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Determination of source term for the fractional Rayleigh–Stokes equation with random data
Journal of Inequalities and Applications volume 2019, Article number: 308 (2019)
Abstract
In this article, we consider the problem of finding a source term of a Rayleigh–Stokes equation. Our problem is not well-posed in the sense of Hadamard. The sought solution does not depend continuously on the given data. Using the truncation method and some new techniques on trigonometric estimators, we give the regularized solution. Moreover, the mean square error and convergence rates are established.
1 Introduction
In the future, there will be a large number of applications of fractional diffusion equations, especially in physics, environment, and some other areas [1,2,3,4,5]. Not only do these applications appear in fluid flow and heat conduction, but they are also appropriate to other topics in mathematics. There are many authors who have studied fractional partial differential equations and ordinary differential equations. Now, we can describe some interesting papers as follows. In [6], the authors considered a class of time-fractional reaction-diffusion equations with nonlocal boundary condition. In 2018, Yong et al. [7] studied Duhamel’s formula for time-fractional Schrödinger equations. The existence and Hölder continuity of solutions for time-fractional Navier–Stokes equations have been studied by Zhou, Peng, and Huang [8]. In this paper, we study the following Rayleigh–Stokes problem:
Here, \(\varOmega\subset\mathbb{R}^{d}\) (\(d=1,2,3\)) is a smooth domain with the boundary ∂Ω, and \(\mathcal {T}_{0}>0\) is a given time. \(u(\mathbf {t},\mathbf {x})\) is the velocity and d is constant with respect to x and t, where x is the distance, t is the time, and \(\partial_{\mathbf {t}}^{\gamma}\) denotes the Riemann–Liouville derivative of order \(\gamma\in(0,1)\) [3, 9]. Our main goal is identifying the source term F if we know the following final value data:
where the source function \(F=\mathbf{F}(\mathbf {t},\mathbf {x}) = \mathcal{Q}(\mathbf {t})f(\mathbf {x})\), and \(\mathcal{Q}(\mathbf {t})\) is known in advance. It is clear that our problem as above is ill-posed by the meaning of Hadamard; in other words, there is not always the existence of a solution. In case the problem has a solution, then a small noise of an exact measurement can imply that the sought solution has large error. Hence, numerical computation is troublesome. So it is essential to have a regularization.
The Rayleigh–Stokes equation (1) is a principal part in the description of dynamic fluids [10]. More applications for such an equation can be found in [10, 11]. The initial and boundary value problems for the Rayleigh–Stokes problem, called direct problems, have already been researched in [10]. In some previous papers, Dehghan et al. [12,13,14,15] considered some numeral solutions of the Rayleigh–Stokes problem. The initial value problem for the Rayleigh–Stokes equation has been studied by applying plenty of numeral methods such as the finite element method, etc. [10, 16].
If the errors coming from unmanageable causes such as wind, rain, humidity, etc. appear, then the model will be considered to be random. The random situations can not directly use the approaches applying for deterministic cases. Sometimes, it is difficult to understand in calculating owing to the random noise. In some real cases, empirical measurements result in function \(\mathbf{h}(\mathbf {x})\), which can be examined with many errors. When the function \(\mathbf{h}(\mathbf {x})\) is measured at fixed points \(\mathbf {x}_{\mathbf {k}}\in\varOmega\), we can collect a set of \(\mathbf{G}(\mathbf {x}_{\mathbf {k}})\), where \(\mathbf{G}(\mathbf {x}_{\mathbf {k}}) \approx \mathbf{h}(\mathbf {x}_{\mathbf {k}}) \). The errors regularly appear in observing practical measurements. Hence
where \(\varepsilon_{\mathbf {k}}\), \(\mathbf {k}=1,\ldots, \mathfrak {M}\), are unknown independent random errors.
