Error Estimates of Spectral Galerkin Methods for a Linear Fractional Reaction–Diffusion Equation

Abstract

We consider a fractional diffusion equation with a reaction term in one dimensional space. We first establish the regularity in weighted Sobolev spaces. Then we present an optimal error estimate for a spectral Galerkin method for the equation and a sub-optimal error estimate for a spectral Petrov–Galerkin method. Numerical results suggest that the convergence order in a weighted $L^2$-norm is $2\alpha +1$ for smooth inputs where $\alpha$ is the order of the fractional Laplacian.

Appendices

Appendix A: Proof of Lemma 2.3

Lemma A.1

[16] Let $u_p= (1-x^2)^p,\, \left| x\right| \le 1$, $p>-1$ and $u_p(x)=0$ when $\left| x\right| >1$, then for $x\in (-1,1)$

$$\begin{aligned} -(-\Delta )^{\alpha /2} u_p(x) = c_{1, \alpha } B(-\alpha /2, \,p+1) {}_{2}F_{1}\left( \frac{\alpha +1}{2},-p+\frac{\alpha }{2}; \frac{1}{2}; x^2\right) . \end{aligned}$$

(A.1)

Let $v_p(x)= (1-x^2)^p x,\, \left| x\right| \le 1$ and $v_p(x)=0$ when $\left| x\right| >1$, $p>-1$. Then for $x\in (-1,1)$

$$\begin{aligned} - (-\Delta )^{\alpha /2} v_p(x) = (\alpha +1) c_{1, \alpha } B(-\alpha /2, \,p+1) {}_{2}F_{1}\left( \frac{\alpha +3}{2},-p+\frac{\alpha }{2}; \frac{3}{2}; x^2\right) x.\nonumber \\ \end{aligned}$$

(A.2)

Here $B(\cdot , \cdot )$ is the Beta function and the hypergeometric function ${}_2F_1(a,b;c;z)$ is defined for $|z| < 1$ by the power series

$$\begin{aligned} {}_2F_1(a,b;c;z) = \sum _{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}.\end{aligned}$$

Here $(q)_n$ is the (rising) Pochhammer symbol, which is defined by:

$$\begin{aligned} (q)_n = {\left\{ \begin{array}{ll} 1 &{} n = 0 \\ q(q+1) \cdots (q+n-1) = \frac{\Gamma (q+n)}{\Gamma (q)} &{} n > 0 \end{array}\right. }.\end{aligned}$$

Proof of Lemma 2.3

The conclusion can be shown by induction. By Lemma A.1, for $n=0,$$(-\Delta )^{\alpha /2}[(1-x^2)^{\alpha /2} ] = \Gamma (\alpha +1),$ and for $n=1$, $(-\Delta )^{\alpha /2}[(1-x^2)^{\alpha /2} x] = \Gamma (\alpha +2) x.$

Now suppose that $k\le n$ the relation holds then when $k=n+1$, we want

$$\begin{aligned} (-\Delta )^{\alpha /2}\left[ (1-x^2)^{\alpha /2}P_{n+1}^{\alpha /2}(x)\right] = D_{n+1,\alpha }P_{n+1}^{\alpha /2}(x). \end{aligned}$$

By integration-by-parts formula [22] (or Lemma 2.1), for any $j\le n$, we have

$$\begin{aligned}&\int _{-1}^{1} (-\Delta )^{\alpha /2}[(1-x^2)^{\alpha /2}P_{n+1}^{\alpha /2}(x)] P_j^{\alpha /2}(1-x^2)^{\alpha /2} ) \,dx \\&\quad = \int _{-1}^{1}(1-x^2)^{\alpha /2}P_{n+1}^{\alpha /2}(x) (-\Delta )^{\alpha /2}[P_j^{\alpha /2}(1-x^2)^{\alpha /2} ])\,dx \\&\quad = D_{j,\alpha } \int _{-1}^{1} (1-x^2)^{\alpha /2} P_{n+1}^{\alpha /2} P_j^{\alpha /2} \,dx=0. \end{aligned}$$

Here we have used the induction assumption. By the orthogonality of Jacobi polynomials and also by Lemma A.1 that $(-\Delta )^{\alpha /2}[(1-x^2)^{\alpha /2}P_{n+1}^{\alpha /2}(x)]$ is a polynomial of order $n+1$, then we have

$$\begin{aligned} (-\Delta )^{\alpha /2}[(1-x^2)^{\alpha /2}P_{n+1}^{\alpha /2}(x)]= C_{n+1,\alpha }P_{n+1}^{\alpha /2}(x). \end{aligned}$$

(A.3)

where $C_{n+1,\alpha }$ is a constant to be determined, depending on $n+1$ and $\alpha$. Then we compare the coefficients of leading-order term over both sides and obtain the conclusion at $k=n+1$.

