# The Bifurcation Curves of a Category of Dirichlet Boundary Value Problems

Authors: Huizeng Qin; Youmin Lu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

We study the Dirichlet boundary value problemu″t+λfut=0,−1<t<1,u−1=u1=0, generally and develop a schema for determining the relationship between the values of its parameters and the number of positive solutions. Then, we focus our attention on the special cases when fu=σ−uexp−K/1+u and fu=∏i=1mai—u, respectively. We prove first that all positive solutions of the first problem are less than or equal to σ, obtain more specific lower and upper bounds for these solutions, and compute a curve in the σK -plane with accuracy up to 10−6, below which the first problem has a unique positive solution and above which it has exactly three positive solutions. For the second problem, we determine its number of positive solutions and find a formula for the value of λ that separates the regions of λ, in which the problem has different numbers of solutions. We also computed the graphs for some special cases of the second problem, and the results are consistent with the existing results. Our code in Mathematica is available upon request.


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## Body

## 1. Introduction

As is well known, Dirichlet boundary value problems have a variety of applications in physics, chemistry, and mathematics [1–7]. In recent years, there has been considerable work on the study of the number of positive solutions and bifurcation diagrams of various special cases of the following Dirichlet boundary value problem (DBVP) [1–21].(1)u″t+λfut=0,−1<t<1,u−1=u1=0,where, λ is a positive parameter. In this paper, we study DBVP (1) generally first and then develop a schema that can be used to determine the exact number of positive solutions of DBVP (1). Then, we focus our attention on two specific problems of DBVP (1) for fu=σ−uexp−K/1+u and fu=∏i=1mai—u. The first problem was derived from a chemical reaction by Aris R. [7], and the second one is a generalization of the problems studied in [2, 4]. We apply our general schema to the first problem, establish an algorithm for efficient computation, and improve the results obtained by K. J. Brown, M.M.A. Ibrahim, and R. Shivaji [4] significantly. More specifically, K. J. Brown, M.M.A. Ibrahim, and R. Shivaji show that the first problem has a unique solution when K≤4 or K>4 and σ<4/K−4, we use a new idea that converts the corresponding boundary value problem into its equivalent integral equation and establishes some precise upper and lower bounds for its solutions, and then use our algorithm to compute a curve in the σK -plane with accuracy up to 10−6, below which the first problem has a unique positive solution and above, which it has exactly three positive solutions. To our knowledge, this curve is completely new. The second problem is a generalization of the problems studied in [2, 4]. While Korman et al. [4] studied a special case of our second problem for fu to be a third degree polynomial, and Kirichuka and Sadyrbaev studied the special case for fu to be a seventh degree polynomial, we apply our schema to polynomials of any degree and give a formula of the parameter λ that separates the regions of the values of λ, in which the problem has different numbers of solutions. Our results recover the ones obtained in [2, 4].


## 2. General Theorem

It has been proved [8] that all the positive solutions of DBVP (1) are even functions with u′t<0 for 0<t<1 and hence, any positive solution is uniquely identified by the value α=u0. Now, we consider the initial value problem(2)u″t+λfut=0,0<t<1,u0=α,u′0=0.Integrating (2) once, we get the following equation:(3)u′2t=2λ∫utαfvdv,0<t<1,u0=α.Becauseu′2t≥0 and λ>0, it is necessary to have(4)∫utαfvdv≥0,0<u<α.Sinceu′t<0 over (0,1), one may get(5)u′t∫utαfvdv−1/2=−2λ,0<t<1,u0=α.With one more integration, we get part of an implicit solution of problem (2)(6)∫utα∫uαfvdv−1/2du=−2λt,0<t<1.Becauseut is an even function, we may set (7)t=12λ∫utα∫uαfvdv−1/2du=1,t>0,−1,t<0.As a solution of DBVP (2), (6) satisfies(8)∫0α∫uαfvdv−1/2du=2λ.It is also easy to see that, for any givenα, if λ>0 is determined by (8), the function ut determined by (7) is a positive solution of DBVP (1) with its maximum α at t=0.Theorem 1.
The number of positive solutions of DBVP (1) is equal to the number of roots α of the equation(9)Hfα=2λ,where(10)Hfx=∫0x∫uxfvdv−1/2du,(11)∫uxfvdv≥0,0<u<x.
Furthermore, the corresponding positive solutions of DBVP (1) are given by (7).Remark 1.
If we get the rootsαii=1,2,⋯,N of (9), we get N implicit solutions by (7) (12)t=12λ∫utαi∫uαifvdv−1/2du=1,t>0,−1,t<0.for i=1,2,⋯,N. Since (10) is an improper double integral, direct computation is very inefficient. We need to develop an algorithm to improve its computation efficiency and increase its accuracy.
In the following sections, we focus our attention on the specific problems whenfu=σ−uexp−K/1+u and fu=∏i=1mai−u, respectively.


