# Global and Local Structures of Bifurcation Curves of ODE with Nonlinear Diffusion

Authors: Tetsutaro Shibata
Journal: International Journal of Differential Equations
 (2018)

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## Abstract

We consider the nonlinear eigenvalue problemDuu′′+λfu=0, u(t)>0, t∈I≔(0,1), u(0)=u(1)=0, where D(u)=uk, f(u)=u2n-k-1+sin⁡u, and λ>0 is a bifurcation parameter. Here, n∈N and k (0≤k<2n-1) are constants. This equation is related to the mathematical model of animal dispersal and invasion, and λ is parameterized by the maximum norm α=uλ∞ of the solution uλ associated with λ and is written as λ=λ(α). Since f(u) contains both power nonlinear term u2n-k-1 and oscillatory term sin⁡u, it seems interesting to investigate how the shape of λ(α) is affected by f(u). The purpose of this paper is to characterize the total shape of λ(α) by n and k. Precisely, we establish three types of shape of λ(α), which seem to be new.


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## Body

## 1. Introduction

This paper is concerned with the following nonlinear eigenvalue problems:(1)Dutut′′+λfut=0,t∈I≔0,1,(2)ut>0,t∈I,(3)u0=u1=0, where D(u)=uk, f(u)=u2n-k-1+sin⁡u, and λ>0 is a bifurcation parameter. Here, n∈N and k (0≤k<2n-1) are constants. Bifurcation problems have a long history and there are so many results concerning the asymptotic properties of bifurcation diagrams. We refer to [1–8] and the references therein. Moreover, bifurcation problems with nonlinear diffusion have been proposed in the field of population biology, and several model equations of logistic type have been considered. We refer to [9] and the references therein. In particular, the case D(u)=uk (k>0) has been derived from a model equation of animal dispersal and invasion in [10, 11]. In this situation, λ is a parameter which represents the habitat size and diffusion rate. On the other hand, there are several papers which treat the asymptotic behavior of oscillatory bifurcation curves. We refer to [7, 12–19] and the references therein. Our equation (1) contains both nonlinear diffusion term and oscillatory nonlinear terms. The purpose of this paper is to find the difference between the structures of bifurcation curves of the equations with only oscillatory term and those with both nonlinear diffusion term and the oscillatory term in (1). To clarify our intention, let k=2 and n=2. Then (1) is given as(4)u2u′′+λu+sin⁡u=0. The corresponding equation without nonlinear diffusion is the case k=0 and n=1, namely,(5)u′′+λu+sin⁡u=0. It should be mentioned that, by using a generalized time-map argument in [9], for any given α>0, there exists a unique classical solution pair (λ,uα) of (1)–(3) satisfying α=uα∞. Furthermore, λ is parameterized by α as λ=λ(α) and is continuous in α>0. For (5), the following asymptotic formula for λ(α) as α→∞ has been obtained.Theorem 1 . (see [12]).
Consider (5) with (2)–(3). Then as α→∞,(6)λα=π2-4παπ2αsin⁡α-π4+oα-3/2.For (5) with (2)–(3), the asymptotic behavior of λ(α) as α→0 is as follows. For a solution pair (λ(α),uα) satisfying uα∞=α, put vα(t)≔uα(t)/α and let α→0. Then we easily obtain the function v0∈C2(I) which satisfies -v0′′(t)=2λ(0)v0(t), v0(t)>0 for t∈I with v0(0)=v0(1)=0. This implies λ(0)=π2/2. By this fact and Theorem 1, the bifurcation curve λ(α) starts from π2/2 and tends to π2 with oscillation and intersects the line λ=π2 infinitely many times for α≫1.Since (4) includes both the nonlinear diffusion function and oscillatory term, it seems interesting how the nonlinear diffusion functions give effect to the structures of bifurcation curves.Now we state our main results.Theorem 2.
Consider (1) with (2)–(3). Then as α→∞,(7)λα=4nα2k+2-2nAk,n2-2Ak,nπ2nαk+1/2-2nsin⁡α-π4+oαk+1/2-2n,where(8)Ak,n=∫01sk1-s2nds.By Theorem2, we obtain the global behavior of λ(α) as α→∞ for n=k=2 and see that the asymptotic behavior of λ(α) is completely different from that for k=0,n=1 by comparing Figures 1 and 2.Figure 1
The graph ofλ(α) for (5) (k=0,n=1).Figure 2
The graph ofλ(α) for k=n=2.Now we establish the asymptotic behavior ofλ(α) as α→0 to obtain a complete understanding of the structure of λ(α). Let(9)B0≔∫01sk1-sk+2ds,(10)B1≔k+212k+4∫01sk1-sk+41-sk+23/2ds,(11)B2=k+22n∫01sk1-s2n1-sk+23/2ds,(12)B3=nk+2∫01sk1-sk+21-s2n3/2ds.Theorem 3.
Consider (1) with (2)–(3). Then the following asymptotic formulas hold as α→0.(i)
Assume thatk+4<2n. Then(13)λα=2k+2αkB02+2B0B1α2+oα2.(ii)
Assume that2n=k+4. Then(14)λα=2k+2αkB02-10B0B1α2+oα2.(iii)
Assume thatk+2<2n<k+4. Then(15)λα=2k+2αkB02-B0B2α2n-k-2+oα2n-k-2.(iv)
Assume that2n=k+2. Then(16)λα=k+2αkB02+B0B1α2+oα2.(v)
Assume thatk+1<2n<k+2. Then(17)λα=4nα2k+1-nAk,n2-2Ak,nB3αk+2-2n+oαk+2-2n.The rough images of the graphs ofλ(α) for k=1,n=2, n=k=2, and k=1,n=3 are given in Figures 3, 4, and 5.Figure 3
The graph ofλ(α) for k=1, n=2.Figure 4
The graph ofλ(α) for k=n=2.Figure 5
The graph ofλ(α) for k=1, n=3.The proofs depend on the generalized time-map argument in [9] and stationary phase method (cf. Lemma 4). It should be mentioned that if we apply Lemma 4 to our situation, careful consideration about the regularity of the functions is necessary.


