This paper was converted on www.awesomepapers.org from LaTeX by an anonymous user.
Want to know more? Visit the Converter page.

Implementation of high-order,
discontinuous Galerkin time stepping
for fractional diffusion problems

William McLean
Abstract

The discontinuous Galerkin (dG) method provides a robust and flexible technique for the time integration of fractional diffusion problems. However, a practical implementation uses coefficients defined by integrals that are not easily evaluated. We describe specialised quadrature techniques that efficiently maintain the overall accuracy of the dG method. In addition, we observe in numerical experiments that known superconvergence properties of dG time stepping for classical diffusion problems carry over in a modified form to the fractional-order setting.

1 Introduction

The discontinuous Galerkin (dG) method provides an effective numerical procedure for the time integration of diffusion problems. In the mid-1980s, Eriksson, Johnson and Thomée [2] provided the first detailed error analysis, which has been subsequently extended and refined by numerous authors [[]and references therein]MakridakisNochetto2006,SchmutzWihler2019,SchotzauSchwab2001. The dG method has also proved effective for time stepping of fractional diffusion problems [9, 12] of the form

tu+t1αAu=f(t)for 0<tT, with u(0)=u0.\partial_{t}u+\partial_{t}^{1-\alpha}Au=f(t)\quad\text{for $0<t\leq T$, with $u(0)=u_{0}$.} (1)

Here, AA is a linear, second-order, elliptic partial differential operator over a spatial domain Ω\Omega, subject to a homogeneous Dirichlet boundary condition u=0u=0 on Ω\partial\Omega. (Our notation suppresses the dependence of uu and ff on the spatial variables.) The fractional diffusion exponent is assumed to satisfy 0<α<10<\alpha<1 (the sub-diffusive case), and the fractional time derivative is understood in the Riemann–Liouville sense: for t>0t>0 and μ>0\mu>0,

tμv=t0tωμ(ts)v(s)𝑑swhereωμ(t)=tμ1Γ(μ).\partial_{t}^{\mu}v=\frac{\partial}{\partial t}\int_{0}^{t}\omega_{\mu}(t-s)v(s)\,ds\quad\text{where}\quad\omega_{\mu}(t)=\frac{t^{\mu-1}}{\Gamma(\mu)}.

The partial integro-differential equation (1) arises in a variety of physical models [3, 11] of diffusing particles whose behaviour is described by a continuous-time random walk for which the waiting-time distribution is a power law that decays like 1/t1+α1/t^{1+\alpha}. The expected waiting time is therefore infinite, and the mean-square displacement is proportional to tαt^{\alpha}. Standard Brownian motion is recovered in the limit as α1\alpha\to 1, when (1) reduces to the classical diffusion equation.

Our main concern in the present work is with the practical implementation of dG time stepping for (1), and in particular with the accurate evaluation of certain coefficients Hijn,nH^{n,n-\ell}_{ij} used during the nnth step. Section 2 introduces the dG method for the fractional ODE case of (1), in which the operator AA is replaced by a scalar λ>0\lambda>0. We will see in the simplest, lowest-order scheme, when the dG solution is piecewise-constant in time, that

H11n,0=tn1tnddt(tn1tωα(ts)𝑑s)𝑑tH^{n,0}_{11}=\int_{t_{n-1}}^{t_{n}}\frac{d}{dt}\biggl{(}\int_{t_{n-1}}^{t}\omega_{\alpha}(t-s)\,ds\biggr{)}\,dt

and

H11n,n=tn1tnddt(t1tωα(ts)𝑑s)𝑑tfor 1n1,H^{n,n-\ell}_{11}=\int_{t_{n-1}}^{t_{n}}\frac{d}{dt}\biggl{(}\int_{t_{\ell-1}}^{t_{\ell}}\omega_{\alpha}(t-s)\,ds\biggr{)}\,dt\quad\text{for $1\leq\ell\leq n-1$,}

where 0=t0<t1<t2<0=t_{0}<t_{1}<t_{2}<\cdots are the discrete time levels. We easily verify that H11n,0=ωα+1(kn)=knα/Γ(α+1)H^{n,0}_{11}=\omega_{\alpha+1}(k_{n})=k_{n}^{\alpha}/\Gamma(\alpha+1), for a step-size kn=tntn1k_{n}=t_{n}-t_{n-1}, and

H11n,n\displaystyle H^{n,n-\ell}_{11} =ωα+1(tnt1)ωα+1(tnt)\displaystyle=\omega_{\alpha+1}(t_{n}-t_{\ell-1})-\omega_{\alpha+1}(t_{n}-t_{\ell}) (2)
ωα+1(tn1t1)+ωα+1(tn1t),\displaystyle\qquad{}-\omega_{\alpha+1}(t_{n-1}-t_{\ell-1})+\omega_{\alpha+1}(t_{n-1}-t_{\ell}),

but for higher-order schemes the coefficients become progressively more complicated. Although the Hijn,nH^{n,n-\ell}_{ij} can always be evaluated via repeated integration by parts, the resulting expressions are likely to suffer from roundoff when evaluated in floating-point arithmetic if nn-\ell is large. Consider just the lowest order case (2) with uniform time steps tn=nkt_{n}=nk, so that

H11n,n=kα[ωα+1(n+1)2ωα+1(n)+ωα+1(n1)].H^{n,n-\ell}_{11}=k^{\alpha}\bigl{[}\omega_{\alpha+1}(n-\ell+1)-2\omega_{\alpha+1}(n-\ell)+\omega_{\alpha+1}(n-\ell-1)\bigr{]}.

Since the factor in square brackets is a second-difference of ωα+1\omega_{\alpha+1}, its magnitude decays like (n)α2(n-\ell)^{\alpha-2} as nn-\ell increases, but the individual terms grow like (n)α(n-\ell)^{\alpha}.

We are therefore led to evaluate the coefficients Hijn,nH^{n,n-\ell}_{ij} via quadratures with positive weights. No special techniques are needed for n2\ell\leq n-2, but when =n\ell=n or n1n-1 we must deal with weakly singular integrands. In Section 3, we show how certain substitutions reduce the problem to dealing with integrands that are either smooth, or are products of smooth functions and standard Jacobi weight functions. Similar substitutions, known as Duffy transformations [1], have long been used to compute singular integrals arising in the boundary element method.

Section 4 introduces a spatial discretisation for the fractional PDE (1) and describes the structure of the linear system that must be solved at each time step. In Section 5, we specialise the expressions for the coefficients by choosing Legendre polynomials as the shape functions employed in the dG time stepping.

Section 6 describes a post-processing technique that, when applied to the dG solution UU, produces a more accurate approximate solution U^\widehat{U}, known as the reconstruction [6] of UU. If UU is a piecewise polynomial of degree at most r1r-1, then U^\widehat{U} is a piecewise polynomial of degree at most rr. For a classical diffusion problem, both UU and U^\widehat{U} are known to be quasi-optimal, that is, accurate of order krk^{r} and kr+1k^{r+1}, respectively. Thus, it is natural to ask what happens in the fractional-order case, and we investigate this question in numerical experiments reported in Section 7.

2 A fractional ODE

Our central concern is present already in the zero-dimensional case when we replace the elliptic operator AA with a scalar λ0\lambda\geq 0, so that the solution u(t)u(t) is a real-valued function satisfying the fractional ODE

u+λt1αu=f(t)for 0<tT, with u(0)=u0.u^{\prime}+\lambda\partial_{t}^{1-\alpha}u=f(t)\quad\text{for $0<t\leq T$, with $u(0)=u_{0}$.} (3)

For the time discretisation, we introduce a grid

0=t0<t1<t2<<tN=T,0=t_{0}<t_{1}<t_{2}<\cdots<t_{N}=T,

and form the vector 𝒕=(t0,t1,,tN)\boldsymbol{t}=(t_{0},t_{1},\ldots,t_{N}). Let kn=tntn1k_{n}=t_{n}-t_{n-1} denote the length of the nnth (open) subinterval In=(tn1,tn)I_{n}=(t_{n-1},t_{n}). We form the disjoint union

I=I1I2IN,I=I_{1}\cup I_{2}\cup\cdots\cup I_{N},

and for any function v:Iv:I\to\mathbb{R} write

v+n=limϵ0v(t+ϵ),vn=limϵ0v(tϵ),vn=v+nvn,v^{n}_{+}=\lim_{\epsilon\downarrow 0}v(t+\epsilon),\qquad v^{n}_{-}=\lim_{\epsilon\downarrow 0}v(t-\epsilon),\qquad\llbracket v\rrbracket^{n}=v^{n}_{+}-v^{n}_{-},

provided the one-sided limits exist.

Given a vector 𝒓=(r1,r2,,rN)\boldsymbol{r}=(r_{1},r_{2},\ldots,r_{N}) of integers rn0r_{n}\geq 0, the trial space 𝒳=𝒳(𝒕,𝒓)\mathcal{X}=\mathcal{X}(\boldsymbol{t},\boldsymbol{r}) consists of the functions X:IX:I\to\mathbb{R} such that X|Inrn1X|_{I_{n}}\in\mathbb{P}_{r_{n}-1} for 1nN1\leq n\leq N. Here, m\mathbb{P}_{m} denotes the space of polynomials of degree at most m0m\geq 0, with real coefficients. The dG solution U𝒳U\in\mathcal{X} of (3) is then defined by [9, 12]

Un1X+n1+In(U+λt1αU)Xdt=InfXdt\llbracket U\rrbracket^{n-1}X^{n-1}_{+}+\int_{I_{n}}(U^{\prime}+\lambda\partial_{t}^{1-\alpha}U)X\,dt=\int_{I_{n}}fX\,dt (4)

for Xrn1X\in\mathbb{P}_{r_{n}-1} and 1nN1\leq n\leq N, where, in the case n=1n=1, we set U0=u0U^{0}_{-}=u_{0} so that U0=U+0U0=U+0u0\llbracket U\rrbracket^{0}=U^{0}_{+}-U^{0}_{-}=U^{0}_{+}-u_{0}. (The monograph of Thomée [15, Chapter 12] is a standard reference providing a general introduction to dG time stepping for classical diffusion problems.)

