Abstract
We develop a sparse spectral method for a class of fractional differential equations, posed on \(\mathbb {R}\), in one dimension. These equations may include sqrt-Laplacian, Hilbert, derivative, and identity terms. The numerical method utilizes a basis consisting of weighted Chebyshev polynomials of the second kind in conjunction with their Hilbert transforms. The former functions are supported on \([-1,1]\) whereas the latter have global support. The global approximation space may contain different affine transformations of the basis, mapping \([-1,1]\) to other intervals. Remarkably, not only are the induced linear systems sparse, but the operator decouples across the different affine transformations. Hence, the solve reduces to solving K independent sparse linear systems of size \(\mathcal {O}(n)\times \mathcal {O}(n)\), with \(\mathcal {O}(n)\) nonzero entries, where K is the number of different intervals and n is the highest polynomial degree contained in the sum space. This results in an \(\mathcal {O}(n)\) complexity solve. Applications to fractional heat and wave equations are considered.
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For reproducibility, an implementation of the spectral method as well as scripts to generate the plots and solutions can be found at https://github.com/ioannisPApapadopoulos/SumSpaces.jl. The version of the software used in this paper is archived on Zenodo [53, 59, 60]. Any timings provided were achieved on computer with a 2.90GHz Intel(R) Core(TM) i7-10700 CPU and 16GB of RAM.
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Acknowledgements
The authors would like to thank the anonymous reviewers for their insightful comments.
Funding
This work was completed with the support of the EPSRC grant EP/T022132/1 “Spectral element methods for fractional differential equations, with applications in applied analysis and medical imaging” and the Leverhulme Trust Research Project Grant RPG-2019-144 “Constructive approximation theory on and inside algebraic curves and surfaces.” IP was also supported by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy – The Berlin Mathematics Research Center MATH+ (EXC-2046/1, project ID: 390685689).
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Appendices
Appendix 1: Approximate inverse Fourier transform via the FFT
The setup of the spectral method requires four inverse Fourier transforms per interval. Although this can be reduced if the intervals are translations on \(\mathbb {R}\) and are of the same size as resulting solutions are simply the equivalent translations. In this section we describe how to use the FFT to approximate the inverse Fourier transform. The FFT implements the discrete Fourier transform (DFT) or its inverse (IDFT) with \(\mathcal {O}(N \log N)\) complexity where N is the size of the input vector. Consider a vector \(\varvec{\textbf{u}} \in \mathbb {R}^N\). We denote the IDFT as \(\mathcal {F}_D^{-1}\) and define it as (as implemented in ifft of FFTW.jl):
Suppose that N is even. Then, we also define the shifted IDFT, \(\hat{\mathcal {F}}_D^{-1}\) such that the components of \(\mathcal {F}_D^{-1}[\varvec{\textbf{u}}]\) are reordered from \(j=N/2+1,\dots ,N,1,\dots , N/2 \),
This is implemented as iffshift(ifft(u)) in FFTW.jl. Consider the approximation, \(j=1,\dots ,N\):
Here, we choose the parameters \(W \gg 1\) and \(N \gg 1\), N even. The points \(\omega _n:= n \delta -W\), \(n=0,\dots ,N-1\), where \(\delta := 2W/N\). Substituting in the definition of \(\omega _n\), we see that
Define the vector \(\varvec{\textbf{u}}_n = u(\omega _{n-1})\), \(n=1,\dots ,N\). Moreover, we fix \(x_j\) as
Then, the right-hand side of A4 is equal to
By noting that \(\textrm{e}^{-\textrm{i}\pi (n-1) } = \textrm{e}^{\textrm{i}\pi (n-1) }\) for \(n \in \mathbb {Z}\), then for \(1 \le j \le N/2\), A6 is equal to
and for \(N/2+1 \le j \le N\), A6 is equal to
Hence, by the definition of the shifted IDFT in A2, we see that, for \(j,n=1,\dots ,N\),
where \(\varvec{\textbf{u}}_n = u(2W(n-1)/N - W)\), \(n = 1,\dots ,N\).
