# Blow-Up Solution of Modified-Logistic-Diffusion Equation

Authors: P. Sitompul; Y. Soeharyadi
Journal: International Journal of Differential Equations
 (2019)

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## Abstract

Modified-Logistic-Diffusion Equationut=Duxx+u|1-u| with Neumann boundary condition has a global solution, if the given initial condition ψ satisfies ψ(x)≤1, for all x∈[0,1]. Other initial conditions can lead to another type of solutions; i.e., an initial condition that satifies ∫01ψ(x)dx>1 will cause the solution to blow up in a finite time. Another initial condition will result in another kind of solution, which depends on the diffusion coefficient D. In this paper, we obtained the lower bound of D, so that the solution of Modified-Logistic-Diffusion Equation with a given initial condition will have a global solution.


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## Body

## 1. Introduction

Logistic-Difussion Equation was first introduced by Fisher at 1937. It describes the growth of mutant gene on long habitatR [1]. The equation is given by(1)ut=Duxx+u1-uwith u(x,t) being the number of mutant genes at time t>0, at location x∈R with initial condition ψ∈C2(R). Lots of researches had been done on this type of partial differential equation (PDE), including how to approximate the solution with various methods [2–5]. This PDE has been applied on different disciplines of knowledge, such as in chemical reaction and economic growth [6–9]. This research will focus on the behavior of solution if the nonlinear factor in (1) is modified with absolute function; i.e., the rate of growth is always positive, (2)ut=Duxx+u1-u.PDE (2) explains that the rate of growth is slowing down around u=1 and is always positive. The state u=1 for all x is the steady state of (1) and (2). In (1), u=1 is a stable equilibrium, while in (2) it is a semi-stable equilibrium. This research is conducted on a bounded domain [0,1]⊂R with Neumann boundary condition.Consider the modified logistic ordinary differential equation(3)u′=u1-u.Equation (3) has a global solution if 0≤u0≤1. Completely, the solution of (3) for initial condition u0>0 is given by(4)ut=u0u0+1-u0e-t,u0<11,u0=1u0u0-u0-1et,u0>1.The behavior of the solution for the three types of initial conditions can be seen in Figure1. Equation (4) shows that the interval of the solution of (3) for u0>1 is [0,t∗) with(5)t∗=log⁡u01-u0.On the contrary, (3) has a global solution for u0<1. According to the solution (4), we can see that, for given initial condition ψ with 0≤ψ(x)<1 or 1<ψ(x) for all x∈R, the diffusion factor does not play many roles in the behavior of solution.Figure 1
The solution of the equationut=u|1-u| for the initial conditions u0<1, u0=1, and u0>1.Now, we will discuss the behavior of solution of (2) for other initial conditions ψ, that is,(6)∫01ψxdx<1,ψx>1for  somexThe effect of the diffusion coefficient D will be observed to guarantee the existence of global solution.Definition 1.
Letψ be the initial condition for (2). ψ is called D-initial condition if ψ satisfies (6).