To compute UU, we choose for each nn a basis ψn1\psi_{n1}, ψn2\psi_{n2}, …, ψnrn\psi_{nr_{n}} for rn1\mathbb{P}_{r_{n}-1} and write

U(t)=j=1rnUnjψnj(t)for tIn.U(t)=\sum_{j=1}^{r_{n}}U^{nj}\psi_{nj}(t)\quad\text{for $t\in I_{n}$.} (5)

When X=ψniX=\psi_{ni}, we find that

U+n1X+n1+InUX𝑑t=j=1rnGijnUnjandUn1X+n1=j=1rn1Kijn,n1Un1,j,U^{n-1}_{+}X^{n-1}_{+}+\int_{I_{n}}U^{\prime}X\,dt=\sum_{j=1}^{r_{n}}G^{n}_{ij}U^{nj}\quad\text{and}\quad U^{n-1}_{-}X^{n-1}_{+}=\sum_{j=1}^{r_{n-1}}K^{n,n-1}_{ij}U^{n-1,j},

with coefficients given by

Gijn=ψnj(tn1)ψni(tn1)+Inψnjψni𝑑tG^{n}_{ij}=\psi_{nj}(t_{n-1})\psi_{ni}(t_{n-1})+\int_{I_{n}}\psi_{nj}^{\prime}\psi_{ni}\,dt

and

Kijn,n1=ψn1,j(tn1)ψni(tn1).K^{n,n-1}_{ij}=\psi_{n-1,j}(t_{n-1})\psi_{ni}(t_{n-1}).

Owing to the convolutional structure of the fractional derivative, it is convenient to introduce the notation

¯=n\bar{\ell}=n-\ell

and define, if tInt\in I_{n},

ρjn¯(t)=ρjn,n(t)=Iωα(ts)ψj(s)𝑑sfor 1n1,\rho^{n\bar{\ell}}_{j}(t)=\rho^{n,n-\ell}_{j}(t)=\int_{I_{\ell}}\omega_{\alpha}(t-s)\psi_{\ell j}(s)\,ds\quad\text{for $1\leq\ell\leq n-1$,}

with

ρjnn¯(t)=ρjn0(t)=tn1tωα(ts)ψnj(s)𝑑s.\rho^{n\bar{n}}_{j}(t)=\rho^{n0}_{j}(t)=\int_{t_{n-1}}^{t}\omega_{\alpha}(t-s)\psi_{nj}(s)\,ds.

We find that

t1αU==1nj=1rUj(ρjn¯)(t)for tIn,\partial_{t}^{1-\alpha}U=\sum_{\ell=1}^{n}\sum_{j=1}^{r_{\ell}}U^{\ell j}(\rho^{n\bar{\ell}}_{j})^{\prime}(t)\quad\text{for $t\in I_{n}$,}

and thus

In(t1αU)X𝑑t==1nj=1rHijn¯UjwhereHijn¯=Hijn,n=In(ρjn)ψni𝑑t.\int_{I_{n}}(\partial_{t}^{1-\alpha}U)X\,dt=\sum_{\ell=1}^{n}\sum_{j=1}^{r_{\ell}}H^{n\bar{\ell}}_{ij}U^{\ell j}\quad\text{where}\quad H^{n\bar{\ell}}_{ij}=H^{n,n-\ell}_{ij}=\int_{I_{n}}(\rho^{n\ell}_{j})^{\prime}\psi_{ni}\,dt.

Hence, putting

Fni=Infψni𝑑t,F^{ni}=\int_{I_{n}}f\psi_{ni}\,dt,

the dG method (4) requires

j=1rn(Gijn+λHijn0)Unj=Fni=1n1j=1rλHijn,nUj+{ψ1i(0)u0,n=1,j=1rn1Kijn,n1Un1,j,2nN.\sum_{j=1}^{r_{n}}\bigl{(}G^{n}_{ij}+\lambda H^{n0}_{ij}\bigr{)}U^{nj}=F^{ni}-\sum_{\ell=1}^{n-1}\sum_{j=1}^{r_{\ell}}\lambda H^{n,n-\ell}_{ij}U^{\ell j}\\ +\begin{cases}\psi_{1i}(0)u_{0},&n=1,\\ \sum_{j=1}^{r_{n-1}}K^{n,n-1}_{ij}U^{n-1,j},&2\leq n\leq N.\end{cases} (6)

At the nnth time step, this rn×rnr_{n}\times r_{n} linear system must be solved to determine Un1U^{n1}, Un2U^{n2},…, UnrnU^{nr_{n}} and hence U(t)U(t) for tInt\in I_{n}.

Remark 1.

If we send α1\alpha\to 1, so that the fractional ODE in (3) reduces to the classical ODE u+λu=f(t)u^{\prime}+\lambda u=f(t), then Hijn¯=0H^{n\bar{\ell}}_{ij}=0 for 1¯n11\leq\bar{\ell}\leq n-1. Indeed, since ω1(t)=1\omega_{1}(t)=1, we see that ρjn¯(t)=Iψj(s)𝑑s\rho^{n\bar{\ell}}_{j}(t)=\int_{I_{\ell}}\psi_{\ell j}(s)\,ds is constant and so (ρjn¯)(t)=0(\rho^{n\bar{\ell}}_{j})^{\prime}(t)=0 for tInt\in I_{n}. Moreover, (ρjn0)(t)=ψnj(t)(\rho^{n0}_{j})^{\prime}(t)=\psi_{nj}(t) so Hijn0=Inψnjψni𝑑tH^{n0}_{ij}=\int_{I_{n}}\psi_{nj}\psi_{ni}\,dt.

Remark 2.

Later we will show certain symmetry properties of Hijn0H^{n0}_{ij} using the identity

ab(tatωα(ts)u(s)𝑑s)v(t)𝑑t=abu(s)(ssbωα(ts)v(t)𝑑t)𝑑s.\int_{a}^{b}\biggl{(}\frac{\partial}{\partial t}\int_{a}^{t}\omega_{\alpha}(t-s)u(s)\,ds\biggr{)}v(t)\,dt\\ =-\int_{a}^{b}u(s)\biggl{(}\frac{\partial}{\partial s}\int_{s}^{b}\omega_{\alpha}(t-s)v(t)\,dt\biggr{)}\,ds. (7)

In fact, the substitution x=tsx=t-s gives

tatωα(ts)u(s)𝑑s\displaystyle\frac{\partial}{\partial t}\int_{a}^{t}\omega_{\alpha}(t-s)u(s)\,ds =t0taωα(x)u(tx)𝑑x\displaystyle=\frac{\partial}{\partial t}\int_{0}^{t-a}\omega_{\alpha}(x)u(t-x)\,dx
=ωα(ta)u(a)+0taωα(x)u(tx)𝑑x\displaystyle=\omega_{\alpha}(t-a)u(a)+\int_{0}^{t-a}\omega_{\alpha}(x)u^{\prime}(t-x)\,dx
=ωα(ta)u(tn1)+atωα(ts)u(s)𝑑s,\displaystyle=\omega_{\alpha}(t-a)u(t_{n-1})+\int_{a}^{t}\omega_{\alpha}(t-s)u^{\prime}(s)\,ds,

and (7) follows after reversing the order of integration and then integrating by parts. Similarly, for n1\ell\leq n-1,

ab(tabωα(ts)u(s)𝑑s)v(t)𝑑t=abu(s)(sabωα(ts)v(t)𝑑t)𝑑s.\int_{a}^{b}\biggl{(}\frac{\partial}{\partial t}\int_{a}^{b}\omega_{\alpha}(t-s)u(s)\,ds\biggr{)}v(t)\,dt=-\int_{a}^{b}u(s)\biggl{(}\frac{\partial}{\partial s}\int_{a}^{b}\omega_{\alpha}(t-s)v(t)\,dt\biggr{)}\,ds. (8)

3 Evaluation of the coefficients

To compute GijnG^{n}_{ij}, HijnH^{n\ell}_{ij} and Kijn,n1K^{n,n-1}_{ij} it is convenient to map each closed subinterval I¯n=[tn1,tn]\bar{I}_{n}=[t_{n-1},t_{n}] to the reference element [1,1][-1,1]. We therefore define the affine function 𝗍n:[1,1]I¯n\mathsf{t}_{n}:[-1,1]\to\bar{I}_{n} by

𝗍n(τ)=12[(1τ)tn1+(1+τ)tn]for 1τ1,\mathsf{t}_{n}(\tau)=\tfrac{1}{2}\bigl{[}(1-\tau)t_{n-1}+(1+\tau)t_{n}\bigr{]}\quad\text{for $-1\leq\tau\leq 1$,}

and let

Ψnj(τ)=ψnj(t)for t=𝗍n(τ) and 1τ1.\Psi_{nj}(\tau)=\psi_{nj}(t)\quad\text{for $t=\mathsf{t}_{n}(\tau)$ and $-1\leq\tau\leq 1$.}

In this way,

Gijn=Ψnj(1)Ψni(1)+11Ψnj(τ)Ψni(τ)𝑑τG^{n}_{ij}=\Psi_{nj}(-1)\Psi_{ni}(-1)+\int_{-1}^{1}\Psi_{nj}^{\prime}(\tau)\Psi_{ni}(\tau)\,d\tau (9)

and

Kijn,n1=Ψn1,j(+1)Ψni(1).K^{n,n-1}_{ij}=\Psi_{n-1,j}(+1)\Psi_{ni}(-1). (10)

Both of these coefficients are readily computed; the remainder of this section is devoted to Hijn¯H^{n\bar{\ell}}_{ij}. The formulae in the next lemma allow us to compute Hijn0H^{n0}_{ij} to machine precision via Gauss–Legendre and Gauss–Jacobi quadrature.

Lemma 3.

If we define the polynomial

Φijn(y)=1211Ψnj(12(1y)(1+z)1)Ψni(112(1y)(1z))𝑑z,\Phi^{n}_{ij}(y)=\frac{1}{2}\int_{-1}^{1}\Psi_{nj}\bigl{(}\tfrac{1}{2}(1-y)(1+z)-1\bigr{)}\Psi^{\prime}_{ni}\bigl{(}1-\tfrac{1}{2}(1-y)(1-z)\bigr{)}\,dz,

then

Hijn0=(kn/2)αΓ(α)\displaystyle H^{n0}_{ij}=\frac{(k_{n}/2)^{\alpha}}{\Gamma(\alpha)} (Ψni(1)11(1σ)αΨnj(σ)dσ\displaystyle\biggl{(}\Psi_{ni}(1)\int_{-1}^{1}(1-\sigma)^{\alpha}\Psi_{nj}(\sigma)\,d\sigma
11(1+y)α1(1y)Φijn(y)dy).\displaystyle\qquad{}-\int_{-1}^{1}(1+y)^{\alpha-1}(1-y)\Phi^{n}_{ij}(y)\,dy\biggr{)}.

