^{1}

^{2}

In this paper, we investigate the dynamics of a stochastic predator-prey model with ratio-dependent functional response and disease in the prey. Firstly, we prove the existence and uniqueness of the positive solution for the stochastic model by using conventional methods. Then we obtain the threshold

The research of eco-epidemiology involving ecological and epidemiological models is a significant field in mathematical biology. To our knowledge, Anderson and May were the first to study the spread and persistence of infectious diseases by formulating an eco-epidemiological prey-predator model [

In predator-prey system, functional response plays an important role in the population dynamics. Holling types functional response functions, namely Holling types I, II, III and IV, have been extensively used and investigated [

{ d S d t = r S ( 1 − S k ) − α S P m P + S + I − b S I , d I d t = b S I − d 1 I − β I P m P + S + I , d P d t = c α S P m P + S + I + c β I P m P + S + I − d 2 P , (1)

where S ( t ) , I ( t ) and P ( t ) denote the densities of the susceptible prey, infected prey and predator respectively. Here, the susceptible prey is subject to the logistic growth, r is the intrinsic growth rate, and r k denotes the interspecific competition rate. The transmission of the disease in the prey is governed by the bilinear incidence rate b S I , where b represent the incidence rate of infected prey to susceptible prey, S ( t ) and I ( t ) denote the densities of the susceptible prey, infected prey. Moreover, the parameters α and β represent the capturing rates of predator to the susceptible and infected prey, respectively; m is the so-called half saturation constant; d 1 and d 2 are the natural death rates of the infected prey and predator. All coefficients mentioned are positive.

As a matter of a fact, most realistic ecosystems are affected by environmental noise (see [

{ d S = [ r S ( 1 − S k ) − α S P m P + S + I − b S I ] d t + σ 1 S d B 1 , d I = [ b S I − d 1 I − β I P m P + S + I ] d t + σ 2 I d B 2 , d P = [ c α S P m P + S + I + c β I P m P + S + I − d 2 P ] d t + σ 3 P d B 3 , (2)

where { B i ( t ) } t ≥ 0 ( i = 1 , 2 , 3 ) are mutually independent standard Brownian motions, and σ i ( i = 1 , 2 , 3 ) denote the intensities of the white noise.

Throughout this article, let ( Ω , F , { F t } t ≥ 0 , ℙ ) be a complete probability space with a filtration { F t } t ≥ 0 satisfying the usual conditions (i.e., it is increasing and right continuous while { F 0 } contains all ℙ -null sets) and let B i ( t ) be defined on the complete probability space, i = 1 , 2 , 3 . Denote ℝ + d = { x = ( x 1 , ⋯ , x d ) ∈ ℝ d : x i > 0 , 1 ≤ i ≤ d } .

In this article, in order to better study the spread of infectious diseases among interacting populations, it is more practical to establish a more accurate random ecological infectious disease model. We will concentrate on the dynamics of the stochastic model (2). The rest of the article is organized as follows. In Section 2, the existence and uniqueness of the positive solution is proved for system (2). In Section 3, we analyze the extinction and persistence of the infected prey. In Section 4, we obtain the conditions on the existence of stationary distribution for model (2). In Section 5, numerical simulations are conducted to support the theoretical results. A conclusion is given in the last section.

To begin with, we recall some basic notations in stochastic differential equation. let X ( t ) be a regular time-homogeneous Markov process in ℝ d described by the stochastic differential equation

d X ( t ) = f ( X ( t ) ) d t + g ( X ( t ) ) d B ( t ) . (3)

The diffusion matrix of the process X ( t ) is defined as A ( x ) = ( a i j ( x ) ) , a i j ( x ) = g i ( x ) g j ( x ) . Furthermore, the differential operator L is defined by

L V ( x ) = ∑ i = 1 d f i ( x ) ∂ V ( x ) ∂ x i + 1 2 ∑ i , j = 1 d [ g T ( x ) g ( x ) ] i j ∂ 2 V ( x ) ∂ x i ∂ x j ,

where V ∈ C 2 ( ℝ d , ℝ + ) .

To investigate the dynamical behavior of the model, the first concern is whether the solution is global and positive. In this section, we show that there exists a unique global positive solution of system (2) by constructing an appropriate Lyapunov function.

Theorem 2.1. For any initial value ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) ∈ ℝ + 3 , there exists a unique positive solution ( S ( t ) , I ( t ) , P ( t ) ) of system (2) on t ≥ 0 and the solution will remain in ℝ + 3 with probability one, that is to say, ( S ( t ) , I ( t ) , P ( t ) ) ∈ ℝ + 3 for all t ≥ 0 almost surely (a.s.).

