Analysis of a class of completely non-local elliptic diffusion operators

Abstract

This work explores the possibility of develo** the analog of some classic results from elliptic PDEs for a class of fractional ODEs involving the composition of both left- and right-sided Riemann-Liouville (R-L) fractional derivatives, ${D^\alpha _{a+}}{D^\beta _{b-}}$, $1<\alpha +\beta <2$. Compared to one-sided non-local R-L derivatives, these composite operators are completely non-local, which means that the evaluation of ${D^\alpha _{a+}}{D^\beta _{b-}}u(x)$ at a point x will have to retrieve the information not only to the left of x all the way to the left boundary but also to the right up to the right boundary, simultaneously. Therefore, only limited tools can be applied to such a situation, which is the most challenging part of the work. To overcome this, we do the analysis from a non-traditional perspective and eventually establish elliptic-type results, including Hopf’s Lemma and maximum principles. As $\alpha \rightarrow 1^-$ or $\alpha ,\beta \rightarrow 1^-$, those operators reduce to the one-sided fractional diffusion operator and the classic diffusion operator, respectively. For these reasons, we still refer to them as “elliptic diffusion operators", however, without any physical interpretation.

1 Introduction and motivation

Non-local equations have been a vibrant area of research in recent years. Compared with classical equations, which usually require solutions to be smooth, the framework of fractional integrals and derivatives assumes much less regularity or smoothness, making it suitable in various applications such as anomalous diffusion processes, image processing, biology models, underground water movement, dynamic fracture propagation and so on. While communities in applied sciences have been using non-local models successfully, their rigorous mathematical analysis still requires the development of fundamental results.

This work is a continuation of the article [8] on the one-sided 1-D fractional diffusion operator. We would like to study the possibility of develo** the parallel between the classical elliptic theory (n-dimensional) and the fractional-order equations and apply such theory to other scientific areas in the future. To do so, it is essential to know what kind of fractional operators (fractional derivatives) are pre-qualified for modeling diffusion and being “elliptic". Furthermore, we would like to discuss the detailed differences and similarities between the “fractional elliptic theory" and the classic elliptic theory. At this stage, our work only focuses on examining the 1-dimensional case.

In previous work [8], it has been shown that the one-sided fractional Riemann-Liouville derivative ${D_{a+}^{\alpha }}, \,1<\alpha <2$ maintains many classic elliptic properties except for the regularity results, such as the maximum principle and principal eigenvalue results, etc. In this paper, we focus our attention on (1-D) double-sided fractional operators ${D^\alpha _{a+}}{D^\beta _{b-}}$, $1<\alpha +\beta <2$ that involve mixed Riemann-Liouville derivatives, that is, the composition between left-sided R-L derivative ${D_{a+}^{\alpha }}$ and right-sided R-L derivative ${D^\beta _{b-}}$ (see Definition 1), and successfully establish the counterpart of Hopf’s Lemma and maximum principle for the homogeneous boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} Au=f \quad \text {in} \quad \Omega ,\quad f\in L^p(\Omega ),\, p\ge 1,\\ u|_{\partial \Omega }=0,\\ (Au)(x):={D_{a+}^{\alpha }}{D_{b-}^{\beta }}u, \quad 0\le \alpha , \beta \le 1, \quad 1<\alpha +\beta <2. \end{array}\right. } \end{aligned}$$

(1.1)

(Theorem 1, Theorem 2 and Theorem 3). These results reveal the similarity between the operator A and the usual elliptic operators. Regarding the regularity results, as we will point out in Remark 1, the solution u usually lacks regularity at the boundary points even for a smooth source function f, which is significantly different from the classic elliptic regularity results. On the other hand, as $\alpha \rightarrow 1^-$ or $\alpha ,\beta \rightarrow 1^-$, ${D^\alpha _{a+}}{D^\beta _{b-}}$ reduces to the one-sided fractional diffusion operator or the classic diffusion operator, respectively. For these reasons, we still refer to operator A as “non-local elliptic diffusion operator" in this paper without any physical interpretation. It is interesting to mention that the maximum principle does not hold for the operator ${^CD^\alpha _{a+}}{D^\beta _{b-}}$ involving the Caputo fractional derivative (it will be presented in a different work instead of the current paper). Therefore, it is still unknown to us whether ${^CD^\alpha _{a+}}{D^\beta _{b-}}$ has other elliptic results, such as principal eigenvalue, etc, due to the loss of maximum principles. Some related spectral analysis on ${^CD^\alpha _{a+}}{D^\beta _{b-}}$ can be found in references such as [5, 11, 12], and will not be discussed further here.

Compared to the one-sided non-local operator ${D_{a+}^{\alpha }}$, we say these double-sided operators ${D^\alpha _{a+}}{D^\beta _{b-}}$ are completely (globally) non-local, in the sense that the evaluation of ${D^\alpha _{a+}}{D^\beta _{b-}}u(x_0)$ at a point $x_0\in (a,b)$ will require the information both from a to $x_0$ and from b to $x_0$, i.e., over the whole domain. The Figure 1 summarizes the comparison between classic, one-sided, and double-sided derivatives.

Fig. 1

Comparison of local classical, non-local one-sided, and completely non-local double-sided operators

This feature brings significant challenges to our analysis simply because many traditional techniques and methods (see Chapter 6 in [3]) are not quite applicable, and in order to achieve the desired results, we have to develop our own. This mainly requires looking at the fractional Sobolev spaces from a different angle to fit the analysis of fractional derivatives in Section 2 and develo** several auxiliary lemmas in Section 4 (such as Lemma 4). To develop the theory of $Au=f$ that is as applicable as possible in future applications, we will do the analysis for the arbitrary bounded interval $\Omega =(a, b)$ instead of (0, 1). This is because, unlike the classic differential operators, fractional derivatives usually lose the simple translation and scaling properties when the function $u(x), \,x\in (0,1)$ is replaced by $u((x-a)/(b-a)), \, x\in (a,b)$, that is, $(Au)(x)\ne \frac{1}{(b-a)^{\alpha +\beta }}Au((x-a)/(b-a))$ in general (see, e.g., Chapter 5, [9]). This means that when restricting from $\Omega =(a, b)$ to $\Omega =(0, 1)$, the problems $Au=f$, $x\in \Omega$ may lose some interesting information. Although the special case $\Omega =(0,1)$ may simplify the calculation and justification substantially, we are more interested in the general bounded interval $\Omega =(a,b)$ throughout this work, where a, b are arbitrary finite real numbers. Meanwhile, unless otherwise specified, we confine ourselves to the discussion of real-valued functions.

Let us notice that if $\alpha =0$ or $\beta =0$ (similarly, if $\alpha =1$ or $\beta =1$), the problem (1.1) essentially reduces to the one-sided problem and has been justified in [8]. Therefore, in this paper, we focus only on the following situation where $\alpha$ and $\beta$ are strictly bigger than 0 and less than 1:

$$\begin{aligned} {\left\{ \begin{array}{ll} Au=f \quad \text {in} \quad \Omega =(a, b),\quad f\in L^1(\Omega ),\\ u(a)=u(b)=0,\\ (Au)(x):={D_{a+}^{\alpha }} {D_{b-}^{\beta }}u, \quad 0<\alpha , \beta<1, \quad 1<\alpha +\beta <2. \end{array}\right. } \end{aligned}$$

(1.2)

The organization of this paper is as follows. In Section 2, we first present some preliminary knowledge on the fractional Sobolev spaces from a non-traditional angle and introduce some useful facts and properties related to hypergeometric functions that will be employed later. Section 3 discusses the properties of the true solution of problem (1.2). In Section 4, we conduct intense analysis toward develo** elliptic-type properties, plot the pictures to illustrate the ideas, and eventually establish the maximum principles and Hopf’s lemma and other related properties, which are our main results.

2 Preliminary knowledge

This section lists basic definitions and properties that are necessary for subsequent analysis and will be used in later calculations and justifications.

2.1 Definitions of Riemann-Liouville integrals and derivatives

Definition 1

Let $s\ge 0$, the left and right Riemann-Liouville integrals of $\psi \in L^1(\Omega )$ are formally defined as

$$\begin{aligned} {I_{a+}^{s}}\psi (x)=\frac{1}{\Gamma (s)}\int _{a}^{x} \frac{\psi (t)\, dt}{(x-t)^{1-s}}\quad \text {and}\quad {I_{b-}^{s}}\psi (x)=\frac{1}{\Gamma (s)}\int _{x}^{b} \frac{\psi (t)\, dt}{(t-x)^{1-s}} \end{aligned}$$

(if $s=0$, it assumes automatically the identity operators) and the left and right Riemann-Liouville derivatives are formally defined as

$$\begin{aligned} {D_{a+}^{s}}\psi (x)=\left( \frac{d}{dx}\right) ^{1+[s]}{I_{a+}^{1-\{s\}}}\psi \quad \text {and}\quad {D_{b-}^{s}}\psi (x)=\left( -\frac{d}{dx}\right) ^{1+[s]}{I_{b-}^{1-\{s\}}}\psi , \end{aligned}$$

where [s] represents the integral part of s and $\{s\}$ represents the fractional part of s.

In particular, when $a=-\infty$ or $b=+\infty$, we shall denote the left- and right-sided R-L derivatives by ${D^{s}_+}$ and ${D^{s}_-}$, respectively.

2.2 Equivalent definitions of fractional Sobolev spaces

Fractional Sobolev spaces $W^{s,p}(\mathbb {R})$ are natural functional spaces to study fractional-order differential equations, especially for investigation of the regularity of solutions. When $p=2$, fractional Sobolev spaces $W^{s,2}(\mathbb {R}):=\widehat{H}^s(\mathbb {R})$, $s \ge 0$ become Hilbert spaces and can be defined in different ways with equal or equivalent (semi)norms. One popular way is via Fourier transform:

Definition 2

Let $\sigma >0$,

$$\begin{aligned} \widehat{H}^\sigma (\mathbb {R}) := \left\{ w \in L^2(\mathbb {R}) : \int _{\mathbb {R}} (1 + |2\pi \xi |^{2\sigma }) |\widehat{w}(\xi ) |^2 \, \textrm{d} \xi < \infty \right\} , \end{aligned}$$

(2.1)

with semi-norm and norm

$$\begin{aligned} |w|_{\widehat{H}^\sigma (\mathbb {R})}:=\Vert (2\pi \xi )^\sigma \widehat{w}\Vert _{L^2(\mathbb {R})}, \quad \Vert w\Vert _{\widehat{H}^\sigma (\mathbb {R})}:=\left( \Vert w\Vert ^2_{L^2(\mathbb {R})} +|w|^2_{\widehat{H}^\sigma (\mathbb {R})}\right) ^{1/2}. \end{aligned}$$

Although Definition 2 plays an important role in the analysis, sometimes, it is not quite applicable in fractional calculus. To fit the analysis of fractional-order differential equations, $\widehat{H}^\sigma (\mathbb {R})$ can also be characterized by the left- or right-sided fractional-order weak derivatives, which can be regarded as a generalization of integer-order weak derivatives:

Definition 3

( [4], Section 3) Given $s\ge 0$ and assume $u\in L^2(\mathbb {R})$, then $u \in \widehat{H}^s(\mathbb {R})$ if and only if there exists a unique $\psi _1 \in L^2(\mathbb {R})$ such that

$$\begin{aligned} \int _\mathbb {R} u \cdot {D_-^s} \psi \, dx =\int _\mathbb {R} \psi _1 \cdot \psi \, dx \end{aligned}$$

(2.2)

for any $\psi \in C^\infty _0(\mathbb {R})$. Similarly, $u \in \widehat{H}^s(\mathbb {R})$ if and only if there exists a unique $\psi _2 \in L^2(\mathbb {R})$ such that

$$\begin{aligned} \int _\mathbb {R} u \cdot {D_{+}^s} \psi \, dx =\int _\mathbb {R} \psi _2 \cdot \psi \, dx \end{aligned}$$

(2.3)

for any $\psi \in C^\infty _0(\mathbb {R})$.

$\psi _1, \psi _2$ are called the left- and right-sided weak R-L derivatives of u, denoted by symbols $\psi _1={D_{+}^s} u$, $\psi _2={D_-^s} u$, respectively, which are understood in the weak sense as in (2.2) and (2.3).

Property 1

( [4], Section 3) Assume $u\in \widehat{H}^s(\mathbb {R})$, $s\ge 0$, then

$$\begin{aligned} |u|_{\widehat{H}^s(\mathbb {R})} = \Vert {D_-^s} u\Vert _{L^2(\mathbb {R})}=\Vert {D_{+}^s} u\Vert _{L^2(\mathbb {R})}, \end{aligned}$$

(2.4)

where ${D_{+}^s} u$ and ${D_-^s} u$ are understood in the weak sense.