## 3. Bounds and the Bifurcation Diagram of the Solutions of DBVP (1) for fu=σ−uexp−K/1+u

In this section, we consider the special case of DBVP (1):(13)u″t+λσ−utexp−K1+u=0,−1<t<1,u−1=u1=0,where σ and K are positive constants, which is a problem arising in chemical reaction theory [7].First, we establish some bounds for all the positive solutions to this problem by proving the following theorem.Theorem 2.
Ifu is a positive solution of DBVP (13), then u has only one extreme point that has a maximum value at t=0 and satisfies(1)
0<ut≤σfor−1<t<1;(2)
λe−Kσ/2+λe−K<u0≤λσ/2e−K/1+σ; and(3)
for−1<t<1.Proof.
First, we prove thatut≤σ for all −1≤t≤1 by dividing it into two cases.(a)
ut has a local maximum ut0<σ. Assume that t1 is the first value of t such that t0<t1<1 and u′t1=0. Then, ut is decreasing for all t0<t<t1 and u″t=−λσ−utexp−K/1+u<0 for all t0<t<t1. Hence, u′t is decreasing over t0,t1. This is a contradiction to the assumption u′t1=0 and therefore, ut has ut0 as its only extremum and t0=0 by its even property.(b)
ut does not have a local maximum before it reaches σ at t0. Now, one has u″t0=0. If ut<σ for some small interval t>t0, then ut will never be able to get back to σ as we discussed in case 1. If ut<σ for some small interval after t0, then ut will never be able to get back to σ and cannot satisfy the boundary condition since u″t=−λσ−utexp−K/1+u<0 in this interval and ut cannot have a maximum.
To prove the rest of our result in this theorem, we need to convert DBVP (13) into its equivalent integral equation.(14)ut=λ2∫−11σ−usGs,texp−K1+usds,t∈−1,1,where, Green’s function Gs,t is calculated as follows:(15)Gs,t=1−t1+s,−1<s≤t<1,1+t1−s,−1<t≤s<1.
It is clear thatGs,t≥0 for all −1<s<1 and −1<t<1, and ∫−11Gs,tds=1−t2. Since e−K/1+x is an increasing function with respect to its variable x, one has(16)ut≤λ2∫−11σGs,te−K/1+σds=λσ2e−K/1+σ1−t2for−1<t<1.
In particular, we have(17)u0≤λσ2e−−K/1+σ.
By using (14) again, we get(18)ut>λ2∫−11σ−u0Gs,te−Kds≥λ2σ−u0e−K1−t2for−1<t<1.
Settingt=0 in (18), one can get(19)u0>λe−Kσ2+λe−K.
Combining (17) and (18), we have(20)ut>λσ21−λ2e−K/1+σe−K1−t2for−1<t<1.
The theorem is proved.
In [4], Brown et al. proved that DBVP (13) has a unique solution if K≤4 or K>4 and σ<4/K−4. To explain our approach below, we first recall the idea of their proof. Let fx,σ,K=σ−xe−K/1+x and Fx,σ,K=∫0xfv,σ,Kdv, equation (9) can be written as follows:(21)x∫01Fx,σ,K−Fxs,σ,K−1/2ds=2λ.
Now, we denote the left side of (21) by Gx,σ,K and take its derivative with respect to x.(22)Gx′x,σ,K=∫01Lx,σ,K−Lxs,σ,KFx,σ,K−Fxs,σ,K3/2ds,where, Lx,σ,K=Fx,σ,K−1/2xfx,σ,K. For the solution of DBVP (13) to be unique, we need Gx,σ,K to be monotone. We take the derivative of Lx,σ,K with respect to x.(23)Lx′x,σ,K=12fx,σ,K−12xfx′x,σ,K=−x22ddxfx,σ,Kx=−x22e−K1+x−σ+Kx−σK−2/2σ+K2−σK4−σK+4σ/4σ+Kx21+x2.
