Nonlocality in Partial Differential Equations

Keng Deng

Department of Mathematics

University of Louisiana at Lafayette

Lafayette, LA 70504, USA

Abstract

In the development of mathematical models for topics of practical importance,many nonlocal problems for partial differential equations have arisen from engineering,physics and life sciences. In this series of lectures, we will introduce various methodsto study the existence/uniqueness and asymptotic behavior of solutions for such modelproblems.

1 Introduction

Locality is an assumption which is central to many mathematical models currently in use.

By locality we mean that an effect at some point (spatial or temporal) depends only on

the physical conditions at the same point. For example, Newton’s second law expresses the

acceleration of a body at a particular time as being dependent only on outside forces at the

same time. Other examples are evident in the study of heat flow (Fourier’s law of conduction)

and in the study of ground water flow (Darcy’s law). Even the nonlinear versions of these

contain only local dependencies. However, it seems quite reasonable to suggest that the flow

of water at a point underground may depend on the water content in some region containing

the point. Furthermore, the phenomenon of hysteresis occurs because of dependence upon

the history of a state. It is similarly reasonable to suggest that the tendency for a plant or

animal to move from one point to another may depend on the number of individuals within

a surrounding acre (crowding effect), or that population growth should depend upon the

population at previous times (gestation period). The phenomenon of chemotaxis, whereby

certain cells are caused to move towards regions of higher concentration of a chemical given

off by other cells of that same species, provides yet another example where such nonlocal

behavior is exhibited.

Nonlocal dependencies have been well studied in the case of ordinary differential equations

where they are more commonly known as delays. However, in the instance that dependen-

cies are nonlocal in both space and time, the subsequent models include initial-boundary

value problems for partial differential equations, and useful general theories, even basic exis-

tence/uniqueness results, are therefore much more difficult to develop. What is in fact needed

are new methods which are effective in analyzing the solutions of a wide variety of model

problems. In this series of lectures, we will introduce various methods to investigate the

solvability and long-time behavior of solutions to several model problems. The motivation

1

for a study of these particular models will be discussed specifically in the context. Further,

as has already been observed in recent investigations on models involving nonlocality, these

methods should be applicable in many situations beyond those to be investigated here.

2

2 Semilinear parabolic equations with nonlocal initial

conditions

We consider the following nonlocal initial-boundary value problem [2]

Lu+ g(x, t, u) = 0 x ∈ Ω, 0 < t < T,

u(x, t) = ϕ(x, t) x ∈ ∂Ω, 0 < t < T,

u(x, 0) =

∫ T0

0

h(x, t)u(x, t)dt+ f(x) x ∈ Ω,

(2.1)

where Ω is a bounded domain in Rn, 0 < T0 ≤ T , and L is a uniformly parabolic operator

with continuous and bounded coefficients,

L =

n∑

i,j=1

aij(x, t)∂2

∂xi∂xj+

n∑

i=1

bi(x, t)∂

∂xi− ∂

∂t.

g(x, t, u) is C0 in (x, t) and C1 in u with gu(x, t, u) ≤ 0, ϕ(x, t) is continuous, h(x, t) is

continuous in x and integrable in t, and f ∈ C(Ω).

This problem is closely related to a mathematical model arising from the physical moti-

vation,

Lu+ c(x, t)u = g(x, t) x ∈ Ω, 0 < t < T,

u(x, t) = ϕ(x, t) x ∈ ∂Ω, 0 < t < T,

u(x, 0) +N∑

k=1

βk(x)u(x, Tk) = f(x) x ∈ Ω

(2.2)

with Tk ∈ (0, T ] (k = 1, . . . , N).

The above problem can be formulated in terms of the theory of diffusion and heat con-

duction. For example, it can be applied to the description of diffusion phenomenon of a

small amount of gas in a transparent tube. We assume that the diffusion is observed via

the surface of the tube. If there is too little gas at the initial time, then the measurement

u(x, 0) of the amount of the gas in this instant may be less precise than the measurement

u(x, 0)+

N∑

k=1

βk(x)u(x, Tk) of the sum of the amounts of this gas. Under the hypotheses that

−1 ≤N∑

k=1

βk(x) ≤ 0 and c(x, t) ≤ −c0 < 0, the existence and uniqueness of solutions of (2.2)

were established in [3, 4] for N = 1 and in [1] for N > 1.

It is easily seen that the initial condition in (2.2) is a special case of that in (2.1).

Moreover, we are concerned here with the behavior of solutions of problem (2.1). To this

end, we first obtain an a priori estimate.

3

Lemma 2.1 There exist a positive function ψ(x) on Ω and a constant γ > 0 such that for

any B > 0 and c(x, t) ≤ 0 on Ω× (0,∞), if u(x, t) satisfies

Lu+ c(x, t)u = 0 x ∈ Ω, t > 0,

u(x, t) = 0 x ∈ ∂Ω, t > 0,

|u(x, 0)| ≤ Bψ(x) x ∈ Ω,

(2.3)

then

|u(x, t)| ≤ Bψ(x)e−γt on Ω× (0,∞).

We then have the following existence/uniqueness and decay results.

Theorem 2.2 Suppose that

∫ ∞

0

|h(x, t)|dt ≤ 1. Then problem (2.1) has a unique solution

u ∈ C2,1(Ω× (0, T ))∩C(Ω× [0, T ]). Furthermore, if ϕ(x, t) = 0, T0 = ∞, and g(x, t, 0) = 0,

then U(t) = maxx∈Ω

|u(x, t)| strictly decreases at exponential rate.

Finally, we indicate that when

∫ ∞

0

|h(x, t)|dt > 1, in general, problem (2.1) is not well-

posed. For instance, consider

ut = ∆u x ∈ Ω, t > 0,

u(x, t) = 0 x ∈ ∂Ω, t > 0,

u(x, 0) =

∫ ∞

0

h(x, t)u(x, t)dt x ∈ Ω.

(2.4)

u ≡ 0 is a solution. On the other hand, choose λ0 as the smallest eigenvalue and φ0(x) as

the associated eigenfunction of

∆φ+ λφ = 0 x ∈ Ω,

φ = 0 x ∈ ∂Ω,(2.5)

and let

h(x, t) =

λ0/(1− e−λ0T0) if t ≤ T0,

0 if t > T0,(2.6)

for some constant T0 < ∞. Note that

∫ ∞

0

h(x, t)dt = λ0T0/(1 − e−λ0T0) > 1. Then u =

φ0(x)e−λ0t is another nontrivial solution.

References

[1] J. Chabrowski, On non-local problems for parabolic equations, Nagoya Math. J. 93

(1984), 109–131.

4

[2] K. Deng, Exponential decay of solutions of semilinear parabolic equations with nonlocal

initial conditions, J. Math. Anal. Appl. 179 (1993), 630–637.

[3] A. A. Kerefov, Non-local boundary value problems for parabolic equations, Differ-

entsial’nye Uravneniya 15 (1979), 74–78.

[4] P. N. Vabishchevich, Non-local parabolic problems and the inverse heat-conduction

problem, Differentsial’nye Uravneniya 17 (1981), 1193–1199.

5

3 Solvability of a nonlocal hyperbolic model

The study of repairable systems is an important topic in reliability, and there are many ap-

plications of reliability models in computer systems, telecommunication systems, and manu-

facturing systems etc. Among the reliability models, there is also an extensive literature on

availability characteristics of repairable systems with two or three components under vari-

ous assumptions on the failures and repairs. For details, see [4, 5] and the references cited

therein.

To make it clear, we consider a two-unit series reliability system with shut-off rule [1].

Generally speaking, a series system operates if and only if each unit operates. In practice,

however, one special unit failure may always shut off some other units but not vice versa.

For example, failure of the power supply may shut down an electronic device but not vice

versa.

We first impose the following assumptions:

• The system consists of two units in series. The system operates if and only if both

units operate.

• Unit 1 upon failure shuts off unit 2, but not vice versa.

• All random variables of the lifetime and the repair time for each unit are mutually

independent.

• The lifetime distribution Fi(t) and the repair time distribution Gi(t) of unit i (i = 1, 2)

are arbitrary, and they are denoted by

Fi(t) =

∫ t

0

fi(s)ds = 1− exp

(

−∫ t

0

λi(s)ds

)

= 1− F i(t),

Gi(t) =

∫ t

0

gi(s)ds = 1− exp

(

−∫ t

0

µi(s)ds

)

= 1− Gi(t),

(3.1)

which implies that for unit i, fi(t) is the density function of the lifetime, λi(t) is the

hazard function of the lifetime, gi(t) is the density function of the repair time, and

µi(t) is the hazard function of the repair time.

• The repaired unit is as good as a new unit. There is only one repair facility, and the

repair discipline for the two units is “first-fail, first-repaired”.

We define the system’s states as follows:

State 0: two units are operating, and the ages of the two units are equal;

State 01: two units are operating, and the age of unit 1 is greater than that of unit 2;

State 02: two units are operating, and the age of unit 2 is greater than that of unit 1;

6

State 1: unit 1 is being repaired, and unit 2 is shut off;

State 2: unit 2 is being repaired, and unit 1 is still operating;

State 3: unit 2 is still being repaired, and unit 1 is waiting for repair;

State 4: unit 1 is being repaired, and new unit 2 is in “suspended animation”.

We then introduce the following notations.

S(t) is the system state at time t;

X(t) is the age of unit 1 at time t when S(t) = 0, 01, 02, 2;

Y (t) is the age of unit 2 at time t when S(t) = 0, 01, 02, 1;

Z1(t) is the elapsed repair time of unit 1 at time t when S(t) = 1, 4;

Z2(t) is the elapsed repair time of unit 2 at time t when S(t) = 2, 3.

Now we can obtain a Markov process S(t), X(t), Y (t), Z1(t), Z2(t) which takes values

on the set:

J∗ = (0, x), (01, x, y), (02, x, y), (1, y, z), (2, x, z), (3, z), (4, z)| 0 ≤ x, y, z <∞.

Next, we define the following state probabilities:

P0(t, x)dx = PS(t) = 0, x ≤ X(t) = Y (t) < x+ dx;P01(t, x, y)dxdy = PS(t) = 0, x ≤ X(t) < x+ dx, y ≤ Y (t) < y + dy, x > y;

P02(t, x, y)dxdy = PS(t) = 0, x ≤ X(t) < x+ dx, y ≤ Y (t) < y + dy, x < y;

P1(t, y, z)dz = PS(t) = 1, z ≤ Z1(t) < z + dz, Y (t) = y;P2(t, x, z)dxdz = PS(t) = 2, x ≤ X(t) < x+ dx, z ≤ Z2(t) < z + dz, x > z;

P3(t, z)dz = PS(t) = 3, z ≤ Z2(t) < z + dz;P4(t, z)dz = PS(t) = 4, z ≤ Z1(t) < z + dz.

Let

P0(x) = limt→∞

P0(t, x), P01(x, y) = limt→∞

P01(t, x, y), P02(x, y) = limt→∞

P02(t, x, y),

P1(y, z) = limt→∞

P1(t, y, z), P2(x, z) = limt→∞

P2(t, x, z), P3(z) = limt→∞

P3(t, z),

P4(z) = limt→∞

P4(t, z).

By using Markov properties, probability considerations, and continuity arguments, we then

obtain the following differential equations governing the behavior of the system:[

d

dx+ λ1(x) + λ2(x)

]

P0(x) = 0,

[

∂

∂x+

∂

∂y+ λ1(x) + λ2(y)

]

P01(x, y) = 0,

[

∂

∂x+

∂

∂y+ λ1(x) + λ2(y)

]

P02(x, y) = 0,

7

[

d

dz+ µ1(z)

]

P1(y, z) = 0,

[

∂

∂x+

∂

∂z+ λ1(x) + µ2(z)

]

P2(x, z) = 0,

[

d

dz+ µ2(z)

]

P3(z) =

∫ ∞

z

P2(x, z)λ1(x)dx,

[

d

dz+ µ1(z)

]

P4(z) = 0,

(3.2)

the boundary conditions:

P0(0) =

∫ ∞

0

P4(z)µ1(z)dz,

P01(x, 0) =

∫ x

0

P2(x, z)µ2(z)dz,

P02(0, y) =

∫ ∞

0

P1(y, z)µ1(z)dz,

P1(y, 0) =

∫ y

0

P02(x, y)λ1(x)dx+

∫ ∞

y

P01(x, y)λ1(x)dx+ P0(y)λ1(y),

P2(x, 0) =

∫ x

0

P01(x, y)λ2(y)dy +

∫ ∞

x

P02(x, y)λ2(y)dy + P0(x)λ2(x),

P3(0) = 0,

P4(0) =

∫ ∞

0

P3(z)µ2(z)dz,

(3.3)

and the regular condition∫ ∞

0

P0(x)dx+

∫ ∞

0

∫ x

0

P01(x, y)dydx+

∫ ∞

0

∫ y

0

P02(x, y)dxdy

+

∫ ∞

0

∫ ∞

0

P1(y, z)dzdy +

∫ ∞

0

∫ x

0

P2(x, z)dzdx +

∫ ∞

0

[P3(z) + P4(z)] dz = 1.

(3.4)

Problem (3.2)-(3.4) is a nonlocal hyperbolic model. To discuss its solvability, however, at

least two difficulties will be encountered. First, in the xy-plane, the boundary condition at

y = 0 can be viewed as an nonlocal initial condition, which usually causes complication (e.g.,

see [3]). Second, (3.3) implies that instead of local existence, global solvability of (3.2) must

be shown. In the existing literature, four methods are often used to establish existence and

uniqueness results: 1) the characteristic method with fixed point argument; 2) the semigroup

of linear operators theoretic approach; 3) the implicit finite difference approximation; 4) the

iterative monotone method based on the comparison principle. None of the above methods

appear readily applicable to problem (3.2)-(3.4). To overcome this obstacle, in [4] Shi and

Li derived a system of integral equations involving three undetermined functions by the

8

method of variation of parameters under the assumption that the solution of (3.2)-(3.4)

is continuously differentiable, i.e., in the classical sense. However, such an assumption is

not necessarily valid, for example, in the case that the hazard functions λi(t) or µi(t) are

only bounded. In contrast to that, we will formulate the system of integral equations under

a weaker assumption that problem (3.2)-(3.4) only admits a mild solution, i.e., a solution

satisfying an equation obtained by integration along the characteristic curves.

