Citation: Dongfu Tong, Yongli Cai, Bingxian Wang, Weiming Wang. Bifurcation structure of nonconstant positive steady states for a diffusive predator-prey model[J]. Mathematical Biosciences and Engineering, 2019, 16(5): 3988-4006. doi: 10.3934/mbe.2019197
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Historically, hypergeometric functions emerged during the 18th century in the work of mathematicians such as Euler and Gauss, who developed theories around hypergeometric series and associated integrals. These functions were first considered as generalizations of geometric series and solutions of higher-order linear differential equations. Today, hypergeometric functions are at the center of an entire field of research, that of special functions [1]. Moreover, it is now well known that most special functions can be expressed in terms of hypergeometric functions, and these functions can be represented using either series or integrals. This dual representation makes it an excellent tool for evaluating series and integrals or solving differential equations. Hypergeometric functions have a wide range of applications. From our literature review, we found that these functions are utilized to express solutions in various fields, including probability and statistics [2,3,4], combinatorics and number theory [5,6,7], random walks [8,9,10], random graphs [11], quantum mechanics (see [12], p. 89, 96, 127 and [13], p. 235, 290, 333), conformal mapping [14], and fractional hypergeometric differential equations [15,16], among many other problems.
Hypergeometric functions usually have no explicit expression and are only represented by power series or integrals, which makes their evaluation time-consuming. Research on hypergeometric functions is divided into two main branches: continuous and discrete. The continuous branch focuses on the analytical study of these functions, treating their arguments as continuous variables (see [17,18,19,20,21,22,23,24]). In contrast, the discrete branch examines these functions by substituting their arguments with integers or specific values. Research in the discrete branch has surged in the past decade, driven by advancements in powerful computer algebra software and the wide range of problems that can be solved using hypergeometric functions. This study belongs to the discrete branch and aims to provide new results for different families of the generalized hypergeometric function 3F2(1). The literature includes several explicit forms of 3F2(a1,a2,a3;b1;b2;1), for particular choices of the parameters (a1,a2,a3,b1,b2). Thus, in [25], the authors used the Gamma function to derive explicit forms for 3F2(a,b,c;1+a−b,1+a−c;1) and 3F2(a,b,c;1+a−b,a+2b−c−1;1). Moreover, in [26], the authors used specialized software and managed to generate, without any mathematical proof, about thirty explicit formulas of 3F2(a1,a2,a3;b1;b2;1), for particular choices of the parameters (a1,a2,a3,b1,b2). Furthermore, in [27], the authors exhibited the explicit expressions of 3F2(−2n,a,1+d;2a+1,d;2) and 3F2(−2n−1,a,1+d;2a+1,d;2) for n∈N. Besides, in [28], the authors succeeded in determining the explicit expressions of the following sequences:
3F2(16,56,12+n;1,32+n;1),3F2(16,56,−12−n;1,12−n;1),3F2(16,56,13+n;1,43+n;1),3F2(16,56,−13−n;1,23−n;1),3F2(16,56,23+n;1,53+n;1),3F2(16,56,−23−n;1,13−n;1),3F2(16,56,14+n;1,54+n;1),3F2(16,56,−14−n;1,34+n;1),3F2(16,56,34+n;1,74+n;1),3F2(16,56,−14−n;1,14−n;1), |
for all n∈N. Finally, in [29], the authors provided the explicit forms of the sets 3F2(2x,2x+12,x;12,1+x;1) and 3F2(2x,2x−12,x;32,1+x;1), for all x∈(−∞,14).
The main objective of our study is to find explicit forms for the sets:
K(α)=3F2(1−α,1,α+1;α+1,α+2;1),α∈(12,+∞),G(α)=3F2(1−α,1,2+α;1+α,3+α;1),α∈(12,+∞),H(p)=3F2(12−p,1,1+2p;32+p,2p+2;1),p∈N∗. | (1.1) |
These sets emerge naturally from an exact evaluation of certain classes of fractional integrals, as we show in Section 4.
In order to understand the notation used above and throughout, we present in this section some basic notations, definitions, and intermediate results, which will be useful to justify certain passages in the proofs of this manuscript. Let us now present a set of symbols and notations.
● N, Z, R, and C denote the sets of non-negative integers, integers, real numbers, and complex numbers, respectively,
● X∗ denotes any set X∖{0},
● Z−0 denotes set −N={…,−n,…,−1,0},
● B(0,1)={z∈C||z|<1},
● ¯B(0,1)={z∈C||z|≤1},
● R(z) denotes the real part of z.
◼ Now we present some basic notations, definitions, and intermediate results related to the so-called the Gauss hypergeometric function.
Definition 2.1. [30] The Euler gamma function Γ(z) is defined by
Γ(z)=∫∞0tz−1e−tdt, ∀z∈C|R(z)>0. | (2.1) |
Using integration by parts, one sees that
Γ(z+1)=zΓ(z), ∀R(z)>0. | (2.2) |
The extension of the Euler gamma function to the half-plane R(z)≤0 is given by
Γ(z)=Γ(z+k)(z)k, (R(z)>−k; k∈N∗;z∉Z−0), |
where (z)k is the Pochhammer symbol defined for all z∈C and k∈N∗ by
(z)0=1 and (z)k=z(z+1)⋯(z+k−1), ∀k∈N. | (2.3) |
Relations (2.2) and (2.3) give
Γ(k+1)=(1)k=k!, ∀k∈N. | (2.4) |
Definition 2.2. [30] The Gauss hypergeometric function 2F1(a,b;c;z) is defined in the unit disk as the sum of the hypergeometric series as follows:
2F1(a,b;c;z)=∞∑k=0(a)k(b)k(c)kzkk!,(a,b∈C;c∈C∖Z−0;z∈¯B;R(c−b−a)>0). | (2.5) |
Furthermore, if 0<R(b)<R(c) and |arg(1−z)|<π, then 2F1(a,b;c;z) is given by the following Euler integral representation:
2F1(a,b;c;z)=Γ(c)Γ(b)Γ(c−b)∫10xb−1(1−x)c−b−1(1−zx)−adx. | (2.6) |
If z=1 with R(c−b−a)>0, the Gauss hypergeometric function has the following property:
2F1(a,b;c;1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b). | (2.7) |
A natural extension of 2F1 to 3F2 is defined by
3F2(a,b,c;d,e;z)=+∞∑k=0(a)k(b)k(c)k(d)k(e)kzkk!, |
∀(z∈¯BandR(d+e−a−b−c)>0). |
In [31] Theorem 38, Rainville proves a general integral representation for p+kFq+k, but here we state the following three special cases,
Case 1: Let p=2, q=k=1, and choose a=1−α, b=1, c=α+1, d=α+1, e=α+2, and z=1. Then, 3F2 is given by the following integral representation:
3F2(1−α,1,α+1;α+1,α+2;1)=(α+1)∫10xα2F1(1−α,1;α+1;x)dx. | (2.8) |
Case 2: Let p=2, q=k=1, and choose a=1−α, b=1, c=α+2, d=α+1, e=α+3, and z=1. Then, 3F2 is given by the following integral representation:
3F2(1−α,1,α+2;α+1,α+3;1)=(α+2)∫10xα+12F1(1−α,1;α+1;x)dx. | (2.9) |
Case 3: Let p=2, q=k=1, and choose a=1−α, b=1, c=2α, d=α+1, e=2α+1, and z=1. Then, 3F2 is given by the following integral representation:
3F2(1−α,1,2α;α+1,2α+1;1)=2α∫10x2α−12F1(1−α,1;α+1;x)dx. | (2.10) |
◼ The following is the definition of the Riemann-Liouville fractional integral Iαf of order α.
Definition 2.3. [30] Let Ω=[τ,η]. The Riemann-Liouville fractional integral Iαf of order α∈C (R(α)>0) is defined by
Iαf(t)=1Γ(α)∫tτ(t−s)α−1f(s)ds,∀t>τandR(α)>0. | (2.11) |
In this section, we establish some results which will play an important role herein. We believe that some of these results may be new.
◼ To justify the interchangeability between the integral and the sum, or to rewrite certain integrals, we give the following two lemmas that we will refer to several times in our work.
Lemma 2.1. Let a,b,c∈C, c∉Z−0, and R(c−a−b)>0. Then, the series
∞∑k=0(a)k(b)k(c)kzk, |
is normally convergent on the interval [−1,1].
Proof of Lemma 2.1. Since a,b,c∈C, c∉Z−0, and R(c−a−b)>0, the Gauss hypergeometric function
2F1(a,b;c;z)=∞∑k=0(a)k(b)k(c)kzkk!, | (2.12) |
is defined for any complex number z∈¯B(0,1). Moreover, the series (2.12) is absolutely convergent for all z=1. Therefore, the series ∑uk is convergent, where uk is defined by
uk=|(a)k(b)k(c)k|,∀k∈N. |
Furthermore, it is clear that for all z∈[−1,1],k∈N, we have
|(a)k(b)k(c)kzk|≤uk. | (2.13) |
The convergence of the series ∑uk together with the relation (2.13) leads to the normal (therefore uniform) convergence of the series ∑(a)k(b)k(c)kzk on the interval [−1,1]. The proof is complete.