Proof:

Since ρjn0(tn1)=0\rho^{n0}_{j}(t_{n-1})=0, integration by parts gives

Hijn0\displaystyle H^{n0}_{ij} =ρjn0(tn)ψni(tn)Inρjn0(t)ψni(t)𝑑t\displaystyle=\rho^{n0}_{j}(t_{n})\psi_{ni}(t_{n})-\int_{I_{n}}\rho^{n0}_{j}(t)\psi^{\prime}_{ni}(t)\,dt
=ρjn0(tn)Ψni(1)11ρjn0(𝗍n(τ))Ψni(τ)𝑑τ,\displaystyle=\rho^{n0}_{j}(t_{n})\Psi_{ni}(1)-\int_{-1}^{1}\rho^{n0}_{j}\bigl{(}\mathsf{t}_{n}(\tau)\bigr{)}\Psi_{ni}^{\prime}(\tau)\,d\tau,

and since 𝗍n(τ)𝗍n(σ)=(τσ)kn/2\mathsf{t}_{n}(\tau)-\mathsf{t}_{n}(\sigma)=(\tau-\sigma)k_{n}/2, the substitution s=𝗍n(σ)s=\mathsf{t}_{n}(\sigma) yields

ρjn0(𝗍n(τ))\displaystyle\rho^{n0}_{j}\bigl{(}\mathsf{t}_{n}(\tau)\bigr{)} =kn21τωα(𝗍n(τ)𝗍n(σ))Ψnj(σ)𝑑σ\displaystyle=\frac{k_{n}}{2}\int_{-1}^{\tau}\omega_{\alpha}\bigl{(}\mathsf{t}_{n}(\tau)-\mathsf{t}_{n}(\sigma)\bigr{)}\Psi_{nj}(\sigma)\,d\sigma
=(kn/2)αΓ(α)1τ(τσ)α1Ψnj(σ)𝑑σ.\displaystyle=\frac{(k_{n}/2)^{\alpha}}{\Gamma(\alpha)}\int_{-1}^{\tau}(\tau-\sigma)^{\alpha-1}\Psi_{nj}(\sigma)\,d\sigma.

Thus,

Hijn0=(kn/2)αΓ(α)(Ψni(1)11(1σ)α1Ψnj(σ)𝑑σBijn),H^{n0}_{ij}=\frac{(k_{n}/2)^{\alpha}}{\Gamma(\alpha)}\biggl{(}\Psi_{ni}(1)\int_{-1}^{1}(1-\sigma)^{\alpha-1}\Psi_{nj}(\sigma)\,d\sigma-B^{n}_{ij}\biggr{)},

where

Bijn=111τ(τσ)α1Ψnj(σ)𝑑σΨni(τ)𝑑τ.B^{n}_{ij}=\int_{-1}^{1}\int_{-1}^{\tau}(\tau-\sigma)^{\alpha-1}\Psi_{nj}(\sigma)\,d\sigma\,\Psi_{ni}^{\prime}(\tau)\,d\tau.

We make the substitution 1+y=τσ1+y=\tau-\sigma, which results in a fixed singularity at y=1y=-1, and then reverse the order of integration:

Bijn\displaystyle B^{n}_{ij} =111τ(1+y)α1Ψnj(τy1)𝑑yΨni(τ)𝑑τ\displaystyle=\int_{-1}^{1}\int_{-1}^{\tau}(1+y)^{\alpha-1}\Psi_{nj}(\tau-y-1)\,dy\,\Psi_{ni}^{\prime}(\tau)\,d\tau
=11(1+y)α1y1Ψnj(τy1)Ψni(τ)𝑑τ𝑑y.\displaystyle=\int_{-1}^{1}(1+y)^{\alpha-1}\int_{y}^{1}\Psi_{nj}(\tau-y-1)\Psi_{ni}^{\prime}(\tau)\,d\tau\,dy.

The substitution τ=12[(1z)y+(1+z)]\tau=\tfrac{1}{2}\bigl{[}(1-z)y+(1+z)\bigr{]} then yields

y1Ψnj(τy1)Ψni(τ)𝑑τ𝑑y=(1y)Φijn(y),\int_{y}^{1}\Psi_{nj}(\tau-y-1)\Psi_{ni}^{\prime}(\tau)\,d\tau\,dy=(1-y)\Phi^{n}_{ij}(y),

implying the desired formula for Hijn0H^{n0}_{ij}.   \spadesuit

To deal with Hijn,nH^{n,n-\ell}_{ij} for n1\ell\leq n-1, we introduce the notation

tn1/2=𝗍n(0)=12(tn1+tn)andDn¯=Dn,n=tn1/2t1/2,t_{n-1/2}=\mathsf{t}_{n}(0)=\tfrac{1}{2}(t_{n-1}+t_{n})\quad\text{and}\quad D_{n\bar{\ell}}=D_{n,n-\ell}=t_{n-1/2}-t_{\ell-1/2},

with

Δn¯(τ,σ)=Δn,n(τ,σ)=τknσk2Dn¯,\Delta_{n\bar{\ell}}(\tau,\sigma)=\Delta_{n,n-\ell}(\tau,\sigma)=\frac{\tau k_{n}-\sigma k_{\ell}}{2D_{n\bar{\ell}}},

so that

𝗍n(τ)𝗍(σ)=Dn¯(1+Δn¯(τ,σ)).\mathsf{t}_{n}(\tau)-\mathsf{t}_{\ell}(\sigma)=D_{n\bar{\ell}}\bigl{(}1+\Delta_{n\bar{\ell}}(\tau,\sigma)\bigr{)}.
Lemma 4.

If 1n11\leq\ell\leq n-1, then

Hijn¯=Dn¯α1Γ(α)k2(Ψni(1)𝒜jn¯Ψni(1)jn¯𝒞ijn¯),H^{n\bar{\ell}}_{ij}=\frac{D_{n\bar{\ell}}^{\alpha-1}}{\Gamma(\alpha)}\,\frac{k_{\ell}}{2}\bigl{(}\Psi_{ni}(1)\mathcal{A}^{n\bar{\ell}}_{j}-\Psi_{ni}(-1)\mathcal{B}^{n\bar{\ell}}_{j}-\mathcal{C}^{n\bar{\ell}}_{ij}\bigr{)},

where

𝒜jn¯\displaystyle\mathcal{A}^{n\bar{\ell}}_{j} =11(1+Δn¯(1,σ))α1Ψj(σ)𝑑σ,\displaystyle=\int_{-1}^{1}\bigl{(}1+\Delta_{n\bar{\ell}}(1,\sigma)\bigr{)}^{\alpha-1}\Psi_{\ell j}(\sigma)\,d\sigma,
jn¯\displaystyle\mathcal{B}^{n\bar{\ell}}_{j} =11(1+Δn¯(1,σ))α1Ψj(σ)𝑑σ,\displaystyle=\int_{-1}^{1}\bigl{(}1+\Delta_{n\bar{\ell}}(-1,\sigma)\bigr{)}^{\alpha-1}\Psi_{\ell j}(\sigma)\,d\sigma,
𝒞ijn¯\displaystyle\mathcal{C}^{n\bar{\ell}}_{ij} =11Ψni(τ)11(1+Δn¯(τ,σ))α1Ψj(σ)𝑑σ𝑑τ.\displaystyle=\int_{-1}^{1}\Psi_{ni}^{\prime}(\tau)\int_{-1}^{1}\bigl{(}1+\Delta_{n\bar{\ell}}(\tau,\sigma)\bigr{)}^{\alpha-1}\Psi_{\ell j}(\sigma)\,d\sigma\,d\tau.

Proof:

Integrating by parts, we find that

Hijn¯\displaystyle H^{n\bar{\ell}}_{ij} =ρjn¯(tn)ψni(tn)ρjn¯(tn1)ψni(tn1)Inρjn¯(t)ψni(t)𝑑t\displaystyle=\rho^{n\bar{\ell}}_{j}(t_{n})\psi_{ni}(t_{n})-\rho^{n\bar{\ell}}_{j}(t_{n-1})\psi_{ni}(t_{n-1})-\int_{I_{n}}\rho^{n\bar{\ell}}_{j}(t)\psi_{ni}^{\prime}(t)\,dt
=ρjn¯(tn)Ψni(1)ρjn¯(tn1)Ψni(1)11ρjn¯(𝗍n(τ))Ψni(τ)𝑑τ.\displaystyle=\rho^{n\bar{\ell}}_{j}(t_{n})\Psi_{ni}(1)-\rho^{n\bar{\ell}}_{j}(t_{n-1})\Psi_{ni}(-1)-\int_{-1}^{1}\rho^{n\bar{\ell}}_{j}\bigl{(}\mathsf{t}_{n}(\tau)\bigr{)}\Psi^{\prime}_{ni}(\tau)\,d\tau.

The substitution s=𝗍(σ)s=\mathsf{t}_{\ell}(\sigma) gives

ρjn¯(𝗍n(τ))=Dn¯α1Γ(α)k211(1+Δn¯(τ,σ))α1Ψj(σ)𝑑σ,\rho^{n\bar{\ell}}_{j}\bigl{(}\mathsf{t}_{n}(\tau)\bigr{)}=\frac{D_{n\bar{\ell}}^{\alpha-1}}{\Gamma(\alpha)}\,\frac{k_{\ell}}{2}\int_{-1}^{1}\bigl{(}1+\Delta_{n\bar{\ell}}(\tau,\sigma)\bigr{)}^{\alpha-1}\Psi_{\ell j}(\sigma)\,d\sigma,

and the formula for Hijn¯H^{n\bar{\ell}}_{ij} follows at once.   \spadesuit

Notice that

1+Δn¯(1,σ)=2(tnt)+(1σ)kkn+2(tn1tl)+k>0for 1n1,1+\Delta_{n\bar{\ell}}(1,\sigma)=\frac{2(t_{n}-t_{\ell})+(1-\sigma)k_{\ell}}{k_{n}+2(t_{n-1}-t_{l})+k_{\ell}}>0\quad\text{for $1\leq\ell\leq n-1$,}

so the integrand of 𝒜ijn¯\mathcal{A}^{n\bar{\ell}}_{ij} is always smooth. However,

1+Δn¯(1,σ)=2(tn1t)+(1σ)kkn+2(tn1tl)+k,1+\Delta_{n\bar{\ell}}(-1,\sigma)=\frac{2(t_{n-1}-t_{\ell})+(1-\sigma)k_{\ell}}{k_{n}+2(t_{n-1}-t_{l})+k_{\ell}},

so the integrands of jn¯\mathcal{B}^{n\bar{\ell}}_{j} and 𝒞jn¯\mathcal{C}^{n\bar{\ell}}_{j} are weakly singular if ¯=1\bar{\ell}=1 (i.e., if =n1\ell=n-1). The next lemma provides alternative expressions that are amenable to Gauss–Jacobi and Gauss–Legendre quadrature.

Lemma 5.

