, the authors have adapted the approach
described in [30], and have assumed that mutations in two TGSs are sufficient to give rise to a tumor.
The mathematical model we consider in this paper consists of four ordinary differential equations describing
the stepwise mutations from a healthy breast stem cell (pre-cancerous cell) to a tumor cell [28]. Then, for
the dynamics of the tumor cell, the model presents basically a partial differential equation modeling the solid
tumor growth and invasion in the sense of examining the invasion of the growing solid tumor and its interaction
with its surrounding environment; by releasing matrix degradative enzymes (MDEs). These MDEs degrade the
tissue upon contact to make space for the solid tumor to develop and grow into (see e.g. [23]). The partial
differential equation modeling the dynamics of the MDEs will constitute the last equation of the mathematical
model considered in this paper.
Now, we give an illustrative diagram for the stepwise mutations from a normal breast stem cell to a tumor
+/+
+/+
cell. The two TGSs considered in [8] are labelled TSG1 and TSG2 , and we denote by TSG1
and TSG2
when both alleles are un-mutated. The stepwise mutation pathway is assumed as follows:
+/+
TSG1
+/+ ρ1
TSG2
+/−
+/−
−→ TSG1
+/+ ρ2
TSG2
−/−
−→ TSG1
+/+ ρ3
TSG2
−/−
−→ TSG1
+/− ρ4
TSG2
−/−
−→ TSG1
−/−
TSG2
,
−/−
where TSGi
represents TSG with LOH and TSGi
represents an inactivated TSG. In the final step of the
pathway when both of the TSGs are inactivated, it is assumed the cell to be a cancer cell (see [8] for more details). The superscript ρi represents the probability of mutating one allele at every step of the mutation diagram.
In this paper, we take the same values for these probabilities as taken in [8] from the studies by Tomlinson et
al. [31] and Nowak et al. [25]. The variation of these values would only result in a faster or delayed mutation
acquisition.
The intention of this paper is to numerically investigate the early stage of the breast cancer development
using a finite volume scheme.
The rest of this paper is organized as follows. In Section 2, we introduce the mathematical system modeling the
breast cancer progression. This system consists of ordinary differential equations and of degenerate parabolic
partial differential equations. We give the main assumptions on this system in order to define a weak solution.
In Section 3, we define the space and time discretization of the space. Then, we introduce the discretization of
the mathematical model using a finite volume method under an orthogonal mesh. Next, we derive the discrete
properties of the scheme and prove the discrete maximum principle and finally show the existence of a discrete
solution to our scheme. In Section 4, we give estimates on differences of time and space translates for the
approximate solutions. In Section 5, we apply the Kolmogorov relative compactness criterion in order to prove
the convergence of a subsequent of the sequence of discrete solutions towards a limit that we identify as a weak
solution to the continuous problem. Finally, in Section 6, a numerical simulation is carried out to investigate
and capture the progression of the breast cancer development.
2 The degenerate nonlinear breast cancer model
Let Ω be an open bounded polygonal and connected subset of Rd , d = 2, 3 and let T > 0 be a fixed finite
time. We denote by QT = Ω × (0, T ) and ΣT = ∂ Ω × (0, T ), where ∂ Ω is the boundary of the domain Ω .
The mechanical interactions of tumor cells with healthy tissue have been modeled in several ways, including
continuum equations [13, 22, 27, 34], multi-phase flow models [1], transport equations [18], and individual
based models [16, 17]. In this paper, we are interested in a modified degenerate nonlinear system [8] modeling
Numerical analysis for the simulation of a degenerate breast cancer model
3
the breast cancer development given by the set of equations

∂t f = −η f v f − ρ1 f





∂t q = λq q(1 − A) − ηq uq + ρ1 f − ρ2 q



∂ r = λ r(1 − A) − η ur + ρ q − ρ r
t
r
r
2
3

∂
s
=
λ
s(1
−
A)
−
η
us
+
ρ
r
−
ρ
t
s
s
3
4s




∂t u − div (a(u)∇u − χ (u) ∇ f ) = λu u(1 − A) + ρ4 s



∂t v − div (dv ∇v) = g(u, v)
in QT ,
in QT ,
in QT ,
in QT ,
in QT ,
in QT ,
(1)
We assume that tumor cells, and consequently the MDEs, remain within the domain of tissue under consideration and therefore no-flux boundary conditions are imposed on ΣT given by
(a(u)∇u − χ (u) ∇ f ) · n = 0,
∇v · n = 0,
(2)
where n is the unit normal vector to ∂ Ω outward to Ω . Finally, we consider the initial conditions on Ω given
by:
ω (x,t) = ω0 (x) ,
in Ω ,
ω = f , q, r, s, u, v.
(3)
In the above model, we represent by f a fraction of healthy breast stem cells in the breast tissue, by q the cells
with LOH in TSG1 , by r the cells with inactivated TSG1 , and by s the cells with LOH in TSG2 . The density of
the tumor cells and the enzymes concentration which they produce are represented by u = u(x,t) and v = v(x,t)
respectively. Next, A = u+ f +q+r +s represents the total tissue and cell population including tumor cells, a(u)
is a density-dependent diffusion coefficient. Furthermore, the function χ(u) represents the enzymes sensitivity
to the healthy breast stem cells, and the function g(u, v) describes the production and degradation of the enzymes
by the tumor cells ; here, we assume it is a nonlinear function given by
g(u, v) = αu(1 − v) − β v,
α, β ≥ 0,
(4)
where α is the production rate and β is the degradation rate. Finally, we denote by λω ≥ 0 the production rate
of ω for ω = q, r, s, u and by ηω ≥ 0 the death rate of ω for ω = f , q, r, s.
We give the main assumptions made about the system:
(A1) The cell-density diffusion a : [0, 1] −→ R is a continuous function such that, a (u) ≥ 0 for 0 ≤ u ≤ 1.
(A2) The sensitivity χ : [0, 1] −→ R is a continuous function such that, χ (0) = χ (1) = 0, and χ (u) > 0 for
0 < u < 1.
(A3) We assume that the production rate λu of u is greater than the final mutation probability ρ4 , that is: λu ≥ ρ4 .
(A4) The initial functions u0 , v0 , and f0 satisfy: u0 , v0 ∈ L2 (Ω ) and f0 ∈ H 1 (Ω ). In addition, we assume that
f0 , q0 , r0 , s0 , u0 , v0 ≥ 0,
in Ω ,
u0 , v0 , f0 + q0 + r0 + s0 ≤ 1,
in Ω .
In the sequel, we use the Lipschitz continuous nondecreasing function A : [0, 1] −→ R defined by
A (u) :=
Z u
a (s) ds,
∀u ∈ [0, 1] .
(5)
0
Definition 1 (Weak solution). Under assumptions (A1)–(A4), we say that the set of measurable functions
( f , q, r, s, u, v) is a weak solution of system (1)–(3) if
0 ≤ ω (x,t) ≤ 1,
ω = f , q, r, s, u, v
A (u) , f , v ∈ L 0, T ; H 1 (Ω ) ,
2
and for all ϕi ∈ D Ω × [0, T ) , i = 1, . . . , 6, one has
for a.e. in QT ,
4
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
−
ZZ
QT
−
ZZ
QT
−
ZZ
QT
−
ZZ
QT
−
ZZ
QT
f ∂t ϕ1 dx dt −
Z
Ω
q ∂t ϕ2 dx dt −
Z
Ω
r ∂t ϕ3 dx dt −
Z
Ω
s ∂t ϕ4 dx dt −
Z
Ω
u ∂t ϕ5 dx dt −
Z
Ω
f0 (x) ϕ10 dx =
ZZ
q0 (x) ϕ20 dx =
ZZ
r0 (x) ϕ30 dx =
ZZ
s0 (x) ϕ40 dx =
ZZ
u0 (x) ϕ50 dx +
ZZ
QT
QT
QT
QT
QT
ZZ
=
QT
−
ZZ
QT
v ∂t ϕ6 dx dt −
Z
Ω
v0 (x) ϕ60 dx +
ZZ
QT
ZZ
=
QT
−η f v f − ρ1 f ϕ1 dx dt,
(λq q (1 − A) − ηq uq + ρ1 f − ρ2 q) ϕ2 dx dt,
(λr r (1 − A) − ηr ur + ρ2 q − ρ3 r) ϕ3 dx dt,
(λs s (1 − A) − ηs us + ρ3 r − ρ4 s) ϕ4 dx dt,
(∇A (u) − χ (u) ∇ f ) · ∇ϕ5 dx dt
(λu u (1 − A) + ρ4 s) ϕ5 (x,t) dx dt,
dv ∇v · ∇ϕ6 dx dt
(αu (1 − v) − β v) ϕ6 (x,t) dx dt,
where ϕi0 = ϕi (x, 0) for all i = 1, . . . , 6.
3 Finite volume approximation
3.1 Space-time discretization and discrete functions
We are placed in the case where the boundary of the domain occupied by the cells and the chemicals is fixed
over the time. We consider that Ω ⊂ Rd is the domain occupied by the cells, and that Ωh ⊂ Ω is the approached
polygonal domain of Ω .
In order to discretize system (1)–(3), we introduce some definitions and notations.