The points \(\mathbf {x}_{\mathbf {k}}\), \(\mathbf {k}= \overline{1,\mathfrak {M}}\), which are non-random, are called design points. We choose
and \(\mathcal{D}= ( \mathbf{G}(\mathbf {x}_{1}),\mathbf{G}(\mathbf {x}_{2}),\ldots, \mathbf {G}(\mathbf {x}_{\mathfrak {M}}) )\), which is the measure of
Let us consider the random model as follows:
Let us assume that \(\varepsilon_{\mathbf {k}}\sim\mathcal{N} (0,1) \) is normally random variables, and the unknown errors \(\sigma_{\mathbf {k}}\), \(\mathbf {k}=1,\ldots, \mathfrak {M}\), are unknown independent noises. As we know, these unknown errors can come from many troubles as measuring environment or instrument, where \(\sigma_{\mathbf {k}}\le R_{\mathrm{max}} \), \(\mathbf {k}=1,\ldots, \mathfrak {M}\), with \(R_{\mathrm{max}}\) being the greatest possible error bound when measuring.
To determine the source value \(f(\mathbf {x})\), we get the following assumptions:
- (a)
\(f(\mathbf {x}) \in L^{2}(\varOmega)\) and \(\mathbf{h}(\mathbf {x}) \in L^{2}(\varOmega)\).
- (b)
P is a positive constant which is the a priori bound of function \(f(\mathbf {x})\)
$$\begin{aligned} \Vert f \Vert _{\mathcal{H}^{\beta}(\varOmega )}\leq\mathbf{P} , \quad\beta> 0. \end{aligned}$$(4) - (c)
The measured data \(\mathcal{G}(x_{k})\) and the function \(\mathbf {h}(\mathbf {x})\) have a relation
$$\begin{aligned} \mathcal{G}(\mathbf {x}_{\mathbf {k}})=\mathbf{h}(\mathbf {x}_{\mathbf {k}})+ \sigma_{\mathbf {k}}\varepsilon_{\mathbf {k}}. \end{aligned}$$
Now, we mention the aim and our methods of this paper. Section 2 and Sect. 3 provide the results to be used in the sequel. In Sect. 4, we obtain some estimations for regularity.
2 Preliminaries
First, we present the definitions of fractional operators and some notations.
Definition 2.1
For given function f, the following formula
is called the Riemann–Liouville integral of order \(\gamma>0\). Here, \(\varGamma(\cdot)\) stands for the gamma function.
Next we remind the \(L^{2} ( \varOmega) \) space. The Neumann–Laplacian operator is defined by
By the spectral theory of the positive elliptic operator, the eigenvalues of \(\mathcal{A}\) are \(\lambda_{\mathfrak {n}}= \mathfrak {n}^{2} \). We denote the corresponding eigenfunctions by \(\phi_{\mathfrak {n}} ( \mathbf {x})=\sqrt{\frac{2}{\pi}}\sin(\mathfrak {n}\mathbf {x})\). Thus the eigenpairs \(( \lambda_{\mathfrak {n}},\phi_{\mathfrak {n}} )\), \(\mathfrak {n}\in\mathbb{Z}^{+}\), satisfy
The sequence \(\{ \phi_{\mathfrak {n}} \}_{\mathfrak {n}\in\mathbb{Z}^{+}}\) is an orthonormal basis of \(L^{2}(\varOmega)\).
It admits the eigenvalues
with \(\lambda_{\mathfrak {n}}\to\infty\) as \(\mathfrak {n}\to\infty\) (see [12]). The corresponding eigenfunctions \(\phi_{\mathfrak {n}}\in H_{0}^{1}(\varOmega)\).
For any \(p\geq0\), denote the space
We recall that \(\mathcal{H}(\varOmega)\) is a Banach space with the following norm:
Now, we get the following lemmas.
