Chebyshev documentation

docs/src/chebyshev.md

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+Chebyshev tranform via Fourier transform
+===============================================
+
+We express a function $f$ as a sum of Chebyshev polynomials
+```math 
+\begin{equation}
+f(x) = \sum^N_{n=0} a_{n}T_n(x) \tag{1}
+\end{equation}
+```
+The Chebyshev polynomials are defined by 
+```math
+\begin{equation}
+T_n(\cos \theta) = \cos n \theta, {\rm~with~}x = \cos \theta.
+\end{equation}
+```
+We can see how to find $\{a_{n}\}$ given $\{f(x_j)\}$ via Fourier transform. 
+The Fourier series representation of $f$ on a uniform grid indexed by $j$ is defined by 
+```math
+\begin{equation}
+f_j = \sum_{k=0}^{M-1} b_{k}\exp\left[i \frac{2\pi k j}{M}\right].\tag{2}
+\end{equation}
+```
+
+Gauss-Chebyshev-Lobotto points
+===============================================
+
+We pick points 
+```math
+\begin{equation}
+x_j = \cos \theta_j, \quad \theta_j = \frac{j \pi}{N} \quad 0 \leq j \leq N.
+\end{equation}
+```
+Then 
+```math
+\begin{equation}
+T_n(x_j) = \cos \frac{n j \pi}{N}.
+\end{equation}
+```
+Assuming that $M = 2N$, with $N$ an integer, and $b_{k} = b_{M-k}$ for $k>0$, we have that 
+```math
+\begin{equation}
+f_j = b_{0} + b_{N}(-1)^j + \sum_{n=1}^{N-1}
+b_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right).
+\end{equation}
+```
+Comparing this to the expression for $f(x_j)$ in the Chebyshev representation,
+```math
+\begin{equation}
+f_j = a_{0} + a_{N}(-1)^j + \frac{1}{2}\sum_{n=1}^{N-1}
+a_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right),
+\end{equation}
+```
+we find that the Chebyshev representation on the Chebyshev points is equivalent 
+to the Fourier representation on the uniform grid points, if we identify
+```math
+\begin{equation}
+b_{0} = a_{0}, \quad  b_{N} = a_{N}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N-1.
+\end{equation}
+```
+This fact allows us to carry out the Chebyshev tranform by Fourier transforming the $\{f_j\}$ data
+and carrying out the correct normalisation of the resulting coefficients. 
+
+Gauss-Chebyshev-Radau points
+===============================================
+
+The last subsection dealt with grids which contain both endpoints on the $[-1,1]$ domain. 
+Certain problems require domains which contain a single endpoint, i.e., $x \in (-1,1]$. For 
+these cases we choose the points 
+```math
+\begin{equation}
+x_j = \cos \theta_j, \quad \theta_j = \frac{2 j \pi}{2 N + 1} \quad 0 \leq j \leq N.
+\end{equation}
+```
+Writing out the Chebyshev series (1),
+we have that 
+```math
+\begin{equation}
+\begin{split}f(x_j) = & \sum^N_{n=0} a_{n} \cos \frac{2 n j \pi}{2 N + 1} \\ &
+= a_{0} + \sum^N_{n=1} \frac{a_{n}}{2}\left(\exp\left[i \frac{2\pi n j}{2N +1}\right] + \exp\left[-i \frac{2\pi n j}{2N +1}\right]\right).\end{split}
+\tag{3}
+\end{equation}
+```
+The form of the series (3) is identical to the form of 
+a Fourier series on an odd number of points, i.e., taking $M = 2 N + 1$ in equation (2),
+and assuming $b_{k} = b_{M -k}$ for $k>1$,
+we have that 
+```math
+\begin{equation}
+f_j = b_{0} + \sum_{k=1}^{N} b_{k}\left(\exp\left[i \frac{2\pi k j}{2N+1}\right] + \exp\left[-i \frac{2\pi k j}{2N+1} \right]\right).
+\end{equation}
+```
+We can thus take a Chebyshev transform using a Fourier transform on Gauss-Chebyshev-Radau points if we identify 
+```math
+\begin{equation}
+b_{0} = a_{0}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N.
