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A nonlinear fractional Rayleigh–Stokes equation under nonlocal integral conditions

Abstract

In this paper, we study the fractional nonlinear Rayleigh–Stokes equation under nonlocal integral conditions, and the existence and uniqueness of the mild solution to our problem are considered. The ill-posedness of the mild solution to the problem recovering the initial value is also investigated. To tackle the ill-posedness, a regularized solution is constructed by the Fourier truncation method, and the convergence rate to the exact solution of this method is demonstrated.

Introduction

Most of fluids in the real world, such as in food products (mayonnaise, mustard, chocolate, ketchup, butter, cheese, yogurt, etc.), in natural substances (honey, magma, lava, gums, etc.), in biology (blood, semen, synovia, mucus, etc.), in industry (paint, glue, lubricant, ink, molten polymer, etc.), in cosmetics (soap solution, toothpaste, cream, silicone, nail polish, etc.) are treated as non-Newtonian fluids. Therefore, the study on non-Newtonian fluids is a substantial subject in science and industrial applications. As the Rayleigh–Stokes problem for an edge, the first problem of Stokes for a non-Newtonian fluid flow past an impulsively started flat plate has received much attention because of its practical importance [1, 2]. For a second grade fluid, the equation of motion is of higher order than the Navier–Stokes equation, because it exhibits all properties of viscoelastic fluids.

Recently, fractional calculus has encountered much success in the description of constitutive relations of viscoelastic fluids. The starting point of the fractional derivative model of a viscoelastic fluid is usually a classical differential equation which is modified by replacing the time derivative of an integer order with a fractional calculus operator. This generalization allows one to define precisely noninteger order integrals or derivatives.

Moreover, there has currently been a considerable increase in examining fractional partial differential equations (FPDEs). We can list several current remarkable research studies; for instance, Abdolrazaghi and Razani [3], Behboudi et al. [4], Ding and Neito [5], Agarwal et al. [6], Baleanu [7], Adiguzel et al. [810], Afshari et al. [11], Alqahtani et al. [12], Karapinar et al. [13], Abdeljawad et al. [14], Baitiche et al. [15], Ardjouni [16] and the references therein.

In this research study, we focus on the Rayleigh–Stokes problem as follows:

$$ \textstyle\begin{cases} \partial _{t} u-\Delta u- \mu \partial _{t}^{\alpha }\Delta u = F(u), & (x,t) \in \Omega \times (0,T), \\ u(x,t)=0, & x \in \partial {\Omega }, \end{cases} $$
(1.1)

associated with the nonlocal integral condition

$$\begin{aligned} \xi _{1} u(x,0)+\xi _{2} \int _{0}^{T}v(t) u(x,s)\,ds = g(x), \quad x \in \Omega , \end{aligned}$$
(1.2)

where \(\xi _{1}, \xi _{2}>0\) and \(\xi _{1}^{2}+\xi _{2}^{2}>0\). Here, \(u(x,t)\) is fluid velocity, μ is viscosity, F is a source function, Δ is the Laplacian, \(\Omega \subset \mathbb{R}^{d}\) (\(d=1,2,3\)) is a smooth domain with the boundary Ω, and \(T>0\) is a given time, g is the final data in \(L^{2}(\Omega )\), \(\partial _{t}=\partial /\partial t\), and \(\partial _{t}^{\alpha }\) is the Riemann–Liouville fractional derivative of order \(\alpha \in (0,1)\) defined in [17, 18]:

$$ \partial _{t}^{\alpha }u(x,t)= \frac{1}{\Gamma (1-\alpha )} \frac{\partial }{\partial t} \biggl( \int _{0}^{t} \frac{u(x,s)}{ (t-s)^{\alpha }}\,ds \biggr), $$

where \(\Gamma (.)\) is the gamma function.

The forward problems for equation (1.1) have been examined in a plenty of studies. For instance, Zierep and Fetecau [19] discussed the energetic balance in the Rayleigh–Stokes problem for a Maxwell fluid for several initial and/or boundary conditions. Fetecau and Zierep [20] found the exact solutions both for the Stokes’ problem and for the Rayleigh–Stokes problem within the context of the fluids of second grade. Bazhlekova et al. [21] presented an introduction about an analysis of the Rayleigh–Stokes problem for a generalized second-grade fluid. The exact solution of the Rayleigh–Stokes problem for a generalized second grade fluid in a porous half-space with a heated flat plate was considered by Xue et al. [22]. Exact solutions of the Rayleigh–Stokes equation in the case of homogeneous initial and boundary conditions was considered by Zhao and Yang [23]. The backward problems have currently been studied by many mathematicians. Ngoc et al. [24], for instance, pondered the inverse problem for the nonlinear fractional Rayleigh–Stokes equations. Equation (1.1) associated with Gaussian random noise was examined by Triet et al. [25], and so forth.

To the best of our knowledge, there are still very few studies on the Rayleigh–Stokes equation accompanied with nonlocal integral conditions. In comparison with the initial condition \(u(x,0)=f(x)\) or the final condition \(u(x,T)=g(x)\), the nonlocal conditions are of more significant complexity. Furthermore, it is emphasized that we cannot apply Parseval’s equality to obtain stable estimates in the \(L^{p}\) space. To tackle this limitation, we need to develop additional techniques and Sobolev embedding in our study.

This manuscript is structured as follows. An introduction of preliminary results is described in Sect. 2; the regularity of the mild solution to the problem in a linear case is illustrated in Sect. 3. In Sect. 4, the problem recovering the initial value and the convergence of the regularized solution are detailed. The regularity of the mild solution to the problem in the nonlinear case is mentioned in Sect. 5. Eventually, the conclusion is presented in Sect. 6.

Preliminary results

We recall the spectral problem as follows:

$$\begin{aligned} \textstyle\begin{cases} \Delta \phi _{n}(x)=-\lambda \phi _{n}(x), & x \in \Omega , \\ \phi _{n}(x)=0, & x \in \partial \Omega , \end{cases}\displaystyle \end{aligned}$$

admitting the eigenvalues \(0 < \lambda _{1} \leq \lambda _{2} \leq \cdots \leq \lambda _{n} \leq \dots \) with \(\lambda _{n} \to \infty \) as \(n\to \infty \) (see [26]). The corresponding eigenfunctions \(\phi _{n} \in H_{0}^{1}(\Omega )\). For all \(s\geq 0\), the operator \(A^{s}\) (here \(A=-\Delta \)) also possesses the following representation:

$$ A^{s}h=\sum_{n=1}^{\infty } (h,\phi _{n} )\lambda _{n}^{s} \phi _{n},\quad h\in D\bigl(A^{s}\bigr)= \Biggl\{ h\in L^{2}(\Omega ): \sum _{n=1}^{\infty } \bigl\vert ( h,\phi _{n} ) \bigr\vert ^{2} \lambda _{n}^{2s}< \infty \Biggr\} . $$

Consider on \(D(A^{s})\) the norm (noting that \(\lambda _{1}>0\))

$$ \Vert h \Vert _{D(A^{s})}= \Biggl( \sum_{n=1}^{\infty } \bigl\vert ( h,\phi _{n} ) \bigr\vert ^{2}\lambda _{n}^{2s} \Biggr) ^{\frac{1}{2}},\quad h\in D \bigl(A^{s}\bigr). $$

By duality, we can set \(D(A^{-s})=(D(A^{s}))^{\ast }\) by identifying \((L^{2}(\Omega ))^{\ast }=L^{2} ( \Omega ) \) and using the so-called Gelfand triple (see [27]). Then \(D(A^{-s})\) is a Hilbert space with the following norm:

$$ \Vert h \Vert _{D(A^{-s})}= \Biggl( \sum_{n=1}^{\infty } \bigl\vert \langle h,\phi _{j} \rangle \bigr\vert ^{2} \lambda _{n}^{-2s} \Biggr) ^{\frac{1}{2}}, $$

wherein \(\langle \cdot ,\cdot \rangle \) denotes the duality bracket between \(D(A^{-s})\) and \(D(A)\). For any \(p\geq 0\), we define the space

$$\begin{aligned} \mathcal{H}^{p}(\Omega )= \Biggl\{ v\in L^{2}(\Omega );\sum _{n=1}^{\infty }\lambda _{n}^{p} \bigl\vert \bigl\langle v(x),\phi _{n}(x) \bigr\rangle \bigr\vert ^{2}< + \infty \Biggr\} , \end{aligned}$$

where \(\langle \cdot ,\cdot \rangle \) is the inner product in \(L^{2}(\Omega )\), then \(\mathcal{H}(\Omega )\) is a Hilbert space with the norm

$$\begin{aligned} \|v\|_{\mathcal{H}^{p}(\Omega )}= \Biggl( \sum_{n=1}^{\infty } \lambda _{n}^{p} \bigl\vert \bigl\langle v(x),\phi _{n}(x) \bigr\rangle \bigr\vert ^{2} \Biggr)^{\frac{1}{2}}. \end{aligned}$$

