Theoretical and Numerical Aspects of a Third-order Three-point Nonhomogeneous Boundary Value Problem

ABSTRACT

In this paper we are considering a third-order three-point equation with nonhomogeneous conditions in the boundary. Using Krasnoselskii’s Theorem and Leray-Schauder Alternative we provide existence results of positive solutions for this problem. Nontrivials examples are given and a numerical method is introduced.

Key words
numerical solutions; third-order; boundary value problem; Krasnoselskii’s Theorem

RESUMO

Neste artigo, consideramos uma equação com três pontos de fronteira de terceira ordem com condições de contorno não homogêneas. Com uso do Teorema de Krasnoselskii e da Alternativa de Leray-Schauder, apresentamos resultados de existência para soluções positivas. Exemplos não triviais são fornecidos e um método numérico é introduzido.

Palavras-chave
soluções numéricas; terceira-ordem; problema de valor de contorno; Teorema de Krasnoselskii

1 INTRODUCTION

Multi-point boundary value problems there has been attention of several studies mainly focused on the existence of solutions with qualitative and quantitative aspects, we recommend^[2,³,⁴,⁵,⁶,⁷,⁸,⁹,¹⁰,¹¹,¹²,¹⁴,^15] and the references therein. It is well known that the Krasnoselskii’s fixed point theorem, Avery-Peterson and Leggett-Williams theorems are massively used in this line.

In this paper, motived by^[13], we discuss the existence of a positive solution for the third-order boundary value problem:

$$u‴+f(t,u,u\prime )=0,0t1,$$(1.1)

$$u(0)=u^{\prime }(0)=0,u^{\prime }(1)-\alpha u^{\prime }(\eta )=\lambda ,$$(1.2)

where $\eta \in (0,1),\alpha \in [0,\frac{1}{\eta })$ are constants and $\lambda \in (0,\infty )$ is a parameter. Essentially, we combine Leray-Schauder Alternative and Krasnoselskii’s theorem to show the existence of a positive solution for (1.1)-(1.2) without supposing superlinearity on $f$. Numerical solutions are poorly explored, thus we complement this work presenting a numerical study for (1.1)-(1.2) based on Banach’s Contraction Principle.

2 BACKGROUND MATERIAL

We begin this section by stating the following results.

Theorem 1. Let $E$ be a Banach space, $C\subset E$ a closed and convex set, $\Omega$ an open set in $C$ and $p\in \Omega$. Then each completely continuous mapping $T:\overline{\Omega }\to C$ has at least one of the following properties:

• (A1) T has a fixed point in $\overline{\Omega }$.

• (A2) There are $u\in \partial \Omega$ and $\lambda \in (0,1)$ such that $u=\lambda T(u)+(1-\lambda )p.$

Theorem 2. Let $E$ be a Banach space and let $K\subset E$ be a cone in $E$. Assume ${\Omega }_1$, ${\Omega }_2$ are bounded open subsets of $E$ with $0\in {\Omega }_1,{\overline{\Omega }}_1\subset {\Omega }_2$, and let $T:K\cap ({\overline{\Omega }}_2\backslash {\Omega }_1)\to K$ be a completely continuous operator such that, either

• (B1)$\Vert Tu\Vert \le \Vert u\Vert ,u\in K\cap \partial {\Omega }_1$ , and $\Vert Tu\Vert \ge \Vert u\Vert ,u\in K\cap \partial {\Omega }_2$, or

• (B2)$\Vert Tu\Vert \ge \Vert u\Vert ,u\in K\cap \partial {\Omega }_1$ , and $\Vert Tu\Vert \le \Vert u\Vert ,u\in K\cap \partial {\Omega }_2$.

Then $T$ has a fixed point in $K\cap ({\overline{\Omega }}_1\backslash {\Omega }_1)$.

The first theorem is a well-known Leray-Schauder alternative and the second theorem is due to Krasnoselskii, see^[1].