3.1.1 Space discretization of Ω
Definition 2 (Admissible mesh of Ω ). An admissible finite volume mesh of Ω is a triplet (T , E, P), where T
is a finite family of disjoint open bounded polygonal convex subsets of Ω called control volumes, E is a finite
family of subsets of Ω contained in hyperplanes of Rd with strictly positive (d − 1)-dimensional measure, called
the edges of the control volumes, and P = {xK , K ∈ T } is a finite family of points of Ω , called the “centers” of
the control volumes.
The triplet (T , E, P) satisfies the following properties:
S
1. The closure of the union of all the control volumes is Ω , i.e. Ω = K∈T
K.
S
S
2. For any K ∈ T , there exists a subset EK of E such that ∂ K = K\K = σ ∈EK σ . Furthermore, E = K∈T EK .
3. For any (K, L) ∈ T 2 with K 6= L, either the “length” (i.e. the (d − 1) Lebesgue measure) of K ∩ L is 0 or
K ∩ L = σ for some σ ∈ E, which then is denoted by σKL i.e. the interface between K and L.
4. The family P = (xK )K∈T is such that xK ∈ K (for all K ∈ T ) and, if σ = σKL , we assumed that xK 6= xL ,
and the straight line SKL going through xK and xL is orthogonal to σKL .
5. For any σ ∈ E such that σ ⊂ ∂ Ω , let K the control volume such that σ ∈ EK . If xK ∈
/ σ , let SK,σ be
the straight line going through xK and orthogonal to σ , then the condition SK,σ ∩ σ 6= 0/ is assumed; let
yσ = SK,σ ∩ σ .
In the sequel, we use the following notations. The size of the mesh T is defined by:
h = size (T ) = max diam (K) ,
K∈T
Numerical analysis for the simulation of a degenerate breast cancer model
σKL
5
xL
L
K
xK
TKL
Fig. 1 Finite volume mesh T : control volumes, centers and diamonds.
where diam (K) represents the least upper bound of the set of all distances between pairs of vertices of every
control volume K ∈ T .
The set of neighbors of K is denoted by N (K), i.e. N (K) = {L ∈ T ; ∃σ ∈ EK , σ = K ∩ L}; a generic neighbor
of K is often denoted by L, furthermore we denote by ηKL and dKL the unit normal vector to σKL outward to K
and the distance |xK − xL | respectively.
For any K ∈ T and σ ∈ E, we denote by |K| the d-dimensional Lebesgue measure of K. If L ∈ N (K), |σKL | will
|σKL |
.
denote the measure of the edge σKL , τKL will denote the “transmissibility” through K|L defined by τKL =
dKL
3.1.2 Discrete control volumes space HT
Let us define the discrete finite volume space HT of piecewise constant functions on the mesh T by
HT = {φ : Ω −→ R; φ|K ∈ R is constant, ∀K ∈ T }.
Every function uT ∈ HT is characterized by its numerical values (uK )K∈T such that uT |K = uK for every K ∈ T .
In other words, given a vector (uK )K∈T ∈ R#T ; then, there exists a unique discrete function uT ∈ HT such that
∀x ∈ K, ∀K ∈ T .
uT (x) = uT (xK ) = uK ,
The discrete space HT is a linear subspace of L2 (Ω ), the usual inner scalar product becomes
Z
(uT , vT )L2 (Ω ) =
uT (x) vT (x) dx =
Ω
∑
|K| uK vK ,
∀uT , vT ∈ HT .
K∈T
Therefore, the associated norm is given by
!1/2
kuT kL2 (Ω ) =
∑
2
|K| |uK |
.
K∈T
We define a discrete equivalent to the semi-norm on H 1 (Ω ), |u|1,Ω =
|uT |1,T =
|σKL |
∑ dKL |uL − uK |2
σKL ∈E
R
2
Ω |∇u|
1/2
, by
!1/2
,
∀uT ∈ HT .
Definition 3. Let σ ∈ E be an interface for a control volume K ∈ T . A “diamond” Tσ constructed from the
neighbor centers xK , xL and the interface σKL is defined by
6
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
(
S
TKL= V ∈{K,L} {cxV + (1 − c) y, c ∈ [0, 1[, y ∈ σKL },
TKσ = {cxK + (1 − c) y, c ∈ [0, 1[, y ∈ σ },
if σ = σKL ∈ E\∂ Ω ,
if σ ∈ ∂ K ∩ ∂ Ω .
|σKL | dKL
, where d > 0.
d
Using definition 3, we define the discrete gradient over every “diamond” Tσ . Specifically, we have the following
definition.
The d-dimensional Lebesgue measure of an interior diamond TKL is |TKL | =
Definition 4 (Discrete gradient). The discrete gradient ∇T is a correspondence that maps a function uT ∈ HT
into a piecewise constant function over the “diamonds” such that

d uL − uK η ,
if σ = σKL ∈ E\∂ Ω ,
KL
dKL
∇T uT |Tσ =
0,
if σ ∈ ∂ K ∩ ∂ Ω .
Using this definition, we remark that there exists a relationship between the semi-norm |·|1,T and the L2 (Ω )
norm of the discrete gradient given by: k∇T uT k2L2 (Ω ) = d |uT |21,T , for all uT ∈ HT .
3.1.3 Time discretization
The time discretization is considered to be uniform (we do not impose any restriction on the time step), it is
given by the sequence of discrete time tn = n∆t for every n ∈ N with a fixed time step ∆t such that there exists
an integer N ∈ N∗ verifying tN+1 = (N + 1) ∆t = T .
3.1.4 Space-time discretization of QT
We define the space and time discrete space HT ,∆t as the set of piecewise constant functions in time with values
in HT .
HT ,∆t = {ϕ ∈ L2 0, T ; H 1 (Ω ) , ϕ (x,t) = ϕ (x,tn+1 ) ∈ HT , ∀t ∈ (tn ,tn+1 ]}.
Remark 1. Hereafter and for simplicity, we consider the nondegenerate version of system (1)–(3) and assume
that the diffusivity a(u) ≡ du . The same analysis is possible and can be adapted for the degenerate case (see
e.g. [2]) and by adding the following assumption on the diffusivity
a (0) = a (1) = 0.
3.2 Discretization of system (1)–(3)
We are now in a position to discretize system (1)–(3). We consider an admissible discretization of QT , which
consists of an admissible mesh T of Ω and a uniform time step ∆t > 0.
A finite volume scheme for the discretization of system (1)–(3) is given by the following set of equations: for
all K ∈ T
w0K =
1
|K|
Z
K
w0 (x) dx,
where w = f , q, r, s, u, v,
and for all K ∈ T and n∈ {0, . . . , N}
(6)
Numerical analysis for the simulation of a degenerate breast cancer model
fKn+1 − fKn
∆t
n
qn+1
K − qK
∆t
rKn+1 − rKn
∆t
n+1
sK − snK
∆t
un+1
− unK
|K| K
∆t
n+1
= −η f vn+1
− ρ1 fKn+1 ,
K fK
n+ 21
= λq qn+1
K (1 − AK
n+ 12
= λr rKn+1 (1 − AK
n+ 12
= λs sn+1
K (1 − AK
− du
) − ηq unK qKn+1 + ρ1 fKn+1 − ρ2 qKn+1 ,
n+1
) − ηr unK rKn+1 + ρ2 qn+1
K − ρ3 rK ,
n+1
n+1
) − ηs unK sn+1
K + ρ3 rK − ρ4 sK ,
n+1
τKL un+1
+
L − uK
∑
L∈N(K)
|K|
∆t
− dv
n+1
n+1
F un+1
K , uL ; dFKL
∑
(7)
L∈N(K)
= |K|
n
vn+1
K − vK
7
n+1
n+1
λu un+1
K (1 − AK ) + ρ4 sK
,
n+1
n+1
τKL vn+1
= |K| αunK (1 − vn+1
,
L − vK
K ) − β vK
∑
L∈N(K)
n+ 1
n+1
n+1
n+1
n+1
n+1
n+1
n+1
n
+ qn+1
where AK 2 = fKn+1 + qn+1
K + rK + sK + uK and AK = f K
K + rK + sK + uK .
3
In the above scheme, we use a numerical flux function F of arguments (a, b, c) ∈ R for
the approximation of
n+1
n+1
n+1
n+1
which are available in
−
f
:=
τ
f
,
and
dF
,
u
the convection term by means of the values un+1
KL
K
L
KL
L
K
the neighborhood of the common interface σKL .
The numerical flux function F is required to satisfy the following properties:
(H1)
(H2)
(H3)
(H4)
(H5)
F (·, b, c) is nondecreasing for all b, c ∈ R, and F (a, ·, c) is nonincreasing for all a, c ∈ R.
F (a, b, c) = −F (b, a, −c) for all a, b, c ∈ R.
F (a, a, c) = χ (a) c for all a, c ∈ R.
There exists a constant CF > 0 such that: |F (a, b, c)| ≤ CF (|a| + |b|) |c| ∀a, b, c ∈ R.
Locally Lipschitz continuity: there exists a constant C > 0 such that:
F (a, b, c) − F a0 , b0 , c ≤ C |c| a − a0 − b − b0 , ∀a, a0 , b, b0 , c ∈ R.
We give now an example on the construction of the numerical flux F satisfying assumptions (H1)–(H5) is to
split χ into the nondecreasing part χ↑ and the nonincreasing part χ↓ (see e.g. [26])
Z z
χ↑ (z) :=
+
χ 0 (s) ds,
0
χ↓ (z) := −
Z z
−
χ 0 (s) ds.
