Multiple Solutions for a Sixth Order Boundary Value Problem

ABSTRACT

This work presents conditions for the existence of multiple solutions for a sixth order equation with homogeneous boundary conditions using Avery Peterson’s theorem. In addition, non-trivial examples are presented and a new numerical method based on the Banach’s Contraction Principle is introduced.

Keywords:
numerical solutions; sixth-order; boundary value problem and multiple solutions

RESUMO

Este trabalho apresenta condições para existência de múltiplas soluções para uma equação de sexta ordem com condições de contorno homogêneas usando o teorema de Avery Peterson. Além disso, exemplos não triviais são apresentados e um novo método numérico baseado no Princípio de Contração de Banach é introduzido.

Palavras-chave:
soluções numéricas; sexta ordem; problema de valor de contorno e múltiplas soluções

In this manuscript we address conditions for the existence of multiple solutions for the sixth order limit value problem:

$u^{\left(6\right)}+f\left(t,u\right)=0,0<t<1,$ (0.1)

$u\left(0\right)=u\text{'}\left(0\right)=u\text{'}\text{'}\left(0\right)=0,u\text{'}\left(1\right)=u\text{'}\text{'}\text{'}\left(1\right)=u^{\left(5\right)}\left(1\right)=0.$ (0.2)

where $f:R^2\to R$ is a continue function.

In the literature, there are several studies mainly focused only on the existence of solutions with qualitative and quantitative aspects. Among them, we recommend ^{1), (2), (3), (5), (13), (6), (7), (8), (12), (4} and the references therein.

Some specific studies, as ^{5), (8} and ¹⁴, have analyzed conditions for the existence of solutions for this class of problems. In ¹⁴, the authors approach a simplified version of problem, in which they consider the dependence of f only on t, the authors apply the Krasnoselskii’s fixed point theorem to determine sufficient conditions for the existence of a positive solution.

Few papers present numerical studies related to the sixth order problem. Numerical solutions are poorly explored, thus we complement this work presenting a numerical study for (0.1)-(0.2) based on Banach’s Contraction Principle.

1 POSITIVE SOLUTIONS

As presented in ¹⁴, we can represent the problem (0.1)-(0.2) as a fixed point of the operator $T:C^1[0,1]\to C^1[0,1]$ defined by:

$Tu\left(t\right)={\int }_0^{1}G(t,s)f(s,u)ds$ (1.1)

where G is the Green’s function:

$G(t,s)=\left(\frac{t^3}{2}-\frac{t^4}{8}\right)\frac{(1-s{)}^4}{24}+\left(-\frac{t^3}{12}+\frac{t^4}{16}\right)\frac{(1-s{)}^2}{2}+\frac{t^3}{48}-\frac{5t^4}{192}+\frac{t^5}{120}-\frac{(t-s{)}^5}{120}H(t-s),$ (1.2)

and

$H\left(\zeta \right)=\left\{\begin{array}{l}1,\zeta \ge 0\\ 0,\zeta <0\end{array}\right..$ (1.3)

In the sequence, some properties that will be useful related to G are listed.

Propriety 1 How $G(1,s)=\frac{s^3}{960}(20-25s+8s^2)\ge 0$ following as presented in ¹⁴ there are polynomials p(t) and q(t) such that:

$p\left(t\right)G(1,t)\le G(t,s)\le q\left(t\right)G(1,s),$ (1.4)

where

$p\left(t\right)=4t^2-4t+t^4,q\left(t\right)=\frac{t^3}{3}(20-25t+8t^2).$

The polynomials p and q are illustrated in Figure 1 .

Figure 1
Illustration of polynomials p and q for $t\in \left[0,1\right]$.

To determine multiple solutions, consider the cone

$E=\{u\in C^1[0,1]:u(0)=0,u(t)\ge 0,\forall t\in [0,1\left]\right\},$

where $C^1\left[0,1\right]$ is the Banach space of continuously differentiable functions in [0,1] equipped with

$\Vert u{\Vert }_E=\Vert u{\Vert }_{\infty }.$

In order, as T is an integral operator, this is continuous and completely continuous as shown in the proposition (1)

Proposition 1. The operator T is continuous and completely continuous.