## 2. Proof of Theorem2

We put(18)Λ≔α>0∣fα>0,∫uαftDtdt>0for  allu∈0,α.It was shown in [9, (2.7)] that if α∈Λ, then λ(α) is well defined. In our situation, it is clear that, for t>0, D(t)>0, f(t)>0, so f(t)D(t)>0. Therefore, Λ≡R+. By this and the generalized time-map obtained in [9] (cf. (24)) and the time-map argument in [8, Theorem 2.1], we see that, for any given α>0, there exists a unique classical solution pair (λ,uα) of (1)–(3) satisfying α=uα∞. Furthermore, λ is parameterized by α as λ=λ(α) and is continuous in α>0. For u≥0, we put(19)Gu≔∫0ufyDydy=12nu2n+G1u≔12nu2n+∫0uyksin⁡ydy. It is known from [9] that if (uα,λ(α))∈C2(I¯)×R+ satisfies (1)–(3), then(20)uαt=uα1-t,0≤t≤1,(21)uα12=max0≤t≤1⁡uαt=α,(22)uα′t>0,0<t<12. In what follows, we denote by C various positive constants independent of α≫1. For 0≤s≤1 and α≫1, we have(23)G1α-G1αsα2n1-s2n=∫αsαwksin⁡wdwα2n1-s2n≤Cαk+11-sk+1α2n1-s2n≤Cαk+1-2n≪1. By this, (19), and Taylor expansion, we have from [9, (2.5)] that(24)λα2=∫0αDuGα-Gudu=∫0αuk1/2nα2n-u2n+G1α-G1udu=2nαk+1-n∫01sk1-s2n+2n/α2nG1α-G1αsds=2nαk+1-n∫01sk1-s2n1-nα2nG1α-G1αs1-s2n1+o1ds=2nαk+1-n∫01sk1-s2nds-nα2nLα1+o1, where(25)Lα≔∫01sk1-s2n3/2G1α-G1αsds. We see from (24) and (25) that if we obtain the precise asymptotic formula for L(α) as α→∞, then we obtain Theorem 2. To do this, we apply the stationary phase method to our situation. By combining [13, Lemma 2] and [7, Lemmas 2.24], we have the following equality.Lemma 4 (see [13, Lemma 2 and 10, Lemma 2.24]).
Assume that the functionf(r)∈C2[0,1], w(r)∈C3[0,1], and(26)w′r<0,r∈0,1,w′0=0,w′′0<0.Then as μ→∞(27)∫01freiμwrdr=12eiμw0-π/42πμw′′0f0+O1μ.In particular, by taking the imaginary part of (27), as μ→∞,(28)∫01frsin⁡μwrdr=122πμw′′0f0sin⁡w0μ-π4+O1μ.We note that, to obtain (27), we have to be careful about the regularity of f and w.Lemma 5.
Asα→∞,(29)Lα=π21n3/2αk+1/2sin⁡α-π4+Oαk.Proof.
We puts=sin⁡θ and(30)Yθ≔Y1θG1α-G1αsin⁡θ≔sinkθ1+sin2θ+⋯+sin2n-2θ3/2G1α-G1αsin⁡θ. By integration by parts, (25) and (30), we have(31)Lα=∫01skG1α-G1αs1-s23/21+s2+⋯+s2n-23/2ds=∫0π/21cos2θsinkθG1α-G1αsin⁡θ1+sin2θ+⋯+sin2n-2θ3/2dθ≔L1α-L2α=tan⁡θYθ0π/2-∫0π/2tan⁡θY1θG1α-G1αsin⁡θ′dθ.By l’Hôpital’s rule, we obtain(32)limθ→π/2G1α-G1αsin⁡θcos⁡θ=limθ→π/2αcos⁡θαsin⁡θksin⁡αsin⁡θsin⁡θ=0. This implies that L1(α)=0. Next,(33)L2α=∫0π/2tan⁡θY1′θG1α-G1αsin⁡θ-Y1θαcos⁡θαsin⁡θksin⁡αsin⁡θdθ.≔L21α-L22α. We first calculate L21(α). Assume that k>0. Then(34)Y1′θ=sink-1θcos⁡θ1+sin2θ+⋯+sin2n-2θ3/2×k-3sin2θ+2sin4⁡θ+⋯+n-1sin2n-2θ1+sin2θ+⋯+sin2n-2θ.This implies that, for α≫1,(35)tan⁡θY1′θ≤Csinkθ≤C.By direct calculation, we also obtain (35) for the case where k=0. By integration by parts, we obtain(36)G1α-G1αsin⁡θ=∫αsin⁡θαwksin⁡wdw≤-wkcos⁡wαsin⁡θα+∫αsin⁡θαkwk-1cos⁡wdw≤Cαk. By (35) and (36), for α≫1, we obtain(37)L21α=tan⁡θY1′θG1α-G1αsin⁡θ≤Cαk. Since(38)L22α=αk+1∫0π/2Y1αsink+1θsin⁡αsin⁡θdθ, by putting θ=(π/2)(1-r), we obtain(39)L22α=π2αk+1∫01cos2k+1π/2r1+cos2π/2r+⋯+cos2n-2π/2r3/2sin⁡αcos⁡π2rdr.Let(40)fr≔cos2k+1π/2r1+cos2π/2r+⋯+cos2n-2π/2r3/2,wr≔cos⁡π2r,μ≔α.Case 1. Assume that k>1/2 or k=0. Then clearly f(r)∈C2[0,1], and we are able to apply Lemma 4 to (39). Then we obtain(41)L22α=π21n3/2αk+1/2sin⁡α-π4+Oαk. By this, (33), and (37), we obtain (29).
Case 2. Assume that 0<k<1/2. Then f(r)∈C1+2k[0,1] with 0<2k<1. Therefore, f(r) does not satisfy the condition in Lemma 4. However, we found in [14] that we can still apply Lemma 4 to (39) in this situation and obtain (41). For completeness, the reason will be explained in the Appendix. By this, (33), and (41), we obtain (29). Thus the proof is complete.By (24) and Lemma 5, we obtain Theorem 2 immediately. Thus the proof is complete.