Proof. Since the coefficients of system (2) are locally Lipschitz continuous, then for any initial value ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) ∈ ℝ + 3 there is a unique local solution

( S ( t ) , I ( t ) , P ( t ) ) on t ∈ [ 0, τ e ) , where τ e is the explosion time. Now, let us

show that this solution is global, i.e., τ e = ∞ a.s.. Let n 0 > 0 be sufficiently large such that S ( 0 ) , I ( 0 ) and P ( 0 ) lie within the interval [ 1 n 0 , n 0 ] . For each integer n ≥ n 0 , define stopping-times

τ n = inf { t ∈ [ 0 , τ e ) : min { S ( t ) , I ( t ) , P ( t ) } ≤ 1 n 0 or max { S ( t ) , I ( t ) , P ( t ) } ≥ n } ,

set inf ∅ = ∞ ( ∅ represents the empty set). It is clear that τ n is increasing as n → ∞ . Let τ ∞ = lim n → ∞ τ n , then τ ∞ ≤ τ e a.s.. In the following, it only needs to show that τ ∞ = ∞ a.s.. If this statement is violated, then there exists a constant T > 0 and ε ∈ ( 0,1 ) such that ℙ { τ ∞ ≤ T } > ε . Consequently, there exists an integer n 1 ≥ n 0 such that

ℙ { τ n ≤ T } ≥ ε for all n ≥ n 1 . (4)

Define a C 2 -function V : ℝ + 3 → ℝ + by

V ˜ ( t ) = V ˜ ( S , I , P ) = ( S − a − a ln S a ) + ( I − 1 − ln I ) + 1 c ( P − 1 − ln P ) ,

where a is a positive constant to be determined later. The nonnegativity of the function can be obtained from u − 1 − log u ≥ 0 for any u > 0 .

Let n ≥ n 0 and T > 0 be arbitrary. Applying Itô’s formula to V ˜ , we obtain that

d V ˜ = ( 1 − a S ) d S + a σ 1 2 2 d t + ( 1 − 1 I ) d I + σ 2 2 2 d t + 1 c ( 1 − 1 P ) d P + σ 3 2 2 c d t : = L V ˜ d t + ( S − a ) σ 1 d B 1 + ( I − 1 ) σ 2 d B 2 + ( P − 1 ) σ 3 d B 3 c ,

where

L V ˜ = ( S − a ) ( r − r k S − α P m P + S + I − b I ) + ( I − 1 ) ( b S − d 1 − β P m P + S + I ) + P − 1 c ( c α S m P + S + I + c β I m P + S + I − d 2 ) + a σ 1 2 2 + σ 2 2 2 + σ 3 2 2 c = r S − r k S 2 − α S P m P + S + I − b S I − r a + r a k S + a α P m P + S + I + a b I + b S I − d 1 I − β P I m P + S + I − b S + d 1 + β P m P + S + I + α S P m P + S + I + β P I m P + S + I − d 2 c P − α S m P + S + I − β I m P + S + I + d 2 c + a σ 1 2 2 + σ 2 2 2 + σ 3 2 2 c

≤ r S + r a k S − r k S 2 + a α P m P + S + I + ( a b − d 1 ) I + β P m P + S + I − r a + d 1 + d 2 c + a σ 1 2 2 + σ 2 2 2 + σ 3 2 2 c .

Choose a = d 1 b such that a b − d 1 = 0 , then

L V ˜ ≤ r S + r k d 1 b S − r k S 2 + α m d 1 b + β m − r d 1 b + d 1 + d 2 c + a σ 1 2 2 + σ 2 2 2 + σ 3 2 2 c ≤ k ( r + r d 1 k b ) 2 4 r + α m d 1 b + β m − r d 1 b + d 1 + d 2 c + d 1 b σ 1 2 2 + σ 2 2 2 + σ 3 2 2 c : = N ,

where N is a positive parameter. Thus, we can obtain

d V ˜ ≤ L V ˜ d t + ( S − a ) σ 1 d B 1 + ( I − 1 ) σ 2 d B 2 + ( P − 1 ) σ 3 d B 3 c . (5)

Interacting and taking the expectation of both sides of (5) yield

E ( V ( S ( τ n ∧ T ) , I ( τ n ∧ T ) , P ( τ n ∧ T ) ) ) ≤ V ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) + N T . (6)

Set Ω n = τ n ≤ T for n ≥ n 1 , then by (4), we have ℙ ( Ω n ) ≥ ε . Noting that for every ω ∈ Ω n , there is at least one of S ( τ n , ω ) , I ( τ n , ω ) , P ( τ n , ω ) that equals to either n or 1 n , and

V ( S ( τ n ) , I ( τ n ) , P ( τ n ) ) ,

is no less than

n − 1 − ln n or 1 n − 1 − ln 1 n .

