Spectral convergence for orthogonal polynomials on triangles

Abstract

We study the behavior of orthogonal polynomials on triangles and their coefficients in the context of spectral approximations of partial differential equations. In these spectral approximations one studies series expansions $u=\sum _{k=0}^{\infty } \hat{u}_k \phi _k$ where the $\phi _k$ are orthogonal polynomials. Our results show that for any function $u\in C^{\infty }$ the series expansion converges faster than with polynomial order.

Appendix

First of all we repeat some of the properties of the Jacobi polynomials which we need in this paper. For further properties see also [16]. The Jacobi polynomials $P_n^{\alpha ,\beta }(x)$ satisfy following properties:

• $P_n^{\alpha ,\beta }(x)$ are orthogonal on $[-1,1]$ with respect to the weight function $w(x)=(1-x)^\alpha (1+x)^\beta , \alpha >-1$ and $\beta >-1$. It holds

  $$\begin{aligned} \int \limits _{-1}^1w(x)P_n^{\alpha ,\beta }(x)P_k^{\alpha ,\beta }(x)dx=\frac{\delta _{n,k}2^{\alpha +\beta +1}}{2n+\alpha +\beta +1}\frac{\Gamma (n+\alpha +1)\Gamma (n+\beta +1)}{n!\Gamma (n+\alpha +\beta +1)}. \end{aligned}$$

• $P_n^{\alpha ,\beta }(1)=\small \left(\begin{array}{c}{n+\alpha }\\ {n}\end{array}\right)$

• $P_n^{\alpha ,\beta }(x)=(-1)^n P_n^{\beta ,\alpha }(-x)$

• If $q=\max \{\alpha ,\beta \}\ge -\frac{1}{2}$, then $\max \nolimits _{-1\le x\le 1}|P_n^{\alpha ,\beta }(x)|=\small \left(\begin{array}{c}{n+q}\\ {n}\end{array}\right)$ holds.

Now we prove the estimates of the orthogonal polynomials which are stated in Sect. 2.

Proof

(of Lemma 2.2)

With the inner product

$$\begin{aligned} (A_{m,l},A_{m,l})= \frac{1}{(2l+\gamma -\alpha )(2(m+l)+\gamma )} \frac{(l+1)_p}{(l+\beta )_p} \frac{(m+a_l)(m)_{\alpha }}{m(m+a_l)_\alpha } \end{aligned}$$

it follows

$$\begin{aligned} \frac{1}{||A_{m,l}||_{L^2(\mathbb{T },h)}}&= \sqrt{\frac{(2l+\gamma -\alpha )(2(m+l)+\gamma )}{1}\frac{m}{(m+a_l)}\frac{(l+\beta )_{p}(m+a_l)_\alpha }{(l+1)_{p}(m)_{\alpha }}}\\&\le \sqrt{\frac{(2m+2l+\gamma )(2m+2l+\gamma )(l+\beta )_p m (m+a_l)_\alpha }{(l+1)_p(m+a_l)(m)_\alpha }}\\&= (2m+2l+\gamma ) \kappa _{l,m}\\&\le 2(m+l+\gamma ) \kappa _{l,m}. \end{aligned}$$

$\square$

In the proof of Lemma 2.3 we prove only estimate (4) and equation (8). The inequality (5) follows directly from the fact that Jacobi polynomials take their maximum at the boundary. The proof of (6) and (7) is a combination of the approach taken in in (4) and (5).

