Logistic model with Allee effects
Consider the case when diffusion is ignorable; that is \(d \rightarrow 0\) (for historical interest, see [42]), then the species birth (or death) is only dependent on growth rate (or death rate). If we allow Allee effects in this type of model, we get the following reduction model of (11) with \(d=0\):
$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \frac{\displaystyle d u}{\displaystyle d t}= ru(t)\left( 1\frac{\displaystyle u(t)}{\displaystyle K}\right) \left( \frac{\displaystyle u(t)}{\displaystyle M}1 \right) ,&{} t>0,\\ u(0)=u_0,&{} \end{array}\right. } \end{aligned}$$
(12)
where \(M < K\). The parametric values, r, is defined as the intrinsic growth rate, K is the carrying capacity, and M is the Allee threshold.
Let \(g(u)=r u(t)\left( 1\displaystyle \frac{\displaystyle u(t)}{\displaystyle K}\right) \left( \displaystyle \frac{ u(t)}{\displaystyle M}1 \right)\). To find the stationary solutions, \(u^*\) of (12), directly we set \(\frac{\displaystyle du^*}{\displaystyle dt}=0\) such that
$$\begin{aligned}&\therefore ru^* \left( 1\frac{\displaystyle u^*}{\displaystyle K}\right) \left( \frac{\displaystyle u^*}{\displaystyle M}1 \right) =0\\&\Rightarrow u^*= \left\{ 0,\; M,\; K \right\} . \end{aligned}$$
Also we get
$$\begin{aligned} \frac{d g}{d u}=r\left\{ \left( 1\frac{\displaystyle u}{\displaystyle K}\right) \left( \frac{\displaystyle u}{\displaystyle M}1 \right) + \frac{u}{M} \left( 1\frac{\displaystyle u}{\displaystyle K}\right)  \frac{u}{K} \left( \frac{\displaystyle u}{\displaystyle M}1 \right) \right\} . \end{aligned}$$
(13)
Since parameters r and K, except M, are strictly positive, and M is the Allee threshold (positive or negative threshold) and \(M<K\), then depending on the choice of M, we have the following two cases:
Case 1 When \(0<M<K\) (weak Allee effect):
Using \(u^*= \left\{ 0,\; M,\; K \right\}\) in (13), we get,
$$\begin{aligned} \frac{d g}{d u}(0)= & {} r<0;\\ \frac{d g}{d u}(M)= & {} r\left( 1\frac{M}{K} \right) >0;\;\text {and}\\ \frac{d g}{d u}(K)= & {} r\left( \frac{K}{M}1 \right) <0. \end{aligned}$$
Therefore, \(u^*=\left\{ 0,\; K \right\}\) are stable solution of (12) and \(u^*=M\) is the unstable state of (12) when the species are facing weak Allee effects.
Case 2 When \(M<0<K\) (strong Allee effect):
Putting \(u^*= \left\{ 0,\; M,\; K \right\}\) in (13), where \(M=M\), we get that the equilibrium 0 does not change its stability. But when, \(M=M\),
$$\begin{aligned} \frac{d g}{d u}(M)= & {} \frac{d g}{d u}(M)=r\left( 1+\frac{M}{K} \right)>0;\;\text {and}\\ \frac{d g}{d u}(K)= & {} r\left( 1+\frac{K}{M} \right) >0. \end{aligned}$$
Therefore, \(u^*=0\) is the only stable solution of (12) and \(u^*=M, \; K\) are unstable state of (12), when the species are facing strong Allee effects.
For a single species population in ecology, the biological significance of these stableunstable strategies is that either the species repelled or converge to the carrying capacity. The present study employs that the stable situation is either \(u=0\) or \(u(t)=K\), i.e., the population will die out or survive, respectively, and the maximum population density is K. A bifurcation point or threshold level can be found in an unstable solution, \(u(t)=M\).
For the sake of diversity and to cover the wide range of studies, in the following section, we introduce the harvesting term in absence of diffusion.