Lemma 2.2. Let Ω=[τ,η] (−∞<τ<η<∞) be a finite interval on the real axis R, and α>1/2. Then,
∫ητ+∞∑k=0(1−α)k(α+1)k(t−τη−τ)kdt=+∞∑k=0(1−α)k(α+1)k∫ητ(t−τη−τ)kdt. | (2.14) |
Proof of Lemma 2.2. If we take a=1−α, b=1, and c=α+1, then c∉Z−0 and R(c−a−b)=2α−1>0, since α>1/2. Then by Lemma 2.1, the series ∑(a)k(b)k(c)kzk converges normally on the interval [0,1]⊂[−1,1]. Consequently, for all α>1/2, we have
∫10+∞∑k=0(1−α)k(α+1)kzkdz=+∞∑k=0(1−α)k(α+1)k∫10zkdz. | (2.15) |
By using the change of variable with z=t−τη−τ, for relation (2.15) we have
1(η−τ)∫ητ+∞∑k=0(1−α)k(α+1)k(t−τη−τ)kdt=+∞∑k=0(1−α)k(α+1)k1(η−τ)∫ητ(t−τη−τ)kdt. | (2.16) |
Thus, we have obtained (2.14). The proof is complete.
◼ We now establish some results, which we believe are new. These results give the limit of some series.
◼ The following lemma gives the sum of the series Sα,0 defined by the left-hand side of relation (2.17).
Lemma 2.3. For all α>1/2, we have
Sα,0=+∞∑k=0(1−α)k(α+1)k(k+α+1)=12α. | (2.17) |
Proof of Lemma 2.3. For all k∈N, let uk be the general term of the series Sα,0 given by the relation (2.17). If we take a=1−α, b=1, and c=α+1, then c∉Z−0 and R(c−a−b)=2α−1>0, since α>1/2. Then, by Lemma 2.1 the series ∑(1−α)k(α+1)k is absolutely convergent, and since 0<uk≤(1−α)k(α+1)k for all k∈N, we then deduce that the series Sα,0 is absolutely convergent. Thus, multiplying and dividing the term uk by (0−α), we obtain
Sα,0=+∞∑k=0(0−α)(1−α)k(0−α)(α+1)k(k+α+1)=+∞∑k=0(−α)k+1−α(α+1)k+1=−1α[+∞∑k=1(−α)k(α+1)k]=−1α[+∞∑k=0(−α)k(α+1)k−1]=1α[1−2F1(−α,1;α+1;1)]. | (2.18) |
From the property of 2F1, given by (2.7) we have
2F1(−α,1;α+1;1)=12, | (2.19) |
and so substituting (2.19) into (2.18), we obtain (2.17). This completes the proof.
◼ The following lemma gives the sum of the series Sα,1 defined by the left-hand side of relation (2.20).
Lemma 2.4. For all α>1/2, we have
Sα,1=+∞∑k=0(1−α)k(α+1)k(α+2+k)=1α+1+12α−22α+1. | (2.20) |
Proof of Lemma 2.4. The proof is similar to the proof of Lemma 2.3. If we take a=1−α, b=1, and c=α+1, then c∉Z−0 and R(c−a−b)=2α−1>0, since α>1/2. We deduce that the series Sα,1 is absolutely convergent. Thus, multiplying and dividing the term uk by the same quantity (k+α+1), we obtain
uk=(1−α)k(k+α+1)(α+1)k(α+1+k)(α+1+k+1)=(1−α)k(1−α+k+2α)(α+1)k+2=(1−α)k+1(α+1)k+2+2α(1−α)k(α+1)k+2. | (2.21) |
Applying the identity (a)k+1=a(1+a)k to α+1, we obtain the following two relations:
(α+1)k+2=(α+1)(α+2)k+1,(α+1)k+2=(α+1)(α+2)(α+3)k, | (2.22) |
and using the above relations and the fact that (a)0=1, the series Sα,1 can be rewritten as follows:
Sα,1=1α+1+∞∑k=0(1−α)k+1(α+2)k+1+2α(α+1)(α+2)+∞∑k=0(1−α)k(α+3)k=1α+1+∞∑k=1(1−α)k(α+2)k+2α(α+1)(α+2)+∞∑k=0(1−α)k(α+3)k=1α+1[+∞∑k=0(1−α)k(α+2)k−1]+2α(α+1)(α+2)+∞∑k=0(1−α)k(α+3)k=1α+1[2F1(1−α,1;α+2;1)−1]+2α(α+1)(α+2)2F1(1−α,1;α+3;1). |
From the property of 2F1 given by (2.7), we have
2F1(1−α,1;α+2;1)=α+12α, | (2.23) |
and
2F1(1−α,1;α+3;1)=α+22α+1. | (2.24) |
Thus, we obtain
Sα,1=1α+1[α+12α−1]+2α(α+1)(α+2)α+22α+1=1α+1+12α−2(2α+1). | (2.25) |
The proof is complete.
◼ The following lemma gives the limit of the series Sα,2 defined by the left-hand side of relation (2.26).
Lemma 2.5. For all α>1/2, we have
Sα,2=+∞∑k=0(1−α)k(α+1)k(k+2α)=12α3F2(1−α,1,2α;α+1,2α+1;1). | (2.26) |
Proof of Lemma 2.5. If a=1−α, b=1, and c=α+1, then c∉Z−0, and R(c−a−b)=2α−1>0, since α>1/2. Then, expressing the rightside of (2.10) as a series and changing the order of integration and summation which is justified by Lemma 2.1 (due to the uniform convergence of the series) gives
2α∫10x2α−12F1(1−α,1;α+1;x)dx=2α+∞∑k=0(1−α)k(1)k(α+1)k∫10x2α−1xkk!dx=2α+∞∑k=0(1−α)k(α+1)k1(k+2α). |
Thus, we obtained (2.26). The proof is complete.
◼ The following lemma gives the sum of the series Sα,3 defined by the left-hand side of relation (2.27).
Corollary 2.1. For all α>1, we have
Sα,3=+∞∑k=0(2−α)k(α+1)k(2α+k)=α(1−α)(2α−1)+[1−3α2α(1−α)]3F2(1−α,1,2α;α+1,2α+1;1). | (2.27) |
Proof of Corollary 2.1. The proof is similar to the proof of Lemma 2.4, and so we just sketch the basic idea. If we take a=2−α, b=1, and c=α+1, then c∉Z−0 and R(c−a−b)=2α−2>0, since α>1. Then, by Lemma 2.1 we deduce that the series Sα,3 is absolutely convergent. Thus, multiplying and dividing the term uk by the same quantity (1−α), we obtain
Sα,3=+∞∑k=0(2−α)k(α+1)k(2α+k)=+∞∑k=0(1−α)(2−α)k(α+1)k(1−α)(2α+k)=+∞∑k=0(1−α)k+1(1−α)(α+1)k(2α+k)=11−α[+∞∑k=0(1−α)k(k+1−α)(α+1)k(2α+k)]=11−α[+∞∑k=0(1−α)k(k+2α+1−3α)(α+1)k(2α+k)]=11−α[+∞∑k=0(1−α)k(α+1)k+(1−3α)+∞∑k=0(1−α)k(α+1)k(2α+k)]=11−α[2F1(1−α,1;α+1;1)+(1−3α)+∞∑k=0(1−α)k(α+1)k(2α+k)]. |
Above we use the property of 2F1 (2.7) and the result of Lemma 2.5 to obtain (2.27). The proof is complete.
◼ In the following corollary, we give the calculation of 3F2(−1,1,4;3,5;1), which we believe may be new.
Corollary 2.1.
3F2(−1,1,4;3,5;1)=1115. | (2.28) |
Proof of Corollary 2.1. Note that when α=2, the numerical series introduced in Lemmas 2.4 and 2.5 coincide. Therefore, the right-hand terms of relations (2.20) and (2.26) are equal when α=2. Thus,
3F2(−1,1,4;3,5;1)=4(13+14−25)=73−85=1115. | (2.29) |
The proof is complete.
This section explores specific subfamilies of 3F2(1) and is structured into two subsections. The first subsection presents the explicit forms of the subfamilies {3F2(1−α,1,α+1;α+1,α+2;1)} and {3F2(1−α,1,α+2;α+1,α+3;1)} for all α>12, while the second subsection provides the explicit form of the subfamily {3F2(12−p,1,1+2p;32+p,2p+2;1)} for all p∈N∗.
This part aims to find the explicit form of the function K(α)=:3F2(1−α,1,α+1;α+1,α+2;1) (Theorem 3.1). To do this, we write K in the form of a series of functions (Lemma 3.1), and then we use the result of Lemma 2.3 to prove our main result of this section, Theorem 3.1.
◼ In the following lemma, we express the function K(α) as a series.
Lemma 3.1. For all α>1/2, we have
3F2(1−α,1,α+1;α+1,α+2;1)=(α+1)+∞∑k=0(1−α)k(α+1)k(k+α+1). | (3.1) |
Proof of Lemma 3.1. If a=1−α, b=1, and c=α+1, then c∉Z−0 and R(c−a−b)=2α−1>0, since α>1/2. Then, expressing the rightside of (2.8) as a series, changing the order of integration and summation, which is justified by Lemma 2.1, and applying the steps of the proof of Lemma 2.5, we can readily derive the proof of Lemma 3.1.
◼ Now we state and prove our main result of this section, Theorem 3.1.
Theorem 3.1. For all α>1/2, we have
3F2(1−α,1,α+1;α+1,α+2;1)=α+12α. | (3.2) |
Proof of Theorem 3.1. By identifying relations (2.17) and (3.1), we obtain
3F2(1−α,1,α+2;α+1,α+3;1)=(α+1)(12α)=α+12α. |
The proof is complete.
This part aims to find the explicit form of the function G(α)=:3F2(1−α,1,α+2;α+1,α+3;1) (Theorem 3.2). To do this, we write G in the form of a series of functions (Lemma 3.2), then we use the result of Lemma 2.4 to prove our main result of this section, Theorem 3.2. Our argument in this section is similar to the previous section.
◼ In the following lemma, we express the function G(α) as a series.