For simplicity, from now on we will use (3.2.i) for the ith equation in (3.2) and (3.3.i)

for the ith boundary condition in (3.3). In view of (3.1) and (3.2.1), we first have

P0(x) = P0(0) exp

(

−∫ x

0

(λ1(t) + λ2(t))dt

)

= CF 1(x)F 2(x), (3.5)

where C = P0(0). Then integrating (3.2.2) along the characteristic curves, we have

P01(x, y) = P01(x− y, 0) exp

(

−∫ y

0

(λ1(x− y + t) + λ2(t))dt

)

= P01(x− y, 0) exp

(

−∫ x

0

λ1(t)dt +

∫ x−y

0

λ1(t)dt

)

F 2(y)

= (P01(x− y, 0)/F 1(x− y))F 1(x)F 2(y)

= Ch1(x− y)F 1(x)F 2(y),

(3.6)

where h1(x) = P01(x, 0)/CF 1(x). Similarly, we find

P02(x, y) = P02(0, y − x) exp

(

−∫ x

0

(λ1(t) + λ2(y − x+ t))dt

)

= (P02(0, y − x)/F 2(y − x))F 1(x)F 2(y)

= Ch2(y − x)F 1(x)F 2(y),

(3.7)

where h2(x) = P02(0, x)/CF 2(x). Now by (3.3.4) and (3.5)-(3.7) we find

P1(y, z) = P1(y, 0)G1(z)

= C

[∫ y

0

h2(y − x)F 2(y)F 1(x)λ1(x)dx

+

∫ ∞

y

h1(x− y)F 1(x)F 2(y)λ1(x)dx

+F 1(y)F 2(y)λ1(y)]

G1(z)

= C

[∫ y

0

h2(t)f1(y − t)dt+

∫ ∞

0

h1(t)f1(y + t)dt+ f1(y)

]

F 2(y)G1(z).

(3.8)

Proceeding as in (3.6) we obtain

P2(x, z) = P2(x− z, 0) exp

(

−∫ z

0

(µ2(t) + λ1(x− z + t))dt

)

= (P2(x− z, 0)/F 1(x− z))F 1(x)G2(z)

= Ch3(x− z)F 1(x)G2(z),

(3.9)

9

where h3(x) = P2(x, 0)/CF 1(x). Then (3.9), together with (3.2.6) yields

P3(z) = G2(z)

∫ z

0

∫ ∞

t

[P2(x, t)λ1(x)/G2(t)]dxdt

= CG2(z)

∫ z

0

∫ ∞

t

[h3(x− t)F 1(x)G2(t)λ1(x)/G2(t)]dxdt

= CG2(z)

∫ z

0

∫ ∞

0

h3(s)f1(t + s)dsdt

= CG2(z)

∫ ∞

0

h3(s)[F 1(s)− F 1(z + s)]ds.

(3.10)

Since P4(z) = P4(0)G1(z), by (3.3.1) we find

P0(0) = P4(0)

∫ ∞

0

µ1(z)G1(z)dz = P4(0),

thus we have P4(z) = CG1(z).

We now establish the relationship between h1, h2, and h3. Making use of (3.3.2) and

(3.9), we first have

h1(x) = P01(x, 0)/CF 1(x)

= (1/CF 1(x))

∫ x

0

Ch3(x− z)F 1(x)G2(z)µ2(z)dz

=

∫ x

0

h3(x− z)g2(z)dz.

(3.11)

By virtue of (3.3.3) and (3.8), we then have

h2(x) = P02(0, x)/CF 2(x)

= (1/CF 2(x))

∫ ∞

0

C

[∫ x

0

h2(t)f1(x− t)dt+

∫ ∞

0

h1(t)f1(x+ t)dt+ f1(x)

]

·F 2(x)G1(z)µ1(z)dz

=

∫ x

0

h2(t)f1(x− t)dt +

∫ ∞

0

h1(t)f1(x+ t)dt + f1(x).

(3.12)

Taking (3.3.5) and (3.5)-(3.7) into account, we now have

h3(x) = P2(x, 0)/CF 1(x)

= (1/CF 1(x))

[∫ x

0

Ch1(x− y)F 1(x)F 2(y)λ2(y)dy

+

∫ ∞

x

Ch2(y − x)F 1(x)F 2(y)λ2(y)dy + CF 1(x)F 2(x)λ2(x)

]

=

∫ x

0

h1(y)f2(x− y)dy +

∫ ∞

0

h2(y)f2(x+ y)dy + f2(x).

(3.13)

10

Therefore, the solution of problem (3.2)-(3.4) can be represented by the following:

P0(x) = CF 1(x)F 2(x),

P01(x, y) = Ch1(x− y)F 1(x)F 2(y),

P02(x, y) = Ch2(y − x)F 1(x)F 2(y),

P1(y, z) = CG1(z)F 2(y)h2(y),

P2(x, z) = Ch3(x− z)F 1(x)G2(z),

P3(z) = CG2(z)

∫ ∞

0

h3(t)[F 1(t)− F 1(z + t)]dt,

P4(z) = CG1(z),

where in view of (3.4) the constant C is determined by

C−1 =

∫ ∞

0

F 1(x)F 2(x)dx+

∫ ∞

0

∫ x

0

h1(x− y)F 1(x)F 2(y)dydx

+

∫ ∞

0

∫ y

0

h2(y − x)F 1(x)F 2(y)dxdy +

∫ ∞

0

∫ ∞

0

G1(z)F 2(y)h2(y)dzdy

+

∫ ∞

0

∫ x

0

h3(x− z)F 1(x)G2(z)dzdx+

∫ ∞

0

G1(z)dz

+

∫ ∞

0

∫ ∞

0

G2(z)h3(t)[F 1(t)− F 1(z + t)]dtdz.

(3.14)

It is worth noting that C−1 is the mean renewal period of the system. Moreover, the limiting

availability of the system, denoted by A, is given by

A = C

[∫ ∞

0

F 1(x)F 2(x)dx+

∫ ∞

0

∫ x

0

h1(x− y)F 1(x)F 2(y)dydx

+

∫ ∞

0

∫ y

0

h2(y − x)F 1(x)F 2(y)dxdy

](3.15)

and the stationary failure frequency of the system, denoted by W , is given by

W = C

1 +

∫ ∞

0

∫ x

0

h1(x− y)[f1(x)F 2(y) + F 1(x)f2(y)]dydx

+

∫ ∞

0

∫ y

0

h2(y − x)[f1(x)F 2(y) + F 1(x)f2(y)]dxdy

.

(3.16)

Here the undetermined functions h1, h2, and h3 are governed by a system of integral equations

h1(x) =

∫ x

0

h3(t)g2(x− t)dt,

h2(x) =

∫ x

0

h2(t)f1(x− t)dt+

∫ ∞

0

h1(t)f1(x+ t)dt+ f1(x),

h3(x) =

∫ x

0

h1(t)f2(x− t)dt+

∫ ∞

0

h2(t)f2(x+ t)dt+ f2(x).

(3.17)

11

Clearly, the solvability of system (3.17) is very important for the performance evaluation of

the reliability system.

We now establish the existence and uniqueness results for system (3.17). In view of the

second and third equations in (3.17), we must show the global solvability of (3.17). To this

end, we first introduce a hypothesis.

(H). There exist positive constants σ and η such that

σ ≤ λ1(x), λ2(x), µ2(x) ≤ η, 2σ > η, and (2σ − η)2(2σ + η) > η3.

Remark 3.1 Restrictions on the model parameters have been introduced for other integral

equations. For example, consider the following equation arising from population dynamics

[6]:

u(x) =

∫ ∞

0

β(t)u(t)dt−∫ x

0

m(t)u(t)dt. (∗)

It is easily seen that the solution of (∗) exists only if

∫ ∞

0

β(x) exp

(

−∫ x

0

m(t)dt

)

dx = 1.

Theorem 3.2 Suppose that the hypothesis (H) holds. Then system (3.17) has a continuous

solution (h1, h2, h3).

Proof. Since we seek a global solution of (3.17), the theory of [2] does not apply. For this

reason, we construct a sequence hn1 , hn2 , hn3∞n=0 as follows:

h01(x) = 0, h02(x) = f1(x), h03(x) = f2(x). For n = 1, 2, · · ·

hn1 (x) =

∫ x

0

hn−13 (t)g2(x− t)dt,

hn2 (x) =

∫ x

0

hn−12 (t)f1(x− t)dt+

∫ ∞

0

hn−11 (t)f1(x+ t)dt+ f1(x),

hn3 (x) =

∫ x

0

hn−11 (t)f2(x− t)dt+

∫ ∞

0

hn−12 (t)f2(x+ t)dt+ f2(x).

(3.18)

We first show that the sequence is monotonically increasing. We then show that hni (x) are

bounded from above by ki(x) (i = 1, 2, 3), where (k1, k2, k3) satisfies an auxiliary system of

integral equations:

k1(x) =

∫ x

0

ηk3(t)e−σ(x−t)dt,

k2(x) =

∫ x

0

ηk2(t)e−σ(x−t)dt+

∫ ∞

0

ηk1(t)e−σ(x+t)dt+ ηe−σx,

k3(x) =

∫ x

0

ηk1(t)e−σ(x−t)dt+

∫ ∞

0

ηk2(t)e−σ(x+t)dt+ ηe−σx.

12

Therefore, (hn1 , hn2 , h

n3 ) converges pointwise to (h1, h2, h3) as n→ ∞. Since hn−1

i (t)fi(x+t) ≤ηki(t)e

−σ(x+t) and ηki(t)e−σt is integrable on (0,∞) (i = 1, 2), by the Dominated Convergence

Theorem, (h1, h2, h3) is a continuous solution of (3.17).

Remark 3.3 Here we only assume that λ1, λ2, µ2 are bounded. If they are continuous on

[0,∞), then (h1, h2, h3) ∈ C1[0,∞).

Theorem 3.4 Suppose that the hypothesis (H) holds. Then the solution of system (3.17) is

unique.

Proof. Suppose that (h1, h2, h3) and (h1, h2, h3) are two solutions of (3.17). Letting wi(x) =

|hi(x)− hi(x)| (i = 1, 2, 3), (w1, w2, w3) satisfies the following:

w1(x) ≤∫ x

0

ηw3(t)e−σ(x−t)dt,

w2(x) ≤∫ x

0

ηw2(t)e−σ(x−t)dt+

∫ ∞

0

ηw1(t)e−σ(x+t)dt,

w3(x) ≤∫ x

0

ηw1(t)e−σ(x−t)dt+

∫ ∞

0

ηw2(t)e−σ(x+t)dt.

(3.19)

For simplicity, we again use (3.19.i) for the ith inequality in (3.19). Let J(x) = w2(x)eσx

and M = η

∫ ∞

0

w1(t)e−σtdt. Rewriting (3.19.2) we have

J(x) ≤∫ x

0

J(t)dt +M,

which by the Gronwall’s inequality yields

J(x) ≤Meηx, i.e., w2(x) ≤Me(η−σ)x. (3.20)

Now using (3.19.3), together with (3.20) we obtain

w3(x) ≤∫ x

0

ηw1(t)e−σ(x−t)dt+ ηM

∫ ∞

0

e−σxe−(2σ−η)tdt

=

∫ x

0

ηw1(t)e−σ(x−t)dt+ ηMe−σx/(2σ − η).

(3.21)

Substituting (3.21) into (3.19.1) we find

w1(x) ≤∫ x

0

∫ t

0

η2w1(s)e−σ(t−s)e−σ(x−t)dsdt+ (η2M/(2σ − η))

∫ x

0

e−σxdt

=

∫ x

0

∫ x

s

η2w1(s)e−σ(x−s)dtds+ (η2M/(2σ − η))xe−σx

= η2∫ x

0

w1(s)e−σ(x−s)(x− s)ds+ (η2M/(2σ − η))xe−σx.

(3.22)

13

Set K(x) = η2∫ x

0

w1(s)eσs(x− s)ds+ (η2M/(2σ − η))x. Then K satisfies

K(0) = 0, K ′(x) = η2∫ x

0

w1(s)eσsds+ η2M/(2σ − η), K ′(0) = η2M/(2σ − η),

and

K ′′(x) = η2w1(x)eσx ≤ η2K(x). (3.23)

Multiplying (3.23) by K ′ and integrating the resulting inequality from 0 to x, we find

(K ′(x))2 − (K ′(0))2 ≤ η2K2(x),

or equivalently,

K ′(x) ≤ η√

γ2 +K2(x), (3.24)

where γ = ηM/(2σ − η). From (3.24) it follows that∫ K(x)

0

(γ2 + τ 2)−1/2dτ ≤∫ x

0

ηdt,

which gives

log((

K(x) +√

γ2 +K2(x))

/γ)

≤ ηx. (3.25)

Rewriting (3.25) we have

K(x) +√

γ2 +K2(x) ≤ γeηx,

that is,√

γ2 +K2(x) ≤ γeηx −K(x). (3.26)

The above inequality is equivalent to 2K(x)eηx ≤ γ(

e2ηx − 1)

. Thus we find

K(x) ≤ (γ/2)(

eηx − e−ηx)

. (3.27)

A combination of (3.23) and (3.27) then yields

w1(x) ≤ (γ/2)(

e(η−σ)x − e−(η+σ)x)

.

Hence we have

M =

∫ ∞

0

ηw1(x)e−σxdx

≤ (ηγ/2)

∫ ∞

0

(

e−(2σ−η)x − e−(2σ+η))

dx

= η2γ/(4σ2 − η2).

Since γ = ηM/(2σ − η), by (H) we further have

M ≤ (η3/(2σ − η)2(2σ + η))M < M,

which implies that M ≡ 0.

Since w1(x) ≥ 0 and w1 ∈ C[0,∞), it follows that w1(x) ≡ 0. Meanwhile, by virtue of

(3.20), w2(x) ≡ 0. Moreover, by means of (3.19.3), w3(x) ≡ 0. The proof is completed.

14

References

[1] A. S. Ackleh, K. Deng, and W. Li, Solvability of a non-local hyperbolic model arising

from a reliability system, IMA J. Appl. Math. 68 (2003), 135-148.

[2] C. Corduneanu, Integral Equations and Applications, Cambridge University Press, New

York, 1991.

[3] Y. Lin, Analytical and numerical solutions for a class of nonlocal nonlinear parabolic

differential equations, SIAM J. Math. Anal. 25 (1994), 1577-1594.

[4] D. Shi and W. Li, Availability analysis of a two-unit series system with shut-off rule

and “first-fail, first-repaired”, Acta Math. Appl. Sinica 9 (1993), 88-91.

[5] D. Shi and L. Liu, Availability analysis of a two-unit series system with a priority

shut-off rule, Naval Res. Logis. 43 (1996), 1009-1024.

[6] G. F. Webb, Theory of Nonlinear Age-Dependent Population Dynamics, Marcel Dekker,

New York, 1985

15

4 Asymptotic behavior for a reaction-diffusion popu-

lation model with delay

In recent years intensive studies have been conducted on reaction-diffusion population mod-

els with time delay. Among them, one kind of studies focuses on the linear stability and

existence of traveling wavefronts for models on unbounded domains (see, e.g. [1, 9]. Another

kind of studies focuses on the asymptotic behavior for models on bounded domains with ho-

mogeneous Neumann boundary conditions (see, e.g. [4, 6, 8]). Here, we study the following

reaction-diffusion population model with time delay [3]

ut −∆u = u

[

f(u)− α

∫ t

−∞

∫

Rn

g(x− y, t− s)u(y, s)dyds

]

(x, t) ∈ Rn × (0,∞),

u(x, t) = ψ(x, t) (x, t) ∈ Rn × (−∞, 0],

(4.1)

where α is a positive constant, g, ψ are nonnegative continuous functions, and f is a contin-

uously differentiable function on [0,∞).