Using the fact that $C_c^\infty (\Omega )$ is a dense set in $\widehat{H}^s_0(\Omega )$, we can define the following analogue:

Definition 4

Given $s\ge 0$.

$$\begin{aligned} \widehat{H}^s_0(\Omega ):=\{\text {Closure of}\, \, u\in C_c^\infty (\Omega ) \, \text {with respect to norm}\, \Vert \tilde{u}\Vert _{\widehat{H}^s(\mathbb {R})} \}, \end{aligned}$$

(2.5)

where symbol $\tilde{u}$ denotes the trivial continuation of u(x) by 0 outside $\Omega$. It is endowed with semi-norm and norm

$$\begin{aligned} |u|_{\widehat{H}^s_0(\Omega )}:=|\tilde{u}|_{\widehat{H}^s(\mathbb {R})}, \quad \Vert u\Vert _{\widehat{H}^s_0(\Omega )}:=\Vert \tilde{u}\Vert _{\widehat{H}^s(\mathbb {R})}. \end{aligned}$$

With Definition 4 and Definition 3, it is not difficult to derive the following:

Property 2

( [7]; Theorems 2.10, 2.13, [2]) Given $\frac{1}{2}<s<1$. Then

• 1. $u\in \widehat{H}^s_0(\Omega )$ can be represented as

  $$\begin{aligned} u(x)={I_{a+}^{s}}\psi _1={I_{b-}^{s}}\psi _2 \end{aligned}$$

  (2.6)

  for certain $\psi _1$, $\psi _2 \in L^2(\Omega )$. As a consequence, the R-L fractional derivatives of elements of $\widehat{H}_0^s(\Omega )$, ${D_{a+}^{s}}u$, and ${D_{b-}^{s}}u$ exist a.e. and coincide with $\psi _1$, $\psi _2$, respectively.

• 2. there exists a positive constant C such that, for any $u\in \widehat{H}^s_0(\Omega )$

  $$\begin{aligned} \Vert u\Vert _{\widehat{H}_0^{s}(\Omega )}\le C \,\Vert {D_{b-}^{s}}u\Vert _{L^2(\Omega )} \quad \text {and} \quad \Vert u\Vert _{\widehat{H}_0^{s}(\Omega )}\le C \,\Vert {D_{a+}^{s}}u\Vert _{L^2(\Omega )}. \end{aligned}$$

  (2.7)

It is worth pointing out that each element $u\in \widehat{H}_0^s(\Omega )$, $s>1/2$ admits a representation that belongs to H$\ddot{\text {o}}$lder space $C^{0, s-1/2}(\overline{\Omega })$, hence it makes sense to talk about u point-wise such as $u\equiv 0$, $u\not \equiv 0$ or $\lim _{x\rightarrow b^-}u$ in subsequent analysis in later sections, and we shall not specify it every time when there is no chance to cause confusion under the context. We shall also need the following fact, which provides a convenient criterion for determining whether a given function belongs to $\widehat{H}_0^s(\Omega )$.

Property 3

(Lemma 3.3, [8]) Given $\frac{1}{2}<s<1$. If u has a fractional integral representation $u={{I_{a+}^{s}}\psi _1}$ for a certain $\psi _1 \in L^2(\Omega )$, and

$$\begin{aligned} u(b)=0,\quad \int _{a}^{b} (b-t)^{-\frac{1}{2}} |\psi _1(t)|\,dt<\infty , \end{aligned}$$

(2.8)

then $u\in \widehat{H}_0^s(\Omega )$.

Similarly, if $u={I_{b-}^{s}}\psi _2$ for a certain $\psi _2 \in L^2(\Omega )$, and

$$\begin{aligned} u(a)=0,\quad \int _{a}^{b} (t-a)^{-\frac{1}{2}} |\psi _2(t)|\,dt<\infty , \end{aligned}$$

(2.9)

then $u\in \widehat{H}_0^s(\Omega )$.

2.3 Map** properties and hypergeometric functions

Now, we point out several map** properties of R-L integrals, as well as some facts about hypergeometric functions.

Property 4

(Property B.5, [7]) Given $t\in (0,1)$ and bounded open interval $\Omega \subset \mathbb {R}$, the fractional operators ${I_{a+}^{t}}, {I_{b-}^{t}}$ are continuous map**s

• 1. from $L^1(\Omega )$ into $L^q(\Omega )$ with $\displaystyle 1\le q <\frac{1}{1-t}$,

• 2. from $L^p(\Omega )$ into $L^q(\Omega )$ with $1<p<1/t$ and $\displaystyle q=\frac{p}{1-tp}$.

Property 5

(Theorem 11.5, p. 206, [10]) Let $0<t<1$ and

$$\begin{aligned} r_a(x):=x-a, \, r_b(x):=b-x,\, x\in \Omega , \, (S\psi )(x):=P.V.\frac{1}{\pi }\int _{a}^{b}\frac{\psi (t)\,dt}{t-x}, x\in \Omega .\nonumber \\ \end{aligned}$$

(2.10)

Then the fractional integrals ${I_{a+}^{t}}\psi , {I_{b-}^{t}}\psi$ and the singular operator S are connected with each other by the relations:

$$\begin{aligned} {I_{b-}^{t}}\psi&= \cos t\pi \,{I_{a+}^{t}}\psi +\sin t\pi \,{I_{a+}^{t}}r_a^{-t}Sr_a^t \psi , \quad \psi \in L^p(\Omega ), \quad p>1, \end{aligned}$$

(2.11)

$$\begin{aligned} {I_{a+}^{t}}\psi&= \cos t\pi \, {I_{b-}^{t}}\psi -\sin t\pi \,{I_{b-}^{t}}r_b^{-t}Sr_b^t\psi , \quad \psi \in L^p(\Omega ),\quad p>1, \end{aligned}$$

(2.12)

$$\begin{aligned} {I_{b-}^{t}}\psi&= \cos t\pi \,{I_{a+}^{t}}\psi +\sin t\pi \,r_b^{t}Sr_b^{-t} {I_{a+}^{t}} \psi , \quad \psi \in L^p(\Omega ), \quad p\ge 1, \end{aligned}$$

(2.13)

$$\begin{aligned} {I_{a+}^{t}}\psi&= \cos t\pi \,{I_{b-}^{t}}\psi -\sin t\pi \,r_a^{t}Sr_a^{-t} {I_{b-}^{t}}\psi , \quad \psi \in L^p(\Omega ),\quad p\ge 1, \end{aligned}$$

(2.14)

which hold almost everywhere if $pt\le 1$ and for each $x\in \Omega$, if $pt>1$.

Property 6

(Theorem 11.3, p. 200, [10]) The operator S, defined in (2.10), is bounded in the weighted space $L^p(\rho , \Omega )$, $1<p<\infty$, $\rho (x)=(x-a)^{\mu _1}(b-x)^{\mu _2}$, if $-1<\mu _1, \mu _2<p-1$.

The following results describe the asymptotic behavior and integral and differential properties of the well-known Gauss hypergeometric function ${{}_2F_1(t_1,t_2;t_3; x)}$ under specific conditions.

Property 7

(Theorem 2.1.3, p. 63, Theorem 2.2.2, p. 66, [1])

• 1. If $\Re (t_3-t_2-t_1)<0$, then

  $$\begin{aligned} \lim _{x\rightarrow 1^-}\frac{{}_2F_1(t_1,t_2;t_3; x)}{(1-x)^{t_3-t_2-t_1}}=\frac{\Gamma (t_3)\Gamma (t_1+t_2-t_3)}{\Gamma (t_1)\Gamma (t_2)}. \end{aligned}$$

  (2.15)

• 2. If $\Re (t_3-t_2-t_1)>0$, then

  $$\begin{aligned} {{}_2F_1(t_1,t_2;t_3; 1)}=\frac{\Gamma (t_3)\Gamma (t_3-t_2-t_1)}{\Gamma (t_3-t_1)\Gamma (t_3-t_2)}. \end{aligned}$$

  (2.16)
• 3.
  $$\begin{aligned} \frac{d}{dx}{{}_2F_1(t_1,t_2;t_3; x)}=\frac{t_1\, t_2}{t_3}{{}_2F_1(t_1+1,t_2+1;t_3+1; x)}. \end{aligned}$$

  (2.17)
• 4.
  $$\begin{aligned} {{}_2F_1(t_1,t_2;t_3; 0)}=1. \end{aligned}$$

  (2.18)

Property 8

((2.19) and (2.46), [10]) For $\Re (t_1)\ge 0, \Re (t_2)>0, t_3\in \mathbb {C}$,

$$\begin{aligned} \begin{aligned}&{I_{b-}^{t_1}} (b-x)^{t_2-1}(x-a)^{t_3-1}\\&=\frac{(b-a)^{t_3-1}\Gamma (t_2)}{\Gamma (t_1+t_2)}(b-x)^{t_1+t_2-1}{{}_2F_1}\left( 1-t_3,t_2;t_1+t_2; \frac{b-x}{b-a}\right) . \end{aligned} \end{aligned}$$

(2.19)

3 The true solution of the BVP

In this section, we establish some useful facts about the true solution of problem (1.2), which will be used later in the paper. In order to interpret the operator A and the boundary condition in (1.2) correctly, we first specify the definition of the solution.

Definition 5

For a given $f\in L^1(\Omega )$, we say that u is a solution of problem (1.2) if

• 1. ${I_{a+}^{1-\alpha }}{D_{b-}^{\beta }}u, \, {I_{b-}^{1-\beta }}u\in AC(\overline{\Omega })$ and $Au=f$ a.e.

• 2. $u\in C(\overline{\Omega })$ and $u(a)=u(b)=0$.

Since $f\in L^1(\Omega )$, the equality $Au=f$ can hold only almost everywhere. In the following, for simplicity we shall sometimes write $Au=f$ instead of $Au=f$ a.e.

Lemma 1

Given $f\in L^1(\Omega )$. If u is a solution of (1.2), then

• 1.
  $$\begin{aligned} u(x)={I_{b-}^{\beta }}\,({I_{a+}^{\alpha }}f+c\,{D_{a+}^{1-\alpha }}\,1), \, x\in \Omega , \end{aligned}$$

  (3.1)

  where

  $$\begin{aligned} c=-\frac{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}{(b-a)^{\alpha +\beta -1}}\cdot {I_{b-}^{\beta }}{I_{a+}^{\alpha }}f \big |_{x=a}. \end{aligned}$$

  (3.2)

• 2. $u \in \widehat{H}_0^{\frac{\alpha +\beta }{2}}(\Omega )$.

• 3. there exists a positive constant $\tilde{C}_{\alpha ,\beta }$ such that

  $$\begin{aligned} \Vert u\Vert _{L^1(\Omega )}\le \tilde{C}_{\alpha ,\beta } \Vert f\Vert _{L^1(\Omega )} \end{aligned}$$

  (3.3)

  for any $f\in L^1(\Omega )$.

• 4. there exists a positive constant $C_{\alpha ,\beta }$ such that

  $$\begin{aligned} \Vert u\Vert _{\widehat{H}_0^{(\alpha +\beta )/2}(\Omega )}\le C_{\alpha ,\beta } \Vert {I_{a+}^{\frac{\alpha +\beta }{2}}}f\Vert _{L^2(\Omega )} \end{aligned}$$

  (3.4)

  for any $f\in L^1(\Omega )$.

Proof

1. Recall Definition 5. Since f is integrable and ${I_{a+}^{1-\alpha }}{D_{b-}^{\beta }}u\in AC(\overline{\Omega })$, it is valid to integrate both sides of $Au=f$ to obtain

$$\begin{aligned} {I_{a+}^{1-\alpha }}{D_{b-}^{\beta }}u={I_{a+}^{1}}f+c \end{aligned}$$

for a certain constant c. Then differentiating both sides by ${D_{a+}^{1-\alpha }}$, we have

$$\begin{aligned} {D_{b-}^{\beta }}u={I_{a+}^{\alpha }}f+c\,{D_{a+}^{1-\alpha }}\,1. \end{aligned}$$

Again, it is valid to integrate both sides to have

$$\begin{aligned} {I_{b-}^{1-\beta }}u={I_{b-}^{1}}({I_{a+}^{\alpha }}f+c\,{D_{a+}^{1-\alpha }}\,1)+\tilde{c} \end{aligned}$$

for another $\tilde{c}$. Hence, differentiating both sides by ${D_{b-}^{1-\beta }}$ yields

$$\begin{aligned}{} & {} u={I_{b-}^{\beta }}\,({I_{a+}^{\alpha }}f+c\,{D_{a+}^{1-\alpha }}\,1)+\tilde{c}\,{D_{b-}^{1-\beta }}\,1\nonumber \\{} & {} \quad ={I_{b-}^{\beta }}\,({I_{a+}^{\alpha }}f+c\,{D_{a+}^{1-\alpha }}\,1) \nonumber \\{} & {} \qquad +\frac{\tilde{c}}{\Gamma (\beta )}\frac{1}{(b-x)^{1-\beta }}.\nonumber \\ \end{aligned}$$

(3.5)

By the map** properties of fractional integrals ${I_{a+}^{\alpha }}, {I_{b-}^{\beta }}$ (Property 4) together with our assumption $\alpha +\beta >1$, we know

$$\begin{aligned} {I_{b-}^{\beta }}\,{I_{a+}^{\alpha }}f \big |_{x=b}={I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}\,1\big |_{x=b}=0. \end{aligned}$$

So, this last line implies that $\tilde{c}$ in (3.5) has to be zero, which gives (3.1).