0
Herein, s+ = max (s, 0) and s− = max (−s, 0). Then we take
F (a, b; c) = c+ χ↑ (a) + χ↓ (b) − c− χ↑ (b) + χ↓ (a) .
(8)
3.3 Main result
Let (Tm )m≥1 be a sequence of admissible meshes of Ω such that
hm = max diam (K) −→ 0 as m → +∞.
K∈Tm
Let (Nm )m be an increasing sequence of integers, then we define the corresponding sequence of time steps
(∆tm )m such that ∆tm → 0 as m → ∞. The intention of this paper is to prove the following main result.
Theorem 1. Let fTm ,∆tm , qTm ,∆tm , rTm ,∆tm , sTm ,∆tm , uTm ,∆tm , vTm ,∆tm m be a sequence of solutions to the scheme
(6)–(7), such that 0 ≤ ωTm ,∆tm ≤ 1 for almost everywhere in QT where ω = f , q, r, s, u, v, then
fTm ,∆tm → f
and
qTm ,∆tm → q
and
rTm ,∆tm → r
a.e. in QT as m → +∞,
sTm ,∆tm → s
and
uTm ,∆tm → u
and
vTm ,∆tm → v
a.e. in QT as m → +∞,
8
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
where the set ( f , q, r, s, u, v) is a weak solution to the system (1)–(3) in the sense of Definition 1.
3.4 Discrete properties and positiveness of the scheme
In this section, we first give a preliminary lemma necessary to prove the positiveness of the finite volume scheme
(6)–(7).
In the sequel, we denote by ζ a continuous real-valued function defined on R+ such that ζ (0) ≥ 0, and
we would like to consider its continuous extension by ζ (0) on R∗− and we denote by U the vector defined by
U = ( f , q, r, s, u, v).
Lemma 1. Let (wnK )K∈T , n∈{0, ..., N+1} be a solution to the following equation
n
wn+1
n+1
K − wK
= ζ (wn+1
K ) − γn wK + σn ,
∆t
in QT ,
(9)
where γn ≥ 0, σn ≥ 0 for all n ∈ {0, . . ., N + 1}.
Assume that w0K ≥ 0 for all K ∈ T . Then, for all K ∈ T and all n ∈ {0, . . ., n + 1}, we have wnK ≥ 0.
Proof. We show this property using an induction on n. The property is true for n = 0 thanks to the assumption
on the nonnegativity of w0K . Now, assume that the claim is true up to time step n, and consider a control volume
n+1 −
n+1
n+1
= 0, where a− = max(−a, 0) =
K such that wn+1
K = min{wL }, we want to show that wK ≥ 0 i.e. wK
L∈T
−
|a| − a
, one has
. Multiplying equation (9) by − wn+1
K
2
n
−
−
−
wn+1
n+1 −
K − wK
wn+1
= −ζ (wn+1
+ γn wn+1
wn+1
− σn wn+1
.
K
K ) wK
K
K
K
∆t
−
+
, the extension by ζ (0) ≥ 0 of the contiuous
= wKn+1 − wn+1
Now, using the following identity wn+1
K
K
function ζ for w ≤ 0 and the nonnegativity of wnK , we get
n+1 − 2
n+1 − 2
n
n+1 −
n+1
n+1 −
n+1 −
− ∆tζ (wK ) wK
− ∆t γn wK
≤ 0.
wK
= −wK wK
+ σn wK
−
According to the choice of the control volume K, then we deduce that {wnK }K∈T is nonnegative for all n ∈
{0, . . ., n + 1}.
3.4.1 Nonnegativity of the solution of scheme (6)–(7)
To prove the discrete maximum principle, we begin by showing the nonnegativity of the solution of the finite
volume scheme (6)–(7).
Proposition 1. Let UKn+1 K∈T , n∈{0,...,N} be a solution to the scheme (6)–(7). Then, for all K ∈ T , and all
n ∈ {0, . . . , N + 1}, we have wnK ≥ 0, where w = f , q, r, s, u, v.
Proof. Let us begin by showing that fKn is nonnegative for all n ∈ {0, . . . , N + 1}. To do this, we replace vn+1
K
+
by vn+1
in the first equation of scheme (7). This substitution has no impact on the scheme because, we will
K
prove later that vn+1
is nonnegative.
K
Taking into account the aforementioned substitution, the first equation of scheme (7) can be rewritten as
follows
fKn+1 − fKn
= ζ fKn+1 − γn fKn+1 + σn ,
∆t
n+1 +
where ζ ≡ 0, γn = η f vK
+ ρ1 , and σn = 0. Thanks to Lemma 1, one can deduce that fKn ≥ 0 for all
n ∈ {0, . . . , N + 1}.
In the same manner, we have
Numerical analysis for the simulation of a degenerate breast cancer model
9
n
ωi n+1
n+ 21
n
n+1
n+1
n+1
K − ωi K
= λωi ωi n+1
K (1 − AK ) − ηωi uK ωi K + ρi ωi−1 K − ρi+1 ωi K ,
∆t
where


q,
ω0 = f , and ωi = r,


s,
for all i ∈ {1, 2, 3}
(10)
if i = 1,
if i = 2,
if i = 3.
Now, we rewrite every equation of system (10) as follows
n
ωi n+1
K − ωi K
= ζ (ωi Kn+1 ) − γi ωi n+1
K + σi ,
∆t
n+ 21
n+1
where ζ (ωi n+1
K ) = λωi ωi K (1 − AK
),
for all i ∈ {1, 2, 3},
σi = ρi ωi−1 n+1
K .
γi = ρi+1 + ηωi unK , and
We have γi ≥ 0, σi ≥ 0, ζ (0) ≥ 0 and for all n ∈ {0, . . . , N + 1}, therefore one can deduce using Lemma 1
that wnK ≥ 0 for all n ∈ {0, . . . , N + 1}, where w = q, r, s.
Now, we want to show that unK ≥ 0 for all n ∈ {0, . . . , N + 1}. To prove this property, we use an induction on
n. The property is true for n = 0 thanks to the assumption (A4) on u0K . Now, assume that the claim is true up to
n+1
n+1
time step n, and consider a control volume K such that un+1
K = min{uL }, we want to show that uK ≥ 0 i.e.
L∈T
−
−
uKn+1 = 0. Multiplying the second-last equation of scheme (7) by − uKn+1 , we have
− |K|
n
−
n+1 −
un+1
n+1
K − uK
un+1
+ du ∑ τKL un+1
uK
K
L − uK
∆t
L∈N(K)
n+1 −
n+1 −
n+1
n+1
n+1
= ∑ F un+1
uK
− |K| ζ (un+1
uK
,
K , uL ; dFKL
K ) + ρ4 sK
L∈N(K)
n+1
where ζ : R+ −→ R is the continuous function defined by ζ (uKn+1 ) = λu un+1
K (1 − AK ) and verifying ζ (0) ≥ 0.
n+1
n+1
Now, since uK ≤ uL due to the choice of the control volume K, one deduces that
∑
n+1
τKL un+1
L − uK
un+1
K
−
≥ 0.
(11)
L∈N(K)
Now, we use the assumptions on the numerical flux F. Recall that the function F is a continuous nonincreasing
function with respect to the second variable un+1
L . Therefore, using the extension by zero of the continuous
function χ (we have χ (u) = 0 for u ≤ 0), we get
n+1
n+1
F un+1
K , uL ; dFKL
un+1
K
−
n+1
≤ F uKn+1 , un+1
K ; dFKL
un+1
K
−
n+1
≤ dFKL
χ un+1
K
uKn+1
−
= 0.
(12)
Using estimates (11)–(12) and the extension by zero of the continuous function ζ , one gets
|K|
−
n+1 − 2
uK
+ unK uKn+1
∆t
n+1 −
Consequently, we deduce that uK
is nonnegative; this implies that
n+1
≤ − |K| ζ (un+1
K ) + ρ4 sK
un+1
K
−
≤ 0.
= 0. According to the choice of the control volume K, then min {un+1
L }L∈T
unK ≥ 0 for all n ∈ {0, . . . , N + 1} and all K ∈ T .
It remains to show that vnK ≥ 0 for all n ∈ {0, . . . , N + 1}. To do that, we argue by induction that for all K ∈ T ,
the claim vnK ≥ 0 is true. We take the control volume K such that vKn+1 = min{vn+1
L }L∈T , we want to show that
vn+1
≥
0.
K
−
For the mentioned claim, we multiply the last equation of scheme (7) by − vn+1
, one gets
K
10
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
|K|
−
n+1 − 2
vK
+ vnK vn+1
K
∆t
+ dv
∑
n+1
τKL vn+1
L − vK
vn+1
K
=
L∈N(K)
− |K| αunK vn+1
K
Since vKn+1 ≤ vn+1
for all L ∈ T , then
L
−
∑
n+1
τKL vn+1
L − vK
−
vn+1
K
− 2
− |K| (αunK + β ) vn+1
.
K
−
≥ 0. Consequently, one gets
L∈N(K)
−
n+1 − 2
vK
+ vnK vn+1
K
∆t
≤ 0.
Finally, we deduce that vn+1
K ≥ 0. This ends the proof of this proposition.