The constant $C_{n+1,\alpha }$ can be found as follows. Suppose that $m=n+1=2k$. We only need to compare the coefficients of the leading order term of both sides of (A.3). By Lemma A.1, it only requires to check the coefficient of leading-order term of $c_{1,\alpha } B(-\alpha /2, \,p+1) {}_{2}F_{1}(\frac{\alpha +1}{2},-p+\frac{\alpha }{2}; \frac{1}{2}; x^2)$, where $p=k+\alpha /2$. By the definition of Pochhammer symbol, the coefficient is

$$\begin{aligned}&\frac{2^\alpha \Gamma (\frac{\alpha +1}{2})}{\pi ^{1/2} \left| \Gamma (-\alpha /2)\right| }\frac{\Gamma (-\alpha /2)\Gamma (p+1)}{\Gamma (p-\alpha /2+1)} \frac{(\frac{(\alpha +1)}{2})_k (-p+\alpha /2)_k }{(1/2)_k k!} \\&\quad = -\frac{2^\alpha \Gamma (\frac{\alpha +1}{2}) }{\pi ^{1/2} }\frac{\Gamma (k+\frac{\alpha }{2}+1)}{\Gamma (k+1)} \frac{\Gamma (\frac{\alpha +1}{2}+k) \Gamma (\frac{1}{2})}{\Gamma (\frac{\alpha +1}{2}) \Gamma (\frac{1}{2}+k) }\frac{(-k)_k}{k!}\\&\quad = (-1)^{k+1}\frac{2^\alpha }{\pi ^{1/2} }\frac{\Gamma (k+\frac{\alpha }{2}+1)}{\Gamma (k+1)} \frac{\Gamma (\frac{\alpha +1}{2}+k) \Gamma (\frac{1}{2})}{ \Gamma (\frac{1}{2}+k) } \\&\quad =\frac{(-1)^{k+1}2^\alpha \Gamma (\frac{1}{2}) }{ \pi ^{1/2} }\frac{ \Gamma (k+\frac{\alpha }{2}+1) \Gamma (\frac{\alpha +1}{2}+k)}{\Gamma (k+1) \Gamma (\frac{1}{2}+k) }. \end{aligned}$$

Then by the the duplication formula, $\Gamma (z) \Gamma \left( z + \tfrac{1}{2}\right) = 2^{1-2z} \; \sqrt{\pi } \; \Gamma (2z),$ and the fact that $\Gamma (1/2)=\sqrt{\pi }$, we have that the coefficient is

$$\begin{aligned} \frac{(-1)^{k+1}2^\alpha \Gamma (\frac{1}{2}) }{ \pi ^{1/2} }\frac{ \Gamma (k+\frac{\alpha }{2}+1) \Gamma (\frac{\alpha +1}{2}+k)}{\Gamma (k+1) \Gamma (\frac{1}{2}+k) }= & {} \frac{(-1)^{k+1}2^\alpha \Gamma (\frac{1}{2}) }{ \pi ^{1/2} }\frac{ \sqrt{\pi } 2^{-\alpha -2k} \Gamma ( \alpha +1 +2k)}{2^{-2k}\sqrt{\pi } \Gamma (2k+1) } \\= & {} (-1)^{k+1}\frac{\Gamma (\alpha +1+ m)}{m! }. \end{aligned}$$

Comparing the coefficients of leading order terms of both sides of (A.3), we have $(-1)^{k} C_{m,\alpha } = (-1)^{k+2}\frac{\Gamma (\alpha +1+ m)}{m!}$ and thus $C_{m,\alpha }= \frac{\Gamma (\alpha +1+ m)}{m!}$. Similarly, we can have the same conclusion when $m=n+1=2k+1$. $\square$

Appendix B: Some Useful Relations of Jacobi Polynomials

The following relations hold for Jacobi polynomials, see e.g. [3, Chapter 2],

$$\begin{aligned} \frac{\text {d}^k}{\text {d} z^k} P_n^{\alpha ,\beta } (z)= & {} d_{n,k}^{\alpha ,\beta } P_{n-k}^{\alpha +k, \beta +k} (z),\quad d_{n,k}^{\alpha ,\beta }=\frac{\Gamma (n+k+\alpha +\beta +1)}{2^k \Gamma (n+\alpha +\beta +1)}. \end{aligned}$$

(B.1)

The Jacobi polynomial can be represented using Rodrigue’s formula:

$$\begin{aligned} \omega ^{\alpha ,\beta }P_n^{\alpha ,\beta } = \frac{(-1)^n}{2^n n!}\frac{d^n}{dx^n}(\omega ^{n+\alpha ,n+\beta }). \end{aligned}$$

(B.2)

The Jacobi polynomial $P_n^{\alpha ,\beta } (x)$ is a solution of the second order linear homogeneous differential equation

$$\begin{aligned} \partial _x \left( (1-x)^{\alpha +1} (1+x)^{\beta +1}\partial _x u \right) + n(n+\alpha +\beta +1) (1-x)^{\alpha } (1+x)^{\beta } u = 0. \end{aligned}$$

The following is an application of Rodrigue’s formula (B.2) and this equation.

