Dynamics of Lotka-Volterra Competition Systems with Fokker-Planck Diffusion

We consider competition systems of two species which have different dispersal strategies and interspecific competitive strengths. One of the dispersal strategies is random dispersal and the other is a Fokker-Planck diffusion whose motility is piecewise constant and jumps upwhen the resource is not enough. In this paper, first we show the Fokker-Planck diffusion allows ideal free distribution. Next we show the linear stability of semitrivial steady states is determined exactly by a threshold on the interspecific competitive strengths. Some conditions for coexistence and global asymptotic stability are also provided.


Introduction and Main Result
In this paper, we study Lotka-Volterra competition systems with Fokker-Planck diffusion of the following form: ( 1 ( 1 ) ) =  ] ( 2 ( 2 ) V) = 0 on Ω,  (, 0) =  0 () ≥ 0, where Ω is a smooth bounded domain in R  and a given resource distribution () is a nonconstant  2 -function satisfying min Ω  > 0. The motility functions  1 ,  2 of Fokker-Planck diffusion are assumed to depend only on respectively.High values of  1 ,  2 mean a scarcity of resources; the difference between them is that  1 depends on the competitive strength of other species while  2 does not.The nonnegative constants  1 ,  2 are interspecific competitive strengths.The intraspecific competitive strengths are assumed to be one.We consider two types of motility functions.The first one is the random dispersal (RD) where   's are constant.The other is based on the following step function: where  and ℎ are positive constants satisfying  < ℎ.For a single species case, if we let  = /, the motility function  (0) models the situation where organisms drastically change their departing probability in random walk from  to ℎ exactly when the resource becomes insufficient.This motility function is called starvation-driven diffusion (SDD) [1].
Unfortunately, due to the jump discontinuity, the Laplacian of  (0) () is not well-defined.So we consider a regularized function  () which is a nondecreasing smooth function such that  ≤  () ≤ ℎ and for given  > 0. For example, we may choose  () fl  (0) *  ()  where  () is a standard mollifier supported in (0, ).
The second motility function is given by this  () .In such cases we may choose  1 ( 1 ) fl  () ( 1 ) or  2 ( 2 ) fl  () ( 2 ).For simplicity we will call the first one SDD 1 and the second one SDD 2 .
Our first main result shows why the SDD is worthy of notice; it is an efficient dispersal strategy helping the steady state be the ideal free distribution (IFD).

𝜕 𝜕]
( () () ) = 0 on Ω, where If the resource distribution  satisfies The uniqueness assumption in Theorem 1 can be replaced by some sufficient conditions.For example, a condition on the resource distribution, max guarantees the uniqueness [2, Theorem 3(i)].There are other sufficient conditions restricting the shape of  () [2, Theorem 2] or the shape of  [2, Theorem 3(ii)].Condition (7) seems not to be a real restriction because it is necessary even when there is no logistic growth term.In [2, Theorem 5], with no logistic growth term, it was shown that
With the logistic term the same IFD property (10) is proved in [2, Theorem 6].But it needs a restrictive condition on  [2, equation (62)] and relies on the explicit form of  () .Although our decay order in (8) is worse than the order in (10), our theorem does not have such restrictions.
In the sense of Theorem 1, the motility function  () gives a best possible dispersal strategy in single species case; the steady state population distribution   () is very close to the resource distribution and furthermore the steady state is globally asymptotically stable [2, Theorem 2].
Next we consider competitions between two dispersal strategies among RD, SDD 1 , and SDD 2 .For instance, to observe the competition between SDD 1 and RD, we let  1 fl  () ( 1 ) and  2 fl  for some positive constant  in (1).Then system (1) can have several steady states such as the trivial steady state ( ≡ 0, V ≡ 0) and semitrivial steady states (  () , 0) and (0,   ).Here   () is a positive solution of ( 5) and   is a unique positive solution of Inspecting the linear stability of each steady state, we can understand which dispersal strategy is more advantageous.
In the competition between SDD 1 and RD, the linear stability of the semitrivial steady state (0,   ), which means RD prevails and SDD 1 goes extinct, changes exactly across a threshold  * 2 > 1 on RD's interspecific competitive strength.Such a threshold exists for the linear stability of (  () , 0) but it is one so less than  * 2 .This suggests that SDD 1 gains an advantage over RD since the RD needs more interspecific competitive strength to prevail.In the competition between RDs, the RD could prevail with smaller interspecific competitive strength  2 < 1 [3, Theorem In the competition between RDs, the one with smaller diffusivity always prevails [4,5] if  1 =  2 = 1.Also even if  2 is slightly less than one (V has weak strength), a lower diffusivity of V (V has strong dispersal strategy) can overcome the disadvantage and make V prevail [3].
But in the competition between SDD 1 and RD, Theorem 2 (i) shows that RD cannot prevail no matter how small diffusivity it has if  2 <  * 2 .When  1 =  2 = 1, the same observation was given in [6, Theorem 1].Furthermore Theorem 2 (ii) claims that SDD 1 loses its advantage as soon as  1 is smaller than one.This means that low interspecific competitive strength of SDD 1 cannot be overcome by the dispersal strategy, contrary to the RD case.This is unexpected since the SDD 1 seems to be more competitive than RD; the SDD 1 can build up the ideal free distribution while the RD cannot.
In the competition between RD and SDD 2 , the dynamics is similar to the case of Theorem 2 but SDD 2 is less competitive than SDD Then there is a coexistence steady state for all sufficiently small .(iv) If  1 < 1 <  2 and  ≤ , then (0,   () ) is globally asymptotically stable for all sufficiently small .
1 is not necessarily bigger than one.So smaller interspecific competitive strength is enough for RD to prevail against SDD 2 than the strength required against SDD 1 is.By Theorem 3 (iv), we can see that Finally, we consider the competition between SDD 1 and SDD 2 .
Although we expect SDD 1 to be more competitive than SDD 2 , our linear stability analysis does not indicate it.We solve system (1) numerically to observe it in the next section (see Figure 3).
The proofs of the theorems are given in the subsequent sections.