3.4.2 Confinement of the solution of scheme (6)–(7)
Here, we want to show that the discrete solution of the finite volume scheme (6)–(7), if it exists, is bounded in
L∞ (QT ).
Proposition 2. Let UKn+1 K∈T , n∈{0,...,N} be a solution to the finite volume scheme (6)–(7). Then, for all K ∈ T ,
and all n ∈ {0, . . . , N + 1}, we have wnK ≤ 1, where w = f , q, r, s, u, v.
Proof. In order to prove that fKn , qnK , rKn , snK ≤ 1 for all n ∈ {0, . . . , N + 1}, we denote by ωKn = fKn + qnK + rKn + snK
and show by induction on n that for all K ∈ T , ωKn ≤ 1. The claim is true up to order 0. We argue by induction
that for all K ∈ T the claim is true up to order n. Consider a control volume K ∈ T , we want to show that
ωKn+1 ≤ 1.
To do that, we sum the first four equations of scheme (7) corresponding to the control volume K, one gets
n
ωKn+1 − ωKn
n+1
n+1
uK
1 − ωKn+1 − λs sKn+1 + λr rKn+1 + λq qn+1
= λs sn+1
K
K + λr rK + λq qK
∆t
n+1
n n+1
n n+1
n+1
−η f vn+1
− ηq unK qn+1
K fK
K − ηr uK rK − ηs uK sK − ρ4 sK .
(13)
−
+
Multiplying equation (13) by (ωKn+1 − 1)+ and using the identity ωKn+1 − 1 = ωKn+1 − 1 − ωKn+1 − 1 as
well as the nonnegativity of the discrete solution given in Proposition 1, one deduces that the first term on the
right hand side of equation (13) is nonpositive and that
+
+
ωKn+1 − 1 − (ωKn − 1) n+1
ωKn+1 − ωKn
n+1
ωK − 1 =
ωK − 1 ≤ 0.
(14)
∆t
∆t
+
−
Using again the identity ωKn+1 − 1 = ωKn+1 − 1 − ωKn+1 − 1 , and that ωKn ≤ 1, one can deduce from
+
estimate (14) that ωKn+1 − 1 = 0. Consequently, we obtain ωKn+1 ≤ 1, for all n ∈ {0, . . . , N} and all K ∈ T .
Let us now focus on unK and show by induction on n that for all for all n ∈ {0, . . . , N + 1}, unK ≤ 1. The claim
is true up to order 0. We argue by induction that for all K ∈ T , the claim is true up to order n. Consider the
+
control volume K ∈ T such that un+1
= maxL∈T {uLn+1 }, we want to show that un+1
≤ 1 i.e. un+1
= 0.
K
K
K −1
We consider the second-last equation of scheme (7) corresponding to the control volume K and multiply it by
+
uKn+1 − 1 . This yields
|K|
+
n+1
+
uKn+1 − unK n+1
uK − 1 − du ∑ τKL uLn+1 − un+1
uK − 1
K
∆t
L∈N(K)
n+1
+
n+1
+
n+1 n+1
n+1
n+1
n+1
+ ∑ F uK , uL ; dFKL
uK − 1 = |K| λu un+1
uK − 1 .
K (1 − AK ) + ρ4 sK
L∈N(K)
Since uKn+1 ≥ un+1
because of the choice of the control volume K, then one has
L
(15)
Numerical analysis for the simulation of a degenerate breast cancer model
du
n+1
τKL un+1
L − uK
∑
11
+
uKn+1 − 1 ≤ 0.
(16)
L∈N(K)
Next, we use assumptions (H1) and (H3) on the numerical flux function F to deduce thanks to the extension by
zero of the continuous function χ for u ≥ 1, that
n+1
n+1
F un+1
K , uL ; dFKL
+
n+1
+
n+1
n+1
un+1
≥ F un+1
uK − 1
K −1
K , uK ; dFKL
n+1 n+1
+
= χ uKn+1 dFKL
uK − 1 ≥ 0.
(17)
Now, we want to show that the term on the right-hand side of equation (15) is nonpositive. Indeed, thanks to the
nonnegativity of the discrete solution, remark that
n+1
n+1
n+1
n+1
n+1 n+1
n+1
λu un+1
K (1 − AK ) + ρ4 sK ≤ λu uK (1 − uK ) − λu uK sK + ρ4 sK .
n+1
n+1 n+1
n+1
Denote by ϒ the differentiable function defined on R+ by ϒ (uKn+1 ) := λu un+1
K (1 − uK ) − λu uK sK + ρ4 sK
n+1
n+1
n+1
for a given sK ∈ [0,
ϒ (1) = (ρ4 − λu )sK ≤ 0 (due to assumption (A3)) and ϒ 0 (uK ) =
1]. We have,
n+1
n+1
λu 1 − 2un+1
−
s
≤
−λ
for
u
≥ 1 i.e. ϒ is nonincreasing for un+1
≥ 1 and ϒ (un+1
u
K
K
K
K
K ) ≤ ϒ (1) ≤ 0 for
n+1
uK ≥ 1. Consequently, one deduces that
n+1
n+1
|K| λu un+1
K (1 − AK ) + ρ4 sK
+
n+1
+
un+1
≤ |K|ϒ un+1
uK − 1 ≤ 0.
K −1
K
(18)
−
+
n+1
− uKn+1 − 1 ,
Plugging estimates (16)–(18) into equation (15), and using the identity un+1
K − 1 = uK − 1
and that unK ≤ 1, one can deduce that
n
n
+
+
un+1
un+1
K − 1 − (uK − 1)
n+1
K − uK
|K|
uK − 1 = |K|
un+1
≤ 0.
(19)
K −1
∆t
∆t
−
+
n+1
− uKn+1 − 1 , and that unK ≤ 1, one can deduce from estiUsing again the identity un+1
K − 1 = uK − 1
+
= 0. Consequently, we obtain un+1
mate (19) that un+1
K ≤ 1, for all n ∈ {0, . . . , N} and all K ∈ T .
K −1
It remains to show that vnK ≤ 1 for all n ∈ {0, . . . , N + 1}. Similarly, we argue by induction that for all K ∈ T
the claim is true up to order n. Consider a control volume K ∈ T such that vnK = max{vnL }L∈T . We want to show
that vn+1
K ≤ 1. For that, we consider thelast equation of scheme (7) corresponding to the aforementioned control
+
volume K and multiply it by vn+1
K − 1 . This yields
|K|
n
+
n+1
+
vn+1
K − vK
vn+1
− dv ∑ τKL vLn+1 − vn+1
vK − 1
K −1
K
∆t
L∈N(K)
= |K| αunK (1 − vKn+1 ) vKn+1 − 1
(20)
+
− β vKn+1
+
vn+1
.
K −1
n+1
According to the choice of the control volume K, we have vn+1
K ≥ vL . This implies that
−dv
∑
τKL vLn+1 − vn+1
K
+
vKn+1 − 1 ≥ 0.
L∈N(K)
+
−
n+1
Finally, we use the identity vn+1
− vn+1
as well as the nonnegativity of the discrete
K − 1 = vK − 1
K −1
solution given in Proposition 1, one deduces that
n
n
+
+
vn+1
vn+1
K − 1 − (vK − 1)
n+1
K − vK
|K|
vK − 1 = |K|
vn+1
≤ 0.
(21)
K −1
∆t
∆t
−
+
n+1
Using again the identity vn+1
− vKn+1 − 1 , and that vnK ≤ 1, one can deduce from estimate
K − 1 = vK − 1
+
(21) that vn+1
= 0. Consequently, we obtain vKn+1 ≤ 1, for all n ∈ {0, . . . , N} and all K ∈ T . This ends
K −1
the proof of this proposition.
12
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
3.5 A priori estimates and existence
In this section, we establish the a priori estimates necessary to prove the existence of at least one solution to the
discrete problem (6)–(7) and the convergence of the scheme towards the weak solution.
The first step consists of deriving the main uniform estimates on the discrete gradient of the tumor cell density u
and the discrete gradient of the chemical concentration v. However, in order to estimate the discrete gradient of
the tumor cell density u, we have to estimate the discrete gradient of the fraction of healthy breast stem cells f .
Therefore, the major difficulty in the proof consists of obtaining an estimation on the discrete spatial derivatives
on f because there is no spatial terms in the equation on f .
3.5.1 Energy estimate on fT
Here, the goal is to derive an estimation on the discrete gradient of f . We have the following proposition.
Proposition 3. Let UKn+1 K∈T , n∈{0,...,N} be a solution to the scheme (6)–(7). Then for all δ > 0 and for all
n ∈ {0, . . . , N}, one has
n+1 2
f
T
1,T
2
− fTn 1,T
2∆t
2
2
η2 + f f n+1 2 .
+ ρ1 fTn+1 1,T ≤ δ vn+1
T
T
1,T
1,T
4δ
(22)
Proof. In order to obtain an estimation on the discrete spatial derivative on f , we multiply the first equation of scheme (7) by minus the discrete laplacian of f namely the piecewise constant function defined by
1
−
∑ τKL fLn+1 − fKn+1 on every control volume K ∈ T and then we perform a sum over K ∈ T . This
|K| L∈N(K)
yields
∑
K∈T
fKn+1 − fKn
∆t
τKL fKn+1 − fLn+1 + η f
∑
∑
n+1
vn+1
K fK
K∈T
L∈N(K)
+ ρ1
∑
∑
τKL fKn+1 − fLn+1
L∈N(K)
fKn+1
K∈T
∑
τKL fKn+1 − fLn+1 = 0.