Lemma B.1

For $\alpha ,\beta >-1$, we have

$$\begin{aligned} \partial _x \left( (1-x)^{\alpha +1} (1+x)^{\beta +1} P_{n-1}^{\alpha +1,\beta +1} \right) = -2n (1-x)^{\alpha } (1+x)^{\beta } P_n^{\alpha ,\beta }. \end{aligned}$$

(B.3)

Lemma B.2

The following relation holds, for $\beta >-1$

$$\begin{aligned} P_{n}^{\beta ,\beta } = {\widehat{A}}_n^{\beta ,\beta } P_{n-2}^{\beta +1,\beta +1} + {\widehat{C}}_{n}^{\beta ,\beta } P_{n}^{\beta +1,\beta +1}, \end{aligned}$$

(B.4)

where ${\widehat{A}}_{n}^{\beta ,\beta } = -\frac{ n+\beta }{2(2n+2\beta +1)}$ and ${\widehat{C}}_{n}^{\beta ,\beta } = \frac{ (n+2\beta +1)(n+2\beta +2)}{2(2n+2\beta +1)(n+\beta +1)}$.

Lemma B.3

For $\alpha >0$, there is a constant C independent of $n,\,l,\,\alpha$ such that

$$\begin{aligned} (h_{n}^{\alpha /2-1})^{-1} \left| (P_{2l+n}^{\alpha /2},P_{n}^{\alpha /2-1})_{\omega ^{\alpha /2-1}}\right| \le C. \end{aligned}$$

(B.5)

Proof

By the Rodrigue’s representation of Jacobi polynomials and integration by parts, we have

$$\begin{aligned} (P_{2l+n}^{\alpha /2},P_{n}^{\alpha /2-1})_{\omega ^{\alpha /2-1}}= & {} \frac{(-1)^n}{n!2^n} (P_{2l+n}^{\alpha /2}, \partial _x^n \omega ^{\alpha /2+n-1}) =\frac{1}{n!2^n}d_{2l+n,n}^{\alpha /2} (P_{2l}^{\alpha /2+n }, \omega ^{\alpha /2+n-1}). \end{aligned}$$

Now we compute the integral $(P_{2l}^{\alpha /2+n }, \omega ^{\alpha /2+n-1}).$ From Lemma B.2, we have

$$\begin{aligned}&{\widehat{C}}_{2l}^{\alpha /2+n-1 } (P_{2l}^{\alpha /2+n }, \omega ^{\alpha /2+n-1}) + {\widehat{A}}_{2l}^{\alpha /2+n-1 } (P_{2l-2}^{\alpha /2+n }, \omega ^{\alpha /2+n-1})\\&\quad = (P_{2l}^{\alpha /2+n-1 }, \omega ^{\alpha /2+n-1}) = h_{0}^{\alpha /2+n-1 } \delta _{2l,0}. \end{aligned}$$

Then by induction, we have for $l\ge 1$

$$\begin{aligned}&(P_{2l}^{\alpha /2+n}, \omega ^{\alpha /2+n-1}) \\&\quad = (-1)^{l-1} \frac{ {\widehat{A}}_{2l}^{\alpha /2+n-1 }}{ {\widehat{C}}_{2l}^{\alpha /2+n-1 }} \cdots \frac{ {\widehat{A}}_{4}^{\alpha /2+n-1 }}{ {\widehat{C}}_{4}^{\alpha /2+n-1 }} \left( \frac{ h_{0}^{\alpha /2+n-1 }}{ {\widehat{C}}_{2}^{\alpha /2+n-1 }}-\frac{ {\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{C}}_{2}^{\alpha /2+n-1 }}(P_{0}^{\alpha /2+n}, \omega ^{\alpha /2+n-1}) \right) \\&\quad = (-1)^{l-1} \frac{ {\widehat{A}}_{2l}^{\alpha /2+n-1 }}{ {\widehat{C}}_{2l}^{\alpha /2+n-1 }} \cdots \frac{ {\widehat{A}}_{4}^{\alpha /2+n-1 }}{ {\widehat{C}}_{4}^{\alpha /2+n-1 }} \frac{ 1-{\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{C}}_{2}^{\alpha /2+n-1 }} h_{0}^{\alpha /2+n-1 }. \end{aligned}$$

