Consider the following initial value problem (IVP) with Caputo-type fractional differential equation:

$$\begin{aligned} \textstyle\begin{cases} {}^{c}D^{p}x(t)=f(t,x(t)),\\ x^{(k)}(0)=x_{k},\quad k=0,1,\ldots, n-1, \end{cases}\displaystyle \end{aligned}$$

(3)

where \(f\in C([0,T]\times R,R)\) and \(0\leq n-1< p\leq n\). The IVP (3) is equivalent to the following Volterra fractional integral equation:

$$\begin{aligned} x(t)=\sum_{k=0}^{n-1} \frac{x_{k}t^{k}}{\Gamma(k+1)}+\frac{1}{\Gamma(p)} \int _{0}^{t}(t-s)^{p-1}f\bigl(s,x(s)\bigr) \,\mathrm{d}s. \end{aligned}$$

(4)

We need the following useful lemma.

### Lemma 1

*Let*
\(m(t): R_{+} \rightarrow R\)
*be locally Lipschitz continuous such that*, *for*
\(t_{1}\in(0,+\infty)\),

$$\begin{aligned} m(t_{1})=0 \quad \textit{and} \quad m(t)\leq0 \quad\textit{for } 0\leq t \leq t_{1}. \end{aligned}$$

*Then*
\(D^{p}m(t_{1})\geq0\)
*for*
\(p\in(0,1)\).

The result of Lemma 1 was presented in [5] for a locally Hölder continuous function \(m(t)\). However, the proof in [5] is wrong since it requires the condition that \(\lambda+p-1>0\), where *λ* is the Hölder constant for \(m(t)\), but \(\lambda+p-1>0\) is not always true for \(p\in(0,1)\). If the Hölder continuity is replaced by the Lipschitz continuity, that is, \(\lambda=1\), then \(\lambda+p-1>0\), and then the proof of Lemma 1 follows from [5].

Now, we present a comparison principle for fractional differential equation with the Caputo derivative under strict inequalities when \(p\in(0,1)\).

### Theorem 1

*Assume that*
\(f(t,x)\)
*and*
\(F(t,x)\)
*are two continuous functions defined on*
\(G= [0,T]\times R\)
*satisfying the inequality*

$$\begin{aligned} f(t,x)< F(t,x) \quad \textit{for } (t,x)\in G. \end{aligned}$$

(5)

*Suppose that*
\(x=\varphi(t)\)
*and*
\(x=\phi(t)\)
*are solutions of the following initial value problems* (\(0< p<1\)):

$$\begin{aligned}& (E_{1}):\quad {}^{c}D^{p}x=f(t,x), \quad x(0)=x_{0}, \\& (E_{2}):\quad {}^{c}D^{p}x=F(t,x), \quad x(0)=x_{0}, \end{aligned}$$

*respectively*. *Then we have*

$$\begin{aligned} \varphi(t)< \phi(t), \quad t\in(0,T]. \end{aligned}$$

(6)

### Proof

Denote \(\psi(t)= \varphi(t)-\phi(t)\), \(t\in[0,T]\). We have

$$\begin{aligned} \begin{aligned} &\psi(0)=\varphi(0)-\phi(0)=0, \\ &{}^{c}D^{p}\psi(0)= {}^{c}D^{p} \varphi(0)- {}^{c}D^{p}\phi(0)=f(0,x_{0})-F(0,x_{0})< 0. \end{aligned} \end{aligned}$$

Then \({}^{c}D^{p}\psi(t)=D^{p}\psi(t)\), and there exists \(\sigma>0\) such that \(\psi(t)<0\) for \(0< t<\sigma\). Suppose that inequality (6) is incorrect. Then there exists at least one \(t_{1}\) (>0) such that \(\psi(t_{1})=0\). Denote

$$\begin{aligned} \beta=\inf\bigl\{ t\mid\psi(t)=0, t>0\bigr\} . \end{aligned}$$

Then \(\psi(\beta)=0\) and \(\psi(t)<0\) for \(0< t<\beta\).

Since \({}^{c}D^{p} \psi(t)=f(t,x)-F(t,x)\), \(t\in[0,T]\), and \(f(t,x)\), \(F(t,x)\) are continuous in *G*, it follows that \(\psi'(t)\) is a locally bounded function by Definition 2. Therefore \(\psi(t)\) is locally Lipschitz continuous in \(0\leq t\leq T\). Thus \(D^{p}\psi(\beta )\geq0\) by Lemma 1.

On the other hand, since \(\psi(\beta)=0\), denote \(\gamma= \varphi (\beta)=\phi(\beta)\). Then

$$\begin{aligned} 0\le D^{p}\psi(\beta)= {}^{c}D^{p} \psi(\beta)= {}^{c}D^{p}\varphi(\beta)- {}^{c}D^{p} \phi(\beta)=f(\beta,\gamma)-F(\beta,\gamma). \end{aligned}$$

(7)

It is obvious that (7) is in contradiction with (5). Hence (6) is proved. □

### Remark 2

When \(p=1\), Theorem 1 is also true as the well-known comparison theorem for ordinary differential equations with initial value condition.

Next, we extend Theorem 1 to the case for fractional differential equations with the Caputo derivative of order \(n-1< p\leq n\) with initial value conditions by employing Theorem 1.

