On solvability of some $p$-Laplacian boundary value problems with Caputo fractional derivative

1.   Introduction

Due to the wide application in many disciplines such as the fields of physics, chemistry, automatic control, signal processing, soft matter research, aerodynamics, etc., the fractional calculus have been widely studied recently. An extensive literature can be found related to the fractional differential equation systems ^[19,22].The solvability of boundary value problems(BVPs) with fractional derivative are studied with various tools, specially, with topological degree, fixed-point theory, the continuation theorems, and other nonlinear functional analysis method. For example, see fractional two-point BVPs ^[8], fractional BVPs at resonance, fractional multi-point problems at nonresonance ^[12,14], fractional initial value problems, fractional impulsive problems ^[14], fractional integral BVPs ^[9], fractional $p$-Laplace problems ^{[4,10,13,15,16,17,18]}, fractional BVPs with the Caputo-Fabrizio derivative ^[1,2,3,5], etc.

Zhang ^[20] has given the solvability results to the BVPs with Caputo fractional derivative

by using some fixed point Theorems.

Salem ^[12] has researched the solvability of the following nonlinear $m$-point fractional BVPs

where $0 < x_1 < x_2 < \cdots < x_{m-2} < 1, \; a_i > 0$ and $\sum\limits_{i = 1}^{m-2}a_i x_i^{\alpha-1} < 1$.

Based upon the above research, some authors studied the differential equations with $p$-Laplacian by the use of the topological method, Leray-Schauder Continuation theorems, fixed-point index theory, etc. For example, in ^[15], the solvability of BVPs of RL-fractional differential equations with $p$-Laplacian

is discussed, where $D^\alpha$ is the RL-fractional differentiation and the function $f: [0, 1]\times[0, +\infty) \to [0, +\infty)$ is continuous. Using a monotone iterative technique, some solvability results are given. We refer the readers to ^{[4,10,13,15,16]} for details.

This work is devoted to the study of the solvability of some $p$-Laplacian $BVPs$ with Caputo fractional derivative

where $2 < \alpha < 3, 0 < \beta < 1, 0 < \xi < 1, \phi_p(t) = |t|^{p-2}t, p > 1, f\in C([0, 1]\times[0, +\infty))$ is nonnegative, ${^C}D^q$ is the Caputo fractional derivative. With the use of some fixed point theorems and properties of the corresponding completely continuous operator, some solvability results of the problem considered are obtained.

2.   Preliminaries and lemmas

Definition 2.1. ^[11] The Caputo fractional derivative of order $\gamma$ of the function $f:[0, \infty)\rightarrow {R}$ is defined as

here $[\gamma]$ is the integer part of real number $\gamma$.

Definition 2.2. ^[11] The RL fractional integral of order $\gamma$ of the function $f:[0, \infty)\rightarrow {R}$ is defined as

provided the integral exists.

Lemma 2.1. ^[11] Let $\gamma > 0, n = [\gamma]+1$, the equation

has solution $f(x) = c_0+c_1x+c_2x^2+\cdots+c_{n-1}x^{n-1}, \; c_i\in{R}, \; i = 0, 1, 2, \cdots, n-1\;$.

Lemma 2.2. ^[11] Let $\gamma > 0, n = [\gamma]+1$, then there holds

for $c_i\in{R}, \; i = 0, 1, 2, \cdots, n-1\;$.

Lemma 2.3 Given $h\in C[0, 1], \; 2 < \alpha < 3, 0 < \beta < 1, 0 < \xi < 1$. A function $y$ solves the BVPs

iff $y$ solves the following equation

where $\phi_q = (\phi_p)^{-1}, \frac{1}{p}+\frac{1}{q} = 1$, and

Proof By Lemma 2.2, we have

for some $c\in{R}.$ Note that $^CD^{\alpha}y(0)) = 0$, we have $c = 0$. So, we obtain

By the use of the Lemma 2.2, we holds

for some $d_1, d_2, d_3\in{R}$. From above formula, one has

By $y^{\prime\prime}(0) = 0$, we obtain $d_3 = 0$, and further the boundary conditions $y(0)+y^\prime(0) = 0$ and $y(1)+ y^\prime(\xi) = 0$ yields that

and

Hence,

So, we have

Thus, the Lemma hold.

Lemma 2.4 The function $G(x, t)$ satisfy:

$(1)\; G(x, t)$ is continuous on $C([0, 1]\times[0, 1])$, and positive for $x, t\in (0, 1)$;

$(2)\; \min\limits_{1/4\leq x \leq 3/4}G(x, t)\geq\varphi(t), \; \max\limits_{0\leq x\leq 1}G(x, t)\leq\omega(t),$

where

Proof By the definition of the function $G(x, t)$, the property (1) is clear.

Setting

for $0\leq t\leq x\leq 1, t\leq\xi$;

for $0 < \xi\leq t\leq x\leq 1$;

for $0\leq x\leq t\leq\xi < 1$;

for $0\leq x\leq t\leq 1$.

By the expression of the $g_i(x, t), i = 1, 2, 3, 4,$ one can check that

So, we obtain

The Lemma hold now.