Lemma 3.2. For all α>1/2, we have
3F2(1−α,1,α+2;α+1,α+3;1)=(α+2)+∞∑k=0(1−α)k(α+1)k(k+α+2). | (3.3) |
Proof of Lemma 3.2. If a=1−α, b=1, and c=α+1, then c∉Z−0 and R(c−a−b)=2α−1>0, since α>1/2. Then, expressing the rightside of (2.9) as a series, changing the order of integration and summation, which is justified by Lemma 2.1, and applying the steps of the proof of Lemma 2.5, we can readily derive the proof of Lemma 3.2.
◼ Now we state and prove our main result of this section, Theorem 3.2.
Theorem 3.2. For all α>1/2, we have
3F2(1−α,1,α+2;α+1,α+3;1)=12+1α+1α+1−32α+1. | (3.4) |
Proof of Theorem 3.2. By identifying relations (2.20) and (3.3), we obtain
3F2(1−α,1,α+2;α+1,α+3;1)=(α+2)(1α+1+12α−22α+1)=12+1α+1α+1−32α+1. |
The proof is complete.
This section deals with the family of functions F(α)=3F2(1−α,1,2α;α+1,2α+1;1). It is easy to find a representation for it by a series of functions ∑+∞k=0fk(α) (see Lemma 2.5). However, it is not obvious to find an explicit one unless α∈N∗ or a rational number of the form α=2p+12 and p∈N∗. Note that when α=p∈N∗, then ∑+∞k=0fk(α) is equal to ∑p−1k=0fk(α). Therefore, this case will be excluded from this study. In summary, this part aims to determine the explicit form of F(α) when α=2p+12, which coincides with 3F2(12−p,1,1+2p;32+p,2p+2;1) and which we will designate by H(p). To do this, we first calculate H(1) in Section 3.3.1 to make the calculation of H(p) easy to follow in Section 3.3.2.
◼ The main result of this section is summarized in the following theorem. The proof of this theorem is postponed to the end of this section.
Theorem 3.3. For all p∈N∗, we have
3F2(12−p,1,1+2p;32+p,2p+2;1)=−(2p+1)Kp(2a2p+1ln(2)+2a2p+1(Ep+Bp)+2p∑j=1ajBj), |
where
Kp=(1−2p)(3−2p)(5−2p)⋯(−1+2p)(1+2p)a2p+1=−22p(2p)!(3p)!(6p+1)!(p)!Ep=2p∑k=112kBj=j−1∑k=012k−2p+1,j=1,…,2paj=2(−1)j(2)2p(j)!(2p−j)!(6p−2j+1),j=1,…,2p. | (3.5) |
This part aims to find the explicit form of the function H(1). To do this, we write H(1) in the form of a numerical series. Then, we determine the limit of this series.
◼ In Lemma 3.3, we give the exact value of 3F2(−12,1,3;52,4;1).
Lemma 3.3.
3F2(−12,1,3;52,4;1)=3335−635ln(2). | (3.6) |
Proof of Lemma 3.3. For p=1 (i.e., α=32), after simplifications, relation (2.26) gives
−3+∞∑k=01(2k−1)(2k+1)(2k+3)(k+3)=133F2(−12,1,3;52,4;1), | (3.7) |
that is, 3F2(−12,1,3;52,4;1)=−9S, where
S=+∞∑k=01(2k−1)(2k+1)(2k+3)(k+3). | (3.8) |
△ The decomposition into partial fractions of the general term uk, of the series S=∑uk, gives
uk=1(2k−1)(2k+1)(2k+3)(k+3)=128(2k−1)−110(2k+1)+112(2k+3)−1105(k+3). | (3.9) |
△ Let n be a fixed positive integer. Then, (Sn)n≥0 and (Tn)n≥1 are sequences defined by
Sn=n∑k=01(2k−1)(2k+1)(2k+3)(k+3)Tn=n∑k=112k−1. | (3.10) |
△ We will simplify the partial sum Sn in order to find its limit when n tends to infinity. To do this, we will express the partial sums,
Un=n∑k=012k−1;Vn=n∑k=012k+1;Wn=n∑k=012k+3, |
as a function of Tn. It is easy to check the following equalities:
Un=−1+TnVn=Tn+12n+1Wn=−1+Tn+12n+1+12n+3. | (3.11) |
△ We then deduce Sn as a function of Tn:
Sn=Un28−Vn10+Wn12−1105n∑k=01k+3=−542+An+Bn, | (3.12) |
where
An=2105Tn−1105n∑k=01k+3Bn=−160(2n+1)+112(2n+3). | (3.13) |
△ Furthermore, we have
n∑k=01k+3=2n∑k=012(k+3)=2n+3∑k=312k=2(−12−14+n∑k=112k+12(n+1)+12(n+2)+12(n+3))=2(−34+n∑k=112k+12(n+1)+12(n+2)+12(n+3))=2(−34+n∑k=112k+Cn), | (3.14) |
where
Cn=12(n+1)+12(n+2)+12(n+3). |
Consequently,
An=2105Tn−2105(−34+n∑k=112k+Cn)=2105(n∑k=112k−1−n∑k=112k+34−Cn)=2105(n∑k=1(−1)k+1k+34−Cn)=2105(Dn+34−Cn), | (3.15) |
where
Dn=n∑k=1(−1)k+1k. |
△ Based on relations (3.12) and (3.15), we obtain
Sn=−542+2105(Dn+34−Cn)+Bn=−11105+2105(Dn−Cn)+Bn. | (3.16) |
△ Furthermore, we know that for all real numbers x≠−1 such that |x|≤1, we have
limn→+∞n∑k=0(−1)k+1kxk=ln(1+x). | (3.17) |
Consequently,
limn→+∞Dn=ln(2). | (3.18) |
△ Moreover, it is easy to verify that
limn→+∞Bn=0,limn→+∞Cn=0. | (3.19) |
△ Finally, we have
limn→+∞Sn=−11105+2105ln(2). | (3.20) |
△ By grouping relations (3.7) and (3.20), we deduce that
3F2(−12,1,3;52,4;1)=−9(−11105+2105ln(2))=3335−635ln(2). | (3.21) |
The proof is complete.
This section explores specific subfamilies of 3F2(1) and is structured into two subsections. The first subsection presents the explicit forms of the subfamilies F(1) and F(2), while the second subsection provides the explicit form of the subfamily F(3). This part aims to find the explicit form of the function H(p), p∈N∗. To do this, we will follow the same approach as that used in Section 3.3.1 to calculate H(1).
◼ In Lemma 3.5, we give the explicit expression of (1−α)k1+α when α=2p+12.
Lemma 3.4. For all p∈N∗,k∈N, we have
(1−2p+12)k(1+2p+12)k=(1−2p)(1−2p+1))⋯(−1+2p)(1+2p)(2k+1−2p)(2k+3−2p)⋯(2k−1+2p)(2k+1+2p). | (3.22) |
Proof of Lemma 3.5. If α=2p+12, then we have
(1−α)k=(12−p)(32−p)(52−p)⋯(k−32−p)(k−12−p),(α+1)k=(32+p)(52+p)⋯(k−12+p)(k+12+p). | (3.23) |
◼ Note that when k≥2p+2, there are j=k−2p−1 terms in common between (1−α)k and (α+1)k. Indeed, we observe the first number in common if k−12−p=32+p, that is k=2p+2. Therefore, if k=2p+3, then there are two terms in common, and so on. Thus, if k≥2p+2, then (1−α)k and (1+α)k are written as follows:
(1−α)k=(12−p)⋯(12−p+2p)(32+p)⋯(k−12−p),(α+1)k=(32+p)(52+p)⋯(k−p−12)(k−p+12)⋯(k+12+p). | (3.24) |
Therefore, their quotient can be simplified as follows:
(1−α)k(α+1)k=(12−p)(12+1−p)⋯(12+2p−p)(k−p+12)(k−p+1+12)⋯(k−p+2p+12)=(12−(p−0))(12−(p−1))⋯(12−(p−2p))(k+12−(p−0))(k+12−(p−1))⋯(k+12−(p−2p))=22p+1(1−2(p−0))(1−2(p−1))⋯(1−2(p−2p))22p+1(2k+1−2(p−0))(2k+1−2(p−1))⋯(2k+1−2(p−2p))=(1−2p)(1−2p+2))⋯(−1+2p)(1+2p)(2k+1−2p)(2k+3−2p)⋯(2k−1+2p)(2k+1+2p)=(1−2p)(3−2p))⋯(−1+2p)(1+2p)(2k+1−2p)(2k+3−2p)⋯(2k−1+2p)(2k+1+2p). | (3.25) |
◼ When k≤2p+1, we will show that the quotient (1−α)k(α+1)k can be reduced to the form given by the last line of relation (3.25).
△ When k=0, we have
(1−α)0(α+1)0=11=(1−2p)(3−2p)⋯(−1+2p)(1+2p)(1−2p)(3−2p))⋯(−1+2p)(1+2p)=(1−2p)(3−2p)⋯(−1+2p)(1+2p)(2×0+1−2p)(2×0+3−2p))⋯(2×0−1+2p)(2×0+1+2p). | (3.26) |
△ When k=1, we have
(1−α)1(α+1)1=1−2p(3+2p)=(1−2p)[(3−2p)⋯(−1+2p)(1+2p)][(3−2p)⋯(−1+2p)(1+2p)](3+2p)=(1−2p)(3−2p)⋯(−1+2p)(1+2p)(2×1+1−2p)(2×1+3−2p)⋯(2×1−1+2p)(2×1+1+2p). | (3.27) |
△ More generally, when k is an integer such that 0≤k≤2p+1, we have
(1−α)k(α+1)k=(1−2p)(3−2p)⋯(2k−3−2p)(2k−1−2p)(3+2p)(5+2p)⋯(2k−1+2p)(2k+1+2p)=(1−2p)(3−2p)⋯(2k−1−2p)[(2k+1−2p)⋯(1+2p)][(2k+1−2p)⋯(1+2p)](3+2p)(5+2p)⋯(2k+1+2p)=(1−2p)(3−2p)⋯(−1+2p)(1+2p)(2k+1−2p)⋯(1+2p)(3+2p)(5+2p)⋯(2k+1+2p). | (3.28) |
Thus, we have shown that (1−α)k(α+1)k is written in the form (3.22) for all k∈N. The proof is complete.