Let us set an auxiliary problem that will be useful in our context.

$$u‴+f(t,x,x\prime )=0,0t1,$$(2.1)

$$u(0)=u\prime (0)=0,u\prime (1)-\alpha u\prime (\eta )=\lambda .$$(2.2)

Related to this problem we have an important lemma.

Lemma 3.Let $x\in C^1[0,1]:=\{x\in C^1[0,1],t\in [0,1]\}$, then we have a unique solution for (2.1)-(2.2). Moreover, this solution is expressed by

$$u(t)={\int }_0^{1}G(t,s)f(s,x(s),x\prime (s))ds+\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^1{G}_1(\eta ,s)f(s,x(s),x\prime (s))ds+$$

$$+\frac{\lambda t^2}{2(1-\alpha \eta )},$$(2.3)

where $G$ is the Green’s function:

$$G(t,s)=\frac{1}{2}\{\begin{array}{c}(2t-t^2-s)s,s\le t\hfill \\ (1-s)t^2,t\le s\hfill \end{array}$$(2.4)

and

$$G_1(t,s)=\frac{\partial G(t,s)}{\partial t}=\{\begin{array}{c}(1-t)s,s\le t\hfill \\ (1-s)t,t\le s\hfill \end{array}.$$(2.5)

Proof. If $u(t)$ is solution of (2.1) , we can suppose that

$$u(t)=-\frac{1}{2}{\int }_0^t(t-s{)}^{2}f(s,x,x\prime )ds+A{t}^2+Bt+C.$$

From condition (2.2), we have $B=C=0$ and

$$A=\frac{1}{2(1-\alpha \eta )}{\int }_0^1(1-s)f(s,x,x\prime )ds-\frac{\alpha }{2(1-\alpha \eta )}{\int }_0^{\eta }(\eta -s)f(s,x,x\prime )ds+\frac{\lambda }{(1-\alpha \eta )}$$

Thus (2.1)-(2.2) has a unique solution. Furthermore $u(t)=-\frac{1}{2}{\int }_0^t(t-s{)}^{2}f(s,x,x\prime )ds+\frac{t^2}{2(1-\alpha \eta )}{\int }_0^1(1-s)f(s,x,x\prime )ds$

$-\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^{\eta }(\eta -s)f(s,x,x\prime )ds+\frac{\lambda t^2}{2(1-\alpha \eta )}$

$=-\frac{1}{2}{\int }_0^t(t-s{)}^{2}f(s,x,x\prime )ds+\frac{t^2}{2}{\int }_0^1(1-s)f(s,x,x\prime )ds$

$+\frac{\alpha \eta t^2}{2(1-\alpha \eta )}{\int }_0^1(1-s)f(s,x,x\prime )ds-\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^{\eta }(\eta -s)f(s,x,x\prime )ds+\frac{\lambda t^2}{2(1-\alpha \eta )}$

$=\frac{1}{2}{\int }_0^t(-t^2+2st-s^2)f(s,x,x\prime )ds+\frac{1}{2}{\int }_0^t(1-s)t^{2}f(s,x,x\prime )ds$

$+\frac{1}{2}{\int }_t^1(1-s)t^{2}f(s,x,x\prime )ds+\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^{\eta }(1-s)\eta f(s,x,x\prime )ds$

$+\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_{\eta }^1(1-s)\eta f(s,x,x\prime )ds-\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^{\eta }(\eta -s)f(s,x,x\prime )ds+\frac{\lambda t^2}{2(1-\alpha \eta )}$

$=\frac{1}{2}{\int }_0^t(2t-t^2-s)sf(s,x,x\prime )ds+\frac{1}{2}{\int }_t^1(1-s)t^{2}f(s,x,x\prime )ds$

$+\frac{\alpha t^2}{2(1-\alpha \eta )}({\int }_0^{\eta }(1-\eta )sf(s,x,x\prime )ds+{\int }_{\eta }^1\eta (1-s)f(s,x,x\prime )ds)+\frac{\lambda t^2}{2(1-\alpha \eta )}$

$={\int }_0^{1}G(t,s)f(s,x,x\prime )ds+\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^1{G}_1(\eta ,s)f(s,x,x\prime )ds+\frac{\lambda t^2}{2(1-\alpha \eta )}.$