The motivation of our study comes from a single biological population model developed

in [1], wherein f(u) = 1+ au− bu2 and α = 1+ a− b. The term au represents an advantage

in local aggregation, the term −bu2 mitigates local crowding, and the convolution term

represents a disadvantage in global population levels.

Our main objective here is to investigate the asymptotic behavior for the model (4.1). To

this end, it is worth mentioning that in [5], convergence to the positive equilibrium is shown

for a reaction-diffusion population model on an infinite one-dimensional spatial domain. The

authors there first prove a comparison principle and then establish nonlinear stability of the

positive uniform equilibrium state. There are two main differences between the model in

[5] and our model (4.1): 1) The spatio-temporal term in [5] is positive while ours does not

possess such positivity. 2) The kernel in [5] decays like e−(|x|2+|t|) while ours is a general

L1 function. To overcome the difficulties due to these differences, we first introduce the

definition of coupled upper and lower solutions. We then are able to establish a comparison

principle and thus construct successive improved monotone sequences of upper and lower

solutions which yield the convergence to the uniform positive equilibrium of the model (4.1).

Throughout our discussion, we impose the following hypotheses.

(H1) The kernel g(x, t) is a continuous, nonnegative function on Rn × [0,∞) with g ∈

L1(Rn × [0,∞)). Moreover,

∫ ∞

0

∫

Rn

g(x− y, s)dyds = 1.

(H2) ψ(x, t) is a bounded continuous, nonnegative function on Rn × (−∞, 0].

For simplicity, we let DT = Rn × (0, T ) and Γ0 = R

n × (−∞, 0], and sometimes we may

use g ∗ u(x, t) to denote

∫ t

−∞

∫

Rn

g(x− y, t− s)u(y, s)dyds for (x, t) ∈ DT .

16

We then introduce the definition of coupled upper and lower solutions of problem (4.1)

as follows.

Definition 4.1 A pair of bounded continuous functions u(x, t) and u(x, t) are called an

upper solution and a lower solution of (4.1) in DT , respectively, if they satisfy the following

conditions.

(i) u, u ∈ C2,1(DT ).

(ii) u(x, t) ≥ ψ(x, t) ≥ u(x, t) on Γ0.

(iii) For any (x, t) ∈ DT ,

ut −∆u ≥ u (f(u)− αg ∗ u) , (4.2)

ut −∆u ≤ u (f(u)− αg ∗ u) . (4.3)

In [7] a comparison principle was established for problems on bounded domains, and it

relies on the fact that a continuous function attains its extremum on compact sets. On the

other hand, for problems on unbounded domains, in [2] a comparison principle was proved

by assuming the initial data integrable. Furthermore, in [5] a comparison principle was

shown by introducing a transformation, which requires the kernel g decaying like e−(|x|2+|t|).

Because our kernel g is a general L1 function, we establish the following comparison result

by constructing a different auxiliary function.

Theorem 4.2 Suppose that (H1)-(H2) hold and f is a continuously differentiable function

on [0,∞). Let u and u be a pair of nonnegative upper and lower solutions of (4.1) in DT ,

respectively. Then u ≥ u in DT .

Proof. Letting w = u− u, we show that w ≤ 0 in DT . For any (x, t) ∈ DT , w satisfies

wt −∆w ≤ uf(u)− uf(u)− αug ∗ u+ αug ∗ u= (θf ′(θ) + f(θ))w − αwg ∗ u+ αug ∗ w= c(x, t)w + αug ∗ w,

where θ is between u and u and c(x, t) = θf ′(θ) + f(θ)− αg ∗ u. Thus, we find

wt −∆w ≤ c(x, t)w + αug ∗ w in DT ,

w(x, t) ≤ 0 on Γ0.(4.4)

By Definition 4.1, u and u are nonnegative and bounded. So there exists M > 0 such that

0 ≤ u(x, t) ≤M and 0 ≤ u(x, t) ≤M for (x, t) ∈ DT ∪ Γ0. Consequently, 0 ≤ θ ≤M . Since

17

f ∈ C1[0,∞), we can choose σ > 0 such that σ > max0≤η≤M

(ηf(η) + f ′(η)). Let u = e−σtw. By

(4.4), we have

ut −∆u+ (σ − c)u ≤ αe−σtu

∫ t

−∞

∫

Rn

g(x− y, t− s)eσsu(y, s)dyds in DT , (4.5)

u(x, t) ≤ 0 on Γ0. (4.6)

It suffices to prove that u ≤ 0 in DT . To this end, we let v =u

1 + |x|2 + γ|t| , where γ is a

positive constant to be determined. By (4.5), we find

(1 + |x|2 + γt)(vt −∆v + (σ − c)v)− 4x · ∇v + (γ − 2n)v

≤ αue−σt

∫ t

−∞

∫

Rn

g(x− y, t− s)eσsu(y, s)dyds in DT .(4.7)

Assume to the contrary that u > 0 somewhere in DT . Then usup = sup(x,t)∈DT∪Γ0

u(x, t) is a

positive number because u is bounded. Since u ≤ 0 on Γ0, there exists (x∗, t∗) ∈ DT such

that u(x∗, t∗) ≥ usup2

. On the other hand, because v(x, t) ≤ 0 on Γ0 and lim|x|→∞

v(x, t) = 0, v

attains its positive maximum vmax at (x, t) in DT . Thus we have

vmax = max(x,t)∈DT

u

1 + |x|2 + γt≥ u(x∗, t∗)

1 + |x∗|2 + γt∗≥ usup

2(1 + |x∗|2 + γt∗). (4.8)

Since vmax is positive, it follows that

usupvmax

≤ 2(1 + |x∗|2 + γt∗). (4.9)

Moreover, at the point (x, t), vt ≥ 0, ∇v = 0 and ∆v ≤ 0. Hence, by (4.7) and (H1) we have

that(σ − c)(1 + |x|2 + γt)vmax + (γ − 2n)vmax

≤ αu(x, t)e−σt

∫ t

−∞

∫

Rn

g(x− y, t− s)eσsu(y, s)dyds

≤ αu(x, t)usup

∫ t

−∞

∫

Rn

g(x− y, t− s)e−σ(t−s)dyds

≤ αu(x, t)usup.

Since vmax > 0 and σ − c ≥ 0, by (4.9) we further have that

γ − 2n ≤ αu(x, t)usupvmax

≤ 2αu(x, t)(1 + |x∗|2 + γt∗)

≤ 2αM(1 + |x∗|2 + γT ).

(4.10)

Since x∗ is independent of γ, if T < 1/2αM , we can choose γ large enough so that (4.10)

yields a contradiction, which means that u ≤ 0 in DT . If T ≥ 1/2αM , we can first prove that

18

u(x, t) ≤ 0 for (x, t) ∈ Rn × [0, 1/2αM) and then prove it for (x, t) ∈ R

n × [1/2αM, 1/αM).

Continuing this process, we can then prove that u(x, t) ≤ 0 in DT .

We now establish the existence and uniqueness results for problem (4.1). We first present

the uniqueness result.

Theorem 4.3 Suppose that (H1)-(H2) hold and f is continuously differentiable on [0,∞).

Then problem (4.1) has at most one bounded solution.

Proof. Suppose that u1 and u2 are two bounded solutions of (4.1) in DT . Then each solution

ui (i = 1, 2) satisfies

ui(x, t) =

∫

Rn

Φ(x− y, t)ψ(y, 0)dy

+

∫ t

0

∫

Rn

Φ(x− y, t− s)ui(y, s)[f(ui(y, s))− αg ∗ ui(y, s)]dyds for t ∈ [0, T ],

where g∗ui(y, s) =∫ s

−∞

∫

Rn

g(y−z, s−τ)ui(z, τ)dzdτ and Φ(x, t) is the fundamental solution

of the heat equation. Let w = u1 − u2, then w satisfies

w(x, t) =

∫ t

0

∫

Rn

Φ(x− y, t− s)[w(y, s)(f(θ) + θf ′(θ)− αg ∗ u2(y, s))

− αu1(y, s)g ∗ w(y, s)]dyds,where θ is between u1 and u2. Since u1 and u2 are bounded, there exists M > 0 such

that |u1| < M and |u2| < M . Moreover, in view of (H1), there exists M1 > 0 such that

|f(θ) + θf ′(θ)− αg ∗ u2| ≤M1. Thus, taking

∫

Rn

Φ(x, t)dx = 1 into account, we have that

‖w(·, t)‖∞ ≤∫ t

0

‖Φ(·, t− s)‖1(M1 ‖w(·, s)‖∞ + αM ‖g ∗ w(·, s)‖∞)ds

≤M1

∫ t

0

‖w(·, s)‖∞ds+ αM

∫ t

0

max0≤τ≤s

‖w(·, τ)‖∞ ds

≤M1

∫ t

0

max0≤τ≤s

‖w(·, τ)‖∞ ds+ αM

∫ t

0

max0≤τ≤s

‖w(·, τ)‖∞ ds

for all t ∈ (0, T ]. Hence, we find

max0≤s≤t

‖w(·, s)‖∞ ≤ (M1 + αM)

∫ t

0

max0≤τ≤s

‖w(·, τ)‖∞ ds.

Applying Gronwall’s inequality, we then obtain

max0≤s≤t

‖w(·, s)‖∞ = 0 for t ∈ (0, T ].

Since w is continuous, w = 0, i.e., u1 = u2 in DT .

We now establish the existence result. To this end, we construct two monotone sequences

of lower and upper solutions and show their convergence to the unique solution of (4.1).

19

Theorem 4.4 Suppose that (H1)-(H2) hold and f is continuously differentiable on [0,∞).

If u(x, t) and u(x, t) are a pair of nonnegative lower and upper solutions of problem (4.1) in

DT , then there exists a unique solution u(x, t) of (4.1) in DT , and u satisfies

u(x, t) ≤ u(x, t) ≤ u(x, t), for (x, t) ∈ DT ∪ Γ0. (4.11)

Proof. By Definition 4.1, we can choose a positive constant M such that 0 ≤ u(x, t) ≤ M

and 0 < u(x, t) ≤ M for (x, t) ∈ DT ∪ Γ0. Since f is continuously differentiable, we can

further choose L > 0 such that ηf(η) + f ′(η) + L > 0 for any η ∈ [0,M ]. Let v0 = u and

w0 = u. We then set up two sequences vk∞k=0 and wk∞k=0 in DT ∪ Γ0 by the following

procedure: For k = 1, 2, . . . , let vk and wk satisfy the system

vkt −∆vk = vk−1f(vk−1)− αvkg ∗ wk−1 − L(vk − vk−1) in DT ,

vk(x, t) = ψ(x, t) on Γ0

(4.12)

and

wkt −∆wk = wk−1f(wk−1)− αwkg ∗ vk−1 − L(wk − wk−1) in DT ,

wk(x, t) = ψ(x, t) on Γ0.(4.13)

The existence of solutions to system (4.12)-(4.13) follows from the fact that (4.12)-(4.13) is

a linear system. We first show that v0 ≤ v1 ≤ w1 ≤ w0 in DT ∪ Γ0. Let v = v0 − v1. Then

v satisfies

vt −∆v ≤ −(αg ∗ w0 + L)v in DT ,

v(x, t) ≤ 0 on Γ0.

Thus, by the comparison principle for parabolic equations, v(x, t) ≤ 0 for (x, t) ∈ DT , and

so v0 ≤ v1 in DT ∪ Γ0. Similarly, one can see that w1 ≤ w0 in DT ∪ Γ0.

We then show that v1 ≤ w1 in DT ∪ Γ0. Indeed, letting u = v1 − w1, u satisfies

ut −∆u = −αv1g ∗ (w0 − v0)− (αg ∗ v0 + L)u+ (v0 − w0)(θf ′(θ) + f(θ) + L)

≤ −(αg ∗ v0 + L)u in DT ,

u(x, t) ≤ 0 on Γ0,

where θ is between v0 and w0.

Furthermore, we claim that v1 and w1 are a pair of lower and upper solutions of problem

(4.1). To this end, we note that since v0 ≤ v1 and w1 ≤ w0, v1 satisfies

v1t −∆v1 = v0f(v0)− αv1g ∗ w0 − L(v1 − v0)

≤ v1f(v1)− αv1g ∗ w1 + (v0 − v1)(θf ′(θ) + f(θ) + L)

≤ v1f(v1)− αv1g ∗ w1 in DT ,

20

where θ is between v0 and v1. On the other hand, w1 satisfies

w1t −∆w1 = w0f(w0)− αw1g ∗ v0 − L(w1 − w0)

≥ w1f(w1)− αw1g ∗ v1 + (w0 − w1)(θf ′(θ) + f(θ) + L)

≥ v1f(v1)− αv1g ∗ w1 in DT ,

where θ is between w0 and w1.

We now assume that for some k > 1, vk and wk are a pair of lower and upper solutions to

(4.1), respectively. Continuing the above process, we can prove that vk ≤ vk+1 ≤ wk+1 ≤ wk.

Thus by induction, we obtain two monotone sequences satisfying

v0 ≤ v1 ≤ · · · ≤ vk ≤ wk ≤ · · · ≤ w1 ≤ w0 in DT ∪ Γ0

for k = 0, 1, 2, . . . . It follows from the monotonicity of these sequences that there exist

functions v and w such that vk → v and wk → w pointwise in DT . Clearly, v ≤ w. On the

other hand, v and w can also be viewed as a pair of upper and lower solutions of (4.1), and

thus by Theorem 4.2 w ≤ v. Hence, we have v = w. Defining this common limit by u and

noticing Theorem 4.3, we find that u is the unique bounded solution of (4.1) in DT .

With an additional condition on f , we have the following global existence result.

Theorem 4.5 Suppose that (H1)-(H2) hold and f is continuously differentiable on [0,∞).

Furthermore, there exists r > 0 such that f(r) = 0 and f(η) < 0 for η > r. Then the

solution u of problem (4.1) is global and bounded by a positive constant K.

Proof. Choose K > 0 such that K > max‖ψ‖∞, r. Then, u = 0 and u = K are a pair

of lower and upper solutions of (4.1) in DT . By Theorem 4.4, problem (4.1) has a unique

solution u(x, t) with 0 ≤ u(x, t) ≤ K in DT . Since K does not depend on T , u is global.

We then investigate the asymptotic behavior for the model (4.1). For this purpose, we

first impose the following hypotheses on the function f .

(H3) There exists h ≥ 0 such that f ∈ C1[0,∞) is increasing on [0, h) and strictly decreasing

on [h,∞), and there exists r > h such that f(r) = 0.

(H4) f(0) > αr.

(H5) If 0 ≤ λ, µ ≤ r, f(λ) = αµ and f(µ) = αλ, then λ = µ.

Remark 4.6 Biologically, (H3) for h > 0 means the introduction of aggregation, that is, the

species cooperates in a small population while competes in a large population; (H4) means

that the natural growth rate of the population is large enough.