That is,

V ( S ( τ n ) , I ( τ n ) , P ( τ n ) ) ≥ ( n − 1 − ln n ) ∧ ( 1 n − 1 − ln 1 n ) .

By (6), we can obtain

∞ > V ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) + N T ≥ E ( 1 Ω n ( ω ) V ( S ( τ n ) , I ( τ n ) , P ( τ n ) ) ) = ℙ ( Ω n ) V ( S ( τ n ) , I ( τ n ) , P ( τ n ) ) ≥ ε V ( S ( τ n ) , I ( τ n ) , P ( τ n ) ) ≥ ε [ ( n − 1 − ln n ) ∧ ( 1 n − 1 − ln 1 n ) ] .

Taking n → ∞ induces ∞ > + ∞ , which is a contradiction. Hence, we have τ ∞ = ∞ , a.s.. The conclusion is confirmed.

According to the theory in [

• If R 0 ≤ 1 , the disease-free equilibria E 0 ( k , 0 , 0 ) is globally asymptotically stable, the disease disappear;

• If R 0 > 1 , the infectious prey population will be persistence in the mean.

In this section, we turn to establish sufficient criteria on the extinction and persistence of infected prey population for the stochastic system (2). Before giving our main results, we first recall the following lemma.

Lemma 3.1. ( [

1) If there exists T > 0 , λ 0 > 0 , λ and n i such that

ln X ( t ) ≤ λ t − λ 0 ∫ 0 t X ( s ) d s + ∑ i = 1 j n i B ( t ) a .s . , for t ≥ T ,

then

{ 〈 X ( t ) 〉 * ≤ λ λ 0 a .s . , if λ ≥ 0 ; l i m t → ∞ X ( t ) = 0 a .s . , if λ < 0.

where 〈 X ( t ) 〉 * = lim sup t → ∞ 〈 X ( t ) 〉 .

2) If there exists T > 0 , λ 0 > 0 , λ > 0 and n i such that

ln X ( t ) ≥ λ t − λ 0 ∫ 0 t X ( s ) d s + ∑ i = 1 j n i B ( t ) a .s . , for t ≥ T ,

then

〈 X ( t ) 〉 * ≥ λ λ 0 a .s .,

where 〈 X ( t ) 〉 * = lim inf t → ∞ 〈 X ( t ) 〉 .

Theorem 3.2. Let ( S ( t ) , I ( t ) , P ( t ) ) be the solution of system (2) with any initial value ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) ∈ ℝ + 3 . If r > σ 1 2 2 , α + β < d 2 + σ 3 2 2 c and R 0 s = b k ( r − σ 1 2 2 ) r ( d 1 + σ 2 2 2 ) < 1 , then

lim sup t → ∞ ln I ( t ) t ≤ ( d 1 + σ 2 2 2 ) ( R 0 s − 1 ) < 0 a .s . ,

lim sup t → ∞ ln P ( t ) t ≤ − [ d 2 + σ 3 2 2 − c ( α + β ) ] < 0 a .s .,

and

l i m t → ∞ 1 t ∫ 0 t S ( s ) d s = k ( r − σ 1 2 2 ) r a .s ..

Proof. By the Itô’s formula, we have

d ln S = ( r − r k S − α P m P + S + I − b I − σ 1 2 2 ) d t + σ 1 d B 1 ( t ) ≤ ( r − σ 1 2 2 − r k S ) d t + σ 1 d B 1 ( t ) , (7)

integrating Equation (7) from 0 to t and dividing it by t, we obtain

1 t ln S ( t ) S ( 0 ) ≤ r − σ 1 2 2 − r k ∫ 0 t S ( s ) d s t + ∫ 0 t σ 1 d B 1 ( θ ) t ,

applying Lemma 3.1, we have

lim sup t → ∞ 1 t ∫ 0 t S ( s ) d s ≤ k ( r − σ 1 2 2 ) r a .s .. (8)

Similarly, by the Itô’s formula, we have

d ln I = ( b S − d 1 − β P m P + S + I − σ 2 2 2 ) d t + σ 2 d B 2 ( t ) ≤ [ b S − ( d 1 + σ 2 2 2 ) ] d t + σ 2 d B 2 ( t ) , (9)

integrate Equation (9) from 0 to t and divide it by t yields, we obtain

1 t ln I ( t ) I ( 0 ) ≤ b ∫ 0 t S ( s ) d s t − ( d 1 + σ 2 2 2 ) + ∫ 0 t σ 1 d B 2 ( θ ) t ,

so

lim sup t → ∞ ln I ( t ) t ≤ − ( d 1 + σ 2 2 2 ) + b k ( r − σ 1 2 2 ) r = ( d 1 + σ 2 2 2 ) ( R 0 s − 1 ) < 0 a .s ..