2. To determine the value of c in (3.2), we first apply the useful identities (2.19) in Property 8 and (2.16) in Property 7 to directly calculate

$$\begin{aligned} \begin{aligned} {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}\,1\big |_{x=a}&=\frac{1}{\Gamma (\alpha )}{I_{b-}^{\beta }}(x-a)^{\alpha -1}\big |_{x=a}\\&=\frac{(b-a)^{\alpha -1}(b-x)^\beta }{\Gamma (\alpha )\Gamma (1+\beta )}{{}_2F_1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \biggl |_{x=a}\\&=\frac{(b-a)^{\alpha +\beta -1}}{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}. \end{aligned} \end{aligned}$$

(3.6)

With (3.6) and the boundary condition $u(a)=0$, evaluating both sides of Equation (3.1) at $x=a$ and solving for c give (3.2).

3. Now let us show that u actually belongs to $\widehat{H}_0^{\frac{\alpha +\beta }{2}}(\Omega )$. In view of Property 3, in order to show $u\in \widehat{H}_0^{\frac{\alpha +\beta }{2}}(\Omega )$ it suffices to show that

$$\begin{aligned} {D_{b-}^{\frac{\alpha +\beta }{2}}}u \in L^q(\Omega ),\quad \text {for a certain}\quad q>2, \end{aligned}$$

(3.7)

and

$$\begin{aligned} u={I_{b-}^{\frac{\alpha +\beta }{2}}}{D_{b-}^{\frac{\alpha +\beta }{2}}}u,\quad x\in \Omega . \end{aligned}$$

(3.8)

To see (3.7), by using expression (3.1), we only need to verify, separately,

$$\begin{aligned} {D_{b-}^{\frac{\alpha +\beta }{2}}}{I_{b-}^{\beta }}{I_{a+}^{\alpha }}f\in L^q(\Omega ), \end{aligned}$$

(3.9)

and

$$\begin{aligned} {D_{b-}^{\frac{\alpha +\beta }{2}}}{I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}\,1\in L^q(\Omega ). \end{aligned}$$

(3.10)

4. Let us first check (3.9). Notice $\displaystyle {I_{b-}^{\beta }}{I_{a+}^{\alpha }}f={I_{b-}^{\beta }}{I_{a+}^{\alpha -\epsilon }}{I_{a+}^{\epsilon }}f$ for any $0<\epsilon <\alpha$. By the first part of Property 4 we know

$$\begin{aligned} {I_{a+}^{\epsilon }}f\in L^{t_0}(\Omega ) \quad \text {for a certain}\quad t_0>1. \end{aligned}$$

(3.11)

Then we apply (2.12) to replace the operator ${I_{a+}^{\alpha -\epsilon }}$, and we have

$$\begin{aligned}{} & {} {I_{b-}^{\beta }}{I_{a+}^{\alpha -\epsilon }}{I_{a+}^{\epsilon }}f \nonumber \\{} & {} \quad =\cos (\alpha \pi -\epsilon \pi ){I_{b-}^{\alpha +\beta -\epsilon }}{I_{a+}^{\epsilon }}f-\sin (\alpha \pi -\epsilon \pi ){I_{b-}^{\alpha +\beta -\epsilon }}r_b^{\epsilon -\alpha }Sr_b^{\alpha -\epsilon } {I_{a+}^{\epsilon }}f.\nonumber \\ \end{aligned}$$

(3.12)

Hence,

$$\begin{aligned} \begin{aligned}&{D_{b-}^{\frac{\alpha +\beta }{2}}}{I_{b-}^{\beta }}{I_{a+}^{\alpha }}f\\&=\cos (\alpha \pi -\epsilon \pi ){I_{b-}^{\frac{\alpha +\beta }{2}-\epsilon }}{I_{a+}^{\epsilon }}f-\sin (\alpha \pi -\epsilon \pi ){I_{b-}^{\frac{\alpha +\beta }{2}-\epsilon }}r_b^{\epsilon -\alpha }Sr_b^{\alpha -\epsilon } {I_{a+}^{\epsilon }}f\\&:=\cos (\alpha \pi -\epsilon \pi )N_1-\sin (\alpha \pi -\epsilon \pi )N_2. \end{aligned} \end{aligned}$$

According to Property 6, it is well-known that the operator $r_b^{-\delta }Sr_b^\delta$ is bounded in $L^t(\Omega )$ for $1<t<\frac{1}{\delta }$. (3.11) tells us that

$$\begin{aligned} r_b^{\epsilon -\alpha }Sr_b^{\alpha -\epsilon } {I_{a+}^{\epsilon }}f\in L^{t_0}(\Omega ). \end{aligned}$$

(3.13)

By the second application of the first part of Property 4 to (3.11) and (3.13), we have

$$\begin{aligned} N_1\in L^t(\Omega )\quad \text {and} \quad N_2\in L^t(\Omega ),\quad 1<t<\frac{2}{2-\alpha -\beta +2\epsilon }. \end{aligned}$$

It is evident that, as long as $\epsilon$ is chosen sufficiently small,

$$\begin{aligned} N_1\in L^q(\Omega ) \quad \text {and}\quad N_2\in L^q(\Omega )\quad \text {for a certain}\quad q>2 \end{aligned}$$

are guaranteed. This yields (3.9).

5. Now let us check (3.10). Rewrite ${D_{a+}^{1-\alpha }}\,1$ in the form

$$\begin{aligned} {D_{a+}^{1-\alpha }}\,1=\frac{1}{\Gamma (\alpha )} (x-a)^{-(1-\alpha )}=\frac{1}{\Gamma (\alpha )} \frac{(x-a)^{\alpha -\epsilon /2}(b-x)^{1-\epsilon /2}}{(x-a)^{1-\epsilon /2}(b-x)^{1-\epsilon /2}}, \, 0<\epsilon <\alpha .\nonumber \\ \end{aligned}$$

(3.14)

On the top, $(x-a)^{\alpha -\epsilon /2}(b-x)^{1-\epsilon /2}$ belongs to the functional space $C_0^{0,\alpha -\epsilon /2}(\Omega )$ (using eq. (29), [6]), which consists of Hölder continuous functions with exponent $\alpha -2\epsilon$ and vanish at the boundary points of $\Omega$. This alternative form (3.14) implies that ${D_{a+}^{1-\alpha }}\,1$ can be represented as

$$\begin{aligned} {D_{a+}^{1-\alpha }}\,1={I_{b-}^{\alpha -\epsilon }}J_\epsilon \quad \text {for a certain} \quad J_\epsilon \in L^1(\Omega ) \,\text {for any}\,\, 0<\epsilon <\alpha . \end{aligned}$$

(3.15)

(see, e.g. Property 8, [6]). Therefore, we have

$$\begin{aligned} {D_{b-}^{\frac{\alpha +\beta }{2}}}{I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}\,1={I_{b-}^{\frac{\alpha +\beta }{2}-\epsilon }}J_\epsilon . \end{aligned}$$

(3.16)

Again, as long as we choose $\epsilon >0$ small enough, by a similar argument as in the previous step, we know that the right-hand side of (3.16) belongs to $L^q(\Omega )$ for some $q>2$. This yields (3.10).

6. It remains to see the validity of (3.8). Recall (3.12) and (3.15), we know that the expression of u in the right-hand side of (3.1) admits an integral representation

$$\begin{aligned} u={I_{b-}^{\frac{\alpha +\beta }{2}}} h\quad \text {for a certain}\quad h\in L^1(\Omega ) \end{aligned}$$

(we can go farther than $\frac{\alpha +\beta }{2}$, but not necessary.) Then equality (3.8) is guaranteed to hold (see, e.g., eq. (2.58), (2.59), p. 45, [10]). Therefore, we arrive at $u\in \widehat{H}_0^{\frac{\alpha +\beta }{2}}(\Omega )$.

7. Let us prove the third part of Lemma 1.

Using expressions (3.1) and (3.2), we readily check that

$$\begin{aligned} \begin{aligned}&\Vert u\Vert _{L^1(\Omega )}\\&=\Vert {I_{b-}^{\beta }}\,({I_{a+}^{\alpha }}f+c\,{D_{a+}^{1-\alpha }}\,1)\Vert _{L^1(\Omega )}\\&\le \Vert {I_{b-}^{\beta }}{I_{a+}^{\alpha }}f\Vert _{L^1(\Omega )}+|c|\cdot \Vert {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}\,1\Vert _{L^1(\Omega )}. \end{aligned} \end{aligned}$$

(3.17)

On one hand, by the boundedness of Riemann-Liouville integrals,

$$\begin{aligned} \Vert {I_{b-}^{\beta }}{I_{a+}^{\alpha }}f\Vert _{L^1(\Omega )}\le C \Vert f\Vert _{L^1(\Omega )}, \end{aligned}$$

where C represents a generic constant. On the other hand,

$$\begin{aligned} |c|=\frac{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}{(b-a)^{\alpha +\beta -1}}\cdot \biggl | {I_{b-}^{\beta }}{I_{a+}^{\alpha }}f \big |_{x=a}\biggl |. \end{aligned}$$

(3.18)

By virtue of (3.12) and choosing $\epsilon =\alpha +\beta -1$ in there,

$$\begin{aligned} \begin{aligned}&\biggl | {I_{b-}^{\beta }}{I_{a+}^{\alpha }}f \big |_{x=a}\biggl |\\&=\biggl |\int _{a}^{b} \left( \cos ( \pi -\beta \pi ){I_{a+}^{\alpha +\beta -1}}f-\sin ( \pi -\beta \pi )r_b^{\beta -1}Sr_b^{1-\beta } {I_{a+}^{\alpha +\beta -1}}f\right) \,dt\biggl |\\&\le |\cos ( \pi -\beta \pi )|\int _{a}^{b} |{I_{a+}^{\alpha +\beta -1}}f|\,dt+|\sin ( \pi -\beta \pi )|\int _{a}^{b}|r_b^{\beta -1}Sr_b^{1-\beta } {I_{a+}^{\alpha +\beta -1}}f|\,dt. \end{aligned} \end{aligned}$$

Utilizing Property 4 and Property 6, this last inequality implies

$$\begin{aligned} \biggl | {I_{b-}^{\beta }}{I_{a+}^{\alpha }}f \big |_{x=a}\biggl |\le C\Vert f\Vert _{L^1(\Omega )}. \end{aligned}$$

Doing back substitution into (3.18), we see

$$\begin{aligned} |c|\le C\Vert f\Vert _{L^1(\Omega )}. \end{aligned}$$

Hence, inequality (3.17) now becomes

$$\begin{aligned} \Vert u\Vert _{L^1(\Omega )}\le \tilde{C}_{\alpha ,\beta }\Vert f\Vert _{L^1(\Omega )}, \end{aligned}$$

where $\tilde{C}_{\alpha ,\beta }$ is a certain suitable positive constant depending on $\alpha , \beta$ only. We, therefore, have proved the third part of Lemma 1.