Appendix 2: Approximating the appended sum space functions
With a naïve approach, it would take four integrals per interval to approximate the additional functions found in the appended sum space. This setup cost may become prohibitive for a large number of intervals. However, the next proposition reveals that the additional functions on intervals that are translated (and not scaled) are simply translations of the additional functions associated to one reference interval.
Proposition B.1
(Translations of the reference interval) Consider the interval \(I = [a, b]\) and its associated reference interval \(I_R = [-(b-a)/2, (b-a)/2]\). Then,
Proof
The result follows from the identity \(\mathcal {F}[f(\diamond + \alpha )](\omega ) = \textrm{e}^{\textrm{i}\alpha \omega } \mathcal {F}[f(\diamond )](\omega )\) and a routine calculation.
The additional functions on the reference interval are approximated with one integral each. Moreover, all intervals of the same width map to the same reference interval. Hence, the setup of the method only requires four integrals per interval of different width. In particular, if all the intervals have the same width, we only require four integrals in total for the setup, irrespective of the number of intervals used.
Appendix 3: Special cases when \(\lambda = 0\)
In this subsection, we discuss the conditioning of \(L^+\) for choices of \(\mu \) and \(\eta \) when \(\lambda = 0\). The ill-conditioning is alleviated by removing the rows of \(L^+\) associated with functions that are no longer in the range of \(\mathcal {L}_{0,\mu ,\eta } S_n^+\).
1.1 C.1 \(\lambda =\mu =\eta =0\)
The first special case that we consider is \(\lambda =\mu =\eta =0\). Here, (1) reduces to finding \(u \in H^{1/2}(\mathbb {R})\) that satisfies
The first issue is that the matrix \(L^+\), as constructed in 66, is singular. This is due to three rows and one column whose entries are all zero. The three rows correspond to the dual sum space functions \(\tilde{U}_{-2}(x)\), \(\tilde{U}_{n+1}(x)\), and \(V_{n+2}(x)\) which do not lie in the range of \((-\Delta )^{1/2} S_n^+(x)\). The column corresponds to the sum space function \(\tilde{T}_0(x)\). Hence, we remove these rows and column from \(L^+\) resulting in an \((2n+4) \times (2n+6)\) matrix. The second issue is that the additional functions computed to form the square matrix \(L^+\) become problematic or redundant. By Proposition 3.5, we have that \(\tilde{u}_{j}(x) = W_j(x)\), \(j \ge 0\) and \(v_j(x) = \tilde{T}_j(x)\), \(j \ge 1\). Hence, two of the additional functions are already included in the sum space and correspond to two columns that should be removed. At first glance, this is advantageous as this reduces \(L^+\) to a \((2n+4) \times (2n+4)\) matrix. However, computing the remaining additional functions \(\tilde{u}_{-1}(x)\) and \(v_0(x)\) poses a numerical difficulty. Although \(\tilde{U}_{-1}, V_0 \in H^{-1/2}(\mathbb {R})\), since \(\lambda = 0\), the Lax–Milgram theorem does not guarantee existence of solutions. Attempting to compute these two solutions by a forward and inverse Fourier transform fails. Hence, as discussed in Sect. 1, there exist solutions that satisfy the Fourier multiplier reformulation 17 but not (6). Ignoring these technical issues and naïvely computing the inverse Fourier transform via (8) does recover the non-decaying solutions:
However, as neither of these tend to zero as \(|x| \rightarrow \infty \), they cannot live in \(H^{1/2}(\mathbb {R})\). Therefore, we choose to remove the columns associated with \(\tilde{u}_{-1}(x)\) and \(v_0(x)\) as well. Since all the additional functions have been removed, we are only required to consider the range of \((-\Delta )^{1/2} S_n(x)\). The range does not include \(\tilde{U}_{-1}(x)\) and \(V_0(x)\). Hence, these functions can be removed from the dual sum space (whilst still preserving the exact map \((-\Delta )^{1/2}: S_n \rightarrow {\textbf{S}}^*_n\)) and their corresponding rows from \(L^+\). In summary, we remove the columns associated with \(\tilde{T}_0(x)\), \(v_0(x)\), \(\tilde{u}_{-1}(x)\), \(v_1(x)\), and \(\tilde{u}_{0}(x)\), and the rows associated with \(\tilde{U}_{-2}(x)\), \(V_0(x)\), \(\tilde{U}_{-1}(x)\), \(V_{n+2}(x)\), and \(\tilde{U}_{n+1}(x)\) reducing \(L^+\) from a \((2n+7)\times (2n+7)\) matrix to a \((2n+2)\times (2n+2)\) matrix. We emphasize that the action of \((-\Delta )^{1/2}\) on \(S_n(x)\) is still represented exactly by the reduced \(L^+\).