By the Itô’s formula, we also have

d ln P = ( c α S m P + S + I + c β I m P + S + I − d 2 − σ 3 2 2 ) d t + σ 3 d B 3 ( t ) ≤ ( c α + c β − d 2 − σ 3 2 2 ) d t + σ 3 d B 3 ( t ) , (10)

integrating Equation (9) from 0 to t and dividing it by t, one can get

1 t ln P ( t ) P ( 0 ) ≤ ( c α + c β − d 2 − σ 3 2 2 ) + ∫ 0 t σ 3 d B 3 ( θ ) t ,

so we can obtain that

lim sup t → ∞ ln P ( t ) t ≤ − [ d 2 + σ 3 2 2 − c ( α + β ) ] < 0 a .s ..

On the other hand,

d ln S = ( r − r k S − α P m P + S + I − b I − σ 1 2 2 ) d t + σ 1 d B 1 ( t ) ,

since lim sup t → ∞ ln P ( t ) t < 0 a .s . , there exists an arbitrarily small constant ε > 0 such that when t > T , we obtain P m P + S + I < ε , so

d ln S ≥ ( r − σ 1 2 2 − r k S − α ε ) d t + σ 1 d B 1 ( t ) , (11)

integrating Equation (11) from 0 to t and dividing it by t, we have

1 t ln S ( t ) S ( 0 ) ≥ r − σ 1 2 2 − r k ∫ 0 t S ( s ) d s t − α ε + ∫ 0 t σ 1 d B 1 ( θ ) t ,

applying Lemma 3.1 and the arbitrariness of ε , we get

lim inf t → ∞ 1 t ∫ 0 t S ( s ) d s ≥ k ( r − σ 1 2 2 ) r a .s .. (12)

From (8) and (12), we obtain

l i m t → ∞ 1 t ∫ 0 t S ( s ) d s = k ( r − σ 1 2 2 ) r a .s ..

Theorem 3.3. Let ( S ( t ) , I ( t ) , P ( t ) ) be the solution of system (2) with any initial value ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) ∈ ℝ + 3 . If r > σ 1 2 2 , α + β < d 2 + σ 3 2 2 c and R 0 s > 1 , then

lim inf t → ∞ ln I ( t ) t ≥ ( d 1 + σ 2 2 2 ) ( R 0 s − 1 ) > 0 a .s ..

Proof. By the Itô’s formula, one we can obtain

d ln I = ( b S − d 1 − β P m P + S + I − σ 2 2 2 ) d t + σ 2 d B 2 ( t ) ,

according to Theorem 3.2, there exists an arbitrarily small constant ε > 0 such

that when t > T , we obtain P m P + S + I < ε , so

d ln I ≥ [ b S − β ε − ( d 1 + σ 2 2 2 ) ] d t + σ 2 d B 2 ( t ) , (13)

integrate Equation (13) from 0 to t and divide it by t yields, we can obtain

1 t ln I ( t ) I ( 0 ) ≥ b ∫ 0 t S ( s ) d s t − β ε − ( d 1 + σ 2 2 2 ) + ∫ 0 t σ 1 d B 2 ( θ ) t = b k ( r − σ 1 2 2 ) r − β ε − ( d 1 + σ 2 2 2 ) ≥ ( d 1 + σ 2 2 2 ) ( R 0 s − 1 ) > 0 a .s ..

According to Theorem 3.2 and 3.3, R 0 s is the threshold for the infected prey population. The disease will go to extinction if R 0 s < 1 , and it will exist in the long time if R 0 s > 1 .

Now we present the following lemma.

Lemma 4.1. ( [^{2}-function V such that LV is negative for any ℝ d \ D . Then

ℙ x { l i m T → ∞ 1 T ∫ 0 T f ( X ( t ) ) d t = ∫ ℝ d f ( x ) μ ( d x ) } = 1,

for all x ∈ ℝ d , where f ( . ) is a function integrable with respect to the measure μ .

Theorem 4.2. Assume that R ^ 0 s = r + d 2 + σ 3 2 2 − σ 1 2 2 − α m − c α − c β r b k ( d 1 + β m + σ 2 2 2 ) > 1 , and

environmental noises are small enough that σ 2 2 < 2 d 1 and σ 3 2 < 2 d 2 . Then for any initial value ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) ∈ ℝ + 3 , there exists a unique stationary distribution for system (2) and it is ergodic.

