Conclusions and Future Work

Abstract

The control theory of fractional diffusion systems constitutes one of the next big challenges of the engineering community. They will require cooperation of multidisciplines, such as mathematical modeling, engineering applications, and information sciences in order to be successful. Among those challenges, the notation of regional anaysis when the system under consideration is only studied on a subset of the whole space has to be one of the most significance to be explored, simply because that focusing on the regional anaysis would allow for a reduction in the number of physical actuators , offer the potential to reduce computational requirements in some cases, and also possible to discuss those systems which are not controllable on the whole domain, etc. So in this book, instead of analyzing a system by purely theoretical viewpoint, we study the time fractional diffusion system by using the notions of sensors and actuators , which allow us to understand the system better and consequently enable us to steer the real-world system in a better way.

Appendix A: Three Different Types of Operators

(1) Fractional Laplacian operator and fractional power of operator

Let us denote $(-\triangle )^{\alpha /2}$ the nonlocal operator (also called the fractional Laplacian operator) defined pointwise by the following Cauchy principal value integral

$$\begin{aligned} (-\triangle )^{\alpha /2} f(x)=C_{\alpha } P.V. \int _{\mathbf {R}}{\frac{f(x)-f(y)}{|x-y|^{1+\alpha }}}dy,~~0<\alpha <2, \end{aligned}$$

(8.2.10)

where $C_{\alpha }=\frac{2^\alpha \varGamma (1/2+\alpha /2)}{\sqrt{\pi } \left| \varGamma \left( -\alpha /2\right) \right| }$ is a constant dependent on the order $\alpha .$ Obviously, the fractional Laplacian $\left( -\triangle \right) ^{\alpha /2}$ is a nonlocal operator which depends on the parameter $\alpha$ and recovers the usual Laplacian as $\alpha \rightarrow 2$. For more information about the fractional Laplacian operator, see [2, 9, 13, 17, 38] and the references cited therein.

Theorem 8.2.1

([25]) Suppose that $(-\triangle )^{\alpha /2}$ is defined in $L^2(0, l)$ for $\alpha \in (0, 2)$. Then, the eigenvalues of the following spectral problem

$$\begin{aligned} (-\triangle )^{\alpha /2} \xi (x)=\lambda \xi (x),~~x\in (0,l), \end{aligned}$$

(8.2.11)

where $\xi \in L^2(0, l)$ is extended to all $\mathbf {R}$ by 0 is

$$\begin{aligned} \lambda _n=\left( \frac{n\pi }{l}-\frac{(2-\alpha )\pi }{4l}\right) ^\alpha +O\left( \frac{1}{n}\right) \end{aligned}$$

(8.2.12)

satisfying

$$\begin{aligned} 0<\lambda _1<\lambda _2\le \cdots \le \lambda _i\le \cdots . \end{aligned}$$

Moreover, the corresponding eigenfunctions $\xi _n$ of $\lambda _n,$ after normalization, form a complete orthonormal basis in $L^2(0,l).$

Note that the constant in the error term $O\left( \frac{1}{n}\right)$ tends to zero when $\alpha$ approaches 2 and in the particular case, when $\alpha =2$, we see that $\lambda _n=\left( n\pi /l\right) ^2$ without the error term.

However, for a positive operator A on bounded domain [0, l]. Suppose $0<\lambda _1\le \lambda _2\le \cdots \le \lambda _n\le \cdots$ are the eigenvalues of A, $\{\xi _1, \xi _2,\ldots ,\xi _n, \ldots \}$ are the corresponding eigenfunctions and $\xi _n~(i=1,2,\ldots )$ form an orthonormal basis of $L^2(0,l)$. Let $(\cdot ,\cdot )$ be the inner product of $L^2(0,l).$ We define the fractional power of operator A as follows:

$$\begin{aligned} A^\beta f(x)=\sum \limits _{n=1}^\infty {\lambda _n^\beta (\xi _n,f)\xi _n(x)},~~f\in L^2(0,l), \end{aligned}$$

(8.2.13)

then $\lambda _n^\beta ~(i=1,2,\ldots )$ are the eigenvalues of $A^\beta$. This implies that the two operators are different.

