Equivalence Transformations for a System
of a Biological Reaction Diusion Model
Zifei Yan
Thesis for the degree Master of Science (two years) in Mathematical Modelling and Simulation 30 credit points (30 ECTS credits)
Apr. 19, 2012
Blekinge Institute of Technology School of Engineering
Abstract
A biological reaction diusion model has gained much attention recently. This model is formulated as a system of nonlinear partial dierential equations that contains an unknown function of one dependent variable. It is complicated to determine this function, but knowing it is useful. This model is considered in this master thesis. The generators of the equivalence groups and invariant solutions are calculated.
Acknowledgements
This work was carried out under the supervision of Prof. Nail H. Ibragimov. I would like to express my deep gratitude to my supervisor for his help and guidance throughout this thesis.
I would also like to express my sincere thanks to my program manager, Dr. Raisa Khamitova, who has taught me a lot.
I am forever grateful to my parents for their endless and unconditional sacri- ces and support.
Contents
1 Introduction . . . 6
1.1 Nonlinear system . . . 6
1.2 Diusion equation . . . 6
1.3 A biological reaction diusion model . . . 7
2 Equivalence transformations . . . 8
3 Application of projections . . . 13
3.1 Theorems on projections of an equivalence Lie algebra . . 13
3.2 Application of projections: Calculation of a principal Lie algebra . . . 14
3.3 Application of projections: Calculation of invariant solutions 15 4 Exact form of the invariant solutions . . . 17
5 Other cases . . . 27
1 Introduction
1.1 Nonlinear system
A nonlinear system is a system which does not satisfy the superposition principle. Its output is not directly proportional to its input. Therefore, for nonlinear problems known solutions cannot be combined with new solutions linearly.
Modern mathematics has over 300 years of history. From the very beginning dierential equations were considered as a major tool for mathematical modelling. Most mathematical models in the academia, such as physics, engineering sciences, and biomathematics, lead to nonlinear dierential equations. However, group analysis is the only universal and eective method for solving nonlinear dierential equations analytically [1].
1.2 Diusion equation
Diusion is a phenomenon in which particles spread from high-concentration re-gions to low-concentration rere-gions. The time dependence of the particle distribu-tion in space is given by a diusion equadistribu-tion. In these circumstances, the power of diusion is ruled by the concentration gradient. Diusion equations can also describe other forms of spread, such as the spread of disease among people, spread of ideas in society, etc.
A diusion equation is a partial dierential equation, which describes the density change in a material undergoing diusion. The equation is usually written as follows:
∂u (r, t)
∂t = ∇ · [D (u, r) ∇u (r, t)] ,
where u (r, t) is the density of the diusing particle at a location r and time t; D (u, r) is the diusion coe cient for the density u at the location r; and the concentration gradient is represented as the ∇ term. The equation is perceived as nonlinear if the diusion coe cient depends on the density; otherwise, it is treated as linear. If D is a constant in the above equation, then the equation can be reduced into the following linear equation:
∂u (r, t)
∂t = D∇
2u (r, t) ,
which is also called the heat equation.
If there are drifts in the diusion phenomenon, one more term is required to describe these drifts. Thus, the diusion equation can be transformed as follows:
It is called the FokkerPlanck equation. Here D1(x, t) denotes the Itô drift, and
D2(x, t) is the diusion coe cient.
1.3 A biological reaction diusion model
In the past years, mathematical modelling in biology has become a growing eld in applied mathematics. One of the active research areas is dealing with mathemat-ical models for microbial growth in a ow reactor and for the diusive epidemic population. The aim of studying these kinds of models is to understand the mechanisms that control the growth and competition of microbes, and to reduce the necessity of time-consuming and complicated experiments.
The autonomous parabolic system modelling microbial growth and the com-petition in a ow reactor have been explained in [2, 3, 4]. The reactor is best considered as the portion, from x = 0 to L, of a long tube, with a ow from the left to the right, carrying fresh nutrient into the reactor at x = 0 and carrying unused nutrient and cells out of the reactor at x = L. By scaling the x-variable the pure culture system is modeled by the following equations:
ut= duxx− αux− f (u) v, (1.1)
vt = vxx− αvx+ (f (u) − k) v, (1.2)
f (0) = 0, f0(u) > 0 for u > 0, lim
u→+∞f (u) > k, (1.3)
where u(x, t) and v(x, t) represent the concentrations of nutrients and microbial population at a position x and time t, respectively, f (u) is the nutrient uptake function, d, α and k are constants.
Based on the discussion in the previous section, the terms −αux and −αvx
are the drift terms, α is the ow velocity, d is the ratio of the diusivity of the nutrient to the random cell motility coe cient of the organism, and k is the cell death rate. The model has the following constrains: α > 0, d > 0, and k > 0. The term −f (u) v in Equation (1.1) shows the constrain in the diusion velocity of nutrient from the growth of the microbial population, and (f (u) − k) v shows the contribution to the diusion velocity of the microbial population from the net population growth.
A typical example of a nutrient uptake function is the Monod function: f (u) = mu
Table 1: Examples of coe cients in the Monod function Culture Substrate m (h−1) b · 103 (kg · m−3) K. aerogenes Glycerol 0.85 9 K. pneumoniae Glucose 0.50 31 A. aerogenes Glucose 1.33 8 E. coli Glucose 0.95 11.4 S. cerevisiae Glucose 0.39 42.7 C. tropicalis Glucose 0.39 198
Equations (1.1) and (1.2) and conditions (1.3) clearly describe the microbial growth system. We obtain the invariant solutions of this system using the equiv-alence transformation method.