Remark 4.7 By (H3) and (H4), we have that f(h) ≥ f(0) > αr > αh and f(r) = 0 < αr.

Thus, from the monotonicity of f there exists a unique d ∈ (h, r) such that f(d) = αd. Note

that d is the positive uniform equilibrium of (4.1).

21

Remark 4.8 Let F (η, ζ) = f(η)−αζ for (η, ζ) ∈ [h,∞)× [0,∞). By (H3) and (H4), there

exists r′(r < r′ ≤ K) such that f(h) ≥ f(0) > αr′. Then for any ζ ∈ [0, r′]

F (h, ζ) = f(h)− αζ > αr′ − αζ ≥ 0 and F (r, ζ) = f(r)− αζ = −αζ ≤ 0.

Thus, by the continuity of F , there exists ϕ(ζ) ∈ (h, r] such that F (ϕ(ζ), ζ) = 0 for any

ζ ∈ [0, r′]. Furthermore, since f is strictly decreasing on [h,∞), ϕ is strictly decreasing

on [0, r′]. And Fη(η, ζ) = f ′(η) < 0 on [h,∞) implies the continuity of ϕ on [0, r′] by the

implicit function theorem.

Remark 4.9 If 0 < λ, µ ≤ r, ϕ(λ) = µ, and ϕ(µ) = λ, from the relation that F (ϕ(ζ), ζ) = 0

for any ζ ∈ [0, r′], it follows that F (µ, λ) = 0 and F (λ, µ) = 0. Then (H5) implies λ = µ.

The following two lemmas will be used in the sequel.

Lemma 4.10 Suppose that all the hypotheses in Theorem 4.5 hold. If ψ(x, 0) ≥ δ > 0 for

all x ∈ Rn, then there exists a positive constant κ such that the solution u(x, t) of problem

(4.1) satisfies u(x, t) ≥ δe−κt for all (x, t) ∈ Rn × [0,∞).

Proof. As in the proof of Theorem 4.5, for any T > 0 let v0 = 0 and w0 = K be a pair of

lower and upper solutions of (4.1) in DT . Consider v1 and w1 defined by (4.12) and (4.13),

respectively. Then, we have that v1 ≤ u ≤ w1 in DT and v1 satisfies

v1t −∆v1 = v0f(v0)− αv1g ∗ w0 − L(v1 − v0) = −(αK + L)v1 in DT ,

v1(x, 0) = ψ(x, 0) on Rn.

Thus,

u(x, t) ≥ v1(x, t) = e−κt

∫

Rn

Φ(x− y, t)ψ(y, 0)dy ≥ δe−κt

for all (x, t) ∈ DT with κ = αK + L. Since κ is independent of T , u(x, t) ≥ δe−κt for all

(x, t) ∈ Rn × [0,∞).

Lemma 4.11 Suppose that (H3)-(H5) hold. For any fixed ζ∗ ∈ [0, r′], η∗ ≡ ϕ(ζ∗) is a

globally asymptotically stable equilibrium to

η = η(f(η)− αζ∗)

on (0,∞).

We now establish the global asymptotic behavior result.

Theorem 4.12 Suppose that (H1)-(H5) hold. If ψ(x, 0) ≥ δ on Rn for some δ > 0, then

the solution u(x, t) of problem (4.1) satisfies limt→∞u(x, t) = u∗ uniformly in x ∈ Rn, where

u∗ = d is the positive uniform steady-state solution of (4.1).

22

Proof. Inspired by [4, 6], We use the method of successive improvement of upper and lower

solutions. To this end, we define

u(t) = infx∈Rn

u(x, t) and u(t) = supx∈Rn

u(x, t) for t > 0,

and I = [lim inft→∞

u(t), lim supt→∞

u(t)]. Then it suffices to show that I = d. As in the proof

of Theorem 4.5, v0 = 0 and w0 = K are a pair of lower and upper solutions of (4.1) in

Rn × (0,∞), respectively. Then 0 ≤ u ≤ K and I ⊆ [0, K].

Let v1 = 0 and w1 satisfy

w1 = w1f(w1), t > 0,

w1(0) = max‖ψ‖∞, h,

and w1 = max‖ψ‖∞, h on Γ0. One can easily see that v1 and w1 are a pair of lower and

upper solutions of (4.1) on Rn × (0,∞), respectively. By Lemma 4.11, w1 → ϕ(0) = r as

t→ ∞. Thus, I ⊆ [0, r].

Let ε > 0 be given such that r+(1+K)ε < r′. Since lim supt→∞

u(t) ≤ r, there exists t1 > 1

such that

u(x, t) ≤ u(t) ≤ r + ε for x ∈ Rn, t ≥ t1 − 1.

In view of (H1), there exists t2 > t1 such that

∫ ∞

t−t1

∫

Rn

g(x− y, s)dyds ≤ ε for x ∈ Rn, t ≥ t2.

Define

δ1 =1

2minu(t) : t2/2 ≤ t ≤ t2.

By Lemma 4.10, δ1 > 0. Let v2 be given by

v2(t) =

0 t ≤ 1

2t2,

δ1t2(2t− t2)

1

2t2 < t ≤ t2,

(4.14)

v2 = v2(f(v2)− αεK − α(r + ε)) for t > t2.

And let w2 be given by

w2(t) =

K t ≤ t1 − 1,r + ε+ (K − r − ε)(t1 − t) t1 − 1 < t ≤ t1,r + ε t > t1.

(4.15)

23

Clearly, v2 ≤ u ≤ w2 on Rn × (−∞, t2]. For t > t2,

g ∗ w2(x, t) =

∫ t

−∞

∫

Rn

g(x− y, t− s)w2(y, s)dyds

=

∫ t1

−∞

∫

Rn

g(x− y, t− s)w2(y, s)dyds+

∫ t

t1

∫

Rn

g(x− y, t− s)w2(y, s)dyds

≤ K

∫ t1

−∞

∫

Rn

g(x− y, t− s)dyds+ (r + ε)

∫ t

t1

∫

Rn

g(x− y, t− s)dyds

≤ εK + (r + ε).

Thus, we have

v2 −∆v2 ≤ v2(f(v2)− αg ∗ w2) for t > t2.

On the other hand, since f(r + ε) < 0, we have

w2 −∆w2 ≥ w2(f(w2)− αg ∗ v2) for t > t2.

Therefore, v2 and w2 are a pair of lower and upper solutions of (4.1) on Rn× (t2,∞), respec-

tively, and hence by the comparison principle, v2 ≤ u ≤ w2 on Rn × [t2,∞). Furthermore,

by Lemma 4.11, v2 → ϕ(r + (1 + K)ε) as t → ∞. Thus, I ⊆ [ϕ(r + (1 +K)ε), r + ε]. By

the continuity of ϕ, letting ε → 0, we have I ⊆ [ϕ(r), r].

We then choose λ3 with h < λ3 < ϕ(r) and µ3 = r, and define two sequences λk∞k=3

and µk∞k=3 by

λk = ϕ(µk−1), µk = ϕ(λk−1), k > 3.

We have that 0 < λ3 < λ4 < µ3 by the definition. Since ϕ(µ3) = λ4, ϕ(d) = d < µ3,

and ϕ is strictly decreasing, we find λ4 < d. On the other hand, since 0 < λ3 < d,

ϕ(d) < ϕ(λ3) < ϕ(0), i.e., d < µ4 < µ3. Thus, we have that 0 < λ3 < λ4 < d < µ4 < µ3.

Continuing the above process, we obtain two monotone sequences that satisfy

0 < λ3 < λ4 < · · · < λk < d < µk < · · · < µ4 < µ3.

We claim that I ⊆ [λk, µk] for all k ≥ 3. Clearly, I ⊆ [ϕ(r), r] ⊆ [λ3, µ3]. Assuming that

I ⊆ [λk−1, µk−1] for some k > 3, we show I ⊆ [λk, µk].

Since λk−1 < λk < µk < µk−1, by the continuity of ϕ, we can choose ε so small that

λk−1 ≤ ϕ(µk−1 + (1 +K)ε) ≤ ϕ((1− ε)(λk−1 − ε)) ≤ µk−1. (4.16)

Since I ⊆ [λk−1, µk−1], there exists tk > 1 such that

λk−1 − ε < u(t) ≤ u(t) < µk−1 + ε for t ≥ tk − 1. (4.17)

Meanwhile, in view of (H1), there exists tk+1 > tk such that∫ ∞

t−tk

∫

Rn

g(x− y, s)dyds ≤ ε for x ∈ Rn, t ≥ tk+1.

24

Let vk be given by

vk(t) =

0 t ≤ tk − 1,(λk−1 − ε)(t− tk + 1) tk − 1 < t ≤ tk,λk−1 − ε tk < t ≤ tk+1,

(4.18)

vk = vk(f(vk)− αεK − α(µk−1 + ε)) for t > tk+1.

And let wk be given by

wk(t) =

K t ≤ tk − 1,(µk−1 + ε−K)(t− tk) + µk−1 + ε tk − 1 < t ≤ tk,µk−1 + ε tk < t ≤ tk+1,

(4.19)

wk = wk(f(wk)− α(1− ε)(λk−1 − ε)) for t > tk+1.

Then, vk ≤ u ≤ wk on Rn × (−∞, tk+1]. Moreover, noticing Remark 4.8 and (4.16), we have

that vk(t) ≥ λk−1 − ε and wk(t) ≤ µk−1 + ε for all t ≥ tk. Thus for t > tk+1,

g ∗ wk(x, t) =

∫ t

−∞

∫

Rn

g(x− y, t− s)wk(y, s)dyds

=

∫ tk

−∞

∫

Rn

g(x− y, t− s)wk(y, s)dyds+

∫ t

tk

∫

Rn

g(x− y, t− s)wk(y, s)dyds

≤ K

∫ tk

−∞

∫

Rn

g(x− y, t− s)dyds+ (µk−1 + ε)

∫ t

tk

∫

Rn

g(x− y, t− s)dyds

≤ εK + (µk−1 + ε)

and

g ∗ vk(x, t) =∫ t

−∞

∫

Rn

g(x− y, t− s)vk(y, s)dyds

=

∫ tk

−∞

∫

Rn

g(x− y, t− s)vk(y, s)dyds+

∫ t

tk

∫

Rn

g(x− y, t− s)vk(y, s)dyds

≥ (λk−1 − ε)

∫ t

tk

∫

Rn

g(x− y, t− s)dyds

≥ (1− ε)(λk−1 − ε).

From the above inequalities, we find that for t > tk+1,

vk −∆vk ≤ vk(f(vk)− αg ∗ wk),

wk −∆wk ≥ wk(f(wk)− αg ∗ vk).

Hence, vk and wk are a pair of lower and upper solutions of (4.1) in Rn × (tk+1,∞), respec-

tively. By the comparison principle,

vk(t) ≤ u(x, t) ≤ wk(t) for (x, t) ∈ Rn × [tk+1,∞). (4.20)

25

Furthermore, by Lemma 4.11, vk → ϕ(µk−1 + (1 +K)ε) and wk → ϕ((1 − ε)(λk−1 − ε)) as

t→ ∞, which implies that I ⊆ [ϕ((µk−1+(1+K)ε)), ϕ((1−ε)(λk−1−ε)]. By the continuity

of ϕ, letting ε→ 0, we obtain

I ⊆ [ϕ(µk−1), ϕ(λk−1)] = [λk, µk].

Therefore, the claim is valid by induction.

Since the sequences λk∞k=3 and µk∞k=3 are monotone and bounded, there exist λ, µ ∈(h, r) such that λk → λ and µk → µ as k → ∞. Then we have

I ⊆ [λk, µk] ⊆ [λ, µ].

Since ϕ(λk−1) = µk and ϕ(µk−1) = λk, by the continuity of ϕ, we further have that ϕ(λ) = µ

and ϕ(µ) = λ. Taking Remark 4.9 into account, we find λ = µ = d. Thus, I = d, and the

proof is completed.

Finally, we apply Theorem 4.12 to several population models.

Example 4.13 We first consider

ut −∆u = u

[

a− bu− α

∫ t

−∞

∫

Rn

g(x− y, t− s)u(y, s)dyds

]

in Rn × (0,∞),

u(x, t) = ψ(x, t) on Γ0.

(4.21)

The equation in (4.21) has been studied on a bounded domain in [6, 8] with g ∗ u being a

temporal convolution, and a global convergence result was established when b > α. Clearly,

hypotheses (H3)-(H5) hold for b > α, and so we conclude that the solution u(x, t) of (4.21)

converges to u∗ =a

α + buniformly in x provided ψ(x, 0) ≥ δ for some δ > 0 when b > α.

Example 4.14 We then consider

ut −∆u = u

[

a− bu

m+ u− α

∫ t

−∞

∫

Rn

g(x− y, t− s)u(y, s)dyds

]

in Rn × (0,∞),

u(x, t) = ψ(x, t) on Γ0.

(4.22)

The equation in (4.22) on a bounded domain has been studied in [6]. The function f is

strictly decreasing and has a unique zero r =am

b− aif b > a. In order for f to satisfy

(H4), we need a > αr which gives b > αm + a. To verify (H5), positive λ, µ must satisfy

am + (a − b)λ = α(m+ λ)µ and am + (a− b)µ = α(m+ µ)λ. A combination of these two

equations yields (a− b+αm)(λ−µ) = 0, which implies λ = µ since a− b+αm < 0. Hence,

(H3)-(H5) hold if b > αm+ a, the same condition as in [6].

26

Example 4.15 We now consider the model developed in [1].

ut −∆u = u

[

1 + au− bu2 − α

∫ t

−∞

∫

Rn

g(x− y, t− s)u(y, s)dyds

]

in Rn × (0,∞),

u(x, t) = ψ(x, t) on Γ0.

(4.23)

The function f satisfies (H3) with h =a

2b, and it has the zero r =

a+√a2 + 4b

2b. The

condition f(0) > αr implies 1 > (1+ a− b)r, which also ensures the validity of (H5). Hence

by our theorem, the solution of (4.23) converges to u∗ = 1 uniformly in x if ψ(x, 0) ≥ δ for

some δ > 0.

References

[1] N.F. Britton, Spatial structures and periodic travelling waves in an integro-differential

reaction-diffusion population model, SIAM J. Appl. Math. 50 (1990), 1663-1688.

[2] K. Deng, On a nonlocal reaction-diffusion population model, Discrete Contin. Dyn.

Syst. Ser. B 9 (2008), 65-73.

[3] K. Deng and Y. Wu, Asymptotic behavior for a reaction-diffusion population model

with delay, Discrete Contin. Dyn. Syst. Ser. B 20 (2015), 385-395.

[4] S.A. Gourley and N.F. Briton, On a modified Volterra population equation with diffu-

sion, Nonlinear Anal. 21 (1993), 389-395.

[5] Y. Kyrychko, S.A. Gourley, and M.V. Bartuccelli, Comparison and convergence to

equlibrium in a nonlocal delayed reaction-diffusion model on an infinite domain, Discrete

Contin. Dyn. Syst. Ser. B 5 (2005), 1015-1028.