$▫$

Defining $x(t)=u(t)$ in Lemma 3 is easy to see that the solution of (1.1)-(1.2) can be expressed as fixed point of the operator $T:C^1[0,1]\to C^1[0,1]$ defined by:

$$Tu(t)={\int }_0^{1}G(t,s)f(s,u,u\prime )ds+\frac{\alpha t^2}{2(1-\alpha \eta )}{\int }_0^1{G}_1(\eta ,s)f(s,u,u\prime )ds+\frac{\lambda t^2}{2(1-\alpha \eta )}.$$(2.6)

Remark 4. Related to $G$ and $G_1$ we have useful properties that will be used in the next section.

• For all$(t,s)\in [0,1]\times [0,1]$:

  $$0\le G_1(t,s)\le (1-s)s$$

• For all$(t,s)\in [0,1]\times [0,1]$:

  $$G(t,s)\le G_1(1,s)=\frac{1}{2}(1-s)s$$

3 POSITIVE SOLUTIONS

Let$E=\{u\in C^1[0,1]:u(0)=0\}$, where $C^1[0,1]$ be the Banach space of continuously differentiable functions in $[0,1]$ equipped with

$$\Vert u{\Vert }_E=max\{\Vert u{\Vert }_{\infty },\Vert u\prime {\Vert }_{\infty }\}.$$

Remark 1. If $u\in E$ then $Tu$ satisfies $Tu(0)=0$. Besides $\Vert (Tu)\prime {\Vert }_{\infty }\ge \Vert Tu{\Vert }_E$.

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

$(H1)$

• (H1) There exist positive constants $A$, $B$ and $\beta$ such that

  • ${max}_{(s,v_1,v_2)\in [0,1]\times [-\beta ,\beta ]\times [-\beta ,\beta ]}\{|f(s,v_1,v_2)|\}\le \frac{\beta (1-\alpha \eta )6B}{1+\alpha (1-\eta )}$

  • $\lambda \le A\beta (1-\alpha \eta )$

  • $A+B\le 1$.

Lemma 2. Suppose that $(H1)$ holds. Thus the problem (1.1)-(1.2) has a solution $u^{*}\in E$ with $\Vert u^{*}{\Vert }_E\le \beta$.

Proof. Let us consider the Theorem 1 with $p=0$ and $\Omega =\{u\in E:\Vert u{\Vert }_E\beta \}$.

We claim that $T$ is continuous and completely continuous. In fact, the continuity follows immediately from the Lebesgue dominated convergence theorem and noting that

$$|T(u)(t)-T(u_n)(t)|\le {\int }_0^{1}G(t,s)|f(s,u(s),u\prime (s))-f(s,u_n(s),u_n\prime (s))|ds+$$

$$+\left|\frac{\alpha t^2}{2(1-\alpha \eta )}\right|{\int }_0^1{G}_1(\eta ,s)|f(s,u(s),u\prime (s))-f(s,u_n(s),u_n\prime (s))|ds$$

$$\le {\int }_0^{1}G1(1,s)|f(s,u(s),u\prime (s))-f(s,u_n(s),u_n\prime (s))|ds+$$

$$+\left|\frac{\alpha }{2(1-\alpha \eta )}\right|{\int }_0^1{G}_1(\eta ,s)|f(s,u(s),u\prime (s))-f(s,u_n(s),u_n\prime (s))|ds,$$

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 $\parallel u\parallel \le {\Lambda }_0$ for each $u\in \Omega$. Now if $u\in \Omega$ we have