Suppose that problem (1.1) has a solution u which admits the form \(u(x,t)=\sum_{n=1}^{\infty } \langle u(x,t), \phi _{n}(x) \rangle \phi _{n}(x) \). From [21, 24], we deduce that the solution of problem (1.1) with the initial condition \(u(x, 0)=u_{0}(x)\) is given by

$$\begin{aligned} \bigl\langle u(x,t),\phi _{n}(x) \bigr\rangle =\mathbf{S}_{n,\alpha }(t) \bigl\langle u_{0}(x),\phi _{n}(x) \bigr\rangle + \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s) \bigl\langle F\bigl(u(x,s)\bigr),\phi _{n}(x) \bigr\rangle \,ds, \end{aligned}$$

where \(\mathbf{S}_{n,\alpha }(t)\) is given by

$$\begin{aligned}& \mathbf{S}_{n,\alpha }(t)= \int _{0}^{\infty }e^{-zt} \mathcal{K} (n, \alpha , z)\,dz, \end{aligned}$$
(2.1)
$$\begin{aligned}& \mathcal{K} (n,\alpha , z) =\frac{\mu }{\pi } \frac{\lambda _{j} \sin (\alpha \pi ) z^{\alpha }}{ (-z+\lambda _{n} \mu z^{\alpha }\cos \alpha \pi +\lambda _{j} )^{2}+ (\lambda _{n} \mu z^{\alpha }\sin \alpha \pi )^{2}}, \end{aligned}$$
(2.2)

and its Laplace transform is given by \(\mathcal{L} (\mathbf{S}_{n,\alpha }(t) )= \frac{1}{t+ \mu \lambda _{n} t^{\alpha }+\lambda _{n}} \). This implies that

$$\begin{aligned} u(x,t)=\sum_{n=1}^{\infty } \biggl[ \mathbf{S}_{n,\alpha }(t) \bigl\langle u_{0}(x),\phi _{n}(x) \bigr\rangle + \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s) \bigl\langle F\bigl(u(x,s)\bigr),\phi _{n}(x) \bigr\rangle \,ds \biggr] \phi _{n}(x). \end{aligned}$$
(2.3)

We can easily see that

$$\begin{aligned} u_{n}(t)=\mathbf{S}_{n,\alpha }(t)u_{n}(0)+ \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds, \end{aligned}$$
(2.4)

where \(F_{n}(u(s))=\langle F(u(\cdot ,s)),\phi (x) \rangle \).

From the equation

$$\begin{aligned} \xi _{1} u(x,0)+\xi _{2} \int _{0}^{T}v(t) u(x,s)\,ds = g(x), \end{aligned}$$
(2.5)

we have that

$$\begin{aligned} \xi _{1} \sum_{1}^{\infty }u_{n}(0) \phi _{n}(x) +\xi _{2} \int _{0}^{T}v(t) \Biggl(\sum _{1}^{\infty }u_{n}(t)\phi _{n}(x) \Biggr)\,dt=\sum_{1}^{ \infty }g_{n}\phi _{n}(x), \end{aligned}$$
(2.6)

where \(g_{n}=\langle g(\cdot ),\phi (x) \rangle \).

Thanks to the uniqueness of the Fourier expansion of a function in \(L^{2}\) space, we get

$$\begin{aligned} g_{n}={}&\xi _{1} u_{n}(0)+\xi _{2} \int _{0}^{T}v(t)u_{n}(t)\,dt \\ ={}&\xi _{1} u_{n}(0) + \xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)u_{n}(0)\,dt \\ &{}+ \xi _{2} \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt. \end{aligned}$$
(2.7)

By a straightforward computation, we obtain

$$\begin{aligned} u_{n}(0)&= \frac{g_{n}-\xi _{2} \int _{0}^{T}v(t)\int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n}(u(s))\,ds\,dt}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \\ &= \frac{1}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggl( g_{n}-\xi _{2} \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr). \end{aligned}$$
(2.8)

Conjoining (2.4) and (2.8), we have

$$\begin{aligned} u_{n}(t)={}& \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggl( g_{n}-\xi _{2} \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr) \\ &{}+ \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds. \end{aligned}$$
(2.9)

As a result, we have the formula or the mild solution of problem (1.1)–(1.2)

$$\begin{aligned} u(x,t)={}&\sum_{n=1}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \\ &{}\times \biggl( g_{n}-\xi _{2} \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr)\phi _{n}(x) \\ &{}+\sum_{n=1}^{\infty } \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds \phi (x). \end{aligned}$$
(2.10)

We posit the terms \(\mathcal{A}_{j=1,2,3}(x,t)\) as follows:

$$\begin{aligned} \mathcal{A}_{1}(x,t)&=\sum_{n=1}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}g_{n} \phi _{n}(x), \\ \mathcal{A}_{2}(x,t)&=\sum_{n=1}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggl(\xi _{2} \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr)\phi _{n}(x), \\ \mathcal{A}_{3}(x,t)&=\sum_{n=1}^{\infty } \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds \phi (x). \end{aligned}$$

Then we have

$$\begin{aligned} u(x,t)=\mathcal{A}_{1}(x,t)-\mathcal{A}_{2}(x,t)+ \mathcal{A}_{3}(x,t). \end{aligned}$$
(2.11)

Lemma 2.1

If \(\alpha \in (0, 1)\), we have that the subsequent estimations hold

$$ \frac{ C_{2} (\mu ,\alpha ,t)}{\lambda _{n}} \leq \mathbf{S}_{n, \alpha }(t) \le \frac{ C_{1} (\mu , \alpha )}{1+\lambda _{n} t^{1-\alpha }},\quad 0 \le t\le T, $$
(2.12)

where

$$ C_{1} (\mu , \alpha )= \frac{\Gamma (1-\alpha )}{\mu \pi \sin (\alpha \pi ) }+1 ,\qquad C_{2} ( \mu , \alpha ,t)= \frac{\mu \sin (\alpha \pi ) e^{-t}}{3\pi (\alpha +1) (\mu ^{2}+1+\frac{1}{\lambda _{1}^{2}})} . $$

Proof

The detailed interpretation can be found in [24]. We briefly present some main steps of the proof. We have

$$\begin{aligned} \mathbf{S}_{n,\alpha }(t)=\frac{\mu }{\pi } \int _{0}^{\infty } \frac{e^{-zt}\lambda _{n}\sin (\alpha \pi )z^{\alpha }}{(-z+\lambda _{n}\mu z^{\alpha }\cos (\alpha \pi ))^{2}+(\lambda _{n}\mu z^{\alpha }\sin (\alpha \pi ))^{2}}\,dz. \end{aligned}$$
(2.13)

In terms of the left-hand side of inequality (2.12), we have

$$\begin{aligned} \mathbf{S}_{n,\alpha }(t)\leq \frac{1}{\mu \pi \sin (\alpha \pi )t^{1-\alpha }\lambda _{n}} \int _{0}^{ \infty }e^{-z}z^{-\alpha }\,dz= \frac{\Gamma (1-\alpha )}{\mu \pi \sin (\alpha \pi )t^{1-\alpha }\lambda _{n}}. \end{aligned}$$
(2.14)

It is worth noting that

$$\begin{aligned} \int _{0}^{\infty }e^{-z}z^{-\alpha }\,dz= \Gamma (1-\alpha ). \end{aligned}$$

With regard to the right-hand side of inequality (2.12), we have

$$\begin{aligned} \mathbf{S}_{n,\alpha }(t) &\geq \frac{\mu \sin (\alpha \pi )}{3\pi \lambda _{n}} \int _{0}^{\infty } \frac{e^{-zt}z^{\alpha }}{\mu ^{2} z^{2\alpha }+1+\frac{z^{2}}{\lambda _{1}^{2}}}\,dz \\ &\geq \frac{\mu \sin (\alpha \pi )}{3\pi \lambda _{n}} \frac{e^{-t}}{\mu ^{2}+1+\frac{1}{\lambda _{1}^{2}}} \frac{1}{\alpha +1}. \end{aligned}$$
(2.15)

 □

For the rest of paper, we give the definition of a mild solution to problem (1.1)–(1.2) in the subsequent notion.

Definition 2.1

The function u is called a mild solution of problem (1.1)–(1.2) if

(i) u belongs to the \(L^{m}(0,T;L^{2}(\Omega ))\) space;

(ii) u satisfies equality (2.9).

The regularity of the mild solution to problem (1.1)–(1.2) in the linear case

In this section, we consider the regularity of the mild solution of problem (1.1)–(1.2) under the condition that the source term is linear.

Theorem 3.1

Suppose that there exist M, N such that \(Nt^{-\gamma }\leq v(t)\leq Mt^{-\theta }\) for \(\gamma , \theta <\frac{1}{2}\).

(i) Let \(1/2 <\alpha <1\) and \(s>0\). If \(\xi _{1}>0\), \(\xi _{2}>0\) and \(g \in D(A^{s}) \), \(F\in L^{2}(0,T;D(A^{s-1}))\), we rest assured that

$$\begin{aligned} \bigl\Vert u(\cdot ,t) \bigr\Vert _{D(A^{s})} \leq \frac{C_{1}(\mu ,\alpha )}{\xi _{1}} \Vert g \Vert _{D(A^{s})}+M_{2} \Vert F \Vert _{L^{2}(0,T;D(A^{s-1}))} + M_{3} \Vert F \Vert _{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.1)

(ii) Let \(1/2 <\alpha <1\) and \(s>0\). If \(\xi _{1}=0\), \(\xi _{2}>0\) and \(g \in D(A^{s}) \), \(F\in L^{2}(0,T;D(A^{s-1}))\), then we can conclude that

$$\begin{aligned} \bigl\Vert u(\cdot ,t) \bigr\Vert _{D(A^{s})} \leq M_{4} \Vert g \Vert _{D(A^{s})}+M_{5} \Vert F \Vert ^{2}_{L^{2}(0,T;D(A^{s-1}))}+M_{3} \Vert F \Vert _{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.2)

Proof

We will estimate the terms \(\mathcal{A}_{1}(x,t)\), \(\mathcal{A}_{2}(x,t)\), and \(\mathcal{A}_{3}(x,t)\) in two cases: \(\xi _{1},\xi _{2} >0\) and \(\xi _{1}=0\), \(\xi _{2}>0 \).