Proof. Continuity follows immediately from the Lebesgue dominated convergence theorem and the fact that

$\begin{array}{c}\begin{array}{ccc}\left|T\left(u\right)\left(t\right)-T\left(u_n\right)\left(t\right)\right|& \le & {\int }_0^{1}G\left(t,s\right)|f\left(s,u\left(s\right)\right)-f\left(s,u_n\left(s\right)\right)|ds,\\ & \le & {\int }_0^{1}G\left(t,s\right)|f\left(s,u\left(s\right)\right)-f\left(s,u_n\left(s\right)\right)|ds,\\ & \le & {\int }_0^{1}q\left(t\right)G\left(1,s\right)|f\left(s,u\left(s\right)\right)-f\left(s,u_n\left(s\right)\right)|ds,\\ & \le & {\int }_0^{1}G\left(1,s\right)|f\left(s,u\left(s\right)\right)-f\left(s,u_n\left(s\right)\right)|ds,\end{array}\end{array}$

with $u_n,u\in E$. To show complete continuity we will use the Arzela-Ascoli’s theorem. Let $\Omega \subseteq E$ be bounded, in other words, there exists ${\Lambda }_0>0$ with $\Vert u\Vert \le {\Lambda }_0$ for each $u\in \Omega$. Now if $u\in \Omega$ we have

$\left|\right(Tu\left)\right(t\left)\right|\le {\int }_0^1\left|G\right(t,s\left)\right|H_{{\Lambda }_0}\left(s\right)ds$

where $H_{{\Lambda }_0}$ is determined by the bounded set and function f. It is easy to check that $H_{{\Lambda }_0}\left(s\right)\in L^1[0,1]$. Then imply that T(Ω) is a bounded equicontinuous family on [0,1]. Consequently the Arzela-Ascoli theorem implies $T:E\to E$ is completely continuous.

To demonstrate the main result of this work, we need to present the main tool to be used.

Avery-Peterson theorem. Now, we need to consider the convex sets

$P(\gamma ,d)=\{x\in P|\gamma \left(x\right)<d\}$

$P(\gamma ,\alpha ,b,d)=\{x\in P|b\le \alpha \left(x\right)and\gamma \left(x\right)<d\}$

$P(\gamma ,\theta ,\alpha ,b,c,d)=\{x\in P|b\le \alpha \left(x\right),\theta \left(x\right)\le cand\gamma \left(x\right)<d\}$

and the closed set

$R(\gamma ,\psi ,a,d)=\{x\in P|a\le \psi \left(x\right)and\gamma \left(x\right)<d\}.$

Theorem 1 Let P be a cone in a real Banach space X. Let γ and θ nonnegative continuous convex functionals on P, α be a nonnegative continuous concave functional on P, and ψ be a nonnegative continuous functional on P satisfying $\psi \left(\lambda x\right)\le \lambda \psi \left(x\right)for0\le \lambda \le 1$ , such that for some positive numbers μ and d,

$\alpha \left(x\right)\le \psi \left(x\right)and\Vert x\Vert \le \mu \gamma \left(x\right),$

for all$x\in \overline{P(\gamma ,d)}$. Suppose

$T:\overline{P(\gamma ,d)}\to \overline{P(\gamma ,d)}$

is completely continuous and there exist positive numbers a, b, c with a<b, such that

$\{u\in P(\gamma ,\theta ,\alpha ,b,c,d\left)\right|\alpha \left(u\right)>b\}\ne \varnothing and$

$u\in P(\gamma ,\theta ,\alpha ,b,c,d)\Rightarrow \alpha \left(Tu\right)>b,$ (1.5)

$\alpha \left(Tu\right)>bforu\in P(\gamma ,\alpha ,b,d)with\theta \left(Tu\right)>c,$ (1.6)

$0\notin R(\gamma ,\psi ,a,d)and\psi \left(Tu\right)<afor$ (1.7)

$u\in R(\gamma ,\psi ,a,d)with\psi \left(u\right)=a.$

Then T has at least three distinct fixed points in $\overline{P(\gamma ,d)}$ .