8. Lastly, to prove (3.4), we consider two different cases.

Case 1: $\beta \ge \alpha$. Using expression (3.1), we compute

$$\begin{aligned} {D_{b-}^{\frac{\alpha +\beta }{2}}}u={I_{b-}^{\frac{\beta -\alpha }{2}}}{I_{a+}^{\alpha }}f+ c\,{I_{b-}^{\frac{\beta -\alpha }{2}}}{D_{a+}^{1-\alpha }}1:=T_1 +T_2, \end{aligned}$$

where c is the same as in (3.2). From the previous justification, i.e. (3.9) and (3.10), we already know that

$$\begin{aligned} T_1\in L^2(\Omega )\quad \text {and}\quad T_2\in L^2(\Omega ). \end{aligned}$$

In view of the second part of Property 2,

$$\begin{aligned} \begin{aligned} \Vert u\Vert _{\widehat{H}_0^{(\alpha +\beta )/2}(\Omega )}&\le C \,\Vert {D_{b-}^{\frac{\alpha +\beta }{2}}}u\Vert _{L^2(\Omega )}\\&\le C\,\left( \Vert T_1\Vert _{L^2(\Omega )}+\Vert T_2\Vert _{L^2(\Omega )}\right) \\&\le C\, \left( \Vert {I_{b-}^{\frac{\beta -\alpha }{2}}}{I_{a+}^{\alpha }}f\Vert _{L^2(\Omega )}+|c|\,\Vert {I_{b-}^{\frac{\beta -\alpha }{2}}}{D_{a+}^{1-\alpha }}1\Vert _{L^2(\Omega )}\right) . \end{aligned} \end{aligned}$$

(3.19)

Observe

$$\begin{aligned} \begin{aligned} |c|&=\frac{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}{(b-a)^{\alpha +\beta -1}}\cdot \biggl | {I_{b-}^{\frac{\beta +\alpha }{2}}}{I_{b-}^{\frac{\beta -\alpha }{2}}}{I_{a+}^{\alpha }}f \big |_{x=a}\biggl |\\&=\frac{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}{(b-a)^{\alpha +\beta -1}\Gamma (\frac{\beta +\alpha }{2})}\cdot \biggl | \int _{a}^{b}(t-a)^{\frac{\beta +\alpha }{2}-1}\, {I_{b-}^{\frac{\beta -\alpha }{2}}}{I_{a+}^{\alpha }}f(t)\,dt \biggl |\\&\le \frac{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}{(b-a)^{\alpha +\beta -1}\Gamma (\frac{\beta +\alpha }{2})}\cdot \Vert (x-a)^{\frac{\beta +\alpha }{2}-1}\Vert _{L^2(\Omega )}\cdot \Vert {I_{b-}^{\frac{\beta -\alpha }{2}}}{I_{a+}^{\alpha }}f\Vert _{L^2(\Omega )}. \end{aligned} \end{aligned}$$

(3.20)

Substituting the last inequality into Equation (3.19), we see that

$$\begin{aligned} \Vert u\Vert _{\widehat{H}_0^{(\alpha +\beta )/2}(\Omega )}\le C \Vert {I_{b-}^{\frac{\beta -\alpha }{2}}}{I_{a+}^{\alpha }}f\Vert _{L^2(\Omega )} \end{aligned}$$

(3.21)

for a certain constant C. Notice $\frac{\beta -\alpha }{2}<\frac{1}{2}$, using (2.13) together with Property 6, it is deduced that

$$\begin{aligned} \Vert {I_{b-}^{\frac{\beta -\alpha }{2}}}{I_{a+}^{\alpha }}f\Vert _{L^2(\Omega )}\le \tilde{C} \Vert {I_{a+}^{\frac{\beta +\alpha }{2}}}f\Vert _{L^2(\Omega )}. \end{aligned}$$

(3.22)

Combining (3.21) and (3.22) gives the desired result (3.4).

Case 2: $\beta <\alpha$. With the expression (3.1), we compute

$$\begin{aligned} \begin{aligned} {D_{b-}^{\frac{\alpha +\beta }{2}}}u&={D_{b-}^{\frac{\alpha -\beta }{2}}}{I_{a+}^{\alpha }}f+ c\,{D_{b-}^{\frac{\alpha -\beta }{2}}}{D_{a+}^{1-\alpha }}1\\&={D_{b-}^{\frac{\alpha -\beta }{2}}}{I_{a+}^{\frac{\alpha -\beta }{2}}}{I_{a+}^{\frac{\alpha +\beta }{2}}}f+ c\,{D_{b-}^{\frac{\alpha -\beta }{2}}}{D_{a+}^{1-\alpha }}1\\&:=M_1+M_2. \end{aligned} \end{aligned}$$

It can be checked in a similar manner that

$$\begin{aligned} {D_{b-}^{\frac{\alpha -\beta }{2}}}{I_{a+}^{\frac{\alpha -\beta }{2}}}\,\text {is bounded, by}\,\, (2.12) \,\,\text {and}\,\, \text {Property}~6, \,\text {in}\,\, L^p(\Omega ), 1<p<\frac{2}{\alpha -\beta }, \end{aligned}$$

and that

$$\begin{aligned} {I_{a+}^{\frac{\alpha +\beta }{2}}}f\in L^2(\Omega )\,\, \text {by}\,\, \text {Property}~4. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert M_1\Vert _{L^2(\Omega )}\le C \Vert {I_{a+}^{\frac{\alpha +\beta }{2}}}f\Vert _{L^2(\Omega )}. \end{aligned}$$

(3.23)

On the other hand, using (3.15) and Property 4, it can be directly verified that

$$\begin{aligned} {D_{b-}^{\frac{\alpha -\beta }{2}}}{D_{a+}^{1-\alpha }}1\in L^2(\Omega ), \end{aligned}$$

and that

$$\begin{aligned}{} & {} |c|=\frac{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}{(b-a)^{\alpha +\beta -1}}\cdot \biggl | {I_{b-}^{\beta }}{I_{a+}^{\frac{\alpha -\beta }{2}}}{I_{a+}^{\frac{\alpha +\beta }{2}}}f \big |_{x=a}\biggl |\nonumber \\{} & {} \qquad =\frac{\Gamma (\alpha )\Gamma (\beta )(\alpha +\beta -1)}{(b-a)^{\alpha +\beta -1}}\cdot M, \end{aligned}$$

(3.24)

where

$$\begin{aligned} \begin{aligned}&M=\\&\biggl |\left( \cos ((\alpha -\beta )\pi /2){I_{b-}^{\frac{\alpha +\beta }{2}}}{I_{a+}^{\frac{\alpha +\beta }{2}}}f-\sin ((\alpha -\beta )\pi /2){I_{b-}^{\frac{\alpha +\beta }{2}}}r_b^{\frac{\beta -\alpha }{2}}Sr_b^{\frac{\alpha -\beta }{2}} {I_{a+}^{\frac{\alpha +\beta }{2}}}f\right) \biggl |_{x=a}\biggl | \end{aligned} \end{aligned}$$

and (2.12) was used in the last equality in (3.24). By triangle inequality and Hölder inequality, we can estimate M to obtain

$$\begin{aligned} \begin{aligned} M&=\biggl |\frac{\cos ((\alpha -\beta )\pi /2)}{\Gamma (\frac{\alpha +\beta }{2})} \int _{a}^{b} (t-a)^{\frac{\alpha +\beta }{2}-1}{I_{a+}^{\frac{\alpha +\beta }{2}}}f(t)\, dt\\&\qquad - \frac{\sin ((\alpha -\beta )\pi /2)}{\Gamma (\frac{\alpha +\beta }{2})}\int _{a}^{b} (t-a)^{\frac{\alpha +\beta }{2}-1} r_b^{\frac{\beta -\alpha }{2}}Sr_b^{\frac{\alpha -\beta }{2}} {I_{a+}^{\frac{\alpha +\beta }{2}}}f(t)\, dt\biggl |\\&\le \frac{\cos ((\alpha -\beta )\pi /2)}{\Gamma (\frac{\alpha +\beta }{2})}\int _{a}^{b}(t-a)^{\frac{\alpha +\beta }{2}-1}\cdot \biggl |{I_{a+}^{\frac{\alpha +\beta }{2}}}f(t)\biggl |\, dt\\&\quad +\frac{\sin ((\alpha -\beta )\pi /2)}{\Gamma (\frac{\alpha +\beta }{2})}\int _{a}^{b}(t-a)^{\frac{\alpha +\beta }{2}-1}\cdot \biggl |r_b^{\frac{\beta -\alpha }{2}}Sr_b^{\frac{\alpha -\beta }{2}} {I_{a+}^{\frac{\alpha +\beta }{2}}}f(t)\biggl |\,dt\\&\le \frac{\cos ((\alpha -\beta )\pi /2)}{\Gamma (\frac{\alpha +\beta }{2})} \Vert (x-a)^{\frac{\beta +\alpha }{2}-1}\Vert _{L^2(\Omega )}\cdot \Vert {I_{a+}^{\frac{\alpha +\beta }{2}}}f\Vert _{L^2(\Omega )}\\&\quad +\frac{\sin ((\alpha -\beta )\pi /2)}{\Gamma (\frac{\alpha +\beta }{2})}\Vert (x-a)^{\frac{\beta +\alpha }{2}-1}\Vert _{L^2(\Omega )}\cdot \Vert r_b^{\frac{\beta -\alpha }{2}}Sr_b^{\frac{\alpha -\beta }{2}} {I_{a+}^{\frac{\alpha +\beta }{2}}}f\Vert _{L^2(\Omega )}. \end{aligned} \end{aligned}$$

Again, since $r_b^{\frac{\beta -\alpha }{2}}Sr_b^{\frac{\alpha -\beta }{2}}$ is bounded in $L^2(\Omega )$, this last inequality yields

$$\begin{aligned} M\le C\Vert {I_{a+}^{\frac{\alpha +\beta }{2}}}f\Vert _{L^2(\Omega )}, \end{aligned}$$

which in turn yields

$$\begin{aligned} \Vert M_2\Vert _{L^2(\Omega )}= |c|\cdot \Vert {D_{b-}^{\frac{\alpha -\beta }{2}}}{D_{a+}^{1-\alpha }}1\Vert _{L^2(\Omega )}\le C \Vert {I_{a+}^{\frac{\alpha +\beta }{2}}}f\Vert _{L^2(\Omega )}. \end{aligned}$$

(3.25)

Therefore, collecting (3.23) and (3.25) and applying (2.7) gives

$$\begin{aligned} \Vert u\Vert _{\widehat{H}_0^{(\alpha +\beta )/2}(\Omega )}\le C \,\Vert {D_{b-}^{\frac{\alpha +\beta }{2}}}u\Vert _{L^2(\Omega )}\le C_{\alpha , \beta }\Vert {I_{a+}^{\frac{\alpha +\beta }{2}}}f\Vert _{L^2(\Omega )} \end{aligned}$$

for a certain suitable constant depending on $\alpha ,\beta$ only. This finishes the whole proof.

$\square$

Remark 1

The regularity result of the fractional diffusion equation is in sharp contrast to the classical diffusion equation. For classic integer-order elliptic problem${\left\{ \begin{array}{ll} -u''=f ,x\in \Omega \\ u(a)=u(b)=0 \end{array}\right. }$, it is well-known that u will automatically become smooth whenever f is smooth. For example, if $f\equiv 1$, it is guaranteed that $u\in W^{n, p}(\Omega )$ for any positive integer n and $p>1$. However, for the fractional case in (3.1), the corresponding solution u does not even belong to $W^{1,p}(\Omega )$ for any $p\ge \max \{1/(1-\beta ),1/(2-\alpha -\beta )\}$ when $f\equiv 1$; this is mainly because it lacks smoothness at both endpoints $x=a$ and $x=b$. As we will see later in Theorem 1, when we take the derivative of solution u, it usually blows up at both boundary points.

4 Maximum principles and Hopf’s Lemma

4.1 The special case when $f(x)=\chi _E$.

This section will develop the elliptic properties for BVP (1.2). We first examine the special case

$$\begin{aligned} {\left\{ \begin{array}{ll} {D_{a+}^{\alpha }}{D_{b-}^{\beta }}u=\chi _E \\ u(a)=u(b)=0, \end{array}\right. } \end{aligned}$$

(4.1)

where $\chi _E$ denotes the step function

$$\begin{aligned} \chi _E= {\left\{ \begin{array}{ll} 0, &{} x\in (a,a_1], \\ 1, &{} x\in (a_1,b_1), \\ 0, &{} x\in [b_1,b), \end{array}\right. } \quad E=(a_1,b_1)\subseteq (a,b). \end{aligned}$$

(4.2)

We have the following auxiliary lemmas.