## 2. Diffusion Time

Letf(u)=u|1-u|. This function is Lipchitz continuous, and so, by the following theorem, (2) has a solution.Theorem 2 (see [10]).
Iff is Lipchitz continuous, there is δ>0 dependent on u0 such that(7)ut=Duxx+fu has a solution at time interval [t0,t0+δ].The functionf(u)=u|1-u|≥0 for all u≥0. Thus, in the absence of diffusion, ut≥0 for all u≥0. This means that the value of u(x,t) will grow for every x∈[0,1]. If there exist x∈[0,1] such that ψ(x)>1, then the solutions interval of (2) is [0,t∗] with(8)t∗=log⁡ψ∗1-ψ∗,ψ∗=maxxψx.On the value of initial condition ψ with ∫01ψ(x)dx>1, the diffusion coefficient is not affected much. It is because the diffusion coefficient is just distributing the concentration from high to low without addition or substraction of the total of concentration on interval [0,1]. As a result, there is any x∈[0,1] such that u(x,t)>1 for all t>0. Furthermore the solution of (2) will blow up at finite time.For further discussion, assume there isx∈[0,1] such that ψ(x)>1 and ∫01ψ(x)dx<1. If the diffusion coefficient D is quite large, then (2) has global solution.Definition 3.
Letu(x,t) be the solution of (2) with initial condition ψ. Diffusion time for (2) with respect to ψ, denoted by TD(ψ), is defined by minimum time t such that u(.,t)≤1.Notice the following usual diffusion equation:(9)ut=Duxx.If u1 and u2 are the solutions of (2) and (9), respectively, with the same initial condition ψ, then u1(x,t)≥u2(x,t) for every (x,t)∈[0,1]×[0,∞). Hence, the diffusion time of (9) with respect to ψ is last or equal to diffusion time of (2) with respect to ψ. Let TD be the diffusion time of (9) with respect to ψ; then TD is a lower bound for diffusion time of (2) with respect to ψ. In [11], the solution of (9) for initial condition ψ is(10)ux,t=12A0+∑k=1∞Ake-kπ2Dtcos⁡kπx.with Ak=∫01ψxcos⁡(kπx)dx. For certain initial condition, the TD value can be obtained easily, while for other initial conditions, we could only determine the lower bound of TD.In this section, we will discuss the diffusion time for two elementary function such as linear and trigonometric functions. A particular linear function family of initial condition,ψ(x)=ax+b,a,b≥0, is a D-initial condition if a+b>1 and a+b/2<1. For this specific family, we found a lower bound of diffusion time with respect to (2).Theorem 4.
Letψ(x)=a+bx be initial condition for (2). If b>0 and 2π2(1-a)/8+π2<b, then(11)TDψ>1π2Dlog⁡4b1-a-b/2π2. Moreover if b<0, and b<2π2(1-a)/π2-8, then(12)TDψ>1π2Dlog⁡-4b1-a-b/2π2.Proof.
The linear functionψ(x)=a+bx, a,b>0, is a monotone function and reaches the maximum at x=1. The volume of ψ is a+b/2, so that, for a+b/2≥1, diffusion time TD(ψ)=∞, and for a+b≤1, diffusion time TD(ψ)=0. The initial condition ψ will satisfy the condition (6); a and b satisfy(13)a<b<11-a<b<2-2a.
The solution of(14)ut=Duxx,ux0,t=ux1,t=0for initial condition ψ(x)=a+bx on x=1 is(15)u1,t=a+b2+∑k=1∞4b2k-1π2e-2k-1π2Dt.
Define(16)u_1,t=a+b2+4be-π2Dtπ2. It is clear that(17)u1,t>u_1,t and if u_(1,T)=1, then T<TD(ψ). The time T with u_(1,T)=1 is(18)T=1π2Dlog⁡4b1-a-b/2π2.
In this caseT>0 for(19)b>2π21-a8+π2.
Forψ(x)=a-bx with a>1, the condition ∫01ψ(x)dx<1 and ψ>1 for some x∈[0,1], satisfied by 2a-2<b<a<2. Assume x=1-y, and substituting to (14), we get(20)ut=Duyy. It means that the diffusion time for ψ(x)=a-bx is equal to diffusion time for φ(y)=a-b+by. Diffusion time for φ (by (18)) is(21)T=1π2Dlog⁡4a1-a+b/2π2. and T>0 is satisfied for(22)b<2π2a-1π2-8.Next, the sinusoidal function class of initial conditionψx=a+bcos⁡(nπx),n∈Z, will be considered. The function ψ will satisfy D-initial condition if a<1 and a+|b|>1. The diffusion time for this class of function is given by this theorem.Theorem 5.
Let the initial conditionψ(x)=a+bcos⁡(nπx),n∈Z, given for (2). If 0<a<1 and a+|b|>1, then(23)TDψ>1nπ2Dlog⁡b1-b.Proof.
Forψx=a+bcos⁡(nπx), where n∈Z, the local maximum or local minimum points of ψ(x) have the same character; that is, they have the same concavity. The volume of ψ is a. So, TD(ψ)=∞ for a>1 and TD(ψ)=0 for a+b<1. We will check the time diffusion for ψ=a+bcos⁡(nπx) with a<1 and a+b>1.
First, we check forn=1. The solution of (14) for initial condition ψx=a+bcos⁡(πx) is given by(24)ux,t=a+be-π2Dtcos⁡πx. For b>0, the solution u(x,t) is decreasing with respect to x for t>0, so u(x,t) reach maximum at x=0. Furthermore, if u(0,T)=1, then u(x,T)≤1 for all x. This time T is(25)T=1π2Dlog⁡b1-a.On the other side, for ψx=a-bcos⁡(πx) with b>0, u(x,t) is increasing with respect to x for t>0. The maximum value of u(x,t) was reached at x=1. By substituting y=1-x, we have the problem (14) with initial condition ψ(x)=a+bcos⁡(πx). The time given by (25) is positive for b>1-a. Therefore, the diffusion time for ψ(x)=a±bcos⁡(πx) is(26)TDψ>1π2Dlog⁡b1-a.
Forn≥2, we can see the problem on interval Ii=[i-1/n,i/n], i=1,…,n. It is clear that ψ(x)=a+bcos⁡(nπx) is monotone on each interval Ii for all i=1,…,n, and u(x,t)x(i/n)=0 for t>0. Therefore, we can see the problem (14) with Neumann Boundary condition on interval Ii=[i-1/n,i/n], and we write it as(27)ut=Duxxx∈i-1n,in,t>0uxi-1n,t=uxin,t=0t>0ux,0=ψxx∈i-1n,in,ψx∈C2i-1n,in.By translation and dilatation(28)x´=nx-i-1n, so, the problem (27) became(29)ut=n2Duxxx∈0,1,t>0ux0,t=ux1,t=0t>0ux,0=ψxx∈0,1,ψx∈C20,1.
The diffusion time forψ(x)=a-bcos⁡(nπx) in (14) is equal to the diffusion time for ψ(x)=a-bcos⁡(πx) in (29). Therefore, the diffusion time for ψ(x)=a-bcos⁡(nπx) in (14) is(30)TDψ>1nπ2Dlog⁡b1-a.