Lemma 2.1
Let \(\mathfrak {n}=1, \ldots, \mathfrak {M}-1\), with \(\mathbf {x}_{\mathbf {k}}=\pi\frac{2\mathbf {k}-1}{2\mathfrak {M}}\)and \(\phi_{\mathfrak {n}} ( \mathbf {x}_{\mathbf {k}})=\sqrt{\frac{2}{\pi}}\sin(\mathfrak {n}\mathbf {x}_{\mathbf {k}})\), then for all \(\mathfrak{m}=1,2,\dots\), we have
If \(\mathfrak{m}= 1, \ldots, \mathfrak {M}-1\), we obtain
and
Lemma 2.2
Let \(\mathfrak {n},\mathfrak {M}\in\mathbb{Z}^{+}\)such that \(\mathfrak {n}=1, \ldots, \mathfrak {M}-1\). Assume thathis piecewise \(C^{1}(\overline{\varOmega})\) \(\mathbf {x}_{\mathbf {k}}=\pi\frac{2\mathbf {k}-1}{2\mathfrak {M}}\)and \(\phi_{\mathfrak {n}} ( \mathbf {x}_{\mathbf {k}})=\sqrt{\frac{2}{\pi}}\sin(\mathfrak {n}\mathbf {x}_{\mathbf {k}})\), then
where
Proof
We will construct the discretization form of the Fourier coefficients. The function h can be written as follows:
where \(\mathbf{h}_{\mathfrak{m}}= \langle\mathbf{h}(\mathbf {x}),\phi _{\mathfrak{m}}(\mathbf {x}) \rangle\). This implies that
Using Lemma 2.1, we get
So the conclusion is completed. □
3 Mild solution of backward in time problem for Rayleigh–Stokes problem
In this part, we establish a representation for the solution of problem (1). The solution u is given by Fourier series as follows:
where
The reference [1] implies that
where \(\mathbf{S}_{\mathfrak {n}}(\mathbf {t},\gamma) \) is as follows:
where \(\mathcal{L}\) is the function with Laplace transform.
This implies that
Lemma 3.1
The function \(\mathbf{S}_{\mathfrak {n}}(\mathbf {t},\gamma)\), \(\mathfrak {n}=1,2,\dots\), is equal to
where
Proof
See the proof in [17]. □
From Lemma 3.1, we derive some estimates.
Lemma 3.2
Assume that \(\gamma\in(0, 1)\). For all \(\mathbf {t}\in[0,\mathcal {T}_{0}]\), the following estimates hold:
There exists \(\mathcal{B}_{1}(d,\mathcal {T}_{0},\gamma)>0\)such that
$$ \mathbf{S}_{\mathfrak {n}}(\mathcal {T}_{0},\gamma) \ge\frac{\mathcal{B}_{1}(d,\mathcal {T}_{0},\gamma )}{\lambda _{\mathfrak {n}}}. $$(13)There exists \(\mathcal{B}_{2}(d,\gamma)>0\)such that
$$ \mathbf{S}_{\mathfrak {n}}(t,\gamma) \le\frac{\mathcal{B}_{2}(d,\gamma)}{1+\lambda _{\mathfrak {n}}\mathbf {t}^{1-\gamma}} ,\quad 0 \le \mathbf {t}\le \mathcal {T}_{0}, $$(14)where
$$ \mathcal{B}_{1}(d,\mathcal {T}_{0},\gamma)=\frac{d \sin(\gamma\pi)e^{-\mathcal {T}_{0}}}{3\pi (\gamma+1)(d^{2} + 1 + \frac{1}{\lambda _{1}^{2}})} ,\qquad \mathcal{B}_{2}(d,\gamma)=\frac{\varGamma(1-\gamma)}{d \pi\sin (\gamma\pi)}+1. $$(15)
Proof
Using the inequality \(({a}_{1} + {a}_{2} + {a}_{3})^{2} \le3({a}_{1}^{2}+ {a}_{2}^{2}+ {a}_{3}^{2})\) for any real numbers \({a}_{1}\), \({a}_{2}\), \({a}_{3}\), we obtain
Hence
Furthermore, we get
This gives that
Note that
This implies that, for \(\mathbf {t}>0\),
where \(\xi=y\mathbf {t}\), and we have the fact that
Hence
Since \(0< \mathbf{S}_{\mathfrak {n}}(\mathbf {t},\gamma) \le1 \) (see Theorem 2.2, [17]), so we imply from that
Therefore
□
Lemma 3.3
Let \(\mathcal{Q}:[\mathcal {T}_{0},0]\rightarrow\mathbb{R}\)be a positive continuous function. Then we have, for all \(\mathfrak {n}\in\mathbb{N}\),
where \(\inf_{\mathbf {t}\in[ 0,\mathcal {T}_{0}]} \vert\mathcal{Q}(\mathbf {t}) \vert =\mathcal{Q}_{0} >0\). Set \(\Vert \mathcal{Q} \Vert _{\infty }=\sup_{\mathbf {t}\in[ 0,\mathcal {T}_{0}]} \vert\mathcal{Q}(\mathbf {t}) \vert\).