+\end{equation}
+```
+
+Chebyshev coefficients of derivatives of a function
+===============================================
+
+Starting from the expression of $f$ as a sum Chebyshev polynomials, equation (1),
+we can obtain an expression for the derivative
+```math
+\begin{equation}
+\frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}.\tag{4}
+\end{equation}
+```
+We note that we must be able to express ${d f}/{d x}$ as a sum 
+of Chebyshev polynomials of up to order $N-1$, i.e.,
+```math
+\begin{equation}
+\frac{d f}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}.
+\end{equation}
+```
+We must determine the set $\{d_{n}\}$ in terms of the set $\{a_{n}\}$.
+First, we equate the two expressions to find that 
+```math
+\begin{equation}
+\sum^N_{k=0} a_{k}\frac{d T_{k}}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}.\tag{5}
+\end{equation}
+```
+We use the Chebyshev polynomials of the second kind $U_n{x}$ to aid us in the calculation of the set $\{d_{n}\}$. 
+These polynomials are defined by 
+```math
+\begin{equation}
+U_{0}(x) = 1, \quad U_{1}(x) = 2x, \quad U_{n+1} = 2 x U_{n}(x) - U_{n-1}(x).
+\end{equation}
+```
+Note the useful relations 
+```math
+\begin{equation}
+\frac{d T_{n}}{d x} = n U_{n-1}, {\rm~for~}n\geq 1, \quad \frac{d T_{0}}{d x} = 0,
+\end{equation}
+```
+```math
+\begin{equation}
+T_{n} = \frac{1}{2}\left(U_{n} - U_{n -2}\right), T_{0} = U_{0}\quad, {\rm ~and~} \quad 2 T_{1} = U_{1}.
+\end{equation}
+```
+Using these identities, which may be obtained from the trigonometric definition of $U_{n}(\cos \theta)$
+```math
+\begin{equation}
+U_{n}(\cos \theta) \sin \theta = \sin \left((n+1)\theta\right),
+\end{equation}
+```
+we find that equation (5) becomes 
+```math
+\begin{equation}
+\begin{split}\sum^N_{n=1} a_{n} n U_{n-1}(x) =& \frac{d_{N-1}}{2}U_{N-1}+\frac{d_{N-2}}{2}U_{N-2} 
+\\ & + \sum^{N-3}_{k=1} \frac{d_{k}-d_{k+2}}{2}U_{k} + \left(d_{0} - \frac{d_{2}}{2}\right)U_{0}. \end{split}\tag{6}
+\end{equation}
+```
+Using the orthogonality relation 
+```math
+\begin{equation}
+\int^1_{-1} U_{m}(x)U_{n}(x)\sqrt{1-x^2} \; d x = 
+\left\{\begin{array}{l} 0 {\rm ~if~} n\neq m  \\ \pi/2 {\rm ~if~} n=m \\ \end{array} \right.
+\end{equation}
+```
+we obtain the (unqiuely-determined) relations 
+```math
+\begin{equation}
+\begin{split} &d_{N-1} = 2Na_{N},\quad d_{N-2} = 2(N-1)a_{N-1}, \\ 
+& d_{k} = 2(k+1) a_{k+1} + d_{k+2}, \quad d_{0} = \frac{d_{2}}{2} + a_{1}.\end{split}\tag{7}
+\end{equation}
+```       
+
+Clenshaw-Curtis integration weights
+===============================================
+
+We require the integration weights for the set of points $\{x_j\}$ chosen 
+in our numerical scheme. The weights $w_{j}$ are defined implicitly by 
+```math
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \sum_{j=0}^N f(x_j) w_{j}.\tag{8}
+\end{equation}
+```
+In the Chebyshev scheme we use the change of variables $x = \cos \theta$
+to write 
+```math
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \int^\pi_0 f(\cos\theta) \sin \theta \; d \theta .\tag{9}
+\end{equation}
+```
+ Using the series expansion (1) in equation (9)
+ we find that 
+```math
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \sum^N_{n=0} a_{n}\int^\pi_0 \cos (n \theta) \sin \theta \; d \theta
+.\tag{10}
+\end{equation}
+```
+Note the integral identity
+```math
+\begin{equation}
+\int^\pi_0 \cos(n \theta) \sin \theta \; d \theta = \frac{\cos(n \pi) +1}{1 - n^2} {\rm~for~} n \geq 0.
+\end{equation}
+```
+Also note that 
+```math
+\begin{equation}
+\frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{array}{l} 0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z}  \\ 2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}  \end{array}\right.