$$\begin{array}{ccc}\hfill |(Tu)\prime (t)|& =\hfill & \hfill |{\int }_0^1{G}_1(t,s)f(s,u,u\prime )ds+\frac{\alpha t}{1-\alpha \eta }{\int }_0^1{G}_1(\eta ,s)f(s,u,u\prime )+\frac{\lambda t}{1-\alpha \eta }|\\ \hfill & \le \hfill & \hfill {\int }_0^1|G_1(t,s)f(s,u,u\prime )ds|+\frac{\alpha t}{1-\alpha \eta }{\int }_0^1|G_1(\eta ,s)f(s,u,u\prime )|+\left|\frac{\lambda t}{1-\alpha \eta }\right|\\ \hfill & \le \hfill & \hfill \underset{t\in [0,1]}{max}\frac{1-\alpha \eta +\alpha t}{1-\alpha \eta }{\int }_0^1|(1-s)sf(s,u,u\prime )|ds+\left|\frac{\lambda t}{1-\alpha \eta }\right|\\ \hfill & \le \hfill & \hfill \frac{1+\alpha (-\eta +1)}{1-\alpha \eta }{\int }_0^1|(1-s)s||f(s,u,u\prime )|ds+\left|\frac{\lambda }{1-\alpha \eta }\right|.\end{array}$$

Then $T\Omega$ is a bounded equicontinuous family on $[0,1]$. Consequently the Arzela-Ascoli theorem implies $T:E\to E$ is completely continuous.

In addition, suppose there are $u\in \partial \Omega$ and $\lambda \in (0,1)$ with $u(x)=\lambda Tu(x)$. According $(H1)$ we have:$\Vert Tu{\Vert }_E\Vert (Tu)\prime {\Vert }_{\infty }={max}_{t\in [0,1]}|(Tu)\prime (t)|,$

$\le {max}_{t\in [0,1]}\frac{1+\alpha (-\eta +1)}{1-\alpha \eta }{\int }_0^1|(1-s)s||f(s,u,u\prime )|ds+\left|\frac{\lambda }{1-\alpha \eta }\right|$

$\le {max}_{(s,v_1,v_2)\in [0,1]\times [-\beta ,\beta ]\times [-\beta ,\beta ]}\frac{1+\alpha (-\eta +1)}{1-\alpha \eta }|f(s,v_1,v_2)|{\int }_0^1(1-s)sds+\left|\frac{\lambda }{1-\alpha \eta }\right|$

$\le {max}_{(s,v_1,v_2)\in [0,1]\times [-\beta ,\beta ]\times [-\beta ,\beta ]}\frac{1+\alpha (-\eta +1)}{1-\alpha \eta }\frac{|f(s,v_1,v_2)|}{6}+\left|\frac{\lambda }{1-\alpha \eta }\right|$

$\le \frac{1}{1-\alpha \eta }[\frac{1+\alpha (1-\eta )}{6}max|f(s,v_1,v_2)|+\lambda ]$

$\le \frac{1}{1-\alpha \eta }[\frac{1+\alpha (1-\eta )}{6}\frac{\beta (1-\alpha \eta )6B}{1+\alpha (1-\eta )}+\lambda ]$

$\le \frac{1}{1-\alpha \eta }[\beta (1-\alpha \eta )B+A\beta (1-\alpha \eta )]$

$\le \beta A+\beta B\le \beta .$ Therefore, $\Vert u{\Vert }_E\beta$ and $(A2)$ in Theorem 1cannot occur. Thus $(A1)$ holds and there is $u^{*}\in E$ such that $\Vert u^{*}{\Vert }_E\le \beta$. $▫$

Theorem 3. Suppose that $(H1)$ holds and $f(s,u,v)\ge 0,\forall (s,u,v)\in [0,1]\times [-\beta ,\beta ]\times [-\beta ,\beta ]$. Then (1.1)-(1.2) has at least one positive solution $u^{*}\in E$.

Proof. We start the proof defining the cone $K\subset E$ by

$$K=\{u\in E:u\ge 0,u(0)=0,u\prime (0)=0\}.$$

From $(H1)$ and the definition of $G$ and $G_1$, we have that $T$ applies $K$ in $K$. As seen in the last result, $T$ is completely continuous.

We shall apply Theorem 2. Thus, we will define ${\Omega }_1=\{u\in E;\Vert u{\Vert }_E\beta \}$, ${\Omega }_2=\{u\in E;\Vert u{\Vert }_E\alpha \}$ and we will show that the following conditions are true for all $u\in K$:

1. if $\Vert u{\Vert }_E=\alpha$ then $\Vert Tu{\Vert }_E\le \alpha$;

2. if $\Vert u{\Vert }_E=\beta$ then $\Vert Tu{\Vert }_E\ge \beta$.