## 3. Reaction Time

The functionf(u)=u|1-u|≥0 for all u≥0. It means the reaction factor is causing the increase of concentration for all time. If f could contribute to the increasing of the concentration such that u(x,t)>1 for time t<∞, the solution will blow up.LetΩ1={x∣ψ(x)≤1} and Ω2={x∣ψ(x)>1}. Then the reaction time for initial condition ψ is defined as follows.Definition 6.
Assumingf is D-initial condition, let u(x,t) be the solution of (2) for D-initial condition ψ. The minimum time T that meets ∫Ω1(u(x,T)-ψ(x))dx=δ is defined as the reaction time for ψ and is notated as TR(ψ).For initial condition that has minimum value at[0,1], the reaction time is given by Theorem 7 below.Theorem 7.
Letψ be the D-initial condition of (2) and have minimum value on [0,1] with m=minx⁡ψ(x); then(31)T=log⁡1+δm∫Ω11-ψxdx-δ is an upper bound for TR(ψ).Proof.
LetTR(ψ)=T and V1[D](t)=∫Ω1u(x,t)dx. So,(32)δ=V1DT-∫Ω1ψxdx>V10T-∫Ω1ψxdx=∫Ω111+1-ψx/ψxe-Tdx-∫Ω1ψxdx=∫Ω11-ψx1+1-ψx/ψxe-T1+1-ψx/ψxe-Tdx=∫Ω1ψx1-ψx1-e-Tψx1-e-T+e-Tdx. Let minx⁡ψ(x)=m; then(33)δ>m1-e-Tm1-e-T+e-T∫Ω11-ψxdx. Furthermore, let(34)PT=m1-e-Tm1-e-T+e-T∫Ω11-ψxdx. If P(T)=δ, then T is an upper bound of TR(ψ). The time T that satisfies this condition is(35)T=log1+δm∫Ω11-ψxdx-δ. Since ψ(x)>1 for every x∈Ω2, then(36)∫Ω11-ψxdx-δ=∫Ω11-ψxdx-∫011-ψxdx=∫Ω11-ψxdx-∫Ω11-ψxdx+∫Ω21-ψxdx=∫Ω2ψx-1dx>0.This shows that T>0 for some D-initial condition ψ.The upper bound of reaction time for initial condition of linear and sinusoidal function class is obtained as follows.Theorem 8.
Let theD-initial condition be given by ψ(x)=a+bx, a>0 with a+b/2<1, a≠b and a≠1.
(1) Ifb>0, then TR(ψ)<log1+2b(1-a-b/2)/a1+a+b2.
(2) Ifb<0, then TR(ψ)<log1+2b(1-a+b/2)/a-b1-a2.Proof.
ForD-initial condition ψ(x)=a+bx, b>0, that is, 0<a<1 and 1-a<b<2-2a, we obtain(37)∫Ω11-ψxdx=12b1-a2,m=a,δ=1-a-b2.Therefore, by Theorem 7 for D-initial condition ψ(x)=a+bx,b>0 we obtain the upper bound of reaction time TR(ψ) as(38)T=log1+2b(1-a-b/2a1-a+b2. In the same way, we obtain the upper bound of reaction time for D-initial condition ψ(x)=a-bx as(39)T=log1+2b(1-a+b/2a-b1-a2.From the first result in Theorem8 we obtain the upper bound of the reaction time for some family of linear functions with positive gradient, while the second result is for the negative gradient.Theorem 9.
Let theD-initial condition be given by ψ(x)=a+bcos⁡(nπx) with n≠0. The reaction time for ψ will satisfy(40)TRψ<log1+πa-bcos-1a-1/b+b/1-a2-11/2-π.Proof.
Letψ(x)=a-bcos⁡(πx) satisfying condition (6); that is, (41)0,5<a<11-a<b<a.We have(42)∫Ω11-ψxdx=1-aπcos-1a-1b+b1-a2-11/2m=a-bδ=1-a.By Theorem 7, we get an upper bound of reaction time for ψ(x)=a-bcos⁡(πx), as(43)T=log1+πa-bcos-1a-1/b+b/1-a2-11/2-π. Since the behavior of solution for initial condition ψ(x)=a-bcos⁡(πx) is the same as the behavior of solution for initial condition φ(x)=a+bcos⁡(πx), TR(ψ)=TR(φ).
Forψ(x)=a+bcos⁡(nπx) with n∈Z,n≥2, we can see the behavior solution on interval [0,1/n]. The behavior solution on each interval Ii=[i-1/n,i/n] is the same for all i=1,…,n. To increase volume as mount of δ on the interval [0,1] is equal to increasing volume as mount of δ/n on each interval Ii. The upper bound of time T to increase the volume as mount of δ/n on interval [0,1/n] is(44)T=log1+πa-bcos-1a-1/b+b/1-a2-11/2-π.This result will be used to show the relationship between diffusion coefficientD and the behavior of the solution of (2) for a given D-initial condition ψ.