Proof. In order to prove Theorem 4.2, first we need to verify Lemma 4.1. To verify B.2, we need to proof there exists a neighborhood D ⊂ ℝ + 3 and a nonnegative C^{2}-function V such that for any ( S , I , P ) ⊂ ℝ + 3 \ D , LV is negative.

Define a C^{2}-function

V ¯ = M ( − log S − r b k log I + b d 1 I ) + 1 2 ( S + I + P c ) 2 ,

where

M = b k r ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) max { 2 + f ¯ 1 + f 2 + f 3 ,2 + f 1 + f ¯ 2 + f 3 ,1 + f 1 + f 2 + f ¯ 3 }

and functions f 1 , f 2 , f 3 , f ¯ 1 , f ¯ 2 and f ¯ 3 will be determined later. There exists a unique minimum point ( S ˜ , I ˜ , P ˜ ) of V ¯ .

Define a nonnegative C^{2}-Lyapunov function

V = M ( − log S − r b k log I + b d 1 I + log P ) + 1 2 ( S + I + P c ) 2 − V ¯ ( S ˜ , I ˜ , P ˜ ) ,

denote

V 1 = M ( − log S − r b k log I + b d 1 I + log P ) , V 2 = 1 2 ( S + I + P c ) 2 .

By Itô’s formula, we get

L V 1 = − 1 S [ r S ( 1 − S k ) − α S P m P + S + I − b S I ] + σ 1 2 2 − r b k 1 I ( b S I − d 1 I − β I P m P + S + I ) + r σ 2 2 2 b k + b d 1 ( b S I − d 1 I − β I P m P + S + I ) + c α S m P + S + I + c β I m P + S + I − d 2 − σ 3 2 2

≤ − ( r − σ 1 2 2 − α m ) + r k S + b I − r k S + r b k ( d 1 + β m + σ 2 2 2 ) + b 2 d 1 S I − b I + c α + c β − d 2 − σ 3 2 2 = − ( r + d 2 + σ 3 2 2 − σ 1 2 2 − α m − c α − c β ) + r b k ( d 1 + β m + σ 2 2 2 ) + b 2 d 1 S I ,

and

L V 2 = ( S + I + P c ) ( r S − r k S 2 − d 1 I − d 2 c P ) + 1 2 ( σ 1 2 S 2 + σ 2 2 I 2 + σ 3 2 c 2 P 2 ) = r S 2 − r k S 3 − d 1 S I − d 2 c S P + r S I − r k I S 2 − d 1 I 2 − d 2 c I P + r c P S − r c k P S 2 − d 1 c P I − d 2 c 2 P 2 + 1 2 ( σ 1 2 S 2 + σ 2 2 I 2 + σ 3 2 c 2 P 2 ) ≤ r S 2 − r k S 3 + r S I − d 1 I 2 + r c P S − d 2 c 2 P 2 + 1 2 ( σ 1 2 S 2 + σ 2 2 I 2 + σ 3 2 c 2 P 2 ) ,

according to Young inequality, x y ≤ f r a c 25 x 5 2 + 3 5 y 5 3 , so

L V 2 ≤ r S 2 − r k S 3 + ( 2 r 5 + 2 r 5 c ) S 5 2 + σ 1 2 2 S 2 − ( d 1 − σ 2 2 2 ) I 2 + 3 r 5 I 5 3 − 1 c 2 ( d 2 − σ 3 2 2 ) P 2 + 3 r 5 c P 5 3 ,

so

L V ≤ M [ − ( r + d 2 + σ 3 2 2 − σ 1 2 2 − α m − c α − c β ) + r b k ( d 1 + β m + σ 2 2 2 ) + b 2 d 1 S I ] − r k S 3 + ( 2 r 5 + 2 r 5 c ) S 5 2 + ( σ 1 2 2 + r ) S 2 − ( d 1 − σ 2 2 2 ) I 2 + 3 r 5 I 5 3 − 1 c 2 ( d 2 − σ 3 2 2 ) P 2 + 3 r 5 c P 5 3 = − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + M b 2 d 1 S I − r k S 3 + ( 2 r 5 + 2 r 5 c ) S 5 2 + ( σ 1 2 2 + r ) S 2 − ( d 1 − σ 2 2 2 ) I 2 + 3 r 5 I 5 3 − 1 c 2 ( d 2 − σ 3 2 2 ) P 2 + 3 r 5 c P 5 3 ,

where

R ^ 0 s = r + d 2 + σ 3 2 2 − σ 1 2 2 − α m − c α − c β r b k ( d 1 + β m + σ 2 2 2 ) , f 1 ( S ) = − r k S 3 + ( 2 r 5 + 2 r 5 c ) S 5 2 + ( σ 1 2 2 + r ) S 2 ,

f 2 ( I ) = − ( d 1 − σ 2 2 2 ) I 2 + 3 r 5 I 5 3 , f 3 ( P ) = − 1 c 2 ( d 2 − σ 3 2 2 ) P 2 + 3 r 5 c P 5 3 .

