1 Introduction

The theory of fractional order derivatives are almost as old as the integer-order [5]. There are many applications, for example in physics [1, 2, 6], finance [8, 9] or biology [3]. Our aim is to prove theoretical mathematical statements.

In this work our goal is to find a solution numerically for the equation A(u) = f. If we assume that u is time-dependent, then one can do this by finding a stationary solution of the equation ∂ tu(t) = −(A(u(t)) − f). The numerical solution of this problem can be highly inaccurate. To avoid this we propose to replace the time derivative with a fractional one. Since the fractional order time derivative is a non-local operator, we expect that this stabilizes the time integration in the numerical solutions. Since the fractional order derivative here is defined as a limit of linear combination of past values, the time discretization will be simple. We also tested our method numerically in a fluid dynamical problem [10].

2 Mathematical Preliminaries

The following theorem is well known, see [11].

Theorem 1

Let H real Hilbert-space, A : H → H nonlinear operator, which satisfies the conditions below with some positive constants M ≥ m:

  1. 1.

    〈A(u) − A(v), u − v〉≥ m∥u − v∥2,

  2. 2.

    ∥A(u) − A(v)∥≤ M∥u − v∥.

Then for any f, u 0 ∈ H there exist a unique solution u ∗of the equation A(u) = f. If \(t\in \mathbb {R}^{+}\)is small enough the following iteration converges to u ∗.

$$\displaystyle \begin{aligned} u_{n+1}=u_n-t \big[ A(u_{n})-f \big]. \end{aligned} $$
(1)

There exist many different definitions of the fractional derivative [4, 7] we will use here the one below which is based on finite differences.

Definition 1

For the exponent β ∈ (0, 1) the fractional order derivative for a given function \(f: \mathbb {R}^{+} \to \mathbb {R}\) is defined as

$$\displaystyle \begin{aligned}\frac{\partial^\beta f(t)}{\partial t^\beta}:= \lim_{N \rightarrow \infty}\Big\{ \sum_{k=0}^{N} \binom{\beta}{k}(-1)^k \frac{f(t-kh)}{h^\beta} \Big\}, \end{aligned} $$

provided that the limit exists.

3 Results

Shortly, our objective is to find a solution for the equation A(u) = f for a given nonlinear operator A, and for a given function f. The solution u is also time-dependent, our goal is to find a stationary solution for

$$\displaystyle \begin{aligned} -(A(u(t))-f)=\partial_t u(t). \end{aligned} $$
(2)

The method in Theorem 1 is one approach to this. Our idea was that to replace the time derivative in (2) with \(\frac {\partial ^\beta }{\partial t^\beta }\) for some β ∈ (0, 1), according to Definition 1, and discretise the equation in time by a natural way.

We need an additional statement before we prove.

Lemma 1 (Pachpatte)

Let \(\left (\alpha _n\right )_{n \in \mathbb {N}}\),\(\left (f_n \right )_{n \in \mathbb {N}}\),\(\left (g_n \right )_{n \in \mathbb {N}}\),\(\left (h_n \right )_{n \in \mathbb {N}}\)nonnegative real sequences with the conditions below:

$$\displaystyle \begin{aligned} \alpha_n \leq f_n + g_n \sum_{s=0}^{n-1}h_s \alpha_s. \end{aligned} $$
(3)

Then the following inequality holds

$$\displaystyle \begin{aligned} \alpha_n \leq f_n + g_n \sum_{s=0}^{n-1}h_s f_s \prod_{\tau=s+1}^{n-1}\big( h_\tau g_\tau +1 \big). \end{aligned} $$
(4)

The main result is a generalisation of Theorem 1. For simplicity, we will not prove the existence of the solution.

Theorem 2

Let H be real Hilbert-space, A : H → H a nonlinear operator, which satisfies the conditions below with some positive constants M ≥ m:

  1. 1.

    〈A(u) − A(v), u − v〉≥ m∥u − v∥2,

  2. 2.

    ∥A(u) − A(v)∥≤ M∥u − v∥.

Let u ∗denote the solution of the equation A(u) = f. For any f, u 0 ∈ H α ∈ (0, 1), and \(t\in \mathbb {R}^{+}\)small enough the following iteration converges to u ∗.

$$\displaystyle \begin{aligned} u_{n+1}=\sum_{j=1}^{n+1}\binom{\alpha}{j}(-1)^{j+1}u_{n+1-j}-t \big[ A(u_{n+1})-f \big]. \end{aligned} $$
(5)

Proof

We first add t[A(u n+1) − f] − u ∗ both sides of the Eq. (5) and taking their norms, we have that

$$\displaystyle \begin{aligned} \left\| u_{n+1}-u^{*}+t \big[ A(u_{n+1}) -A(u^{*}) \big] \right\|= \left\| \sum_{j=1}^{n+1} \binom{\alpha}{j}(-1)^{j+1} u_{n+1-j}-u^{*} \right\|. \end{aligned} $$
(6)

Using the first assumption, we get the lower estimation

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} \| u_{n+1}-u^{*}+t \big[ A(u_{n+1}) -A(u^{*}) \big] \|{}^2 \\ =\| u_{n+1}-u^{*}\|{}^2 +t^2 \|A(u_{n+1})-A(u^{*}) \|{}^2 + 2t \langle A(u_{n+1})-A(u^{*}),u_{n+1}-u^{*}\rangle \\ \geq \| u_{n+1}-u^{*}\|{}^2 +2tm \|u_{n+1}-u^{*} \|{}^2\geq \| u_{n+1}-u^{*} \|{}^2. \end{array} \end{aligned} $$
(7)