### Theorem 2

*Under assumption* (5), *suppose that*
\(x=\varphi(t)\)
*and*
\(x=\phi (t)\)
*are solutions of the following initial value problems* (\(1\leq n-1< p\leq n\)):

$$\begin{aligned} (E_{3}): \quad{}^{c}D^{p}x=f(t,x), \quad x^{(k)}(0)=x_{k},k=0,1,2,\ldots n-1, \\ (E_{4}):\quad{}^{c}D^{p}x=F(t,x), \quad x^{(k)}(0)=x_{k}, k=0,1,2,\ldots n-1, \end{aligned}$$

*respectively*. *Then*
\(\varphi(t)<\phi(t)\)
*for*
\(t\in(0,T]\).

### Proof

Consider \({}^{c}D^{p}x(t)= {}^{c}D^{q}D^{n-1}x(t)\) with \(q=p-n+1\in(0,1]\). Denote \(y(t)=D^{n-1}x(t)\), \(t\in[0,T]\). Then the IVPs (\(E_{3}\)) and (\(E_{4}\)) can be transformed into initial value problems of order \(0< q\leq1\). Indeed, we have

$$\begin{aligned} &{}^{c}D^{p}x(t)= {}^{c}D^{q}D^{n-1}x(t)= {}^{c}D^{q}y(t), \quad y(t)=D^{n-1}x(t), \end{aligned}$$

(8)

$$\begin{aligned} &x(t)=\sum_{k=0}^{n-2}\frac{x_{k}t^{k}}{\Gamma(k+1)}+ \frac{1}{\Gamma (n-1)} \int_{0}^{t}(t-s)^{n-2}y(s)\, \mathrm{d}s. \end{aligned}$$

(9)

Denote

$$\begin{gathered} \hat{f}\bigl(t,y(t)\bigr)=f\bigl(t,x(t)\bigr)=f \Biggl(t,\sum _{k=0}^{n-2}\frac{x_{k}t^{k}}{\Gamma (k+1)}+\frac{1}{\Gamma(n-1)} \int_{0}^{t}(t-s)^{n-2}y(s)\,\mathrm{d}s \Biggr), \\ \hat{F}\bigl(t,y(t)\bigr)=F\bigl(t,x(t)\bigr)=F \Biggl(t,\sum _{k=0}^{n-2}\frac{x_{k}t^{k}}{\Gamma (k+1)}+\frac{1}{\Gamma(n-1)} \int_{0}^{t}(t-s)^{n-2}y(s)\,\mathrm{d}s \Biggr). \end{gathered} $$

By inequality (5) we have \(\hat{f}(t,y(t))<\hat{F}(t,y(t))\) for \(t\in[0,T]\). The original initial value problems (\(E_{3}\)) and (\(E_{4}\)) can be written as

$$\begin{aligned} &(\hat{E_{3}}):\quad {}^{c}D^{q}y(t)=\hat{f} \bigl(t,y(t)\bigr), \quad y(0)=D^{n-1}x(0)=x_{n-1}, \\ &(\hat{E_{4}}):\quad {}^{c}D^{q}y(t)=\hat{F} \bigl(t,y(t)\bigr), \quad y(0)=D^{n-1}x(0)=x_{n-1}. \end{aligned}$$

Assume that \(\hat{\varphi}(t)\) and \(\hat{\phi}(t)\) are solutions of (\(\hat{E_{3}}\)) and (\(\hat{E_{4}}\)), respectively. Then by Theorem 1 and Remark 2 we have \(\hat{\varphi}(t)<\hat{\phi}(t)\) for \(t\in(0,T]\).

Since \(x(t)\) and \(y(t)\) satisfy equation (9), we obtain

$$\begin{aligned} \varphi(t)&=\sum_{k=0}^{n-2} \frac{x_{k}t^{k}}{\Gamma(k+1)}+\frac{1}{\Gamma (n-1)} \int_{0}^{t}(t-s)^{n-2}\hat{\varphi}(s)\, \mathrm{d}s \\ &< \sum_{k=0}^{n-2}\frac{x_{k}t^{k}}{\Gamma(k+1)}+ \frac{1}{\Gamma(n-1)} \int _{0}^{t}(t-s)^{n-2}\hat{\phi}(s)\, \mathrm{d}s=\phi(t), \end{aligned}$$

which finishes the proof. □

### Example 1

Assume that \(1 < p \leq2\) and that \(\varphi(t)\) and \(\phi(t)\) are solutions to the two initial value problems

$$\begin{aligned} &{}^{c}D^{p}x = x,\quad x(0)=1, x'(0) = 2, \\ &{}^{c}D^{p}x = t + x,\quad x(0)=1, x'(0) = 2, \end{aligned}$$

respectively, where \(t\geq0\). Then we have \(x< t+x\) for \(t>0\), and

$$\begin{aligned} &\varphi(t)= E_{p,1}\bigl(t^{p}\bigr)+2E_{p,2} \bigl(t^{p}\bigr), \\ &\phi(t)=E_{p,1}\bigl(t^{p}\bigr)+2E_{p,2} \bigl(t^{p}\bigr)+t^{p+1}E_{p,p+2}\bigl(t^{p} \bigr), \end{aligned}$$

where \(E_{\alpha,\beta}(z)=\sum_{k=0}^{\infty}\frac{z^{k}}{\Gamma(k\alpha +\beta)}\). We know that \(\varphi(t)<\phi(t)\) for \(t>0\). This is consistent with Theorem 2.

### Remark 3

Some researchers [17, 18] attempted to give similar comparison principles for fractional differential equations with the Caputo derivatives by the method of the integral mean value theorem, which, however, does not guarantee the correctness of their proofs. Yu et al. [19] presented comparison theorems for fractional differential equations involving the Riemann–Liouville and Caputo derivatives of order \(0< p<1\) with Lipschitz condition \(f(t,u)-f(t,v)\leq L(u-v)\) for \(u>v\).