In fact, the demonstration of (a) is similar to the proof of the Lemma 2. To prove (b) is necessary to verify that there is $\overline{\gamma }>0$ with

$$\Vert Tu{\Vert }_E\ge \Vert u{\Vert }_E,\forall u\in K\cap \partial {\Omega }_3,$$

where ${\Omega }_3=\{u\in E;\Vert u{\Vert }_E\overline{\gamma }\}$.

Let us assume that the inequality is false, that is, for every $\overline{\gamma }$ such that $\beta >\overline{\gamma }>0$ there exists $u\in E$ with $\Vert u{\Vert }_E=\overline{\gamma }$ and $\Vert Tu{\Vert }_E\overline{\gamma }$. Thus for all $n\in \{1,2,\cdots \}$ with $\frac{1}{n}\alpha$, we can find $u_n\in K$ such that

$$\Vert u_n{\Vert }_E=\frac{1}{n}\mathrm{\text{and}}\Vert T{u}_n{\Vert }_E\frac{1}{n}.$$

Then $\Vert u_n\Vert \to 0$ and $\Vert T{u}_n\Vert \to 0$, when $n\to \infty$. Being $T$ continuous, we have $\Vert T0{\Vert }_E=0$. On the other hand, using $(H1)$ and the definition of $G$ and $G_1$ we have

$$\begin{array}{ccc}\hfill \Vert T0{\Vert }_{\infty }& \ge \hfill & \hfill \underset{t\in [0,1]}{max}\left\{\frac{\lambda t^2}{2(1-\alpha n)}\right\},\\ \hfill & \ge \hfill & \hfill \frac{\lambda }{2(1-\alpha n)}>0\end{array}$$

which is a contradiction. Therefore we have the result. $▫$

Remark 4. Note that the most important step in the proof of Theorem 3 is to impose conditions to conclude that $0$ is not fixed point of $T$.

Example 3.1. Let us consider (1.1)-(1.2) with

$$\begin{array}{ccc}\hfill f(t,u,v)& =\hfill & \hfill \frac{1}{4}t+u^2+v^2\\ \hfill \eta & =\hfill & \hfill \frac{1}{10},\alpha =\frac{1}{3},\lambda =\frac{1}{4}\end{array}$$

Choosing the constants

$$\beta =10,A=0.54,B=0.45,$$

we can easily verify that in these conditions the hypotheses $(H1)$ are satisfied.

Example 3.2.Let us define

$$\begin{array}{ccc}\hfill f(t,u,v)& =\hfill & \hfill \frac{1}{4}t+sin(u)+\frac{1}{4}cos(v)\\ \hfill \eta & =\hfill & \hfill \frac{1}{9},\alpha =\frac{1}{6},\lambda \frac{14}{10}\end{array}$$

As before, choosing the constants

$$\beta =2,A=0.75,B=0.2,$$

we can verify that$(H1)$ is satisfied.

4 NUMERICAL SOLUTIONS

In this section we show the existence and uniqueness for (1.1)-(1.2) 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

$$\begin{array}{ccc}\hfill u^{k+1}& =\hfill & \hfill T(u^k)\end{array}$$

and the basic assumptions$(H2)$ $|f(s,u,u\prime )-f(s,v,v\prime )|\le Amax\{|u(s)-v(s)|,|u\prime (s)-v\prime (s)|\};$ $(H3)$ $\frac{-t^2+t}{2}+\frac{\alpha t\eta (-\eta +1)}{2(1-\alpha \eta )}\le \frac{1}{A}.$

Theorem 1. Suppose that $(H1)$, $(H2)$ and $(H3)$ are satisfied. Then (1.1)- (1.2) has a unique solution $u$ with $\Vert u{\Vert }_E\le \beta$. Moreover, $u^{k+1}=T(u^k)\to u$.