2 Equivalence transformations
The method for determining generators of continuous equivalence groups is de-scribed in [6]. It was suggested by L. V. Ovsyannikov in [7]. The central part in this method is played by a secondary prolongation. In this thesis the method is used frequently.
An equivalence transformation of Equations (1.1) and (1.2) is a change of variables
(t, x, u, v, f, d, α, k) → ¯t, ¯x, ¯u, ¯v, ¯f , ¯d, ¯α, ¯k , carrying the system (1.1), (1.2) into a system of the same form:
¯
u¯t= ¯d¯u¯x¯x− ¯α¯ux¯− ¯f (¯u) ¯v, (2.5)
¯
v¯t= ¯vx¯¯x− ¯α¯v¯x+ f (¯¯ u) − ¯k ¯v. (2.6)
Systems (1.1), (1.2), and (2.5), (2.6) are considered equivalent.
The set of all equivalence transformations forms an equivalence group E . We shall nd a continuous subgroup Ec of the set using the in nitesimal method.
Thus, we have f (u), d (t, x, u, v), α (t, x, u, v), k (t, x, u, v) and introduce the local notation f = f1, d = f2, α = f3, k = f4. We can look for an operator of the
equivalence group Ec having the form
Y = ξ1 ∂ ∂t+ ξ 2 ∂ ∂x + η 1 ∂ ∂u + η 2 ∂ ∂v + µ k ∂ ∂fk (2.7)
where the coordinates ξ1, ξ2, η1 and η2 of the operator (2.7) are considered
f1, f2, f3and f4, k = 1, 2, 3, 4. The operator Y de nes the group E cof equivalence transformations ¯ t = ϕ (t, x, u, v) , ¯ x = ψ (t, x, u, v) , ¯ u = Φ (t, x, u, v) , ¯ v = Ψ (t, x, u, v) , ¯ fk = ¯fk t, x, u, v, fk
for the family of Equations (1.1) and (1.2), if and only if Y satis es the invariance conditions for the the following system:
ut− f2uxx+ f3ux+ f1v = 0,
vt− vxx+ f3vx− f1− f4 v = 0, (2.8)
ft1 = fx1 = fv1 = 0.
The invariance conditions for System (2.8) are e Y ut− f2uxx+ f3ux+ f1v (2.8) = 0, (2.9) e Y vt− vxx+ f3vx− f1− f4 v (2.8) = 0, (2.10) e Y ft1 (2.8) = eY f 1 x (2.8) = eY f 1 v (2.8) = 0, (2.11)
whereYe is the prolongation of the operator (2.7): e Y = Y + ζ11 ∂ ∂ut + ζ21 ∂ ∂ux + ζ12 ∂ ∂vt + ζ22 ∂ ∂vx + ζ221 ∂ ∂uxx + ζ222 ∂ ∂vxx + ω11 ∂ ∂f1 t + ω12 ∂ ∂f1 x + ω14 ∂ ∂f1 v · (2.12) The coe cients ζ in (2.12) are given by the usual prolongation formulae:
where Dx = ∂ ∂x + ux ∂ ∂u + vx ∂ ∂v + uxx ∂ ∂ux + vxx ∂ ∂vx + · · · , and Dt= ∂ ∂t+ ut ∂ ∂u + vt ∂ ∂v + utt ∂ ∂ut + vtt ∂ ∂vt + · · · .
The other coe cient, ω1
1, is obtained by the new prolongation formula:
ωk1 = ˜Dt µk − ftkD˜t ξ1 − fxkD˜t ξ2 − fukD˜t η1 − fvkD˜t η2 , (2.14)
where ˜Dt has the following form:
˜ Dt= ∂ ∂t+ f k t ∂ ∂fk· (2.15)
Recalling Equation (2.8), the operator (2.15) can be reduced to the form ˜
Dt=
∂
∂t· (2.16)
Furthermore, replacing ˜Dt in (2.14) by the operators
˜ Dx = ∂ ∂x + f k x ∂ ∂fk, ˜ Dv = ∂ ∂v + f k v ∂ ∂fk,
we obtain the prolongation formulae for ω1
2 and ω41, respectively. Using (2.8), the
operators ˜Dx, ˜Dv are reduced to
˜ Dx = ∂ ∂x, ˜ Dv = ∂ ∂v· The use of the operator (2.16) in (2.14) gives
ω11 = µ1t − fu1ηt1. Similarly, we obtain the coordinates ω1
The invariance conditions (2.11) yield ω11 = ω12 = ω41 = 0, (2.18) therefore µ1t − fu1η1t = 0, µ1x− fu1η1x= 0, µ1v− f1 uη1v = 0.
Noting that the equations hold for an arbitrary function f1
u, we obtain
µ1t = µ1x = µ1v = 0, ηt1 = ηx1 = η1v = 0.