L∈N(K)
We reorganize the sum over the edges (integration by parts on both sides), we find
E1 + E2 + E3 = 0,
(23)
where
1
∑ τKL fKn+1 − fLn+1 fKn+1 − fLn+1 − ( fKn − fLn ) ,
∆t σKL
∈E
n+1
E2 = η f ∑ τKL fKn+1 − fLn+1 vKn+1 fKn+1 − vn+1
,
L fL
E1 =
σKL ∈E
E3 = ρ1
∑
σKL ∈E
τKL fKn+1 − fLn+1
2
2
= ρ1 fTn+1 1,T .
1 2
a − b2 , ∀a, b ∈ R. This yields
2
f n+1 2 − f n 2
1
T 1,T
T
1,T
n+1
n+1 2
n
n 2
E1 ≥
τKL fK − fL
− ( fK − fL ) =
.
∑
2∆t σKL ∈E
2∆t
For the first term, we use the following inequality: (a − b) a ≥
For the second term of equation (23), we write
n+1
n+1
n+1
n+1
E2 = η f ∑ τKL fKn+1 − fLn+1 vn+1
− vn+1
+ vn+1
− vn+1
= E2,1 + E2,2 ,
K fK
K fL
K fL
L fL
σKL ∈E
(24)
Numerical analysis for the simulation of a degenerate breast cancer model
13
where
E2,1 = η f
2
τKL fKn+1 − fLn+1 vn+1
K ,
∑
and E2,2 = η f
σKL ∈E
∑
τKL fKn+1 − fLn+1
n+1
n+1
vn+1
fL .
K − vL
σKL ∈E
Now, using the weighted Young inequality and the fact that vKn+1 ≥ 0 and fKn+1 ≤ 1 for all K ∈ T , one can deduce
the following estimates
2
η2 + f f n+1 2 .
(25)
E2,1 ≥ 0 and |E2,2 | ≤ δ vn+1
T
T
1,T
1,T
4δ
Plugging estimates (24)–(25) into equation (23), one can deduce that estimate (3) holds.
3.5.2 Energy estimate on uT
In the following, we denote by C a generic constant, which need not have the same value throughout the proofs.
Proposition 4. Let UKn+1 K∈T , n∈{0,...,N} be a solution to the scheme (6)–(7). Then for all 0 < ε < du and for
all n ∈ {0, . . . , N}, one has
n+1 2
u 2 − un 2 2
T L (Ω )
T
L (Ω )
2∆t
2
2
2
C2 ρ4
+ (du − ε) uTn+1 1,T ≤ F fTn+1 1,T + (ρ4 + λu ) uTn+1 L2 (Ω ) + |Ω | .
ε
4
(26)
Proof. We multiply the second-last equation of scheme (7) by un+1
and sum over K ∈ T , this yields
K
∑
|K|
K∈T
n
un+1
n+1
n+1
n+1
K − uK n+1
+ ∑ un+1
F un+1
uK + du ∑ un+1
τKL un+1
∑
∑
K
K , uL ; dFKL
K
K − uL
∆t
K∈T
K∈T
L∈N(K)
L∈N(K)
n+1
n+1
n+1
= ∑ |K| λu uK (1 − AK ) + ρ4 sK un+1
K .
K∈T
We reorganize the sum over the edges, we find
E1 + E2 + E3 = E4 ,
(27)
where
E1 =
∑
|K|
K∈T
E3 =
∑
F
n
un+1
K − uK n+1
uK
∆t
n+1
un+1
K , uL ;
E2 = du
∑
σKL ∈E
n+1
dFKL
n+1
un+1
K − uL
2
n+1 2
τKL un+1
= du uTn+1 1,T
K − uL
E4 =
σKL ∈E
∑
n+1
n+1
n+1
|K| λu un+1
uK .
K (1 − AK ) + ρ4 sK
K∈T
1 2
a − b2 , ∀a, b ∈ R, then the following estimate holds
2
n+1 2
n+1 2
u 2 − un 2 2
u − |un |2
T L (Ω )
T
L (Ω )
K
K
E1 ≥ ∑ |K|
=
.
2∆t
2∆t
K∈T
Using the inequality: (a − b) a ≥
(28)
Let us now focus on the third term of equation (27). Thanks to Proposition 2, assumption (H4) on the numerical
flux function F and the weighted Young inequality, one has
|E3 | ≤ 2CF ∑ dF n+1 un+1 − un+1 = 2CF ∑ τKL f n+1 − f n+1 un+1 − un+1 KL
K
L
K
σKL ∈E
L
σKL ∈E
≤
2 CF
f n+1 2 + ε un+1 2 .
T
T
1,T
1,T
ε
K
L
(29)
Finally, for the last term of equation (27) and thanks to Propositions 1 and 2 and to the weighted Young inequality, one has
14
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
E4 =
∑
n+1
n+1
n+1
|K| λu un+1
uK ≤
K (1 − AK ) + ρ4 sK
∑
n+1
n+1
|K| λu un+1
uK
K + ρ4 sK
K∈T
K∈T
2
2 + ρ4 sn+1 2 2
≤ (ρ4 + λu ) un+1
T
T
L (Ω )
L (Ω )
4
n+1 2
ρ
4
≤ (ρ4 + λu ) uT L2 (Ω ) + |Ω | .
4
(30)
Collecting estimates (28)–(30), and plugging them into equation (27), one deduces that estimate (26) holds.
This ends the proof of this proposition.
3.5.3 Energy estimate on vT
Proposition 5. Let UKn+1 K∈T , n∈{0,...,N} be a solution to the scheme (6)–(7). Then for all n ∈ {0, . . . , N}, one
has
n+1 2
v 2 − vn 2 2
2
2
α
α
T L (Ω )
T
L (Ω )
2 .
≤ kunT k2L2 (Ω ) + vn+1
(31)
+ dv vn+1
T
T
1,T
L (Ω )
2∆t
2
2
and sum over K ∈ T , we get
Proof. We multiply the last equation of scheme (7) by vn+1
K
∑
K∈T
|K|
n
vn+1
n+1
n+1 n+1
K − vK n+1
vK + dv ∑ vn+1
τKL vn+1
= ∑ |K| αunK (1 − vn+1
vK .
∑
K
K − vL
K ) − β vK
∆t
K∈T
K∈T
L∈N(K)
Reorganizing the sum over the edges and using the inequality: (a − b) a ≥
n+1 2
v 2 − vn 2 2
T L (Ω )
T
L (Ω )
2∆t
2
+ dv vn+1
≤
T
1,T
∑
1 2
a − b2 , ∀a, b ∈ R, one has
2
n+1
|K| αunK (1 − vn+1
K ) vK .
(32)
K∈T
Now, thanks to Proposition 1 and 2, and to the weighted Young inequality, one can deduce from estimate (32)
that estimate (31) holds. This ends the proof of the proposition.
Now, we give the main proposition of this section.
Proposition 6 (Discrete energy estimates). Let UKn+1 K∈T , n∈{0,...,N} be a solution to the finite volume scheme
(6)–(7). Under assumption (A4), there exists a constant C > 0 such that for all n ∈ {0, . . . , N}, one has
N
∆t
∆t
∆t
2
≤ C,
(33)
n=0
N 2
≤ C,
(34)
n=0
N 2
≤ C.
(35)
∑ fTn+1 1,T
∑ un+1
T
1,T
∑ vn+1
T
1,T
n=0
dv
du
Proof. In order to prove estimates (33)–(35), we fix δ =
and ε = . Then, we sum estimates (22), (26),
2
2
(31), one gets
f n+1 2 + un+1 2 2 + vn+1 2 2
− | fTn |21,T + kunT k2L2 (Ω ) + kvnT k2L2 (Ω )
T
T
T
1,T
L (Ω )
L (Ω )
2
+ ∆tdv vn+1 2 + 2∆tρ1 f n+1 2
(36)
+∆tdu un+1
T
T
T
1,T
1,T
1,T
n+1 2
n+1 2
n+1 2
≤ C1 ∆t fT 1,T +C2 ∆t uT L2 (Ω ) +C3 ∆t vT L2 (Ω ) +C4 ∆t,
η 2f
2
4CF
ρ4 |Ω |.
, C2 = 2 (ρ4 + λu ), C3 = α, and C4 = α +
dv
du
2
We define the constant C5 as the maximum of C1 , C2 , and C3 . Then, equation (36) implies in particular that
where C1 =
+
Numerical analysis for the simulation of a degenerate breast cancer model
15
yn+1 − yn ≤ C4 ∆t +C5 ∆tyn+1 ,
(37)
n 2
n 2
T 1,T + uT L2 (Ω ) + vT L2 (Ω ) .
Now, we apply the discrete Gronwall’s inequality given by the following Lemma.
2
where yn = f n Lemma 2 (Discrete Gronwall’s inequality). Given K1 ≥ 0 and K2 > 0. Let (yn )n∈N be real sequence verifying
1
and a fixed time T > 0. Then,
0 ≤ yn+1 ≤ yn + K1 ∆t + K2 ∆tyn+1 for all n ∈ N. Given a fixed time-step ∆t0 <
K2
for all 0 < ∆t ≤ ∆t0 , we have
K1
K2 T
n
0
y ≤ y +
exp
, for all n ∈ N such that n∆t ≤ T.
