Least-Squared Mixed Variational Formulation Based on Space Decomposition for a Kind of Variable-Coefficient Fractional Diffusion Problems

Abstract

In this paper, we decompose the fractional derivative space as the direct-sum of a fractional Sobolev space and a singular space spanned by $x^{-\beta }$ and then propose a $x^{-\beta }$-independent mixed type variational formulation over the commonly used Sobolev spaces for a kind of variable-coefficient fractional diffusion equations, based on the least-squared techniques and the merits of the direct-sum decomposition. We then prove the existence and uniqueness of the variational formulation, show the equivalence between the variational formulation and the fractional diffusion equation and discuss the regularity of the solution to the equation with a general right hand side function. As a consequence, an easily-computed and optimal-order-convergent least-squared mixed finite element method is established. The optimal-order numerical analysis with supporting numerical experiments is also conducted.

Appendix

In this appendix, we first introduce the definitions and notations of the negative fractional spaces and their relations, and then prove the $H^{\gamma -\beta }(\varOmega )$-regularity of $x^{-\beta }$, which is used in the second assertion of Theorem 3.4.

Definition 6.1

(Negative fractional derivative space [4, 6]) For any $\beta \ge 0$, define space $J_{L}^{-\beta }(\varOmega )$ as the closure of $C_0^\infty (\varOmega )$ with respect to the norm $\Vert \cdot \Vert _{J_{L}^{-\beta }(\varOmega )}$, defined by

$$\begin{aligned} \Vert \mathbf{v}\Vert _{J_{L}^{-\beta }(\varOmega )}:=\Vert _0I_x^{\beta }\mathbf{v}\Vert _{L^2(\varOmega )}. \end{aligned}$$

Definition 6.2

(Negative Sobolev space [13, 23]) For any $\beta \ge 0$, define space $H^{-\beta }(\varOmega )$ as the dual space of $H_0^{\beta }(\varOmega )$.

Lemma 6.1

([4]) For any $\beta \ge 0$, $\beta \ne n-\frac{1}{2}$ with $n\in \mathbb {N}$. Then, the negative fractional derivative space $J_{L}^{-\beta }(\varOmega )$ and the negative fractional Sobolev space $H^{-\beta }(\varOmega )$ are equal with equivalent norms.

Now we give the main lemma.

Lemma 6.2

Assume that $0\le \beta \le 1.$ Then, there exists a real number $\gamma$ ($0<\gamma <{1\over 2}$), which can be selected as close as possible to $1\over 2$, such that $x^{-\beta }\in H^{\gamma -\beta }(\varOmega ).$

Proof

We consider two cases for $\beta$.

Case (1). As $0\le \beta <\frac{1}{2}$.

Recalling the definition of fractional Sobolev spaces [23, 33], we know that the norm of $\mathbf{v}\in H^s(\varOmega )$ for $0\le s<1$ is defined by

$$\begin{aligned} \begin{array}{lll} \Vert \mathbf{v}\Vert ^2_{H^s(\varOmega )}=\Vert \mathbf{v}\Vert ^2_{L^2(\varOmega )}+\int _0^1\int _0^1\frac{|\mathbf{v}(x)-\mathbf{v}(y)|^2}{|x-y|^{1+2s}}dxdy. \end{array} \end{aligned}$$

(6.1)

Apparently, $x^{-\beta } \in L^2(\varOmega )$ for $0\le \beta <{1\over 2},$ then, it suffices to show that the singular integral in (6.1) is finite to draw the conclusion of $x^{-\beta }\in H^s(\varOmega ).$

For this purpose, we decompose the integral into two parts:

$$\begin{aligned} \begin{array}{lll} &{}\int _0^1\int _0^1\frac{|x^{-\beta }-y^{-\beta }|^2}{|x-y|^{1+2s}}dxdy\\ &{}\quad =\int _0^1\int _0^y\frac{(x^{-\beta }-y^{-\beta })^2}{(y-x)^{1+2s}}dxdy+\int _0^1\int _y^1 \frac{(x^{-\beta }-y^{-\beta })^2}{(x-y)^{1+2s}}dxdy\\ &{}\quad := I_1+I_2. \end{array} \end{aligned}$$

For $I_1$, applying the mean value theorem of differentials, changing variables from x to t, then to $\eta ,$ and tediously calculating, we yield

$$\begin{aligned} I_1= & {} \int _0^1 \int _0^y(y-x)^{-(1+2s)}(\frac{y^\beta -x^\beta }{x^\beta y^\beta })^2dxdy\\= & {} \beta ^2\int _0^1 \int _0^y(y-x)^{-(1+2s)}\{\frac{[ x+\theta (y-x)]^{\beta -1}(y-x)}{x^\beta y^\beta }\}^2 dxdy\\= & {} \int _0^1\frac{\beta ^2 dy }{y^{2s+2\beta +1}}\int _0^y\left( 1-\frac{x}{y}\right) ^{1-2s} \left( \frac{x}{y}\right) ^{-2\beta }\left[ \frac{x}{y}+\theta \left( 1-\frac{x}{y}\right) \right] ^{2\beta -2}dx\\= & {} \int _0^1\frac{\beta ^2 dy}{y^{2(s+\beta )}}\int _0^1(1-t)^{1-2s}t^{-2\beta }[t+\theta (1-t)]^{2\beta -2}dt \left( {\mathrm{let}}\ t=\frac{x}{y}\right) \\= & {} \int _0^1\frac{\beta ^2 dy}{y^{2(s+\beta )}}\int _0^1\eta ^{1-2s}(1-\eta )^{-2\beta } (1-\eta +\theta \eta )^{2\beta -2}d\eta ({\mathrm{let}}\ \eta = 1-t)\\= & {} \int _0^1\frac{\beta ^2 dy}{y^{2(s+\beta )}}\int _0^1\eta ^{1-2s}(1-\eta )^{-2\beta } (1-(1-\theta )\eta )^{2\beta -2}d\eta , \end{aligned}$$