Remark 3.1. For α=2p+12, we deduce from Lemma 3.5 that
(1−α)k(α+1)k(k+2p+1)=Kpuk, | (3.29) |
where
Kp=(1−2p)(3−2p)⋯(−1+2p)(1+2p)uk=1(2k+1−2p)(2k+3−2p)⋯(2k+1+2p)(k+2p+1). | (3.30) |
◼ From the above remark, the decomposition into partial fractions of uk gives
uk=a02k+1−2p+a12k+3−2p+⋯+a2p2k+1+2p+a2p+1k+2p+1=a02k+1−2(p−0)+a12k+1−2(p−1)+⋯+a2p2k+1−2(p−2p)+a2p+1k+2p+1=2p∑i=0ai2k+1−2(p−i)+a2p+1k+2p+1. | (3.31) |
For all i=0,…,2p, to find ai, simply multiply uk by (2k+1−2(p−i)), Then, evaluate the resulting expression at k=2(p−i)−12. For now, let us calculate only the first three coefficients a0,a1, and a2.
△ For a0, we evaluate the resulting expression at k=2p−12 (i.e., 2k=2p−1), so we obtain
1a0=(2p−1+1−2(p−1))⋯(2p−1+1−2(p−2p)12(2p−1+4p+2)=(2×1)(2×2)⋯(2×2p)12(6p+1),=22p(2p)!12(6p+1)=12(−1)0(0!)22p(2p−0)!(6p−2×0+1). | (3.32) |
Consequently,
a0=2(−1)0(0!)22p(2p−0)!(6p−2×0+1). | (3.33) |
△ For a1, we evaluate the resulting expression at k=2p−32 (i.e., 2k=2p−3), so we obtain
1a1=(2p−3+1−2(p−0))⋯(2p−3+1−2(p−2p))12(2p−3+4p+2)=(−2)(2×1)(2×3)⋯(2×(2p−1))12(6p−1)=(−1)12122p−1(2p−1)!12(6p−1)=12(−1)1(1!)(2)2p(2p−1)!(6p−2×1+1). | (3.34) |
Consequently,
a1=2(−1)1(1!)(2)2p(2p−1)!(6p−2×1+1). | (3.35) |
△ For a2, we evaluate the resulting expression at k=2p−52 (i.e., 2k=2p−5), so we obtain
1a2=(2p−5+1−2(p−0))⋯(2p−5+1−2(p−2p))12(2p−5+4p+2)=(−22)(−21)(2×1)(2×3)⋯(2×(2p−2))12(6p−3)=(−1)222(2!)22p−2(2p−2)!12(6p−3)=12(−1)2(2!)(2)2p(2p−2)!(6p−2×2+1). | (3.36) |
Consequently,
a2=2(−1)2(2)2p(2p−2)!(6p−2×2+1). | (3.37) |
◼ The calculation of the first three coefficients a0,a1,a2 allowed us to guess the general expression of ai for all i=0,…,2p, which is stated in the following Lemma 3.5.
Lemma 3.5. Let p∈N∗. Then, for all i=0,…,2p, we have
ai=2(−1)i(2)2p(i)!(2p−i)!(6p−2i+1). | (3.38) |
Proof of Lemma 3.5. We will confirm this expression by a direct calculation of ai. To do this, we evaluate the resulting expression at k=2p−(2i+1)2 (i.e., 2k=2p−(2i+1)), so we obtain
1ai=(2p−(2i+1)+1−2(p−0))(2p−(2i+1)+1−2(p−1))⋯⋯(2p−(2i+1)+1−2(p−(i−1)))(2p−(2i+1)+1−2(p−(i+1)))⋯⋯(2p−(2i+1)+1−2(p−2p))12(2p−(2i+1)+4p+2)=(−2(i−0))(−2(i−1))⋯(−2(i−(i−1)))(2×1)(2×2)⋯⋯(2(2p−i))12(6p−2i+1)=12(−2)i(i)!22p−i(2p−i)!(6p−2i+1)=12(−1)i(2)i(i)!22p−i(2p−i)!(6p−2i+1)=12(−1)i(2)2p(i)!(2p−i)!(6p−2i+1). | (3.39) |
Thus, we find the expression of ai stated in relation (3.38). The proof is then complete.
◼ All that remains is to determine the expression of a2p+1. This will be the subject of Lemma 3.6.
Lemma 3.6. The last coefficient of the decomposition into partial fractions (3.31) is given by
a2p+1=−22p(2p)!(3p)!(6p+1)!(p)!. | (3.40) |
Proof of Lemma 3.6. To find a2p+1, multiply uk by (k+2p+1). Then, evaluate the resulting expression at k=−2p−1 (i.e., 2k=−4p−2). We then obtain
1a2p+1=(−4p−2−2p+1)(−4p−2−2p+3)⋯(−4p−2+2p−1)(−4p−2+2p+1)=(−6p−1)(−6p+1)(−6p+3)⋯(−2p−3)(−2p−1)=(−6p−1)(−6p)(−6p+1)(−6p+2)(−6p+3)⋯(−2p−3)(−2p−2)(−2p−1)(−6p)(−6p+2)⋯(−2p−4)(−2p−2)=(−1)2p+1(6p+1)(6p)(6p−1)(6p−2)(6p−3)⋯(2p+3)(2p+2)(2p+1)(−1)2p(6p)(6p−2)⋯(2p+4)(2p+2)=−(2p+1)(2p+2)⋯(6p+1)2(p+1)2(p+2)⋯2(3p)=−[1×2×⋯×2p](2p+1)(2p+2)⋯(6p+1)[1×2×⋯×p][1×2×⋯×2p]23p−(p+1)+1[1×2×⋯×p](p+1)(p+2)⋯(3p)=−(6p+1)!(p)!22p(2p)!(3p)!. | (3.41) |
The proof is complete.
◼ The following remark shows the validation of formulas (3.38) and (3.40) when p=1.
Remark 3.2. For p=1, we will check the concordance of the coefficients a0,…,a3 given by relations (3.40) and (3.38) with those presented in relation (3.9). When p=1, relations (3.38) and (3.40) give
a0=2(−1)0(2)2(0)!(2)!(7)=128a1=2(−1)1(2)2(1)!(1)!(5)=−110a2=2(−1)2(2)2(2)!(0)!(3)=112a3=−22(2)!(3)!(7)!(1)!=−1105. | (3.42) |
Thus, we find the same coefficients of the decomposition into partial fractions of uk given by (3.9).
◼ Lemma 3.7 presents a relation between the coefficient a2p+1 and the coefficients (ai)0≤i≤2p.
Lemma 3.7. For all p∈N∗, we have
2p∑i=0ai=−2a2p+1. | (3.43) |
Proof of Lemma 3.7. We have shown in relation (3.31) that, for all k>0, the term uk is written as a partial fraction (3.31), where (ai)0≤i≤2p and a2p+1 are given by (3.38) and (3.40).
By reducing all the partial fractions to the same denominator and identifying the numerators, we then obtain 2p+2 equations with unknowns a0, …, a2p, a2p+1. It is easy to see that the equation that relates the coefficients of the monomial k2p+1 is written in the following form ∑2pi=0k(2k)2pai+(2k)2p+1a2p+1=0, or in the equivalent form
2p∑i=0ai=−2a2p+1. | (3.44) |
The proof is complete.
◼ In the following Lemma 3.8, we establish a supporting result that arises from the calculations performed in the previous results of this section. This result provides an explicit form of a finite sum.
Lemma 3.8. For all p∈N∗, we have
2p∑i=0(−1)i(2)2p(i)!(2p−i)!(6p−2i+1)=22p(2p)!(3p)!(6p+1)!(p)!. | (3.45) |
Proof of Lemma 3.8. Based on formula (3.43), we have
2p∑i=0ai=−2a2p+1. | (3.46) |
By replacing in the previous equality the coefficients (ai)0≤i≤2p+1 by their expressions presented in relations (3.38) and (3.40), we directly obtain relation (3.45). The proof is complete.
◼ In the following section, we will prove that the series ∑uk converges and we calculate its sum, where uk is defined by (3.30).
Convergence of the series ∑uk
Let n be a fixed positive integer, and (Sn)n≥0 and (Tn)n≥1 be the sequences defined by
Sn=n∑k=0uk,Tn=n∑k=012k−2p+1. | (3.47) |
◼ We will write Sn as a function of Tn, prove that the sequence Sn converges, and determine its limit.
△ For all j=1,…,2p, we have
n∑k=012k−2(p−j)+1=n∑k=012(k+j)−2p+1=n+j∑k=j12k−2p+1=n∑k=012k−2p+1+n+j∑k=n+112k−2p+1−j−1∑k=012k−2p+1=Tn+Aj,n−Bj, | (3.48) |
where
Aj,n=n+j∑k=n+112k−2p+1,Bj=j−1∑k=012k−2p+1. | (3.49) |
Note that for j=0, we also have
n∑k=012k−2(p−0)+1=n∑k=012k−2p+1=Tn+A0,n−B0, | (3.50) |
where
A0,n=B0=0. | (3.51) |
△ Furthermore, we also have
n∑k=01k+2p+1=2n∑k=012(k+2p+1)=2n+2p+1∑k=2p+112k=2n∑k=112k+2n+2p+1∑k=n+112k−22p∑k=112k=2n∑k=112k+2Dn−2Ep, | (3.52) |
where
Dn=n+2p+1∑k=n+112k,Ep=2p∑k=112k. | (3.53) |
△ Moreover, we have
Tn=n∑k=012k−2p+1=p−1∑k=012k−2p+1+n+p∑k=p12k−2p+1=p−1∑k=012k−2p+1+n∑k=012k+1=Bp+n∑k=012k+1, | (3.54) |
where
Bp=p−1∑k=012k−2p+1. | (3.55) |
◼ After writing Sn as a function of Tn, we will inject the expression of Tn given by (3.54) into Sn to prove that Sn converges and deduce its limit.