Proof
As in the proof of Lemma 3.2, we obtain
This implies that
and
The proof is completed. □
4 Main results
The first result is given as follows.
Lemma 4.1
Assume \(0<\mathcal {N}< \mathfrak {M}\), with \(\mathcal {N}\in\mathbb{N}\). Lethhold in Lemma 2.2. Under the conditions stated above, we have
Proof
First, we have the following equality.
Substituting t by \(\mathcal {T}_{0}\) into equation (12), by the supplementary condition \(u(\mathcal {T}_{0},\mathbf {x})=\mathbf{h}(\mathbf {x})\) and \(u(0,\mathbf {x})=0\), we have
where \(\langle F(\mathfrak {s},\mathbf {x}),\phi_{\mathfrak {n}}(\mathbf {x}) \rangle= \langle\mathcal {Q}(\mathfrak {s})f(\mathbf {x}),\phi_{\mathfrak {n}}(\mathbf {x}) \rangle=\mathcal{Q}(\mathfrak {s}) \langle f(\mathbf {x}),\phi_{\mathfrak {n}}(\mathbf {x}) \rangle\).
The source function f is given by
Using Lemma 3.2, we have
The proof is completed. □
4.1 The ill-posedness of the problem
In order to illustrate the ill-posedness of the backward problem through an example, let \(h=0\) and \(h(\mathbf {x}_{\mathbf {k}})=0\). This implies \(f=0\), and define function \(u_{\mathfrak {M}}(\mathcal {T}_{0}, \mathbf {x}_{\mathbf {k}})=\frac{1}{\sqrt{\mathfrak {M}}}\varepsilon_{\mathbf {k}}\). First, we have
Let \(f_{\mathfrak {M}}\) be the source function of the problem
Moreover, by the Parseval equality, we have
Using \(\mathbb{E} ( \varepsilon_{j} \varepsilon_{l} )=0\) (\(j\neq l \)), \(\mathbb{E} ( \varepsilon_{j} )=0\) (\(j = 1,2,\ldots, \mathfrak {n}\)), and Lemma 2.1, we can deduce that
Then
We have
where
We get that \(\widetilde{\mathcal{G}}_{\mathfrak {n},\mathfrak {M}}=0\) for \(\mathfrak {n}> \mathfrak {M}\). Applying Lemma 2.1, we obtain
By the Parseval equality, we obtain
Applying \(\mathbb{E} ( \varepsilon_{j} \varepsilon_{l} )=0\) (\(j\neq l \)), \(\mathbb{E} ( \varepsilon_{j} )=0\) (\(j = 1,2,\ldots, \mathfrak {n}\)) with Lemma 2.1, we have
Using Lemma 3.3, we get
Then
From the above argument, we can deduce that problem (1) is not well-posed. Hence, a regularization method is necessary.
4.2 Regularization and convergence rate under a priori bounded condition
We impose the following a priori bound on the exact solution \(f(\mathbf {x})\).