Part i: In the case of \(\xi _{1},\xi _{2} >0\). Thanks to Parseval’s equality, we have

$$\begin{aligned} \bigl\Vert \mathcal{A}_{1}(\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})}&=\sum_{n=1}^{\infty } \biggl( \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr)^{2}\lambda ^{2s}_{n} \vert g_{n} \vert ^{2} \\ &\leq \frac{C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1}}\sum_{1}^{\infty } \lambda ^{2s}_{n} \vert g_{n} \vert ^{2} \\ &=\frac{C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1}} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$
(3.3)

This implies

$$\begin{aligned} \bigl\Vert \mathcal{A}_{1}(\cdot ,t) \bigr\Vert _{D(A^{s})}\leq \frac{C_{1}(\mu ,\alpha )}{\xi _{1}} \Vert g \Vert _{D(A^{s})}. \end{aligned}$$
(3.4)

Similarly, the term \(\|\mathcal{A}_{2}(\cdot ,t)\|_{D}(A^{2})\) can be assessed as follows:

$$\begin{aligned} &\bigl\Vert \mathcal{A}_{2}(\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \\ &\quad =\sum_{n=1}^{\infty } \biggl( \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr)^{2} \lambda ^{2s} \biggl( \xi _{2} \int _{0}^{T}v(t) \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr)^{2} \\ &\quad \leq \sum_{n=1}^{\infty }\lambda _{n}^{2s-2} \frac{\xi _{2}^{2}}{\xi _{1}^{2}} \biggl( \int _{0}^{T} v^{2}(t)\,dt \biggr) \biggl( \int _{0}^{T} \biggl( \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds \biggr)^{2}\,dt \biggr) . \end{aligned}$$
(3.5)

Note that \(\theta <\frac{1}{2}\), we have

$$\begin{aligned} \int _{0}^{T}v^{2}(t)\,dt\leq \int _{0}^{T}M^{2}t^{-2\theta }=M_{\theta }. \end{aligned}$$
(3.6)

Put another way, the integral \(\int _{0}^{T}v^{2}(t)\,dt\) is convergent

$$\begin{aligned} &\int _{0}^{T} \biggl( \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds \biggr)^{2}\,dt \\ &\quad \leq \int _{0}^{T} \biggl( \int _{0}^{t} \mathbf{S}^{2}_{n, \alpha }(t-s)\,ds \biggr) \biggl( \int _{0}^{t} \bigl\vert F\bigl(u(s)\bigr) \bigr\vert ^{2}\,ds \biggr)\,dt \\ &\quad \leq \frac{C^{2}_{1}(\mu ,\alpha )}{\lambda _{n}^{2}} \int _{0}^{T} \biggl( \int _{0}^{t} (t-s)^{2\alpha -2}\,ds \biggr) \biggl( \int _{0}^{t} \bigl\vert F\bigl(u(s)\bigr) \bigr\vert ^{2}\,ds \biggr)\,dt \\ &\quad \leq \frac{C^{2}_{1}(\mu ,\alpha )}{\lambda _{n}^{2}} \frac{T^{2\alpha }}{2\alpha -1} \biggl( \int _{0}^{T} \bigl\vert F\bigl(u(s)\bigr) \bigr\vert ^{2}\,ds \biggr). \end{aligned}$$
(3.7)

Conjoining (3.5), (3.6), and (3.7), we arrive at

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(\cdot ,t) \bigr\Vert _{D(A^{s})}^{2} &\leq M_{\theta }C_{1}^{2}( \mu ,\alpha )\frac{\xi _{2}^{2}}{\xi _{1}^{2}} \frac{T^{2\alpha }}{2\alpha -1} \int _{0}^{T}\sum_{n=1}^{\infty } \lambda _{n}^{2s-4} \bigl\vert F\bigl(u(s)\bigr) \bigr\vert ^{2}\,ds \\ &\leq M_{\theta }C_{1}^{2}(\mu ,\alpha ) \frac{\xi _{2}^{2}}{\xi _{1}^{2}}\frac{T^{2\alpha }}{2\alpha -1} \Vert F \Vert ^{2}_{L^{2}(0,T;D(A^{s-2}))}. \end{aligned}$$
(3.8)

In other words, we have

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(\cdot ,t) \bigr\Vert _{D(A^{s})}\leq \sqrt{M_{\theta }C_{1}^{2}( \mu ,\alpha )\frac{\xi _{2}^{2}}{\xi _{1}^{2}} \frac{T^{2\alpha }}{2\alpha -1}} \Vert F \Vert _{L^{2}(0,T;D(A^{s-2}))}=M_{2} \Vert F \Vert _{L^{2}(0,T;D(A^{s-2}))}. \end{aligned}$$
(3.9)

Using Parseval’s equality again, we can estimate the term \(\mathcal{A}_{3}(\cdot ,t)\) as follows:

$$\begin{aligned} \bigl\Vert \mathcal{A}_{3}(\cdot ,t) \bigr\Vert ^{2}_{D(A^{S})}&=\sum_{n=1}^{\infty } \lambda _{n}^{2s} \biggl( \int _{0}^{t} \mathbf{S}_{n}(\alpha ,t-s)F_{n}\bigl(u(s)\bigr)\,ds \biggr)^{2} \\ &\leq C^{2}_{1}(\mu ,\alpha )\sum _{n=1}^{\infty } \lambda _{n}^{2s-2} \biggl( \int _{0}^{t} (t-s)^{2\alpha -2}\,ds \biggr) \biggl( \int _{0}^{t} \bigl\vert F\bigl(u(s)\bigr) \bigr\vert ^{2}\,ds \biggr)\,dt \\ &\leq C^{2}_{1}(\mu ,\alpha ) \frac{T^{2\alpha }}{2\alpha -2} \biggl( \int _{0}^{T}\lambda _{n}^{2s-2} \bigl\vert F\bigl(u(s)\bigr) \bigr\vert ^{2}\,ds \biggr) \\ &\leq C^{2}_{1}(\mu ,\alpha ) \frac{T^{2\alpha }}{2\alpha -2} \Vert F \Vert ^{2}_{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.10)

The latter estimation allows us to deduce that

$$\begin{aligned} \bigl\Vert \mathcal{A}_{3}(\cdot ,t) \bigr\Vert _{D(A^{S})}\leq \sqrt{C^{2}_{1}(\mu , \alpha ) \frac{T^{2\alpha }}{2\alpha -2}} \Vert F \Vert _{L^{2}(0,T;D(A^{s-2}))}=M_{3} \Vert F \Vert _{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.11)

Combining (3.4), (3.9), and (3.11) helps us rest assured that

$$\begin{aligned} \bigl\Vert u(\cdot ,t) \bigr\Vert _{D(A^{s})} \leq \frac{C_{1}(\mu ,\alpha )}{\xi _{1}} \Vert g \Vert _{D(A^{s})}+M_{2} \Vert F \Vert _{L^{2}(0,T;D(A^{s-2}))} + M_{3} \Vert F \Vert _{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.12)

Part ii: In the case of \(\xi _{1}=0\), \(\xi _{2} >0\).

Paserval’s equality gives us

$$\begin{aligned} \bigl\Vert \mathcal{A}_{1}(\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})}&=\sum_{n=1}^{\infty } \biggl( \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr)^{2}\lambda ^{2s}_{n} \vert g_{n} \vert ^{2}. \end{aligned}$$
(3.13)

It should be noted that

$$\begin{aligned} \mathbf{S}_{n,\alpha }(t)\geq \frac{C_{2}(\mu ,\alpha ,t)}{\lambda _{n}} \end{aligned}$$

and

$$\begin{aligned} v(t)\geq Nt^{-\gamma }. \end{aligned}$$

We get

$$\begin{aligned} \bigl\Vert \mathcal{A}_{1}(\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})}&\leq \frac{-2\gamma +2}{\xi _{2}^{2} N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }}\sum_{n=1}^{\infty } \lambda _{n}^{2s} \vert g_{n} \vert ^{2} \\ &\leq \frac{-2\gamma +2}{\xi _{2}^{2} N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$
(3.14)

Therefore, we immediately obtain that

$$\begin{aligned} \bigl\Vert \mathcal{A}_{1}(\cdot ,t) \bigr\Vert _{D(A^{s})} \leq \frac{-\gamma +1}{\xi _{2} NT^{-\gamma +1}} \frac{C_{1}(\mu ,\alpha )}{C_{2}(\mu ,\alpha ,t)} \frac{1}{t^{1-\alpha }} \Vert g \Vert _{D(A^{s})} =M_{4} \Vert g \Vert _{D(A^{s})}. \end{aligned}$$
(3.15)

In terms of \(\mathcal{A}_{2}(\cdot ,t)\), by computations similar to above, we have

$$\begin{aligned} &\bigl\Vert \mathcal{A}_{2}(\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \\ &\quad =\sum_{n=1}^{\infty } \biggl( \frac{\mathbf{S}_{n,\alpha }(t)}{ \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr)^{2}\lambda _{n}^{2s} \biggl( \int _{0}^{T}v(t) \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr)^{2} \\ &\quad \leq \frac{-2\gamma +2}{ N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \\ &\qquad {}\times \sum _{n=1}^{\infty }\lambda _{n}^{2s} \biggl( \int _{0}^{T} v^{2}(t)\,dt \biggr) \biggl( \int _{0}^{T} \biggl( \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds \biggr)^{2}\,dt \biggr) \\ &\quad \leq \frac{M_{\theta }(-2\gamma +2)}{ N^{2}T^{-2\gamma +2}} \frac{C^{4}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{T^{2\alpha -2}}{t^{2-2\alpha }(2\alpha -1)}\sum _{n=1}^{\infty } \lambda _{n}^{2s-2} \biggl( \int _{0}^{T} \bigl\vert F\bigl(u(s)\bigr) \bigr\vert ^{2}\,ds \biggr) \\ &\quad \leq M^{2}_{5} \Vert F \Vert ^{2}_{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.16)

This implies

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(\cdot ,t) \bigr\Vert _{D(A^{s})}\leq M_{5} \Vert F \Vert ^{2}_{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.17)

Connecting (3.15), (3.16), and (3.17) allows us to come to a conclusion that

$$\begin{aligned} \bigl\Vert u(\cdot ,t) \bigr\Vert _{D(A^{s})} \leq M_{4} \Vert g \Vert _{D(A^{s})}+M_{5} \Vert F \Vert ^{2}_{L^{2}(0,T;D(A^{s-1}))}+M_{3} \Vert F \Vert _{L^{2}(0,T;D(A^{s-1}))}. \end{aligned}$$
(3.18)

 □

The problem recovering the initial value

In this section, we are inclined to deliberate on the following problem:

$$ \textstyle\begin{cases} \partial _{t} u-\Delta u- \mu \partial _{t}^{\alpha }\Delta u = F(u),& (x,t) \in \Omega \times (0,T), \\ u(x,t)=0, & x \in \partial {\Omega }, \\ \xi _{2} \int _{0}^{T}v(t) u(x,s)\,ds= g(x),& x \in \Omega . \end{cases} $$
(4.1)

Our primary aim in this part is to recover the initial data \(u(x,0)=V(x)\). To achieve this goal, we firstly go to prove the point that the problem is ill-posed in \(L^{2}(0,T)\). For convenience of readers, we presume that \(F=0\).