Let ρn=kn/kn1\rho_{n}=k_{n}/k_{n-1}. Then,

𝒜jn1\displaystyle\mathcal{A}^{n1}_{j} =(1+ρn)1α11(2ρn+1σ)α1Ψn1,j(σ)𝑑σ,\displaystyle=(1+\rho_{n})^{1-\alpha}\int_{-1}^{1}(2\rho_{n}+1-\sigma)^{\alpha-1}\Psi_{n-1,j}(\sigma)\,d\sigma,
jn1\displaystyle\mathcal{B}^{n1}_{j} =(1+ρn)1α11(1σ)α1Ψn1,j(σ)𝑑σ\displaystyle=(1+\rho_{n})^{1-\alpha}\int_{-1}^{1}(1-\sigma)^{\alpha-1}\Psi_{n-1,j}(\sigma)\,d\sigma

and

𝒞ijn1=(1+ρn)1α(11(1+τ)αΨni(τ)01(ρn+z)α1Ψn1,j(1z(1+τ))𝑑z𝑑τ+11(1σ)αΨn1,j(σ)01(ρnz+1)α1Ψni(z(1σ)1)dzdσ).\mathcal{C}^{n1}_{ij}=(1+\rho_{n})^{1-\alpha}\biggl{(}\\ \int_{-1}^{1}(1+\tau)^{\alpha}\Psi^{\prime}_{ni}(\tau)\int_{0}^{1}(\rho_{n}+z)^{\alpha-1}\Psi_{n-1,j}\bigl{(}1-z(1+\tau)\bigr{)}\,dz\,d\tau\\ +\int_{-1}^{1}(1-\sigma)^{\alpha}\Psi_{n-1,j}(\sigma)\int_{0}^{1}(\rho_{n}z+1)^{\alpha-1}\Psi_{ni}^{\prime}\bigl{(}z(1-\sigma)-1\bigr{)}\,dz\,d\sigma\biggr{)}.

Proof:

Since 1+Δn1(1,σ)=kn1(1σ)/(kn+kn1)1+\Delta_{n1}(-1,\sigma)=k_{n-1}(1-\sigma)/(k_{n}+k_{n-1}), the formula for ijn1\mathcal{B}^{n1}_{ij} follows at once. To deal with 𝒞ijn,1\mathcal{C}^{n,1}_{ij} we begin by mapping [1,1]2[-1,1]^{2} onto [0,2]2[0,2]^{2} with the substitution (τ,σ)=(x1,1y)(\tau,\sigma)=(x-1,1-y). In this way, the singularity at (τ,σ)=(1,1)(\tau,\sigma)=(-1,1) moves to (x,y)=(0,0)(x,y)=(0,0), and

𝒞ijn1=0202(1+Δn1(x1,1y))α1Ψn1,j(1y)Ψni(x1)𝑑x𝑑y\mathcal{C}^{n1}_{ij}=\int_{0}^{2}\int_{0}^{2}\bigl{(}1+\Delta_{n1}(x-1,1-y)\bigr{)}^{\alpha-1}\Psi_{n-1,j}(1-y)\Psi_{ni}^{\prime}(x-1)\,dx\,dy

with

1+Δn1(x1,1y)=xkn+ykn1kn+kn1.1+\Delta_{n1}(x-1,1-y)=\frac{xk_{n}+yk_{n-1}}{k_{n}+k_{n-1}}.

By splitting the integration domain [0,2]2[0,2]^{2} into the triangular halves where x>yx>y and x<yx<y, we obtain

𝒞ijn1\displaystyle\mathcal{C}^{n1}_{ij} =02Ψni(x1)0x(xkn+ykn1kn+kn1)α1Ψn1,j(1y)𝑑y𝑑x\displaystyle=\int_{0}^{2}\Psi_{ni}^{\prime}(x-1)\int_{0}^{x}\biggl{(}\frac{xk_{n}+yk_{n-1}}{k_{n}+k_{n-1}}\biggr{)}^{\alpha-1}\Psi_{n-1,j}(1-y)\,dy\,dx
+02Ψn1,j(1y)0y(xkn+ykn1kn+kn1)α1Ψni(x1)𝑑x𝑑y.\displaystyle\qquad{}+\int_{0}^{2}\Psi_{n-1,j}(1-y)\int_{0}^{y}\biggl{(}\frac{xk_{n}+yk_{n-1}}{k_{n}+k_{n-1}}\biggr{)}^{\alpha-1}\Psi_{ni}^{\prime}(x-1)\,dx\,dy.

The substitution y=zxy=zx tranforms the inner integral in the first term to

xα01(kn+zkn1kn+kn1)α1Ψn1,j(1zx)𝑑z,x^{\alpha}\int_{0}^{1}\biggl{(}\frac{k_{n}+zk_{n-1}}{k_{n}+k_{n-1}}\biggr{)}^{\alpha-1}\Psi_{n-1,j}(1-zx)\,dz,

and the substitution x=zyx=zy transforms that in the second to

yα01(zkn+kn1kn+kn1)α1Ψni(zy1)𝑑z.y^{\alpha}\int_{0}^{1}\biggl{(}\frac{zk_{n}+k_{n-1}}{k_{n}+k_{n-1}}\biggr{)}^{\alpha-1}\Psi_{ni}^{\prime}(zy-1)\,dz.

Thus,

𝒞ijn1\displaystyle\mathcal{C}^{n1}_{ij} =02xαΨni(x1)01(kn+zkn1kn+kn1)α1Ψn1,j(1zx)𝑑z𝑑x\displaystyle=\int_{0}^{2}x^{\alpha}\Psi^{\prime}_{ni}(x-1)\int_{0}^{1}\biggl{(}\frac{k_{n}+zk_{n-1}}{k_{n}+k_{n-1}}\biggr{)}^{\alpha-1}\Psi_{n-1,j}(1-zx)\,dz\,dx
+02yαΨn1,j(1y)01(zkn+kn1kn+kn1)α1Ψni(zy1)𝑑z𝑑y.\displaystyle\qquad{}+\int_{0}^{2}y^{\alpha}\Psi_{n-1,j}(1-y)\int_{0}^{1}\biggl{(}\frac{zk_{n}+k_{n-1}}{k_{n}+k_{n-1}}\biggr{)}^{\alpha-1}\Psi_{ni}^{\prime}(zy-1)\,dz\,dy.

Now make the substitutions x=1+τx=1+\tau and y=1σy=1-\sigma.   \spadesuit

We also have the following alternative representation.

Lemma 6.

If 1n21\leq\ell\leq n-2, then

Hijn¯=1αΓ(α)knk4Dn¯α211Ψni(τ)11(1+Δn(τ,σ))α2Ψj(σ)𝑑σ𝑑τ.H^{n\bar{\ell}}_{ij}=-\frac{1-\alpha}{\Gamma(\alpha)}\,\frac{k_{n}k_{\ell}}{4}\,D_{n\bar{\ell}}^{\alpha-2}\int_{-1}^{1}\Psi_{ni}(\tau)\int_{-1}^{1}\bigl{(}1+\Delta_{n\ell}(\tau,\sigma)\bigr{)}^{\alpha-2}\Psi_{\ell j}(\sigma)\,d\sigma\,d\tau.

Proof:

When n2\ell\leq n-2,

(ρjn¯)(t)=Iωα1(ts)ψj(s)𝑑sfor t>t,(\rho^{n\bar{\ell}}_{j})^{\prime}(t)=\int_{I_{\ell}}\omega_{\alpha-1}(t-s)\psi_{\ell j}(s)\,ds\quad\text{for $t>t_{\ell}$,}

and so

Hijn¯=Inψni(t)Iωα1(ts)ψj(s)𝑑s.H^{n\bar{\ell}}_{ij}=\int_{I_{n}}\psi_{ni}(t)\int_{I_{\ell}}\omega_{\alpha-1}(t-s)\psi_{\ell j}(s)\,ds. (11)

The result now follows via the substitutions t=𝗍n(τ)t=\mathsf{t}_{n}(\tau) and s=𝗍(σ)s=\mathsf{t}_{\ell}(\sigma), noting that Γ(α)=(α1)Γ(α1)\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha-1).   \spadesuit

Remark 7.

If the time levels are uniformly spaced, and if the reference basis functions are the same for each subinterval, say

k=k,r=randΨj=Ψjfor 1n and 1jr,k_{\ell}=k,\quad r_{\ell}=r\quad\text{and}\quad\Psi_{\ell j}=\Psi_{j}\quad\text{for $1\leq\ell\leq n$ and $1\leq j\leq r$,}

then

Dn¯=¯kandΔn¯(τ,σ)=τσ2¯,D_{n\bar{\ell}}=\bar{\ell}k\quad\text{and}\quad\Delta_{n\bar{\ell}}(\tau,\sigma)=\frac{\tau-\sigma}{2\bar{\ell}},

so the formulae of Lemma 4 show that Hijn¯H^{n\bar{\ell}}_{ij} depends on nn and \ell only through the difference ¯=n\bar{\ell}=n-\ell; for further details, see Example 12 below.

4 Spatial discretisation

The initial-boundary value problem (1) is known to be well-posed [5, 8, 10]. Let u,v=Ωuv\langle u,v\rangle=\int_{\Omega}uv denote the usual inner product in L2(Ω)L^{2}(\Omega), and let a(u,v)a(u,v) denote the bilinear form associated with AA via the first Green identity. For example, if A=2A=-\nabla^{2} then a(u,v)=Ωuva(u,v)=\int_{\Omega}\nabla u\cdot\nabla v. In this way, the weak solution u:(0,T]H01(Ω)u:(0,T]\to H^{1}_{0}(\Omega) satisfies

tu,v+a(t1αu,v)=f(t),vfor vH01(Ω) and 0<tT.\langle\partial_{t}u,v\rangle+a(\partial_{t}^{1-\alpha}u,v)=\langle f(t),v\rangle\quad\text{for $v\in H^{1}_{0}(\Omega)$ and $0<t\leq T$.}

We choose a finite dimensional subspace VnH01(Ω)V_{n}\subseteq H^{1}_{0}(\Omega) for 0nN0\leq n\leq N, and form the vector 𝑽=(V1,,VN)\boldsymbol{V}=(V_{1},\ldots,V_{N}). For example, VnV_{n} might be a (conforming) finite element space constructed using a triangulation of Ω\Omega. Our trial space 𝒳=𝒳(𝒕,𝒓,𝑽)\mathcal{X}=\mathcal{X}(\boldsymbol{t},\boldsymbol{r},\boldsymbol{V}) then consists of the functions X:IH01(Ω)X:I\to H^{1}_{0}(\Omega) such that X|Inrn1(In;Vn)X|_{I_{n}}\in\mathbb{P}_{r_{n}-1}(I_{n};V_{n}), that is, the restriction X|InX|_{I_{n}} is a polynomial in tt of degree at most rn1r_{n}-1, with coefficients from VnV_{n}. Generalising (4), the dG solution U𝒳U\in\mathcal{X} of (1) satisfies

Un1,X+n1+IntU,Xdt+Ina(t1αU,X)dt=Inf(t),Xdt\bigl{\langle}\llbracket U\rrbracket^{n-1},X^{n-1}_{+}\bigr{\rangle}+\int_{I_{n}}\langle\partial_{t}U,X\rangle\,dt+\int_{I_{n}}a(\partial_{t}^{1-\alpha}U,X)\,dt=\int_{I_{n}}\langle f(t),X\rangle\,dt (12)

for Xrn1(In;Vn)X\in\mathbb{P}_{r_{n}-1}(I_{n};V_{n}) and 1nN1\leq n\leq N, with U0=U0U^{0}_{-}=U_{0} for a suitable U0V0U_{0}\in V_{0} such that U0u0U_{0}\approx u_{0}.