[6] R. Laister, Global asymptotic behavior in some functional parabolic equations, Nonlin-

ear Anal. 50 (2002), 347-361.

[7] R. Redlinger, Existence theorems for semilinear parabolic systems with functionals,

Nonlinear Anal. 8 (1984), 667-682.

[8] R. Redlinger, On Volterra’s population with diffusion, SIAM J. Math. Anal. 16 (1985),

135-142.

[9] Z-C. Wang, W-T. Li, and S. Ruan, Existence and stability of traveling wave fronts

in reaction advection diffusion equations with nonlocal delay, J. Differential Equations

238 (2007), 153-200.

27

5 Global solvability for the porous medium equation

with boundary flux governed by nonlinear memory

We initiate a study of global solvability for the nonlinear diffusion model [2]

ut = ∆(um) on ΩT

∇ (um) · n = 0 on (∂Ω \ Σ)T∇ (um) · n = uqv on ΣT

u = u0 on Ω× 0,(5.1)

where

vt = up on ΣT

v = 0 on Σ× 0 (5.2)

Here, m, p > 0, q ≥ 0, and ΩT = Ω × (0, T ), where Ω is a bounded domain in RN having

piecewise smooth boundary ∂Ω with outward pointing unit normal n. Σ is a relatively

open subset of ∂Ω. The initial condition u0 is a nonnegative, continuous function on Ω.

We establish local existence, continuation, and subsolution comparison for weak solutions

of (5.1)-(5.2). Our main result is that for m > 1, every solution of (5.1)-(5.2) is global if

0 < p+ q ≤ 1, whereas if p+ q > 1 and u0 > 0, then all maximal solutions blow up in finite

time. The analysis herein also applies in the case of the heat equation, m = 1. However,

this has already been addressed in a previous paper [3].

Motivation for the study of (5.1)-(5.2) comes from both its identification as a subset of

previously introduced models of tumor-induced angiogenesis and presenting a memory-type

flux condition having had limited treatment in the literature. Formally integrating (5.2),

one obtains the boundary condition

∇(um) · n = uq∫ t

0

up(·, s)ds on ΣT

Our result may thus be viewed in analogy to the localized version of (5.2), namely∇(um)·n =

uq+p. As such, we extend results by Wolanski for the localized model [9].

As part of a model for the growth of new capillary networks (“angiogenesis”), initiated

by a developing solid tumor, Levine et. al [5] introduced a general transmission condition,

similar to (5.2), of the form

vt = f(x, t, u, v) +G(u)t on ΣT

v = v0 on Σ× 0

In the application to angiogenesis, u and v are the concentrations of tumor-released

growth factor outside and within the wall of a nearby capillary, respectively. The boundary

condition arises in an effort to correctly represent uptake of growth factor within the capillary

wall and its transport through the wall. Additionally, Σ represents the capillary wall.

28

In the context of characterizing conditions under which these models are globally solvable,

it is helpful to consider possible comparisons to corresponding localized models. As such,

there is the aforementioned work by Wolanski [9], as well as the extension of this work to

the reaction-diffusion model ut = ∆(um) + uα with strictly positive initial data, u0, and

nonlinear flux ∇u · n = uβ at the boundary [6]. These authors have established that the

model is globally solvable if and only if α ≤ 1, β ≤ min1, (m+ 1)/2, via suitable choices

of subsolutions and supersolutions.

We first establish the local existence and comparison of solutions for a slightly more

general model

ut = ∆φ(u) on ΩT

∇φ(u) · n = 0 on (∂Ω \ Σ)T∇φ(u) · n = g(u, v) on ΣT

u = u0 ≥ 0 on Ω× 0(5.3)

where

vt = f(u) on ΣT

v = v0 ≥ 0 on Σ× 0 (5.4)

The nonlinear terms φ, g, and f are assumed to be continuous for u, v ≥ 0, differentiable

for u, v > 0, with φ′(u) > 0 for u > 0, φ(0) = f(0) = 0, and g(0, v) ≥ 0 for v ≥ 0. We

also require monotonicity gv ≥ 0 and f ′ ≥ 0. The initial condition v0 will be assumed to be

continuous for purposes of this work, although it is possible to obtain the same results for

v0 ∈ L∞(Σ).

The notion of weak solvability used herein is a standard approach obtained upon moving

all derivatives to a test function via integration by parts. In order to formulate the definition,

it is convenient to introduce the following notation.

IΩ(u, v, ξ, u0) ≡∫

Ω

uξ(x, t)dx−∫

Ω

u0ξ(x, 0)dx

−∫ t

0

∫

Ω

[uξs + φ(u)∆ξ] dxds

−∫ t

0

∫

Σ

g(u, v)ξdxds+

∫ t

0

∫

∂Ω

φ(u)∇ξ · ndxds

(IΩ)

IΣ(u, v, η, v0) ≡∫

Σ

vη(x, t)dx−∫

Σ

v0η(x, 0)dx

−∫ t

0

∫

Σ

[vηs + f(u)η(x, s)] dxds

(IΣ)

Here, ξ, η are smooth, nonnegative test functions on ΩT , ΣT .

29

Definition 5.1 u, v ≥ 0 is a weak solution (subsolution, supersolution) of (5.3)-(5.4) on ΩT

if u, v are bounded, defined everywhere on ΩT , ΣT , respectively, and IΩ, IΣ = (≤,≥) 0 for

every set of smooth test functions ξ, η ≥ 0.

Clearly, this definition simply incorporates the integral identities (inequalities) result-

ing from formal integration by parts on (5.3)-(5.4) and classical subsolution/supersolution

formulations of the model.

Theorem 5.2 For some T > 0, there exists a solution of (5.3)-(5.4) on ΩT , U, V . Further-

more, if u, v is a nonnegative subsolution of (5.3)-(5.4) on the same set, with u ≤ U v ≤ V

a.e. for t = 0, then u ≤ U , v ≤ V on ΩT , ΣT , respectively.

Either T = ∞ or ‖U(·, t)‖∞,Ω + ‖V (·, t)‖∞,Σ is unbounded for t < T . Here, ‖U(·, t)‖∞,Ω

is the L∞ norm of U(·, t) on Ω, with a similar definition for ‖V (·, t)‖∞,Σ.

Proof. Fix 0 < ǫ < 1 and define

R ≡ ‖u0‖∞,Ω + ‖v0‖∞,Σ + 1

Let C1 ≡ C1(R) ≥ f(u) for all 0 ≤ u ≤ R. Since f ′ ≥ 0, a suitable choice is C1(R) = f(R).

Additionally, define

mg(r) ≡ min

0, min0≤u≤2r

g(u, 0)

We construct a solution U, V of (5.3)-(5.4) as the limit of un, vn, where the sequences

un∞n=1, vn∞n=0 are defined as follows.

v0(x, t) ≡ v0(x) + ǫ+ C1t

Note: In a general case of v0(x) bounded but not necessarily continuous, we instead define

v0(x, t) similarly with a continuous function v0(x) having v0 ≥ v0.

With n ≥ 1 and vn−1 defined, un is the solution of

ut = ∆φ(u) on ΩT

∇φ(u) · n = 0 on (∂Ω \ Σ)T∇φ(u) · n = g(u, vn−1)−mg(ǫ/n) on ΣT

u = u0 + ǫ/n on Ω× 0(5.3ǫ,n)

Given un, vn is then defined as the solution of

vt = f(un) on ΣT

v = v0 + ǫ/n on Σ× 0 (5.4ǫ,n)

Since v0(x, t) ≥ 0, the solution of (3ǫ,1) exists on ΩT1for sufficiently small T1 > 0, with

u1 ≥ ǫ, following arguments in [1]. A necessary technicality invoked in this reference is that

30

φ and g(·, v0) be bounded, with φ′(u) ≥ θ > 0 for some constant θ and u ≥ ǫ/2. This may

be satisfied by modifying φ(u) and g(u, v) appropriately for u, v ≥ R + 1. Namely, we may

replace φ, g with φR, gR, respectively, such that φ = φR and g = gR for 0 ≤ u, v ≤ R + 1.

Furthermore, we set θ(R) ≤ φ′ for ǫ/2 ≤ u ≤ R + 1 and φ, g ≤ α(R) for u, v ≤ R + 1.

The standard maximum principle estimate, e.g., [1, Lemma 3.1], yields

ǫ ≤ u1 ≤ ‖u0‖∞,Ω + ǫ+Kt1/(N+2)

where K = K(Ω, α, θ, N, T ). Selecting T1 = T1(R) > 0 small enough, we have u1 ≤ R on

ΩT1. Hence, u1 is a solution of (5.3ǫ,1).

Upon direct integration of (5.4ǫ,1),

ǫ ≤ v1(x, t)

= v0(x) + ǫ+

∫ t

0

f(u1(x, s))ds

≤ v0(x) + ǫ+ C1(R)t

Subsequently, ǫ ≤ u1, v1 ≤ R for t < T2 = T2(R) with 0 < T2 ≤ T1. From this identity, we

also have v1(x, t) ≤ v0(x, t).

It will be useful for purposes of establishing the continuation of solutions to observe

that, without loss of generality, we can define T2(R) to be decreasing and finite for R ≥ 1.

Specifically, for 1 ≤ R1 ≤ R2, the choices θ(R1) ≡ θ(R2), α(R1) ≡ α(R2) are clearly

acceptable, giving rise to T1(R1) = T1(R2). As more optimal choices of θ(R1), α(R1) are

possible and C1(R1) ≤ C1(R2), we have T2(R1) ≥ T2(R2).

Now, with vn−1 ≥ 0 for n ≥ 1, we may similarly conclude the existence of a local solution

for (5.3ǫ,n). In order to obtain the lower estimate, un ≥ ǫ/n, note that

g(ǫ/n, vn−1) ≥ g(ǫ/n, 0) ≥ mg(ǫ/n)

Thus, ǫ/n is a classical subsolution of (5.3ǫ,n), and hence un ≥ ǫ/n.

Due to the monotonicity of g in v and the fact mg(ǫ/2) ≥ mg(ǫ), v1 ≤ v0 now implies

ǫ/2 ≤ u2 ≤ u1 ≤ R on ΩT2. The monotonicity of f , in turn, implies ǫ/2 ≤ v2 ≤ v1 ≤ R on

ΣT2. Continuing the process, we obtain

ǫ/n ≤ un ≤ un−1 ≤ · · · ≤ u1 ≤ R on ΩT2

and

ǫ/n ≤ vn ≤ vn−1 ≤ · · · ≤ v1 ≤ R on ΣT2

It is now easy to show

IΩ(un, vn−1, ξ, u0 + ǫ/n) +mg(ǫ/n)

∫ t

0

∫

Σ

ξdxds = 0

31

and IΣ(un, vn, η, v0+ǫ/n) = 0. Upon letting n→ ∞, the limiting functions un → U , vn → V

are solutions of (5.3)-(5.4) on ΩT2.

The continuation result follows from the observation that if U(·, t), V (·, t) remain bounded

for 0 ≤ t < T ∗, then the process above may be repeated with R ≥ U + V + 1, u0 ≡U(·, T ∗ − T2(R)/2) and v0 ≡ V (·, T ∗ − T2(R)/2). Since such initial conditions u0, v0 at

T ∗ − T2(R)/2 yield

R1 ≡ ‖u0‖∞,Ω + ‖v0‖∞,Σ + 1 ≤ R,

we have T2(R1) ≥ T2(R) and, thus, obtain a solution of (5.3)-(5.4) on ΩT for T ≥ T ∗ +

T2(R)/2. Therefore, if T∗ <∞ is maximal for which (5.3)-(5.4) is solvable on ΩT ∗ , the result

follows.

If u, v ≥ 0 is a subsolution of (5.3)-(5.4) on ΩT , such that u ≤ u0, v ≤ v0 at t = 0 then∫

Ω

(u− un) ξ(x, t)dx ≤∫ t

0

∫

Ω

(u− un) ξs + φn∆ξ dxds

−∫ t

0

∫

∂Ω

(u− un)φn∇ξ · ndxds

+

∫ t

0

∫

Σ

[

g(u, v)− g(un, vn−1) +mg

( ǫ

n

)]

ξdxds

and∫

Σ

(v − vn) η(x, t)dx ≤∫ t

0

∫

Σ

(v − vn) ηs + [f(u)− f(un)] η dxds

Here, (u− un)φn ≡ φ(u)− φ(un). Note that

g(u, v)− g(un, vn−1) = g(u, v)− g(un, v)+ g(un, v)− g(un, vn−1)

≡ (u− un) gu,n + (v − vn−1) gv,n

We recall the assumption gv ≥ 0, which in the above notation is gv,n ≥ 0.

Choose the test function ξn as the solution of

ξs + φn∆ξ = 0 x ∈ Ω, s < t∇ξ · n = 0 x ∈ ∂Ω \ Σ, s < tgu,nξ − φn∇ξ · n = 0 x ∈ Σ, s < tξ(x, t) = χ(x) x ∈ Ω

where χ ≥ 0. We also choose η(x, t) ≡ χ(x).

This process is technically more correct by introducing appropriate smooth approxima-

tions of φn and gu,n to insure the solution ξ is sufficiently smooth. Then, upon passing to

the limit, we ultimately obtain the same results.

With such choices for test functions, ξn ≥ 0,∫

Ω

(u− un)χ(x)dx ≤∫ t

0

∫

Σ

[

(v − vn−1) gv,n +mg

( ǫ

n

)]

ξndxds

32

and∫

Σ

(v − vn)χ(x)dx ≤∫ t

0

∫

Σ

[f(u)− f(un)]χ(x)dxds

By the definition of subsolution of (5.3)-(5.4), we also have IΣ(u, v, η, v0) ≤ 0, that is,

∫

Σ

vχ(x)dx ≤∫ t

0

∫

Σ

f(u)χ(x)dxds +

∫

Σ

v(·, 0)χ(x)dx

Let R1 ≥ u on ΩT and R1 ≥ R ≡ ‖u0‖∞,Ω + ‖v0‖∞,Σ + 1. Recalling the definitions

above, we have f(u) ≤ C1(R1), and C1(R) ≤ C1(R1). Hence, selecting T1 < T so that

T1(C1(R1)− C1(R)) ≤ ǫ, there follows

∫

Σ

v(x, t)χ(x)dx ≤∫

Σ

v0χdx+

∫ t

0

∫

Σ

f(u)χdx

≤∫

Σ

v0χdx+ C1(R1)t

∫

Σ

χdx

≤∫

Σ

v0χdx+ [(C1(R1)− C1(R))T1 + C1(R)t]

∫

Σ

χdx

≤∫

Σ

(v0 + ǫ+ C1(R)t)χdx

for any smooth χ ≥ 0 and t ≤ T1. Therefore,

v ≤ v0 + ǫ+ C1(R)t ≤ v0(·, t)

on ΣT1.