△ Using relations (3.48), (3.54) (3.52), and (3.43), the partial sum Sn equals
Sn=2p∑j=0ajn∑k=012k−2(p−j)+1+a2p+1n∑k=01k+2p+1=2p∑j=0aj(Tn+Aj,n−Bj)+a2p+1(2n∑k=112k+2Dn−2Ep)=Tn2p∑j=0aj+2a2p+1n∑k=112k+2p∑j=1aj(Aj,n−Bj)+2a2p+1(Dn−Ep)=−2a2p+1Tn+2a2p+1n∑k=112k+2p∑j=1ajAj,n−2p∑j=1ajBj+2a2p+1Dn−2a2p+1Ep=−2a2p+1n∑k=012k+1−2a2p+1Bp+2a2p+1n∑k=112k+2p∑j=1ajAj,n−2p∑j=1ajBj+2a2p+1Dn−2a2p+1Ep=−2a2p+1(n∑k=012k+1−n∑k=112k)−2a2p+1Ep−2p∑j=1ajBj−2a2p+1Bp+2p∑j=1ajAj,n+2a2p+1Dn=−2a2p+1n∑k=1(−1)k+1k−2a2p+1Ep−2p∑j=1ajBj−2a2p+1Bp+2p∑j=1ajAj,n+2a2p+1Dn. | (3.56) |
\vartriangle Since
\begin{equation} \begin{array}{lll} \lim\limits_{n\rightarrow +\infty}\sum\limits_{k = 1}^{n} \dfrac{(-1)^{k+1}}{k}& = &\ln(2)\\ \lim\limits_{n\rightarrow +\infty}A_{j, n}& = &0, \; \forall\, j = 1, \ldots, 2p\\ \lim\limits_{n\rightarrow +\infty}D_n& = &0, \end{array} \end{equation} | (3.57) |
we deduce that S_n converges, and that its limit verifies
\begin{equation} \lim\limits_{n\rightarrow +\infty}S_n = -2a_{2p+1}\ln(2)- 2a_{2p+1}(E_p+B_p)-\sum\limits_{j = 1}^{2p} a_jB_j. \end{equation} | (3.58) |
\blacksquare The following Lemma 3.9 gives the sum of the series \sum u_k .
Lemma 3.9. For all p\in {{ \mathbb{N}}}^* , we have
\begin{equation} \sum\limits_{k = 0}^{+\infty}\dfrac{(\tfrac{1}{2}-p)_k}{(\tfrac{1}{2}+p)_k (k+2p+1)} = -K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right), \end{equation} | (3.59) |
where K_p , a_{2p+1} , E_p , B_j , and a_j are defined in (3.5).
Proof of Lemma 3.9. By grouping relations (3.29), (3.47), and (3.58), we obtain
\begin{equation} \begin{array}{lll} \sum\limits_{k = 0}^{+\infty}\dfrac{(\tfrac{1}{2}-p)_k}{(\tfrac{1}{2}+p)_k (k+2p+1)}& = & K_p\lim\limits_{n\rightarrow +\infty}S_n\\ & = & -K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right). \end{array} \end{equation} | (3.60) |
The proof is complete.
\blacksquare The following remark gives the validation of formula (3.59) when p = 1 .
Remark 3.3. For p = 1 , we will check the concordance of the limit given by relation (3.20) with that presented in relation (3.58). When p = 1 , we easily obtain E_1 = \tfrac{3}{4} , B_1 = -1 , and B_2 = 0 . Moreover, the values of (a_{i})_{0\leq i\leq 3} are given in (3.42). Thus, relation (3.58) gives
\begin{equation} \begin{array}{lll} \lim\limits_{n\rightarrow +\infty}S_n& = &-2(-\dfrac{1}{105})\ln(2)- 2(-\dfrac{1}{105})(\dfrac{3}{4}-1)-[-\dfrac{1}{10}(-1)+\dfrac{1}{12}( 0)]\\ & = & \dfrac{2}{105}\ln(2)+\dfrac{1}{105}(\dfrac{3}{2}-2)-\dfrac{1}{10}\\ & = & \dfrac{2}{105}\ln(2)-\dfrac{1}{210}-\dfrac{21}{210}\\ & = & \dfrac{2}{105}\ln(2)-\dfrac{22}{210}\\ & = & \dfrac{2}{105}\ln(2)-\dfrac{11}{105}. \end{array} \end{equation} | (3.61) |
Thus, we find the same limit as that found in relation (3.20).
\blacksquare With the intermediate results from Section 3.3 now found, we can prove Theorem 3.3 stated at the beginning of this section.
Proof of Theorem 3.3. Based on relation (2.26) and taking \alpha = \tfrac{2p+1}{2} , we obtain
\begin{equation} \begin{array}{lll} \sum\limits_{k = 0}^{+\infty}\dfrac{(\tfrac{1}{2}-p)_k}{(\tfrac{1}{2}+p)_k (k+2p+1)}& = &\dfrac{{ _3F_2(1-\alpha, 1, 2\alpha;\alpha+1, 2\alpha+1;1) }}{2\alpha} \\ & = & \dfrac{{ _3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1) }}{2p+1}. \end{array} \end{equation} | (3.62) |
By identifying the right-hand sides of relations (3.59) and (3.62), we obtain
\begin{equation*} \label{Lem7Eq3bis} -K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right) = \dfrac{{ _3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1) }}{2p+1}, \end{equation*} |
that is,
\begin{equation*} -(2p+1)K_p\left(2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right) = { _3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1) }. \end{equation*} |
The proof is complete.
\blacksquare We now state and prove the following two results of Theorem 3.3 corresponding to the special cases p = 2 and p = 3 .
\vartriangle For p = 2 , Theorem 3.3 yields the first new special result.
Corollary 3.1.
\begin{equation} {_3F_2(-\frac{3}{2}, 1, 5;\frac{7}{2}, 6;1)} = \frac{4045}{6006}+\frac{10}{1001}\ln(2). \end{equation} | (3.63) |
Proof of corollary 3.1. Let p = 2 . Then, from Theorem 3.3, we easily compute the following quantities:
\begin{eqnarray*} && K_2 = 45, \ E_2 = \frac{25}{24}, \ B_1 = -\frac{1}{3}, \ B_2 = -\frac{4}{3}, \ B_3 = -\frac{1}{3}, \ B_4 = 0, \ a_1 = -\frac{1}{528}, \\ &&a_2 = \frac{1}{288}, \ a_3 = -\frac{1}{336}, \ a_4 = \frac{1}{960}, \ a_5 = -\frac{1}{45045}, \ \sum\limits_{j = 1}^{4} a_jB_j = -\frac{25}{8316}. \end{eqnarray*} |
Thus, to derive relation (3.63), we simply substitute the values above into Theorem 3.3. The proof is complete.
\vartriangle Also, for p = 3 , Theorem 3.3 yields the second new special result.
Corollary 3.2.
\begin{equation} {_3F_2(-\frac{5}{2}, 1, 7;\frac{9}{2}, 8;1)} = \frac{221158}{415701}-\frac{70}{138567}\ln(2). \end{equation} | (3.64) |
Proof of Corollary 3.2. The proof follows similar lines of argument to that of Corollary 3.1. Let p = 3 . Then, from Theorem 3.3, we easily compute the following quantities:
\begin{equation} \begin{array}{cccccc} B_1 = -\dfrac{1}{5}, \ B_2 = -\dfrac{8}{15}, \ B_3 = -\dfrac{23}{15}, \ B_4 = -\dfrac{8}{15}, \ B_5 = -\dfrac{3}{15}, \ B_6 = 0, \\ a_1 = -\dfrac{1}{65280}, \ a_2 = \dfrac{1}{23040}, \ a_3 = -\dfrac{1}{14976}, \ a_4 = \dfrac{1}{16896}, \ a_5 = -\dfrac{1}{34560}, \\a_6 = \dfrac{1}{161280}, \ a_7 = -\dfrac{1}{43648605}, \ K_3 = -1575, \ E_3 = \dfrac{49}{40}, \ \sum\limits_{j = 1}^{6} a_jB_j = \dfrac{371}{6563700}. \end{array} \end{equation} | (3.65) |
Thus, to derive relation (3.64), we simply substitute the values above into Theorem 3.3. The proof is complete.
This section is devoted to giving our new evaluation of a certain class of fractional integrals whose values are written in terms of hypergeometric functions _3F_2 which we obtained in the previous sections.
Theorem 4.1. For all \tau\le s\le t\le \eta , and \alpha > 1/2, the following integral representations for the Gauss hypergeometric function hold true.
(1)
\begin{eqnarray} I^\alpha_1(t)&: = &\int_{\tau}^{t} (t-s)^{\alpha-1} (\eta-s)^{\alpha-1} \ ds\\ & = &\left[\dfrac{(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha}}{\alpha}\right] {_2F_1(1-\alpha, 1;\alpha+1;g(t))}, \end{eqnarray} | (4.1) |
where g(t): = \dfrac{t-\tau}{\eta-\tau}.