Therefore,Lx′x,σ,K>0 and then Lx,σ,K−Lxs,σ,K≥0 when K≤4 or K>4 and σ<4/K−4 for 0<s<1. In turn, we have Gx′x,σ,K≥0 when K≤4 or K>4 and σ<4/K−4 for all x≥0. Therefore, Gx,σ,K is monotone and DBVP (13) has a unique solution when K≤4 or K>4 and σ<4/K−4. Hence, the condition K>4 and σ<4/K−4 or K>4+4/σ is necessary for DBVP (13) to have multiple (two or three) solutions. Since the integral Gx′x,σ,K can still be positive when Lx,σ,K−Lxs,σ,K is less than zero in a small interval, there should be some ε>0 such that DBVP (13) has a unique solution when 4<K>4+ε+4/σ.
Let’s first work on this value ofK heuristically first. When σ≫u>0 (this is actually a reasonable assumption when we realize that σ is forced to go to infinity by the condition σ>4/K−4 when K is getting close to 4), DBVP (13) is approaching the DBVP(24)u″t+λσexp−Kα+u=0,−1<t<1,u−1=u1=0,which brings us back to the well-known one-dimensional perturbed Gelfand two-point DBVP(25)u″t+λexpαuα+u=0,−1<t<1,u−1=u1=0.
In [19], we get that DBVP (25) has exactly three solutions when α>α0 and 4.0686722336<α0 < 4.068722344. Using the transformation v=u/α, one may convert problem (25) to(26)u″t+λαexp−α1+v=0,−1<t<1,v−1=v1=0.
Comparing (24) with (26), we can conclude that DBVP (23) or (13) has exactly three positive solutions when K>α0. In fact, this heuristic result is quite close to our result obtained in the following numerical computation:
Through the above discussion, we see that there is a region contained inK>4 and σ>4/K−4 in the σK -plane such that DBVP (13) has exactly three solutions. Now we use the result in Section 2 to develop an efficient algorithm for computing this region.
Let’s introduce a curveK=Sσ defined by(27)Sσ=infK|3.1has multiple roots forK>0,σ>0.
Since it is very inefficient to compute improper double integrals, we apply integration by parts on the integral in (11), expand fu=σ−uexp−Ku+1 into its power series(28)fu=σ−uexp−Ku+1=σ+K−u−2σ+1+KK21+u+∑n=2∞σ+1n+1+K−Knn+1!1+un,and plug back to equation (10) to get(29)Hσ,KC=λ2,where(30)Hσ,Ku=eKσgσ,K0,u+∫0u1σ−w−K1+w2eK/1+wgσ,Kw,uσ−wdw,(31)gσ,Kw,u=∫wuσ−vexp−Kv+1dv=σ+Ku−w−u2−w22−2σ+1+KK2ln1+u1+w+∑n=2∞σ+1n+1+K−Knn−1n+1!1+un11+wn−1−11+un−1.
It follows (28) and (29) that limu⟶0Hσ,Ku=0,limu⟶σ−0Hσ,Ku=∞, and Hσ,Ku is defined in 0,σ. Hence, it is clear that DBVP (13) has positive solutions. By equation (23), we can find that the left side of equation (9) for the problem in this section may have zero or one local maximum, which implies that equation (9) may have one or two roots or three roots for given parameters λ,σ, and K. This recovers the result for this problem given by Brown et al. [9].
Using numerical methods, we look forS=S1σ and S=S2σ so that the bifurcation diagram is monotone when K<S1σ and it is S-shaped when K>S2σ. The following results are obtained by using our code in Mathematica.
Curve fitting of the data in Table1 tells us the following formulas for functions S1σ and S2σ and the figures:(32)S1σ=4.069519+4.1340721σS2σ=4.070043+4.1340804σ.