Note that the work spaces of the two operators (fractional Laplacian operator and fractional power of operator A) are different. Before stating our main results, we first introduce two Banach spaces, which are specified in [12,13,14, 31].

For $\varOmega \subseteq \mathbf {R}^n$ is a bounded domain, $s\in (0,1)$ and $p\in [1,\infty )$, we define the classical Sobolev space $W^{s,p}(\varOmega )$ as follows [12]:

$$\begin{aligned} W^{s,p}(\varOmega ):=\left\{ f\in L^p(\varOmega ): \frac{f(x)-f(y)}{|x-y|^{\frac{n}{p}+s}}\in L^p(\varOmega \times \varOmega )\right\} \end{aligned}$$

(8.2.14)

endowed with the natural norm

$$\begin{aligned}&~&\Vert f\Vert _{W^{s,p}(\varOmega )}:=\left( \int _\varOmega {|f(x)|^p}dx+\int _\varOmega \int _\varOmega {\frac{|f(x)-f(y)|^p}{|x-y|^{n+sp}}}dxdy\right) ^{1/p} \end{aligned}$$

is an intermediary Banach space between $L^p(\varOmega )$ and $W^{1,p}(\varOmega )$. When a noninteger $s>1$, let $s=m+\sigma$ with $m\in \mathbf {N}$ and $\sigma \in (0,1)$. In this case, let $D^\beta f$ with $|\beta |=m$ be the distributed derivative of f, then the classical Sobolev space $W^{s,p}(\varOmega )$ defined by

(8.2.15)

with respect to the norm

$$\begin{aligned} \Vert f\Vert _{W^{s,p}(\varOmega )}:=\left( \Vert f\Vert ^p_{W^{m,p}(\varOmega )}+\Vert D^\beta f\Vert ^p_{W^{\sigma ,p}(\varOmega )}\right) ^{1/p} \end{aligned}$$

(8.2.16)

is a Banach space. Clearly, if $s = m$ is an integer, the space $W^{s,p}(\varOmega )$ coincides with the Sobolev space $W^{m,p}(\varOmega )$.

Besides, let $\rho (x)\sim \frac{1}{\delta ^\alpha (x)}$ with $\delta (x)=\text{ dist }(x,\varOmega ^c)$. Define another space as follows:

$$\begin{aligned} W^{s,p}_\rho (\varOmega ):=\left\{ f\in W^{s,p}(\varOmega ) : \rho (x)f(x)\in L^p(\varOmega )\right\} \end{aligned}$$

(8.2.17)

with the norm

$$\begin{aligned} \begin{array}{l} \Vert f\Vert _{W^{s,p}_\rho (\varOmega )}:=\left( \int _\varOmega {|\rho (x)f(x)|^p}dx+\int _\varOmega \int _\varOmega {\frac{|f(x)-f(y)|^p}{|x-y|^{n+sp}}}dxdy\right) ^{1/p}.\qquad \qquad \end{array}\end{aligned}$$

(8.2.18)

Actually, $W^{s,p}_\rho (\varOmega )$ is called nonlocal Sobolev space and we have $W^{s,p}_\rho (\varOmega ) \subseteq W^{s,p}(\varOmega )$ [13, 14].

By the Remark 2.1 in [31], for the fractional power of operator, we take the classical fractional Sobolev space as its work space. But for fractional Laplacian operator, we must take the nonlocal Sobolev space as its work space, which can be regarded as the weighted fractional Sobolev space. More precisely, for any element $f\in W^{s,p}(\varOmega )$, since $-\triangle$ is a local operator and we do not know how f(x) approaches 0 when $x \rightarrow \partial \varOmega$. Even if considering the space $W_0^{s,p}(\varOmega );=\{f\in W^{s,p}(\varOmega ): f|_{\partial \varOmega }=0\}$, we only know that the function $f= 0$ on the boundary and we do not know how f approaches 0. However, for the fractional Laplacian operator, it is a nonlocal operator and it is defined in the whole space. So, it provides information about how f approaches 0 as $x \rightarrow \partial \varOmega$. In fact, from the definition of nonlocal Sobolev space, we know that $\frac{f(x)}{\delta ^\alpha (x)}\rightarrow 0$ as $x \rightarrow \partial \varOmega$, which dictates how f approaches 0 near the boundary. It coincides with the result of the Theorem 1.2 in [42]. Thus, this is a significant difference between the fractional power of operator $A=-\triangle$ and the fractional Laplacian operator.