(2)
\begin{eqnarray} \int_{\tau}^{\eta} I^\alpha_1(t) dt& = &\left[\dfrac{(\eta-\tau)^{2\alpha}}{\alpha(\alpha+1)} \right] {_3F_2(1-\alpha, 1, \alpha+1;\alpha+1, \alpha+2;1) } \end{eqnarray} | (4.2) |
\begin{eqnarray} & = &\dfrac{(\eta-\tau)^{2\alpha}}{2\alpha^2}. \end{eqnarray} | (4.3) |
Proof of Theorem 4.1.
(1) Let s = \tau+x(t-\tau) . By changing the integration variable from s to x , the integral I^\alpha_1(t) becomes
\begin{eqnarray} I^\alpha_1(t)& = & (t-\tau)\int_{0}^{1} ((t-\tau)-x(t-\tau))^{\alpha-1} \left((\eta-\tau)-x(t-\tau)\right)^{\alpha-1} dx \\ & = & (t-\tau)(t-\tau)^{\alpha-1} \int_{0}^{1} (1-x)^{\alpha-1} \left((\eta-\tau)-\frac{(\eta-\tau)(t-\tau)}{(\eta-\tau)} x\right)^{\alpha-1} dx\\ & = &(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha}\int_{0}^{1} (1-x)^{\alpha-1} \left(1-\frac{(t-\tau)}{(\eta-\tau)} x\right)^{\alpha-1} dx. \end{eqnarray} |
Above, we have the Euler integral representation of _2F_1(a, b; c;z) with a = 1-\alpha , b = 1 , c = \alpha+1 , and z = g(t) = \frac{t-\tau}{\eta-\tau} . Thus,
\begin{eqnarray} I^\alpha_1(t)& = &\frac{(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha}}{\alpha} {_2F_1(1-\alpha, 1;\alpha+1;g(t)) }. \end{eqnarray} |
This completes the proof of (4.1).
(2) Now we are in a position to evaluate the integral (4.2), so denoting the left-hand side of (4.2) by \lambda_1 , we have
\begin{equation} \lambda_1 : = \frac{(\eta-\tau)^{\alpha-1}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{\alpha} {_2F_1(1-\alpha, 1;\alpha+1;g(t))} dt. \end{equation} | (4.4) |
Now, expressing _2F_1 as a series and changing the order of integration and summation, which is justified by Lemma 2.2, we have
\begin{eqnarray} \lambda_1 &: = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \int_\tau^\eta (t-\tau)^{\alpha}\left(\dfrac{t-\tau}{\eta-\tau}\right)^k dt\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+\alpha} dt\\ & = &\left[\frac{(\eta-\tau)^{2\alpha}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+\alpha+1)}. \end{eqnarray} |
Now, by Lemma 3.1, we obtain (4.2), that is
\begin{equation*} \lambda_1 : = \left[\dfrac{(\eta-\tau)^{2\alpha}}{\alpha(\alpha+1)} \right] {_3F_2(1-\alpha, 1, \alpha+1;\alpha+1, \alpha+2;1) }, \end{equation*} |
and Theorem 3.1 gives (4.3). This completes the proof.
Theorem 4.2. For all \tau\le s\le t\le \eta and \alpha > 1/2, the following integral representations for the Gauss hypergeometric function hold true.
(1)
\begin{eqnarray} I^\alpha_2(t)&: = &\int_{\tau}^{t} (t-\tau) (t-s)^{\alpha-1} (\eta-s)^{\alpha-1} \ ds\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}(t-\tau)^{\alpha+1}}{\alpha}\right] {_2F_1(1-\alpha, 1;\alpha+1;g(t))}, \end{eqnarray} | (4.5) |
where g(t): = \frac{t-\tau}{\eta-\tau}.
(2)
\begin{eqnarray} \int_{\tau}^{\eta} I^\alpha_2(t) dt& = &\left[\frac{(\eta-\tau)^{2\alpha+1}}{\alpha(\alpha+2)} \right] {_3F_2(1-\alpha, 1, \alpha+2;\alpha+1, \alpha+3;1) } \end{eqnarray} | (4.6) |
\begin{eqnarray} & = &\frac{(\eta-\tau)^{2\alpha+1}}{\alpha} \left[ \dfrac{1}{\alpha+1}+\dfrac{1}{2\alpha}-\dfrac{2}{2\alpha+1}\right]. \end{eqnarray} | (4.7) |
Proof of Theorem 4.2.
(1) The proof is similar to the proof of Theorem 4.1, and so we just sketch the basic idea. In exactly the same manner, the integral I^\alpha_2(t) can be obtained.
(2) Now denoting the left-hand side of (4.6) by \lambda_2 , we have
\begin{equation} \lambda_2 : = \frac{(\eta-\tau)^{\alpha-1}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{\alpha+1} {_2F_1(1-\alpha, 1;\alpha+1;g(t))} dt. \end{equation} | (4.8) |
Now, expressing _2F_1 as a series and changing the order of integration and summation, which is justified by Lemma 2.2, we have
\begin{eqnarray} \lambda_2 &: = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \int_\tau^\eta (t-\tau)^{\alpha+1}\left(\dfrac{t-\tau}{\eta-\tau}\right)^k dt\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+\alpha+1} dt\\ & = &\left[\frac{(\eta-\tau)^{2\alpha+1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+\alpha+2)}. \end{eqnarray} |
Now, by Lemma 3.2, we obtain (4.6), that is
\begin{eqnarray} \lambda_2 &: = &\left[\frac{(\eta-\tau)^{2\alpha+1}}{\alpha(\alpha+2)} \right] {_3F_2(1-\alpha, 1, \alpha+2;\alpha+1, \alpha+3;1) }, \end{eqnarray} |
and Theorem 3.2 gives (4.7). This completes the proof.
Theorem 4.3. For all \tau\le s\le t\le \eta , and \alpha > 1/2, the following integral representations for the Gauss hypergeometric function hold true.
(1)
\begin{eqnarray} I^\alpha_3(t)&: = &\int_{\tau}^{t} (t-\tau)^{\alpha-1} (t-s)^{\alpha-1} (\eta-s)^{\alpha-1} \, ds\\ & = &\left[\frac{(\eta-\tau)^{\alpha-1}(t-\tau)^{2\alpha-1}}{\alpha}\right] {_2F_1(1-\alpha, 1, \alpha+1;g(t)) }, \end{eqnarray} | (4.9) |
where g(t): = \frac{t-\tau}{\eta-\tau} .
(2)
\begin{equation} \int_{\tau}^{\eta} I^\alpha_3(t) dt = \left[\frac{(\eta-\tau)^{3\alpha-1}}{2\alpha^2}\right] { _3F_2(1-\alpha, 1, 2\alpha;\alpha+1, 2\alpha+1;1) }. \end{equation} | (4.10) |
Proof of Theorem 4.3.
(1) The proof is similar to the proof of Theorem 4.1, and so we just sketch the basic idea. In exactly the same manner, the integral I^\alpha_3(t) can be obtained.
(2) Now, denoting the left-hand side of (4.10) by \lambda_3 , we have
\begin{equation} \lambda_3 : = \frac{(\eta-\tau)^{\alpha-1}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{2\alpha-1} {_2F_1(1-\alpha, 1;\alpha+1;g(t))} dt. \end{equation} | (4.11) |
By expressing _2F_1 as a series and change the order of integration and summation, which is justified by justified by Lemma 2.2, we have
\begin{eqnarray} \lambda_3 & = &\left[\frac{(\eta-\tau)^{\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+2\alpha-1} dt\\ & = &\left[\frac{(\eta-\tau)^{3\alpha-1}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(1-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+2\alpha)}. \end{eqnarray} | (4.12) |
Thus, the result of Lemma 2.5 gives (4.10). This completes the proof.
Remark 4.1. Note that when \alpha = 2 , the results of Theorems 4.2 and 4.3 coincide. Therefore, the right-hand terms of relations (4.6) and (4.10) are equal when \alpha = 2 , which can be justified by Lemma 2.1. Thus,
\begin{eqnarray} \int_{\tau}^{\eta} I^2_2(t) dt = \int_{\tau}^{\eta} I^2_3(t) dt & = & \frac{11(\eta-\tau)^{5}}{120}. \end{eqnarray} | (4.13) |
\blacksquare For the choices \alpha = \tfrac{2p+1}{2} and p\in {{ \mathbb{N}}}^*, we have the following new explicit evaluation of a certain class of integrals as a special case from our Theorem 4.3.
Theorem 4.4. For all \tau\le s\le t\le \eta and p\in {{ \mathbb{N}}}^*, the following integral holds true.
(1)
\begin{eqnarray} \int_{\tau}^{\eta} I^{\tfrac{2p+1}{2}}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{6p+1}{2}}}{(2p+1)^2}\right] {_3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1)} \end{eqnarray} | (4.14) |
\begin{eqnarray} & = &-K_p\left[\frac{2(\eta-\tau)^{\frac{6p+1}{2}}}{(2p+1)}\right] \left( 2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right ), \end{eqnarray} | (4.15) |
where I^{\tfrac{2p+1}{2}}_3 is defined in (4.9) with \alpha = \tfrac{2p+1}{2} , and K_p, a_{2p+1}, E_p, B_j , and a_j are defined in (3.5).
Proof of Theorem 4.4. By letting \alpha = \tfrac{2p+1}{2} and p\in {{ \mathbb{N}}}^* in the integral (4.10), we obtain (4.14) and (4.15) is obtained by replacing the explicit evaluation of the fucntion {_3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1)} given by Theorem 3.3. The proof is complete.
\blacksquare The following result presents the integrals of I^{\alpha}_3(t) for some \alpha when \alpha = 3/2 , \alpha = 5/2 , and \alpha = 7/2 .
Corollary 4.1. For all \tau\le s\le t\le \eta, the following integrals hold true.