Lemma 2

The solution of (4.1) can be represented as

$$\begin{aligned} u(x)=T(x)+\frac{c\, (b-a)^{\alpha -1}}{\Gamma (\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) ,\, x\in \Omega ,\nonumber \\ \end{aligned}$$

(4.3)

where

$$\begin{aligned} \begin{aligned}&T(x):= \\&{\left\{ \begin{array}{ll} \frac{1}{\Gamma (\beta )\Gamma (1+\alpha )}\left( \int _{a_1}^{b}(t-x)^{\beta -1}(t-a_1)^{\alpha }\,dt\right. \\ \quad - \left. \int _{b_1}^{b}(t-x)^{\beta -1}(t-b_1)^{\alpha }\,dt\right) , &{} x\in (a,a_1) \\ \frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-a_1}\right) \\ \quad -\frac{1}{\Gamma (\beta )\Gamma (1+\alpha )}\,\int _{b_1}^{b}(t-x)^{\beta -1}(t-b_1)^{\alpha }\,dt, &{} x\in [a_1,b_1) \\ \frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-a_1}\right) \\ \quad -\frac{(b-b_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-b_1}\right) , &{} x\in [b_1,b) \end{array}\right. } \end{aligned} \end{aligned}$$

(4.4)

and

$$\begin{aligned} \begin{aligned} c&=-\frac{\alpha +\beta -1}{\alpha \,(b-a)^{\alpha +\beta -1}}\left( \int _{a_1}^{b}(t-a)^{\beta -1}(t-a_1)^{\alpha }dt-\int _{b_1}^{b}(t-a)^{\beta -1}(t-b_1)^{\alpha }dt \right) \\&<0. \end{aligned}\nonumber \\ \end{aligned}$$

(4.5)

Proof

This is a direct consequence of (3.1). However, for the sake of rigorousness, careful justification is provided simply because the calculations can be tricky when dealing with hypergeometric functions. Without loss of generality, we assume in the following that $(a_1,b_1)$ is a proper subset of (a, b), namely, $a<a_1<b_1<b$.

1. From (3.1), we know that the solution u is

$$\begin{aligned} u(x)={I_{b-}^{\beta }}\,({I_{a+}^{\alpha }}\,\chi _E+c\,{D_{a+}^{1-\alpha }}\,1),\quad x\in \Omega =(a, b). \end{aligned}$$

(4.6)

First, compute

$$\begin{aligned} {I_{a+}^{\alpha }}\,\chi _E= {\left\{ \begin{array}{ll} 0 &{} ,\quad x\in (a,a_1) \\ \frac{1}{\Gamma (1+\alpha )}(x-a_1)^{\alpha } &{} ,\quad x\in [a_1,b_1) \\ \frac{1}{\Gamma (1+\alpha )}\left( (x-a_1)^{\alpha }-(x-b_1)^{\alpha } \right) &{} ,\quad x\in [b_1,b). \end{array}\right. } \end{aligned}$$

Then apply (2.19) to compute

$$\begin{aligned} {I_{b-}^{\beta }}(c\,{D_{a+}^{1-\alpha }}\,1)=\frac{c\, (b-a)^{\alpha -1}}{\Gamma (\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) , x\in \Omega . \end{aligned}$$

2. For convenience, let us represent ${I_{a+}^{\alpha }}\,\chi _E$ as two pieces, i.e., ${I_{a+}^{\alpha }}\,\chi _E=y_1+y_2$, where

$$\begin{aligned}{} & {} y_1(x)= {\left\{ \begin{array}{ll} 0 &{} ,\quad x\in (a,a_1) \\ \frac{1}{\Gamma (1+\alpha )}(x-a_1)^{\alpha } &{} ,\quad x\in [a_1,b_1) \\ \frac{1}{\Gamma (1+\alpha )}(x-a_1)^{\alpha } &{} ,\quad x\in [b_1,b), \end{array}\right. } \\{} & {} y_2(x)= {\left\{ \begin{array}{ll} 0 &{} ,\quad x\in (a,a_1) \\ 0 &{} ,\quad x\in [a_1,b_1) \\ -\frac{1}{\Gamma (1+\alpha )}(x-b_1)^\alpha &{} ,\quad x\in [b_1,b). \end{array}\right. } \end{aligned}$$

Then continuing calculations with the aid of Property 8, we have

$$\begin{aligned} \begin{aligned}&{I_{b-}^{\beta }}\,{I_{a+}^{\alpha }}\,\chi _E\\&= {I_{b-}^{\beta }}y_{1}+{I_{b-}^{\beta }}y_2 \\&= {\left\{ \begin{array}{ll} \frac{1}{\Gamma (\beta )\Gamma (1+\alpha )}\,\int _{a_1}^{b}(t-x)^{\beta -1}(t-a_1)^{\alpha }\,dt &{} , x\in (a,a_1) \\ \frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-a_1}\right) &{} , x\in [a_1,b) \end{array}\right. } \\&\quad + {\left\{ \begin{array}{ll} -\frac{1}{\Gamma (\beta )\Gamma (1+\alpha )}\,\int _{b_1}^{b}(t-x)^{\beta -1}(t-b_1)^{\alpha }\,dt &{} , x\in (a,b_1) \\ -\frac{(b-b_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-b_1}\right) &{} , x\in [b_1,b) \end{array}\right. } \\&= {\left\{ \begin{array}{ll} \frac{1}{\Gamma (\beta )\Gamma (1+\alpha )}\left( \int _{a_1}^{b}(t-x)^{\beta -1}(t-a_1)^{\alpha }\,dt\right. \\ \quad -\left. \int _{b_1}^{b}(t-x)^{\beta -1}(t-b_1)^{\alpha }\,dt\right) &{}, x\in (a,a_1) \\ \frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-a_1}\right) \\ \quad -\frac{1}{\Gamma (\beta )\Gamma (1+\alpha )}\,\int _{b_1}^{b}(t-x)^{\beta -1}(t-b_1)^{\alpha }\,dt&{}, x\in [a_1,b_1) \\ \frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-a_1}\right) \\ \quad -\frac{(b-b_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-b_1}\right) &{}, x\in [b_1,b) \end{array}\right. } \\&:=T(x),\quad \quad x\in (a,b). \end{aligned} \end{aligned}$$

Substituting the above expressions into (4.6) gives

$$\begin{aligned} u(x)=T(x)+\frac{c\, (b-a)^{\alpha -1}}{\Gamma (\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) ,\quad x\in \Omega . \end{aligned}$$

3. Now, we need to compute the constant c by using the boundary condition $u(a)=0$. That is

$$\begin{aligned} 0=T(a)+\frac{c\, (b-a)^{\alpha +\beta -1}}{\Gamma (\alpha )\Gamma (1+\beta )}\,{}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;1\right) , \end{aligned}$$

where T(a) is understood in the limiting sense. Solving the last equation for c, evaluating T(a) with expression (4.4), and simplifying ${}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;1\right)$ with Equation (2.16) in Property 7, we obtain

$$\begin{aligned} \begin{aligned} c&=-\frac{T(a)\cdot \Gamma (\alpha )\Gamma (1+\beta )}{(b-a)^{\alpha +\beta -1} {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;1\right) }\\&=-\frac{\alpha +\beta -1}{\alpha \,(b-a)^{\alpha +\beta -1}}\left( \int _{a_1}^{b}(t-a)^{\beta -1}(t-a_1)^{\alpha }dt\,-\,\int _{b_1}^{b}(t-a)^{\beta -1}(t-b_1)^{\alpha }dt \right) . \end{aligned} \end{aligned}$$

It is clear that $c<0$ from our assumptions $\alpha +\beta >1$ and $a_1<b_1$. This completes the proof. $\square$

Lemma 3

Let u be the same solution as in Lemma 2, then it is true that

$$\begin{aligned} \lim _{x\rightarrow a^+}u'(x)=+\infty ,\quad \lim _{x\rightarrow b^-}u'(x)=-\infty . \end{aligned}$$

Proof

1. Let $x\in (a,a_1)$. From (4.3) and (2.17), we readily check

$$\begin{aligned} \begin{aligned} u'(x)&= T'(x)+\frac{c\,(b-a)^{\alpha -1}}{\Gamma (\alpha )\Gamma (1+\beta )}\frac{d}{dx}\left( (b-x)^{\beta }\cdot {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \right) \\&= \frac{1-\beta }{\Gamma (\beta )\Gamma (1+\alpha )}\left( \int _{a_1}^{b}(t-x)^{\beta -2}(t-a_1)^{\alpha }dt-\int _{b_1}^{b}(t-x)^{\beta -2}(t-b_1)^{\alpha }dt \right) \\&\quad +\Bigg \{ \frac{-c\beta \,(b-a)^{\alpha -1}}{\Gamma (\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta -1}\cdot {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \\&\quad + \frac{-c(1-\alpha )(b-a)^{\alpha -2}}{(1+\beta )\Gamma (\alpha )\Gamma (1+\beta )}\cdot (b-x)^{\beta }\cdot {}_{2}F_{1}\left( 2-\alpha ,2;2+\beta ;\frac{b-x}{b-a}\right) \Bigg \} \\&:= T_1+T_2+T_3. \end{aligned} \end{aligned}$$

We see that

$$\begin{aligned} \begin{aligned}&\lim _{x\rightarrow a^+}T_1(x)\\&=\frac{1-\beta }{\Gamma (\beta )\Gamma (1+\alpha )}\left( \int _{a_1}^{b}(t-a)^{\beta -2}(t-a_1)^{\alpha }dt-\int _{b_1}^{b}(t-a)^{\beta -2}(t-b_1)^{\alpha }dt \right) >0. \end{aligned} \end{aligned}$$

Notice that $c<0$ and $\alpha +\beta >1$, then

$$\begin{aligned} \begin{aligned}&\lim _{x\rightarrow a^+}T_2(x)\\&=\frac{-c\beta \,(b-a)^{\alpha +\beta -2}}{\Gamma (\alpha )\Gamma (1+\beta )}\cdot {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;1\right) \\&=\frac{-c\beta \,(b-a)^{\alpha +\beta -2}}{(\alpha +\beta -1)\Gamma (\alpha )\Gamma (\beta )}>0. \end{aligned} \end{aligned}$$

Lastly, applying (2.15) in Property 7 and condition $\alpha +\beta <2$ to $T_3$ gives

$$\begin{aligned} \lim _{x\rightarrow a^+}T_3(x)=\lim _{x\rightarrow a^+}\frac{-c(1-\alpha )\Gamma (2-\alpha -\beta )}{\Gamma (\alpha )\Gamma (2-\alpha )}\cdot (x-a)^{\alpha +\beta -2}=+\infty . \end{aligned}$$

Hence,

$$\begin{aligned} \lim _{x\rightarrow a^+}u'(x)=+\infty . \end{aligned}$$

2. Let $x\in (b_1,b)$. From (4.3) and by the product rule, we similarly check

$$\begin{aligned} \begin{aligned}&u'(x)\\&\quad = -\beta (b-x)^{\beta -1}\Biggl ( \frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-a_1}\right) \\&\qquad -\frac{(b-b_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot {}_{2}F_{1}\left( -\alpha ,1;1+\beta ;\frac{b-x}{b-b_1}\right) \\&\qquad +\frac{c\,(b-a)^{\alpha -1}}{\Gamma (\alpha )\Gamma (1+\beta )}\cdot {}_{2}F_{1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \Biggr ) \\&\qquad + (b-x)^{\beta }\Biggl ( \frac{\alpha (b-a_1)^\alpha }{(1+\beta )(b-a_1)\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot {}_{2}F_{1}\left( 1-\alpha ,2;2+\beta ;\frac{b-x}{b-a_1}\right) \\&\qquad -\frac{\alpha (b-b_1)^\alpha }{(1+\beta )(b-b_1)\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot {}_{2}F_{1}\left( 1-\alpha ,2;2+\beta ;\frac{b-x}{b-b_1}\right) \\&\qquad +\frac{-c(1-\alpha )(b-a)^{\alpha -2}}{(1+\beta )\Gamma (\alpha )\Gamma (1+\beta )}\cdot {}_{2}F_{1}\left( 2-\alpha ,2;2+\beta ;\frac{b-x}{b-a}\right) \Biggr ) \\&\quad := -\beta (b-x)^{\beta -1}\cdot T_1(x)+(b-x)^{\beta }\cdot T_2(x). \end{aligned}\nonumber \\ \end{aligned}$$

(4.7)

It is easy to see $T_2(b)$ is finite. We claim that

$$\begin{aligned} T_1(b)\quad \text {is also finite and}\quad T_1(b)\ne 0. \end{aligned}$$

To verify this, apply (2.18) in Property 7 to carry out

$$\begin{aligned} T_1(b)=\frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}-\frac{(b-b_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}+\frac{c\,(b-a)^{\alpha -1}}{\Gamma (\alpha )\Gamma (1+\beta )}. \end{aligned}$$