Theorem 4.1
Suppose \(f \in\mathcal{H}^{\beta}(\varOmega)\) and there exists \(\mathbf{P}>0\) such that
then we have
Proof
We have
Thanks to Lemma 3.3, we obtain
□
4.2.1 Error estimate in \(L^{2}(\varOmega)\)
Theorem 4.2
Let \(\epsilon>0\)and \(\epsilon_{k} \sim N(0,1)\)with \(\mathbf {k}= 1,\ldots ,\mathfrak {M}\)if the function \(f(\mathbf {x})\)satisfies the prior bounded condition (4).
A regularized function \(\widetilde{f}_{\mathfrak {M},\mathcal {N}}\)is as follows:
where \(\mathfrak {M}\), \(\mathcal {N}\)are called regularization parameters. Then the error estimate between the exact source and its regularized source is as follows:
Let \(\mathcal {N}:= \mathcal {N}_{\mathfrak {M}}\)such that \(0<\mathcal {N}:= \mathcal {N}_{\mathfrak {M}}<\mathfrak {M}\)and
then
Remark 4.1
By choosing \(\mathcal {N}:= \mathfrak {M}^{\frac{1}{5+2 \beta}}\) and by (37), we can conclude that
Proof of Theorem 4.2
First, we have the following estimate:
This follows from the Parseval identity
The fact that \(\mathbb{E} ( \varepsilon_{j} \varepsilon_{l} )=0\) (\(l\neq j \)), and \(\mathbb{E} ( \varepsilon_{j} )=0\) (\(j = \overline{1,\mathfrak {n}} \)). So we can deduce that
We have
By equation (25), we know that, for \(\mathfrak {n}\ge1\),
Using the last two equations, we get
By \(1=\mathfrak {n}^{-2\beta} \mathfrak {n}^{2\beta}\), we can rewrite \(\mathbb{I}_{1}\) as follows:
In the last series (42), since \(\mathfrak {n}\geq \mathcal {N}+1 > \mathcal {N}\), we get \(\mathfrak {n}^{-2\beta}\leq \mathcal {N}^{-2\beta}\). Using the last two observations, we obtain
We shall begin with showing that
Recall the definition of \(\mathbb{I}_{2}\) in equation (40)
We invoke Lemma 3.3 to deduce that
where \(\sum_{l=1}^{\infty}\frac{1}{l^{2}}=\frac{\pi^{2}}{6}\). We can now combine the results of Lemma 2.2 and equation (47) to obtain
Since \(\sigma_{k} < R_{\mathrm{max}} \), we estimate \(\mathbb{I}_{2}\) as follows:
Combining equation (45) with equation (50), we obtain
It is shown that our main results are stated and proved. □
4.2.2 Error estimate in \(\mathcal{H}^{\beta}\)
Now, we give error estimate between the source and the regularized source in higher Sobolev spaces. For any \(\beta>0\), the convergence rate in \(\mathcal{H}^{\beta}\) norm is as follows.
Theorem 4.3
Assume that \(f(\mathbf {x}) \in\mathcal{H}^{\beta+ \mathfrak{m}}\)for any \(\mathfrak{m}> 0\). Then
Proof
We have
Take the expectation to both sides and use Theorem 4.2. The fact that \(\mathbb{E} ( \varepsilon_{j} \varepsilon_{l} )=0\) (\(j\neq l \)), and \(\mathbb{E} ( \varepsilon_{j} )=0\) (\(j = \overline{1,\mathfrak {n}} \)). Now we can deduce that
First, using equation (25), we get
We can now proceed analogously to the proof of Theorem 4.2, we continue to estimate the error \(\widetilde{\mathbb{I}}_{2}\):
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Binh, T.T., Baleanu, D., Luc, N.H. et al. Determination of source term for the fractional Rayleigh–Stokes equation with random data. J Inequal Appl 2019, 308 (2019). https://doi.org/10.1186/s13660-019-2262-9
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DOI: https://doi.org/10.1186/s13660-019-2262-9