From Lemma B.2, we have

$$\begin{aligned} \frac{ {\hat{A}}_{2l}^{\alpha /2+n-1 }}{ {\hat{C}}_{2l}^{\alpha /2+n-1 }} = - \frac{(2l+\alpha /2+n-1)(2l+\alpha /2+n) }{(2l+\alpha +2n-1)(2l+\alpha +2n )}. \end{aligned}$$

This leads to

$$\begin{aligned} (P_{2l}^{\alpha /2+n}, \omega ^{\alpha /2+n-1}) = h_{0}^{\alpha /2+n-1 } \frac{\Gamma (2l+ \alpha /2+n+1) \Gamma (\alpha +2n+3)}{\Gamma (2l+ \alpha +2n+1)\Gamma (\alpha /2+n+3)}\frac{ 1-{\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{A}}_{2}^{\alpha /2+n-1 }} .\nonumber \\ \end{aligned}$$

(B.6)

Then we have

$$\begin{aligned}&(P_{2l+n}^{\alpha /2},P_{n}^{\alpha /2-1})_{\omega ^{\alpha /2-1}}\\&\quad = \frac{1}{n!2^n}d_{2l+n,n}^{\alpha 2/,\alpha /2}h_{0}^{\alpha /2+n-1 } \frac{\Gamma (2l+ \alpha /2+n+1) \Gamma (\alpha +2n+3)}{\Gamma (2l+ \alpha +2n+1)\Gamma (\alpha /2+n+3)}\frac{ 1-{\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{A}}_{2}^{\alpha /2+n-1 }} \\&\quad = \frac{\Gamma (2l+2n+\alpha +1)}{n!2^{2n} \Gamma (2l+n+\alpha +1)} \frac{\Gamma (2l+ \alpha /2+n+1) \Gamma (\alpha +2n+3)}{\Gamma (2l+ \alpha +2n+1)\Gamma (\alpha /2+n+3)} h_{0}^{\alpha /2+n-1 }\frac{ 1-{\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{A}}_{2}^{\alpha /2+n-1 }} \\&\quad = \frac{\Gamma (2l+ \alpha /2+n+1)}{2^n \Gamma (2l+n+\alpha +1) } \frac{ \Gamma (\alpha +2n+3)}{ \Gamma (\alpha /2+n+3)n!2^n} h_{0}^{\alpha /2+n-1 } \frac{ 1-{\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{A}}_{2}^{\alpha /2+n-1 }}. \end{aligned}$$

Then by (2.5),

$$\begin{aligned}&(h_{n}^{\alpha /2-1})^{-1}(P_{2l+n}^{\alpha /2},P_{n}^{\alpha /2-1})_{\omega ^{\alpha /2-1}} \\&\quad = \frac{\Gamma (2l+ \alpha /2+n+1)}{n!2^{2n} \Gamma (2l+n+\alpha +1) } \frac{ \Gamma ( \alpha +2n+3) }{ \Gamma (n+\alpha /2+3) } \frac{h_{0}^{\alpha /2+n-1 }}{h_{n}^{\alpha /2-1}} \frac{ 1-{\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{A}}_{2}^{\alpha /2+n-1 }}\\&\quad = \frac{\Gamma (2l+ \alpha /2+n+1)}{ \Gamma (2l+n+\alpha +1) }\frac{ \Gamma (\alpha +2n+3) }{ \Gamma (n+\alpha /2+3) } \frac{ \Gamma (n+\alpha -1)}{ \Gamma (2n+\alpha -1) }\frac{ 1-{\widehat{A}}_{2}^{\alpha /2+n-1 }}{ {\widehat{A}}_{2}^{\alpha /2+n-1 }}. \end{aligned}$$

Then by the fact which can be proved by Stirling’s formula that

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{\Gamma (n+\delta )}{n^{\delta -\gamma }\Gamma (n+ \gamma )} = \lim _{n\rightarrow \infty } (1+\frac{(\delta -\gamma )(\delta +\gamma -1)}{2n}+\mathcal {O}(n^{-2}))=1, \end{aligned}$$

(B.7)

we obtain the conclusion (B.5). $\square$

Proof of Lemma 2.5

By Lemma B.2 and Lemma 2.3, we have

$$\begin{aligned} (-\Delta )^{\alpha /2}[(1-x^2)^{\alpha /2}P_n^{\alpha /2-1}(x)]= & {} (-\Delta )^{\alpha /2}[(1-x^2) ^{\alpha /2}({\widehat{A}}_{n}^{\alpha /2-1} P_{n-2}^{\alpha /2}(x) + {\widehat{C}}_{n}^{\alpha /2-1} P_{n}^{\alpha /2}(x)] \\= & {} {\widehat{A}}_{n}^{\alpha /2-1}D_{n-2,\alpha }P_{n-2}^{\alpha /2}(x) + {\widehat{C}}_{n}^{\alpha /2-1} D_{n,\alpha }P_{n}^{\alpha /2}(x). \end{aligned}$$