From the formula of the mild solution (2.10), we have that

$$\begin{aligned} V(x)=u(x,0)=\sum_{n=0}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}g_{n} \phi _{n}(x). \end{aligned}$$
(4.2)

Theorem 4.1

In the sense of Hadamard, problem (4.1) is ill-posed in the space \(L^{2}(0,T)\) with reference to the case of \(t=0\).

Proof

We ponder on the linear operator \(\mathcal{K}:L^{2}(D)\to L^{2}(D)\) as follows:

$$\begin{aligned} \mathcal{K}V(x)&=\sum_{n=0}^{\infty } \frac{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}{\mathbf{S}_{n,\alpha }(t)} \bigl\langle V(x),\phi _{n}(x) \bigr\rangle \phi _{n}(x) \\ &= \int _{D} m(x,\tau )V(\tau )\,d\tau , \end{aligned}$$
(4.3)

where

$$\begin{aligned} m(x,\tau )=\sum_{n=0}^{\infty } \frac{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}{\mathbf{S}_{n,\alpha }(t)} \phi _{n}(x)\phi _{n}(\tau ). \end{aligned}$$

It is worth mentioning that \(m(x,\tau )=m(\tau ,x)\). Therefore, the operator \(\mathcal{K}\) is self-adjoint and the compactness of \(\mathcal{K}\) is presented as follows.

We define the finite rank operator \(\mathcal{K}_{L}\)

$$\begin{aligned} \mathcal{K}_{L}V=\sum_{n=1}^{L} \frac{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}{\mathbf{S}_{n,\alpha }(t)} \bigl\langle V(x),\phi _{n}(x) \bigr\rangle \phi _{n}(x). \end{aligned}$$
(4.4)

From (4.3) and (4.4), we have

$$\begin{aligned} \Vert \mathcal{K}_{L}V-\mathcal{K}V \Vert ^{2}_{L^{2}(D)}&= \sum_{n=L+1}^{ \infty } \biggl( \frac{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}{\mathbf{S}_{n,\alpha }(t)} \biggr)^{2}\bigl\langle V(x),\phi _{n}(x) \bigr\rangle ^{2} \\ &\leq \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)}t^{2-2 \alpha } \frac{M^{2}T^{-2\theta +2}}{(-\theta +1)^{2}} \sum _{n=L+1}^{ \infty }\bigl\langle V(x),\phi _{n}(x) \bigr\rangle ^{2}. \end{aligned}$$
(4.5)

In that

$$\begin{aligned} \Vert \mathcal{K}_{L}V-\mathcal{K}V \Vert _{L^{2}(D)} \leq \frac{C_{1}(\mu ,\alpha )}{C_{2}(\mu ,\alpha ,t)}t^{1-\alpha } \frac{MT^{-\theta +1}}{(-\theta +1)} \Vert V \Vert _{L^{2}(D)}. \end{aligned}$$
(4.6)

This allows us to conclude that \(\mathcal{K}\) is a compact operator. In conjunction with (4.3), we have

$$\begin{aligned} \mathcal{K}V(x)=g(x). \end{aligned}$$
(4.7)

Combining the latter conclusion and using Kirsch [28], we deduce that the problem of recovering the initial value V from (4.7) is ill-posed. To ensure mathematical clarity, we provide an example as follows. If we choose the input data \(g^{j}(x)=\frac{\phi _{j}(x)}{\lambda _{j}^{1 /2}}\), the \(L^{2}\) norm of \(g^{j}\) is

$$\begin{aligned} \bigl\Vert g^{j} \bigr\Vert _{L^{2}(D)}= \frac{1}{\lambda _{j}^{1/2}}\to 0 \quad \text{when } j\to +\infty , \end{aligned}$$
(4.8)

and the initial data regarding \(g^{j}\) is

$$\begin{aligned} V^{j}(x)&=\sum_{n=1}^{\infty } \frac{\mathbf{S}_{j,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{j,\alpha }(t)\,dt} \bigl\langle g^{j}(x),\phi _{j}(x) \bigr\rangle \phi _{j}(x) \\ &= \frac{\mathbf{S}_{j,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{j,\alpha }(t)\,dt} \frac{\phi _{j}(x)}{\lambda _{j}^{1 /2}}. \end{aligned}$$
(4.9)

In the next step, we assess the initial data \(V^{j}\) in respect of \(L^{2}\) norm

$$\begin{aligned} \bigl\Vert V^{j} \bigr\Vert _{L^{2}(D)}&= \biggl\Vert \frac{\mathbf{S}_{j,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{j,\alpha }(t)\,dt} \frac{\phi _{j}(x)}{\lambda _{j}^{1 /2}} \biggr\Vert _{L^{2}(D)} \\ &\geq \frac{\frac{C_{2}(\mu ,\alpha ,t)}{\lambda _{1}}}{\xi _{2}M_{1,\beta } \frac{C_{1}(\mu ,\alpha )}{1+\lambda _{j}t^{1-\alpha }}} \frac{1}{\lambda _{j}^{1/2}} \\ &\geq \frac{C_{2}(\mu ,\alpha ,t)t^{1-\alpha }}{\xi _{2}M_{1,\beta }\lambda _{1}C_{1}(\mu ,\alpha )} \lambda _{j}^{1/2}\to +\infty \quad \text{when } j\to +\infty . \end{aligned}$$
(4.10)

Conjoining (4.8) with (4.10), we arrive at the conclusion that the solution of problem (4.1) is instable. □

With the aim of putting forward the next theory, we give an assumption as follows. Presume that \(g^{\epsilon } \in L^{p}(D)\) and \(F^{\epsilon } \in L^{\infty }(0,T;L^{p}(D))\) are noisy data, and presume

$$\begin{aligned} \bigl\Vert g-g^{\epsilon } \bigr\Vert _{L^{p}(D)} + \bigl\Vert F-F^{\epsilon } \bigr\Vert _{L^{\infty }(0,T;L^{p}(D))} \leq \epsilon . \end{aligned}$$
(4.11)

Using the Fourier truncation method, we provide a construction of regularized solution to problem (4.1) as follows:

$$\begin{aligned} V^{\epsilon }(x)={}&\sum_{n=1}^{K} \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \bigl\langle g^{\epsilon }(x),\phi _{n}(x)\bigr\rangle \phi _{n}(x), \\ &{}+\sum_{n=1}^{K} \frac{\mathbf{S}_{n,\alpha }(t)}{ \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggl( \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F^{ \epsilon }_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr)\phi _{n}(x). \end{aligned}$$
(4.12)

Theorem 4.2

Let \((g^{\epsilon },F^{\epsilon })\) satisfy (4.11), and let b, δ be such that

$$\begin{aligned} -\frac{d}{4}< b\leq \min \biggl(0,\frac{(q-2)d}{4q}\biggr), \quad 0\leq \delta < \frac{d}{4}. \end{aligned}$$
(4.13)

Suppose that there exists β such that

$$\begin{aligned} V \in D\bigl(A^{\beta }\bigr). \end{aligned}$$

Choose \(K=K(\epsilon )\) such that

$$\begin{aligned} \lim_{\epsilon \to 0}\lambda _{K}^{\delta -b}\epsilon =0,\qquad \lim_{\epsilon \to 0}\lambda _{k}=+\infty . \end{aligned}$$
(4.14)

Then we have that the following estimation holds:

$$\begin{aligned} \bigl\Vert V-V^{\epsilon } \bigr\Vert _{L^{\frac{2d}{d-4\delta }}(\Omega )}\leq C \bigl\Vert V-V^{ \epsilon } \bigr\Vert _{D(A^{\delta })}\leq \lambda _{K}^{\delta -b} \epsilon + \lambda _{K}^{\delta -b}\epsilon +\lambda _{K+1}^{-\gamma } \Vert V \Vert _{D(A^{ \delta +\beta })}. \end{aligned}$$
(4.15)

Remark 4.1

On account of \(\lambda _{K}\sim K^{2/d}\), we need to choose K such that

$$\begin{aligned} K^{\frac{2}{d}(\delta -b)}\epsilon \to 0,\quad \text{when } \epsilon \to 0. \end{aligned}$$

In fact, we choose \(K=\epsilon ^{(r-1)d/2(\delta -b)}\) for \(0< r<1\), and then the error is of order

$$\begin{aligned} \max \bigl( \bigl(\epsilon ^{(r-1)d/2(\eta -b)}+1\bigr)^{\frac{-2\delta }{d}}, \epsilon ^{r} \bigr). \end{aligned}$$

Proof

To begin with, we define the function \(W^{\epsilon }(x)\) in the following way:

$$\begin{aligned} W^{\epsilon }(x)&=\sum_{n=1}^{K} \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \bigl\langle g(x),\phi _{n}(x)\bigr\rangle \phi _{n}(x), \\ &+\sum_{n=1}^{K} \frac{\mathbf{S}_{n,\alpha }(t)}{ \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggl( \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr)\phi _{n}(x). \end{aligned}$$
(4.16)