Lemma 2.5. ^[6] Let $X$ be an order Banach space, $K\subset X$ is a cone, and that $\Omega_1, \Omega_2$ are bounded open subsets of $X$ with $0\in\Omega_1\subset \overline{\Omega}_1\subset \Omega_2$, and let $T:K\rightarrow K$ be a completely continuous operator such that either

$(A_1)\; \|Ty\|\leq \|y\|, y\in K\cap\partial\Omega_1$ and $\|Ty\|\geq \|y\|, y\in K\cap\partial\Omega_2$, or

$(A_2)\; \|Ty\|\geq \|y\|, y\in K\cap\partial\Omega_1$ and $\|Ty\|\leq \|y\|, y\in K\cap\partial\Omega_2$

Then, $T$ has a fixed point in $K\cap\overline{\Omega}_2\backslash\Omega_1$.

Lemma 2.6. ^[7] Let $K$ be a cone in a real $Banach$ space $X, K_c = \{y\in K\mid\|y\|\leq c\}, \theta$ be a nonnegative continuous concave functional on a cone $K$ such that $\theta(y)\leq\|y\|$, for all $y\in\overline{K}_c$, and $K(\theta, b, d) = \{y\in K\mid b\leq\theta(y), \|y\|\leq d\}.$ Suppose $T:\overline{K}_c\rightarrow \overline{K}_c$ is completely continuous and there exist constants $0 < a < b < d\leq c$ such that

$(B_1)\; \{y\in K(\theta, b, d)\mid\theta(y) > b\}\neq\emptyset$ and $\theta(Ty) > b$ for $y\in K(\theta, b, d)$;

$(B_2)\; \|Ty\| < a$ for $\|y\|\leq a$;

$(B_3)\; \theta(Ty) > b$ for $y\in K(\theta, b, c)$ with $\|Ty\| > d$.

Then $T$ has at least three fixed points $y_1, y_2, y_3$ satisfying

$\|y_1\| < a, \; \; b < \theta(y_2)$ and $\|y_3\| > a$ with $\theta(y_3) < b$.

3.   Existence of positive solutions

Let $X = C[0, 1]$ with the norm $\|y\| = \max\limits_{0\leq x\leq 1}|y(x)|$. The cone $K\subset X$ is defined as $K = \{y\in X \mid y(x)\geq 0, 0\leq x\leq 1\}$ and the continuous concave functional $\theta$ on the $K$ defined as

Lemma 3.1 Suppose $T: K\rightarrow X$ be defined by

Then $T: K\rightarrow K$ is completely continuous.

Proof Taking into account that two functions $G(x, t)$ and $f(x, y)$ are all nonnegative and continuity in their domain, one has the operator $T:K\rightarrow K$ and it is continuous. Furthermore, by the use of Lebesgue dominated convergence Theorem and Ascoli-Arzela Theorem, a standard argument can show that $T:K\rightarrow K$ is completely continuous operator (see, for example, ^[21]).

For the convenience, the following notations are introduced.

where $\varphi(t), \omega(t)$ are defined as (6) and $f\in C([0, 1]\times [0, +\infty), [0, +\infty))$.

Theorem 3.1 Suppose that $f: [0, 1]\times [0, +\infty)\rightarrow [0, +\infty)$ is a continuous function, and there exists two constants $0 < k < d$ such that

Then BVPs (1), (2) have at least one solution $y\in K$ satisfying $k\leq\|y\|\leq d$.

Proof We will use Lemma 2.5 to obtain the results. The proof is separated into two steps.

Step 1 Let $\Omega_d = \{y\in K\mid\|y\| < d\}$. For any $y\in \partial\Omega_d$, we have $\|y\| = d$ and $f(x, y(x))\leq\chi(d)\leq \phi_p(dM)$ for $(x, y)\in [0, 1]\times[0, d]$. Hence,

That is to say that,

Step 2 Let $\Omega_k = \{y\in K\mid\|y\| < k\}$. For any $y\in \partial\Omega_k$, we have $\|y\| = k$ and $f(x, y(t))\geq\psi(k)\geq \phi_p(kN)$ for $(x, y)\in [0, 1]\times[0, k]$. Hence,

That is to say that,

Combining step 1, step 2 and Lemma 2.5, one obtains the results that the operator $T$ has at least one fixed point $y\in K\cap\overline{\Omega}_d\backslash\Omega_k$. Consequently, the BVPs (1), (2) have at least one solution $y$ satisfying $k\leq\|y\|\leq d$.

Theorem 3.2 Suppose that $f: [0, 1]\times [0, +\infty)\rightarrow [0, +\infty)$ is a continuous function and there exist constants $0 < a < b < c$ such that the following three assumptions hold

$(C_1)\; f(x, y) < \phi_p(Ma)$, for $(x, y)\in[0, 1]\times[0, a]$;

$(C_2)\; f(x, y) > \phi_p(Nb)$, for $(x, y)\in[\frac{1}{4}, \frac{3}{4}]\times[b, c]$;

$(C_3)\; f(x, y)\leq\phi_p(Mc)$, for $(x, y)\in[0, 1]\times[0, c]$.