Here ${\widehat{A}}_{n}^{\alpha /2-1}$, ${\widehat{C}}_{n}^{\alpha /2-1}$ are from Lemma B.2 and $D_{n,\alpha }, D_{n-2,\alpha }$ are from Lemma 2.3. $\square$

Appendix C: Proof of (3.3)

By (3.2) and Lemma B.2, we have

$$\begin{aligned} \sum _{k=0}^\infty u_k (E_{k,\alpha }P_{k-2}^{\alpha /2} + F_{k,\alpha }P_{k}^{\alpha /2} )= & {} \sum _{k=0}^\infty f_k \left( {\widehat{A}}_{k}^{\alpha /2-1} P_{k-2}^{\alpha /2} + {\widehat{C}}_{k}^{\alpha /2-1} P_{k}^{\alpha /2}\right) . \end{aligned}$$

Here $E_{k,\alpha }$, $F_{k,\alpha }$ are from Lemma 2.5 and ${\widehat{A}}_{k}^{\alpha /2-1}$, ${\widehat{C}}_{k}^{\alpha /2-1}$ are defined in Lemma B.2. Thus, multiplying by $P_{n}^{\alpha /2}$ over both sides of the last equation and by the orthogonality of the Jacobi polynomials, we have

$$\begin{aligned} E_{n+2,\alpha } u_{n+2} + F_{n,\alpha } u_n = {\widehat{A}}_{n+2}^{\alpha /2-1} f_{n+2} + {\widehat{C}}_{n}^{\alpha /2-1} f_n, \quad n\ge 0, \end{aligned}$$

or equivalently for $n\ge 0$,

$$\begin{aligned} {\widehat{A}}_{n+2}^{\alpha /2-1} u_{n+2} + {\widehat{C}}_{n}^{\alpha /2-1} u_n = D_{n,\alpha }^{-1}\big ( {\widehat{A}}_{n+2}^{\alpha /2} f_{n+2} + {\widehat{C}}_{n}^{\alpha /2} f_n\big ). \end{aligned}$$

(C.1)

Multiplying both sides of (C.1) by $P_n^{\alpha /2}$, adding all resulting equations, we have

$$\begin{aligned} \sum _{n=0}^\infty ({\widehat{A}}_{n+2}^{\alpha /2-1} u_{n+2} + {\widehat{C}}_{n}^{\alpha /2-1} u_n) P_{n}^{\alpha /2 } =\sum _{n=0}^\infty D_{n,\alpha }^{-1}\big ( {\widehat{A}}_{n+2}^{\alpha /2} f_{n+2} + {\widehat{C}}_{n}^{\alpha /2} f_n\big )P_{n}^{\alpha /2 }. \end{aligned}$$

Applying Lemma B.2 leads to

$$\begin{aligned} \sum _{n=0}^\infty u_n P_{n}^{\alpha /2-1} =\sum _{n=0}^\infty D_{n,\alpha }^{-1} f_{n} P_{n}^{\alpha /2-1} + \sum _{n=0}^\infty f_{n+2} \left( D_{n,\alpha }^{-1}-D_{n+2,\alpha }^{-1}\right) {\widehat{A}}_{n+2}^{\alpha /2-1} P_{n}^{\alpha /2}. \end{aligned}$$

Thus we have from the orthogonality of the Jacobi polynomials that

$$\begin{aligned} u_n =D_{n,\alpha }^{-1}f_{n} + \left( h_{n}^{\alpha /2-1}\right) ^{-1} \sum _{k=n}^\infty f_{k+2} \left( D_{k,\alpha }^{-1}-D_{k+2,\alpha }^{-1}\right) {\widehat{A}}_{k+2}^{\alpha /2-1} (P_{k}^{\alpha /2},P_{n}^{\alpha /2-1})_{\omega ^{\alpha /2-1}}. \end{aligned}$$

By the fact that $P_n^{\beta ,\beta }(-x)= (-1)^nP_n^{\beta ,\beta }(x)$, we have $(P_{k}^{\alpha /2},P_{n}^{\alpha /2-1})_{\omega ^{\alpha /2-1}}=0$ if $\left| n-k\right|$ is an odd number. Further, the orthogonality of Jacobi polynomials leads to $(P_{k}^{\alpha /2},P_{n}^{\alpha /2-1})_{\omega ^{\alpha /2-1}}=0$ if $n>k$. Then we obtain (3.3).