Proof. Let us consider $u,v\in \Omega$ with $\Vert u{\Vert }_E\le \beta$ and $\Vert v{\Vert }_E\le \beta$. Then$\Vert Tu-Tv{\Vert }_E=\Vert (Tu-Tv)\prime {\Vert }_{\infty }$

$=|{\int }_0^1{G}_1(t,s)[f(s,u,u\prime )-f(s,v,v\prime )]ds+\frac{\alpha t}{1-\alpha \eta }{\int }_0^1{G}_1(t,s)[f(s,u,u\prime )-f(s,v,v\prime )]ds|$

$\le A{max}_s\{|u(s)-v(s)|,|u\prime (s)-v\prime (s)|\}({\int }_0^1{G}_1(t,s)ds+\frac{\alpha t}{1-\alpha \eta }{\int }_0^1{G}_1(\eta ,s)ds)$

$\le A{max}_s\{|u(s)-v(s)|,|u\prime (s)-v\prime (s)|\}(\frac{-t^2+t}{2}+\frac{\alpha t\eta (-\eta +1)}{2(1-\alpha \eta )})$

Using $(H3)$ we obtain

$\le A{max}_s\{|u(s)-v(s)|,|u\prime (s)-v\prime (s)|\}\frac{1}{A}$

$\le {max}_s\{|u(s)-v(s)|,|u\prime (s)-v\prime (s)|\}=\Vert u-v{\Vert }_E$ $▫$

Motivated by the last result we can define Algorithm 1.

Algorithm 1
Fixed-Point

Example 4.1. In this example, we consider

$$\begin{array}{ccc}\hfill f(x,u,u\prime )& =\hfill & \hfill -u\prime \\ \hfill \eta & =\hfill & \hfill \frac{\pi }{4},\alpha =\frac{1}{10},\lambda =0.770760306689242\end{array}$$

The the analytical solution is $u^{*}(x)=1-cos(x)$. The Table 1 contains results of application in Example 4.1.

We can make additional tests. From Theorem 3 we have a solution for Examples3.1 and 3.2 but in both case, we do not know which they are. Let us apply Algorithm 1 in these problems. For this purpose, we can consider the condition

$$\frac{\Vert u^{k+1}-u^k\Vert }{\Vert u^{k+1}{\Vert }_{\infty }}{10}^{-4}$$

as stopping criterion for the algorithm. The results for these examples are presented in Table 2 and 3, respectively. The illustrations of these results are given in Figure 1 and 2.

Table 1
Algorithm 1 considering Example 4.1.

It	${\epsilon }_u^k$	${\epsilon }^k$	${\overline{\epsilon }}^k$
1	0.104585227251908	0.355112466879952	1.000000000000000
2	0.072538564385106	0.032046662866802	0.082773878760819
3	0.069264937033799	0.003273627351307	0.008384612437847
4	0.068925441261629	0.000339495772170	0.000868781674432
5	0.068890166416009	0.000035274845620	0.000090261428243

Table 2
Algorithm 1 considering Example 3.1.

It	${\epsilon }_u^k$	${\epsilon }^k$	${\overline{\epsilon }}^k$
1	-	0.168278823890335	1.000000000000000
2	-	0.007563461402919	0.043012756518181
3	-	0.000744660869474	0.004216964422657
4	-	0.000077049209989	0.000436134198292

Table 3
Algorithm 1 considering Example 3.2.

It	${\epsilon }_u^k$	${\epsilon }^k$	${\overline{\epsilon }}^k$
1	-	0.740971458506793	1.000000000000000
2	-	0.010141509530254	0.013876702158219
3	-	0.000276799190473	0.000378602975785
4	-	0.000007307338952	0.000009995000056

Figure 1
Numerical solution obtained from Example 1 using Algorithm 1.

Figure 2
Figure 2: Numerical solution obtained from Example 2 using Algorithm 1.

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Publication Dates

• Publication in this collection
  13 Dec 2019
• Date of issue
  Sep-Dec 2019

History

• Received
  3 Dec 2018
• Accepted
  21 Feb 2019