Returning to the inequality

∫

Ω

(u− un)χ(x)dx ≤∫ t

0

∫

Σ

[

(v − vn−1) gv,n +mg

( ǫ

n

)]

ξndxds

along with the facts gv,n ≥ 0 and mg(ǫ/n) ≤ 0, it now follows

∫

Ω

(u− u1)χ(x)dx ≤ 0

for any smooth χ ≥ 0. Thus, u ≤ u1 on ΩT1.

We now return to the second inequality

∫

Σ

(v − vn)χ(x)dx ≤∫ t

0

∫

Σ

[f(u)− f(un)]χ(x)dxds

along with f ′ ≥ 0 to conclude∫

Σ

(v − v1)χ(x)dx ≤ 0

for any smooth χ ≥ 0. Thus, v ≤ v1 on ΣT1.

33

Continuing the process, we have u ≤ un and v ≤ vn on ΩT1, ΣT1

, respectively. Passing

to the limit n→ ∞ yields u ≤ U , v ≤ V on the same sets. Continuation of the argument to

ΩT , ΣT is possible due to u(·, T1) ≤ U(·, T1), v(·, T1) ≤ V (·, T1).Short of imposing differentiability of the nonlinearities at u = 0, thus limiting the power

laws allowed, we are not aware of a general uniqueness and supersolution comparison result

for (5.3)-(5.4). As we establish global existence by integral estimates and do not resort

to supersolution constructions, this does not limit the current work. Specifically, letting

φ(u) = um, g(u, v) = uqv, and f(u) = up in (5.3)-(5.4), we have the following result for the

model (5.1)-(5.2).

Theorem 5.3 For m > 1, every solution of (5.1)-(5.2) is global if p + q ≤ 1, whereas if

p+ q > 1 and u0 > 0, then all maximal solutions of the model blow up in finite time.

Proof. We first show that for any finite T > 0, there is a solution of (5.1)-(5.2) on ΩT

in the case p + q ≤ 1, i.e., the model is globally solvable. Due to the continuation result

above and the form of the differential equation for V , this will follow from an upper bound

on the solution U . Our focus is on the power law model (5.1)-(5.2), however some further

generalizations are certainly possible.

We first establish estimates of ‖U‖r+1,ΩT, ‖∇U (r+m)/2‖2,ΩT

, and ‖U‖r+p+q,ΣTfor suffi-

ciently large r > 0 and any finite T > 0 so long as the solution exists on ΩT . These turn

out to be true in all cases m > 0, and the specific result is as follows. Throughout the work,

fairly standard notation for integral norms will be used, e.g.,

‖U‖r+1,ΩT≡

[∫ T

0

∫

Ω

U r+1(x, s)dxds

]

1

r+1

Lemma 5.4 Let U, V denote the maximal solution of (5.1)-(5.2) on ΩT . Assume m > 0

and p + q ≤ min1, (m + 1)/2. For any r > 0 with r ≥ m − 2(p + q) and all T > 0 such

that the solution exists, there is an M > 0 for which

sup0≤t≤T

‖U(·, t)‖r+1,Ω, ‖U‖r+1,ΩT, ‖∇U (r+m)/2‖2,ΩT

, ‖U‖r+p+q,ΣT≤M

A key step in proving Lemma 5.4 is controlling boundary integral terms by using the

standard Sobolev trace estimate∫

Σ

wdx ≤ C

∫

Ω

(w + |∇w|)dx

Resulting constants involved do not permit passing to the limit r → ∞ to obtain a bound on

the L∞ norm. Instead, we are able to slightly revise the maximum principle method invoked

for proving local solvability and obtain the necessary L∞ bound.

34

We now establish the global L∞ estimate on ΩT using estimates in Lr+1(ΩT ), for suffi-

ciently high r established above. To do so, begin with the standard calculation for (un−k)+ ≡maxun − k, 0, with m ≥ 1 and k ≥ ‖u0‖∞ + 1. For fixed t ∈ (0, T ],

1

2

∫

Ω

(un − k)+2(x, t)dx = −∫ t

0

∫

Ω

mum−1n |∇(un − k)+|2dxds

+

∫ t

0

∫

Σ

(un − k)+uqn

[∫ s

0

upn−1dτ +ǫ

n− 1

]

dxds

≤ −mkm−1

∫ t

0

∫

Ω

|∇(un − k)+|2dxds

+

∫ t

0

∫

Σ

(un − k)+uqn

[∫ s

0

upn−1dτ +ǫ

n− 1

]

dxds

Passing to the limit n→ ∞, we have

1

2

∫

Ω

(U − k)+2(x, t)dx+mkm−1

∫ t

0

∫

Ω

|∇(U − k)+|2dxds

≤∫ t

0

∫

Σ

(U − k)+U q

[∫ s

0

Updτ

]

dxds

Select λ > 1 so λ < 1 + 1/N and µ > 1 such that 1/λ+ 1/µ = 1. Then for r ≥ 1

∫ t

0

∫

Σ

(U − k)+U q

[∫ s

0

Updτ

]

dxds

=

∫ t

0

∫

Σ

(U − k)+

U rU r+q

[∫ s

0

Updτ

]

dxds

≤[

∫ t

0

∫

Σ

(

(U − k)+

U r

)λ

dxds

]1/λ[∫ t

0

∫

Σ

(

U r+q

∫ s

0

Updτ

)µ

dxds

]1/µ

We further find∫ t

0

∫

Σ

(

U r+q

∫ s

0

Updτ

)µ

dxds

≤∫ t

0

∫

Σ

Uµr+µq

[

(∫ s

0

Uµpdτ

)1/µ

t1/λ

]µ

dxds

≤ t(λ+µ)/λ

∫ t

0

∫

Σ

U r+q+pdxds

where r + p+ q ≡ µ(r + p+ q), Lemma 5.4 now yields

[∫ t

0

∫

Σ

(

U r+q

∫ s

0

Updτ

)µ

dxds

]1/µ

≤M

for some constant M , which is independent of t ∈ (0, T ].

35

With r ≥ 1 > 1− 1/λ, we have λr > λ− 1. Since k ≥ 1, U ≥ k implies

U ≤ Uλr/(λ−1) ≤ Uλr/(λ−1) + k

Hence, (U − k)+(λ−1) ≤ Uλr, that is,(

(U − k)+

U r

)λ

≤ (U − k)+

It follows by applying the Sobolev trace estimate that

1

2

∫

Ω

(U − k)+2dx+mkm−1

∫ t

0

∫

Ω

|∇(U − k)+|2dxds

≤ M1

[∫ t

0

∫

Σ

(U − k)+dxds

]1/λ

≤ CM1

[∫ t

0

∫

Ω

(

(U − k)+ + |∇(U − k)+|)

dxds

]1/λ

≤ CM1

[∫ t

0

∫

Ω

(U − k)+dxds

]1/λ

+ CM1

[∫ t

0

∫

Ω

|∇(U − k)+|dxds]1/λ

The right hand side of this estimate contains a power which is not normally encountered

in this method of establishing an L∞ bound, e.g., see [1]. We will thus continue to show the

technique still may be successfully applied, due to our choice of λ.

To begin, define Ak(t) ≡ x ∈ Ω : U(x, t) > k and

µ(k) ≡∫ t

0

∫

Ak(s)

dxds

For α > 1,

1

2

∫

Ω

(U − k)+2dx+mkm−1

∫ t

0

∫

Ω

|∇(U − k)+|2dxds

≤ CM1

[

(∫ t

0

∫

Ω

(U − k)+αdxds

)1/α

µ(k)1−1/α

]1/λ

+ CM1

[∫ t

0

∫

Ω

|∇(U − k)+2|dxds]1/2λ

µ(k)1/2λ

Selecting α = (2λ− 1)/(λ− 1) yields 1− 1/α = λ/(2λ− 1), and

1

2

∫

Ω

(U − k)+2dx+mkm−1

∫ t

0

∫

Ω

|∇(U − k)+|2dxds

≤ CM1‖U‖1/λLα(ΩT )µ(k)1/(2λ−1)

+1

2λmkm−1

∫ t

0

∫

Ω

|∇(U − k)+2|dxds

+

(

1− 1

2λ

)[

CM1

(mkm−1)1/2λµ(k)1/2λ

]2λ/(2λ−1)

36

Therefore,

1

2

∫

Ω

(U − k)+2dx+1

2mkm−1

∫ t

0

∫

Ω

|∇(U − k)+|2dxds ≤ C2µ(k)1/(2λ−1)

The standard argument now proceeds to obtain

(h− k)2µ(h)N/(N+2) ≤ C2µ(k)1/(2λ−1),

for h > k > k0, e.g., [1, p.116-117] for the details. It follows

µ(h) ≤[

C2µ(k)1/(2λ−1)

(h− k)2

](N+2)/N

=C3

(h− k)2(N+2)/Nµ(k)(N+2)/N(2λ−1)

Since 1 < λ < 1 + 1/N ,

2λ− 1 < 2(1 + 1/N)− 1 = 1 + 2/N = (N + 2)/N

and thus1

2λ− 1

(

N + 2

N

)

>N

N + 2

(

N + 2

N

)

= 1

Subsequently, [4, p.63] applies to establish µ(k0 + d) = 0 for some d > 0. We thus have

U ≤ k0 + d on ΩT .

We then show the blow-up in finite time for p+ q > 1.

Assuming ∂ΩT = ΣT , we rewrite the model (5.1)-(5.2) as follows

ut = ∆(um) on ΩT

∇(um) · n = uq∫ t

0

up(·, s)ds on ΣT

u = u0 on Ω× 0

(5.5)

Clearly, to establish the blow-up result for (5.5), it suffices to consider the following problem

ut = ∆(um) on ΩT

∇(um) · n = uq∫ t

0

up(·, s)ds on ΣT

u = minΩu0 on Ω× 0

(5.6)

Due to the lack of a supersolution comparison result, we first show that solutions of (5.6)

are monotone in time. Let v(x, t) = u(x, t+ κ) (κ > 0). Then v satisfies

vt = ∆(vm) on ΩT

∇(vm) · n ≥ vq∫ t

0

up(·, s)ds on ΣT

v = u(x, κ) on Ω× 0

(5.7)

37

Since u(x, κ) ≥ minΩu0, from the comparison principle for the porous medium equation

with a localized boundary condition (cf. [1]), it follows that v(x, t) ≥ u(x, t), that is, u is

nondecreasing in t.

Taking into account the monotonicity of the solution of (5.6), we now consider the fol-

lowing problem

ut = ∆(um) on ΩT

∇(um) · n =

∫ t

0

up+q(·, s)ds on ΣT

u = minΩu0 on Ω× 0

(5.8)

Since the solution of (5.8) is a subsolution of (5.6), as in [7, 8], we seek a subsolution of

(5.8) which blows up in finite time. To this end, letting w = um, we rewrite problem (5.8)

as follows

(w1

m )t = ∆w in ΩT

∇w · n =

∫ t

0

wp+q

m (·, s)ds on ΣT

w =

(

minΩu0

)m

on Ω× 0

(5.9)

By comparison, it is easy to see that w(x, t) ≥ σ =

(

minΩu0

)m

> 0 on ΩT . Let h(x) be the

solution of the problem

∆h = 1 on Ω

∇h · n =|Ω||∂Ω| on ∂Ω

(5.10)

Since h(x) + c is also a solution of (5.10) for any positive constant c, we may assume that

h(x) > 0 on Ω. We then consider three cases.

Case 1 (1 < p+ q < m− 1). We choose ϕ(ζ) to satisfy the following

ϕ(ζ) ≡ σ for 0 ≤ ζ ≤ 1 and ϕ′(ζ) =

∫ ζ

1

ϕ−α(η)dη for ζ > 1 (5.11)

where α = 1− p+ q + 1

m. Since p+ q < m− 1, 0 < α < 1. Clearly,

ϕ′′(ζ) ≡ 0 for 0 ≤ ζ < 1 and ϕ′′(ζ) = ϕ−α(ζ) for ζ > 1 (5.12)

Multiplying the second equation in (5.12) by ϕ′(ζ) and integrating over (1, ζ), we obtain

ϕ′(ζ) =

[

2

1− α(ϕ1−α − σ1−α)

]1

2

for ζ > 1

38

Then it follows that ϕ(ζ) exists for 0 ≤ ζ <∞ and ϕ(ζ) → ∞ as ζ → ∞.

Let g(t) be the solution of the following problem

g′(t) = mδϕ1− 1

m (g(t)) t > 0

g(0) =1

2

(5.13)

where δ is chosen so that 0 < δ ≤ |∂Ω||Ω| and δmax

Ωh(x) <

1

2. Since g′(t) ≥ mδσ1− 1

m and∫ ∞

ϕ−1+ 1

m− 1−α

2 dϕ < ∞, g must blow up in finite time. We now construct a subsolution

w(x, t) of (5.9) as follows

w(x, t) = ϕ(g(t) + δh(x)).

By (5.12)-(5.13) and the fact that ϕ(ζ) is nondecreasing for ζ > 0, we have that

(w1

m )t =1

mϕ

1

m−1(g + δh)ϕ′(g + δh)g′(t)

≤ δϕ′(g + δh)

≤ δϕ′(g + δh)∆h + δ2ϕ′′(g + δh)|∇h|2 = ∆w

a.e. on ΩT . On the other hand, by (5.11) and (5.13), we find that

∇w · n = δϕ′(g + δh)∇h · n

≤∫ g(t)+δh(x)

1

ϕ−α(η)dη

≤∫ t

0

ϕ−α(g(s) + δh(x))g′(s)ds

≤∫ t

0

wp+q

m (x, s)ds

on ΣT . Moreover, w(x, 0) = ϕ(g(0) + δh(x)) = σ ≤ w(x, 0). Thus, w(x, t) is indeed a

subsolution of (5.9). Since g(t) blows up in finite time and limζ→∞

ϕ(ζ) = ∞, w(x, t) blows up

in finite time, and so does w(x, t).

Case 2 (m− 1 ≤ p+ q ≤ m). We choose ϕ(ζ) to satisfy the following

ϕ(ζ) ≡ σ for 0 ≤ ζ ≤ 1 and ϕ′(ζ) =

∫ ζ

1

ϕβ(η)dη for ζ > 1 (5.14)

where β =p+ q + 1

m− 1. Since m − 1 ≤ p + q ≤ m, 0 ≤ β ≤ 1

m. We then let g(t) be the

solution of problem (5.13). If 0 < β ≤ 1

m, proceeding as for the case 1 < p+ q < m− 1, we

can show that ϕ(ζ) exists globally, ϕ(ζ) → ∞ as ζ → ∞, and g(t) blows up in finite time.

Thus, w(x, t) = ϕ(g(t)+ δh(x)) is a desired subsolution of (5.9). If β = 0, i.e., m = p+ q+1,

39

ϕ(ζ) =ζ2

2− ζ +

1

2+ σ for ζ > 1, and hence g′(t) = mδ

[

g2(t)

2− g(t) +

1

2+ σ

]1− 1

m

for

t ≥ 1

mδσ1− 1

m

. It then follows that g(t) can only exist locally, and so w(x, t) = ϕ(g(t)+δh(x))

blows up in finite time.