1.2 C.2 \(\lambda = \mu = 0\), \(|\eta | > 0\)
We now consider where \(\lambda = \mu = 0\) but \(|\eta |>0\). This corresponds to finding \(u \in H^1(\mathbb {R})\) that satisfies
This special case is similar to the previous one. Again \(L^+\) is singular due to three rows and a column which contain only zeroes. The rows are associated with the dual sum space functions \(\tilde{U}_{-2}(x)\), \(\tilde{U}_{n+1}(x)\), and \(V_{n+2}(x)\) which do not lie in the range of \((\eta \frac{\textrm{d}}{\textrm{d}x}+ (-\Delta )^{1/2}) S_n^+(x)\) and the column corresponds to \(\tilde{T}_0(x)\). Hence, we remove those three rows and column. Moreover, \(\tilde{u}_{1}(x)\) and \(v_0(x)\) cannot be computed via a forward and inverse Fourier transform due to the same issues as in the previous case. Hence, the columns associated with \(\tilde{u}_{1}(x)\) and \(v_0(x)\) must also be removed. This implies that the rows associated with \(V_0(x)\) and \(\tilde{U}_{-1}(x)\) are now all zero and must also be removed. Hence, as in the previous case, we remove the columns associated with \(\tilde{T}_0(x)\), \(v_0(x)\), \(\tilde{u}_{-1}(x)\), \(v_1(x)\), and \(\tilde{u}_{0}(x)\), and the rows associated with \(\tilde{U}_{-2}(x)\), \(V_0(x)\), \(\tilde{U}_{-1}(x)\), \(V_{n+2}(x)\), and \(\tilde{U}_{n+1}(x)\) reducing \(L^+\) from a \((2n+7)\times (2n+7)\) matrix to a \((2n+2)\times (2n+2)\) matrix. Moreover, the action of \((\frac{\textrm{d}}{\textrm{d}x}+ (-\Delta )^{-1/2})\) on \(S_n(x)\) is still represented exactly by the reduced \(L^+\).
Remark C.1
Since we have removed the additional functions from the appended sum space, the setup cost is minimal when \(\lambda =\mu =0\) and \(\eta \in \mathbb {R}\) as no integrals are required.
1.3 C.3 \(\lambda = 0\), \(|\mu |>0\), \(\eta \in \mathbb {R}\)
The final case to consider is when \(\lambda = 0\) but \(\mu \ne 0\). In this case there is only one zero row and column associated with \(\tilde{U}_{-2}(x)\) and \(\tilde{T}_0(x)\), respectively. Removing that row and column results in an invertible matrix that still represents the mapping exactly. Moreover, unlike the previous two cases, we recover \(v_0(x), \tilde{u}_{-1}(x) \in H^{1/2}(\mathbb {R})\). However, we note that with increasing n (truncation degree) \(L^+\) has two singular values that quickly decrease to zero which impact the conditioning of \(L^+\). If the problem setup requires high truncation degree n, and small parameter \(|\mu | \ll 1\), then \(L^+\) may become numerically singular. In this case, we suggest using the additional functions \(v_{n+2}(x)\) and \(\tilde{u}_{n+1}(x)\) instead of \(v_1(x)\) and \(\tilde{u}_{0}(x)\) as suggested in Remark 4.3. This results in a well-conditioned \(L^+\) irrespective of the choices of n and \(\mu \).
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Papadopoulos, I.P.A., Olver, S. A sparse spectral method for fractional differential equations in one-spatial dimension. Adv Comput Math 50, 69 (2024). https://doi.org/10.1007/s10444-024-10164-1
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DOI: https://doi.org/10.1007/s10444-024-10164-1