We choose a basis {ϕnp}p=1Pn\{\phi_{np}\}_{p=1}^{P_{n}} for VnV_{n}. In the expansion (5), the coefficient UnjU^{nj} is now a function in VnV_{n}, so there exist real numbers UqnjU^{nj}_{q} such that

Unj(x)=q=1PnUqnjϕnq(x)for xΩ;U^{nj}(x)=\sum_{q=1}^{P_{n}}U^{nj}_{q}\phi_{nq}(x)\quad\text{for $x\in\Omega$;}

for example, Uqnj=Unj(xnq)U^{nj}_{q}=U^{nj}(x_{nq}) if xnqx_{nq} is the qqth free node of a finite element mesh and if ϕnq\phi_{nq} is the corresponding nodal basis function. Similarly, for the discrete initial data, there are real numbers U0qU_{0q} such that

U0(x)=q=1P0U0qϕ0q(x)for xΩ.U_{0}(x)=\sum_{q=1}^{P_{0}}U_{0q}\phi_{0q}(x)\quad\text{for $x\in\Omega$.}

Choosing X(x,t)=ψni(t)ϕnq(x)X(x,t)=\psi_{ni}(t)\phi_{nq}(x) in (12), we find that the equations (6) for time stepping the scalar problem generalise to

j=1rnq=1Pn(GijnMpqnn+Hijn0Apqnn)Uqnj=Fpni=1n1j=1rq=1PHijn,nApqnUqj+{ψ1i(0)q=1P0Mpq10U0q,n=1,j=1rn1q=1Pn1Kijn,n1Mpqn,n1Uqn1,j,2nN,\sum_{j=1}^{r_{n}}\sum_{q=1}^{P_{n}}\bigl{(}G^{n}_{ij}M^{nn}_{pq}+H^{n0}_{ij}A^{nn}_{pq}\bigr{)}U^{nj}_{q}=F^{ni}_{p}-\sum_{\ell=1}^{n-1}\sum_{j=1}^{r_{\ell}}\sum_{q=1}^{P_{\ell}}H^{n,n-\ell}_{ij}A^{n\ell}_{pq}U^{\ell j}_{q}\\ +\begin{cases}\psi_{1i}(0)\sum_{q=1}^{P_{0}}M^{10}_{pq}U_{0q},&n=1,\\[6.0pt] \sum_{j=1}^{r_{n-1}}\sum_{q=1}^{P_{n-1}}K^{n,n-1}_{ij}M^{n,n-1}_{pq}U^{n-1,j}_{q},&2\leq n\leq N,\end{cases} (13)

where

Mpqn=ϕq,ϕnp,Apqn=a(ϕq,ϕnp),Fpni=Inf(t),ϕnpψni(t)𝑑t.M^{n\ell}_{pq}=\langle\phi_{\ell q},\phi_{np}\rangle,\qquad A^{n\ell}_{pq}=a(\phi_{\ell q},\phi_{np}),\qquad F^{ni}_{p}=\int_{I_{n}}\langle f(t),\phi_{np}\rangle\psi_{ni}(t)\,dt.

By introducing the Pn×PP_{n}\times P_{\ell} mass matrix 𝑴n=[Mpqn]\boldsymbol{M}^{n\ell}=[M^{n\ell}_{pq}] and stiffness matrix 𝑨n=[Apqn]\boldsymbol{A}^{n\ell}=[A^{n\ell}_{pq}], and forming the column vectors

𝑼nj=[U1njU2njUPnnj],𝑭ni=[F1niF2niFPnni],𝑼0=[U01U02U0P0],\boldsymbol{U}^{nj}=\begin{bmatrix}U^{nj}_{1}\\ U^{nj}_{2}\\ \vdots\\ U^{nj}_{P_{n}}\end{bmatrix},\qquad\boldsymbol{F}^{ni}=\begin{bmatrix}F^{ni}_{1}\\ F^{ni}_{2}\\ \vdots\\ F^{ni}_{P_{n}}\end{bmatrix},\qquad\boldsymbol{U}_{0}=\begin{bmatrix}U_{01}\\ U_{02}\\ \vdots\\ U_{0P_{0}}\end{bmatrix},

we can rewrite the equations (13) as

j=1rn(Gijn𝑴nn+Hijn0𝑨nn)𝑼nj=𝑭ni=1n1j=1rHijn,n𝑨n𝑼j+{ψ1i(0)𝑴10𝑼0,n=1,j=1rn1Kijn,n1𝑴n,n1𝑼n1,j,2nN.\sum_{j=1}^{r_{n}}\bigl{(}G^{n}_{ij}\boldsymbol{M}^{nn}+H^{n0}_{ij}\boldsymbol{A}^{nn}\bigr{)}\boldsymbol{U}^{nj}=\boldsymbol{F}^{ni}-\sum_{\ell=1}^{n-1}\sum_{j=1}^{r_{\ell}}H^{n,n-\ell}_{ij}\boldsymbol{A}^{n\ell}\boldsymbol{U}^{\ell j}\\ +\begin{cases}\psi_{1i}(0)\boldsymbol{M}^{10}\boldsymbol{U}_{0},&n=1,\\[6.0pt] \sum_{j=1}^{r_{n-1}}K^{n,n-1}_{ij}\boldsymbol{M}^{n,n-1}\boldsymbol{U}^{n-1,j},&2\leq n\leq N.\end{cases} (14)

To write (14) even more compactly, define the rn×rnr_{n}\times r_{n} matrix 𝑮n=[Gijn]\boldsymbol{G}^{n}=[G^{n}_{ij}] and the rn×rr_{n}\times r_{\ell} matrix 𝑯n¯=[Hijn¯]\boldsymbol{H}^{n\bar{\ell}}=[H^{n\bar{\ell}}_{ij}], together with the (block) column vectors

𝑼n=[𝑼n1𝑼n2𝑼nrn]and𝑭n=[𝑭n1𝑭n2𝑭nrn].\boldsymbol{U}^{n}=\begin{bmatrix}\boldsymbol{U}^{n1}\\ \boldsymbol{U}^{n2}\\ \vdots\\ \boldsymbol{U}^{nr_{n}}\end{bmatrix}\quad\text{and}\quad\boldsymbol{F}^{n}=\begin{bmatrix}\boldsymbol{F}^{n1}\\ \boldsymbol{F}^{n2}\\ \vdots\\ \boldsymbol{F}^{nr_{n}}\end{bmatrix}.

We also form the rn×rn1r_{n}\times r_{n-1} matrix 𝑲n,n1=[Kijn,n1]\boldsymbol{K}^{n,n-1}=[K^{n,n-1}_{ij}] and the column vector

𝝍+𝟎=[ψ11(0)ψ12(0)ψ1rn(0)].\boldsymbol{\psi^{0}_{+}}=\begin{bmatrix}\psi_{11}(0)\\ \psi_{12}(0)\\ \vdots\\ \psi_{1r_{n}}(0)\\ \end{bmatrix}.

Utilising the Kronecker product, the linear system (14) takes the form

(𝑮n𝑴nn+𝑯n0𝑨nn)𝑼n=𝑭n=1n1(𝑯n,n𝑨n)𝑼+{(𝝍+0𝑴10)𝑼0,n=1,(𝑲n,n1𝑴n,n1)𝑼n1,j,2nN.\bigl{(}\boldsymbol{G}^{n}\otimes\boldsymbol{M}^{nn}+\boldsymbol{H}^{n0}\otimes\boldsymbol{A}^{nn}\bigr{)}\boldsymbol{U}^{n}=\boldsymbol{F}^{n}-\sum_{\ell=1}^{n-1}\bigl{(}\boldsymbol{H}^{n,n-\ell}\otimes\boldsymbol{A}^{n\ell}\bigr{)}\boldsymbol{U}^{\ell}\\ +\begin{cases}\bigl{(}\boldsymbol{\psi}^{0}_{+}\otimes\boldsymbol{M}^{10}\bigr{)}\boldsymbol{U}_{0},&n=1,\\[6.0pt] \bigl{(}\boldsymbol{K}^{n,n-1}\otimes\boldsymbol{M}^{n,n-1}\bigr{)}\boldsymbol{U}^{n-1,j},&2\leq n\leq N.\end{cases} (15)

5 Legendre polynomials

Let P0P_{0}, P1P_{1}, P2P_{2}, …denote the Legendre polynomials with the standard normalisation Pj(1)=1P_{j}(1)=1 for all j0j\geq 0. By choosing

Ψnj(τ)=Pj1(τ),\Psi_{nj}(\tau)=P_{j-1}(\tau), (16)

we obtain a convenient and well-conditioned basis for rn1\mathbb{P}_{r_{n}-1} with the properties

11Ψnj(τ)Ψni(τ)𝑑τ=2δij2j1andΨnj(τ)=(1)j1Ψnj(τ)\int_{-1}^{1}\Psi_{nj}(\tau)\Psi_{ni}(\tau)\,d\tau=\frac{2\delta_{ij}}{2j-1}\quad\text{and}\quad\Psi_{nj}(-\tau)=(-1)^{j-1}\Psi_{nj}(\tau)

for ii, j{1,2,,rn}j\in\{1,2,\ldots,r_{n}\}.

Lemma 8.

With the choice (16) of basis functions,

Ψnj(1)=1andΨnj(1)=(1)j1,\Psi_{nj}(1)=1\quad\text{and}\quad\Psi_{nj}(-1)=(-1)^{j-1}, (17)

and the coefficients (9) and (10) are given by

Gijn={(1)i+j,if ij,1,if i<j,G^{n}_{ij}=\begin{cases}(-1)^{i+j},&\text{if $i\geq j$,}\\ 1,&\text{if $i<j$,}\end{cases}

and

Kijn,n1=(1)i1.K^{n,n-1}_{ij}=(-1)^{i-1}.

Proof:

The properties (17) follow from Pj(1)=1P_{j}(1)=1 and Pj(1)=(1)jP_{j}(-1)=(-1)^{j}. Hence, the formula for Kijn,n1K^{n,n-1}_{ij} follows from (10), and by (9),

Gijn=(1)i+j+EijwhereEij=11Pj1(τ)Pi1(τ)𝑑τ.G^{n}_{ij}=(-1)^{i+j}+E_{ij}\quad\text{where}\quad E_{ij}=\int_{-1}^{1}P_{j-1}^{\prime}(\tau)P_{i-1}(\tau)\,d\tau.