K2
1 − K2 ∆t0
The proof of this lemma can be obtained using an induction on n (see e.g. [32]). As a consequence of Lemma
2, one can deduce thanks to inequality (37) that
N
∆t
∑ yn+1 ≤ C6 ,
n=0
C5 T
y0 + CC4 . This proves in particular estimates (33) with C = C6 .
where C6 = T exp 1−C
∆t0
5
5
Let us now focus on estimate (34). Indeed, estimate (36) implies that
N
yN+1 − y0 + du ∆t
2
+ dv ∆t
∑ un+1
T
1,T
n=0
This proves estimate (34) with C =
y0 + TC4 +C5C6
.
du
In the same manner, we prove estimate (35) with C =
N
2
∑ vn+1
T
1,T
≤ TC4 +C5C6 .
n=0
y0 +C4 T +C5C6
. This ends the proof of the Proposition.
dv
3.5.4 Existence of a discrete solution
The existence of a discrete solution to the finite volume scheme (6)–(7) will be obtained with the help of
Brouwer fixed-point theorem proved for example in [9, 21, 29]. Specifically, we have the following proposition.
Proposition 7. Consider an admissible mesh of QT as described in § 3.1.1. Given (UKn )K∈T , then there exists at
least one solution UKn+1 K∈T to the scheme (6)–(7).
Proof. We first provide the existence of vTn+1 a unique solution to the discrete last equation of system (7). Note
that this equation is a standard time-implicit finite volume discretization of a uniformly parabolic equation,
where the contribution of u in the right-hand side is discretized in an explicit way. Hence, the aforementioned
discrete equation can be written as a finite dimensional linear system Avn+1
= B with respect to the unknown
T
n+1
n+1
vT (which is characterized by its numerical values vK K∈T ). The coefficients (Ai, j )i, j∈{1,...,#T } of the matrix
A and the components (Bi )i∈{1,...,#T } of the vector B are given, for all i, j ∈ {1, . . . , #T }, i 6= j, by
Ai,i = |K| (1 + ∆t (β + αunK )) + dv ∆t
∑
τKL ,
Ai, j = −dv ∆tτKL ,
Bi = |K| (α∆tunK + vnK ) .
L∈N(K)
It is clear that the resulting matrix A of this system is symmetric and strictly diagonally dominant with nonnegative diagonal coefficients. As a consequence, A is invertible and definite positive. Thus the last equation of
system (7) admits a unique solution vn+1
T .
Using this unique solution, one can deduce the existence of a uniquesolution fTn+1 to the
discrete first equation
n+1
n+1 +
n
of system (7). Indeed, this solution is given by: fK = fK / 1 + ∆t η f vK
+ ρ1 .
16
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
The remained proof of this proposition follows the same lines as in [5] where, we introduce the truncated version of the discrete system (6)–(7). Then,
by an application of the Brouwer fixed-point theorem, we deduce the
existence of a discrete solution UKn+1 K∈T ∈ (HT )6 to the finite volume scheme (6)–(7).
Remark 2. We can prove by induction on n that the discrete solutions vnT and fTn ≥ 0 for all n = {0, . . . , N + 1}
under the assumption that the given vector (UKn )K∈T is nonnegative. Indeed, one can remark that the strictly
diagonally dominant matrix A has nonpositive off-diagonal and positive diagonal entries, so that A is a nonsingular M-matrix and vnT , fTn ≥ 0.
4 Compactness estimates on discrete solutions
We give below the time translate estimate for the family ωTm ,∆tm
Ω × (0, T − τ), for all τ ∈ (0, T ).
m
, where ω = f , u, v. We denote by QT −τ :=
Proposition 8 (Time translate estimates). Let ∆t0 > 0 small enough. Given a time-step ∆tm ≤ ∆t0 , then there
exists a constant C > 0 independent of m and τ such that,
ZZ
QT −τ
ωT ,∆t (x,t + τ) − ωT ,∆t (x,t)2 dx dt ≤ C (τ + ∆tm ) ,
m
m
m
m
ω = f , u, v
(38)
for all τ ∈ (0, T ).
Proof. Let us prove estimate (38) for ω = u, the proof being similar for ω = f , v. We consider the quantity
Am (t) defined by
Am (t) =
Z uT ,∆t (x,t + τ) − uT ,∆t (x,t)2 dx,
m
m
m
m
for all t ∈ (0, T − τ) .
Ω
For t ∈ (0, T ], we denote by ξ (t) ∈ {0, . . . , N} the unique positive integer such that tξ (t) < t ≤ tξ (t)+1 , so that,
we can rewrite Am (t) as follows
Am (t) =
ξ (t+τ)+1
ξ (t)+1 2
|K| =
uK
− uK
∑
K∈Tm
ξ (t+τ)
∑
∑
n
|K| un+1
K − uK
ξ (t+τ)+1
ξ (t)+1
uK
− uK
, ∀t ∈ (0, T − τ)
K∈Tm n=ξ (t)+1
Now, using the second-last equation of scheme (7) then gathering by edges and using the weighted Young
inequality as well as Proposition 1 and 2, we obtain
ξ (t+τ)
Am (t) =
∑
n=ξ (t)+1
∆tm
∑
h
n+1
n+1
du τKL un+1
+ F uKn+1 , un+1
K − uL
L ; dFKL
σKL ∈Em
×
i
ξ (t)+1
ξ (t)+1
ξ (t+τ)+1
ξ (t+τ)+1
− uK
− uL
uK
− uL
ξ (t+τ)
+
∑
n=ξ (t)+1
∆tm
∑
ξ (t+τ)+1
ξ (t)+1
n+1
n+1
|K| λu un+1
1
−
A
+
ρ
s
u
−
u
4
K
K
K
K
K
K∈Tm
≤ A1,m (t) + A2,m (t) + A3,m (t) + A4,m (t) + A5,m (t),
where, we have set
Numerical analysis for the simulation of a degenerate breast cancer model
ξ (t+τ)
A1,m (t) =
∑
σKL ∈Em
n=ξ (t)+1
A2,m (t) =
A3,m (t) =
n+1 2
τKL un+1
K − uL
∑
∆tm
ξ (t+τ)
2
du2 +CF
∑
2
∆tm
∑
∆tm
σKL ∈Em
n=ξ (t)+1
∑
∆tm
2
τKL fKn+1 − fLn+1 σKL ∈Em
n=ξ (t)+1
ξ (t+τ)
A5,m (t) = C
∑
ξ (t+τ)+1
ξ (t+τ)+1 2
− uL
τKL uK
ξ (t+τ)
ξ (t+τ)
∑
∑
ξ (t)+1
ξ (t)+1 2
− uL
τKL uK
σKL ∈Em
n=ξ (t)+1
2
du2 +CF
2
A4,m (t) =
17
∑
∑
∆tm
K∈Tm
n=ξ (t)+1
ξ (t+τ)+1
ξ (t)+1 |K| uK
− uK
Now, we introduce the characteristic function ρ (n,t, τ) = 1 if t < n∆tm ≤ t + τ and ρ (n,t, τ) = 0 otherwise. Let
(an )n∈{0,...,N} be a family of non negative real values, we have the following properties
Z T −τ
tn −τ
0
Z T −τ ξ (t+τ)
∑
0
ξ (t+τ)
Z tn
ρ (n,t, τ) dt =
dt = τ,
∆tm an+1 dt =
n=ξ (t)+1
∆tm =
∑
∑
Z T −τ N
0
tn+1 − tn ≤ τ + ∆tm , and
n;t≤tn <t+τ
n=ξ (t)+1
N
∑ ∆tm an+1 ρ (n,t, τ) dt = τ ∑ ∆tm an+1 .
n=0
n=0
Using these properties and the a priori estimates (33)–(35) on fTm , uTm and vTm , one can deduce that
Z T −τ
0
A1,m (t) dt ≤ C (τ + ∆tm ),
Z T −τ
0
A4,m (t) dt ≤ C (τ + ∆tm ),
for some constant C > 0 independent of m and τ.
Now, we consider the function ζ defined by ζ (n,t) = 1 if ξ (t) = n and ζ (n,t) = 0 otherwise. We have the
following result (see e.g. [12])
!
Z
Z
ξ (t+τ)
T −τ
∑
0
∆tm aξ (t+τ)+1 dt ≤ (τ + ∆tm )
n=ξ (t)+1
T −τ N
0
N
∑ am+1 ζ (m,t + τ) dt = (τ + ∆tm ) ∑ am+1 ∆tm .
m=0
m=0
One can conclude the proof using again the a priori estimates (33)–(35). This ends the proof of this proposition.
We give now the space translate estimate for the family ωTm ,∆tm m , where ω = f , u, v.
Proposition 9 (Space translate estimates). Let ∆t0 > 0 small enough. Given a time-step ∆tm ≤ ∆t0 , then there
exists a constant C > 0 independent of m and τ such that, for all y ∈ Rd
Z TZ
0
Ωy
ωT ,∆t (x + y,t) − ωT ,∆t (x,t)2 dx dt ≤ C kyk (kyk + 2 (KΩ − 1) hm ) ,
m
m
m
m
2
2
ω = f , u, v
(39)
where Ωy = {x ∈ Ω | x + y ∈ Ω } and KΩ is the number of sides of Ω .