As the next step, we ponder the term \(\|V^{\epsilon }-V\|_{D(A^{\delta })}\) and present an estimation of it with regard to the \(D(A^{\delta })\) norm for \(\delta >0\). Thanks to the triangle inequality, we obtain

$$\begin{aligned} \bigl\Vert V^{\delta }-V \bigr\Vert _{D(A^{\eta })}\leq \bigl\Vert W^{\delta }-V^{\delta } \bigr\Vert _{D(A^{ \eta })}+ \bigl\Vert W^{\delta }-V \bigr\Vert _{D(A^{\eta })}. \end{aligned}$$
(4.17)

In terms of \(\|W^{\delta }-V^{\delta }\|_{D(A^{\eta })}\), we have

$$\begin{aligned} &\bigl\Vert W^{\epsilon }-V^{\epsilon } \bigr\Vert ^{2}_{D(A^{\delta })} \\ &\quad \leq 2 \sum_{n=1}^{K} \biggl\vert \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr\vert ^{2}\lambda _{n}^{2\delta } \bigl\langle g^{\epsilon }-g,\phi _{n} \bigr\rangle ^{2} \\ &\qquad {}+\sum_{n=1}^{K} \biggl\vert \frac{\mathbf{S}_{n,\alpha }(t)}{ \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr\vert ^{2}\lambda _{n}^{2\delta } \\ &\qquad {}\times \biggl( \int _{0}^{T}v(t) \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s) \bigl(F_{n}^{\epsilon }\bigl(u(s)\bigr)-F_{n}\bigl(u(s) \bigr)\bigr)\,ds\,dt \biggr)^{2}. \end{aligned}$$
(4.18)

From the Sobolev embedding \(L^{q}(\Omega )\hookrightarrow D(A^{b})\) for \(-\frac{d}{4}\leq b<0\) and \(q\geq \frac{2d}{d-4b}\), it should be noted that there exists a positive constant \(C(b,q)\) such that

$$\begin{aligned} \bigl\Vert g^{\epsilon }-g \bigr\Vert _{D(A^{b})} \leq C(b,q) \bigl\Vert g^{\epsilon }-g \bigr\Vert _{L^{q}(D)} \leq C(b,q)\epsilon . \end{aligned}$$
(4.19)

We denote the two terms \(\mathcal{B}_{1}\) and \(\mathcal{B}_{2}\) as follows:

$$\begin{aligned} &\mathcal{B}_{1}=\sum_{n=1}^{K} \biggl\vert \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr\vert \bigl\langle g^{\epsilon }-g, \phi _{n}\bigr\rangle \phi _{n}(x), \\ &\mathcal{B}_{2}=\sum_{n=1}^{K} \biggl\vert \frac{\mathbf{S}_{n,\alpha }(t)}{ \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \\ &\hphantom{\mathcal{B}_{2}=}{}\times \biggr\vert \biggl( \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s) \bigl(F_{n}^{\epsilon }\bigl(u(s)\bigr)-F_{n}\bigl(u(s) \bigr) \bigr)\,ds\,dt \biggr)\phi _{n}(x). \end{aligned}$$
(4.20)

It is worth noting that \(\delta >b\), the term \(\mathcal{B}_{1}\) can be assessed in the following way:

$$\begin{aligned} \Vert \mathcal{B}_{1} \Vert _{D(A^{\delta })}&=\sqrt {\sum_{n=1}^{K} \biggl\vert \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr\vert ^{2}\lambda _{n}^{2\delta -2b} \lambda _{n}^{2b}\bigl\langle g^{ \epsilon }-g,\phi _{n}\bigr\rangle ^{2}} \\ &\leq \frac{-\gamma +1}{\xi _{2} NT^{-\gamma +1}} \frac{C_{1}(\mu ,\alpha )}{C_{2}(\mu ,\alpha ,t)} \frac{1}{t^{1-\alpha }}\lambda _{K}^{\delta -b}\sqrt{\sum_{n=1}^{K} \lambda _{n}^{2b}\bigl\langle g^{\epsilon }-g,\phi _{n}\bigr\rangle ^{2}} \\ &\leq \frac{-\gamma +1}{\xi _{2} NT^{-\gamma +1}} \frac{C_{1}(\mu ,\alpha )}{C_{2}(\mu ,\alpha ,t)} \frac{1}{t^{1-\alpha }}\lambda _{K}^{\delta -b} \bigl\Vert g^{\epsilon }-g \bigr\Vert _{D(A^{b})} \\ &\leq \frac{-\gamma +1}{\xi _{2} NT^{-\gamma +1}} \frac{C_{1}(\mu ,\alpha )}{C_{2}(\mu ,\alpha ,t)} \frac{1}{t^{1-\alpha }}C(b,q)\lambda _{K}^{\delta -b}\epsilon =M_{6} \lambda _{K}^{\delta -b}\epsilon . \end{aligned}$$
(4.21)

Similarly, the term \(\mathcal{B}_{2}\) can be estimated as follows:

$$\begin{aligned} \Vert \mathcal{B}_{2} \Vert ^{2}_{D(A^{\delta })}={}& \sum_{n=1}^{K} \biggl\vert \frac{\mathbf{S}_{n,\alpha }(t)}{ \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr\vert ^{2} \lambda _{n}^{2\delta } \\ &{}\times \biggl( \int _{0}^{T}v(t) \int _{0}^{t} \mathbf{S}_{n,\alpha }(t) \bigl(F_{n}^{\epsilon }\bigl(u(s)\bigr)-F_{n}\bigl(u(s) \bigr) \bigr)\,ds\,dt \biggr)^{2} \\ \leq{}& \frac{-2\gamma +2}{ N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2} (\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \\ &{}\times\sum _{n=1}^{K} \lambda _{n}^{2\delta -2b} \lambda _{n}^{2b} \biggl( \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s) \bigl(F_{n}^{\epsilon }\bigl(u(s)\bigr)-F_{n}\bigl(u(s) \bigr) \bigr)\,ds\,dt \biggr)^{2} \\ \leq{}& \frac{-2\gamma +2}{ N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \bigl\Vert F^{\epsilon }-F \bigr\Vert _{L^{\infty }(0,T;D(A^{b}))} \\ &{}\times \sum _{n=1}^{K} \lambda _{n}^{2\delta -2b} \biggl( \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)\,ds\,dt \biggr)^{2} \\ \leq{}& \frac{-2\gamma +2}{ N^{2}T^{-2\gamma +1}} \frac{C^{4}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \bigl\Vert F^{\epsilon }-F \bigr\Vert _{L^{\infty }(0,T;D(A^{b}))} \sum _{n=1}^{K} \lambda _{n}^{2\delta -2b} \biggl( \int _{0}^{T}v(t)\,dt \biggr)^{2} \\ \leq{}& \frac{-2\gamma +2}{ N^{2}T^{-2\gamma +1}} \frac{C^{4}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \Vert v \Vert ^{2}_{L^{2}(0,T)}\lambda _{K}^{2\delta -2b-2} \bigl\Vert F^{\delta }-F \bigr\Vert _{L^{\infty }(0,T;D(A^{b}))} \\ \leq{}& M^{2}_{7} \lambda _{K}^{2 \delta -2b}\epsilon ^{2}. \end{aligned}$$
(4.22)

The latter estimation and the previous one allow us to have that

$$\begin{aligned} \bigl\Vert W^{\epsilon }-V^{\epsilon } \bigr\Vert _{D(A^{\delta })}\leq \Vert \mathcal{B}_{1} \Vert _{D(A^{ \delta })}+ \Vert \mathcal{B}_{2} \Vert _{D(A^{\delta })}\leq M_{6} \lambda _{K}^{ \delta -b}\epsilon +M_{7} \lambda _{K}^{\delta -b}\epsilon . \end{aligned}$$
(4.23)

Regarding the term \(\|W^{\epsilon }-V\|_{D(A^{\delta })}\), we have

$$\begin{aligned} \bigl\Vert W^{\epsilon }-V \bigr\Vert _{D(A^{\delta })}= \sqrt{\sum_{n=K+1}^{\infty } \lambda _{n}^{2\delta }V_{n}^{2}}=\sqrt{\sum _{n=K+1}^{\infty }\lambda _{n}^{-2 \beta } \lambda _{n}^{2\delta +2\beta }V_{n^{2}}}\leq \lambda _{K+1}^{- \beta } \Vert V \Vert _{D(A^{\delta +\beta })}. \end{aligned}$$
(4.24)

Conjoining (4.23) and (4.24), we come to the conclusion that

$$\begin{aligned} \bigl\Vert V-V^{\epsilon } \bigr\Vert _{D(A^{\delta })}&\leq \bigl\Vert W^{\epsilon }-V^{\epsilon } \bigr\Vert _{D(A^{ \delta })}+ \bigl\Vert W^{\epsilon }-V \bigr\Vert _{D(A^{\delta })} \\ &\leq M_{6}\lambda _{K}^{\delta -b}\epsilon +M_{7} \lambda _{K}^{ \delta -b}\epsilon +\lambda _{K+1}^{-\beta } \Vert V \Vert _{D(A^{\delta +\beta })}. \end{aligned}$$
(4.25)

The Sobolev embedding \(D(A^{\delta })\hookrightarrow L^{\frac{2d}{d-4\delta }}(\Omega ) \) allows us to rest assured that

$$\begin{aligned} \bigl\Vert V-V^{\epsilon } \bigr\Vert _{L^{\frac{2d}{d-4\delta }}(\Omega )}\leq C \bigl\Vert V-V^{ \epsilon } \bigr\Vert _{D(A^{\delta })} \leq \lambda _{K}^{\delta -b}\epsilon + \lambda _{K}^{\delta -b} \epsilon +\lambda _{K+1}^{-\beta } \Vert V \Vert _{D(A^{ \delta +\beta })}. \end{aligned}$$
(4.26)

 □

The regularity of the mild solution to problem (1.1)–(1.2) in the nonlinear case

In this section, we concentrate on examining the subsequent nonlinear problem

$$ \textstyle\begin{cases} \partial _{t} u-\Delta u- \mu \partial _{t}^{\alpha }\Delta u = F(u), & (x,t) \in \Omega \times (0,T), \\ u(x,t)=0, & x \in \partial {\Omega }, \\ \xi _{1}u(x,0) +\xi _{2} \int _{0}^{T}v(t) u(x,s)\,ds = g(x), & x \in \Omega . \end{cases} $$
(5.1)

Theorem 5.1

Suppose that there exists \(Nt^{-\gamma }\leq v(t)\leq Mt^{-\theta }\) for \(\theta <1/2\).