Proof

(of Lemma 2.3)

In the proof we use the estimate

$$\begin{aligned} |(1-x^2)^{\frac{1}{4}}g_n^{\alpha ,\beta }(x)| \le \frac{C}{(2n+\alpha +\beta +1)^{\frac{1}{4}}} \quad \forall x \in [-1,1], \end{aligned}$$

(24)

with $\alpha , \beta \ge 0, n\in \mathbb{N }_0, g_n^{\alpha ,\beta }(x)$ defined by

$$\begin{aligned} g_n^{\alpha ,\beta }(x)=\left(\frac{\Gamma (n+1)\Gamma (n+\alpha +\beta +1)}{\Gamma (n+\alpha +1)\Gamma (n+\beta +1)} \right)^{\frac{1}{2}} \left(\frac{1-x}{2} \right)^{\frac{\alpha }{2}}\left(\frac{1+x}{2} \right)^{\frac{\beta }{2}}P_n^{\alpha ,\beta }(x) \end{aligned}$$

and $C$ is a positive constant less than 12, see [5, p.1]. With this inequality and the relation

$$\begin{aligned} \frac{\Gamma (m+\alpha )}{\Gamma (m)}=(m)_{\alpha } \end{aligned}$$

we estimate the factors of the APK polynomials in ${\stackrel{\circ }{\mathbb{T }}}$. We have

$$\begin{aligned}&|P_m^{\alpha -1,a_l}(1-2x)| \le \frac{C}{(2m+a_l+\alpha )^{\frac{1}{4}}(1-(1-2x))^{\frac{1}{4}}(1+(1-2x))^{\frac{1}{4}} }\\&\quad \times \frac{2^{\frac{\alpha -1+a_l}{2}}}{(1-(1-2x))^{\frac{\alpha -1}{2}}(1+(1-2x))^{\frac{a_l}{2}}} \left( \frac{\Gamma (m+\alpha )\Gamma (m+a_l+1)}{\Gamma (m+1)\Gamma (m+\alpha +a_l)} \right)^\frac{1}{2}\\{}&=\frac{C \cdot 2^{\frac{\alpha -1+a_l}{2}}}{(2(m+l)+\gamma )^{\frac{1}{4}}(2x)^{\frac{1}{4}+\frac{\alpha -1}{2}}(2(1-x))^{\frac{1}{4}+\frac{a_l}{2}}} \left( \frac{(m)_\alpha (m+a_l)}{(m+a_l)_\alpha m} \right)^{\frac{1}{2}}\\&=\frac{C}{(2(m+l)+\gamma )^{\frac{1}{4}}(x)^{\frac{1}{4}+\frac{\alpha -1}{2}}((1-x))^{\frac{1}{4}+\frac{a_l}{2}}\sqrt{2}}\left( \frac{(m)_\alpha (m+a_l)}{(m+a_l)_\alpha m} \right)^{\frac{1}{2}} \end{aligned}$$

and

$$\begin{aligned}&\left|P_l^{p,\beta -1}\left(\frac{2y}{1-x}-1\right)(1-x)^l\right|\\&\quad \le \frac{C(1-x)^l}{\left[(2l+p+\beta ) \left(1-\frac{2y}{1-x}+1\right)\left(1+\frac{2y}{1-x}-1\right)\right]^{\frac{1}{4}}}\\&\qquad \times \left( \frac{\Gamma (l+p+1)\Gamma (l+\beta )2^{p+\beta -1}}{(\Gamma (l+1) \Gamma (l+p+\beta )\left(2-\frac{2y}{1-x} \right)^p \left( \frac{2y}{1-x} \right)^{\beta -1}} \right)^\frac{1}{2}\\&\quad = \frac{C(1-x)^{l+\frac{p+\beta -1}{2}+\frac{1}{2}} 2^{\frac{p+\beta -1}{2}}}{\sqrt{2}(2l+\beta +p)^{\frac{1}{4}}2^{\frac{p+\beta -1}{2}} (1-x-y)^{\frac{1}{4}+\frac{p}{2} } y^{\frac{1}{4}+\frac{\beta -1}{2}} } \left(\frac{(l+1)_p}{(l+\beta )_p} \right)^{\frac{1}{2}}\\&\quad = \frac{C(1-x)^{l+\frac{p+\beta }{2}}}{\sqrt{2}(2l+\beta +p)^{\frac{1}{4}}(1-x-y)^{\frac{1}{4}+\frac{p}{2} } y^{\frac{1}{4}+\frac{\beta -1}{2}} } \left(\frac{(l+1)_p}{(l+\beta )_p} \right)^{\frac{1}{2}}. \end{aligned}$$