In order to prove the existence of solutions, we need to consider some basic assumptions.

(H1) For problem (0.1)-(0.2) there is a positive constant d such that:

$·Forall(s,v)\in [0,1]\times [0,d]then0\le f(s,v)\le \frac{d}{r_1}$

$·r_1={\int }_0^{1}G\left(1,s\right)ds.$

The lemma presented will be fundamental for demonstrating our main result.

Lemma 2.Suppose that(H1)holds and$P=Eand\gamma (.)=\Vert .{\Vert }_E$, then T defined in(1.1)fulfills$T:\overline{P(\gamma ,d)}\to \overline{P(\gamma ,d)}$.

Proof. Let us consider $u\in Ewith\Vert u{\Vert }_E\le d$, so from (H1) we can obtain:

$\begin{array}{ccc}{||Tu||}_E& =& \underset{t\in \left[0,1\right]}{max}\left|\left(Tu\right)\left(t\right)\right|,\\ & \le & \underset{t\in \left[0,1\right]}{max}{\int }_0^1\left|G\left(t,s\right)\right|\left|f\left(s,u\right)\right|ds\\ & \le & \underset{t\in \left[0,1\right]}{max}{\int }_0^{1}q\left(t\right)G\left(1,s\right)\left|f\left(s,u\right)\right|ds\\ & \le & \frac{{\displaystyle d}}{{\displaystyle r_1}}\left[{\int }_0^{1}G\left(1,s\right)ds\right]\underset{t\in \left[0,1\right]}{max}q\left(t\right)\\ & \le & d\underset{t\in \left[0,1\right]}{max}q\left(t\right)\\ & \le & d.\end{array}$

Therefore $T:\overline{P(\gamma ,d)}\to \overline{P(\gamma ,d)}$.

Theorem 2 presents conditions under which the problem defined in (0.1)-(0.2) has at least three positive solutions.

Theorem 2 Suppose that the hypothesis (H1) is satisfied. Suppose, in addition, that there exist a, 0 < a < d such that f satisfies the following conditions:

(H2) $f(s,u)>\frac{2a}{r_2},\forall (s,u)\in [0,1]\times [2a,8a],where{r}_2=\frac{423}{2048}{\int }_{\frac{3}{8}}^{\frac{5}{8}}G(1,s)ds$

(H3) $f(s,u)<\frac{a}{r_1},\forall (s,u)\in [0,1]\times [0,a]$

Then, the Problem (0.1) - (0.2) has at least three positive solutions.

Proof. We will apply Avery-Peterson theorem, let us consider T and P as defined before. Furthermore, we need define the following functionals:

$\begin{array}{c}\gamma \left(u\right)={||u||}_E,\\ \psi \left(u\right)=\underset{t\in \left[0,1\right]}{max}\left|u\left(t\right)\right|,\\ \theta \left(u\right)=\underset{t\in \left[\frac{3}{8},\frac{5}{8}\right]}{max}\left|u\left(t\right)\right|\\ \alpha \left(u\right)=\underset{t\in \left[\frac{3}{8},\frac{5}{8}\right]}{min}\left|u\left(t\right)\right|.\end{array}$

Therefore, from Lemma 2 we obtain

$T:\overline{P(\gamma ,d)}\to \overline{P(\gamma ,d)}$

and T is completely continuous and there exist positive numbers b and c with a<b, such that

$\{u\in P(\gamma ,\theta ,\alpha ,b,c,d)\left|\alpha \right(u)>b\}\ne \varnothing and$

$u\in P(\gamma ,\theta ,\alpha ,b,c,d)\Rightarrow \alpha \left(Tu\right)>b$ (1.8)

$\alpha \left(Tu\right)>bforu\in P(\gamma ,\alpha ,b,d)with\theta \left(Tu\right)>c,$ (1.9)

$0\notin R(\gamma ,\psi ,a,d)and\psi \left(Tu\right)<aforu\in R(\gamma ,\psi ,a,d)with\psi \left(u\right)=a.$ (1.10)

Now, we consider the constants b and c as follows:

$b=2a$

and

$c=8a.$

Clearly, we have $\{u\in P(\gamma ,\theta ,\alpha ,b,c,d\left)\right|\alpha \left(u\right)>b\}\ne \varnothing$. Let us demonstrate (1.8).