+\end{equation}
+```
+We define 
+```math
+\begin{equation}
+J_{n} = \frac{\cos(n \pi) +1}{1 - n^2}.
+\end{equation}
+```
+Using this definition, we can write the integral of $f(x)$ can be written 
+in terms of a sum over of the Chebyshev coefficients:
+```math
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N J_{n} a_{n}.\tag{11}
+\end{equation}
+```
+To avoid computing the set of coefficients $\{a_{n}\}$ every time we wish to integrate $f(x_j)$,
+we use the inverse transforms. This transform allows us to rewrite equation (11) in the form (8).
+Since the inverse transform differs between the Gauss-Chebyshev-Lobotto and Gauss-Chebyshev-Radau cases, we treat each 
+case separately below. 
+
+Weights on Gauss-Chebyshev-Lobotto points
+===============================================
+We use the inverse transformation 
+```math
+\begin{equation}
+a_{n} = \frac{q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \tag{12}
+\end{equation}
+```
+where 
+```math
+\begin{equation}
+q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n\neq0,N  \\ 1 {\rm ~if~} n=0,N  \end{array}\right.
+\end{equation}
+```
+and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
+```math
+\begin{equation}
+\hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1.
+\end{equation}
+```
+With this inverse tranformation, we can write 
+```math
+\begin{equation}
+\begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N-1}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
+& = \sum^{2N-1}_{j=0}\sum^{2N-1}_{n=0} \frac{\hat{f}_j J_{n}}{2N} \exp\left[-i \frac{2\pi n j}{2N}\right] \\ 
+& = \sum^{2N-1}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\tag{13}
+\end{equation}
+```
+where in the first step we have extended the sum from $N$ to $2N-1$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
+```math
+\begin{equation}
+J_{j} = J_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1,
+\end{equation}
+```
+```math
+\begin{equation}
+a_{j} = a_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1.
+\end{equation}
+```
+In the second step we use the definition of the inverse transform (14) , and 
+in the third step we define 
+```math
+\begin{equation}
+v_{j} = \sum_{n=0}^{2N-1}\frac{J_{n}}{2N}\exp\left[-i \frac{2\pi n j}{2N}\right].
+\end{equation}
+```
+Finally, we can compare equations (8) and (13) and deduce that 
+```math
+\begin{equation}
+w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.
+\end{equation}
+```
+We can write $v_{j}$ in terms of a discrete cosine transform, i.e.,
+```math
+\begin{equation}
+v_{j} = \frac{1}{2N}\left(J_{0} + (-1)^jJ_{N} + 2\sum_{n=1}^{N-1}J_{n}\cos\left(\frac{\pi n j}{N}\right)\right).
+\end{equation}
+```
+
+Weights on Gauss-Chebyshev-Radau points
+===============================================
+We use the inverse transformation 
+```math
+\begin{equation}
+a_{n} = \frac{q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right],\tag{14}
+\end{equation}
+```
+where 
+```math
+\begin{equation}
+q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n > 0  \\ 1 {\rm ~if~} n=0  \end{array}\right.
+\end{equation}
+```
+and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
+```math
+\begin{equation}
+\hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j+1}){\rm~for~} N+1 \leq j \leq 2N.
+\end{equation}
+```
+Note that the details of what is the appropriate FFT order depends on the order in which the points $x_j$ are stored.
+The key detail in the Chebyshev-Radau scheme is that (in the notation above)
+$x_0 = 1$ is not a repeated point, and must occupy $\hat{f}_0$. 
+With this inverse tranformation, we can write 
+```math
+\begin{equation}
+\begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
+& = \sum^{2N}_{j=0}\sum^{2N}_{n=0} \frac{\hat{f}_j J_{n}}{2N+1} \exp\left[-i \frac{2\pi n j}{2N+1}\right] \\ 
+& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\tag{15}\end{equation}
+```
+where in the first step we have extended the sum from $N$ to $2N$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
+```math
+\begin{equation}
+J_{j} = J_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N,
+\end{equation}
+```
+```math
+\begin{equation}
+a_{j} = a_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N.
+\end{equation}
+```
+In the second step we use the definition of the inverse transform (14), and 
+in the third step we define 
+```math
+\begin{equation}
+v_{j} = \sum_{n=0}^{2N}\frac{J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].
+\end{equation}
+```
+Finally, we can compare equations (8) and (15) and deduce that 
+```math
+\begin{equation}
+w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.
+\end{equation}
+```