We aim to prove that L V ≤ − 1 , consider the bounded set D

D = { ( S , I , P ) : ε ≤ S ≤ 1 ε , ε ≤ I ≤ 1 ε , ε ≤ P ≤ 1 ε } ,

then ℝ + 3 \ D = D 1 c ∪ D 2 c ∪ D 3 c ∪ D 4 c ∪ D 5 c ∪ D 6 c , with

D 1 c = { ( S , I , P ) ∈ ℝ + 3 : 0 < S < ε } , D 2 c = { ( S , I , P ) ∈ ℝ + 3 : 0 < I < ε } ,

D 3 c = { ( S , I , P ) ∈ ℝ + 3 : 0 < P < ε } , D 4 c = { ( S , I , P ) ∈ ℝ + 3 : S > 1 ε } ,

D 5 c = { ( S , I , P ) ∈ ℝ + 3 : I > 1 ε } , D 6 c = { ( S , I , P ) ∈ ℝ + 3 : P > 1 ε } ,

and ε is a sufficiently small positive constant satisfying the following conditions

− M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + b 2 M ε d 1 + f 1 ( S ) + f ¯ 2 ( I ) + f 3 ( P ) ≤ − 1 , b 2 M ε d 1 < 1 2 ( d 1 − σ 2 2 2 ) , (14)

− M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + b 2 M ε d 1 + f ¯ 1 ( S ) + f 2 ( I ) + f 3 ( P ) ≤ − 1 , b 2 M ε d 1 < r 2 k , (15)

− M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + f 1 ( S ) + f 2 ( I ) − 1 c 2 ( d 2 − σ 3 2 2 ) P 2 + 3 r 5 c ε 5 3 ≤ − 1 , (16)

− M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + A − r 2 k ε 3 ≤ − 1, (17)

− M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + B − 1 2 ε 2 ( d 1 − σ 2 2 2 ) ≤ − 1, (18)

− M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + C − 1 2 ε 2 c 2 ( d 2 − σ 3 2 2 ) ≤ − 1, (19)

where

A = sup ( S , I , P ) ∈ ℝ + 3 { f ¯ 1 ( S ) + f 2 ( I ) + f 3 ( P ) + 2 5 b 2 M d 1 S 5 2 + 3 5 b 2 M d 1 I 5 3 } < ∞ ,

B = sup ( S , I , P ) ∈ ℝ + 3 { f 1 ( S ) + f ¯ 2 ( I ) + f 3 ( P ) + 2 5 b 2 M d 1 S 5 2 + 3 5 b 2 M d 1 I 5 3 } < ∞ ,

C = sup ( S , I , P ) ∈ ℝ + 3 { f 1 ( S ) + f 2 ( I ) + f ¯ 3 ( P ) + 2 5 b 2 M d 1 S 5 2 + 3 5 b 2 M d 1 I 5 3 } < ∞ .

Case 1. If ( S , I , P ) ∈ D 1 c , S I < ε I < ε ( 1 + I 2 ) , we have

L V ≤ − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + b 2 M ε d 1 + f 1 ( S ) + [ − d 1 − σ 2 2 2 2 + b 2 M ε d 1 ] I 2 + f ¯ 2 ( I ) + f 3 ( P ) ≤ − 1 ,

which follows from (14), where

f ¯ 2 ( I ) = − 1 2 ( d 1 − σ 2 2 2 ) + 3 r 5 I 5 3 .

Case 2. If ( S , I , P ) ∈ D 2 c , S I < ε S < ε ( 1 + S 3 ) , we have

L V ≤ − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + b 2 M ε d 1 + f ¯ 1 ( S ) + ( − r 2 k + b 2 M ε d 1 ) S 3 + f 2 ( I ) + f 3 ( P ) ≤ − 1 ,

which follows from (15), where

f ¯ 1 ( S ) = − r 2 k S 3 + ( 2 r 5 + 2 r 5 c ) S 5 2 + ( σ 1 2 2 + r ) S 2 .