Figure1 shows that the fitting between (32) and the data in Table 1 work very well. Functions in (32) also show that S=S1σ and S=S2σ almost coincide. Using the fitting functions S=S1σ and S=S2σ determined by (32), we get the following values Figure 2
For the data in Table2, we draw Hσ,Ku to verify the correctness of the above conclusion. As one can see from the second figure, when σ,K is on K=S1σ, y=Hσ,Ku is monotonically increasing overall, so (29) has only one root. One can also see from Figure 3, when σ,K is on K=S2σ and y=Hσ,Ku has a maximum and a minimum overall, so (29) has three roots.
To refine the curves, we use our code in Mathematica to get the following data. Curve fitting of the data in Table3 provides us the following functions S1σ and S2σ:(33)S1σ=4.069896776+4.133808594σ,σ<5,4.068807657+4.137501617σ,5≤σ<14,4.068687154+4.138886208σ,14≤σ,S2σ=4.069896942++4.133808514σ,σ<5,4.068807664+4.137502175σ,5≤σ<14,4.068687389+4.138883103σ,14≤σ.
Since the functionsS1σ and S2σ in (33) are so close to each other, we can use either K=S1σ or K=S2σ for K=Sσ if we limit our accuracy up to 10−6. Now, let’s take K=Sσ=S1σ, we get the following values for some σ:
If we add1/50000 to each piece of the data in Table 4, the corresponding graph of Hσ,Ku must have a minimax value. Their images are shown in Figure 4 below.
Because both the extremum interval and the difference between the extrema are very small, the curve drawing is very time-consuming and error-prone. To avoid these problems, we take 41 points on the curve. All the curves in Figure4 have clear maximum and minimum, so DBVP (13) has exactly three solutions in this case. The curve K=Sσ separates the first quarter of the σK -plane such that DBVP (13) has exactly one solution when σ,K is below the curve K=Sσ and has exactly three solutions when σ,K is above the curve K=Sσ.Table 1
The numerical results forSσ.
σ1258341322468101420S1σ123361000267125095819100008204110000341315000613710005103110000237950022935001793400174540011371000S2σ987802136920005989625410235000170672500491801275925009517200091732000448310004464100049180Figure 1
Graphs ofS1σ and S2σ.Figure 2
Graphics ofHσ,Ku for S1σ in Table 2.Table 2
The numerical results forSσ.
σ1116943579S1σ10.08275.723155.447544.896334.660104.52886S2σ10.08325.723685.448054.896856.660614.52937Figure 3
Graphics ofHσ,Ku for S2σ in Table 2.Table 3
More numerical results forSσ.
σ1234124681014203040S1σ25872685320971520313980851332768000013441748911638400003142200751200002612821151200002436307751200002935047640000459014131024000071505027163840004378249102400053845131280000213614475120000S2σ12936342710485760156990426316384000053766995676553600003927751640000522564231024000012181539256000046960753102400007344226116384000286020116553600140103973276800861522120480002670181640000Table 4
The numerical results forSσ.
σ1116916356080Sσ10.082714.528534.327374.186944.137674.12042Figure 4
Graphics ofHσ,Ku for S2σ in Table 2.