If jij\leq i, then Eij=0E_{ij}=0 because Pj1P_{j-1}^{\prime} is orthogonal to Pi1P_{i-1}. Otherwise, if j>ij>i, then Pj1P_{j-1} is orthogonal to Pi1P_{i-1}^{\prime} so integration by parts gives

Eij=[Pj1(x)Pi1(x)]1111Pj1(x)Pi1(x)𝑑x=1(1)i+jE_{ij}=\bigl{[}P_{j-1}(x)P_{i-1}(x)\bigr{]}_{-1}^{1}-\int_{-1}^{1}P_{j-1}(x)P_{i-1}^{\prime}(x)\,dx=1-(-1)^{i+j}

and hence Gijn=1G^{n}_{ij}=1.   \spadesuit

Example 9.

If rn=4r_{n}=4 and rn1=3r_{n-1}=3, then the matrices 𝑮n=[Gijn]\boldsymbol{G}^{n}=[G^{n}_{ij}] and 𝑲n,n1=[Kijn,n1]\boldsymbol{K}^{n,n-1}=[K^{n,n-1}_{ij}] are

𝑮n=[1111111111111111]and𝑲n,n1=[111111111111].\boldsymbol{G}^{n}=\left[\begin{array}[]{rrrr}1&1&1&1\\ -1&1&1&1\\ 1&-1&1&1\\ -1&1&-1&\phantom{-}1\end{array}\right]\quad\text{and}\quad\boldsymbol{K}^{n,n-1}=\left[\begin{array}[]{rrr}1&1&1\\ -1&-1&-1\\ 1&1&1\\ -1&-1&-1\end{array}\right].

We have no analogous, simple formula for the remaining coefficients HijnH^{n\ell}_{ij}. However, when ¯=0\bar{\ell}=0 (=n\ell=n) the following parity property holds.

Lemma 10.

With the choice (16) of basis functions,

Hjin0=(1)i+jHijn0.H^{n0}_{ji}=(-1)^{i+j}H^{n0}_{ij}. (18)

Proof:

Using (7), we find that

Hjin0=11(BPi1)(τ)Pj1(τ)𝑑τ=11Pi1(σ)(BPj1)(σ)𝑑σ,H^{n0}_{ji}=\int_{-1}^{1}(BP_{i-1})^{\prime}(\tau)P_{j-1}(\tau)\,d\tau=-\int_{-1}^{1}P_{i-1}(\sigma)(B^{*}P_{j-1})^{\prime}(\sigma)\,d\sigma,

where

(Bv)(τ)=1τωα(τσ)v(σ)𝑑σand(Bv)(σ)=σ1ωα(τσ)v(τ)𝑑τ.(Bv)(\tau)=\int_{-1}^{\tau}\omega_{\alpha}(\tau-\sigma)v(\sigma)\,d\sigma\quad\text{and}\quad(B^{*}v)(\sigma)=\int_{\sigma}^{1}\omega_{\alpha}(\tau-\sigma)v(\tau)\,d\tau.

Let (RV)(τ)=V(τ)(RV)(\tau)=V(-\tau). A short calculation shows that RB=BRRB^{*}=BR, so

(BPj1)(σ)\displaystyle(B^{*}P_{j-1})^{\prime}(-\sigma) =ddσ[(BPj1)(σ)]=ddσ(RBPj1)(σ)\displaystyle=-\frac{d}{d\sigma}\bigl{[}(B^{*}P_{j-1})(-\sigma)\bigr{]}=-\frac{d}{d\sigma}(RB^{*}P_{j-1})^{\prime}(\sigma)
=(BRPj1)(σ)=(1)j(BPj1)(σ),\displaystyle=-(BRP_{j-1})^{\prime}(\sigma)=(-1)^{j}(BP_{j-1})^{\prime}(\sigma),

and therefore, using the substitution σ=x\sigma=-x,

Hjin0\displaystyle H^{n0}_{ji} =(1)j+111Pi1(x)(BPj1)(x)𝑑x\displaystyle=(-1)^{j+1}\int_{-1}^{1}P_{i-1}(-x)(BP_{j-1})^{\prime}(x)\,dx
=(1)i+j11(BPj1)(x)Pi1(x)𝑑x=(1)i+jHijn0,\displaystyle=(-1)^{i+j}\int_{-1}^{1}(BP_{j-1})^{\prime}(x)P_{i-1}(x)\,dx=(-1)^{i+j}H^{n0}_{ij},

as claimed.   \spadesuit

Remark 11.

In the limit as α1\alpha\to 1, we see from Remark 1 that

Hijn0Inψnj(t)ψni(t)𝑑t=kn211Ψj(τ)Ψi(τ)𝑑τ=knδij2j1.H^{n0}_{ij}\to\int_{I_{n}}\psi_{nj}(t)\psi_{ni}(t)\,dt=\frac{k_{n}}{2}\int_{-1}^{1}\Psi_{j}(\tau)\Psi_{i}(\tau)\,d\tau=\frac{k_{n}\delta_{ij}}{2j-1}.
Example 12.

Consider the uniform case kn=kk_{n}=k, rn=rr_{n}=r and Ψnj=Ψj\Psi_{nj}=\Psi_{j} for 1nN1\leq n\leq N (as in Remark 7), with Ψj(τ)=Pj1(τ)\Psi_{j}(\tau)=P_{j-1}(\tau) as above. We then have

Hijn¯=kαHij¯for 1nN and ij{1,2,,r},H^{n\bar{\ell}}_{ij}=k^{\alpha}H^{\bar{\ell}}_{ij}\quad\text{for $1\leq\ell\leq n\leq N$ and $i$, $j\in\{1,2,\ldots,r\}$,}

where, by Lemma 3,

Hij0=12αΓ(α)\displaystyle H^{0}_{ij}=\frac{1}{2^{\alpha}\Gamma(\alpha)} (11(1σ)αPj1(σ)dσ\displaystyle\biggl{(}\int_{-1}^{1}(1-\sigma)^{\alpha}P_{j-1}(\sigma)\,d\sigma (19)
11(1+y)α1(1y)Φij(y)dy),\displaystyle\qquad{}-\int_{-1}^{1}(1+y)^{\alpha-1}(1-y)\Phi_{ij}(y)\,dy\biggr{)},

with

Φij(y)=1211Pj1(12(1y)(1+z)1)Pi1(112(1y)(1z))𝑑z,\Phi_{ij}(y)=\frac{1}{2}\int_{-1}^{1}P_{j-1}\bigl{(}\tfrac{1}{2}(1-y)(1+z)-1\bigr{)}P^{\prime}_{i-1}\bigl{(}1-\tfrac{1}{2}(1-y)(1-z)\bigr{)}\,dz, (20)

and by Lemma 4,

Hij¯=¯α12Γ(α)(𝒜j¯+(1)ij¯𝒞ij¯)for 1,H^{\bar{\ell}}_{ij}=\frac{\bar{\ell}^{\alpha-1}}{2\Gamma(\alpha)}\bigl{(}\mathcal{A}^{\bar{\ell}}_{j}+(-1)^{i}\mathcal{B}^{\bar{\ell}}_{j}-\mathcal{C}^{\bar{\ell}}_{ij}\bigr{)}\quad\text{for $\ell\geq 1$,}

with, letting Δ¯(τ)=τ/(2¯)\Delta_{\bar{\ell}}(\tau)=\tau/(2\bar{\ell}),

𝒜j¯\displaystyle\mathcal{A}^{\bar{\ell}}_{j} =11(1+Δ¯(1σ))α1Pj1(σ)𝑑σ,\displaystyle=\int_{-1}^{1}\bigl{(}1+\Delta_{\bar{\ell}}(1-\sigma)\bigr{)}^{\alpha-1}P_{j-1}(\sigma)\,d\sigma,
j¯\displaystyle\mathcal{B}^{\bar{\ell}}_{j} =11(1Δ¯(1+σ))α1Pj1(σ)𝑑σ,\displaystyle=\int_{-1}^{1}\bigl{(}1-\Delta_{\bar{\ell}}(1+\sigma)\bigr{)}^{\alpha-1}P_{j-1}(\sigma)\,d\sigma,
𝒞ij¯\displaystyle\mathcal{C}^{\bar{\ell}}_{ij} =11Pi1(τ)11(1+Δ¯(τσ))α1Pj1(σ)𝑑σ𝑑τ.\displaystyle=\int_{-1}^{1}P_{i-1}^{\prime}(\tau)\int_{-1}^{1}\bigl{(}1+\Delta_{\bar{\ell}}(\tau-\sigma)\bigr{)}^{\alpha-1}P_{j-1}(\sigma)\,d\sigma\,d\tau.

Moreover, Lemma 5 provides alternative expressions when ¯=1\bar{\ell}=1:

𝒜ij1\displaystyle\mathcal{A}^{1}_{ij} =21α11(3σ)α1Pj1(σ)𝑑σ,\displaystyle=2^{1-\alpha}\int_{-1}^{1}(3-\sigma)^{\alpha-1}P_{j-1}(\sigma)\,d\sigma,
ij1\displaystyle\mathcal{B}^{1}_{ij} =21α11(1σ)α1Pj1(σ)𝑑σ\displaystyle=2^{1-\alpha}\int_{-1}^{1}(1-\sigma)^{\alpha-1}P_{j-1}(\sigma)\,d\sigma

and

𝒞ij1\displaystyle\mathcal{C}^{1}_{ij} =21α(11(1+τ)αPi1(τ)01(1+z)α1Pj1(1z(1+τ))dzdτ\displaystyle=2^{1-\alpha}\biggl{(}\int_{-1}^{1}(1+\tau)^{\alpha}P_{i-1}^{\prime}(\tau)\int_{0}^{1}(1+z)^{\alpha-1}P_{j-1}\bigl{(}1-z(1+\tau)\bigr{)}\,dz\,d\tau
+11(1σ)αPj1(σ)01(z+1)α1Pi1(z(1σ)1)dzdσ).\displaystyle{}+\int_{-1}^{1}(1-\sigma)^{\alpha}P_{j-1}(\sigma)\int_{0}^{1}(z+1)^{\alpha-1}P_{i-1}^{\prime}\bigl{(}z(1-\sigma)-1\bigr{)}\,dz\,d\sigma\biggr{)}.