Using (H2) we obtain

$\begin{array}{ccc}\alpha \left(Tu\right)& =& \underset{t\in \left[\frac{3}{8},\frac{5}{8}\right]}{min}\left(Tu\right)\left(t\right)\\ & =& \underset{t\in \left[\frac{{\displaystyle 3}}{{\displaystyle 8}},\frac{{\displaystyle 5}}{{\displaystyle 8}}\right]}{min}\left({\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & \underset{t\in \left[\frac{{\displaystyle 3}}{{\displaystyle 8}},\frac{{\displaystyle 5}}{{\displaystyle 8}}\right]}{min}\left({\int }_0^{1}p\left(t\right)G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & p\left(0.375\right){\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\\ & \ge & \frac{423}{2048}{\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\\ & \ge & \frac{423}{2048}{\int }_{\frac{3}{8}}^{\frac{5}{8}}G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\\ & \ge & \frac{423}{2048}\frac{2a}{r_2}{\int }_{\frac{3}{8}}^{\frac{5}{8}}G\left(1,s\right)ds\\ & \ge & 2a=b.\end{array}$

Let us demonstrate (1.9). Let $u\in P(\gamma ,\alpha ,b,d)with\theta \left(Tu\right)>c$. Then

$\begin{array}{ccc}\alpha \left(Tu\right)& =& \underset{t\in \left[\frac{3}{8},\frac{5}{8}\right]}{min}\left(Tu\right)\left(t\right)\\ & =& \underset{t\in \left[\frac{{\displaystyle 3}}{{\displaystyle 8}},\frac{{\displaystyle 5}}{{\displaystyle 8}}\right]}{min}\left({\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & \underset{t\in \left[\frac{{\displaystyle 3}}{{\displaystyle 8}},\frac{{\displaystyle 5}}{{\displaystyle 8}}\right]}{min}\left({\int }_0^{1}p\left(t\right)G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & p\left(0.375\right)\left({\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & q\left(0.625\right)\frac{p\left(0.375\right)}{q\left(0.625\right)}\left({\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & \frac{p\left(0.375\right)}{q\left(0.625\right)}\underset{t\in \left[\frac{{\displaystyle 3}}{{\displaystyle 8}},\frac{{\displaystyle 5}}{{\displaystyle 8}}\right]}{max}\left({\int }_0^{1}q\left(t\right)G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & \frac{1}{4}\underset{t\in \left[\frac{{\displaystyle 3}}{{\displaystyle 8}},\frac{{\displaystyle 5}}{{\displaystyle 8}}\right]}{max}\left({\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right)\right)ds\right)\\ & \ge & \frac{1}{4}\theta \left(Tu\right)\\ & \ge & \frac{1}{4}c=b.\end{array}$

Now, to show (1.10) let us consider $u\in R(\gamma ,\psi ,a,d)with\psi \left(u\right)=a$. From (H3) we have,

$\begin{array}{ccc}\psi \left(Tu\right)& =& \underset{t\in \left[0,1\right]}{max}\left|\left(Tu\right)\left(t\right)\right|\\ & \le & \underset{t\in \left[0,1\right]}{max}{\int }_0^1\left|G\left(t,s\right)\right|\left|f\left(s,u\right)\right|ds\\ & \le & \underset{t\in \left[0,1\right]}{max}{\int }_0^{1}q\left(t\right)G\left(1,s\right)\left|f\left(s,u\right)\right|ds\\ & \le & \frac{a}{r_1}\left[{\int }_0^{1}G\left(1,s\right)ds\right]\underset{t\in \left[0,1\right]}{max}q\left(t\right)\\ & \le & a.\end{array}$

Applying Avery-Peterson theorem we obtain that the problem has at least three distinct solutions in the set $\overline{P(\gamma ,d)}$, so these solutions are non-negative. On the other hand, they must satisfy the hypothesis (H2) so they cannot be null. Therefore, the Problem (0.1) - (0.2) has at least three positive.