## 4. Number of Solutions of DBVP (1) for fu=∏i=1mai−u

In this section, we consider the special case of DBVP (1): (34)u″t=−λ∏i=1mai−u=−λ∑i=0mAiui,−1<t<1,u−1=u1=0,where, a1,a2,⋯,am are constants satisfying 0<ai<aj for 0≤i<j≤m,Am=−1m and Am−k=−1m−k∑1≤i1<i2<⋯<ik≤mai1ai2⋯aik for k=1,2,⋯,m. Let gx=∑i=0mAi/i+1xi+1, then(35)∫uxfvdv=gx−gu=x−uhx,u,where(36)hx,u=∑i=0mAii+1∑j=0ixi−juj.By (34), we see that(37)fu<0,u∈a2k+1,a2k+2,k=0,1,⋯,m−32,>0,u∈a2k,a2k+1,k=1,2,⋯,m−12,where, a0=0 and(38)fu<0,u>amfor evenm,>0,u>amfor oddm.Whenα∈a2k+1,a2k+2 or α>am for odd m, by (37) and (38), it is clear that the solution of the initial value problem (2) cannot satisfy DBVP (1). However, under some conditions, there are still multiple solutions for DBVP (1).Lemma 1.
Ifga2k+1−ga2k−1>0,k=1,2,⋯,m−1/2,m=2,3,4,⋯, let(39)ak∗=minu|gu−ga2k−1≥0,u>a2k.
Then, the function(40)Hx=∫0xdvgx−gv=∫0xdvx−vhx,v=x∫01dv1−vhx,xv.
Satisfies(41)limx⟶a2k+1−0Hx=+∞,limu⟶ak∗+0Hx=+∞.Proof.
By the mean value theorem of integrals,(42)∫ua2k+1fvdv=∫ua2k+1a2k+1−v∏i=1,i≠2k+1mai−vdv=a2k+1−u22∏i=1,i≠2k+1mai−ξ,where, ξ∈u,a2k+1 for k=0,1,⋯,m−2/2.. Thus,(43)limx⟶a2k+1Hx=limx⟶a2k+1∫0xdvgx−gv=+∞.
One can do it similarly to get(44)limu⟶ak∗+0Hx=+∞.Theorem 3.
Under the assumption of Lemma1, if m=2,4,6,⋯, DBVP (34) has at least two positive solutions for all λ>0 and Hx has m−2/2 minimum values Ha^k,k=1,⋯,m−2/2, such that(45)Ha^1≤⋯≤Ha^m−2/2,where, a^k is the minimum point, α^k∈αk∗,a2k+1,k=1,⋯,m−2/2. Further, the following result holds:(1)
DBVP (34) has m positive solutions(46)t=12λ∫uαidvαi−vhαi,v1,t>0−1,t<0,i=1,2,⋯,m.forλ>1/2H2a^m−2/2 and m>2, where αii=1,2,⋯,m are the m roots of the equation(47)Hx=2λ.(2)
There are 3 tom−1 positive solutions of DP (1) for 1/2H2a^1≤λ≤1/2H2a^m−2/2.Proof.
By Lemma1, (45) holds. By (36), we have(48)hx,xu=∑i=0mAii+1xi∑j=0iuj.
By (43) and (47), we have(49)limx⟶0Hx=0,limx⟶∞Hx=limx⟶∞x∫01∑i=0mAii+1xi∑j=0ivj−1/2dv1−v=0.
By (44) and (49) and the continuity of Hx over 0,a1 and am,∞, the solution of DBVP (34) exists on each of these two intervals, i.e.,(50)t=12λ∫uαidvαi−vhαi,v1,t>0−1,t<0,i=1,2,for α1<a1,Hα1=2λ and α2>a2,Hα2=2λ.
Without loss of generality, supposeHa^1<⋯<Ha^m−2/2.. If λ=Hα^1, in addition to the two solutions (50), the function(51)t=12λ∫uα^1dvα^1−vhα^1,v1,t>0,−1,t<0,is also a solution of DBVP (34). If Hα^1<λ<Hα^2, in addition to the two solutions (50), the functions(52)t=12λ∫uαidvαi−vhαi,v1,t>0−1,t<0,i=3,4,are also two solution of DBVP (1), where α3 and α4 are two roots of (47) in a^1,a3. Working up similarly for other pairs of ai, we complete the prove of the theorem.Theorem 4.
Under the assumption of Lemma1, if m=3,5,7,⋯, the function(53)Hx=∫0xdvgx−gv=∫0xdvx−vhx,v.
hasm−1/2 minimum values Ha^k,k=1,⋯,m−1/2, such that(54)Ha^1≤⋯≤Ha^m−1/2,where, α^k is the minimum point with α^k∈αk∗,a2k+1 for k=1,⋯,m−1/2. So we have(1)