Hence we get an upper bound for all points $(x,y)\in {\stackrel{\circ }{\mathbb{T }}}$ as

$$\begin{aligned} |A_{m,l}(x,y)|&= \left|P_m^{\alpha -1,a_l}(1-2x)P_l^{p,\beta -1}\left(\frac{2y}{1-x}-1\right)(1-x)^l\right| \\&= \left|P_m^{\alpha -1,a_l}(1-2x)\right| \left|P_l^{p,\beta -1}\left(\frac{2y}{1-x}-1\right)(1-x)^l\right|\\&\le \frac{C}{(2(m+l)+\gamma )^{\frac{1}{4}}(x)^{\frac{1}{4}+\frac{\alpha -1}{2}}(1-x)^{\frac{1}{4}+\frac{a_l}{2}}\sqrt{2}}\left( \frac{(m)_\alpha (m+a_l)}{(m+a_l)_\alpha m} \right)^{\frac{1}{2}}\\&\quad \times \frac{C(1-x)^{l+\frac{p+\beta }{2}}}{\sqrt{2}(2l+\beta +p)^{\frac{1}{4}}(1-x-y)^{\frac{1}{4}+\frac{p}{2} } y^{\frac{1}{4}+\frac{\beta -1}{2}} } \left(\frac{(l+1)_p}{(l+\beta )_p} \right)^{\frac{1}{2}}\\&= \tilde{D}(x,y)\frac{1}{(2l+\beta +p)^{\frac{1}{4}}(2(m+l)+\gamma )^{\frac{1}{4}}\kappa _{l,m}}, \end{aligned}$$

with

$$\begin{aligned} \tilde{D}(x,y)= \frac{C^2}{2(1-x-y)^{\frac{p}{2}+\frac{1}{4}} y^{\frac{\beta -1}{2}+\frac{1}{4}}x^{\frac{\alpha -1}{2}+ \frac{1}{4}}(1-x)^{\frac{1}{4}}}. \end{aligned}$$

Finally we have to demonstrate equation (8). To this end we examine the factor

$$\begin{aligned} P_l^{p,\beta -1}\left(\frac{2y}{1-x}-1\right)(1-x)^l. \end{aligned}$$

The series expansion of the Jacobi polynomials yields

$$\begin{aligned}&P_l^{p,\beta -1}\left(\frac{2y}{1-x}-1\right)(1-x)^l\\&\quad = \frac{(1\!-\!x)^l\Gamma (p\!+\!l\!+\!1)}{l!\Gamma (p+\beta \!-\!1\!+\!l\!+\!1)} \sum \limits _{n=0}^l \left(\begin{array}{c}{l}\\ {n}\end{array}\right) \frac{\Gamma (p+\beta -1+l+n+1)}{\Gamma (p+n+1)} \left(\frac{2(y-1+x)}{2(1-x)} \right)^n\\&\quad =\frac{\Gamma (p+l+1)}{l!\Gamma (p+\beta -1+l+1)} \sum \limits _{n=0}^l \left(\begin{array}{c}{l}\\ {n}\end{array}\right) \frac{\Gamma (p+\beta -1+l+n+1)}{\Gamma (p+n+1)}\\&\qquad \times \left(y-1+x \right)^n(1-x)^{l-n}. \end{aligned}$$

If $l\ne 0$ and $(x,y)=(1,0)$ then the value of the series is $0$ and so part I is proven. If $l=0$ we get

$$\begin{aligned} \displaystyle |A_{m,0}(1,0)|=\left|P_m^{\alpha -1,a_l}(-1) \cdot \frac{\Gamma (p+1)}{\Gamma (p+\beta )} \frac{\Gamma (p+\beta )}{\Gamma (p+1)}\right|= \left(\begin{array}{c}{m+a_l}\\ {m}\end{array}\right). \end{aligned}$$

$\square$