Logistic model with Allee effects and harvesting
If harvesting is involved then the model can be formulated in the following way from (12):
$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \frac{\displaystyle d u}{\displaystyle d t}= ru(t)\left( 1\frac{\displaystyle u(t)}{\displaystyle K}\right) \left( \frac{\displaystyle u(t)}{\displaystyle M}1 \right) Hu(t),&{} t>0,\\ u(0)=u_0&{}. \end{array}\right. } \end{aligned}$$
(14)
Here the harvesting function H is constant, \(H \ge 0\) and \(H < r\), and in case of Allee effects, \(M < K\). Let us define the following function
$$\begin{aligned} g^*(u)= u(t)\left[ r \left( 1\frac{ u(t)}{ K}\right) \left( \frac{ u(t)}{ M}1 \right) H\right] , \end{aligned}$$
and
$$\begin{aligned} \frac{d g^*}{d u}=r\left\{ \left( 1\frac{\displaystyle u}{\displaystyle K}\right) \left( \frac{\displaystyle u}{\displaystyle M}1 \right) + \frac{u}{M} \left( 1\frac{\displaystyle u}{\displaystyle K}\right)  \frac{u}{K} \left( \frac{\displaystyle u}{\displaystyle M}1 \right) \right\} H. \end{aligned}$$
(15)
Now letting \(\frac{\displaystyle d u}{\displaystyle d t}=0\), we get
$$\begin{aligned}&u\left[ r \left( 1\frac{ u}{ K}\right) \left( \frac{ u}{ M}1 \right) H\right] =0\\&\quad \Rightarrow ru^2r(K+M)u+KM(r+H)=0; \; \text {and} \; u=0\\&\quad \Rightarrow u=\frac{\displaystyle r(K+M) \pm \sqrt{r^2(KM)^24rKMH}}{\displaystyle 2r}; \; \text {and} \; u=0. \end{aligned}$$
Therefore, the three equilibrium states of (14) are
$$\begin{aligned} u_1&=0,\\ u_2&=\frac{\displaystyle 1}{\displaystyle 2r}\left[ r(K+M) + \sqrt{r^2(KM)^24rKMH}\right] ,\;\text {and} \\ u_3&=\frac{\displaystyle 1}{\displaystyle 2r}\left[ r(K+M)  \sqrt{r^2(KM)^24rKMH}\right] . \end{aligned}$$
The equilibrium states \(u_2\) and \(u_3\) exist as long as
$$\begin{aligned} r^2(KM)^24rKMH>0. \end{aligned}$$
Let us define \(D:= \sqrt{r^2(KM)^24rKMH}\) which modify \(u_2\) and \(u_3\) in the following mathematical structures
$$\begin{aligned} u_2= & {} \frac{\displaystyle 1}{\displaystyle 2}\left[ (K+M)+\frac{\displaystyle D}{\displaystyle r}\right] ,\\ u_3= & {} \frac{\displaystyle 1}{\displaystyle 2}\left[ (K+M)\frac{\displaystyle D}{\displaystyle r}\right] . \end{aligned}$$
Now from Eq. (15), we get
$$\begin{aligned} \frac{\displaystyle d g^*}{\displaystyle d u}(u_1=0)=(r+H) <0, \;\;\text {since}\;r>0,\;H\ge 0. \end{aligned}$$
Thus we conclude that \(u_1=0\) is a stable solution either there is a strong Allee effect or weak Allee effect. The rest two equilibria states are \(u_2\) and \(u_3\), and in terms of H and M, the following two situations arise:
When \(0<M<K\), i.e., weak Allee effect, then we get the following three cases in terms of H:
$$\begin{aligned} H<\frac{\displaystyle r(KM)^2}{\displaystyle 4KM},\;\; H=\frac{\displaystyle r(KM)^2}{\displaystyle 4KM},\;\text {and}\;\; H>\frac{\displaystyle r(KM)^2}{\displaystyle 4KM}. \end{aligned}$$
Case I When \(H>\displaystyle \frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\), that is if \(D^2<0\) then \(u_2, u_3 \in {\mathbb {C}}\setminus {\mathbb {R}}\) hence their stability is irrelevant and we have no conclusion.
Case II If \(H=\displaystyle \frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\) then \(D=0\) and the system (14) has one identical solution, \(u_2=u_3=\frac{\displaystyle K+M}{\displaystyle 2}\) and from (15) it is seen that
$$\begin{aligned} \frac{\displaystyle d g^*}{\displaystyle d u}\left( \frac{\displaystyle K+M}{\displaystyle 2}\right) =0. \end{aligned}$$
Therefore, in this situation, the system has a unique positive equilibrium, which is semiunstable.
Case III If \(H<\displaystyle \frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\) then \(D>0\) and the problem (14) has two distinct solutions \(u_2 \; \text {and} \; u_3\), where
$$\begin{aligned} u_2=\frac{\displaystyle 1}{\displaystyle 2}\left[ (K+M)+\frac{\displaystyle D}{\displaystyle r}\right] \; \text {and} \; u_3=\frac{\displaystyle 1}{\displaystyle 2}\left[ (K+M)\frac{\displaystyle D}{\displaystyle r}\right] . \end{aligned}$$
Since \(D>0\), \(u_2\) is always positive. We recall (15) and get the simplified form as follows:
$$\begin{aligned} \frac{d g^*}{d u}&=r\left[ \frac{\displaystyle 3}{\displaystyle KM} u^2+ \frac{\displaystyle 2(K+M)}{\displaystyle KM}u 1 \right] H, \end{aligned}$$
such that
$$\begin{aligned} \frac{d g^*}{d u}(u_2)&=r\left[ \frac{\displaystyle 3}{\displaystyle KM} u_2^2+ \frac{\displaystyle 2(K+M)}{\displaystyle KM}u_2 1 \right] H. \end{aligned}$$