Thus,
µ1 = µ1(u, f ) , η1 = η1(u) . (2.19) We now rewrite the invariance conditions (2.9) and (2.10):
f2 ζ221 + µ2uxx− f3ζ21− µ 3 ux− f1η2− µ1v − ζ11 (2.8) = 0, (2.20) ζ2 22− f3ζ22− µ3vx+ f1− f4 η2+ µ1v − µ4v − ζ12 (2.8) = 0. (2.21)
Now if we look into the expression of ζ1
22, we can obtain ζ221 =Dx ζ21 − utDx ξ1 − uxDx ξ2 =Dx2 η1 − utD2x ξ 1 − u txDx ξ1 | {z } −uxxDx ξ2 − uxD2x ξ 2 − u tDx ξ1 − uxDx ξ2 . After substituting uxx = 1 f2 (ut+ f 3u
x+ f1v) , one can notice that −utxDx(ξ1)
will be the only term with utxin equation (2.20). So it is obvious that Dx(ξ1) = 0,
which leads to ξ1 = ξ1(t), and also can reduce the expressions (2.13) to
Substituting these expressions and uxx = 1 f2 (ut+ f 3u x+ f1v) in (2.20), we ob-tain f2ηuu1 u2x− ξxx2 ux− 2ξxu2 u 2 x− ξ 2 uuu 3 x− 2ξ 2 uvu 2 xvx− 2ξxv2 uxvx− ξvv2 uxvx2− ξ 2 vuxvt −f3ξ2 xux+ 2ξu2u 2 x+ ξ 2 vuxvx + ξt1ut+ ξt2ux+ ξv2uxvt− µ3ux −f1η2− η1 uv + 2ξ 2 xv + 3ξ 2 uuxv + 2ξv2vxv −2ξ2xut− 2ξu2uxut− 2ξv2utvx −µ1v − f2f3ξ2 vuxvx+ f2ξv2ux f1− f4 v+ µ2 f2ut+ µ2f3 f2 ux+ µ2f1 f2 v = 0 (2.22) The terms with ut in Equation (2.22) give
uxut: ξu2 = 0; utvx: ξv2 = 0; ut: ξ1t − 2ξx2+
µ2 f2 = 0.
Because f2 canbe an arbitrary positive constant, so µ2 = 0 and ξ1
t − 2ξ2x= 0. We
already know that ξ1 = ξ1(t), so ξ1
tx = 2ξxx2 = 0. Now we can reduce Equation
(2.22) to f2ηuu1 u2x− f3 ξx2ux+ ξt2ux | {z } −f1η2− η1 uv + 2ξ 2 xv −µ 3 ux | {z } −µ1 v = 0 (2.23) The terms with ux in Equation (2.23) give
ux2 : ηuu1 = 0; ux : ξt2− f3ξx2− µ3 = 0.
The determining equation reduces to −f1η2 − η1
uv + 2ξ 2
xv − µ
1v = 0.
Substituting the expressions (2.13) and vxx = vt+f3vx−(f1− f4) vin (2.21), and
simplifying the determining equation with the results we got above, we obtain uxx : ηu2 = 0, ux : 2ηxu2 + η 2 uuux = 0, whence ηxu2 = η2uu= 0; vx : 2ηvx2 + η2vvvx+ f3ξx2+ ξt2− µ3− f3ξt1 = 0, whence ηvv2 = 0, 2ηvx2 + ξt2 = 0, µ3 = ξx2− ξt1 f3 ; and the remaining terms give
After integrating the resulting equations, we obtain: ξ1 = 2C6t + C1, ξ2 = C6x + C2, η1 = C4u + C3, η2 = C5v, (2.24) µ1 = (C4− 2C6− C5) f, µ2 = 0, µ3 = −C6α, µ4 = (C4− C5) f − 2C6k,
where C1, C2, C3, C4, C5 and C6 are arbitrary constants.
Substituting (2.24) in (2.7), one arrives the following result.
The equivalence Lie algebra LE for the family of nonlinear diusion-reaction
equations (1.1) and (1.2) is a six-dimensional Lie algebra spanned by Y1 = ∂ ∂t, Y2 = ∂ ∂x, Y3 = ∂ ∂u, Y4 = u ∂ ∂u + f ∂ ∂f + f ∂ ∂k, Y5 = v ∂ ∂v − f ∂ ∂f − f ∂ ∂k, (2.25) Y6 = 2t ∂ ∂t+ x ∂ ∂x − 2f ∂ ∂f − α ∂ ∂α − 2k ∂ ∂k·
3 Application of projections
3.1 Theorems on projections of an equivalence Lie algebra
The method of preliminary group classi cation has been hinted by the observa-tions made in applicaobserva-tions of group analysis. Most of the extensions of a principal Lie algebra Lp (the algebra admitted by every equation of the family of equations
under consideration) are taken from an equivalence algebra LE. These extensions
are called E extensions.
We use the equivalence algebra to simplify the calculation of symmetry groups without solving the determining equation. If we derive the equivalence algebra, E extensions for the equations of a given family are obtained by solving only simple algebraic equations [8, 9].
of a continuous group of equivalence transformations, we introduce the following projections of this equivalence operator:
X ≡ pr(t,x,u,v)(Y ) = ξ1 ∂ ∂t+ ξ 2 ∂ ∂x + η 1 ∂ ∂u + η 2 ∂ ∂v, Z ≡ pr(u,f,d,α,k)(Y ) = η1 ∂ ∂u + µ k ∂ ∂fk·
The signi cance of these projections is de ned by the following simple state-ments [8].
Theorem 3.1. An operator X belongs to the principal Lie algebra Lp of a system
if and only if
X = pr(t,x,u,v)(Y )
with an equivalence generator Y , such that
pr(u,f,d,α,k)(Y ) = 0. (3.26)
Theorem 3.2. Let Y be an equivalence operator. The operator X = pr(t,x,u,v)(Y )
is a symmetry operator for a system with equations
fk = fk(u), k = 1, 2, 3, 4, (3.27) if and only if Equations (3.27) are invariant under the group generated by
Z = pr(u,f,d,α,k)(Y ).