Proof. The proof of this proposition is developed by Eymard and al. in [10, Lemma 9.3]. Indeed, following the
same lines of their proof, one gets
Z TZ
0
ωT ,∆t (x + y,t) − ωT ,∆t (x,t)2 dx dt
m
m
m
m
Ωy
N
≤ kyk2 (kyk2 + 2 (KΩ − 1) hm ) ∆tm
∑ ∑
n=0 σKL ∈E
2
τKL ωKn+1 − ωLn+1 ,
ω = f , u, v.
18
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
Thanks to estimates (33)–(35), one can deduce space translate estimates. This ends the proof of the proposition.
5 Convergence of the finite volume scheme
In this section, we give the main result in this paper. Specifically, we have the following Proposition.
Proposition 10 (Convergence in L2 (QT )). Let UTm ,∆tm m>0 be a solution to the finite volume scheme (6)–(7),
for T = Tm . Then, there exists f , u, v ∈ L2 0, T ; H 1 (Ω ) and q, r, s ∈ L2 (Ω ) such that, up to a subsequence
ωTm ,∆tm −→ ω
ξTm ,∆tm −→ ξ ,
strongly in L p (QT ) and a.e. in QT for all 1 ≤ p < +∞,
2
weakly in L (QT ),
ω = f , u, v,
ξ = q, r, s.
(40)
(41)
Furthermore,
∇Tm ωTm ,∆tm −→ ∇ω,
d
weakly in L2 (QT )
ω = f , u, v.
(42)
Proof. Let us prove the first convergence (40) for ω = f , since the proof is similar for ω = u, v. We want to
apply the Kolmogorov’s compactness criterion (see [14], [10, Theorem 3.9, p: 93]) for the study of the sequence
fTm ,∆tm m∈N . The first and the second items of the Kolmogorov compactness criterion are satisfied with N =
d + 1, q = 2, and p fTm ,∆tm = f˜Tm ,∆tm with f˜ defined by f˜Tm ,∆tm = fTm ,∆tm on QT and f˜Tm ,∆tm = 0 outside of
QT . It suffices to prove the third item of the Kolmogorov theorem; indeed, using the triangle inequality, one has
for any y ∈ Rd and τ ∈ R,
f˜T
m ,∆tm
(· + y, · + τ) − f˜Tm ,∆tm (·, ·)L2 (Rd+1 ) ≤ f˜Tm ,∆tm (· + y, ·) − f˜Tm ,∆tm (·, ·)L2 (Rd+1 )
+ f˜Tm ,∆tm (·, · + τ) − f˜Tm ,∆tm (·, ·)L2 (Rd+1 ) .
Propositions 8 and 9 ensure that f˜Tm ,∆tm (· + y, · + τ) − f˜Tm ,∆tm (·, ·)L2 (Rd+1 ) → 0, as y → 0 and τ → 0. This
yields the compactness of the sequence fTm ,∆tm m in L2 (QT ). Therefore, by application of the Kolmogorov
theorem, there exists a subsequence still denoted by fTm ,∆tm m , and there exists f ∈ L2 (QT ) such that
fTm ,∆tm −→ f
strongly in L2 (QT ) and a.e. in QT .
This convergence holds also in L p (QT ) for all 1 ≤ p < +∞ since the sequence fTm ,∆tm m∈N is bounded in
L∞ (QT ).
Let us now prove the second convergence (41). Indeed, the sequence qTm ,∆tm m (resp. rTm ,∆tm m , sTm ,∆tm m ) is
uniformly bounded in L2 (QT ). Consequently, the sequence qTm ,∆tm m (resp. rTm ,∆tm m , sTm ,∆tm m ) converges
weakly in L2 (QT ), up to unlabeled subsequence, to a function q in L2 (QT ) (resp. r, s in L2 (QT )).
It remains to show the last convergence (42) for ω = f , the proof for ω = u, v being similar. Thanks to the
d
discrete energy estimate (33), one has that the sequence ∇Tm fTm ,∆tm m is uniformly bounded in L2 (QT ) .
As a consequence, the sequence ∇Tm fTm ,∆tm m converges weakly, up to unlabeled subsequence, to a function
d
p? ∈ L2 (QT ) . It remains to identify ∇ f by p? in the sense of distributions. For that, it is enough to use the
following convergence (see e.g. [3, 4] for more details)
Z TZ
0
∇Tm fTm ,∆tm · φ + fTm ,∆tm div (φ ) dx dt −→ 0,
as m → +∞,
∀φ ∈ D (QT )d .
Ω
5.1 Identification as a weak solution
In this section, we want to show that the set of functions ( f , q, r, s, u, v) defined in Proportion 10 constitutes a
weak solution to the continuous problem (1)–(3) in the sense of Definition 1. We will show the convergence for
the first and the second-last equation of the scheme (7). The convergence for the other equations being similar.
Numerical analysis for the simulation of a degenerate breast cancer model
19
A. Convergence of the discrete health breast stem
cells equation
We consider a test function ψ ∈ D Ω × [0, T ) , and denote by ψKn = ψ (xK ,tn ), for all K ∈ Tm and all n ∈
{0, . . . , Nm }. Multiplying the first equation of the scheme (7) by ∆tm ψKn and summing over n ∈ {0, . . . , Nm } and
K ∈ Tm yields
Am = Bm + Cm ,
(43)
where
Nm
Am =
∑ ∑
Nm
|K| fKn+1 − fKn ψKn ,
Bm = − ∑ ∆tm
n=0 K∈Tm
n=0
Nm
Cm = − ∑ ∆tm
n=0
∑
n+1 n
|K| vn+1
K f K ψK ,
K∈Tm
|K| ρ1 fKn+1 ψKn .
∑
K∈Tm
Note that ψKNm +1 = 0 for all K ∈ Tm , then, performing a summation by parts in time, the first term Am can be
rewritten
Nm
Am =
Nm
∑ ∑
n=0 K∈Tm
|K| fKn+1 ψKn − ∑
∑
|K| fKn ψKn −
n=1 K∈Tm
K∈Tm
Nm
ψ n+1 − ψKn
=−
∆tm
fKn+1 K
∆tm
n=0
K∈Tm
∑
=−
∑
ZZ
|K| −
∑
|K| fK0 ψK0
K∈Tm
fTm ,∆tm (x,t) ∂t ψTm ,∆tm (x,t) dx dt −
QT
|K| fK0 ψK0
∑
Z
fTm ,∆tm (x, 0) ψTm ,∆tm (x, 0) dx.
Ω
p
Thanks to
the regularity of ψ, and the strong convergence in L (QT ), for all 1 ≤ p < +∞, of the sequence
fTm ,∆tm m towards f , it follows that
Am −→ −
ZZ
QT
f (x,t) ∂t ψ (x,t) dx dt −
Z
f (x, 0) ψ (x, 0) dx,
as m → +∞.
Ω
We now focus on the second term of equation (43) and prove that
lim Bm = lim −
m→∞
m→∞
ZZ
QT
η f vTm ,∆tm fTm ,∆tm ψTm ,∆tm (·,t − ∆tm ) dx dt = −
ZZ
QT
η f v (x,t) f (x,t) ψ (x,t) dx dt. (44)
Thanks to Hölder’s inequality as well as the triangle inequality, one has
vT ,∆t fT ,∆t ψT ,∆t − v f ψ 1
m
m
m
m
m
m
L (QT )
≤ vTm ,∆tm L∞ (Q ) fTm ,∆tm ψTm ,∆tm − f ψ L1 (Q ) + k f kL2 (QT ) vTm ,∆tm − vL2 (Q ) kψkL∞ (QT ) .
T
T
T
Using the regularity of the test function ψ, the boundedness of the sequence vTm ,∆tm in L∞ (QT ) and its strong
convergence in L2 (QT ) towards v, one can deduce that
vT ,∆t fT ,∆t ψT ,∆t − v f ψ 1
−→ 0, as m → +∞,
m
m
m
m
m
m
L (Q )
T
and consequently, convergence (44) holds.
In the same manner, we prove that the last term of equation (43) is convergent and that
lim Cm = −
m→∞
ZZ
QT
ρ1 f (x,t) ψ (x,t) dx dt.
B. Convergence of the discrete tumor cells equation
In the same manner, we consider a test function ψ ∈ D Ω × [0, T ) , and denote by ψKn = ψ (xK ,tn ), for all
K ∈ Tm and all n ∈ {0, . . . , Nm }. Multiplying the second-last equation of the scheme (7) by ∆tm ψKn and summing
over n ∈ {0, . . . , Nm } and K ∈ Tm yields, after a reorganization of the sum (see e.g. [11])
20
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
Am + Bm + Cm = Dm ,
(45)
where
Nm
Am =
∑ ∑
Nm
n
n
|K| un+1
K − uK ψK ,
Bm = du
n=0 K∈Tm
n=0
Nm
Cm =
n+1
n+1
F un+1
K , uL ; dFKL
∑ ∆tm ∑
n=0
n
n+1
τKL un+1
(ψK − ψLn ),
K − uL
σKL ∈Em
(ψKn − ψLn ),
σKL ∈Em
Nm
Dm =
∑ ∆tm ∑
n
|K| λu un+1
1 − An+1
+ ρ4 vn+1
ψK .