Besides, it is well known that

$$\begin{aligned} \Vert \triangle f\Vert _{W^{s,p}(\varOmega )}=\Vert f\Vert _{W^{s+2,p}(\varOmega )}.\end{aligned}$$

(8.2.19)

However, for the fractional Laplacian operator $(-\triangle )^\alpha$,

$$\begin{aligned} \Vert (-\triangle )^\alpha f\Vert _{W^{s,p}(\varOmega )}=\Vert f\Vert _{W^{s+2\alpha ,p}(\varOmega )}\end{aligned}$$

(8.2.20)

will not hold. By using Fourier transform, we have

$$\begin{aligned} \Vert (-\triangle )^\alpha f\Vert _{W_\rho ^{s,p}(\varOmega )}=\Vert f\Vert _{W_\rho ^{s+2\alpha ,p}(\varOmega )}.\end{aligned}$$

(8.2.21)

This also indicates that the fractional power of operator $-\triangle$ and the fractional Laplacian operator are different.

(2) Fractional Laplacian operator and fractional derivative

According to [44], the fractional Laplacian is the operator with symbol $|x|^\alpha .$ In other words, the following formula holds:

$$\begin{aligned} (-\triangle )^{\alpha /2}f(x)= \mathscr {F}^{-1}|x|^\alpha \mathscr {F} f(x), \end{aligned}$$

(8.2.22)

where $\mathscr {F}$ and $\mathscr {F}^{-1}$ denote the Fourier transform and inverse Fourier transform of f(x), respectively. We refer the readers to [26] for a detailed proof of the equivalence between the two definitions (8.2.10) and (8.2.22) of fractional Laplacian operator.

The Riesz fractional operator is defined as follows:

Definition 8.2.1

([19]) The Riesz fractional operator for $n-1<\alpha \le n$ on a finite interval $0\le x\le l$ is defined as

$$\begin{aligned} \frac{\partial ^\alpha }{\partial |x|^\alpha }f(x) =\frac{-1}{2\cos (\frac{\alpha \pi }{2})}\left[ {}_0D_x^\alpha +{}_xD_l^\alpha \right] f(x), \end{aligned}$$

(8.2.23)

where $_0D_x^\alpha$ and $_xD_l^\alpha$ are the left-sided and right-sided Riemann–Liouville fractional derivative, respectively.

By using Luchko’s theorem in [30], we obtain the following result on the equivalent relationship between the Riesz fractional derivative $\frac{\partial ^\alpha }{\partial |x|^\alpha }$ and the fractional Laplacian operator $-(-\triangle )^{\alpha /2}$:

Lemma 8.2.1

([22]) For a function f(x) defined on the finite domain [0, l], $f(0)=f(l)=0$ and $\alpha \in (1,2)$, the following equality holds:

$$\begin{aligned} -\left( -\triangle \right) ^{\alpha /2} f(x)=\frac{-1}{2\cos (\frac{\alpha \pi }{2})} \left[ {}_0D_x^\alpha f(x)+{}_xD_l^\alpha f(x)\right] =\frac{\partial ^\alpha }{\partial |x|^\alpha }f(x).\qquad \qquad \end{aligned}$$

(8.2.24)

For more information on the analytical solution of the generalized multi-term time and space fractional partial differential equations with Dirichlet nonhomogeneous boundary conditions, we refer the readers to [22]. For more information on the numerical solution of the fractional partial differential equation with Riesz space fractional derivatives on a finite domain, consult [28, 54].