Appendix D: Estimates of Products of Functions in Weighted Spaces

Lemma D.1

Let $u=v\omega ^{\alpha /2}$ with $v\in B^{\beta }_{ {\alpha /2}}(I)$, $0<\beta \le \alpha \le 2$. Then $u\in B^{\beta }_{ {\alpha /2}}(I)$.

Lemma D.2

If $v\in B^{\beta }_{ {\alpha /2-1}}(I)$, then $u=v\omega ^{\alpha /2}\in B^{\beta }_{ {\alpha /2-1}}(I)$, when $0<\beta \le \alpha \le 2$ and $\beta \ne \alpha /2$.

Lemma D.3

Let $u=v\omega ^{\alpha /2}$ with $v\in B^{\beta }_{ {\alpha /2-1}}(I)$, $0\le \beta \le \alpha +1$, $0<\alpha \le 2$. Then $u\in B^{\beta }_{ {\alpha /2}}(I)$ when $\beta \ne \alpha /2$.

The proofs of these lemmas are similar and we only present the proof of Lemma D.1.

We need an equivalent definition of the weighted Sobolev space. In [17], it is shown that the norm in $B_{\theta }^s(I)$ ($s=m+\sigma$, $0<\sigma <1$ and $s\ne 1+\theta$ if $-1<\theta <0$) is equivalent to the following

$$\begin{aligned} \left\| u\right\| _{s,\omega ^{\theta },B}= \big ( \left\| u\right\| _{m,\omega ^{\theta },B}^2 + \iint _{\Omega _{I,a}} \omega ^{\theta +s}\frac{\left| u^{(m)}(x) - u^{(m)}(y)\right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy \big )^{1/2}, \end{aligned}$$

(D.1)

where for $a>1$, $\Omega _{I,a}$ is a subset of $I\times I$ and specifically

$$\begin{aligned} \Omega _{I,a} = \left\{ (x,y)\in I \times I | a^{-1}(1-\left| x\right| )< 1 -{{\text {sgn}}(x)}y < a (1-\left| x\right| ) \right\} .\end{aligned}$$

Here a can be any number larger than 1 and we take $a=2$.

We also need the following lemma to prove Lemma D.1.

Lemma D.4

Let $v\in L^{2}_{\omega ^{\gamma +1-\beta +2\theta }}(I)$, where $\beta <3$, $\gamma ,\theta$ are real numbers. Then

$$\begin{aligned}&\iint _{\Omega _{I,a}} \omega ^{\gamma }(x)v^{2}(x) \frac{\left| \omega ^{\theta }(x) - \omega ^{\theta }(y) \right| ^2}{\left| x-y\right| ^{ \beta }}\,dx\,dy\\&\quad \le C\int _{I} \omega ^{\gamma +1-\beta }(x) v^2(x) \int _{a^{-1}}^a [\partial _t g (t,\left| x\right| ,\theta )]^2 \left| 1-t\right| ^{ 2-\beta } \,dt \,dx \le C \left\| v\right\| ^2_{\omega ^{\gamma +1-\beta +2\theta }}. \end{aligned}$$

Here $g(t,x,\theta )= t^{-2\theta } (1-x)^{\theta }(2t-(1-x))^{\theta }$.

Proof

Note that the integration domain can be split into two parts that $\Omega _{I,a}\cap \left\{ x>0\right\}$ and $\Omega _{I,a}\cap \left\{ x<0\right\} .$

Let consider the case when $x>0$. Let $1-x=(1-y)t$. Then $a^{-1}\le t\le a$, $\frac{dy}{dt}=\frac{1-x}{t^2}$ and $\left| x-y\right| = \left| 1-y -(1-x)\right| = \frac{\left| 1-t\right| }{t} (1-x)$. Let $g(t,x,\theta )= t^{-2\theta } (1-x)^{\theta }(2t-(1-x))^{\theta }=\omega ^{\theta }(y)$. Observe that $g(1,x,\theta )= \omega ^{\theta }(x)$. The double integral $\Omega _{I,a}\cap \left\{ x>0\right\}$ becomes

$$\begin{aligned}&\iint _{\Omega _{I,a},x>0} \omega ^{\gamma }(x)v^{2}(x) \frac{\left| \omega ^{\theta }(x) - \omega ^{\theta }(y) \right| ^2}{\left| x-y\right| ^{ \beta }}\,dx\,dy\\&\quad = \int _{0}^1 \omega ^{\gamma }(x)(1-x)^{1-\beta } v^2(x) \int _{a^{-1}}^a \frac{ t^{\beta -2} (g(t,x,\theta ) -g(1,x,\theta ))^2 }{ \left| 1-t\right| ^{ \beta }} \,dt\,dx \\&\quad \le C \int _{0}^1 \omega ^{\gamma +1-\beta }(x) v^2(x) \int _{a^{-1}}^a (\partial _tg(t,x,\theta ) )^2 \left| 1-t\right| ^{2- \beta } \,dt \,dx. \end{aligned}$$

In the last step, we have applied Hardy’s inequality on [1, a] and $[a^{-1},1]$.