Numerical Simulation and Conjecture
In this section, by numerically solving the competition system (1), we observe how competitive the dispersal strategies RD, SDD 1 , and SDD 2 are.
When the interspecific competitive strengths are small ( 1 ,  2 ≪ 1), the opponents are relatively weak so two species can coexist (see [3,7,11,12] for other models).Numerical simulation shows in our case the theoretically claimed constants  * 2 ,  * 1 or one play as the "smallness" thresholds of the interspecific competitive strengths for coexistence.For example, in Figure 1, two species coexist if  1 < 1 and  2 <  * 2 .Combining our analysis and numerical observation, we give some conjectures.with the parameters in (14).In the red and blue region SDD 1 and RD prevail, respectively.In any cases the SDD 1 survives near Conjecture.For simplicity, we ignore unstable coexistence steady states.Assume  is sufficiently small.Then we conjecture the following: (1) When SDD 1 competes with RD as in Theorem 2 (see Table 1).

Proof of Theorem 1
To prove Theorem 1, we need some lemmas.In the rest of the paper, we omit the -dependence in proofs if there is no confusion.
Lemma 6 (approximation from below).Assume the hypotheses in Theorem 1 and fix a number  such that 0 <  < 1.Then the Lebesgue measure of the set for some positive constant  independent of .
Proof.By [2, Lemma 1(ii)], we know By definition of Ω () and Lemma 5, the right-hand side is estimated by (1 + ) . (28) Combining the above inequalities, we have < where  is a positive constant independent of .Because   has a uniform bound independent of , so does ∇(()  ).Furthermore, from the computation we have From this relation, we can observe that ∇  also has a uniform bound as claimed; if we divide the both sides by  +   (  /), then we can easily see that the gradient has a bound. ( for any  ≥ 1. Applying the previous claims, we have The exponent of  is maximized when  = √ and the proof is complete.

Proof of Theorem 2
First we introduce the diffusion pressure  for the first species : Then for given , V, and , the function  can be completely determined by the equation G ≡ 0 since  is a nondecreasing function.Hence, we may write  = (,V, ), and ( 1) is rewritten as where  0 fl (( 0 +  2 V 0 )/) 0 .Let (  , V  ) be a steady state solution of (1).The stability of (  , V  ) is equivalent to that of the steady state (  , V  ) of (40) for   fl ((  +  2 V  )/)  .Let  =   + Φ and V = V  +  with |Φ| and || small.Then,  =  (, V  + ,   + Φ) =   +  V  +   Φ + higher order terms (41) and we have the following linearized eigenvalue problem: At the semitrivial steady state (0,   ), where So the eigenvalue value problem for (40) at (0,   ) is Also consider the other semitrivial steady state (  , 0) and define Then, similarly, the eigenvalue problem linearized at (  , 0) is We will use the following stability criteria for semitrivial steady states.The proof is similar to that of [6, Lemma 1], but we give it here for completeness.Lemma 7. Let   be unique positive steady states of ( 5).
Proof.(i) First suppose  1 > 0. For a given , consider the eigenvalue problem and denote its first eigenvalue by   1 .Then  0 1 =  1 > 0 by (48).Moreover,   1 will be negative when  is large enough.Thus, there exists  1 > 0 such that   1 1 = 0 and the corresponding eigenfunction Φ 1 satisfies the first equation in (45) with  =  1 .Now due to (11), it is easy to see that the operator is invertible.Hence there is a solution of L = ( 1   / ( 2   ))Φ 1 and it satisfies the second equation in (45).This yields that  1 > 0 is an eigenvalue of the linearized problem (45), which implies that (0,   ) is linearly unstable.
(ii) Consider the linearized eigenvalue problem of (5) at   , which is written as Since   is a unique positive steady state, it is linearly neutrally stable (in fact it is globally asymptotically stable [2, Theorem 2]).Hence every eigenvalue of the linear operator Δ + ( − 2  )(/) is nonpositive.Thus, if ] 1 > 0, because of G/ = (  ) +   (  )  > 0, the operator has strictly negative eigenvalues and thus it is invertible.Let {] 1 ,  1 } be an eigenpair of (49) and Φ 1 be the solution of Then, {] 1 , ( 1 , Ψ 1 )} is an eigenpair of the linearized problem (47), which implies that (  , 0) is linearly unstable.Since an eigenvalue of (47) is also an eigenvalue of (49), eigenvalues of (47) are all strictly negative if ] 1 < 0 and hence (  , 0) is linearly stable.Now we are ready to prove Theorem 2. Due to Lemma 7, it is sufficient to compute the sign of the first eigenvalues  1 and ] 1 .
Similarly, for a given  1 < 1, there is a  > 0 such that for any sufficiently small .
Then with  1 fl 1, (61) (iii) Suppose there is a positive coexistence steady state (, V).Then by (40), Considering the Rayleigh quotient of this equation, we have (63) Similarly we also have by the assumption and (64) .
Now observe the sign of the integrand.If  1 = ( +  2 V)/ ≤ 1, ( 1 ) =  so the integrand vanishes.Hence the integral is actually taken over the set where ( +  2 V)/ > 1, and then the last integral is positive.This is a contradiction so there is no coexistence steady state.

Proof of Theorem 3
In this section, we consider the following system:

Proof of Theorem 4
We consider the following system: where respectively.