(i) Let \(1/2 < \alpha < 1\), and presume \(0< s<1\). If \(\xi _{1}, \xi _{2}>0\), g belongs to \(D(A^{s})\), and F is a global Lipschitz source function satisfying

$$\begin{aligned} \bigl\Vert F(u)-F(v) \bigr\Vert _{L^{2}(D)}\leq K_{F} \Vert u-v \Vert _{L^{2}(D)} \end{aligned}$$
(5.2)

for T is small enough, then problem (5.1) has a unique solution

$$\begin{aligned} u\in L^{\infty }\bigl(0,T;D\bigl(A^{s}\bigr)\bigr). \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \|u\|_{L^{\infty }(0,T;D(A^{s}))}\leq \frac{C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1} ( 1-M_{9}\sqrt{T}-M_{10}T^{\alpha -1} )} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$

(ii) Let \(1/2 < \alpha < 1\), and presume \(0< s<1\). If \(\xi _{1}=0\), \(\xi _{2}>0\), \(g\in D(A^{s+\alpha })\), and F is a global Lipschitz source function satisfying

$$\begin{aligned} \bigl\Vert F(u)-F(v) \bigr\Vert _{D(A^{\alpha })}\leq K_{F} \Vert u-v \Vert _{D(A^{\alpha })} \end{aligned}$$
(5.3)

for T is small enough, then problem (5.1) has a unique solution

$$\begin{aligned} u \in L^{\infty }\bigl(0,T;D\bigl(A^{s}\bigr)\bigr). \end{aligned}$$

Moreover, we have

$$\begin{aligned} \|u\|_{L^{\infty }(0,T;D(A^{s}))}&\leq \frac{M_{4}}{1-M_{11}\sqrt{T}-M_{10}T^{\alpha -1}} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$

Proof

Part i. Using (2.11) in Sect. 2, we obtain

$$\begin{aligned} u(x,t)=\mathcal{A}_{1}(x,t)-\mathcal{A}_{2}(u) (x,t)+\mathcal{A}_{3}(u) (x,t). \end{aligned}$$
(5.4)

For convenience of calculations in the next steps, we posit

$$\begin{aligned} &\mathcal{A}_{1}(x,t)=\sum_{n=1}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}g_{n} \phi _{n}(x), \\ &\mathcal{A}_{2}(u) (x,t)=\sum_{n=1}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \\ &\hphantom{\mathcal{A}_{2}(u) (x,t)=}{}\times \biggl(\xi _{2} \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(s,u(s)\bigr)\,ds\,dt \biggr)\phi _{n}(x), \\ &\mathcal{A}_{3}(u) (x,t)=\sum_{n=1}^{\infty } \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)F_{n} \bigl(s,u(s)\bigr)\,ds \phi (x), \end{aligned}$$

where

$$\begin{aligned} F_{n}\bigl(x,u(s)\bigr)=\bigl\langle F\bigl(x,s,u(s)\bigr),\phi _{n}(x)\bigr\rangle . \end{aligned}$$
(5.5)

With the purpose of proving the point that nonlinear equation (5.4) has a unique solution in \(L^{\infty }(0,T;D(A^{s}))\), we posit the following mapping:

$$\begin{aligned} &\mathcal{I}:L^{\infty }\bigl(0,T;D\bigl(A^{s}\bigr)\bigr)\to L^{\infty }\bigl(0,T;D\bigl(A^{s}\bigr)\bigr) , \\ &\mathcal{I}w=\mathcal{A}_{1}(x,t)-\mathcal{A}_{2}(w) (x,t)+ \mathcal{A}_{3}(w) (x,t). \end{aligned}$$
(5.6)

Using the triangle inequality again, we have

$$\begin{aligned} \Vert \mathcal{I}w_{1}-\mathcal{I}w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))} \leq{}& \bigl\Vert \mathcal{A}_{2}(w_{1})- \mathcal{A}_{2}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))} \\ &{}+ \bigl\Vert \mathcal{A}_{3}(w_{1})- \mathcal{A}_{3}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.7)

We will show that the mapping \(\mathcal{I}\) is a contraction mapping in \(L^{\infty }(0,T;D(A^{s}))\). Now, we give estimations of the terms \(\|\mathcal{A}_{2}(w_{1})-\mathcal{A}_{2}(w_{2})\|_{L^{\infty }(0,T;D(A^{s}))}\) and \(\|\mathcal{A}_{3}(w_{1})-\mathcal{A}_{3}(w_{2})\|_{L^{\infty }(0,T;D(A^{s}))}\) as follows.

Step 1: In terms of the term \(\|\mathcal{A}_{2}(w_{1})-\mathcal{A}_{2}(w_{2})\|_{L^{\infty }(0,T;D(A^{s}))}\).

Firstly, we have

$$\begin{aligned} & \bigl\Vert \mathcal{A}_{2}(w_{1}) (\cdot ,t)- \mathcal{A}_{2}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \\ &\quad =\sum_{n=1}^{\infty } \biggl( \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr)^{2} \lambda _{n}^{2s} \\ &\qquad {}\times \biggl(\xi _{2} \int _{0}^{T}v(t) \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s) \bigl(F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds\,dt \biggr)^{2}. \end{aligned}$$
(5.8)

It is noteworthy that \(\frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \leq \frac{C_{1}(\mu ,\alpha )}{\xi _{1}}\), we obtain

$$\begin{aligned} &\leq \sum_{n=1}^{\infty } \frac{\xi _{2}^{2}C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1}} \lambda _{n}^{2s} \int _{0}^{T}v^{2}(t)\,dt \\ &\quad {}\times \int _{0}^{T} \biggl( \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s) \bigl(F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds \biggr)^{2}\,dt \\ & \leq \sum_{n=1}^{\infty } \frac{\xi _{2}^{2}C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1}} \lambda _{n}^{2s}M_{ \theta } \\ &\quad {}\times \int _{0}^{T} \biggl( \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s) \bigl(F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds \biggr)^{2}\,dt. \end{aligned}$$
(5.9)

Thanks to the inequality \(\mathbf{S}_{n,\alpha }(t)\leq \frac{C_{1}(\mu ,\alpha )}{1+\lambda _{n}(t-s)^{1-\alpha }}\) (see Lemma (2.1)), we get

$$\begin{aligned} & \int _{0}^{T} \biggl( \int _{0}^{t}\mathbf{S}_{n,\alpha }(t) \bigl(F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds \biggr)^{2}\,dt \\ &\quad \leq \int _{0}^{T} \biggl( \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)\,ds \biggr) \biggl( \int _{0}^{t} \bigl(F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \biggr)\,dt \\ &\quad \leq \frac{C_{1}^{2}(\mu ,\alpha )T^{2\alpha }}{2\alpha -1}\lambda _{n}^{-2} \int _{0}^{T} \bigl(F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds. \end{aligned}$$
(5.10)

On account of \(s<1\), we have that

$$\begin{aligned} &\bigl\Vert \mathcal{A}_{2}(w_{1}) (\cdot ,t)- \mathcal{A}_{2}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \\ &\quad \leq \sum_{n=1}^{\infty } \frac{\xi _{2}^{2}C^{4}_{1}(\mu ,\alpha )T^{2\alpha }M_{\theta }}{\xi ^{2}_{1}(2\alpha -1)} \lambda _{n}^{2s-2} \int _{0}^{T} \bigl(F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \\ &\quad \leq M_{8} \int _{0}^{T}\sum_{n=1}^{\infty } \lambda _{n}^{2s-2} \bigl(F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \\ &\quad \leq M_{8} \lambda _{1}^{2s-2} \int _{0}^{T} \bigl\Vert F\bigl(s,w_{1}(s) \bigr)-F\bigl(s,w_{2}(s)\bigr) \bigr\Vert ^{2}_{L^{2}(D)}\,ds. \end{aligned}$$
(5.11)

Using the global Lipschitz property of F, we arrive at

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(w_{1}) (\cdot ,t)- \mathcal{A}_{2}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} &\leq M_{8} \lambda _{1}^{2s-2}K_{F}^{2} \int _{0}^{T} \bigl\Vert w_{1}(s)-w_{2}(s) \bigr\Vert ^{2}_{L^{2}(D)}\,ds. \end{aligned}$$
(5.12)

The Sobolev embedding \(D(A^{s})\hookrightarrow L^{2}(D)\) allows us to deduce that

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(w_{1}) (\cdot ,t)- \mathcal{A}_{2}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \leq M_{8} \lambda _{1}^{2s-2}K_{F}^{2}C_{s}^{2} \int _{0}^{T} \bigl\Vert w_{1}(s)-w_{2}(s) \bigr\Vert ^{2}_{D(A^{s})}\,ds. \end{aligned}$$
(5.13)

This implies that, for any t, \(0\leq t \leq T\), we have

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(w_{1}) (\cdot ,t)-\mathcal{A}_{2}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \leq M_{8} \lambda _{1}^{2s-2}K_{F}^{2}C_{s}^{2}T \Vert w_{1}-w_{2} \Vert ^{2}_{L^{ \infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.14)

In other words,

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(w_{1}) (\cdot ,t)- \mathcal{A}_{2}(w_{2}) (\cdot ,t) \bigr\Vert _{D(A^{s})} \leq M_{9}\sqrt{T} \Vert w_{1}-w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.15)

We can see that the right-hand side of (5.14) is independent of t, as a result we reach the conclusion that

$$\begin{aligned} \bigl\Vert \mathcal{A}_{2}(w_{1})- \mathcal{A}_{2}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))} \leq M_{9}\sqrt{T} \Vert w_{1}-w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.16)

Step 2: In terms of the term \(\|\mathcal{A}_{3}(w_{1})(\cdot ,t)-\mathcal{A}_{3}(w_{2})(\cdot ,t) \|_{L^{\infty }(0,T;D(A^{s}))}\).