(3) Fractional derivative and fractional power of operator

In this part, we first show the following definition of the positive operator.

Definition 8.2.2

([1]) The operator A is said to be positive if its spectrum $\sigma (A)$ lies in the interior of the sector of angle $\varphi \in (0,\pi )$, symmetric with respect to the real axis, and if on the edges of this sector, $S_1=\{\rho e^{i\varphi }: 0\le \rho <\infty \}$ and $S_2=\{\rho e^{-i\varphi }: 0\le \rho <\infty \}$, and outside it the resolvent $(\lambda I-A)^{-1}$ is subject to the bound

$$\begin{aligned} \left\| (\lambda I-A)^{-1}\right\| \le \frac{M(\varphi )}{1+|\lambda |}. \end{aligned}$$

(8.2.25)

For a positive operator A, any $\alpha >0$, one can define the negative fractional power of operator A by the following formula [24]:

$$\begin{aligned} \begin{array}{l} A^{-\alpha }=\frac{1}{2\pi i}\int _\varGamma {\lambda ^{-\alpha } R(\lambda )}d\lambda ,\\ \\ \left( R(\lambda )=(A-\lambda I)^{-1}, \varGamma =S_1 \cup S_2\right) . \end{array} \end{aligned}$$

(8.2.26)

It is then quite easy to see that $A^{-\alpha }$ is a bounded operator, which is an entire function of $\alpha$, satisfying $A^{-\alpha }=A^{-n}$ if $\alpha$ is an integer n, and $A^{-(\alpha +\beta )}=A^{-\alpha }A^{-\beta }$ for all $\alpha ,\beta \in \mathbf {C}$. Using (8.2.26), we have

$$\begin{aligned} A^{-\alpha }=\frac{1}{2\pi i}\int _{-\infty }^0{\lambda ^{-\alpha } R(\lambda )}d\lambda + \frac{1}{2\pi i}\int _0^{-\infty }{\lambda ^{-\alpha } R(\lambda )}d\lambda . \end{aligned}$$

(8.2.27)

Then taking the integration along the lower and upper sides of the cut respectively: $\lambda =se^{-\pi i}$ and $\lambda =se^{\pi i}$, it follows that

$$\begin{aligned} A^{-\alpha }=&\frac{e^{\alpha \pi i}}{2\pi i}\int _0^\infty {s^{-\alpha } R(-s)}ds- \frac{e^{-\alpha \pi i}}{2\pi i}\int _0^\infty {s^{-\alpha } R(-s)}ds\\ =&\frac{\cos (\alpha \pi )+i\sin (\alpha \pi )}{2\pi i}\int _0^\infty {s^{-\alpha } R(-s)}ds\\&~- \frac{\cos (\alpha \pi )-i\sin (\alpha \pi )}{2\pi i}\int _0^\infty {s^{-\alpha } R(-s)}ds\\ =&\frac{\sin (\alpha \pi )}{\pi }\int _0^\infty {s^{-\alpha } R(-s)}ds\\ =&\frac{1}{\varGamma (\alpha )\varGamma (1-\alpha )}\int _0^\infty {s^{-\alpha } R(-s)}ds. \end{aligned}$$

(a) Moreover, for any $\alpha \in (n-1,n)$, if we set

$$\begin{aligned} A^{\alpha }f=A^{\alpha -n}A^{n}f = \frac{\int _0^\infty {s^{\alpha -n} R(-s)A^{n}f}ds}{\varGamma (n-\alpha )\varGamma (1+\alpha -n)}, \end{aligned}$$

(8.2.28)

we get that

Theorem 8.2.2

Let the absolutely space

$$\begin{aligned} AC^n[0,l]: =\{f: f^{(n-1)}(x) \in C[0, l],~ f^{(n)}(x) \in L^2[0, l]\} \end{aligned}$$

and let A be the operator defined by the formula $Af(x) = f'(x)$ with the domain

$$\begin{aligned} D(A)=\{f: f\in AC^n[0,l], ~f^{(n)}(0)=0\}. \end{aligned}$$

Then A is a positive operator in the Banach space $AC^n[0,l]$ and

$$\begin{aligned} A^\alpha f(x)={}_0^CD_x^\alpha f(x),~~n-1<\alpha <n\end{aligned}$$

(8.2.29)

for all $f (x) \in D(A)$.