## 4. The Lower Bound ofD for Global Solution

The solution of (2) with D-initial condition ψ will blow up for t∗<∞ if TR(ψ)<TD(ψ). This is because the reaction term is growing much faster to supply the volume than the suppressing of diffusion term. From Theorems 4, 5, 8, and 9, we obtained the lower bound for diffusion coefficient D such that the global solution for linear and sinusoidal function family of initial condition exists.Theorem 10.
Let the Modified-Logistic-Diffusion be(45)ut=Duxx+u1-u,x∈0,1ux0,t=ux1,t=0,t>0ux,0=a+bx,x∈0,1.with ∫01ψ(x)<1 and a+b>1. For b>0 and 2π2(1-a)/8+π2<b, the solution u(x,t) will blow up if(46)D<log4b/1-a-b/2π2π2log⁡1+2b1-a-b/2/a1+a+b2. and for b<0, 0<2π2(a-1)/π2-8+b and 0<b+a, the solution u(x,t) will blow up if(47)D<log-4b/1-a+b/2π2π2log⁡1+2b1-a-b/2/a+b1+a2Theorem10 gives the lower bound of D such that (2) on [0,1] with the D-initial conditions of the increasing linear function and the decreasing linear function has a global solution. In this case(48)D=log4b/1-a-b/2π2π2log⁡1+2b1-a-b/2/a1+a+b2.is lower bound of D so the system with D-initial condition of the increasing linear function ψ(x)=a+bx,b>0 has a global solution. As for the decreasing linear function ψ(x)=a+bx,b<0, the lower bound of D is(49)D=log-4b/1-a+b/2π2π2log⁡1+2b1-a-b/2/a+b1+a2.The next theorem discusses some family of trigonometric function of initial condition.Theorem 11.
Let the Modified-Logistic-Difussion be(50)ut=Duxx+u1-u,x∈0,1ux0,t=ux1,t=0,t>0ux,0=a-bcos⁡nπx,x∈0,1.with n≠0, 1/2<a<1, and 1-a<b<a. The solution u(x,t) will blow up if(51)D<logb/1-anπ2log⁡1+π/a-bcos-1a-1/b+b/1-a2-11/2-π.


## 5. Discussion

The diffusion-reaction equation with modified logistic function of reaction term and Neumann’s boundary condition at[0,1] can have a global or blow-up solution. If the initial conditions given are D-initial conditions, then the diffusion term plays an important role in determining whether the system will have a global solution or a blow-up solution. In this study, we obtain the lower bound of the diffusion coefficient D such that the system has a global solution. The objective for further investigation is determining the limiting value of the diffusion coefficient D such that, for D-initial condition, the solution has a global soluton or a blow-up solution.



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