Likewise, Lemma 6 provides an alternative expression for ¯2\bar{\ell}\geq 2:

Hij¯=1α4Γ(α)¯α211Pi1(τ)11(1+Δ¯(τσ))α2Pj1(σ)𝑑σ𝑑τ.H^{\bar{\ell}}_{ij}=-\frac{1-\alpha}{4\Gamma(\alpha)}\,\bar{\ell}^{\alpha-2}\int_{-1}^{1}P_{i-1}(\tau)\int_{-1}^{1}\bigl{(}1+\Delta_{\bar{\ell}}(\tau-\sigma)\bigr{)}^{\alpha-2}P_{j-1}(\sigma)\,d\sigma\,d\tau. (21)

Finally, by arguing as in the proof of Lemma 10, we can show that

Hji¯=(1)i+jHij¯for all ¯0.H^{\bar{\ell}}_{ji}=(-1)^{i+j}H^{\bar{\ell}}_{ij}\quad\text{for all $\bar{\ell}\geq 0$.} (22)

6 Reconstruction

Throughout this section, we continue to use the Legendre basis (16). Some insight into the dG method can be had by considering the trivial case of (1) when A=0A=0, that is, tu=f(t)\partial_{t}u=f(t) for 0<tT0<t\leq T, with u(0)=u0u(0)=u_{0}. The dG scheme (12) then reduces to

Un,X+n+IntU,Xdt=Intu,Xdt\bigl{\langle}\llbracket U\rrbracket^{n},X^{n}_{+}\bigr{\rangle}+\int_{I_{n}}\langle\partial_{t}U,X\rangle\,dt=\int_{I_{n}}\langle\partial_{t}u,X\rangle\,dt (23)

for Xrn1(In;Vn)X\in\mathbb{P}_{r_{n}-1}(I_{n};V_{n}) and 1nN1\leq n\leq N, with U0=U0U^{0}_{-}=U_{0}. To state our next result, let 𝒫n\mathcal{P}_{n} denote the orthoprojector from L2(Ω)L_{2}(\Omega) onto VnV_{n}, and define

𝒬n=𝒫n𝒫n1𝒫+1.\mathcal{Q}_{n\ell}=\mathcal{P}_{n}\mathcal{P}_{n-1}\cdots\mathcal{P}_{\ell+1}.
Lemma 13.

If A=0A=0 and U0=𝒫0u0U_{0}=\mathcal{P}_{0}u_{0}, then for 1nN1\leq n\leq N the dG solution U𝒳U\in\mathcal{X} satisfies

Un=𝒫nu(tn)+=0n1𝒬n(𝒫I)u(t)U^{n}_{-}=\mathcal{P}_{n}u(t_{n})+\sum_{\ell=0}^{n-1}\mathcal{Q}_{n\ell}(\mathcal{P}_{\ell}-I)u(t_{\ell}) (24)

and

InUu,tX𝑑t=0for all Xn(In;V).\int_{I_{n}}\langle U-u,\partial_{t}X\rangle\,dt=0\quad\text{for all $X\in\mathbb{P}_{n}(I_{n};V)$.} (25)

Proof:

Integrating by parts in (23), we find that

Unu(tn),Xn=Un1u(tn1),X+n+InUu,tX𝑑t.\langle U^{n}_{-}-u(t_{n}),X^{n}_{-}\rangle=\langle U^{n-1}_{-}-u(t_{n-1}),X^{n}_{+}\rangle+\int_{I_{n}}\langle U-u,\partial_{t}X\rangle\,dt.

Given vVnv\in V_{n}, by choosing the constant function X(t)=vX(t)=v for tInt\in I_{n} we deduce that Unu(tn),v=Un1u(tn1),v\langle U^{n}_{-}-u(t_{n}),v\rangle=\langle U^{n-1}_{-}-u(t_{n-1}),v\rangle and so (25) is satisfied. Moreover,

𝒫n(Unu(tn))=𝒫n(Un1u(tn1)),\mathcal{P}_{n}\bigl{(}U^{n}_{-}-u(t_{n})\bigr{)}=\mathcal{P}_{n}\bigl{(}U^{n-1}_{-}-u(t_{n-1})\bigr{)},

and, by the choice of initial condition, we see that (24) is satisfied for n=1n=1:

U1𝒫1u(t1)\displaystyle U^{1}_{-}-\mathcal{P}_{1}u(t_{1}) =𝒫1(U1u(t1))=𝒫1(I𝒫0+𝒫0)(U0u(t0))\displaystyle=\mathcal{P}_{1}\bigl{(}U^{1}_{-}-u(t_{1})\bigr{)}=\mathcal{P}_{1}(I-\mathcal{P}_{0}+\mathcal{P}_{0})\bigl{(}U^{0}_{-}-u(t_{0})\bigr{)}
=𝒫1(𝒫0I)u(t0)+𝒫1(U0𝒫0u0)=𝒬11(𝒫0I)u(t0).\displaystyle=\mathcal{P}_{1}(\mathcal{P}_{0}-I)u(t_{0})+\mathcal{P}_{1}(U_{0}-\mathcal{P}_{0}u_{0})=\mathcal{Q}_{11}(\mathcal{P}_{0}-I)u(t_{0}).

Letting n2n\geq 2, we make the induction hypothesis

Un1=𝒫n1u(tn1)+=0n1𝒬n1,(𝒫I)u(t),U^{n-1}_{-}=\mathcal{P}_{n-1}u(t_{n-1})+\sum_{\ell=0}^{n-1}\mathcal{Q}_{n-1,\ell}(\mathcal{P}_{\ell}-I)u(t_{\ell}),

and observe that

Un𝒫nu(tn)\displaystyle U^{n}_{-}-\mathcal{P}_{n}u(t_{n}) =𝒫n(Unu(tn))=𝒫n(I𝒫n1+𝒫n1)(Un1u(tn1))\displaystyle=\mathcal{P}_{n}\bigl{(}U^{n}_{-}-u(t_{n})\bigr{)}=\mathcal{P}_{n}(I-\mathcal{P}_{n-1}+\mathcal{P}_{n-1})\bigl{(}U^{n-1}_{-}-u(t_{n-1})\bigr{)}
=𝒫n(𝒫n1I)u(tn1)+𝒫n=0n1𝒬n1,(𝒫I)u(t),\displaystyle=\mathcal{P}_{n}(\mathcal{P}_{n-1}-I)u(t_{n-1})+\mathcal{P}_{n}\sum_{\ell=0}^{n-1}\mathcal{Q}_{n-1,\ell}(\mathcal{P}_{\ell}-I)u(t_{\ell}),

which gives the desired formula (24).   \spadesuit

For the remainder of this section, we will assume that the subspaces VnV_{n} are nested, as follows:

V0V1V2VN.V_{0}\supseteq V_{1}\supseteq V_{2}\supseteq\cdots\supseteq V_{N}. (26)

It follows that 𝒫+1(𝒫I)=0\mathcal{P}_{\ell+1}(\mathcal{P}_{\ell}-I)=0 for 0N10\leq\ell\leq N-1 and so

Un=𝒫nu(tn).U^{n}_{-}=\mathcal{P}_{n}u(t_{n}). (27)

The following explicit representation for UU holds.

Lemma 14.

If A=0A=0, U0=𝒫0u0U_{0}=\mathcal{P}_{0}u_{0} and the subspaces satisfy (26), then

U(t)=j=1rn1anjψnj(t)+a~nψnrn(t)for tIn,U(t)=\sum_{j=1}^{r_{n}-1}a_{nj}\psi_{nj}(t)+\tilde{a}_{n}\psi_{nr_{n}}(t)\quad\text{for $t\in I_{n}$,} (28)

where

anj=2j1knIn𝒫nu(t)ψnj(t)𝑑ta_{nj}=\frac{2j-1}{k_{n}}\int_{I_{n}}\mathcal{P}_{n}u(t)\psi_{nj}(t)\,dt

are the local Fourier–Legendre coefficients of 𝒫nu\mathcal{P}_{n}u, but

a~n=𝒫nu(tn)j=1rn1anj.\tilde{a}_{n}=\mathcal{P}_{n}u(t_{n})-\sum_{j=1}^{r_{n}-1}a_{nj}.

Proof:

By definition, U|Inrn1(In;Vn)U|_{I_{n}}\in\mathbb{P}_{r_{n}-1}(I_{n};V_{n}) so there exist coefficients anja_{nj} and a~n\tilde{a}_{n} in VnV_{n} such that UU has the desired expansion. The formula for anja_{nj} follows at once from the orthogonality property of the ψnj\psi_{nj} (see Remark 11). The formula for a~n\tilde{a}_{n} follows from (27) because ψnj(tn)=Pj1(1)=1\psi_{nj}(t_{n})=P_{j-1}(1)=1 for all jj.   \spadesuit

We have a Peano kernel 𝖦r\mathsf{G}_{r} for the Fourier–Legendre expansion of degree rr,

f(τ)=j=1r+1bjΨj(τ)+11𝖦r(τ,σ)f(r+1)(σ)𝑑σfor 1τ1,f(\tau)=\sum_{j=1}^{r+1}b_{j}\Psi_{j}(\tau)+\int_{-1}^{1}\mathsf{G}_{r}(\tau,\sigma)f^{(r+1)}(\sigma)\,d\sigma\quad\text{for $-1\leq\tau\leq 1$,}

assuming f:[1,1]f:[-1,1]\to\mathbb{R} is Cr+1C^{r+1}, and also a Peano kernel 𝖬j(τ)\mathsf{M}_{j}(\tau) for the jjth coefficient:

bj=2j1211f(τ)Ψj(τ)𝑑τ=11𝖬j(τ)f(j1)(τ)𝑑τ.b_{j}=\frac{2j-1}{2}\int_{-1}^{1}f(\tau)\Psi_{j}(\tau)\,d\tau=\int_{-1}^{1}\mathsf{M}_{j}(\tau)f^{(j-1)}(\tau)\,d\tau.

Thus, if t=𝗍n(τ)t=\mathsf{t}_{n}(\tau) and s=𝗍n(σ)s=\mathsf{t}_{n}(\sigma), and if we define the local Peano kernels

𝗀nr(t,s)=(kn/2)r𝖦r(τ,σ)and𝗆nj(t)=(kn/2)j2𝖬j(τ),\mathsf{g}_{nr}(t,s)=(k_{n}/2)^{r}\mathsf{G}_{r}(\tau,\sigma)\quad\text{and}\quad\mathsf{m}_{nj}(t)=(k_{n}/2)^{j-2}\mathsf{M}_{j}(\tau),

then

𝒫nu(t)=j=1rn+1anjψnj(t)+In𝗀rn(t,s)𝒫nu(rn+1)(s)𝑑sfor tIn,\mathcal{P}_{n}u(t)=\sum_{j=1}^{r_{n}+1}a_{nj}\psi_{nj}(t)+\int_{I_{n}}\mathsf{g}_{r_{n}}(t,s)\mathcal{P}_{n}u^{(r_{n}+1)}(s)\,ds\quad\text{for $t\in I_{n}$,} (29)

and

anj=In𝗆nj(s)𝒫nu(j1)(s)𝑑s.a_{nj}=\int_{I_{n}}\mathsf{m}_{nj}(s)\mathcal{P}_{n}u^{(j-1)}(s)\,ds.

It follows that anj=O(knj1)a_{nj}=O(k_{n}^{j-1}) provided uu is Cj1C^{j-1} on I¯n\bar{I}_{n}.