(1)
\begin{eqnarray} \int_{\tau}^{\eta} I^{3/2}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{7}{2}}}{3^2}\right] \left( \dfrac{33}{35}-\dfrac{6}{35}\ln(2)\right), \end{eqnarray} | (4.16) |
where I^{3/2}_3 is defined in (4.9) with \alpha = 3/2.
(2)
\begin{eqnarray} \int_{\tau}^{\eta} I^{5/2}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{13}{2}}}{5^2}\right] \left( \frac{4045}{6006}+\frac{10}{1001}\ln(2)\right), \end{eqnarray} | (4.17) |
where I^{5/2}_3 is defined in (4.9) with \alpha = 5/2.
(3)
\begin{eqnarray} \int_{\tau}^{\eta} I^{7/2}_3(t) dt & = & \left[\frac{2(\eta-\tau)^{\frac{19}{2}}}{7^2}\right] \left( \frac{221158}{415701}-\frac{70}{138567}\ln(2)\right), \end{eqnarray} | (4.18) |
where I^{7/2}_3 is defined in (4.9) with \alpha = 7/2.
Proof of corollary 4.1.
(1) Let p = 1 . Then, from Theorem 4.4 (4.14), we obtain
\begin{equation} \int_{\tau}^{\eta} I^{3/2}_3(t) dt = \left[\frac{2(\eta-\tau)^{\frac{7}{2}}}{3^2}\right] {_3F_2(-\dfrac{1}{2}, 1, 3;\dfrac{5}{2}, 4;1)}. \end{equation} | (4.19) |
From Lemma 3.3, we have
\begin{equation} _3F_2(-\dfrac{1}{2}, 1, 3;\dfrac{5}{2}, 4;1) = \dfrac{33}{35}-\dfrac{6}{35}\ln(2), \end{equation} | (4.20) |
and so substituting (4.20) into (4.19), we obtain (4.31). This completes the proof of (4.31).
(2) Similarly, let p = 2 . Then, from Theorem 4.4 (4.14), we obtain
\begin{equation} \int_{\tau}^{\eta} I^{5/2}_3(t) dt = \left[\frac{2(\eta-\tau)^{\frac{13}{2}}}{5^2}\right] {_3F_2(-\frac{3}{2}, 1, 5;\frac{7}{2}, 6;1)}. \end{equation} | (4.21) |
From Corollary 3.1, we have
\begin{equation} {_3F_2(-\frac{3}{2}, 1, 5;\frac{7}{2}, 6;1)} = \frac{4045}{6006}+\frac{10}{1001}\ln(2), \end{equation} | (4.22) |
and so substituting (4.22) into (4.21), we obtain (4.32). This completes the proof of (4.32).
(3) Also, if we let p = 2 , then from Theorem 4.4 (4.14), we obtain
\begin{equation} \int_{\tau}^{\eta} I^{7/2}_3(t) dt = \left[\frac{2(\eta-\tau)^{\frac{19}{2}}}{7^2}\right]{_3F_2(-\frac{5}{2}, 1, 7;\frac{9}{2}, 8;1)}. \end{equation} | (4.23) |
From Corollary 3.2, we have
\begin{equation} {_3F_2(-\frac{5}{2}, 1, 7;\frac{9}{2}, 8;1)} = \frac{221158}{415701}-\frac{70}{138567}\ln(2), \end{equation} | (4.24) |
and so substituting (4.24) into (4.23), we obtain (4.33). This completes the proof.
Theorem 4.5. For all \tau\le s\le t\le \eta and \alpha > 1, the following integral representations for the Gauss hypergeometric functions hold true.
(1)
\begin{eqnarray} I^\alpha_4(t) &: = &\int_{\tau}^{t} (t-\tau)^{\alpha-1} (t-s)^{\alpha-1} (\eta-s)^{\alpha-2} \, ds\\ & = &\left[\frac{(\eta-\tau)^{\alpha-2}(t-\tau)^{2\alpha-1}}{\alpha}\right] {_2F_1(2-\alpha, 1;\alpha+1;g(t)) }, \end{eqnarray} | (4.25) |
where g(t): = \frac{t-\tau}{\eta-\tau} .
(2)
\begin{equation} \int_{\tau}^{\eta} I^\alpha_4(t) dt = \dfrac{(\eta-\tau)^{3\alpha-2}}{(1-\alpha)(2\alpha-1)}+\left[ \dfrac{(1-3\alpha)(\eta-\tau)^{3\alpha-2}}{2\alpha^2(1-\alpha)} \right] {_3F_2(1-\alpha, 1, 2\alpha;\alpha+1, 2\alpha+1;1)}. \end{equation} | (4.26) |
Proof of Theorem 4.5.
(1) The proof follows similar lines of argument to that of the above theorems and so we just sketch the basic idea. In exactly the same manner, we have
\begin{eqnarray} I^\alpha_4(t)& = & (\eta-\tau)^{\alpha-2}(t-\tau)^{2\alpha-1}\int_{0}^{1} (1-x)^{\alpha-1} \left(1-\frac{(t-\tau)}{(\eta-\tau)} x\right)^{\alpha-2} dx. \end{eqnarray} |
Above, we have the Euler integral representation of _2F_1(a, b; c;z) with a = 2-\alpha , b = 1 , c = \alpha+1 , and z = g(t) = \frac{t-\tau}{\eta-\tau} , thus
\begin{equation} I^\alpha_4(t) = \frac{(\eta-\tau)^{\alpha-2}(t-\tau)^{2\alpha-1}}{\alpha} {_2F_1(2-\alpha, 1;\alpha+1;g(t)) }. \end{equation} | (4.27) |
This completes the proof of (4.25).
(2) Now denoting the left-hand side of (4.26) by \lambda_4 , we have
\begin{equation} \lambda_4 : = \frac{(\eta-\tau)^{\alpha-2}}{\alpha} \int_{\tau}^{\eta} (t-\tau)^{2\alpha-1} {_2F_1(2-\alpha, 1;\alpha+1;g(t))} dt, \end{equation} | (4.28) |
and by expressing _2F_1 as a series and change the order of integration and summation, which is justified by justified by Lemma 2.2, we have
\begin{eqnarray} \lambda_4 & = &\left[\frac{(\eta-\tau)^{\alpha-2}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(2-\alpha)_k}{(\alpha+1)_k} \left(\dfrac{1}{\eta-\tau}\right)^k \int_\tau^\eta \left(t-\tau\right)^{k+2\alpha-1} dt\\ & = &\left[\frac{(\eta-\tau)^{3\alpha-2}}{\alpha} \right] \sum\limits_{k = 0}^{\infty} \dfrac{(2-\alpha)_k}{(\alpha+1)_k} \dfrac{1}{(k+2\alpha)}. \end{eqnarray} | (4.29) |
Thus, the result of Lemma 2.1 gives (4.26). This completes the proof.
\blacksquare For the choices \alpha = \tfrac{2p+1}{2} and p\in {{ \mathbb{N}}}^*, we have also the following new explicit evaluation of a certain class of Integrals as special case from our Theorem 4.5.
Theorem 4.6. For all \tau\le s\le t\le \eta and p\in {{ \mathbb{N}}}^*, the following integral holds true.
\begin{equation} \begin{array}{lll} \int_{\tau}^{\eta} I^{\tfrac{2p+1}{2}}_4(t) dt & = & \dfrac{(\eta-\tau)^{\frac{6p-1}{2}}}{p(1-2p)}+\left[ \dfrac{(6p+1)(\eta-\tau)^{\frac{6p-1}{2}}}{(2p-1)(2p+1)^2} \right] {_3F_2(\tfrac{1}{2}-p, 1, 1+2p;\tfrac{3}{2}+p, 2p+2;1)}\\ & = &\dfrac{(\eta-\tau)^{\frac{6p-1}{2}}}{p(1-2p)}-\left[ \dfrac{K_p(6p+1)(\eta-\tau)^{\frac{6p-1}{2}}}{(2p-1)(2p+1)} \right]\left( 2a_{2p+1}\ln(2)+ 2a_{2p+1}(E_p+B_p)+\sum\limits_{j = 1}^{2p} a_jB_j\right ), \end{array} \end{equation} | (4.30) |
where I^{\tfrac{2p+1}{2}}_4 is defined in (4.25) with \alpha = \tfrac{2p+1}{2} , and K_p, a_{2p+1}, E_p, B_j, and a_j are defined in (3.5)
Proof of Theorem 4.6. The proof is similar to the proof of Theorem 4.4, so we omit the proof for brevity.
\blacksquare The following result presents the values of I^{\alpha}_4 for some \alpha that are when \alpha = 3/2, \alpha = 5/2 , and \alpha = 7/2.
Corollary 4.2. For all \tau\le s\le t\le \eta, the following integrals hold true.
(1)
\begin{eqnarray} \int_{\tau}^{\eta} I^{3/2}_4(t) dt & = & -(\eta-\tau)^{\frac{5}{2}}+\frac{7(\eta-\tau)^{\frac{5}{2}}}{9} \left( \dfrac{33}{35}-\dfrac{6}{35}\ln(2)\right), \end{eqnarray} | (4.31) |
where I^{3/2}_3 is defined in (4.25) with \alpha = 3/2.
(2)
\begin{eqnarray} \int_{\tau}^{\eta} I^{5/2}_4(t) dt & = & -\dfrac{(\eta-\tau)^{\frac{11}{2}}}{6}+\frac{12(\eta-\tau)^{\frac{11}{2}}}{75} \left( \frac{4045}{6006}+\frac{10}{1001}\ln(2)\right), \end{eqnarray} | (4.32) |
where I^{5/2}_3 is defined in (4.25) with \alpha = 5/2.
(3)
\begin{eqnarray} \int_{\tau}^{\eta} I^{7/2}_4(t) dt & = & -\dfrac{(\eta-\tau)^{\frac{17}{2}}}{15}+\frac{19(\eta-\tau)^{\frac{17}{2}}}{245} \left( \frac{221158}{415701}-\frac{70}{138567}\ln(2)\right), \end{eqnarray} | (4.33) |
where I^{7/2}_3 is defined in (4.25) with \alpha = 7/2.