Case 3. If ( S , I , P ) ∈ D 3 c , S I ≤ 2 5 S 5 2 + 3 5 I 5 3 , we have

L V ≤ − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + f 1 ( S ) + 2 5 b 2 M d 1 S 5 2 + f 2 ( I ) + 3 5 b 2 M d 1 I 5 3 − 1 c 2 ( d 2 − σ 3 2 2 ) P 2 + 3 r 5 c ε 5 3 ≤ − 1 ,

which follows from (16).

Case 4. If ( S , I , P ) ∈ D 4 c , S I ≤ 2 5 S 5 2 + 3 5 I 5 3 , − r 2 k S 3 < − r 2 k ε 3 , we have

L V ≤ M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + f ¯ 1 ( S ) + 2 5 b 2 M d 1 S 5 2 + f 2 ( I ) + 3 5 b 2 M d 1 I 5 3 + f 3 ( P ) − r 2 k ε 3 = − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + A − r 2 k ε 3 ≤ − 1 ,

which follows from (17).

Case 5. If ( S , I , P ) ∈ D 5 c , S I ≤ 2 5 S 5 2 + 3 5 I 5 3 , − 1 2 ( d 1 − σ 2 2 2 ) I 2 < − 1 2 ε 2 ( d 1 − σ 2 2 2 ) , we have

L V ≤ − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + f 1 ( S ) + 2 5 b 2 M d 1 S 5 2 + f ¯ 2 ( I ) + 3 5 b 2 M d 1 I 5 3 + f 3 ( P ) − r 2 k ε 3 = − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + B − 1 2 ε 2 ( d 1 − σ 2 2 2 ) ≤ − 1 ,

which follows from (18).

Case 6. If ( S , I , P ) ∈ D 6 c , S I ≤ 2 5 S 5 2 + 3 5 I 5 3 , − 1 2 c 2 ( d 2 − σ 3 2 2 ) P 2 < − 1 2 c 2 ε 2 ( d 2 − σ 3 2 2 ) , we have

L V ≤ − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + f 1 ( S ) + 2 5 b 2 M d 1 S 5 2 + f 2 ( I ) + 3 5 b 2 M d 1 I 5 3 + f ¯ 3 ( P ) − 1 2 c 2 ε 2 ( d 2 − σ 3 2 2 ) = − M [ r b k ( d 1 + β m + σ 2 2 2 ) ( R ^ 0 s − 1 ) ] + C − 1 2 ε 2 c 2 ( d 2 − σ 3 2 2 ) ≤ − 1 ,

which follows from (19), where f ¯ 3 ( P ) = − 1 2 c 2 ( d 2 − σ 3 2 2 ) P 2 + 3 r 5 c P 5 3 .

The proof of B.2 in Lemma 4.1 is completed. We get the conclusion that L V ≤ 1 on D ⊂ ℝ + 3 .

Direct computation shows that the diffusion matrix of system (2) is given by

A = ( σ 1 2 S 2 0 0 0 σ 2 2 I 2 0 0 0 σ 3 2 P 2 ) .

It is clearly that the matrix A is positive definite for any compact subset of ℝ + 3 . The proof of B.1 in Lemma 4.1 is verified. According to Lemma 4.1, we know system (2) admits a stationary distribution.

In this section, we conduct Example 1 - 3 numerical simulations to show the effect of noise on the dynamics of the system. Applying Milstein’s higher-order method [

{ S k + 1 = S k + ( r S k ( 1 − S k k ) − α S k P k m P k + S k + I k − b S k I k ) Δ t + σ 1 S k Δ t ξ k + σ 1 2 2 S k ( Δ t ξ k 2 − Δ t ) , I k + 1 = I k + ( b S k I k − d 1 I k − β I k P k m P k + S k + I k ) Δ t + σ 2 I k Δ t η k + σ 2 2 2 I k ( Δ t η k 2 − Δ t ) , P k + 1 = P k + ( c α S k P k m P k + S k + I k + c β I k P k m P k + S k + I k − d 2 P k ) Δ t + σ 3 P k Δ t ς k + σ 3 2 2 P k ( Δ t ς k 2 − Δ t ) , (20)

where the time increment Δ t > 0 , ξ k , η k and ς k , k = 1 , 2 , 3 , ⋯ , n , are independent Gaussian random variables with normal distribution N ( 0,1 ) , and σ i , 1 ≤ i ≤ 3 , are noise intensities. The unit of all time is days, the unit of the population S ( t ) , I ( t ) , P ( t ) are the individual.

Example 1. Choose the initial value S ( 0 ) = 5 , I ( 0 ) = 5 , P ( 0 ) = 2 , and r = 0.1 , k = 9 , b = 0.007 , m = 1 , α = 0.02 , β = 0.02 , c = 0.8 , d 1 = 0.1 , d 2 = 0.05 .