Substituting for c by using the expression (4.5), this becomes

$$\begin{aligned} \begin{aligned}&T_1(b)\\&\quad =\frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}-\frac{(b-b_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}- \frac{\alpha +\beta -1}{(b-a)^{\beta }\Gamma (1+\beta )} \times \\&\quad \left( \frac{1}{\Gamma (1+\alpha )}\int _{a_1}^{b}(t-a)^{\beta -1}(t-a_1)^{\alpha }dt-\frac{1}{\Gamma (1+\alpha )}\int _{b_1}^{b}(t-a)^{\beta -1}(t-b_1)^{\alpha }dt \right) \\&\quad =\frac{(b-a_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}-\frac{(b-b_1)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}- \frac{\alpha +\beta -1}{(b-a)^{\beta }\Gamma (1+\beta )} \times \\&\quad \left( {I_{b-}^{\alpha +1}}\,(t-a)^{\beta -1}\big |_{t=a_1}-{I_{b-}^{\alpha +1}}\,(t-a)^{\beta -1}\big |_{t=b_1} \right) \\&\quad :=F(a_1)-F(b_1), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} F(x):=\frac{(b-x)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}-\frac{\alpha +\beta -1}{(b-a)^{\beta }\Gamma (1+\beta )}\cdot {I_{b-}^{\alpha +1}}\,(t-a)^{\beta -1}\big |_{t=x},\quad x\in \Omega . \end{aligned}$$

3. Therefore, it can be seen that $T_1(b)$ is finite since F(x) takes finite values in $\overline{\Omega }$. In order to further show that $T_1(b)\ne 0$, let us find the sign of $F(a_1)-F(b_1)$ by examining the derivative $\displaystyle \frac{dF}{dx}$.

$$\begin{aligned} \begin{aligned} F'(x)&=-\frac{\alpha \, (b-x)^{\alpha -1}}{\Gamma (1+\alpha )\Gamma (1+\beta )}+\frac{\alpha +\beta -1}{(b-a)^{\beta }\Gamma (1+\beta )}\cdot {I_{b-}^{\alpha }}\,(t-a)^{\beta -1}\big |_{t=x}\\&{=}\frac{(b-x)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\left( \frac{\alpha +\beta -1}{b-a} \cdot {}_{2}F_{1}\left( 1-\beta ,1;1+\alpha ;\frac{b-x}{b-a}\right) {-}\frac{\alpha }{b-x}\right) \\&=\frac{(b-x)^\alpha }{\Gamma (1+\alpha )\Gamma (1+\beta )}\cdot G(x), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} G(x):=\frac{\alpha +\beta -1}{b-a} \cdot {}_{2}F_{1}\left( 1-\beta ,1;1+\alpha ;\frac{b-x}{b-a}\right) -\frac{\alpha }{b-x}. \end{aligned}$$

Utilizing Property 7, it can be directly checked that for $x\in \Omega$

$$\begin{aligned} G'(x)=-\frac{(\alpha +\beta -1)(1-\beta )}{(b-a)^2(1+\alpha )}\cdot {}_{2}F_{1}\left( 2-\beta ,2;2+\alpha ;\frac{b-x}{b-a}\right) -\frac{\alpha }{(b-x)^2}<0\nonumber \\ \end{aligned}$$

(4.8)

and

$$\begin{aligned} G(a)=\frac{\alpha +\beta -1}{b-a} \cdot {}_{2}F_{1}\left( 1-\beta ,1;1+\alpha ;1\right) -\frac{\alpha }{b-a}=\frac{\alpha }{b-a}-\frac{\alpha }{b-a}=0.\nonumber \\ \end{aligned}$$

(4.9)

Hence, (4.8) and (4.9) imply that $F'(x)<0, x\in \Omega$, which in turn yields $F(a_1)-F(b_1)>0$. So, $T_1(b)>0$, and our claim is verified.

4. Going back to the last line in (4.7), we finally see that

$$\begin{aligned} \lim _{x\rightarrow b^-}u'(x){} & {} =-\lim _{x\rightarrow b^-} \beta (b-x)^{\beta -1}\cdot T_1(x)+\lim _{x\rightarrow b^-}(b-x)^{\beta }\cdot T_2(x)\\ {}{} & {} = -\infty +0=-\infty . \end{aligned}$$

This completes the proof. $\square$

Remark 2

As we can see, the derivatives of the solution u near the boundary points all blow up, which is an interesting difference from the classic elliptic theory. The following Figure 2, Figure 3, Figure 4 and Figure 5 summarize some differences between the solutions of the classic (local), one-sided (non-local), and double-sided (completely non-local) diffusion operators at the boundary points.

Fig. 2

For simplicity of illustration, we take $f=1$ and $(a,b)=(0,1)$. The red dashed lines in each figure represent the tangent lines of solutions u at the boundary points $x=0$ and $x=1$. We can see that, for the classic diffusion operator in (2a), $u'(0)$ and $u'(1)$ are finite. Meanwhile, for one-sided non-local operators in (2b), (2c) and(2d), $u'(0)$ always blows up and $u'(1)$ stays finite. In particular, as $\mu$ increases, the maximum point of u is shifting from left to right and u always stays concave down

Fig. 3

${\left\{ \begin{array}{ll} D_{a+}^\alpha D_{b-}^\beta u=1\\ u(0)=u(1)=0 \end{array}\right. }$ with $\alpha +\beta =1.2$. For two-sided completely non-local operators, we can see that both $u'(0)$ and $u'(1)$ blow up for general $\alpha , \beta$. We can also see how the weights $\alpha , \beta$ affect the shape of the graph: if $\alpha$ and $\beta$ are very close, such as (3c) and (3d), it is easy to see the vertical tangent lines (in red) at the boundary, however, if $\alpha$ and $\beta$ are not sufficiently close, such as (3a) and (3e), the behavior of $u'$ at one of the boundary points is hidden and can only be observed when zooming in by taking the $\log$ scale

Fig. 4

${\left\{ \begin{array}{ll} D_{a+}^\alpha D_{b-}^\beta u=1\\ u(0)=u(1)=0 \end{array}\right. }$ with $\alpha +\beta =1.4$. This figure together with Figure 3 above and Figure 5 below shows that as the sum $\alpha +\beta$ increase, the graph of u becomes more and more regular as the classic case in (2a). Finally, let us observe that, similar to the one-sided operator, when the total derivative is between 1 and 2, i.e. $1<\alpha +\beta <2$, the graph is always concave down as well

Fig. 5

${\left\{ \begin{array}{ll} D_{a+}^\alpha D_{b-}^\beta u=1\\ u(0)=u(1)=0 \end{array}\right. }$ with $\alpha +\beta =1.6$ and $\alpha +\beta =1.8$

Lemma 4

Let u be the same solution as in Lemma 2. Then ${D_{b-}^{\beta }}u$ can be expressed as

$$\begin{aligned} ({D_{b-}^{\beta }}u)(x)= {\left\{ \begin{array}{ll} f_1(x), &{} x\in (a,\xi ) \\ f_2(x), &{} x\in [\xi ,b], \end{array}\right. } \end{aligned}$$

(4.10)

where $\xi$ is a certain real number satisfying $a<\xi <b_1$, $f_1(x)$ is negative and $f_1'(x)>0$ on $(a,\xi )$, and $f_2(x)$ is positive on $(\xi ,b]$.

Proof

1. Use expression (4.6) and compute

$$\begin{aligned} \begin{aligned} {D_{b-}^{\beta }}u&= {I_{a+}^{\alpha }}\chi _E+c\,{D_{a+}^{1-\alpha }}\,1 \\&= {\left\{ \begin{array}{ll} \frac{c}{\Gamma (\alpha )}(x-a)^{-(1-\alpha )} &{} , x\in (a,a_1) \\ \frac{1}{\Gamma (1+\alpha )}(x-a_1)^\alpha +\frac{c}{\Gamma (\alpha )}(x-a)^{-(1-\alpha )} &{} , x\in [a_1,b_1) \\ \frac{1}{\Gamma (1+\alpha )}\left( (x-a_1)^\alpha -(x-b_1)^\alpha \right) +\frac{c}{\Gamma (\alpha )}(x-a)^{-(1-\alpha )} &{} , x\in [b_1,b] \end{array}\right. } \\&:= {\left\{ \begin{array}{ll} T_1(x), &{} x\in (a,a_1) \\ T_2(x), &{} x\in [a_1,b_1) \\ T_3(x), &{} x\in [b_1,b]. \end{array}\right. } \end{aligned} \end{aligned}$$

Then, continue to compute the first derivatives of $T_1$, $T_2$ to obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} T_1'(x)=\frac{-c\,(1-\alpha )}{\Gamma (\alpha )} (x-a)^{-(2-\alpha )}&{}>0 \\ T_2'(x)=\frac{1}{\Gamma (\alpha )}(x-a_1)^{\alpha -1}-\frac{c\,(1-\alpha )}{\Gamma (\alpha )} (x-a)^{-(2-\alpha )}&{}>0. \end{array}\right. } \end{aligned}$$

(4.11)

This means that ${D_{b-}^{\beta }}u$ increases on $(a,b_1]$ from $-\infty$ to ${D_{b-}^{\beta }}u\big |_{x=b_1}$. Let us note that ${D_{b-}^{\beta }}u$ is continuous on (a, b]. Therefore, if we can show that $T_3(x)>0$ on $(b_1,b]$, then Lemma 4 is verified, and the existence of such a point $\xi$ and such functions $f_1$ and $f_2$ is guaranteed automatically.

2. To see that $T_3(x)>0$, we first write out $T_3$ by substituting the constant c in (4.5) into its expression.

$$\begin{aligned} \begin{aligned}&T_3(x)\\&\quad = \frac{1}{\Gamma (1+\alpha )}\left( (x-a_1)^\alpha -(x-b_1)^\alpha \right) -\frac{(\alpha +\beta -1)(x-a)^{-(1-\alpha )}}{(b-a)^{\alpha +\beta -1}} \times \\&\quad \left( \frac{1}{\Gamma (1+\alpha )}\int _{a_1}^{b}(t-a)^{\beta -1}(t-a_1)^{\alpha }dt-\frac{1}{\Gamma (1+\alpha )}\int _{b_1}^{b}(t-a)^{\beta -1}(t-b_1)^{\alpha }dt \right) \\&\quad = \frac{1}{\Gamma (1+\alpha )}\left( (x-a_1)^\alpha -(x-b_1)^\alpha \right) -\frac{(\alpha +\beta -1)(x-a)^{-(1-\alpha )}}{(b-a)^{\alpha +\beta -1}} \times \\&\quad \left( {I_{b-}^{\alpha +1}}\,(t-a)^{\beta -1}\big |_{t=a_1}-{I_{b-}^{\alpha +1}}\,(t-a)^{\beta -1}\big |_{t=b_1} \right) \\&\quad = F(x,a_1)-F(x, b_1), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&F(t_1,t_2)\\&:=\frac{1}{\Gamma (1+\alpha )}(t_1-t_2)^\alpha -\frac{(\alpha +\beta -1)(t_1-a)^{-(1-\alpha )}}{(b-a)^{\alpha +\beta -1}}\cdot {I_{b-}^{\alpha +1}}(t-a)^{\beta -1}\big |_{t=t_2}, \\&\quad (t_1,t_2)\in [b_1,b]\times [a_1,b_1]. \end{aligned} \end{aligned}$$

Then, let us note that, to obtain $T_3(x)>0$ for $x\in (b_1,b]$, it suffices to show $\displaystyle \frac{\partial {F}}{\partial {t_2}}<0$ for any $(t_1,t_2)\in (b_1,b]\times [a,b_1] \supset (b_1,b]\times [a_1,b_1]$.

3. To obtain $\displaystyle \frac{\partial {F}}{\partial {t_2}}<0$, we examine

$$\begin{aligned} \begin{aligned} \frac{\partial {F}}{\partial {t_2}}&= -\frac{1}{\Gamma (\alpha )}(t_1-t_2)^{\alpha -1}+\frac{(\alpha +\beta -1)(t_1-a)^{-(1-\alpha )}}{(b-a)^{\alpha +\beta -1}}\cdot {I_{b-}^{\alpha }}(t-a)^{\beta -1}\big |_{t=t_2} \\&= -\frac{1}{\Gamma (\alpha )}(t_1-t_2)^{\alpha -1}+ \nonumber \\&\frac{\alpha +\beta -1}{(b-a)^\alpha \Gamma (1+\alpha )}(b-t_2)^\alpha \cdot {}_{2}F_{1}\left( 1-\beta ,1;1+\alpha ;\frac{b-t_2}{b-a}\right) \cdot (t_1-a)^{-(1-\alpha )}. \end{aligned}\nonumber \\ \end{aligned}$$

(4.12)

Based on the expression (4.12), it is not immediately clear that $\displaystyle \frac{\partial {F}}{\partial {t_2}}<0$.