where $\theta \in (0,1)$ is a mean value constant. Denoting $z=1-\theta ,$ we find that the inner singular integral is closely related to the well-known Gauss hypergeometric function ${_2F_1}(a,b,c;z)$ with $a=2-2\beta ,b=2-2s,c=3-2\beta -2s$ and $z=1-\theta ,$ expressed by [30],

$$\begin{aligned} \begin{array}{lll} \int _0^1\eta ^{1-2s}(1-\eta )^{-2\beta }(1-(1-\theta )\eta )^{2\beta -2}d\eta =\frac{\varGamma (b)\varGamma (c-b)}{\varGamma (c)}{_2F_1}(a,b,c;z). \end{array} \end{aligned}$$

The Gauss hypergeometric function ${_2F_1}(a,b,c;z)$ is convergent since $0<b<c$ and $0<z<1,$ and thus the inner singular integral in $I_1$. As a consequence of the convergence of ${_2F_1}(a,b,c;z)$ , $I_1$ is finite if and only if $\int _0^1\frac{1}{y^{2s+2\beta }}dy$ is convergent, which implies equivalently that $2(s+\beta )<1$ or $s<\frac{1}{2}-\beta$.

A similar calculation for $I_2$ shows that

$$\begin{aligned} \begin{array}{ll} I_2&{}=\int _0^1\int _y^1(x-y)^{-(1+2s)}\left( \frac{y^\beta -x^\beta }{x^\beta y^\beta }\right) ^2dxdy\\ &{}=\beta ^2\int _0^1dy\int _y^1(x-y)^{-(1+2s)} \left\{ \frac{[y+\theta (x-y)]^{\beta -1}(x-y)}{x^\beta y^\beta }\right\} ^2dx\\ &{}=\int _0^1\frac{\beta ^2dy}{y^{2s+2\beta +1}} \int _y^1\left( 1-\frac{y}{x}\right) ^{1-2s}\left( \frac{y}{x}\right) ^{1+2s} \left[ \frac{y}{x}+\theta \left( 1-\frac{y}{x}\right) \right] ^{2\beta -2}dx\\ &{}=\int _0^1\frac{\beta ^2dy}{y^{2(s+\beta )}} \int _y^1(1-t)^{1-2s}t^{2s-1}[t+\theta (1-t)]^{2\beta -2}dt \quad \left( {\mathrm{let}}\ t=\frac{y}{x}\right) \\ &{}=\int _0^1\frac{\beta ^2dy}{y^{2(s+\beta )}} \int _0^{1-y}\eta ^{1-2s}(1-\eta )^{2s-1}[1-\eta +\theta \eta ]^{2\beta -2}d\eta \quad ({\mathrm{let}}\ \eta =1-t)\\ &{}\le \int _0^1\frac{\beta ^2dy}{y^{2(s+\beta )}} \int _0^1\eta ^{1-2s}(1-\eta )^{2s-1}[1-(1-\theta )\eta ]^{2\beta -2}d\eta . \end{array} \end{aligned}$$

In the last step, the conditions $0<1-y<1$ and that the integrand is non-negative are used. Analogous to $I_1$, we involve the inner singular integral in $I_2$ into the convergent Gauss hypergeometric function ${_2F_1}(a,b,c;z),$

$$\begin{aligned} \begin{array}{lll} \int _0^1\eta ^{1-2s}(1-\eta )^{2s-1}[1-(1-\theta )\eta ]^{2\beta -2}d\eta =\frac{\varGamma (b)\varGamma (c-b)}{\varGamma (c)}{_2F_1}(a,b,c;z), \end{array} \end{aligned}$$

with $a=2-2\beta ,b=2-2s,c=2$ and $z=1-\theta .$ So, the finiteness of $I_2$ is equivalent to requiring $s<\frac{1}{2}-\beta .$

Collecting the finiteness of $I_1$ and $I_2$, we conclude that $\Vert x^{-\beta }\Vert _{H^s(\varOmega )}$ is finite for $s<{1\over 2}-\beta$. Taking $s=\gamma -\beta$, we obtain the stated result.

Case (2). As $\frac{1}{2}\le \beta <1$.

In this case $\beta -\gamma >0$. The fact $_0I_x^{\beta -\gamma }x^{-\beta }=\frac{\varGamma (1-\beta )}{\varGamma (1-\gamma )}x^{-\gamma }\in L^2(\varOmega )$ implies $x^{-\beta }\in J_L^{-(\beta -\gamma )}$. Considering the equivalence between space $J_L^{-(\beta -\gamma )}(\varOmega )$ and space $H^{-(\beta -\gamma )}(\varOmega )$ (see Lemma 6.1), we obtain the result. $\square$