3.2 Application of projections: Calculation of a principal
Lie algebra
Let us determine the principal Lie algebra using Theorem 3.1. For the general operator of the equivalence algebra of (1.1) and (1.2),
Z =pr(u,f,d,α,k)(Y ) = (C4u + C3) ∂ ∂u + (C4− 2C6− C5) f ∂ ∂f − C6α ∂ ∂α + [(C4− C5) f − 2C6k] ∂ ∂k· (3.30)
Therefore, from Equation (3.26) we get
C3 = C4 = C5 = C6 = 0. Hence, Y = C1 ∂ ∂t + C2 ∂ ∂x.
The principal Lie algebra Lp is two-dimensional and spanned by
Xp1 = ∂ ∂t, X 2 p = ∂ ∂x· (3.31)
3.3 Application of projections: Calculation of invariant
so-lutions
Consider the operator
Z1 = u ∂ ∂u + 2f ∂ ∂f + α ∂ ∂α + 2k ∂ ∂k, (3.32)
which corresponds to the equivalence operator (3.28) with C3 = 0, C4 = 1, C5 = 1, C6 = −1.
Hence, we look for Equations (1.1) and (1.2) admitting an extension of the principal Lie algebra Lp by one-dimensional algebra with the basis
X1 = u ∂ ∂u + v ∂ ∂v − 2t ∂ ∂t− x ∂ ∂x· (3.33)
According to Theorem 3.2, we have to nd invariant equations with func-tions (3.27) for algebra L1 with the basis (3.32). For this algebra we can derive
functionally independent invariants and invariant equations by doing as follows. Let us consider the equations f = F (u), d = D, α = A, k = K. The invariance conditions for the operator Z1 have the form
Z1(f − F (u))|f =F (u) = 0,
Z1(d − D)|d=D = 0,
Z1(α − A)|α=A = 0,
whence
2F = udF du, A = 0, K = 0.
The latter equations give us the invariant equations with the functions f = Cu2, C = const., C > 0
α = 0, (3.34)
k = 0. Thus, the particular form,
ut= duxx− Cu2v, (3.35)
vt= vxx+ Cu2v, (3.36)
of the system (1.1), (1.2) admits the algebra L3 spanned by the operators (3.31)
and (3.33).
The characteristic system for X1,
dt −2t = dx −x = du u = dv v , provides three functionally independent invariants:
K1 =
x2
t , K2 = xu, K3 = xv.
Consequently, we seek for invariant solutions having the form: u = 1 xϕ x2 t , v = 1 xψ x2 t · (3.37)
Substitution of the functions (3.37) in Equations (3.35) and (3.36) leads us to the following system of ordinary dierential equations:
ϕ0 = −4dϕ00+2d λ ϕ 0− 2d λ2ϕ + C λ2ϕ 2ψ, (3.38) ψ0 = −4ψ00+ 2 λψ 0− 2 λ2ψ − C λ2ϕ 2ψ, (3.39) where λ = x2
4 Exact form of the invariant solutions
We can rewrite the system (3.38), (3.39) as ϕ + 4dϕ0− 2d λϕ 0 = C λ2ϕ 2ψ, ψ + 4ψ0− 2 λψ 0 = −C λ2ϕ 2ψ
which leads to the equation
ϕ + 4dϕ0− 2d
λ ϕ = −ψ − 4ψ
0+ 2
λψ + L, (4.40) where L is an arbitrary constant.
We can simplify the process of solving the system (3.38), (3.39) by choosing the rst order dierential equation (4.40) instead of the second order equation (3.39). Hence, the new system is
1 −2d λ ϕ0+ 4dϕ00+ 2d λ2ϕ − C λ2ϕ 2ψ = 0, (4.41) 1 −2d λ ϕ + 4dϕ0+ 1 − 2 λ ψ + 4ψ0− L = 0. (4.42) Based on the method used before, we look for an operator of the equivalence group Ec for Equations (3.38) and (4.40) having the following form:
X = ξ ∂ ∂λ + η 1 ∂ ∂ϕ + η 2 ∂ ∂ψ + µ ∂ ∂d, (4.43) where ξ = ξ (λ, ϕ, ψ), η1 = η1(λ, ϕ, ψ), η2 = η2(λ, ϕ, ψ) and µ = µ (λ, ϕ, ψ, d).
The invariance conditions for Equations (4.41) and (4.42) are e X 1 − 2d λ ϕ0+ 4dϕ00+2d λ2ϕ − C λ2ϕ 2ψ (4.41) = 0, (4.44) e X 1 − 2d λ ϕ + 4dϕ0+ 1 − 2 λ ψ + 4ψ0− L (4.42) = 0, (4.45) whereXe is the prolongation of the operator (4.43):
The coe cients ζ in (4.46) are given by using the usual prolongation formulae: ζ11 = Dλ η1 − ϕ0Dλ(ξ) , ζ12 = Dλ η2 − ψ0Dλ(ξ) , ζ21 = Dλ ζ11 − ϕ 00 Dλ(ξ) , (4.47) ζ22 = Dλ ζ12 − ψ 00 Dλ(ξ) , where Dλ = ∂ ∂λ + ϕ 0 ∂ ∂ϕ + ψ 0 ∂ ∂ψ + ϕ 00 ∂ ∂ϕ0 + ψ 00 ∂ ∂ψ0 + · · · ,
whereas the coe cients ωk are obtained from the following prolongation formulae:
ω1 = ˜Dλ(µ) − dλD˜λ ξ1 − dϕD˜λ η1 − dψD˜λ η2 , ω2 = ˜Dϕ(µ) − dλD˜ϕ ξ1 − dϕD˜ϕ η1 − dψD˜ϕ η2 , (4.48) ω3 = ˜Dψ(µ) − dλD˜ψ ξ1 − dϕD˜ψ η1 − dψD˜ψ η2 , where ˜ Dλ = ∂ ∂λ + dλ ∂ ∂d, ˜ Dϕ = ∂ ∂ϕ + dϕ ∂ ∂d, (4.49) ˜ Dψ = ∂ ∂ψ + dψ ∂ ∂d·
From Equations (4.41) it follows that the operators (4.49) can be reduced to the form ˜ Dλ = ∂ ∂λ, ˜ Dϕ = ∂ ∂ϕ, ˜ Dψ = ∂ ∂ψ, (4.50)
therefore Equations (4.48) reduce to the system
ω1 = µλ, ω2 = µϕ, ω3 = µψ.