K
K
K
∑ ∆tm ∑
n=0
K∈Tm
For the time evolution term Am , we follow the same guideline as before and get the following convergence
Am −→ −
ZZ
u (x,t) ∂t ψ (x,t) dx dt −
QT
Z
u (x, 0) ψ (x, 0) dx,
as m → +∞.
Ω
Let us now focus on the diffusion term Bm , and show that
lim Bm =
ZZ
m→∞
QT
du ∇u · ∇ψ (x,t) dx dt.
(46)
Indeed, using the definition of the discrete gradient (Definition 4), we have
Nm
Bm =
∑ ∆tm
n=0
∑
|TKL | d
σKL ∈Em
n+1
un+1
ψ (xL ,tn ) − ψ (xK ,tn )
L − uK
dKL
dKL
ZZ
=
QT
du ∇Tm uTm ,∆tm · (∇ψ)Tm ,∆tm (·,t − ∆tm ) dx dt,
where (∇ψ)Tm ,∆tm (x,t) = ∇ψ (tn , xKL ), for all (x,t) ∈ TKL × (tn−1 ,tn ] and xKL = θ xK + (1 − θ )xL , 0 < θ < 1 is
some point on the segment ]xK , xL [.
The weak convergence (42) of the discrete gradient of the tumor cells uTm ,∆tm and the regularity of the test
function ψ allow us to deduce that
Bm =
ZZ
QT
du ∇Tm uTm ,∆tm · (∇ψ)Tm ,∆tm (·,t − ∆tm ) dx dt −→
ZZ
QT
du ∇u · ∇ψ (x,t) dx dt,
as m → +∞.
Now, we show the convergence for the convective term
lim Cm = −
m→∞
ZZ
χ (u) ∇ f · ∇ψ (x,t) dx dt.
(47)
QT
To do that, we follow the same steps as in [2]. First we prove that |Cm − Cm? | → 0 as m → ∞, where
Cm? = −
ZZ
QT
χ uTm ,∆tm ∇Tm fTm ,∆tm · (∇ψ)Tm ,∆tm (·,t − ∆tm ) dx dt,
where uTm ,∆tm is defined by uTm ,∆tm |(t ,t ]×T = un+1
:= min un+1
, uLn+1 .
KL
K
n n+1
KL
From the assumptions (H3)–(H5) on the consistency and the local Lipschitz continuity of the numerical flux
function F, one has
F un+1 , un+1 , dF n+1 − χ un+1 dF n+1 = F un+1 , un+1 , dF n+1 − F un+1 , un+1 , dF n+1 KL
KL
K
L
KL
KL
KL
KL
K
L
KL
≤ C dF n+1 un+1 − un+1 .
KL
Define uTm ,∆tm by uTm ,∆tm |(tn ,tn+1 ]×TKL
L
K
n+1
:= max un+1
. Then, from the above inequality, one has
K , uL
Numerical analysis for the simulation of a degenerate breast cancer model
|Cm − Cm? | ≤ C
ZZ
QT
21
uT ,∆t − uT ,∆t ∇T fT ,∆t · (∇ψ)
(·,t
−
∆t
)
m dx dt.
m
m
m
m
m
Tm ,∆tm
m
m
Note that uTm ,∆tm − uTm ,∆tm converges to zero almost everywhere in QT due to estimate (34). Applying the
Cauchy-Schwarz inequality and thanks the uniform bound on ∇Tm fTm ,∆tm , one can deduce that |Cm − Cm? | → 0
as m → ∞. It remains to show the desired limit given in (47). Indeed, we have uTm ,∆tm ≤ uTm ,∆tm ≤ uTm ,∆tm and
uTm ,∆tm → u a.e.
on QT . Consequently, and due to the continuity of the enzymes sensitivity function χ, one
gets χ uTm ,∆tm → χ (u) a.e. on QT and in L p (QT ) for 1 ≤ p < +∞. Thanks to the weak convergence (42), we
deduce that
ZZ
lim Cm = lim Cm? = −
χ (u) ∇ f · ∇ψ (x,t) dx dt.
m→∞
m→∞
QT
Finally, for the reaction term Dm , we rely on the boundedness of the discrete solution given in Propositions 1
and 2 and on the convergence results given in Proportion 10 to deduce that
lim Dm = lim
m→∞
m→∞
ZZ
QT
λu uTm ,∆tm 1 − ATm ,∆tm + ρ4 sTm ,∆tm ψTm ,∆tm (·,t − ∆tm ) dx dt
ZZ
=
QT
(λu u (1 − A) + ρ4 s) ψ (x,t) dx dt.
6 Numerical experiment in two-dimensional space
In this section, we show a two-dimensional numerical result provided by scheme (6)–(7) for the simulation
of the breast cancer development model (1)–(3). Newton’s algorithm is employed to compute the solution of
the nonlinear equations of system (7). This algorithm is coupled with a bigradient method to solve the linear
systems arising from the Newton algorithm as well as the linear system given by the last equation of system (7).
For the numerical test, we have chosen η f to be 1 since the MDE which is produced by all cells should
easily degrade the tissue. We took smaller values for death parameters to reflect the idea that cell death due
to environmental acidosis occurs at a slower rate; for instance, we took ηq = 0.5, ηr = 0.5, and ηs = 0.5,
since the change in these parameters will not alter the outcome of the simulations significantly. As estimated
in [7, 8], we take λu = 0.75, λq = 7.5 × 10−3 , λr = 7.575 × 10−3 , and λs = 0.0375. We also define as in [7, 8]
the mutation rates ρi in line with the mutation probabilities stated in the introduction ρ1 = 2 × 10−7 , ρ2 =
5 × 10−6 , ρ3 = 10−3 , and ρ4 = 3 × 10−2 . We fix ∆t = 0.01, α = 0.1, β = 0, a (u) = du u (1 − u), du = 0.0005,
χ (u) = 0.005 × (u (1 − u))2 ,and dv = 0.005. In the definition
of the numerical
flux F defined in equation (8),
we consider χ↑ (z) = χ min z, 21
and χ↓ (z) = χ max z, 12 − χ 21 .
In this test, we consider an orthogonal mesh for the shape of a breast consisting of 9307 triangles. We have
verified that this mesh verifies the orthogonality condition despite the presence of triangles with obtuse angles (
Fig. 2).
Fig. 2 Mammography of a breast (left), and an associated orthogonal mesh for the breast with 9307 triangles (right).
22
Françoise Foucher, Moustafa Ibrahim and Mazen Saad
In Figure 3, we show the initial conditions for healthy stem cells, cells with LOH in TSG1 and the other cell
densities. We assume in the model that as few as one of the breast stem cells has acquired the first mutation and
thus this mutation has been spread through a certain part of the breast when the breast was formed by clonal
expansion via the stem cells [20]. Therefore, we assume that the initial condition of the healthy stem cells f is
given by f0 (x, y) = 0.7 in the square region defined by (x, y) ∈ [2; 2.45] × [2.85; 3.25] and 1 otherwise. For the
+/−
initial condition for the cells with LOH in TSG1 i.e. TSG1 , we take q0 (x, y) = 0.3 in the same square region
defined by (x, y) ∈ [2; 2.45] × [2.85; 3.25] and 0 otherwise. Finally, for the other density cells, we consider that
r0 = s0 = u0 = 0 on the whole breast, and since we assume the absence of tumor cells initially, then we take
v0 = 0 as an initial condition for the chemicals.
Fig. 3 Initial condition for the healthy stem cells f : 0.7 ≤ f0 ≤ 1(left), and for the cells with LOH in TSG1 : 0 ≤ q0 ≤ 0.3 (center),
and for the other cell densities (right).
In Figure 4, we show the evolution of each of the healthy stem cells, the tumor cells and the chemicals
densities at time t = 41.4. We see that the tumor begins to form into the same region where there was initially
cells with LOH. As a consequence of the formation of tumor cells, we see the production of the chemicals by
the later which have an essential role to degrade the healthy stem cells in order to make place for the tumor to
spread into the breast.
Fig. 4 Evolution at t = 41.5 of the healthy stem cells f : 0.18 ≤ f ≤ 1(left), and for the tumor cell density u: 8.24×10−9 ≤ u ≤ 0.4
(center), and for the chemicals density v: 7.75×10−8 ≤ v ≤ 0.182 (right).
In Figure 5, we show the evolution of each of the healthy stem cells, the tumor cells and the chemicals
densities at time t = 44. We see that the tumor cells u invade the healthy stem cells f which are degraded by the
chemicals v, this phenomenon shows the spread of the tumor cells in gradient of the healthy stem cells.
In Figure 6, we show the evolution of each of the healthy stem cells, the tumor cells and the chemicals
densities at time t = 58. We see that the tumor spreads quickly and form an aggregation which continues in
invading the healthy stem cells through the chemicals produced by the tumor cells.
Numerical analysis for the simulation of a degenerate breast cancer model
23
Fig. 5 Evolution at t = 44 of the healthy stem cells f : 1.08×10−7 ≤ f ≤ 1(left), and for the tumor cell density u: 9.97×10−9 ≤ u ≤ 1
(center), and for the chemicals density v: 10−7 ≤ v ≤ 0.85 (right).
Fig. 6 Evolution at t = 58 of the healthy stem cells f : 0 ≤ f ≤ 1(left), and for the tumor cell density u: 2.4×10−8 ≤ u ≤ 1 (center),
and for the chemicals density v: 3.35×10−7 ≤ v ≤ 1 (right).
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