## 3. Proof of Theorem3

In this section, let0<α≪1. The proofs of Theorem 3 (i)-(v) are similar. Therefore, we only prove (i) and (iv).Proof of Theorem3 (i).
We assume that2n>k+4. Then by Taylor expansion, for 0≤s≤1, we have(42)Gα-Gαs=12nα2n1-s2n+1k+2αk+21-sk+2-16k+4αk+41-sk+41+o1. By this, (24), Taylor expansion, and putting u=αs, we obtain(43)λα2=∫0αuk1/2nα2n-u2n+1/k+2αk+2-uk+2-1/6k+4αk+4-uk+41+o1du=k+2αk/2∫01sk1-sk+21-k+2/6k+41-sk+4/1-sk+2α2+oα2ds=k+2αk/2∫01sk1-sk+21+k+212k+41-sk+41-sk+2α2+oα2ds=k+2αk/2B0+B1α2+oα2. This implies (13). Thus the proof is complete.Proof of Theorem3 (iv).
We assume that2n=k+2. Then by (42), for 0≤s≤1, we have(44)Gα-Gw=2k+2αk+2-wk+2-16k+4αk+4-wk+41+o1. By this, (24), and putting w=αs, we obtain(45)λα2=∫0αwk2/k+2αk+2-wk+2-1/6k+4αk+4-wk+41+o1dw=k+22αk/2∫01sk1-sk+21-k+212k+41-sk+41-sk+2α2+oα2-1/2ds=k+22αk/2∫01sk1-sk+21+k+224k+41-sk+41-sk+2α2+oα2ds.This implies(46)λ=k+2αk/2B0+12B1α2+oα2.This implies (16). Thus the proof is complete.



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