Thus, we obtain µλ = µϕ = µψ = 0, whence µ = µ (d) . Now the operator Xe has the form e X =ξ (λ, ϕ, ψ) ∂ ∂λ + η 1(λ, ϕ, ψ) ∂ ∂ϕ + η 2(λ, ϕ, ψ) ∂ ∂ψ + µ (d) ∂ ∂d + ζ11 ∂ ∂ϕ0 + ζ 2 1 ∂ ∂ψ0 + ζ 1 2 ∂ ∂ϕ00 + ζ 2 2 ∂ ∂ψ00, where ζ11 = Dλ η1 − ϕ0Dλ(ξ) , ζ12 = Dλ η2 − ψ0Dλ(ξ) , ζ22 = Dλ ζ12 − ψ 00 Dλ(ξ) , and ζ21 = Dλ ζ11 − ϕ 00 Dλ(ξ) = ηλλ1 + ηϕ1ϕ00+ 2η1λϕ− ξλλ ϕ0+ η1ϕϕ− 2ξλϕ ϕ02 + 2ηϕψ1 − 2ξλψ ϕ0ψ0− ξϕϕϕ03− 3ξϕϕ0ϕ00 + 2ηλψ1 ψ0+ ηψψ1 ψ02+ η1ψψ00− 2ξϕψϕ02ψ0 − ξψψϕ0ψ02− 2ξψϕ00ψ0− ξψϕ0ψ00− 2ξλϕ00.
Substituting (4.47) and (4.41) in (4.51). We obtain 2d λ2ϕ 0− 4d λ3ϕ + 2C λ3ϕ 2ψ ξ + 2d λ2 − 2C λ2 ϕψ η1− C λ2ϕ 2η2 + 1 −2d λ η1 λ + η 1 ϕ− ξλ ϕ0− ξϕϕ02+ η1ψψ 0− ξ ψϕ0ψ0 +4dη1 λλ+ 2η 1 λϕ− ξλλ ϕ0+ η1ϕϕ− 2ξλϕ ϕ02− ξϕϕϕ03 + 2ηϕψ1 − 2ξλψ ϕ0ψ0 + 2η1λψψ 0+ η1 ψψψ 02− 2ξ ϕψϕ02ψ0− ξψψϕ0ψ02 −ηϕ1 − 2ξλ−3ξϕϕ0 1 − 2d λ ϕ0+ 2d λ2ϕ− C λ2ϕ 2ψ +dηψ2 − 2ξψψ0−dξψϕ0 2 − λ λ ψ 0− 2 λ2ψ − C λ2ϕ 2ψ = 0 (4.53)
The terms with ϕ0ψ in Equation (4.51) give −3ξ ϕ C λ2ϕ 2 + 2d λ2ξψ + dC λ2ϕ 2ξ ψ = 0,
where C and d are arbitrary positive constants, so we got ξϕ = ξψ = 0. Now
Equation (4.53) can be reduced to 2d λ2ϕ 0− 4d λ3ϕ + 2C λ3 ϕ 2ψ ξ + 2d λ2 − 2C λ2 ϕψ η1− C λ2ϕ 2η2 + 1 −2d λ h ηλ1+ η1ϕ+ ξλ ϕ0+ ηψ1ψ 0i− η1 ϕ− 2ξλ 2d λ2ϕ − C λ2ϕ 2ψ +4dhηλλ1 + 2ηλϕ1 − ξλλ ϕ0+ η1ϕϕϕ 02 + 2ηϕψ1 ϕ0ψ0+ 2ηλψ1 ψ0+ ηψψ1 ψ02i + dηψ2 − 2ξψψ0 2 − λ λ ψ 0− 2 λ2ψ − C λ2ϕ 2ψ = 0. (4.54) The terms with ϕ0ψ0 in Equation (4.54) give η1
ϕψ = 0. The terms with ϕ 02 give
ηϕϕ1 = 0. The terms with ϕ0 give 2d λ2ξ + 1 − 2d λ η1ϕ+ ξλ + 4d 2ηλϕ1 − ξλλ = 0.
Because d is an arbitrary positive constant, so we got ξ = 0 and η1
ϕ = 0. Now
The terms with ψ0 in Equation (4.55) give 1 −2d λ η1 ψ+ dη2ψ 2 − λ λ = 0, whence η1
ψ = ηψ2 = 0. Now Equation (4.55) is reduced to
2d λ2 − 2C λ2 ϕψ η1− C λ2ϕ 2η2+ 1 −2d λ η1λ+ 4dηλλ1 = 0.
Substituting (4.47) and (4.42) in (4.52). We obtain ϕ0 : 2d
λ2ξ = 0, whence ξ = 0;
and the remaining terms give 2η1 λ2 − 2ηλ1 λ + 4η 1 λλ= 0, − 2C λ2ϕψη 1− C λ2ϕ 2η2+ η1 λ = 0.