So, we continue to calculate the following auxiliary expressions. First,

$$\begin{aligned}{} & {} \frac{\partial ^2{F}}{\partial {t_2^2}}= \frac{\alpha -1}{\Gamma (\alpha )}(t_1-t_2)^{\alpha -2}\nonumber \\{} & {} \qquad -\frac{\alpha +\beta -1}{(b-a)^\alpha \Gamma (\alpha )}(b-t_2)^{\alpha -1}\cdot {}_{2}F_{1}\left( 1-\beta ,1;1+\alpha ;\frac{b-t_2}{b-a}\right) \cdot (t_1-a)^{-(1-\alpha )} \nonumber \\{} & {} \quad -\frac{(\alpha +\beta -1)(1-\beta )}{(b-a)^{\alpha +1}\Gamma (2+\alpha )}(b-t_2)^\alpha \cdot {}_{2}F_{1}\left( 2-\beta ,2;2+\alpha ;\frac{b-t_2}{b-a}\right) \cdot (t_1-a)^{-(1-\alpha )}\nonumber \\{} & {} <0, \quad \forall (t_1,t_2)\in (b_1,b]\times [a, b_1].\nonumber \\ \end{aligned}$$

(4.13)

Second, by evaluating (4.12) at $t_2=a$, we find

$$\begin{aligned}{} & {} \frac{\partial {F}}{\partial {t_2}}(t_1,a)\nonumber \\{} & {} \quad = -\frac{1}{\Gamma (\alpha )}(t_1-a)^{\alpha -1}+\frac{\alpha +\beta -1}{\Gamma (1+\alpha )}\cdot {}_{2}F_{1}\left( 1-\beta ,1;1+\alpha ;1\right) \cdot (t_1-a)^{-(1-\alpha )} \nonumber \\{} & {} \quad = -\frac{1}{\Gamma (\alpha )}(t_1-a)^{\alpha -1}+\frac{1}{\Gamma (\alpha )}\cdot (t_1-a)^{\alpha -1} \nonumber \\{} & {} \quad = 0.\nonumber \\ \end{aligned}$$

(4.14)

Combining above (4.13) and (4.14), we deduce that

$$\begin{aligned} \frac{\partial {F}}{\partial {t_2}}<0\quad \text {for any}\quad (t_1,t_2)\in (b_1,b]\times [a_1,b_1]. \end{aligned}$$

Therefore, $F(x,a_1)-F(x,b_1)>0$ for $x\in (b_1,b]$, i.e., $T_3(x)>0,x\in (b_1,b]$. This confirms Lemma 4. $\square$

Using Lemma 4, we obtain the following, which is the preliminary version of the Maximum principle.

Lemma 5

In $\Omega =(a, b)$, $u(x)>0$, and $\underset{\overline{\Omega }}{\min u}=\underset{\partial {\Omega }}{\min u}$.

Proof

1. From (4.10) in Lemma 4, the solution u can be written as

$$\begin{aligned} u(x)= {I_{b-}^{\beta }}{D_{b-}^{\beta }}u={I_{b-}^{\beta }}\tilde{f_1}+{I_{b-}^{\beta }}\tilde{f_2}:=T_1(x)+T_2(x), x\in (a,b), \end{aligned}$$

where $\tilde{f_1}, \tilde{f_2}$ denote the trivial extensions of $f_1, f_2$ by zero outside $(a, \xi )$ and $[\xi , b]$, respectively.

2. Let us examine the behavior of u as x goes from b to a (from right to left). On one hand, by the definition of R-L integral,

$$\begin{aligned} T_2(x)=\frac{1}{\Gamma (\beta )}\int _{x}^{b}(t-x)^{\beta -1}\tilde{f_2}(t)\,dt>0, \, \,x\in [\xi , b), \end{aligned}$$

(4.15)

and it is not difficult to see that

$$\begin{aligned} T_2(x)\text { decreases as }x\rightarrow a^+\text { in }(a, \xi ). \end{aligned}$$

(4.16)

On the other hand,

$$\begin{aligned} T_1(x)=\frac{1}{\Gamma (\beta )}\int _{x}^{b}(t-x)^{\beta -1}\tilde{f_1}(t)\,dt<0, \, \, x\in (a,\xi ), \end{aligned}$$

(4.17)

and

$$\begin{aligned} T_1(x)\text { also decreases as }x\rightarrow a^+\text { on }(a, \xi ), \end{aligned}$$

(4.18)

because

$$\begin{aligned} \begin{aligned} T_1'(x)&=\left( \frac{1}{\Gamma (\beta )}\int _{x}^{b}(t-x)^{\beta -1}\tilde{f_1}(t)\,dt\right) ' =\left( \frac{1}{\Gamma (\beta )}\int _{x}^{\xi }(t-x)^{\beta -1}f_1(t)\,dt\right) '\\&=\frac{1}{\Gamma (\beta )}\int _{x}^{\xi }(t-x)^{\beta -1}f'_1(t)\,dt\\&>0,\quad \forall x\in (a,\xi ) \end{aligned} \end{aligned}$$

by noting from (4.11) that $f'_1\in L^1((a+\epsilon , \xi ))$ for any small $0<\epsilon <\xi -a$. (4.15), (4.16), (4.17) and (4.18) imply that

$$\begin{aligned} u(x)>0, x\in [\xi , b), \text { and }u(x)\hbox { decreases as }x\rightarrow a^+\hbox { on } (a, \xi ). \end{aligned}$$

(4.19)

3. Now, suppose there exists a $t_0\in (a, b)$ such that $u(t_0)\le 0$. From (4.19), we know $t_0<\xi$ and therefore u(x) strictly decreases from $t_0$ to the left boundary a, which contradicts the fact that $u(a)=0$. Therefore, $u(x)>0, x\in \Omega$, and $\underset{\overline{\Omega }}{\min u}=\underset{\partial {\Omega }}{\min u}$. Lemma 5 is proved. $\square$

4.2 The general case when $f\in L^1(\Omega )$.

It is well-known that the step functions defined in (4.2) for arbitrary open interval $(a_1,b_1)\subset (a,b)=\Omega$ are dense in $L^p(\Omega )$, $p\ge 1$. Therefore, we have the following lemmas for general real-valued functions $f\in L^p(\Omega )$, which are analogies of the Maximum principle and Hopf’s Lemma.

Lemma 6

Suppose $f\in L^1(\Omega )$ is a real-valued, non-zero function, and $f(x)\ge 0$ a.e. in $\Omega$. Let u be a solution of problem (1.2). Then

• 1.
  $$\begin{aligned} \lim _{x\rightarrow a^+}\frac{u(x)}{x-a}=+\infty ,\quad \lim _{x\rightarrow b^-}\frac{u(x)}{x-b}=-\infty . \end{aligned}$$

  (4.20)

• 2. In $\Omega =(a,b)$, $u(x)> 0$, and $\underset{\overline{\Omega }}{\min u}=\underset{\partial {\Omega }}{\min u}$.

Proof

1. Since $f\in L^1(\Omega )$ is non-zero and $f(x)\ge 0$ a.e. in $\Omega$, it is well-known from real analysis that there must exist a sequence of $\{f_n\}_{n=1}^\infty$ such that

$$\begin{aligned} f_n=\sum _{i=1}^{n}c_i\psi _i, \text { each }\psi _i\hbox { is a step function as defined in (4.2)}, \,c_1>0, c_i\ge 0, \end{aligned}$$

and

$$\begin{aligned} f_n(x)\le f(x)\, \text {a.e.}, \quad f_n\rightarrow f\quad \text {in}\quad L^1(\Omega ). \end{aligned}$$

2. By Lemma 1, we have a sequence of corresponding solutions $\{\phi _i\}$ for $\{c_i\psi _i\}$, satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} {D_{a+}^{\alpha }}{D_{b-}^{\beta }}\phi _i=c_i\psi _i\\ \phi _i(a)=\phi _i(b)=0. \end{array}\right. } \end{aligned}$$

(4.21)

The fractional derivative ${D_{a+}^{\alpha }}{D_{b-}^{\beta }}$ is a linear operator. Summing up (4.21) gives a sequence of corresponding solutions $\{u_n\}$ for $\{f_n\}$, satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} {D_{a+}^{\alpha }}{D_{b-}^{\beta }}u_n=f_n\\ u_n(a)=u_n(b)=0. \end{array}\right. } \end{aligned}$$

Notice each $f_n$ is non-zero and the sequence $\{f_n\}$ is non-decreasing for each fixed $x\in \Omega$. According to Lemma 3 and Lemma 5, we know that

$$\begin{aligned} \lim _{x\rightarrow a^+}u'_n(x)=+\infty ,\quad \lim _{x\rightarrow b^-}u'_n(x)=-\infty , \end{aligned}$$

(4.22)

and

$$\begin{aligned} u_n(x)>0\quad x\in \Omega ,\quad \{u_n\}_{n=1}^{\infty }\quad \text {is non-decreasing for each given}\quad x\in \Omega . \end{aligned}$$

(4.23)

3. According to the norm inequality (3.3), $\Vert u_n-u\Vert _{L^1(\Omega )}\le \tilde{C}_{\alpha ,\beta } \Vert f_n-f\Vert _{L^1(\Omega )}$, we know that

$$\begin{aligned} u_n\rightarrow u \quad \text {in}\quad L^1(\Omega )\quad \text {as}\quad f_n\rightarrow f \quad \text {in}\quad L^1(\Omega ). \end{aligned}$$

Therefore, there must exists a subsequence $\{u_{n_i}\}\subset \{u_n\}$ and a set $E\subset \Omega$ with zero measure such that

$$\begin{aligned} \lim _{i\rightarrow +\infty }u_{n_i}(x)=u(x) \quad \text {for every}\quad x\in \Omega \backslash E. \end{aligned}$$

(4.24)

4. The last key step is to notice that u is continuous in $\Omega$. We derive that

$$\begin{aligned} \underset{x\in \Omega }{\lim _{x\rightarrow a^+}}\frac{u(x)}{x-a}=\underset{x\in \Omega \backslash E}{\lim _{ x\rightarrow a^+}}\frac{u(x)}{x-a} \ge \underset{x\in \Omega \backslash E}{\lim _{ x\rightarrow a^+}}\frac{u_{n_1}(x)}{x-a}=\underset{x\in \Omega \backslash E}{\lim _{ x\rightarrow a^+}}u'_{n_1}(x)=+\infty . \end{aligned}$$

(4.25)

Similarly,

$$\begin{aligned} \underset{x\in \Omega }{\lim _{x\rightarrow b^-}}\frac{u(x)}{x-b}=\underset{x\in \Omega \backslash E}{\lim _{ x\rightarrow b^-}}\frac{u(x)}{x-b} \le \underset{x\in \Omega \backslash E}{\lim _{ x\rightarrow b^-}}\frac{u_{n_1}(x)}{x-b} = \underset{x\in \Omega \backslash E}{\lim _{ x\rightarrow b^-}}u'_{n_1}(x)=-\infty . \end{aligned}$$

(4.26)

(4.25) and (4.26) give the first part of Lemma 6.

5. It remains to prove the second part of Lemma 6. (4.23) and (4.24) imply that

$$\begin{aligned} u(x)>0,\quad x\in \Omega \backslash E. \end{aligned}$$

Then the continuity of u on $\Omega$ and the boundary condition $u(a)=u(b)=0$ further yield that

$$\begin{aligned} u(x)>0,\quad x\in \Omega \quad \text {and}\quad \underset{\overline{\Omega }}{\min u}=\underset{\partial {\Omega }}{\min u}. \end{aligned}$$

The proof is complete. $\square$

Let us close this subsection by pointing out that if ${I_{a+}^{\alpha }}f\in C^1(\overline{\Omega })$, then u is differentiable at every point $x\in \Omega$ by using expression (3.1). This means that by imposing a stronger condition on f, ${I_{a+}^{\alpha }}f\in C^1(\overline{\Omega })$, rather than $f\in L^1(\Omega )$, we will have a more compact statement for (4.20), which is usually more useful in applications. We directly state it now without repeating the proof.

Lemma 7

Suppose ${I_{a+}^{\alpha }}f\in C^1(\overline{\Omega })$, f is a real-valued, non-zero function, and $f(x)\ge 0$ a.e. in $\Omega$. Let u be a solution of problem (1.2). Then

• 1.
  $$\begin{aligned} \lim _{x\rightarrow a^+}u'(x)=+\infty ,\quad \lim _{x\rightarrow b^-}u'(x)=-\infty . \end{aligned}$$

  (4.27)

• 2. In $\Omega =(a,b)$, $u(x)> 0$, and $\underset{\overline{\Omega }}{\min u}=\underset{\partial {\Omega }}{\min u}$.