Case 3 (p+ q > m). We choose ϕ(ζ) to satisfy the following

ϕ(ζ) ≡ σ for 0 ≤ ζ ≤ 1 and ϕ′(ζ) =

∫ ζ

1

ϕp+q

m (η)dη for ζ > 1 (5.15)

Thus we have that

ϕ′′(ζ) ≡ 0 for 0 ≤ ζ < 1 and ϕ′′(ζ) = ϕp+q

m (ζ) for ζ > 1 (5.16)

Multiplying the second equation in (5.16) by ϕ′(ζ) and integrating over (1, ζ), we find

ϕ′(ζ) =

[

2m

p+ q +m(ϕ

p+q

m+1 − σ

p+q

m+1)

]1

2

for ζ > 1

Sincep+ q

m> 1, ϕ(ζ) must blow up in finite time. We now construct a subsolution w(x, t)

of (9) as follows

w(x, t) = ϕ(εt+ δh(x))

where ε is chosen so that 0 < ε ≤ min1, mδσ1− 1

m.By (5.16) and the monotonicity of ϕ(ζ), we have that

(w1

m )t =ε

mϕ

1

m−1(εt+ δh)ϕ′(εt+ δh)

≤ δϕ′(εt+ δh)∆h + δ2ϕ′′(εt+ δh)|∇h|2 = ∆w

a.e. on ΩT . On the other hand, in view of (5.15), we find that

∇w · n = δϕ′(εt+ δh)∇h · n

≤∫ εt+δh(x)

1

ϕp+q

m (η)dη

≤∫ t

0

ϕp+q

m (εs+ δh(x))ds

=

∫ t

0

wp+q

m (x, s)ds

on ΣT . Moreover, w(x, 0) = ϕ(δh(x)) = σ ≤ w(x, 0). Thus, w(x, t) of (5.9) blows up in

finite time.

40

References

[1] J. Anderson, Local existence and uniqueness of solutions of degenerate parabolic equa-

tions. Commun. Partial Differential Equations 16 (1991), 105-143.

[2] J. Anderson and K. Deng, Global solvability for the porous medium equation with

boundary flux governed by nonlinear memory, J. Math. Anal. Appl. 423 (2015), 1183-

1202.

[3] J. Anderson, K. Deng, and Z. Dong, Global Solvability for the heat equation with

boundary flux governed by nonlinear memory, Quart. Appl. Math. 69 (2011), 759-770.

[4] D. Kinderlehrer and G. Stampacchia, An Introduction to Variational Inequalities and

Their Application, Academic Press, New York, NY, 1980.

[5] H. Levine, S. Pamuk, B. Sleeman, and M. Nilsen-Hamilton, Mathematical modeling

of capillary formation and development in tumor angiogenesis: penetration into the

stroma, Bulletin Math. Biol. 63 (2001), 801-863.

[6] X. Song and S. Zheng, Blow-up and blow-up rate for a reaction-diffusion model with

multiple nonlinearities, Nonlinear Anal. 54 (2003), 279-289.

[7] W. Walter, On existence and nonexistence in the large of solutions of parabolic differ-

ential equations with a nonlinear boundary condition, SIAM Journal on Mathematical

Analysis 6 (1975), 85-90.

[8] M.-X. Wang and Y.-H. Wu, Global existence and blow up problems for quasilinear

parabolic equations with nonlinear boundary conditions, SIAM J. Math. Anal. 24

(1993), 1515-1521.

[9] N. Wolanski, Global behavior of positive solutions to nonlinear diffusion problems with

nonlinear absorption through the boundary, SIAM J. Math. Anal. 24 (1993), 317-326.

41

6 A nonlocal elliptic equation arising from an SIS model

Over the past few years, several SI-type epidemic reaction-diffusion models have been de-

veloped to study the impact of spatial heterogeneity of environment and movement rates of

individuals on the dynamics of the models. In [1], an SIS (susceptible-infected-susceptible)

reaction-diffusion model with homogeneous Neumann boundary conditions has been consid-

ered:

St = dS∆S − β(x)SI

S + I+ γ(x)I, x ∈ Ω, t > 0,

It = dI∆I +β(x)SI

S + I− γ(x)I, x ∈ Ω, t > 0,

(6.1)

under the condition∫

Ω

(S(x, 0) + I(x, 0)) dx ≡ N > 0,

where Ω is a bounded domain in Rm and N is the total number of individuals at t = 0.

In [1], a basic reproduction number R0 is defined, and it is shown that the disease-free

equilibrium is globally asymptotically stable if R0 ≤ 1, while a unique endemic equilibrium

exists if R0 > 1. Later in [5, 6], the global attractivity of the endemic equilibrium of the

model is proved for two cases: the case dS = dI and the case γ(x) = rβ(x), although the

global attractivity of the endemic equilibrium for general cases remains open.

It is worth pointing out that the well-known SIS model, due to Kermack and McKendrick

(see [2]), takes the form of a system of ordinary differential equations:

S ′ = −βSI + γI, t > 0,

I ′ = βSI − γI, t > 0(6.2)

with initial data satisfying

S(0) + I(0) = N > 0,

where N represents the total population. A basic reproduction number can be defined as

R0 = Nβ/γ. It is proved that if R0 ≤ 1, the solution (S(t), I(t)) of (6.2) approaches the

disease-free equilibrium (N, 0), while if R0 > 1, a unique endemic equilibrium exists:

S∗ =γ

β, I∗ = N − γ

β,

and it is globally asymptotically stable.

Our main objective here is to generalize model (6.2) to an epidemic reaction-diffusion

model and then study the existence of the disease-free equilibrium and the endemic equi-

librium and their global attractivity. Even though our model bears a resemblance to model

(6.1), there is a main difference between our model and model (6.1): The authors in [1]

consider the SIS reaction-diffusion model with frequency dependent interaction, while we

42

focus on mass action type nonlinearity. Consequently, our arguments for proving the global

existence and boundedness of the model, the existence of the endemic equilibrium, and the

global attractivity of the endemic equilibrium are quite different.

Let Ω be a bounded domain in Rm with smooth boundary ∂Ω. Let S(x, t) and I(x, t)

be the density of susceptible and infected individuals at location x and time t, respectively.

We assume that the individuals randomly move in the domain Ω with diffusion rates dS and

dI for susceptible and infected individuals, respectively. If all the infected individuals at the

same location have the same rate to recover and become susceptible immediately, an SIS

epidemic reaction-diffusion model can be formulated as follows [4]:

St = dS∆S − β(x)SI + γ(x)I, x ∈ Ω, t > 0,

It = dI∆I + β(x)SI − γ(x)I, x ∈ Ω, t > 0,(6.3)

where the disease transmission rate function β(x) describes the effective interaction between

susceptible and infected individuals at location x, and the function γ(x) represents the

recovery rate of the infected individuals at location x. Both β and γ are positive Holder

continuous functions in Ω. Furthermore, we assume that there is no flux across the boundary

∂Ω, that is,∂S

∂n=∂I

∂n= 0, x ∈ ∂Ω, t > 0, (6.4)

where ∂/∂n is the outward normal derivative to ∂Ω. We also assume that the initial data

satisfy

(H1) S(x, 0) and I(x, 0) are nonnegative continuous functions in Ω, and initially the number

of infected individuals is positive, i.e.,

∫

Ω

I(x, 0) dx > 0.

Let∫

Ω

(S(x, 0) + I(x, 0)) dx ≡ N

be the total number of individuals at t = 0. Adding up the two equations in (6.3) and then

integrating over the domain Ω, we find

∂

∂t

∫

Ω

(S + I) dx = 0, t > 0,

which implies that the total population size is a constant given by∫

Ω

(S(x, t) + I(x, t)) dx = N. (6.5)

We then consider the equilibria of problem (6.3)-(6.4), that is, the solutions of the fol-

lowing semilinear elliptic system:

dS∆S − βSI + γI = 0, x ∈ Ω,

dI∆I + βSI − γI = 0, x ∈ Ω(6.6)

43

with boundary conditions∂S

∂n=∂I

∂n= 0, x ∈ ∂Ω. (6.7)

Here S(x) and I(x) are the density of susceptible and infected individuals at location x,

respectively. In view of (6.5), we impose an additional condition:∫

Ω

(S + I) dx = N. (6.8)

And we are only interested in nonnegative solutions of (6.6)-(6.8). As considered in other

epidemic models, we will focus on the disease-free equilibrium (DFE) and the endemic equi-

librium (EE). A DFE is a solution of (6.6)-(6.8) with I(x) = 0 for all x ∈ Ω, while an EE is a

solution with I(x) > 0 for some x ∈ Ω. To distinguish between these two types of equilibria,

we will denote a DFE by (S, 0) and an EE by (S∗, I∗). Let |Ω| be the measure of Ω. We first

show that the disease-free equilibrium exists uniquely.

Proposition 6.1 Problem (6.6)-(6.8) has a unique DFE given by (S, 0) = (N/|Ω|, 0).

We then consider the existence of the endemic equilibrium. To this end, we define a basic

reproduction number R0. The variational formula suggests that we can define R0 as follows

R0 = sup

N|Ω|

∫

Ωβϕ2 dx

∫

Ω(dI |∇ϕ|2 + γϕ2) dx

: ϕ ∈ H1(Ω) and ϕ 6= 0

.

On the other hand, there exists a least eigenvalue λ∗ with its corresponding eigenvector

ψ∗, where λ∗ is a real number, ψ∗ is strictly positive on Ω, and (λ∗, ψ∗) satisfies

dI∆ψ∗ +

(

N

|Ω|β − γ

)

ψ∗ + λ∗ψ∗ = 0, x ∈ Ω and∂ψ∗

∂n= 0, x ∈ ∂Ω. (6.9)

Moreover, the eigenvalue λ∗ is given by the variational formula

λ∗ = inf

∫

Ω

(

dI |∇ϕ|2 +(

γ − N

|Ω|β)

ϕ2

)

dx : ϕ ∈ H1(Ω) and

∫

Ω

ϕ2 dx = 1

.

We now state a result which is similar to Lemmas 2.2 and 2.3 in [1].

Proposition 6.2 The following statements about R0 and λ∗ hold.

(a) R0 > 1 when λ∗ < 0, R0 = 1 when λ∗ = 0, and R0 < 1 when λ∗ > 0;

(b) If

∫

Ω

N

|Ω|β dx ≥∫

Ω

γ dx, then λ∗ ≤ 0 for all dI > 0;

(c) IfN

|Ω|β − γ changes sign on Ω and if

∫

Ω

N

|Ω|β dx <

∫

Ω

γ dx, then there exists d∗I > 0

such that λ∗ = 0 when dI = d∗I , λ∗ < 0 when dI < d∗I , and λ

∗ > 0 when dI > d∗I .

44

Remark 6.3 Clearly, by the variational formula, if

∫

Ω

N

|Ω|β dx >∫

Ω

γ dx, then λ∗ < 0.

We then study the existence of the endemic equilibrium. We first convert problem (6.6)-

(6.8) to a more approachable problem.

Lemma 6.4 The pair (S, I) is a nonnegative solution of problem (6.6)-(6.8) if and only if

it is a nonnegative solution of the following problem

dI∆I + I

(

N

|Ω|β − γ −(

1− dIdS

)

β

|Ω|

∫

Ω

I dx− dIβ

dSI

)

= 0, x ∈ Ω, (6.10)

S =N

|Ω| −(

1− dIdS

)

∫

ΩI dx

|Ω| − dIdSI , x ∈ Ω, (6.11)

∂I

∂n= 0, x ∈ ∂Ω. (6.12)

Proof. Through routine calculations, one can easily check that (S, I) is a nonnegative

solution of problem (6.6)-(6.8) if and only if it solves the following problem

dSS + dI I = K, x ∈ Ω, (6.13)

dI∆I + I(βS − γ) = 0, x ∈ Ω, (6.14)

∂S

∂n=∂I

∂n= 0, x ∈ ∂Ω, (6.15)

∫

Ω

(S + I) dx = N, (6.16)

where K is some positive constant that is independent of x ∈ Ω. Thus we only need to

show the equivalence between problem (6.13)-(6.16) and problem (6.10)-(6.12). On one

hand, suppose that (S, I) is a nonnegative solution of (6.13)-(6.16). By (6.13), we have

S = (K−dI I)/dS. Substituting it into (6.16), we find K =

(

dSN − (dS − dI)

∫

Ω

I dx

)

/|Ω|.It then follows from (6.13) that

S =K − dI I

dS=

N

|Ω| −(

1− dIdS

)

∫

ΩI dx

|Ω| − dIdSI ,

which is the equation (6.11). Substituting this S into (6.14), we obtain the equation (6.10).

On the other hand, suppose that (S, I) is a nonnegative solution of problem (6.10)-(6.12).

Taking a normal derivative on both sides of (6.11) and using (6.12), we find ∂S/∂n = 0 which

verifies (6.15). Furthermore, by (6.11), we have that

dIdSI =

N

|Ω| −(

1− dIdS

)

∫

ΩI dx

|Ω| − S

and substitution of this in (6.10) gives (6.14). We then integrate both sides of (6.11) over

Ω to obtain (6.16). Applying the Laplace operator to both sides of (6.11), we find that

45

dS∆S + dI∆I = ∆(dSS + dI I) = 0. Since ∂/∂n(dS S + dI I) = 0, the maximum principle

implies that dSS+dI I is a constant. In view of (6.16), this constant must be positive, which

yields (6.13).

Problem (6.10)-(6.12) is more approachable since (6.10) and (6.12) are independent of S.

In addition, the following result indicates that we can actually focus on a nonlocal elliptic

problem that involves only I.

Lemma 6.5 If I ∈ C2(Ω)∩C1(Ω) is a nonnegative solution of the nonlocal elliptic problem:

dI∆I + I

(

N

|Ω|β − γ −(

1− dIdS

)

β

|Ω|

∫

Ω

I dx− dIβ

dSI

)

= 0, x ∈ Ω, (6.17)

∂I

∂n= 0, x ∈ ∂Ω, (6.18)

then we have that(

1− dIdS

)

∫

ΩI dx

|Ω| +dIdSI(x) ≤ N

|Ω| for all x ∈ Ω. (6.19)

Proof. If I is trivial, then the claim holds. If I is not identically zero on Ω, we assume to

the contrary that the claim is false. Since I is continuous on Ω, it attains its maximum value

on Ω, say, I(x0) = maxx∈Ω

I(x) > 0 for some x0 ∈ Ω. Under the assumption, one must have

that(

1− dIdS

)

∫

ΩI dx

|Ω| +dIdSI(x0) >

N

|Ω| . (6.20)

If x0 ∈ Ω, we can choose a closed ball B centered at x0 such that B ⊂ Ω. By (6.17) and

(6.20), we can make the ball so small that dI∆I > 0 in B. Since I attains its maximum at

an interior point x0 of B, by the strong maximum principle, I must be a constant in B. But

this is impossible since dI∆I > 0 in B. So x0 ∈ ∂Ω, and I(x0) > I(x) for all x ∈ Ω. Then

we can find a closed ball B ⊂ Ω such that B ∩ Ω = x0. Again by (6.17) and (6.20), we

can make the ball so small that dI∆I > 0 in the interior of B. It then follows from the Hopf

lemma that ∂I/∂n(x0) > 0, which is also impossible by virtue of (6.18).