After integrating the resulting equations, we obtain: ξ = 0, µ = 0, η1 = C1 √ λ, (4.56) η2 = −C1 √ λ + C3 √ λe−14λ−C2, (4.57) − 2C λ2ϕψη 1− C λ2ϕ 2η2+ η1 λ = 0, (4.58)
where C1, C2, and C3 are arbitrary constants, and C is the arbitrary positive
constant mentioned in (3.34). The calculation that leads to the result (4.57) can be found in [1] on pages 113-114.
Now we can use Equations (4.56) and (4.57) to form the operator (4.43). Thus, the family of non-linear ordinary dierential Equations (4.41) has a three-dimensional equivalence Lie algebra. One of the three operators is:
X1 = √ λ ∂ ∂ϕ − √ λ ∂ ∂ψ· (4.59)
The characteristic system of X1 is
dϕ √ λ = − dψ √ λ, which provides one invariant
Inserting (4.56), (4.57) and (4.60) in (4.58), we obtain: ϕ =2CC1C4± p 4C2C2 1C42− CC1λ(6C1 − 2C3e−λ/4−C2) 6CC1− 2CC3e−λ/4−C2 , ψ =4CC1C4− 2CC3C4e −λ/4−C2 6CC1− 2CC3e−λ/4−C2 ∓ p 4C2C2 1C42− CC1λ(6C1− 2C3e−λ/4−C2) 6CC1− 2CC3e−λ/4−C2 = C4− ϕ.
Based on (3.37), it is obvious that
u (t, x) = 2CC1C4± r 4C2C2 1C42− CC1 x2 t (6CC − 2C3e −C2−x2/(4t)) (6CC1− 2CC3e−C2−x2/(4t))x , v (t, x) =4CC1C4− 2CC3C4e −C2−x2/(4t) (6CC1− 2CC3e−C2−x2/(4t))x ∓ r 4C2C2 1C42− CC1 x2 t (6C1− 2C3e −C2−x2/(4t)) (6CC1− 2CC3e−C2−x2/(4t))x = C4 x − u, where C1, C2, and C3 are arbitrary constants, and C is an arbitrary positive
constant. According to the physical meaning of u and v, C4 should be a positive
constant as well.
Based on operator (3.33), we only used the rst part of (4.57): η2 = −C1
√ λ,
which indicates C3 = 0. Thus, we can simplify our results to the following forms:
u (t, x) = 2C4± r 4C2 4 − 6x2 Ct 6x , (4.61) v (t, x) = 4C4∓ r 4C2 4 − 6x2 Ct 6x = C4 x − u. (4.62) According to (4.61) and (4.62), we obtain:
ux = − 1 Ct r 4C2 4 − 6x2 Ct − 2C4± r 4C2 4 − 6x2 Ct 6x2 , uxx = ± 1 Ctx r 4C2 4 − 6x2 Ct − 6x C2t2 r 4C2 4 − 6x2 Ct 3 + 2C4± r 4C2 4 − 6x2 Ct 3x3 and vt= − ut= ∓ x 2Ct2 r 4C2 4 − 6x2 Ct , vx = − C4 x2 − ux= 1 Ct r 4C2 4 − 6x2 Ct + −4C4± r 4C2 4 − 6x2 Ct 6x2 , vxx = 2C4 x3 − uxx = 6x C2t2 r 4C42− 6x 2 Ct 3 ∓ 1 Ctx r 4C2 4 − 6x2 Ct − −4C4± r 4C2 4 − 6x2 Ct 3x3 ·
where we set A = 4C2
4 −
6x2
Ct . Equating to zero the coe cients for dierent powers of x and t, we obtain
C4 → 0, C → ∞, (4.63)
which means our results (4.61) and (4.62) are approximate solutions to our sys-tem, with conditions (4.63) which agree with reality for the following reason.
According to (3.34), our calculation leading to the results (4.61) and (4.62) is based on the speci c form of f (u):
f (u) = Cu2.
Figure 3 shows the graphs of the Monod function (1.4) and f (u) = Cu2. One
can notice that these two lines can be closer to each other at small values of u when C goes to in nity.
Based on Equation (4.60), we can obtain C4 = x (u + v) , whence it follows
that values of u and v have to be small.
Figures 1 and 2 show the functions (4.61) and (4.62) for medium values of C and C4, while Figures 4-6 show the functions (4.61) and (4.62) for a large
value of C and a small value of C4. As we can see from these gures, based on
constrain (4.63), the values of u and v are really small and f (u) = Cu2 can t
the Monod function better. These gures can give us a breif view of how the microbe population distribution will look like under a very low nutrient control.
0 5 10 15 0 5 10 15 0 0.2 0.4 0.6 0.8 x t u
0 5 10 15 0 5 10 15 0 0.1 0.2 0.3 0.4 x t v
Figure 2: v (t, x) when C = 2 and C4 = 1
0 0.5 1 1.5 2 2.5 3 3.5 4 0 20 40 60 80 100 120 140 160 u f
Figure 3: The Monod function (blue) when m = 100, b = 1, and f (u) = Cu2
0 5 10 15 0 5 10 15 0 0.02 0.04 0.06 0.08 x t u
Figure 4: u (t, x) when C = 2000 and C4 = 0.1
0 5 10 15 0 5 10 15 0 0.002 0.004 0.006 0.008 0.01 x t v
0 0.02 0.04 0.06 0.08 0.1 0 5 10 15 20 25 u f
Figure 6: The Monod function (blue) when m = 100, b = 1, and f (u) = Cu2
(red) when C = 2000
5 Other cases
In Chapter 4 we had got some meaningful, but not general results. To obtain a general result, the rst thing to do is to get a reasonable expression for f (u). The Monod function (1.4) is an empirical function which can be the criterion for our choice of f (u).