By simple computations,

R 0 = 0.63 < 1 , α + β = 0.04 < 0.075 = d 2 + σ 3 2 2 c , R 0 s = 0.5925 < 1.

The numerical simulation is shown in

Example 2. Choose the initial value S ( 0 ) = 5 , I ( 0 ) = 5 , P ( 0 ) = 2 , and r = 0.1 , k = 9 , b = 0.015 , m = 1 , α = 0.02 , β = 0.02 , c = 0.8 , d 1 = 0.1 , d 2 = 0.05 .

By simple computations,

R 0 = 1.35 > 1 , α + β = 0.04 < 0.075 = d 2 + σ 3 2 2 c , R 0 s = 1.2214 > 1.

The conclusion of Theorem 3.3 holds, and the numerical simulation is shown in

Example 3. Choose the initial value S ( 0 ) = 5 , I ( 0 ) = 5 , P ( 0 ) = 5 , and r = 0.1 , k = 9 , b = 0.007 , m = 1 , α = 0.02 , β = 0.02 , c = 0.8 , d 1 = 0.01 , d 2 = 0.01 , the (a) of

Under this condition, by simple computations, (a) and (b) of

R ^ 0 s = 1.2099 > 1 , σ 2 2 = 0.0004 < 0.02 = 2 d 1 , σ 3 2 = 0.0004 < 0.02 = 2 d 2 .

R ^ 0 s = 1.2160 > 1 , σ 2 2 = 0.0001 < 0.02 = 2 d 1 , σ 3 2 = 0.0001 < 0.02 = 2 d 2 .

The numerical simulation is shown in

A stochastic predator-prey model with ratio-dependent functional response and disease in the prey population is investigated. We have shown that the existence of a unique positive global solution of the stochastic model (2). The threshold R 0 s between stochastic persistence in the mean and extinction for infectious

prey population is given. If r > σ 1 2 2 , α + β < d 2 + σ 3 2 2 c and R 0 s < 1 , we obtained that the population of the infected prey and the predator will die out in time mean sense. If r > σ 1 2 2 , α + β < d 2 + σ 3 2 2 c and R 0 s > 1 , we note that the susceptible prey and the infectious prey are persistent and predator is going to die out. By constructing some suitable Lyapunov function, the existence of stationary distribution for both populations is established under certain parametric restrictions. If R ^ 0 s > 1 , and environmental noises are small enough that σ 2 2 < 2 d 1 and σ 3 2 < 2 d 2 , then for any initial value ( S ( 0 ) , I ( 0 ) , P ( 0 ) ) ∈ ℝ + 3 , there exists a unique stationary distribution for system (2) and it is ergodic.

The bell for graduation sounded immediately, and three years passed unconsciously. Looking back on the three years, I find that there are sweet and bitter, tension, anxiety, joy, joy and tears. I am very grateful to my tutor, friends and family for their concern, care and tolerance to me. During my postgraduate period, my tutor Ma Jiying was strict in study and gave careful guidance. She also cared about my life and gave me meticulous care. I am very grateful to my mentor Ma Jiying. From I began writing my paper, horse the teacher give me a clear research direction are pointed out, in many times in the discussion class for my proposed amendments, encounter problems ma teacher and I discuss research, finally during the period of thesis writing finalized, teacher ma for all aspects of the draft, including writing format, paper typesetting, English grammar, logic of paper, proof of modified earnestly repeatedly. Every time we met, Ms. Ma urged me and my sister to understand the principle of each theorem and to derive it by ourselves. We should not only know how it is, but also know why it is so that we can successfully complete my graduation thesis. I have also learned from Mr. Ma a lot of hard research, serious study of the spirit of scientific research, here, I would like to express the most sincere thanks to my respected mentor.

Second, thank Wei Guoliang dean, professor held, Liu Xiping professor, the original three led, professor jammeh, an associate professor in the school and they taught me a very rich academic professional theory knowledge, really thank ginger teachers and Wang Ting counselor, would also like to thank the teacher elder sister of my horse sasha and yi qing, each encounter problems they were positive and I explore together, thanks to pool Lin Wei accompany and encouragement, I thank my roommate for the care and help.

Finally, I would like to express my heartfelt thanks to my parents and my boyfriend for their care and support and help in my study. Thank you all.

The authors declare no conflicts of interest regarding the publication of this paper.

Ma, J.Y. and Ren, H.M. (2021) On the Dynamics of a Stochastic Ratio-Dependent Predator-Prey System with Infection for the Prey. Open Journal of Applied Sciences, 11, 440-457. https://doi.org/10.4236/ojapps.2021.114032