Then the BVPs $(1), (2)$ possess three non-negative solutions $y_1, y_2$ and $y_3$ such that

Proof We will use Lemma $2.6$ to obtain the multiplicity results.

Step 1. Firstly, for any $y\in\overline{K}_c$, there is $\|y\|\leq c$. Then the assumption $(C_3)$ yields that $f(x, y(x))\leq\phi_p(Mc)$ for $0\leq x\leq 1$. So,

Hence the operator $T:\overline{K}_c\rightarrow \overline{K}_c$.

Step 2 Secondly, We claim that the condition $(B_1)$ of Lemma 2.6 hold. Choosing $y_{0}(x) = (b+c)/2$ aconstant function. It is clear that $y_{0}(x)\in K(\theta, b, c), \theta(y_{0}) > b$. That is to say that the set $\{y\in K(\theta, b, c)\mid\theta(y) > b\}$ is not empty. Moreover, for $y\in K(\theta, b, c)$, there holds $b\leq y(x)\leq c$ for $\frac{1}{4}\leq x\leq \frac{3}{4}$. Then, the assumption $(C_2)$ yields that

$f(x, y(x)) > \phi_p(Nb)$, for $y\in K(\theta, b, c)$

So

Then, we have

The above arguments implies that the condition $(B_1)$ of Lemma 2.6 holds.

Step 3 Thirdly, for $y\in\overline{K}_a$, there holds $\|y\|\leq a$. Then, the Assumption $(C_1)$ implies that $f(x, y(x)) < \phi_p(Ma)$ for all $0\leq x\leq 1$. Thus

That is to say that

$\|Ty\| < a$, for all $y\in\overline{K}_a$,

i.e., the condition $(B_2)$ of Lemma 2.6 is also satisfied.

Step 4 At last, we claim that the condition $(B_3)$ of Lemma 2.6 holds too. If $y\in K(\theta, b, c)$, taking into account Step 2, there holds $\theta(Ty) > b$. Consequently, the condition $(B_3)$ of Lemma 2.6 holds.

The above four steps show that all the conditions By Lemma 2.6 hold. Thus, the BVPs (1), (2) possess three non-negative solutions $y_1, y_2$ and $y_3$ such that

The Theorem is proven.

4.   Some examples

In the section, we give two examples to illustrate the main results obtained for BVPs (1), (2) in the section 3.

$\; \; \; \;$Let $p = \frac{3}{2}, \; \alpha = \frac{5}{2}, \; \beta = \frac{1}{2}, \; \xi = \frac{1}{2}$. By simple computation, we have

Example 4.1 Consider the following BVPs:

It is clear that the real function

is nonnegative continuous on $[0, 1]\times[0, +\infty)$. Choosing $d = 3, \; k = \frac{1}{39}$, one can check that $\chi(d)\leq 2 < \phi_p(dM)\approx 2.0275$ and $\psi(k)\geq \frac{38}{39}\approx 0.9744 > \phi_p(kN)\approx0.9680$. By theorem $3.1$, the BVPs $(4.1)$ possess one positive solution $y$ such that $\frac{1}{39}\leq\|y\|\leq 3$.

Example 4.2 Consider the following BVPs:

where

Choosing $a = \frac{1}{36}, \; b = 1, \; c = 60$, then the function $f(x, y)$ satisfies the following three conditions

$(1)\; f(x, y) = \frac{1}{6}x^2+7y^{2} < 0.1721 < \phi_p(Ma)\approx 0.1951, \; \; \;$ for $(x, y)\in[0, 1]\times[0, \frac{1}{36}]$;

$(2)\; f(x, y) = \frac{1}{6}x^2+\frac{1}{60}y+\frac{419}{60} > 7.0104 > \phi_p(Nb)\approx 6.0449, \; \; \;$ for $(x, y)\in[\frac{1}{4}, \frac{3}{4}]\times[1,60]$;

$(3)\; f(x, y) = \frac{1}{6}x^2+\frac{1}{60}y+\frac{419}{60} < 8.1500 < \phi_p(Mc)\approx 9.0674, \; \; \;$ for $(x, y)\in[0, 1]\times[0, 60]$.

That is to say that all conditions of Theorem 3.2 hold. Then, Theorem 3.2 yields that the BVPs $(4.2)$ possess three non-negative solutions $y_1, y_2$ and $y_3$ such that

5.   Conclusions

The present paper concentrated the solvability of some $p$-Laplace boundary value problems with Caputo fractional derivative. In this work, we obtained some existence results of one or three non-negative solutions by using the fixed-point theory. At the last we yielded two examples which fulfills our findings.

Acknowledgments

The authors would like to express their deep gratitude to the journal editor and referees for their careful reviews and valuable comments which helped to improve the paper, and thank Professor Y.Tian for his help with this research. This research is supported by Chenzhou Science and Technology Planning Project of Hunan Province (ZDYF2020164).

Conflict of interest

The authors declare that they have no competing interests.