The example presented below illustrates the hypotheses assumed in Theorem 2.

Example 1.1. Let us consider (0.1) - (0.2) with

$f\left(t,u\right)=\left\{\begin{array}{l}6e^t+6561+5\frac{{\left(u-2a\right)}^2}{a}u\ge 2a\\ 6e^t+{\left(\frac{9u}{2a}\right)}^{4}u<2a\end{array}\right.$

Choosing the constants

$d=10,a=1,$

we can easily verify that in these conditions the hypotheses (H1) and hypotheses of Theorem 2 are satisfied.

2 NUMERICAL SOLUTIONS

In this section, we show the existence and uniqueness for (0.1)-(0.2) using Banach Fixed Point Theorem. This approach is classical but very important to define numerical methods for our problem. Let us consider the iterative sequence

$\begin{array}{ccc}{u}^{k+1}\left(t\right)& =& \left(T{u}^k\right)\left(t\right)\\ & =& {\int }_0^{1}G\left(t,s\right)f\left(s,u^k\left(s\right)\right)ds.\end{array}$

and the basic assumptions

(H4) $\left|f\right(s,u)-f(s,v\left)\right|\le \frac{\beta }{r_1}\left|u\right(s)-v(s\left)\right|;\forall u,v\in [0,d],s\in [0,1]and\beta \in (0,1).$

Theorem 3.Suppose that(H1)and(H4)are satisfied. Then(0.1)-(0.2)has an unique solution u with$\Vert u{\Vert }_E\le d$. Moreover, $u^{k+1}=T\left(u^k\right)\to u^{*}$.

Proof. We will prove that the operator T is a contraction. For this, consider $u,v\in Ewith\Vert u{\Vert }_E\le dand\Vert v{\Vert }_E\le d$. Then

$\begin{array}{ccc}{||Tu-T||}_E& =& {||\left(Tu-Tv\right)||}_{\infty }\\ & =& \underset{t\in \left[0,1\right]}{max}\left|{\int }_0^{1}G\left(t,s\right)\left[f\left(s,u\right)-f\left(s,v\right)\right]ds\right|\\ & \le & \underset{t\in \left[0,1\right]}{max}{\int }_0^{1}G\left(t,s\right)\left|f\left(s,u\right)\left(s,v\right)\right|ds\\ & \le & \underset{t\in \left[0,1\right]}{max}{\int }_0^{1}q\left(t\right)G\left(1,s\right)\left|f\left(s,u\right)-f\left(s,v\right)\right|ds\\ & \le & \left(\frac{\beta }{r_1}\underset{s}{max}\left|u\left(s\right)-v\left(s\right)\right|\right)\left(\underset{t\in \left[0,1\right]}{max}q\left(t\right)\right){\int }_0^{1}G\left(1,s\right)ds\\ & \le & \beta \underset{s}{max}\left|u\left(s\right)-v\left(s\right)\right|\\ & \le & \beta {||u-v||}_E.\end{array}$

Therefore, by the principle of contraction there is only one solution that can be obtained as a limit of the sequence $u^{k+1}=T\left(u^k\right)\to u^{*}$.

Motivated by the last result we can define Algorithm 1.

Algorithm 1
Fixed-Point

1: Define an uniformly distributed mesh {x j } in [0, 1];

2: Define an initial approximation $u_j^0=u^0\left(x_j\right)$, tolerance ε > 0;

3: k = 0;

4: while$\Vert u^{k+1}-u^k{\Vert }_{\infty }>\epsilon ork=0$do

5: Compute $u_j^{k+1}u\sin g{u}^{k+1}=T\left(u^k\right)$ and Trapezoidal Rule

6: K = k + 1;

7: end while

8: output: u k .

In sequence, examples are presented in order to establish the effectiveness of Algorithm 1. In the Table 1, ${\epsilon }_u^{k}denotes\Vert u^{*}-u^k{\Vert }_{\infty }where{u}^{*}$ is the exact solution, ${\epsilon }_{k}denotes{||u^{k+1}-u^k||}_{\infty }and{\overline{\epsilon }}^k=\frac{{||u^{k+1}-u^k||}_{\infty }}{{||u^{k+1}||}_{\infty }}$. Still, “It” denotes “iteration”.