It is also known that \(\sum _{j=1}^{\infty } \binom {\alpha }{j}(-1)^{j+1}=1\) and \(\binom {\alpha }{j}(-1)^{j+1}>0\). Using this, the triangle inequality and (6) for the inequality in (7) we get

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} \| u_{n+1}-u^{*} \|\leq \left\| \sum_{j=1}^{n+1} \binom{\alpha}{j}(-1)^{j+1} u_{n+1-j}-u^{*} \right\| \\ = \left\| \sum_{j=1}^{n+1} \binom{\alpha}{j}(-1)^{j+1} u_{n+1-j}-\sum_{j=1}^{\infty} \binom{\alpha}{j}(-1)^{j+1} u^{*} \right\| \\ \leq \sum_{j=1}^{n+1} \binom{\alpha}{j}(-1)^{j+1} \|u_{n+1-j}-u^{*} \|+ \sum_{j=n+2}^{\infty} \binom{\alpha}{j}(-1)^{j+1} \| u^{*} \|. \end{array} \end{aligned} $$
(8)

Let α n := ∥u n − u ∗∥, \(f_n :=\sum _{j=n+1}^{\infty } \binom {\alpha }{j}(-1)^{j+1} \| u^{*} \|\) and \(\beta _n=\binom {\alpha }{n}(-1)^{n+1}\). With these, we can rewrite (8) as

$$\displaystyle \begin{aligned}{{}} \alpha_{n+1} \leq f_{n+1}+\sum_{j=1}^{n+1} \beta_j \alpha_{n+1-j}. \end{aligned} $$
(9)

Also using the notation h j instead of β n+1−j, (9) can be recognised as

$$\displaystyle \begin{aligned}{{}} \alpha_{n+1} \leq f_{n+1} + \sum_{j=0}^{n} h_j \alpha_j. \end{aligned} $$
(10)

Therefore, with g n := 1 we can apply Lemma 1.

$$\displaystyle \begin{aligned} \alpha_{n+1} \leq f_{n+1} +\sum_{s=0}^{n}h_s f_s \prod_{\tau=s+1}^{n} (h_{\tau}+1). \end{aligned} $$
(11)

Estimate \(\prod _{\tau =s+1}^{n} (h_{\tau }+1)\) as

$$\displaystyle \begin{aligned} \begin{array}{rcl} \prod_{\tau=s+1}^{n} (h_{\tau}+1) = \prod_{\tau=s+1}^{n} (\beta_{n+1-\tau}+1) \\ \leq \prod_{\tau=1}^{n} (\beta_{n+1-\tau}+1) \leq \Big( \frac{n+\sum_{j=1}^{n}\beta_j}{n}\Big)^{n} \leq \Big(1+\frac{1}{n} \Big)^n \leq e. \end{array} \end{aligned} $$

Consequently, for (11) the following holds.

$$\displaystyle \begin{aligned}\alpha_{n+1} \leq f_{n+1} +\sum_{s=0}^{n}h_s f_s \prod_{\tau=s+1}^{n} (h_{\tau}+1) \leq f_{n+1}+e \sum_{s=0}^n h_s f_s.\end{aligned}$$

It is clear that if n →∞ then f n+1 → 0. We prove that \(\sum _{s=0}^n h_s f_s \rightarrow 0\).

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} \sum_{s=0}^n h_s f_s = \|u^{*}\| \beta_{n+1} + \|u^{*}\| \sum_{s=1}^n \beta_{n+1-s} \sum_{j=s+1}^{\infty} \beta_j \\= \|u^{*}\| \beta_{n+1} + \|u^{*}\| \sum_{s=1}^n \beta_{n+1-s} \Big( 1-\sum_{j=1}^{s} \beta_j \Big)\\ = \|u^{*}\| \beta_{n+1}+\|u^{*}\| \sum_{s=1}^{n}\beta_{n+1-s}- \|u^{*}\| \sum_{s=1}^{n} \sum_{j=1}^{n} \beta_{n+1-s}\beta_j. \end{array} \end{aligned} $$
(12)

Observe first, that the last term in (12) is a Cauchy product.

$$\displaystyle \begin{aligned}\lim_{n\rightarrow \infty} \Big(\sum_{s=1}^{n} \sum_{j=1}^{n} \beta_{n+1-s}\beta_j\Big)=\Big( \sum_{j=1}^{\infty} \beta_j \Big)^2=1. \end{aligned}$$

Therefore, the first term in (12) tends to zero, the second and the third term to ∥u ∗∥, since \(\sum _{j=1}^{\infty } \beta _j=1\). This means that α n+1 → 0 if n →∞, which has been stated. □

4 Discussion

In this work, we solved nonlinear time-independent equations of type A(u) = f, where the operator A is on a Hilbert space. We assumed that it is monotone and Lipschitz-continuous and we proved that the algorithm is convergent.

Our numerical experiences show that if we replace the time-derivative operator in the equation ∂ tu = −[A(u) − f] with a fractional derivative, then it stabilizes the time integration in the numerical solutions. We have tested our method numerically in a fluid dynamical problem previously [10].