Proof

By [16], the operator $A+sI$ $(s\ge 0)$ has a bounded inverse and the resolvent of A is given by

$$\begin{aligned} \left( \left( A+sI\right) ^{-1}f\right) (x)=\int _0^x{e^{-s(x-y)}f(y)}dy. \end{aligned}$$

(8.2.30)

Then the operator A is a positive operator in $AC^n[0,l]$ and Eq. (8.2.31) gives

$$\begin{aligned} A^{\alpha }f(x)= & {} \frac{\int _0^\infty {s^{\alpha -n} R(-s)A^{n}f(x)}ds}{\varGamma (n-\alpha )\varGamma (1+\alpha -n)}\\= & {} \frac{\int _0^\infty {s^{\alpha -n} \left( A+sI\right) ^{-1}f^{(n)}(x)}ds}{\varGamma (n-\alpha )\varGamma (1+\alpha -n)}\\= & {} \frac{\int _0^\infty {s^{\alpha -n} \int _0^x{e^{-s(x-y)}f^{(n)}(y)}dy}ds}{\varGamma (n-\alpha )\varGamma (1+\alpha -n)}\\= & {} \frac{\int _0^x\left[ \int _0^\infty {s^{\alpha -n}e^{-s(x-y)}}ds\right] f^{(n)}(y)dy}{\varGamma (n-\alpha )\varGamma (1+\alpha -n)}. \end{aligned}$$

Let $s(x-y)=\lambda$. we get that

$$\begin{aligned} \int _0^\infty {s^{\alpha -n}e^{-s(x-y)}}ds=(x-y)^{n-\alpha -1}\int _0^\infty { \lambda ^{\alpha -n}e^{-\lambda }}d\lambda =(x-y)^{n-\alpha -1}\varGamma (\alpha -n+1). \end{aligned}$$

Then

$$\begin{aligned} A^{\alpha }f(x)= & {} \frac{ \int _0^x{(x-y)^{n-\alpha -1}\varGamma (\alpha -n+1)f^{(n)}(y)}dy}{\varGamma (n-\alpha )\varGamma (1+\alpha -n)}\\= & {} \frac{1}{\varGamma (n-\alpha )} \int _0^x{(x-y)^{n-\alpha -1}f^{(n)}(y)}dy \\= & {} {}_0^CD_x^{\alpha } f(x). \end{aligned}$$

This completes the proof.

(b) Similarly, for any $\alpha \in (n-1,n)$, let

$$\begin{aligned} A^{\alpha }f=A^{n}A^{\alpha -n}f = A^{n}\frac{\int _0^\infty {s^{\alpha -n} R(-s)f}ds}{\varGamma (n-\alpha )\varGamma (1+\alpha -n)}. \end{aligned}$$

(8.2.31)

It is not difficult to obtain the following result and we omit the detail proof.

Theorem 8.2.3

Let A be the operator defined by the formula $Af(x) = f'(x)$ with the domain

$$\begin{aligned} D(A)=\{f: f\in L^2[0,l]\}. \end{aligned}$$

Then A is a positive operator in the Banach space $AC^n[0,l]$ and

$$\begin{aligned} A^\alpha f(x)={}_0D_x^\alpha f(x),~~n-1<\alpha <n\end{aligned}$$

(8.2.32)

for all $f (x) \in D(A)$.

What’s more, it is worth noting that in the more recent monograph [51] and articles [20, 21, 50], the theory of pseudo-differential operators with singular symbols, and the connections between them and those three types of operators are explored. So that we can study the system (8.2.1)–(8.2.3) by using those theory of pseudo-differential operators.