Proof of Corollary 4.2. The proof is similar to the proof of Theorem 4.1, so we omit the proof for brevity.
In this study, we expressed four families of fractional integrals, denoted as {{ \mathcal{F} }}_1^{\alpha} = \{I^\alpha_1(\alpha), I^\alpha_2(\alpha)\; ;\; \alpha > \tfrac{1}{2}\} , {{ \mathcal{F} }}_2^{\alpha} = \{I^\alpha_3(\alpha), \; ;\; \alpha > \tfrac{1}{2}\} and {{ \mathcal{F} }}_3^{\alpha} = \{ I^\alpha_4(\alpha)\; ;\; \alpha > 1\} , using the class of hypergeometric functions _3F_2(1) . The series representation of the hypergeometric functions allowed us to derive explicit forms for the integrals of the family {{ \mathcal{F} }}_1^{\alpha} for all \alpha > \tfrac{1}{2} , as well as for the integrals of the subfamily {{ \mathcal{F} }}_2^{\tfrac{2p+1}{2}} and {{ \mathcal{F} }}_3^{\tfrac{2p+1}{2}} for all p\in {{ \mathbb{N}}}^* .
For the integrals of the family {{ \mathcal{F} }}_2^{\alpha} and {{ \mathcal{F} }}_3^{\alpha} , we have only calculated the explicit forms of those of the subfamily {{ \mathcal{F} }}_2^{\tfrac{2p+1}{2}} and {{ \mathcal{F} }}_3^{\tfrac{2p+1}{2}} . However, by examining other subfamilies of {{ \mathcal{F} }}_2^{\alpha} and {{ \mathcal{F} }}_3^{\alpha} , we could derive more interesting formulas relating fractional integrals to the family of functions _3F_2(1) .
Saleh S. Almuthaybiri: Conceptualization, supervision, validation, investigation, writing original draft preparation, formal analysis, writing review and editing. Abdelhamid Zaidi: Investigation, validation, writing original draft preparation, methodology, formal analysis, writing review and editing, doing the revision, funding acquisition. All authors have read and agreed to the published version of the manuscript.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The Researchers would like to thank the Deanship of Graduate Studies and Scientific Research at Qassim University for financial support (QU-APC-2025).
The authors declare that they have no conflicts of interest.
[1] | J. Jang, W.-M. Ni and M. Tang, Global bifurcation and structure of Turing patterns in the 1-D Lengyel-Epstein model, J. Dyn. Diff. Equa., 2 (2004), 297–320. |
[2] | A. M. Turing, The chemical basis of morphogenesis, Philos. T. R. SOC. B, 237 (1852), 37–72. |
[3] | N. F. Britton, Essential Mathematical Biology, Springer, New York, 2003. |
[4] | L. A. Segel and J. L. Jackson, Dissipative structure: an explanation and an ecological example, J. Theor. Biol., 37 (1972), 545–559. |
[5] | D. Alonso, F. Bartumeus and J. Catalan, Mutual interference between predators can give rise to Turing spatial patterns, Ecology, 83 (2002), 28–34. |
[6] | W.-M. Wang, L. Zhang, H. Wang, et al., Pattern formation of a predator-prey system with Ivlev-type functional response, Ecol. Model., 221 (2008), 131–140. |
[7] | C. Neuhauser, Mathematical challenges in spatial ecology, Notices AMS, 48 (2001), 1304–1314. |
[8] | A. Okubo and S. A. Levin, Diffusion and Ecological Problems: Modern Perspectives, Springer, New York, 2001. |
[9] | A. B. Medvinsky, S. V. Petrovskii, I. A. Tikhonova, et al., Spatiotemporal complexity of plankton and fish dynamics. SIAM Rev., 44 (2002), 311–370. |
[10] | J. D. Murray, Mathematical biology. II: Spatial models and biomedical applications, Springer, New York, 2003. |
[11] | P. Y. H. Pang and M. Wang, Qualitative analysis of a ratio-dependent predator-prey system with diffusion, Proc. R. Soc. Edinb., 133 (2003), 919–942. |
[12] | K. Kuto and Y. Yamada. Multiple coexistence states for a prey-predator system with crossdi ffusion, J. Differ. Equations, 197 (2004), 315–348. |
[13] | K. Kuto, Stability of steady-state solutions to a prey-predator system with cross-diffusion, J. Differ. Equations, 197 (2004), 293–314. |
[14] | X. Zeng and Z. Liu, Non-constant positive steady states of a prey-predator system with cross-diffusions, J. Math. Anal. Appl., 332 (2007), 989–1009. |
[15] | R. Peng and J. Shi, Non-existence of non-constant positive steady states of two Holling type-II predator-prey systems: Strong interaction case, J. Differ. Equations, 247 (2009), 866–886. |
[16] | F. Yi, J. Wei and J. Shi, Bifurcation and spatiotemporal patterns in a homogeneous diffusive predator-prey system, J. Differ. Equations, 246 (2009), 1944–1977. |
[17] | H. Shi,W.-T. Li and G. Lin. Positive steady states of a diffusive predator-prey system with modified Holling-Tanner functional response, Nonl. Anal. Real, 11 (2010), 3711–3721. |
[18] | Y. Cai, M. Banerjee, Y. Kang, et al., Spatiotemporal complexity in a predator-prey model with weak Allee effects, Math. Biosci. Eng., 11 (2014), 1247–1274. |
[19] | S. Li, J. Wu and Y. Dong, Turing patterns in a reaction-diffusion model with the Degn-Harrison reaction scheme, J. Differ. Equations, 259 (2015), 1990–2029. |
[20] | H. Shi and S. Ruan. Spatial, temporal and spatiotemporal patterns of diffusive predator-prey models with mutual interference, IMA J. Appl. Math., 80 (2015), 1534–1568. |
[21] | Y. Cai and W.-M. Wang, Fish-hook bifurcation branch in a spatial heterogeneous epidemic model with cross-diffusion, Nonl. Anal. Real, 30 (2016), 99–125. |
[22] | T. Kuniya and J.Wang, Global dynamics of an SIR epidemic model with nonlocal diffusion, Nonl. Anal. Real, 43 (2018), 262–282. |
[23] | J.Wang, J.Wang and T. Kuniya, Analysis of an age-structured multi-group heroin epidemic model.Appl. Math. Comp., 347 (2019), 78–100. |
[24] | Y. Cai, Z. Ding, B. Yang, et al., Transmission dynamics of Zika virus with spatial structure–A case study in Rio de Janeiro, Brazil. Phys. A, 514 (2019), 729–740. |
[25] | Y. Cai, K. Wang and W.M. Wang, Global transmission dynamics of a Zika virus model, Appl. Math. Lett., 92 (2019), 190–195. |
[26] | Y. Cai, X. Lian, Z. Peng, et al., Spatiotemporal transmission dynamics for influenza disease in a heterogenous environment, Nonl. Anal. Real., 46 (2019), 178–194. |
[27] | Y. Cai, Z. Gui, X. Zhang, et al., Bifurcations and pattern formation in a predator-prey model,. Inter. J. Bifur. Chaos, 28 (2018), 1850140. |
[28] | H. Zhang, Y. Cai, S. Fu, et al., Impact of the fear effect in a prey-predator model incorporating a prey refuge, Appl. Math. Comp., 356 (2019), 328–337. |
[29] | X. Cao, Y. Song and T. Zhang. Hopf bifurcation and delay-induced Turing instability in a diffusive Iac Operon model, Inter. J. Bifur. Chaos, 26 (2016), 1650167. |
[30] | J. Jiang, Y. Song and P. Yu, Delay-induced Triple-Zero bifurcation in a delayed Leslie-type predator-prey model with additive Allee effect, Inter. J. Bifur. Chaos, 26 (2016), 1650117. |
[31] | Y. Song, H. Jiang, Q. Liu, et al., Spatiotemporal dynamics of the diffusive Mussel-Algae model near Turing-Hopf bifurcation. SIAM J. Appl. Dyn. Sys., 16 (2017), 2030–2062. |
[32] | Y. Song and X. Tang, Stability, steady-state bifurcations, and Turing patterns in a predator-prey model with herd behavior and prey-taxis, Stud. Appl. Math., 139 (2017), 371–404. |
[33] | S. Wu and Y. Song, Stability and spatiotemporal dynamics in a diffusive predator Cprey model with nonlocal prey competition, Nonl. Anal. Real, 48 (2019), 12–39. |
[34] | S.-B. Hsu and T.-W. Huang, Global stability for a class of predator-prey systems, SIAM J. Appl. Math., 55 (1995), 763–783. |
[35] | Y. Lou and W.-M. Ni. Diffusion, self-diffusion and cross-diffusion, J. Differ. Equations, 131 (1996), 79–131. |
[36] | W.-M. Ni and M. Tang, Turing patterns in the Lengyel-Epstein system for the CIMA reaction, T. Am. Math. Soc., 357 (2005), 3953–3969. |
[37] | P. H. Rabinowitz, Some global results for nonlinear eigenvalue problems, J. Func. Anal., 7 (1971), 487–513. |
[38] | Y. Nishiura, Global structure of bifurcating solutions of some reaction-diffusion systems, SIAM J. Math. Anal., 13 (1982), 555–593. |
[39] | I. Takagi, Point-condensation for a reaction-diffusion system, J. Differ. Equations, 61 (1986), 208–249. |
[40] | A.Chertock, A. Kurganov, X. Wang, et al., On a chemotaxis model with saturated chemotactic flux, Kinet. Relat. Models, 5 (2012), 51–95. |