Let us consider the operator Z2 = 2au ∂ ∂u + 2f ∂ ∂f + α ∂ ∂α + 2k ∂ ∂k, (5.64)
which corresponds to the equivalence operator (3.28) with C3 = 0, C4 = 2a, C5 = 2a, C6 = −1,
where a is a constant. So here we look for Equations (1.1) and (1.2) admitting an extension of the principal algebra Lp by one-dimensional algebra with the basis
X2 = 2au ∂ ∂u + 2av ∂ ∂v − 2t ∂ ∂t− x ∂ ∂x· (5.65)
According to Theorem 3.2, we have to nd invariant Equations (3.27) for algebra L1 with the basis (5.64). For this algebra we can get functionally independent
The invariance conditions for the operator Z2 have the form
Z2(f − F (u))|f =F (u) = 0,
Z2(d − D)|d=D = 0,
Z2(α − A)|α=A= 0,
Z2(k − K)|k=K = 0.
The characteristic system of Z2 is
F = audF du, A = 0, K = 0,
which gives us the invariant equations with functions f = Cu1/a
, α = 0, k = 0.
As we can see from Figure 7, we can have dierent cases by choosing dierent values of C and a. Our function f (u) can be used for dierent cases very well.
The special form,
ut= duxx− Cu 1/a v, (5.66) vt= vxx+ Cu 1/a v, (5.67)
of the system (1.1) and (1.2) admits the algebra L3 spanned by the operators
(3.31) and (5.65). The characteristic system for the operator X2,
dt −2t = dx −x = du 2au = dv 2av, provides three functionally independent invariants:
K1 = x2 t , K2 = tu 1/a , K3 = tv 1/a .
0 5 10 15 20 0 0.5 1 1.5 u f
Figure 7: The Monod function (blue) with m = 1, a = 1 and f (u) = Cu1/a (red)
with C = 0.5, 1/a= 0.35 whence ut= −a ϕ ta+1 − x2 ta+2ϕ 0 , ux = 2x ta+1ϕ 0 , uxx = 4x2 ta+2ϕ 00 + 2 ta+1ϕ 0 , vt = −a ψ ta+1 − x2 ta+2ψ 0 , vx = 2x ta+1ψ 0 , (5.69) vxx = 4x2 ta+2ψ 00 + 2 ta+1ψ 0 .
Substituting the expressions (5.68) and (5.69) in Equations (5.66) and (5.67), one obtains:
4dλϕ00+ (λ + 2d) ϕ0 + aϕ − Cϕ1/a
ψ = 0, (5.70) 4λψ00+ (λ + 2) ψ0+ aψ + Cϕ1/a
where λ = x2
t is the independent variable, ϕ and ψ are the dependent variables. We can seek for an operator of the equivalence group Ec for Equations (5.70)
and (5.71) having the form: X = ξ ∂ ∂λ + η 1 ∂ ∂ϕ+ η 2 ∂ ∂ψ + D ∂ ∂d + µ 1 ∂ ∂a + µ 2 ∂ ∂C, (5.72) where ξ = ξ (λ, ϕ, ψ), η1 = η1(λ, ϕ, ψ), η2 = η2(λ, ϕ, ψ), D = D (λ, ϕ, ψ, d), µ1 = µ1(λ, ϕ, ψ, d, a) and µ2 = µ2(λ, ϕ, ψ, d, C).
For dierent kinds of microbe, the coe cients in a Monod function, i.e., m and b in function (1.4), will be dierent. So we need to choose suitable sets of values of C and a in Equations (5.66) and (5.67) to get a good shape of f (u) to t dierent Monod functions related to dierent types of microbe. Among all the coe cients, i.e. d, α, and k, in our model (1.1) and (1.2), the diusion coe cient d is the one most likely related to the type of microbe. Thus, I consider our a and C in Equations (5.70) and (5.71) depend on d.
The invariant conditions of Equations (5.70) and (5.71) are: e X 4dλϕ00+ (λ + 2d) ϕ0 + aϕ − Cϕ1/a ψ (5.70) = 0, (5.73) e X 4λψ00+ (λ + 2) ψ0+ aψ + Cϕ1/a ψ (5.71) = 0, (5.74)
where Xe is the prolongation of the operator (5.72): e X = X + ζ11 ∂ ∂ϕ0 + ζ 2 1 ∂ ∂ψ0 + ζ 1 2 ∂ ∂ϕ00 + ζ 2 2 ∂ ∂ψ00 + ω11 ∂ ∂dλ + ω12 ∂ ∂dϕ + ω31 ∂ ∂dψ + ω12 ∂ ∂aλ + ω22 ∂ ∂aϕ + ω32 ∂ ∂aψ (5.75) + ω13 ∂ ∂Cλ + ω23 ∂ ∂Cϕ + ω33 ∂ ∂Cψ ·
The coe cients ζ in (5.75) are given by the usual prolongation formulae (4.47). After similar calculations as we have done before, we obtain:
Now the invariant conditions (5.73) and (5.74) can be written as following: ξ (4dϕ00+ ϕ0) + η1 a −C aϕ 1/a−1 ψ + η2 −Cϕ1/a + D (4λϕ00 + 2ϕ0) +ζ11(λ + 2d) + 4ζ21dλ − µ2ϕ1/a ψ + µ1 ϕ − lnaC aϕ 1/a−1 ψ (5.70) = 0, (5.76) ξ (4ψ00+ ψ0) + η1C aϕ 1/a−1 ψ + η2 a + Cϕ1/a + ζ2 1(λ + 2) +ζ224λ + µ2ϕ1/a ψ + µ1 ψ + lnaC aϕ 1/a−1 ψ (5.71) = 0. (5.77)
Substituting the expressions (4.47) in (5.76) and (5.77), and equating to zero the coe cients for dierent powers of ϕ00, ψ0, ψ00 and ϕ0, we split the equations.