Now consider the case when $x<0$, Let $1+x=(1+y)t$. Then $a^{-1}\le t\le a$, $\frac{dy}{dt}=-\frac{1+x}{t^2}$ and $\left| x-y\right| = \left| 1+x -(1+y)\right| = \frac{\left| 1-t\right| }{t}(1+x)$. Let $f(t,x,\theta )= t^{-2\theta } (1+x)^{\theta }(2t-(1+x))^{\theta }=\omega ^{\theta }(y)$. Observe that $f(1,x,\theta )= \omega ^{\theta }(x)$. The double integral $\Omega _{I,a}\cap \left\{ x<0\right\}$ becomes

$$\begin{aligned}&\iint _{\Omega _{I,a},x<0} \omega ^{\gamma }(x)v^{2}(x) \frac{\left| \omega ^{\theta }(x) - \omega ^{\theta }(y) \right| ^2}{\left| x-y\right| ^{ \beta }}\,dx\,dy\\&\quad = \int _{-1}^0 \omega ^{\gamma }(x)(1+x)^{1-\beta } v^2(x) \int _{a^{-1}}^a \frac{ t^{\beta -2} (f(t,x,\theta ) -f(1,x,\theta ))^2 }{ \left| 1-t\right| ^{ \beta }} \,dt\,dx \\&\quad \le C\int _{0}^1 \omega ^{\gamma +1-\beta }(x) v^2(x) \int _{a^{-1}}^a (\partial _tf(t,x,\theta ) )^2 \left| 1-t\right| ^{2- \beta } \,dt \,dx. \end{aligned}$$

In the last step, we have applied Hardy’s inequality on [1, a] and $[a^{-1},1]$. Note that $f(t,x,\theta ) = g(t,-x,\theta )$. Observe that $\int _{a^{-1}}^a (\partial _tf(t,x,\theta ) )^2 \left| 1-t\right| ^{2- \beta } \,dt \le C (1+x)^{2\theta }$. We then obtain the desired conclusion. $\square$

Proof of Lemma D.1

If $\beta =1$ or 2, we can use direct calculation to get the desired results.

When $0<\beta <1$, $\int _{I}u^2\omega ^{\alpha /2}\,dx = \int _{I}v^2\omega ^{3\alpha /2}\,dx \le \int _{I}v^2\omega ^{\alpha /2}\,dx$. Moreover,

$$\begin{aligned}&\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)\frac{\left| \omega ^{\alpha /2}(x)v(x) - \omega ^{\alpha /2}(y)v(y)\right| ^2}{\left| x-y\right| ^{1+2\beta }}\,dx\,dy \nonumber \\&\quad \le 2\iint _{\Omega _{I,a}} \omega ^{ \alpha /2+\beta } (x)\omega ^{\alpha }(y)\frac{\left| v(x) - v(y)\right| ^2}{\left| x-y\right| ^{1+2\beta }}\,dx\,dy \nonumber \\&\qquad +\,2\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)v^{2}(x) \frac{\left| \omega ^{\alpha /2}(x) - \omega ^{\alpha /2}(y) \right| ^2}{\left| x-y\right| ^{1+2\beta }}\,dx\,dy \nonumber \\&\quad \le 2\left\| v\right\| _{\beta , \omega ^{\alpha /2},B}^2 +\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)v^{2}(x) \frac{\left| \omega ^{\alpha /2}(x) - \omega ^{\alpha /2}(y) \right| ^2}{\left| x-y\right| ^{1+2\beta }}\,dx\,dy. \end{aligned}$$

(D.2)

By Lemma D.4, we have

$$\begin{aligned} \iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)v^{2}(x) \frac{\left| \omega ^{\alpha /2}(x) - \omega ^{\alpha /2}(y) \right| ^2}{\left| x-y\right| ^{1+2\beta }}\,dx\,dy\le & {} C \int _I \omega ^{3\alpha /2-\beta } v^{2} \,dx . \end{aligned}$$

Since $\beta \le \alpha$, then $\int _{I} \omega ^{3\alpha /2- \beta } v^2 \,dx$ is bounded by $\left\| v\right\| _{\omega ^{\alpha /2}}^2$. Then (D.2) can be bounded by $C\left\| v\right\| _{\beta ,\omega ^{\alpha /2},B}^2$.