Using \(u_2=\frac{\displaystyle 1}{\displaystyle 2}\left[ (K+M)+\frac{\displaystyle D}{\displaystyle r}\right]\), we get
$$\begin{aligned} \frac{d g^*}{d u}(u_2)=r\left[ \frac{\displaystyle (K+M)^2}{\displaystyle 4KM} 1 \right] H \left[ \frac{K+M}{2KM}D+\frac{3}{4rKM}D^2 \right] . \end{aligned}$$
(16)
It is noted that \(\frac{\displaystyle d g^*}{\displaystyle d u}\left( \frac{\displaystyle K+M}{\displaystyle 2} \right) =r\left[ \frac{\displaystyle (K+M)^2}{\displaystyle 4KM} 1 \right] H=0\) when \(H=\frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\). Hence there exits at least one set of positive parameters, \(r,\;K,\;M,\;\text {and}\;H\) with \(M<K\) such that we assume \(H=\frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\epsilon ,\;\epsilon >0\) for \(H<\frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\), where \(\epsilon\) is small enough. Using the value of H, it is found that \(D=2\sqrt{r\epsilon KM}>0\). Substituting H and D in (16) yields
$$\begin{aligned} \frac{d g^*}{d u}(u_2)&=r+\epsilon \left[ \frac{K+M}{2KM}2\sqrt{r\epsilon KM}+\frac{3}{4rKM} 4r\epsilon KM \right] \\&=r+\epsilon  \frac{K+M}{KM}\sqrt{r\epsilon KM} 3\epsilon \\&=r2 \epsilon  \frac{(K+M)\sqrt{r\epsilon KM}}{KM}<0. \end{aligned}$$
In case of equilibrium point \(u_3\), it is seen that \(u_3\) is positive only for \(r(K+M)>D\). Then using the fact that, \(\frac{\displaystyle dg^*}{\displaystyle du}\left( \frac{\displaystyle K+M}{\displaystyle 2} \right) =0\) and \(H<\frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\) and from the above constructive manner, we found that \(\frac{d g^*}{d u}(u_3)>0\).
All the results from case III are summarized as follows:

1.
For \(D>0\), it is obvious that \(u_2 >0\). Using the fact that \(\displaystyle \frac{\displaystyle d g^*}{\displaystyle d u}\left( \frac{\displaystyle K+M}{\displaystyle 2}\right) =0\) and \(H<\frac{\displaystyle r(KM)^2}{\displaystyle 4KM}\), we have \(\displaystyle \frac{\displaystyle d g^*}{\displaystyle d u}(u_2) <0.\)

2.
When \(r(K+M)>D\) then \(u_3 >0\), and we have \(\displaystyle \frac{\displaystyle d g^*}{\displaystyle d u}(u_3) >0.\) If \(r(K+M)<D\), the solution \(u_3\) is strictly negative and the solution is not acceptable.
The stable and convergent solutions are presented in Fig. 3 both for weak and strong Allee effects. For large time, t, the solution \(u(t)\rightarrow u_2\) when M is positive (Fig. 3a). Similarly, in case of strong Allee effect, the solution u(t) is converging to \(u_1=0\) (Fig. 3b).
In conclusion, for weak Allee effect, we have two stable equilibria \(u_1=\left\{ 0, \; u_2\right\}\), and one unstable equilibrium state \(u_3\). Hence the solution of the model (14) is decreasing for \(0<u(t)<u_3\), and increasing for \(u_3<u(t)<u_2\).
In case of strong Allee effects, that is when \(M<0<K\), we put \(M\) instead of M, then we get the following equilibria
$$\begin{aligned} u_1= & {} 0,\\ u_2= & {} \frac{\displaystyle 1}{\displaystyle 2} \left[ (KM) + \frac{\displaystyle D^*}{\displaystyle r}\right] ,\;\text {and}\\ u_3= & {} \frac{\displaystyle 1}{\displaystyle 2}\left[ (KM)\frac{\displaystyle D^*}{\displaystyle r}\right] . \end{aligned}$$
Where \(D^*:= \sqrt{r^2(K+M)^2+4rKMH}\). Here we also get three cases in terms of H:
$$\begin{aligned} H<\frac{\displaystyle r(K+M)^2}{\displaystyle 4KM},\;\; H=\frac{\displaystyle r(K+M)^2}{\displaystyle 4KM},\;\text {and}\;\; H>\frac{\displaystyle r(K+M)^2}{\displaystyle 4KM}. \end{aligned}$$
Since \(D^*\) is not valid for this \(H<\displaystyle \frac{\displaystyle r(K+M)^2}{\displaystyle 4KM}\), so there is no equilibrium point, that is \(u_2, u_3 \in {\mathbb {C}}\setminus {\mathbb {R}}\).
On contrary, the value \(\frac{\displaystyle r(K+M)^2}{\displaystyle 4KM}\) is strictly negative since \(r, \; K\) and M are strictly positive. Hence H is negative for either, \(H=\displaystyle \frac{\displaystyle r(K+M)^2}{\displaystyle 4KM}\) or, \(H>\displaystyle \frac{\displaystyle r(K+M)^2}{\displaystyle 4KM}\), which is a contradiction of our earliest assumption that harvesting can not be negative. Considering this cases, we can conclude that there is no condition of \(D^*\) found, such that a positive equilibrium exists when the harvest is positive.
Therefore, if we allow harvesting in a strong Allee effect, the resulting behavior is diverging, and does not exist a stable point except possibly, \(u_1 = 0\).
Let us now introduce the diffusion term in our considered model for further study.