Thanks to Parseval’s equality and Holder’s inequality, we can assess the term \(\|\mathcal{A}_{3}(w_{1})(\cdot ,t)-\mathcal{A}_{3}(w_{2})(\cdot ,t) \|_{L^{\infty }(0,T;D(A^{s}))}\) in the following way:

$$\begin{aligned} &\bigl\Vert \mathcal{A}_{3}(w_{1}) (\cdot ,t)-\mathcal{A}_{3}(w_{2}) (\cdot ,t) \bigr\Vert _{D(A^{s})} \\ & \quad = \sum_{n=1}^{\infty }\lambda _{n}^{2s} \biggl( \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s) \bigl(F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds \biggr)^{2} \\ &\quad \leq \sum_{n=1}^{\infty } \lambda _{n}^{2s} \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)\,ds \int _{0}^{t} \bigl(F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \\ &\quad \leq C_{1}^{2}(\mu ,\alpha ) \sum _{n=1}^{\infty } \lambda _{n}^{2s-2} \int _{0}^{t}(t-s)^{2\alpha -2}\,ds \int _{0}^{t} \bigl(F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds. \end{aligned}$$
(5.17)

It is worth noting that \(\alpha >1/2\), we get

$$\begin{aligned} &\bigl\Vert \mathcal{A}_{3}(w_{1}) (\cdot ,t)- \mathcal{A}_{3}(w_{2}) (\cdot ,t) \bigr\Vert _{D(A^{s})} \\ &\quad \leq C_{1}^{2}(\mu ,\alpha ) \frac{T^{2\alpha -1}}{2\alpha -1}\sum_{n=1}^{ \infty }\lambda _{n}^{2s-2} \int _{0}^{T} \bigl(F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \\ &\quad \leq C_{1}^{2}(\mu ,\alpha ) \frac{T^{2\alpha -1}}{2\alpha -1} \int _{0}^{T} \sum_{n=1}^{\infty } \lambda _{n}^{2s-2} \bigl(F_{n}\bigl(s,w_{1}(s) \bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds. \end{aligned}$$
(5.18)

We can estimate the term \(\int _{0}^{T}\sum_{n=1}^{\infty }\lambda _{n}^{2s-2} (F_{n}(s,w_{1}(s))-F_{n}(s,w_{2}(s)) )^{2}\,ds\) as follows:

$$\begin{aligned} & \int _{0}^{T}\sum_{n=1}^{\infty } \lambda _{n}^{2s-2} \bigl(F_{n}\bigl(s,w_{1}(s) \bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \\ &\quad \leq \lambda _{1}^{2s-2} \int _{0}^{T} \bigl\Vert F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr\Vert ^{2}_{L^{2}(D)}\,ds \\ &\quad \leq \lambda _{1}^{2s-2}K^{2}_{F} \int _{0}^{T} \bigl\Vert w_{1}(s)-,w_{2}(s)\,ds \bigr\Vert ^{2}_{L^{2}(D)}\,ds \\ &\quad \leq \lambda _{1}^{2s-2}K^{2}_{F}T \Vert w_{1}-w_{2} \Vert ^{2}_{L^{\infty }(0,T;D(A^{s}))}, \end{aligned}$$
(5.19)

where we used the global Lipschitz property of F function and note that \(s<1\).

Conjoining the latter estimate and the previous one, we claim that

$$\begin{aligned} \bigl\Vert \mathcal{A}_{3}(w_{1}) (\cdot ,t)- \mathcal{A}_{3}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})}& \leq \frac{T^{2\alpha -2}}{2\alpha -1}\lambda _{1}^{2s-2}K^{2}_{F} \Vert w_{1}-w_{2} \Vert ^{2}_{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$

Put another way,

$$\begin{aligned} \bigl\Vert \mathcal{A}_{3}(w_{1}) (\cdot ,t)- \mathcal{A}_{3}(w_{2}) (\cdot ,t) \bigr\Vert _{D(A^{s})}& \leq M_{10}T^{\alpha -1} \Vert w_{1}-w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}, \end{aligned}$$

this implies that

$$\begin{aligned} \bigl\Vert \mathcal{A}_{3}(w_{1})- \mathcal{A}_{3}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))}& \leq M_{10}T^{\alpha -1} \Vert w_{1}-w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.20)

Combining (5.7), (5.16), and (5.20), we have that

$$\begin{aligned} \Vert \mathcal{I}w_{1}-\mathcal{I}w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}\leq \bigl(M_{9} \sqrt{T}+M_{10}T^{\alpha -1} \bigr) \Vert w_{1}-w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.21)

Suppose that T is small enough such that

$$\begin{aligned} M_{9}\sqrt{T}+M_{10}T^{\alpha -1}< 1, \end{aligned}$$

we can claim that \(\mathcal{I}\) is a contraction mapping in \(L^{\infty }(0,T;D(A^{s})\).

Furthermore, it should be noted that if \(w_{1}=0\), then

$$\begin{aligned} \mathcal{I}w_{1}(x,t)=\mathcal{A}_{1}(x,t). \end{aligned}$$
(5.22)

This implies

$$\begin{aligned} \bigl\Vert \mathcal{I}w_{1}(\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})}&=\sum_{n=1}^{\infty } \biggl( \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{1}+\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggr)^{2}\lambda ^{2s}_{n} \vert g_{n} \vert ^{2} \\ &\leq \frac{C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1}}\sum_{1}^{\infty } \lambda ^{2s}_{n} \vert g_{n} \vert ^{2} \\ &=\frac{C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1}} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$
(5.23)

The latter estimation helps us assert that

$$\begin{aligned} \mathcal{I}w_{1} \in L^{\infty }\bigl(0,T;D\bigl(A^{s} \bigr)\bigr). \end{aligned}$$
(5.24)

On the basis of the Banach fixed point theorem, we claim that problem (5.1) has a unique solution belonging to the \(L^{\infty }(0,T;D(A^{s}))\) space. In addition, if we take \(w=0\), we have that

$$\begin{aligned} \|u\|_{L^{\infty }(0,T;D(A^{s}))}&\leq \Vert \mathcal{I}u-\mathcal{I}w \Vert _{L^{ \infty }(0,T;D(A^{s}))}+ \Vert \mathcal{I}w \Vert _{L^{\infty }(0,T;D(A^{s}))} \\ &\leq \bigl(M_{9}\sqrt{T}+M_{10}T^{\alpha -1} \bigr) \Vert u \Vert _{L^{\infty }(0,T;D(A^{s}))} +\frac{C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1}} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$
(5.25)

It is worth noting that \(M_{9}\sqrt{T}+M_{10}T^{\alpha -1}<1\), we get

$$\begin{aligned} \|u\|_{L^{\infty }(0,T;D(A^{s}))}\leq \frac{C^{2}_{1}(\mu ,\alpha )}{\xi ^{2}_{1} ( 1-M_{9}\sqrt{T}-M_{10}T^{\alpha -1} )} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$
(5.26)

We break the back of the proof of Part i.

Part ii: From (2.11) in Sect. 2, we get

$$\begin{aligned} u(x,t)=\mathcal{B}_{1}(x,t)+\mathcal{B}_{2}(u) (x,t)+\mathcal{B}_{3}(u) (x,t), \end{aligned}$$
(5.27)

where \(\mathcal{B}_{j}\) (\(j=1,2,3\)) are defined as follows:

$$\begin{aligned} &\mathcal{B}_{1}(x,t)=\sum_{n=1}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\xi _{2} \int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt}g_{n} \phi _{n}(x), \\ &\mathcal{B}_{2}(u) (x,t)=\sum_{n=1}^{\infty } \frac{\mathbf{S}_{n,\alpha }(t)}{\int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t)\,dt} \biggl( \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds\,dt \biggr)\phi _{n}(x), \\ &\mathcal{B}_{3}(u) (x,t)=\sum_{n=1}^{\infty } \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s)F_{n} \bigl(u(s)\bigr)\,ds \phi (x), \end{aligned}$$

and

$$\begin{aligned} F_{n}\bigl(x,u(s)\bigr)=\bigl\langle F\bigl(x,s,u(s)\bigr),\phi _{n}(x)\bigr\rangle . \end{aligned}$$

Firstly, we posit the following function:

$$\begin{aligned} &\mathcal{J}:L^{\infty }\bigl(0,T;D\bigl(A^{s} \bigr)\bigr)\to L^{\infty }\bigl(0,T;D\bigl(A^{s}\bigr)\bigr) , \\ &\mathcal{J}w=\mathcal{B}_{1}(x,t)+\mathcal{B}_{2}(w) (x,t)+ \mathcal{B}_{3}(w) (x,t). \end{aligned}$$
(5.28)