Ifλ<1/2H2α^1, let α be such that Hα=2λ,α<a1. Then, a unique solution of DBVP (34) is(55)t=1Hα∫uαdvα−vhα,v1,t>0,−1,t<0.(2)
Ifλ>1/2H2α^m−1/2, let α be such that Hα=2λ,α<a1. There are m positive solutions of DBVP (34), one of the solutions is(56)t=1Hα∫uαdvα−vhα,v1,t>0−1,t<0,and the otherm−1 solutions are(57)t=1Hα∫uα2i−jdvα2i−j−vhα2i−j,v1,t>0−1,t<0,i=1,2,⋯,m−12,j=0,1,where,αii=1,2,⋯,m−1 are the m−1 roots of (47) and α2i−1,α2i∈ai∗,a2i+1, i=1,2,⋯,m−1/2.(3)
There are 2 tom−1 positive solutions of DBVP (13) for 1/2H2α^1≤λ≤1/2H2α^m−1/2.Proof.
Whenm=3,5,7,⋯ notice that Hx is undefined for x>am. The rest of the proof is similar to the proof of Theorem 3, so it is omitted.Example 1.
Ifm=2, then DBVP (34) has at least two positive solutions by Theorem 3 for all λ>0. For a1=1,a2=2, (48) becomes(58)Hx=∫0x2−3x+v2+x2+2v+v23−1/2dvx−v=2λ.
Letλ=3, the two roots are α1≈0.7418993534279 and α2≈3.8120743716586 for (58). The graph of y=Hx and the two corresponding positive solutions of DBVP (34) are shown in Figure 5.Figure 5
Graphs ofy=Hx and 2 positive solutions for DBVP (1) and λ=3.Example 2.
Ifm=3,a2<a1+a3/2, then ga3−ga1>0, α1∗=2a3+2a2−a1−22a3−a1−a2a3+a1−2a2/3 in (4.9) and DBVP (13) has three positive solutions by Theorem 4 for λ>1/2H2α^1. For a1=1/4,a2=1/2,a3=5/4, the minimum value of y=Hx is Hα1≈7.76663974235139, where α1≈1.01627651425078. There are two positive solutions to DBVP (1) for λ0=1/2H2α^1 and the three positive solutions to DBVP (1) for λ=32>1/2H2α^1 respectively. They are shown in Figure 6.Figure 6
The multiple positive solutions for DBVP (1).Remark 2.
Whena1=a,a2=b,a3=1, the condition a2<a1+a3/2 in Example 2 is consistent with [2]. However, the λ0 does not only exist but also is clearly identified by λ0=1/2H2α1.Example 3.
Ifm=4,ga3−ga1>0, then DBVP (1) has at least four positive solutions by Theorem 3 for all λ>1/2H2α^1. However, unlike Example 2, it is difficult to determine a simplified form for the condition of ga3−ga1>0.


## 5. Conclusion

In this paper, we studied the multiplicity and bifurcation diagram of the Dirichlet boundary value problem (1). Our major results are on the problem for fu=σ−uexp−K/1+u and fu=∏i=1mai−u, respectively. We first proved that all positive solutions of the first problem are less than or equal to σ, obtained more specific lower and upper bounds for these solutions, and computed a curve with accuracy up to 10−6, below which the first problem has a unique positive solution and above which it has exactly three positive solutions. This curve is completely new to our knowledge. For the second problem, we determine its exact number of positive solutions and find a formula used to compute the values of λ that separates the regions of λ for different numbers of solutions.



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