4.3 Hopf’s lemma, maximum principle

We next substantially strengthen the foregoing assertions. We first state Hopf’s lemma, which says that the (one-sided) directional derivative of u in the direction of the outward-pointing normal to the boundary $x_0$ is always positive infinity. This differs from the classic elliptic result for integer-order derivatives, which is always positive but finite.

Theorem 1

(Hopf’s Lemma) Let $f\in L^1(\Omega )$. Assume $u\in C(\overline{\Omega })\cap C^1(\Omega )$, ${I_{a+}^{1-\alpha }}{D_{b-}^{\beta }}u$, ${I_{b-}^{1-\beta }}u\in AC(\overline{\Omega })$ and $Au=f, a.e.$ in $\Omega$. Suppose further that $f\ge 0, a.e.$ and there exists a point $x_0\in \partial \Omega$ such that $u(x_0)<u(x)$ for all $x\in \Omega$. Then

• 1. if $x_0$ is the left boundary, i.e., $x_0=a$, then it is true that

  $$\begin{aligned} \lim _{ x\rightarrow a^+}u'(x)=+\infty . \end{aligned}$$

  (4.28)

• 2. if $x_0$ is the right boundary, i.e., $x_0=b$, then it is true that

  $$\begin{aligned} \lim _{ x\rightarrow b^-}u'(x)=-\infty . \end{aligned}$$

  (4.29)

Proof

1. By Lemma 1, there exists a solution $\tilde{u}\in \widehat{H}_0^{\frac{\alpha +\beta }{2}}(\Omega )$ satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} A\tilde{u}=f \quad \text {in} \quad \Omega ,\\ \tilde{u}|_{\partial \Omega }=0. \end{array}\right. } \end{aligned}$$

(4.30)

Hence,

$$\begin{aligned} A(u-\tilde{u})=0,\, \text {a.e.} \quad \text {in} \quad \Omega . \end{aligned}$$

Since ${I_{a+}^{1-\alpha }}{D_{b-}^{\beta }}u$, ${I_{b-}^{1-\beta }}u\in AC(\overline{\Omega })$, it is valid to integrate both sides of $A(u-\tilde{u})=0$ to carry out that

$$\begin{aligned} u-\tilde{u}=c\, {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1+\tilde{c}\, {D_{b-}^{1-\beta }}1, \end{aligned}$$

where c and $\tilde{c}$ are certain real constants. Due to $u\in C(\overline{\Omega })$, $\tilde{c}$ has to be zero, and the last equality becomes

$$\begin{aligned} u=\tilde{u}+c\, {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1,\quad x\in \Omega . \end{aligned}$$

(4.31)

2. Now, let us consider the first case: $x_0=a$. First, note that if f is a zero function, then the solution $\tilde{u}$ is trivially zero everywhere. Otherwise, by Lemma 6 we know that $\tilde{u}(x)>0$ in $\Omega$. Hence, by combining these two situations, we always have

$$\begin{aligned} \tilde{u}(a)=\tilde{u}(b)=0,\quad \text {and}\quad \tilde{u}\ge 0\quad \text {in} \quad \Omega . \end{aligned}$$

(4.32)

Secondly, observe that

$$\begin{aligned} {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1\big |_{x=b}=0\quad \text {and}\quad {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1>0\quad \text {in} \quad [a,b). \end{aligned}$$

(4.33)

Since we assumed that $u(a)<u(x)$ for all $x\in \Omega$, (4.32) and (4.33) imply that c in Equation (4.31) must be non-positive, that is, $c\le 0$. Furthermore, in particular, $c<0$, provided that f is a zero function.

3. Due to the assumption $u\in C^1(\Omega )$ and the fact that ${I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1\in C^1(\Omega )$, we deduce that $\tilde{u}\in C^1(\Omega )$. Therefore,

$$\begin{aligned} \lim _{ x\rightarrow a^+}u'(x)=\lim _{ x\rightarrow a^+}\tilde{u}'(x)+c\lim _{ x\rightarrow a^+} \left( {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1\right) ':=T_1+ c\,T_2. \end{aligned}$$

(4.34)

We continue to carry out $T_1$ and $T_2$ to obtain

$$\begin{aligned} T_1= {\left\{ \begin{array}{ll} 0,\quad \text {if }f\hbox { is a zero function}\\ +\infty , \quad \text {if }f\hbox { is not a zero function} \quad \text {(by Lemma~6)} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} T_2&=\lim _{ x\rightarrow a^+} \left( \frac{(b-a)^{\alpha -1}(b-x)^\beta }{\Gamma (\alpha )\Gamma (1+\beta )}{{}_2F_1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \right) '\\&=\lim _{ x\rightarrow a^+}-\frac{(b-a)^{\alpha -1}(b-x)^{\beta -1}}{\Gamma (\alpha )\Gamma (\beta )}{{}_2F_1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \\&\quad -\lim _{ x\rightarrow a^+} \frac{(1-\alpha )(b-a)^{\alpha -2}(b-x)^\beta }{(1+\beta )\Gamma (\alpha )\Gamma (1+\beta )}{{}_2F_1}\left( 2-\alpha ,2;2+\beta ;\frac{b-x}{b-a}\right) \\&=-\frac{\beta (b-a)^{\alpha +\beta -2}}{(\alpha +\beta -1)\Gamma (\alpha )\Gamma (\beta )}-\lim _{ x\rightarrow a^+}\frac{\Gamma (2-\alpha -\beta )}{\Gamma (\alpha )\Gamma (1-\alpha )}(x-a)^{\alpha +\beta -2}\\&=-\infty \qquad \text {(due to }\alpha +\beta <2). \end{aligned} \end{aligned}$$

Doing back substitution into (4.34), we see

$$\begin{aligned} \lim _{ x\rightarrow a^+}u'(x)=+\infty , \text { no matter whether }f\hbox { is a zero function or not,} \end{aligned}$$

which is the desired result (4.28).

4. The second case: $x_0=b$ remains to be considered. Since we assume that $u(b)<u(x)$ for all $x\in \Omega$, (4.32) and (4.33) imply that c in Equation (4.31) must be non-negative, that is, $c\ge 0$. Furthermore, in particular, $c>0$, provided that f is a zero function.

Then, in a similar fashion, we can compute and find

$$\begin{aligned} \lim _{ x\rightarrow b^-}u'(x)=\lim _{ x\rightarrow b^-}\tilde{u}'(x)+c\lim _{ x\rightarrow b^-} \left( {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1\right) '=-\infty , \end{aligned}$$

(4.35)

no matter whether f is a zero function or not. This is (4.29), and it completes the proof. $\square$

We next develop the weak maximum principle, which characterizes the location of the minimum or maximum of u.

Theorem 2

(Weak maximum principle) Assume $u\in C(\overline{\Omega })\cap C^1(\Omega )$ and ${I_{a+}^{1-\alpha }}{D_{b-}^{\beta }}u,$ $\,{I_{b-}^{1-\beta }}u\in AC(\overline{\Omega })$.

• 1. If $Au \ge 0$ a.e. in $\Omega$, then

  $$\begin{aligned} \underset{\overline{\Omega }}{\min u }=\underset{\partial \Omega }{\min u }. \end{aligned}$$

  (4.36)

• 2. If $Au \le 0$ a.e. in $\Omega$, then

  $$\begin{aligned} \underset{\overline{\Omega }}{\max u }=\underset{\partial \Omega }{\max u }. \end{aligned}$$

  (4.37)

Proof

1. Denote $Au:=f$. In light of the proof in Theorem 1, we already know that u can be written as

$$\begin{aligned} u=\tilde{u}+c\, {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1,\quad x\in \Omega , \end{aligned}$$

(4.38)

where c is a certain real constant, $\tilde{u}$ belongs to $\widehat{H}_0^{\frac{\alpha +\beta }{2}}(\Omega )$ and is the solution of the boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} A\tilde{u}=f \quad \text {in} \quad \Omega ,\\ \tilde{u}|_{\partial \Omega }=0. \end{array}\right. } \end{aligned}$$

(4.39)

2. Let us examine $\tilde{u}$ and ${I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1$ separately in (4.38). First note that $f\in L^1(\Omega )$. If $f\ge 0$ a.e. in $\Omega$, then $\tilde{u}$ either is a zero function (when $f=0$ a.e.), or, otherwise, $\tilde{u}(x)>0$ in $\Omega$ by Lemma 6. Secondly, it is readily verified that ${I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1$ is monotonically decreasing, that is,

$$\begin{aligned} \begin{aligned} \frac{d}{dx}{I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1&=\frac{d}{dx}\frac{(b-a)^{\alpha -1}(b-x)^\beta }{\Gamma (\alpha )\Gamma (1+\beta )}{{}_2F_1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \\&=-\frac{(b-a)^{\alpha -1}(b-x)^{\beta -1}}{\Gamma (\alpha )\Gamma (\beta )}{{}_2F_1}\left( 1-\alpha ,1;1+\beta ;\frac{b-x}{b-a}\right) \\&\quad -\frac{(1-\alpha )(b-a)^{\alpha -2}(b-x)^\beta }{(1+\beta )\Gamma (\alpha )\Gamma (1+\beta )}{{}_2F_1}\left( 2-\alpha ,2;2+\beta ;\frac{b-x}{b-a}\right) \\&<0\quad \text {in} \quad \Omega . \end{aligned} \end{aligned}$$

Therefore, combining the above facts and using (4.38), we see that

$$\begin{aligned} \begin{aligned}&u\text { attains its minimum at }x=a\text { provided }c<0.\\&u\text { attains its minimum at }x=b\text { provided }c>0.\\&u\text { attains its minimum at }x=a,b\text { provided }c=0. \end{aligned} \end{aligned}$$

(4.40)

This proves (4.36).

3. Since $-u$ attains its maximum whenever u attains its minimum, assertion Equation (4.37) follows. $\square$

We continue to state the strong maximum principle by going one step further from the weak maximum principle, which says that u cannot attain its minimum (similarly, maximum) at an interior point unless u is a constant. However, it is somewhat interesting that our proof is independent of Hopf’s lemma, which shows a difference in analytic techniques between classic elliptic equations and fractional-order elliptic equations.

Theorem 3

(Strong maximum principle) Assume $u\in C(\overline{\Omega })\cap C^1(\Omega )$ and ${I_{a+}^{1-\alpha }}{D_{b-}^{\beta }}u, \,$ ${I_{b-}^{1-\beta }}u\in AC(\overline{\Omega })$.

• 1. If $Au \ge 0$ a.e. in $\Omega$ and u attains its minimum over $\overline{\Omega }$ at an interior point, then

  $$\begin{aligned} u \text { must be a zero function}. \end{aligned}$$

  (4.41)

• 2. Similarly, if $Au \le 0$ a.e. in $\Omega$ and u attains its maximum over $\overline{\Omega }$ at an interior point, then

  $$\begin{aligned} u \text { must be a zero function}. \end{aligned}$$

  (4.42)

The proof is like that above in Theorem 2.

Proof

1. Suppose $Au \ge 0$ a.e. in $\Omega$. Recall from (4.38) that u can be decomposed into

$$\begin{aligned} u=\tilde{u}+c\, {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1,\quad x\in \Omega , \end{aligned}$$

(4.43)

where

$$\begin{aligned} \tilde{u}(x)\ge 0, x\in \Omega \quad \text {and}\quad \tilde{u}(x)> 0, x\in \Omega \quad \text {if}\quad A\tilde{u}=0, \,a.e.\quad \text {does not hold},\nonumber \\ \end{aligned}$$

(4.44)

and

$$\begin{aligned} {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1 \quad \text {is a monotonically decreasing function with}\quad {I_{b-}^{\beta }}{D_{a+}^{1-\alpha }}1\big |_{x=b}=0.\nonumber \\ \end{aligned}$$

(4.45)

(4.44) and (4.45) mean that u attains its minimum over $\overline{\Omega }$ at an interior point only when $c=0$ and $A\tilde{u}=0$ a.e. Therefore, $\tilde{u}\equiv 0$ since $\tilde{u}$ is a solution of boundary value problem (4.39). Hence, u is a zero function.

Finally, (4.42) follows immediately, since u attains its minimum over $\overline{\Omega }$ at an interior point if and only if $-u$ attains its maximum over $\overline{\Omega }$ at the same point. $\square$

5 Conclusions

In this work, we have developed elliptic-type results, such as Hopf’s lemma and maximum principles, for a class of fractional ODEs involving the composite operators of both left- and right-sided R-L derivatives. By plotting the pictures, we have also demonstrated the similarity and difference of the asymptotic behaviors at the boundary between the solutions of fractional diffusion equations and the classic diffusion equations. This work shows a possibility of develo** other elliptic-type results for those operators, for example, existence of principal eigenvalue, etc., which will be explored in separate work.