Similar to Part i, we go to prove the point that \(\mathcal{J}\) is a contraction mapping in \(L^{\infty }(0,t;D(A^{s}))\). In view of the triangle inequality, we have

$$\begin{aligned} \Vert \mathcal{J}w_{1}-\mathcal{J}w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))} \leq{}& \bigl\Vert \mathcal{B}_{2}(w_{1})- \mathcal{B}_{2}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))} \\ &{}+ \bigl\Vert \mathcal{B}_{3}(w_{1})- \mathcal{B}_{3}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.29)

The term \(\|\mathcal{B}_{2}(w_{1})-\mathcal{B}_{2}(w_{2})\|_{L^{\infty }(0,T;D(A^{s}))}\) can be estimated in the following way:

$$\begin{aligned} & \bigl\Vert \mathcal{B}_{2}(w_{1}) (\cdot ,t)- \mathcal{B}_{2}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \\ &\quad =\sum_{n=1}^{\infty } \frac{\mathbf{S}^{2}_{n,\alpha }(t)}{ (\int _{0}^{T}v(t)\mathbf{S}_{n,\alpha }(t) )^{2}} \lambda _{n}^{2s} \\ &\qquad {}\times \biggl( \int _{0}^{T}v(t) \int _{0}^{t}\mathbf{S}_{n, \alpha }(t-s) \bigl( F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds\,dt \biggr)^{2}. \end{aligned}$$
(5.30)

It is important to note that \(\mathbf{S}_{n,\alpha }(t)\geq \frac{C_{2}(\mu ,\alpha ,t)}{\lambda _{n}}\) and \(v(t)\geq Nt^{-\gamma }\), we get

$$\begin{aligned} & \bigl\Vert \mathcal{B}_{2}(w_{1}) (\cdot ,t)- \mathcal{B}_{2}(w_{2}) (\cdot ,t) \bigr\Vert ^{2}_{D(A^{s})} \\ &\quad \leq \frac{-2\gamma +2}{ N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }}\sum _{n=1}^{\infty }\lambda _{n}^{2s} \\ &\qquad {}\times \biggl( \int _{0}^{T} v^{2}(t)\,dt \biggr) \biggl( \int _{0}^{T} \biggl( \int _{0}^{t} \mathbf{S}_{n,\alpha }(t-s) \bigl( F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds \biggr)^{2}\,dt \biggr) \\ &\quad \leq \frac{M^{2}_{\theta }(-2\gamma +2)}{ N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1} (\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \\ &\qquad {}\times \sum _{n=1}^{\infty }\lambda _{n}^{2s} \biggl( \int _{0}^{T} \biggl( \int _{0}^{t}\mathbf{S}_{n,\alpha }(t-s) \bigl( F_{n}\bigl(s,w_{1}(s)\bigr)-F_{n} \bigl(s,w_{2}(s)\bigr) \bigr)\,ds \biggr)^{2}\,dt \biggr) \\ &\quad \leq \frac{M^{2}_{\theta }(-2\gamma +2)}{ N^{2}T^{-2\gamma +2}} \frac{C^{4}_{1} (\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{T^{2\alpha -1}}{t^{2-2\alpha }(2\alpha -1)}\lambda _{1}^{-2} \\ &\qquad {}\times \sum_{n=1}^{\infty } \lambda _{n}^{2s} \int _{0}^{T} \bigl( F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \\ &\quad \leq \frac{M^{2}_{\theta }(-2\gamma +2)}{ N^{2}T^{-2\gamma +2}} \frac{C^{4}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{T^{2\alpha -1}}{t^{2-2\alpha }(2\alpha -1)}\lambda _{1}^{-2} \\ &\qquad {}\times \int _{0}^{T}\sum_{n=1}^{\infty } \lambda _{n}^{2s} \bigl( F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr)^{2}\,ds \\ &\quad \leq \frac{M^{2}_{\theta }(-2\gamma +2)}{ N^{2}T^{-2\gamma +2}} \frac{C^{4}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{T^{2\alpha -1}}{t^{2-2\alpha }(2\alpha -1)}\lambda _{1}^{-2} \\ &\qquad {}\times \int _{0}^{T} \bigl\Vert F_{n} \bigl(s,w_{1}(s)\bigr)-F_{n}\bigl(s,w_{2}(s)\bigr) \bigr\Vert ^{2}_{D(A^{s})}\,ds \\ &\quad \leq \frac{M^{2}_{\theta }(-2\gamma +2)}{ N^{2}T^{-2\gamma +2}} \frac{C^{4}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{T^{2\alpha -1}}{t^{2-2\alpha }(2\alpha -1)}\lambda _{1}^{-2}K^{2}_{F} \int _{0}^{T} \bigl\Vert w_{1}(s)-w_{2}(s) \bigr\Vert ^{2}_{D(A^{s})}\,ds \\ &\quad \leq M^{2}_{11}T \bigl\Vert w_{1}(s)-w_{2}(s) \bigr\Vert ^{2}_{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.31)

In that,

$$\begin{aligned} \bigl\Vert \mathcal{B}_{2}(w_{1})- \mathcal{B}_{2}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))} \leq M_{11}\sqrt{T} \bigl\Vert w_{1}(s)-w_{2}(s) \bigr\Vert ^{2}_{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.32)

With regard to the term \(\|\mathcal{B}_{3}(w_{1})-\mathcal{B}_{3}(w_{2})\|_{L^{\infty }(0,T;D(A^{s}))}\), by a similar calculation in Part i, we immediately obtain

$$\begin{aligned} \bigl\Vert \mathcal{B}_{3}(w_{1})- \mathcal{B}_{3}(w_{2}) \bigr\Vert _{L^{\infty }(0,T;D(A^{s}))}& \leq M_{10}T^{\alpha -1} \Vert w_{1}-w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.33)

Conjoining (5.29), (5.32), and (5.33), we have

$$\begin{aligned} \Vert \mathcal{J}w_{1}-\mathcal{J}w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}\leq \bigl(M_{11} \sqrt{T}+M_{10}T^{\alpha -1} \bigr) \Vert w_{1}-w_{2} \Vert _{L^{\infty }(0,T;D(A^{s}))}. \end{aligned}$$
(5.34)

Suppose that there exists T small enough such that

$$\begin{aligned} M_{11}\sqrt{T}+M_{10}T^{\alpha -1}< 1, \end{aligned}$$
(5.35)

we arrive at the conclusion that \(\mathcal{J}\) is a contraction mapping in the space \({L^{\infty }(0,T;D(A^{s}))}\). Moreover, it is noteworthy that if \(w_{1}=0\), then

$$\begin{aligned} \mathcal{J}w_{1}(x,t)=\mathcal{B}_{1}(x,t). \end{aligned}$$

In conjunction with (3.4), we have

$$\begin{aligned} \bigl\Vert \mathcal{J}w_{1}(\cdot ,t) \bigr\Vert ^{2}_{L^{\infty }(0,T;D^{(}A^{s}))} \leq \frac{-2\gamma +2}{\xi _{2}^{2} N^{2}T^{-2\gamma +2}} \frac{C^{2}_{1}(\mu ,\alpha )}{C^{2}_{2}(\mu ,\alpha ,t)} \frac{1}{t^{2-2\alpha }} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$

In other words,

$$\begin{aligned} \bigl\Vert \mathcal{J}w_{1}(\cdot ,t) \bigr\Vert ^{2}_{L^{\infty }(0,T;D^{(}A^{s}))} \leq M_{4} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$

The latter estimation allows us to deduce that if \(w_{1}(x,t)\in L^{\infty }(0,T;D(A^{s}))\), then

$$\begin{aligned} \mathcal{J}w_{1}(x,t) \in L^{\infty }\bigl(0,T;D \bigl(A^{s}\bigr)\bigr). \end{aligned}$$

Thanks to the Banach fixed point theorem, we come to the conclusion that \(\mathcal{J}w=w\) has a unique solution \(w \in L^{\infty }(0,T;D(A^{s}))\). Furthermore, in a similar way to Part i, we take \(w=0\), and we immediately get

$$\begin{aligned} \|u\|_{L^{\infty }(0,T;D(A^{s}))}&\leq \Vert \mathcal{I}u-\mathcal{I}w \Vert _{L^{ \infty }(0,T;D(A^{s}))}+ \Vert \mathcal{I}w \Vert _{L^{\infty }(0,T;D(A^{s}))} \\ &\leq \bigl(M_{11}\sqrt{T}+M_{10}T^{\alpha -1} \bigr) \Vert u \Vert _{L^{\infty }(0,T;D(A^{s}))} +M_{4} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$

Put another way,

$$\begin{aligned} \|u\|_{L^{\infty }(0,T;D(A^{s}))}&\leq \frac{M_{4}}{1-M_{11}\sqrt{T}-M_{10}T^{\alpha -1}} \Vert g \Vert ^{2}_{D(A^{s})}. \end{aligned}$$
(5.36)

We get the proof out of the way. □

Conclusion

In this paper, we examined the fractional nonlinear Rayleigh–Stokes equation under nonlocal integral conditions. The existence and uniqueness are considered using the Fourier truncation method. The convergence rate between the obtained solution and the regularized solution is demonstrated.

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Acknowledgements

The authors would like to thank for the support from the National Research Foundation of Korea under grant number NRF-2020K1A3A1A05101625 and from the Institute of Construction and Environmental Engineering at Seoul National University. The authors also would like to thank the handling editor and two anonymous referees for their valuable and constructive comments to improve our manuscript.

Funding

This research was funded by the National Research Foundation of Korea under grant number NRF-2020K1A3A1A05101625 and received the support from the Institute of Construction and Environmental Engineering at Seoul National University.

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Luc, N.H., Long, L.D., Van, H.T.K. et al. A nonlinear fractional Rayleigh–Stokes equation under nonlocal integral conditions. Adv Differ Equ 2021, 388 (2021). https://doi.org/10.1186/s13662-021-03545-z

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Keywords

  • Fractional Rayleigh–Stokes equation
  • Ill-posed problem
  • Regularization
  • Existence
  • Uniqueness
  • Convergence estimation