Table 1
Algorithm 1 considering Example 2.1.

It	${\epsilon }_u^k$	εk	${\overline{\epsilon }}^k$
1	8.015149 × 10-4	4.746984	0.999687
2	8.015149 × 10-4	0	0

Example 2.1. Consider in problem (0.1) - (0.2) :

$f(t,u)=-\left(32400t\right(t-1{)}^2+14400(t-1{)}^3+6480t^2(2t-2)+720t^3);$

The analytical solution of (0.1) - (0.2) is $u^{*}\left(t\right)=t^3(1-t{)}^6$. Table 1 contains results of application of the Algorithmic 1 in this example and the results are shown in Figure 2.

Figure 2
Graph of the analytical solution u ^k and approximate solution u ^* obtained by the algorithm 1.

Figure 2 shows that the solution provided by algorithm 1 is very close to the analytical solution and the error increases when t tends to 1. This behavior can be justified because in (0.2) does not specify a condition for u(1).

Example 2.2.This example consider the function components of Example 1. We know that, according to theorem 2, the problem of example 1 has at least 3 solutions with a norm less than 1, Algorithm 1 is not the most suitable for determining multiples solutions because it requires that the operator T be in the vicinity of the solution contraction, as seen in Theorem 3. Even so, we performed a test to verify the behavior of Algorithm 1 in an attempt to determine multiple solutions. So inspired by the works^{9), (10} and¹¹, how know that the solutions we are looking for must be continuous and satisfy the condition 0.2. We choose initial approaches that satisfy the conditions $u\left(0\right)=u\text{'}\left(0\right)=0andu\text{'}\left(1\right)=0$. Thus, functions parable approaches are reasonable ways to approach the solution. In this sense, our heuristic methodology is to generate parables about starting points as follows:

$u^0\left(t\right)=\zeta (2t^2-t^4)$

where the constants ζ is a random numbers in [0,d]. For practical purposes, the proposed procedure is defined by Algorithm 2. It is expected that this procedure returns several solutions.

Algorithm 2

1. Choose a vector $\zeta \in [0,d{]}^N$.
for k = 1, ..., N do
	1. Compute $u_{k,i}^0=u_k^0\left(x_i\right)={\zeta }_k\left(2\right(x_i{)}^2-\left(x_i{)}^4\right)$
	2. Run the Algorithm 1 with initial guess $u_k^0$.
end for

Therefore, it is necessary to establish a way to compare these solutions. Note that the magnitude of the solutions may be different. In this sense, we say that the numeric solutions u ^* and u ^** are equivalent if

$\Vert u^{*}-u^{**}\Vert \le \max \{{10}^{-4},{10}^{-2}\min \{\Vert u^{*}\Vert ,\Vert u^{**}\Vert \left\}\right\}.$ (2.1)

is satisfied.

We consider N=50 in Algorithm 2 and $\epsilon ={10}^{-6}$ in Algorithm 1, we get the convergence of Algorithm 1 in all initializations. Of these 32 initializations converged to the solution $u_1^{*}$ the others converged on the $u_2^{*}$ solution illustrated in the figure 3. We can notice that the curves obtained seem to fulfill the hypotheses of Theorem 2 and the conditions (0.2).

Figure 3
Illustration of solutions obtained for Example 1. The left solutions obtained are illustrated on a linear scale, the right for better visualization we present the solutions on a logarithmic scale

3 FINAL REMARKS

This work is restricted to the problem (0.1), (0.2) can have several solutions if the f function meets certain conditions through of the Avery-Peterson theorem. Additionally, conditions are determined for convergence of the interactive sequence $u^{k+1}=T{u}^k$ through the principle of contraction. To complement the analysis, the implementation of this method is performed and non-trivial examples were tested. The results were detailed showing the feasibility of the proposed methods.

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Publication Dates

• Publication in this collection
  05 Apr 2021
• Date of issue
  Jan-Mar 2021

History

• Received
  11 Apr 2020
• Accepted
  18 Nov 2020