When $1<\beta \le \alpha <2$. $\left\| u\right\| _{\omega ^{\alpha /2}}= \left\| \omega ^{\alpha /2}v\right\| _{\omega ^{\alpha /2}}\le \left\| v\right\| _{\omega ^{\alpha /2}}$. As $\partial _x u = \omega ^{\alpha /2} \partial _x v + (-\alpha x)\omega ^{\alpha /2-1}v$, then

$$\begin{aligned} \left\| \partial _x u\right\| _{\omega ^{\alpha /2+1}}\le & {} \left\| \omega ^{\alpha /2}\partial _xv\right\| _{\omega ^{\alpha /2+1}} + \alpha \left\| \omega ^{\alpha /2-1}v\right\| _{\omega ^{\alpha /2+1}} \le \left\| \partial _xv\right\| _{\omega ^{\alpha /2+1}} + \alpha \left\| v\right\| _{\omega ^{ \alpha /2}}. \end{aligned}$$

It remains to check the weighted fractional-order norm. Since $\partial _x u = \omega ^{\alpha /2} \partial _x v + (-\alpha x)\omega ^{\alpha /2-1}v$ and we only need to check that the weighted fractional norms of $\omega ^{\alpha /2} \partial _x v$ and $x\omega ^{\alpha /2-1}v$ are bounded. The boundedness of the weighted fractional norm of $\omega ^{\alpha /2} \partial _x v$ is proved as follows. Let $\beta =1+\sigma$.

$$\begin{aligned}&\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)\frac{\left| \omega ^{\alpha /2}(x)\partial _xv(x) - \omega ^{\alpha /2}(y)\partial _xv(y)\right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy \\&\quad \le 2\iint _{\Omega _{I,a}} \omega ^{ \alpha /2+\beta }(x)\frac{\left| \partial _xv(x) - \partial _x v(y)\right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy\\&\qquad +\,2\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)(\partial _xv)^2\frac{\left| \omega ^{\alpha /2}(x) - \omega ^{\alpha /2}(y) \right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy \\&\quad \le C\left\| v\right\| ^2_{\beta ,\omega ^{\alpha /2},B} +2\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)(\partial _xv)^2\frac{\left| \omega ^{\alpha /2}(x) - \omega ^{\alpha /2}(y) \right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy. \end{aligned}$$

By Lemma D.4, the last term is bounded by $\int _{I} \omega ^{3\alpha /2+\beta -2\sigma }(\partial _x v)^2\,dx$, which is further bounded by $\left\| \partial _x v\right\| ^2_{\omega ^{\alpha /2+1} }$ as $\beta =1+\sigma \le \alpha$.

For $x\omega ^{\alpha /2-1}v$, we only need to show the regularity of $\omega ^{\alpha /2-1}v$. Then by the fact that $\omega ^{\gamma }(y)\le C \omega ^{\gamma }(x)$ on $\Omega _{I,a}$ for any $\gamma$, we have

$$\begin{aligned}&\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)\frac{\left| \omega ^{\alpha /2-1}(x)v(x) - \omega ^{\alpha /2-1}(y)v(y)\right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy \\&\quad \le 2\iint _{\Omega _{I,a}} \omega ^{ \alpha /2 +\beta }(x)\omega ^{ \alpha /2-1}(y)\frac{\left| v(x) - v(y)\right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy \\&\qquad +\,2\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)v^{2}(x) \frac{\left| \omega ^{\alpha /2-1}(x) - \omega ^{\alpha /2-1}(y) \right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy \\&\quad \le 2 \left\| v\right\| _{\sigma , \omega ^{\alpha /2},B}^2 + C\iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x) v^2(x) \frac{\left| \omega ^{\alpha /2-1}(x) - \omega ^{\alpha /2-1}(y) \right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy . \end{aligned}$$

By Lemma D.4, the last term is bounded by $C \int _{I} \omega ^{3\alpha /2+\beta -2\sigma -2 }(x) v^2(x) \,dx$ and thus it is bounded by $C \int _{I} \omega ^{\alpha /2 }(x) v^2(x) \,dx.$ Then we have shown that

$$\begin{aligned} \iint _{\Omega _{I,a}} \omega ^{\alpha /2+\beta }(x)\frac{\left| \omega ^{\alpha /2}(x)\partial _xv(x) - \omega ^{\alpha /2}(y)\partial _xv(y)\right| ^2}{\left| x-y\right| ^{1+2\sigma }}\,dx\,dy \le C \left\| v\right\| _{\beta ,\omega ^{\alpha /2},B}^2. \end{aligned}$$

Then by the definition of $B^{\beta }_{\alpha /2}(I)$, we have $u \in B^{\beta }_{\alpha /2}(I)$. $\square$