In view of Lemma 6.5, if there is a nonnegative solution I of the nonlocal elliptic problem

(6.17)-(6.18), then one can define

S ≡ N

|Ω| −(

1− dIdS

)

∫

ΩI dx

|Ω| − dIdSI for x ∈ Ω,

which is nonnegative by (6.19). Then it follows that the pair (S, I) solves problem (6.10)-

(6.12).

Let Y = z ∈ C2,α(Ω) : ∂z/∂n = 0 on ∂Ω. For simplicity, we introduce

f(τ, I) =N

|Ω|β − γ −(

1− dIdS

)

β

|Ω|τ −dIβ

dSI ,

46

and define a mapping F : R+ × Y → Cα(Ω) by

F (τ, I) = dI∆I + If(τ, I).

We then consider an eigenvalue problem:

dI∆ϕ+ f(τ, 0)ϕ+ λϕ = 0, x ∈ Ω,

∂ϕ

∂n= 0, x ∈ ∂Ω,

(6.21)

and let λτ be the principle eigenvalue of (6.21).

We now state a well-known result about the existence of positive solutions of an elliptic

problem.

Lemma 6.6 Suppose that τ ≥ 0 and consider the problem:

dI∆I + If(τ, I) = 0, x ∈ Ω,

∂I

∂n= 0, x ∈ ∂Ω.

(6.22)

Then the following statements hold.

(a) If λτ ≥ 0, the only nonnegative solution of (6.22) is I = 0;

(b) If λτ < 0, there is a unique positive solution I ∈ Y of (6.22).

Using the implicit function theorem (cf. [3]), we then prove the following result.

Lemma 6.7 Suppose that λ∗ < 0 and dS > dI . Then there exists a smooth curve (τ, Iτ (x))

in R+ × Y such that F (τ, Iτ ) = 0. And there is a T > 0 such that Iτ (x) > 0 for all x ∈ Ω

and τ ∈ [0, T ) and IT = 0. Moreover, Iτ is decreasing and continuously differentiable in τ

on (0, T ).

Proof. Suppose that (τ0, Iτ0) ∈ R+ × Y satisfies F (τ0, Iτ0) = 0 and Iτ0(x) > 0 on Ω. The

Frechet derivative of F with respect to the second variable at (τ0, Iτ0) is Fy(τ0, Iτ0)w =

dI∆w + (f(τ0, Iτ0) − (dI/dS)βIτ0)w for all w ∈ Y . To see that Fy(τ0, Iτ0) is invertible, for

any h ∈ Cα(Ω), consider the following problem:

dI∆w +

(

f(τ0, Iτ0)−dIdSβIτ0

)

w = h x ∈ Ω,

∂w

∂n= 0, x ∈ ∂Ω.

(6.23)

Let στ0 be the principle eigenvalue of the problem:

dI∆ϕ+

(

f(τ0, Iτ0)−dIdSβIτ0

)

ϕ+ σϕ = 0, x ∈ Ω,

∂ϕ

∂n= 0, x ∈ ∂Ω.

(6.24)

47

By the Fredholm alternative, problem (6.23) has a unique solution for every h ∈ Cα(Ω) if

0 is not an eigenvalue of (6.24). To show this, we note that since F (τ0, Iτ0) = 0, Iτ0 is an

eigenvector of the eigenvalue problem

dI∆ϕ + f(τ0, Iτ0)ϕ+ σϕ = 0, x ∈ Ω,

∂ϕ

∂n= 0, x ∈ ∂Ω

(6.25)

for the eigenvalue σ = 0. Then by the Krein-Rutman Theorem, the positivity of Iτ0 implies

that σ = 0 is the principle eigenvalue of (6.25). Since f(τ0, Iτ0) − (dI/dS)βIτ0 < f(τ0, Iτ0),

it follows that στ0 > 0. So all the eigenvalues of problem (6.24) are positive, which yields

the unique solvability of (6.23). The continuity of the inverse of Fy(τ0, Iτ0) follows from the

classical C2,α estimates. Since λ0 = λ∗ < 0, by Lemma 6.6, there exists a unique positive

I0 ∈ Y such that F (0, I0) = 0. Then by the implicit function theorem, there is a unique

Iτ ∈ Y such that F (τ, Iτ) = 0 for τ ∈ [0, τ ′) with τ ′ > 0, and this Iτ is continuously

differentiable with respect to τ .

To show that Iτ is decreasing with respect to τ , we may consider 0 < τ1 < τ2 < τ ′.

Since dS > dI , we have that F (τ1, Iτ2) > 0 and hence Iτ2 is a lower solution of the equation

F (τ1, I) = 0. On the other hand, we can choose a sufficiently large number as an upper

solution. Then the method of upper/lower solutions and the uniqueness of the positive

solution of F (τ2, I) = 0 imply Iτ1 > Iτ2 .

The curve (τ, Iτ ) with Iτ > 0 continues as long as λτ < 0. By the variational formula,

λτ is increasing with respect to τ and λτ > 0 for large τ . Thus by Lemma 6.6, there is no

positive solution of F (τ, I) = 0 if τ is large. Let [0, T ) be the maximal interval of existence

of τ such that Iτ > 0. Then IT = 0.

An analogous result in the case dS < dI can also be proved.

Lemma 6.8 Suppose that λ∗ < 0 and dS < dI . Then there exists a smooth curve (τ, Iτ (x))

in R+ × Y such that F (τ, Iτ) = 0 with Iτ (x) > 0 for all x ∈ Ω and τ ∈ (0,∞). Moreover,

Iτ is increasing and continuously differentiable in τ on (0,∞), and it satisfies the following

estimate∫

Ω

Iτ dx ≤ dSdIN +

(

1− dSdI

)

τ.

Proof. The existence and continuity of the curve (τ, Iτ ) follow from a similar argument as

in the proof of Lemma 6.7. Since dS < dI , one can see that Iτ is increasing with respect to

τ , and thus the curve continues as τ → ∞. It then remains to show the estimate. For any

τ > 0, one can check that

I =dSN

dI |Ω|+

(

1− dSdI

)

τ

|Ω|

48

is an upper solution of F (τ, I) = 0. On the other hand, I = Iτ with τ < τ is a lower solution

of F (τ, I) = 0. Then the method of upper/lower solutions implies that

Iτ ≤ dSN

dI |Ω|+

(

1− dSdI

)

τ

|Ω| ,

which upon integration over Ω yields the estimate.

We are now in a position to prove the existence of the endemic equilibrium.

Theorem 6.9 If R0 > 1, then there exists a unique EE for dS ≥ dI and for dS < dI with

N/|Ω| ≥ γ/β.

Proof. By Lemmas 6.4 and 6.5, it suffices to show that problem (6.17)-(6.18) has a unique

positive solution. The case dS = dI follows directly from Lemma 6.6. We then consider

the case dS > dI . By Lemma 6.7, there exists a smooth curve (τ, Iτ ) for τ ∈ [0, T ) with

F (τ, Iτ ) = 0. By the definition of F , Iτ is a solution of problem (6.17)-(6.18) if τ =

∫

Ω

Iτ dx.

Since 0 <

∫

Ω

I0 dx and T >

∫

Ω

IT dx = 0, the continuity and monotonicity of Iτ in τ implies

that there exists a unique τ0 ∈ [0, T ) such that τ0 =

∫

Ω

Iτ0 dx. Hence problem (6.17)-(6.18)

has a unique positive solution.

We now consider the case dS < dI . By Lemma 6.8, there exists a smooth curve (τ, Iτ )

with F (τ, Iτ) = 0. Since 0 <

∫

Ω

I0 dx, by the continuity and monotonicity in Iτ , and using

the estimate of

∫

Ω

Iτ dx in Lemma 6.8, we can see that there exists at lease a τ0 > 0 such

that τ0 =

∫

Ω

Iτ0 dx. To show the uniqueness of the solution, suppose that I is a positive

solution of (6.17)-(6.18) and let I(x0) = minI(x) : x ∈ Ω. If x0 ∈ Ω, then ∆I(x0) ≥ 0,

and we find

I(x0)

(

N

|Ω|β(x0)− γ(x0)−(

1− dIdS

)

β(x0)

|Ω|

∫

Ω

I dx− dIβ(x0)

dSI(x0)

)

≤ 0. (6.26)

If x0 ∈ ∂Ω, we may assume to the contrary that (6.26) is not valid, then we can find a ball

B ⊂ Ω such that B ∩ Ω = x0 and I(x0) < I(x),∆I(x) < 0 for all x ∈ B. It then follows

from the Hopf lemma that ∂I/∂n(x0) < 0, which is impossible by (6.18). Let d = dI/dS − 1,

(6.26) indicates that

dIdSI(x0) ≥

N

|Ω| −γ(x0)

β(x0)+

d

|Ω|

∫

Ω

Idx ≥ d

|Ω|

∫

Ω

Idx,

which leads todIdSI ≥ d

|Ω|

∫

Ω

Idx.

49

Clearly, the above inequality can never be an equality. Suppose that I1 and I2 are two

different positive solutions. Without loss of generality, we may assume I1 > I2 and define

k = maxk ≥ 0 : kI1 ≤ I2.

We then have that 0 < k < 1, kI1 ≤ I2, and kI1(x1) = I2(x1) for some x1 ∈ Ω.

We now introduce a mapping A : E ⊂ C+(Ω) → C+(Ω) by

A(I) = (a− dI∆)−1I

(

a +N

|Ω|β − γ −(

1− dIdS

)

β

|Ω|

∫

Ω

I dx− dIβ

dSI

)

for I ∈ E , where a is a positive constant, and

E =

I ∈ C+(Ω) :dIdSI ≥ d

|Ω|

∫

Ω

Idx and ‖I‖ ≤ ‖I1‖

.

Note that both I1 and I2 are fixed points of A. Moreover, if a is large enough, then A is

strictly increasing on E . Thus, we have that kA(I) < A(kI) for any k ∈ (0, 1) and I ∈ E

withdIdSI ≥ d

|Ω|

∫

Ω

Idx anddIdSI 6≡ d

|Ω|

∫

Ω

Idx. (6.27)

Since both I1 and I2 satisfy (6.27), we further find

kI1 = kA(I1) < A(kI1) ≤ A(I2) = I2,

which contradicts kI1(x1) = I2(x1).

We then discuss the nonexistence of the endemic equilibrium.

Theorem 6.10 If dS ≥ dI, then the EE does not exist when R0 ≤ 1; if dS < dI, then the

EE does not exist when R0 ≤ dS/dI . Furthermore, if dS < dI and γ(x) = rβ(x) with r

positive constant, then the EE does not exist when R0 ≤ 1.

Proof. The case dS = dI follows directly from Lemma 6.6. We then consider the case

dS > dI with R0 ≤ 1. Assume to the contrary that an EE, (S∗, I∗), exists. Then there

is a τ ∗ > 0 such that τ ∗ =

∫

Ω

I∗ dx and F (τ ∗, I∗) = 0, and it follows from Lemma 6.6

that λτ∗ < 0. Since f(τ, 0) is decreasing in τ when dS > dI , by the variational formula,

λ∗ = λ0 ≤ λτ∗ < 0. This implies R0 > 1 by Proposition 6.2, which is a contradiction.

We now consider the case dS < dI with R0 ≤ dS/dI . Assume to the contrary that an EE

(S∗, I∗) exists. Let τ ∗ =

∫

Ω

I∗ dx. Then F (τ ∗, I∗) = 0, and this implies λτ∗ < 0 by Lemma

6.6. On the other hand, by Lemma 6.5, one can see that for all x ∈ Ω,

(

1− dIdS

)

∫

ΩI∗ dx

|Ω| +dIdSI∗(x) ≤ N

|Ω| .

50

Integrating the above inequality over Ω then gives τ ∗ =

∫

Ω

I∗ dx ≤ N . Since f(τ, 0) is

increasing in τ when dS < dI , λN ≤ λτ∗ < 0. Note that λN is the principle eigenvalue of the

following problem:

dI∆ϕ+

(

dIN

dS|Ω|β − γ

)

ϕ+ λϕ = 0, x ∈ Ω,

∂ϕ

∂n= 0, x ∈ ∂Ω.

Similar to R0, one can define R′0 as follows:

R′0 = sup

dINdS |Ω|

∫

Ωβϕ2 dx

∫

Ω(dI |∇ϕ|2 + γϕ2) dx

: ϕ ∈ H1(Ω) and ϕ 6= 0

.

Then λN < 0 if and only if R′0 > 1. Since R′

0 = dI/dSR0, λN < 0 implies R0 > dS/dI , which

is a contradiction.

We then consider the case dS < dI with R0 ≤ 1 for γ(x) = rβ(x). Assume to the contrary

that an EE (S∗, I∗) exists. Proceeding as in the proof of Lemma 6.5, one can see that for all

x ∈ Ω,(

1− dIdS

)

∫

ΩI∗ dx

|Ω| +dIdSI∗(x) ≤ N

|Ω| − r.

Integrating the above inequality over Ω then gives

∫

Ω

I∗ dx ≤ N − r|Ω|, and it follows that

I∗ is a lower solution of the problem:

dS∆I + I

(

N

|Ω|β − γ − βI

)

= 0, x ∈ Ω,

∂I

∂n= 0, x ∈ ∂Ω.

(6.28)

On the other hand, it is easy to see that M = N/|Ω| is an upper solution of problem (6.28).

Then by the upper/lower solution argument there exists a unique positive solution of (6.28).

However, since R0 ≤ 1, by Proposition 6.2 λ∗ ≥ 0. It then follows from the variational

formula of λ∗ and γ(x) = rβ(x) that γ ≥ Nβ/|Ω|. Thus problem (6.28) has no positive

solution. This leads to a contradiction.

References

[1] L. J. S. Allen, B. M. Bolker, Y. Lou, and A. L. Nevai, Asymptotic profiles of the steady

states for an SIS epidemic reaction-diffusion model, Discrete Contin. Dyn. Syst. 21

(2008), 1-20.

51

[2] F. Brauer and C. Castillo-Chavez, Mathematical Models in Population Biology and

Epidemiology, Springer, New York, 2001.

[3] R. S. Cantrell and C. Cosner, Spatial Ecology Via Reaction-Diffusion Equations, Wiley,

Chichester, West Sussex, UK, 2003.

[4] K. Deng and X. Wu, Dynamics of a susceptible-infected-susceptible epidemic reaction-

diffusion model, Proc. Royal Soc. Edinburgh Sect. A 146 (2016), 929-946.

[5] R. Peng and S. Liu, Global stability of the steady states of an SIS epidemic reaction-

diffusion model, Nonlinear Anal. 71 (2009), 239-247.

[6] R. Peng and X. Zhao, A reaction-diffusion SIS epidemic model in a time-periodic envi-

ronment, Nonlinearity 25 (2012), 1451-1471.

52