In Equation (5.76), the terms with ϕ00 give
ϕ00 : ηϕ2 = 0, ξϕ = 0.
The terms with ψ0 give
ψ0 : ξψ = 0, ξ + (λ + 2) ξλ− ξ λ + 4λ 2ηψλ2 − ξλλ = 0.
The terms with ψ02 give
ψ02 : ηψψ2 = 0. The determining equation (5.76) reduces to
C aη 1ϕ1/a−1 ψ + aη2+ Cη2ϕ1/a + ηλ2(λ + 2) + 4ληλλ2 + 2aξλψ − a ξ λψ − aη 2 ψψ + 2Cξλϕ 1/a ψ − Cξ λϕ 1/a ψ − Cη2 ψϕ 1/a ψ + µ2ϕ1/a ψ + µ1 ψ + lnaC aϕ 1/a−1 ψ = 0.
In Equation (5.77), the terms with ψ00 give
ψ00: ηψ1 = 0.
The terms with ϕ0 give
The terms with ϕ02 give
η1ϕϕ= 0. The remaining terms give
aη1− C aη 1ϕ1/a−1 ψ − Cη2ϕ1/a + η1λ(λ + 2d) + 4dλη1λλ+ Cξ λϕ 1/a ψ − aξ λϕ + D dCϕ 1/a ψ − aD dϕ + Cη 1 ϕϕ 1/a ψ − aηϕ1ϕ − 2Cξλϕ 1/a ψ + 2aξλϕ − µ2ϕ 1/a ψ + µ1 ϕ − lnaC aϕ 1/a−1 ψ = 0.
After integrating the resulting equations, we obtain ξ = 0, η1 = K1ϕ, η2 = K1ψ,
D = 0, µ1 = 0, µ2 = −K1
C
a, (5.78)
where K1 is an arbitrary constant. Substituting (5.78) in (5.72), one arrives at
the following result.
The family of non-linear ODEs (5.70) and (5.71) has a one-dimensional equiv-alence Lie algebra LE spanned by the operator:
X = ϕ ∂ ∂ϕ + ψ ∂ ∂ψ − C a ∂ ∂C· The characteristic system of X is
dϕ ϕ = dψ ψ = −a dC C , which provides two functionally independent invariants
P1 =
ψ
ϕ, P2 = ϕC
a. (5.79)
Inserting ψ = C1ϕ2 in Equations (5.70) and (5.71), we obtain
4dλϕ00+ (λ + 2d) ϕ0+ aϕ − CC1ϕ
1/a+2 = 0,
8C1λϕϕ00+ 8C1λϕ02+ 2C1(λ + 2) ϕϕ0+ C1aϕ2+ CC1ϕ
1/a+2 = 0,
which are much easier for solving. We can choose to exclude term ϕ00 to get
8dλϕ02+ 2 (d − 1) λϕϕ0+ (d − 2) aϕ2+ (2C1ϕ + d) Cϕ
1/a+2 = 0, (5.80)
or exclude term CC1ϕ
1/a+2 which results in the equation
4 (d + 2C1ϕ) λϕ00+ (λ + 2d + 2C1λϕ + 4C1ϕ) ϕ0
+8C1λϕ02+ (C1ϕ + 1) aϕ = 0. (5.81)
6 Summary of results
In this thesis we have investigated a biological reaction diusion model described by the system:
ut= duxx− αux− f (u) v, (1.1)
vt = vxx− αvx+ (f (u) − k) v, (1.2)
f (0) = 0, f0(u) > 0 for u > 0, lim
u→+∞f (u) > k. (1.3)
Firstly, it is shown that the equivalence Lie algebra LE for the system is a
six-dimensional Lie algebra spanned by Y1 = ∂ ∂t, Y2 = ∂ ∂x, Y3 = ∂ ∂u, Y4 = u ∂ ∂u + f ∂ ∂f + f ∂ ∂k, Y5 = v ∂ ∂v − f ∂ ∂f − f ∂ ∂k, (2.25) Y6 = 2t ∂ ∂t+ x ∂ ∂x − 2f ∂ ∂f − α ∂ ∂α − 2k ∂ ∂k·
Secondly, by using projections of the Lie algebra, we have obtained the system's approximate solution: u (t, x) = 2C4± r 4C2 4 − 6x2 Ct 6x , (4.61) v (t, x) = 4C4∓ r 4C2 4 − 6x2 Ct 6x = C4 x − u, (4.62) which can well t the reality under the condition:
C4 → 0,
C → ∞. (4.63)
Finally, by using one of the equations:
8dλϕ02+ 2 (d − 1) λϕϕ0 + (d − 2) aϕ2+ (2C1ϕ + d) Cϕ
1/a+2 = 0, (5.80)
4 (d + 2C1ϕ) λϕ00+ (λ + 2d + 2C1λϕ + 4C1ϕ) ϕ0
+8C1λϕ02+ (C1ϕ + 1) aϕ = 0. (5.81)
values of C and a to make sure that the nutrient uptake function f (u) ts the corresponding Monod function well. Then, depending on the values of C and a, one should solve the simplest equation from Equations (5.80), (5.81). Using the solution, the function ϕ and the relation ψ = C1ϕ2, one can calculate the
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