Theorem 15.

Assume that A=0A=0, U0=𝒫u0U_{0}=\mathcal{P}u_{0} and the subspaces satisfy (26). If u:I¯nL2(Ω)u:\bar{I}_{n}\to L^{2}(\Omega) is Crn+1C^{r_{n}+1}, then an,rn+1=O(knrn)a_{n,r_{n}+1}=O(k_{n}^{r_{n}}) and

𝒫nu(t)U(t)=an,rn+1[ψn,rn+1(t)ψn,rn(t)]+O(knrn+1)for tIn.\mathcal{P}_{n}u(t)-U(t)=a_{n,r_{n}+1}\bigl{[}\psi_{n,r_{n}+1}(t)-\psi_{n,r_{n}}(t)\bigr{]}+O(k_{n}^{r_{n}+1})\quad\text{for $t\in I_{n}$.} (30)

Proof:

Subtracting (28) from (29), we have

𝒫nu(t)U(t)=(an,rna~n)ψnrn(t)+an,rn+1ψn,rn+1(t)+O(knrn+1)\mathcal{P}_{n}u(t)-U(t)=(a_{n,r_{n}}-\tilde{a}_{n})\psi_{nr_{n}}(t)+a_{n,r_{n}+1}\psi_{n,r_{n}+1}(t)+O(k_{n}^{r_{n}+1})

for tInt\in I_{n}. Since Un=𝒫nu(tn)U^{n}_{-}=\mathcal{P}_{n}u(t_{n}) and ψn,rn(tn)=ψn,rn+1(tn)=1\psi_{n,r_{n}}(t_{n})=\psi_{n,r_{n}+1}(t_{n})=1, taking the limit as ttnt\to t_{n} yields an,rna~n=an,rn+1+O(knrn+1)a_{n,r_{n}}-\tilde{a}_{n}=-a_{n,r_{n}+1}+O(k_{n}^{r_{n}+1}).   \spadesuit

Corollary 16.

𝒫nUn1=2(1)rn+1an,rn+1+O(knrn+1)\mathcal{P}_{n}\llbracket U\rrbracket^{n-1}=2(-1)^{r_{n}+1}a_{n,r_{n}+1}+O(k_{n}^{r_{n}+1}).

Proof:

As ttn1+t\to t_{n-1}^{+}, the left-hand side of (30) tends to

𝒫nu(tn1)U+n1\displaystyle\mathcal{P}_{n}u(t_{n-1})-U^{n-1}_{+} =𝒫n(I𝒫n1+𝒫n1)Un1U+n1=𝒫nUn1U+n1\displaystyle=\mathcal{P}_{n}(I-\mathcal{P}_{n-1}+\mathcal{P}_{n-1})U^{n-1}_{-}-U^{n-1}_{+}=\mathcal{P}_{n}U^{n-1}_{-}-U^{n-1}_{+}
=𝒫n(U+n1Un1)=𝒫nUn1,\displaystyle=-\mathcal{P}_{n}(U^{n-1}_{+}-U^{n-1}_{-})=-\mathcal{P}_{n}\llbracket U\rrbracket^{n-1},

and on the right-hand side, ψn,rn+1(t)ψn,rn(t)\psi_{n,r_{n}+1}(t)-\psi_{n,r_{n}}(t) tends to Prn(1)Prn1(1)=(1)rn(1)rn1=2(1)rnP_{r_{n}}(-1)-P_{r_{n}-1}(-1)=(-1)^{r_{n}}-(-1)^{r_{n}-1}=2(-1)^{r_{n}}.   \spadesuit

Figure 1: The polynomials Pr(τ)Pr1(τ)P_{r}(\tau)-P_{r-1}(\tau).
Refer to caption

In light of Theorem 15, we consider the polynomials

ψn,rn+1(t)ψn,rn(t)=Ψrn+1(τ)Ψrn(τ)=Prn(τ)Prn1(τ).\psi_{n,r_{n}+1}(t)-\psi_{n,r_{n}}(t)=\Psi_{r_{n}+1}(\tau)-\Psi_{r_{n}}(\tau)=P_{r_{n}}(\tau)-P_{r_{n}-1}(\tau).

As illustrated in Figure 1, there are r+1r+1 points

1=τr0<τr1<<τrr=1-1=\tau_{r0}<\tau_{r1}<\cdots<\tau_{rr}=1

such that

(PrPr1)(τrj)=0for 1jr.(P_{r}-P_{r-1})(\tau_{rj})=0\quad\text{for $1\leq j\leq r$.}

In fact, the rr zeros τr1\tau_{r1}, τr2\tau_{r2}, …, τrr\tau_{rr} are the points of a right-Radau quadrature rule [4, Chapter 9] on the interval [1,1][-1,1]. We put

tnj=𝗍n(τrnj)for 0jrn,t^{*}_{nj}=\mathsf{t}_{n}(\tau_{r_{n}j})\quad\text{for $0\leq j\leq r_{n}$,} (31)

so that tn1=tn0<tn1<<tnrn=tnt^{*}_{n-1}=t^{*}_{n0}<t_{n1}<\cdots<t^{*}_{nr_{n}}=t_{n} and

ψn,rn+1(tnj)ψn,rn(tnj)=0for 1jrn.\psi_{n,r_{n}+1}(t^{*}_{nj})-\psi_{n,r_{n}}(t^{*}_{nj})=0\quad\text{for $1\leq j\leq r_{n}$.}

From Theorem 15, we see that 𝒫nu(t)U(t)=O(knrn)\mathcal{P}_{n}u(t)-U(t)=O(k_{n}^{r_{n}}) for general tInt\in I_{n}, but 𝒫nu(tnj)U(tnj)=O(knrn+1)\mathcal{P}_{n}u(t^{*}_{nj})-U(t^{*}_{nj})=O(k_{n}^{r_{n}+1}) for 1jrn1\leq j\leq r_{n}. Let 𝒳^\widehat{\mathcal{X}} denote the space obtained from 𝒳\mathcal{X} by increasing the maximum allowed polynomial degree over the subinterval InI_{n} from rnr_{n} to r^n=rn+1\hat{r}_{n}=r_{n}+1, for 1nN1\leq n\leq N. The reconstruction U^𝒳^\widehat{U}\in\widehat{\mathcal{X}} of U𝒳U\in\mathcal{X} is then defined by requiring that

U^(tnj)=U(tnj)for 1jrn1,\widehat{U}(t^{*}_{nj})=U(t^{*}_{nj})\quad\text{for $1\leq j\leq r_{n}-1$,}

and that the one-sided limits at the end points are

U^+n1=𝒫nUn1andU^n=Un.\widehat{U}^{n-1}_{+}=\mathcal{P}_{n}U^{n-1}_{-}\quad\text{and}\quad\widehat{U}^{n}_{-}=U^{n}_{-}.

Since U^|In\widehat{U}|_{I_{n}} is a polynomial of degree at most r^n1=rn\hat{r}_{n}-1=r_{n}, it is uniquely determined by these rn+1r_{n}+1 interpolation conditions. Notice also that U^\widehat{U} is continuous at tn1t_{n-1} if Vn1=VnV_{n-1}=V_{n} because 𝒫nUn1=Un1\mathcal{P}_{n}U^{n-1}_{-}=U^{n-1}_{-}.

Makridakis and Nochetto [6] introduced the reconstruction in their analysis of a posteriori error bounds for parabolic PDEs. Since the polynomial (UU^)|In(U-\widehat{U})|_{I_{n}} has degree at most rnr_{n} and vanishes at tnjt^{*}_{nj} for 1nrn1\leq n\leq r_{n}, it must be a multiple of ψn,rn+1ψnrn\psi_{n,r_{n}+1}-\psi_{nr_{n}}. In fact, by taking limits as ttn1+t\to t_{n-1}^{+}, we see that

U(t)U^(t)=12(1)rn𝒫nUn1[ψn,rn+1(t)ψn,rn(t)]for tIn.U(t)-\widehat{U}(t)=\frac{1}{2}(-1)^{r_{n}}\mathcal{P}_{n}\llbracket U\rrbracket^{n-1}\bigl{[}\psi_{n,r_{n}+1}(t)-\psi_{n,r_{n}}(t)\bigr{]}\quad\text{for $t\in I_{n}$.} (32)

At the same time, by Theorem 15 and Corollary 16,

U(t)𝒫nu(t)=12(1)rn𝒫nUn1[ψn,rn+1(t)ψn,rn(t)]+O(knrn+1)for tIn,U(t)-\mathcal{P}_{n}u(t)=\frac{1}{2}(-1)^{r_{n}}\mathcal{P}_{n}\llbracket U\rrbracket^{n-1}\bigl{[}\psi_{n,r_{n}+1}(t)-\psi_{n,r_{n}}(t)\bigr{]}+O(k_{n}^{r_{n}+1})\quad\text{for $t\in I_{n}$,} (33)

implying that U^𝒫nu\widehat{U}-\mathcal{P}_{n}u is O(knrn+1)O(k_{n}^{r_{n}+1}) on InI_{n}. One of our principal aims in the next section is to investigate numerically the error in the dG solution UU and its reconstruction U^\widehat{U} in non-trival cases where A0A\neq 0. We can hope that something like (33) still holds, because the time derivative in the term t1αAu\partial_{t}^{1-\alpha}Au is of lower order than in tu\partial_{t}u. Notice that (5) and (32) imply

U^(t)=j=1r^nU^njψnj(t)for tIn,\widehat{U}(t)=\sum_{j=1}^{\hat{r}_{n}}\widehat{U}^{nj}\psi_{nj}(t)\quad\text{for $t\in I_{n}$,}

where

U^nj={Unj,1jrn1,Unrn+12(1)rn𝒫nUn1,j=rn,12(1)rn+1𝒫nUn1,j=rn+1=r^n.\widehat{U}^{nj}=\begin{cases}U^{nj},&1\leq j\leq r_{n}-1,\\ U^{nr_{n}}+\tfrac{1}{2}(-1)^{r_{n}}\mathcal{P}_{n}\llbracket U\rrbracket^{n-1},&j=r_{n},\\ \tfrac{1}{2}(-1)^{r_{n}+1}\mathcal{P}_{n}\llbracket U\rrbracket^{n-1},&j=r_{n}+1=\hat{r}_{n}.\end{cases}

7 Numerical experiments

A Julia package [7] provides functions to evaluate the coefficients GijnG^{n}_{ij}, Kijn,n1K^{n,n-1}_{ij} and Hijn¯H^{n\bar{\ell}}_{ij} based on the results of Sections 3 and 5. This package also includes (in the examples directory) the scripts used for the examples below.

7.1 The matrix 𝑯¯{\boldsymbol{H}^{\bar{\ell}}}

Let α=3/4\alpha=3/4, and consider for simplicity the case when kn=kk_{n}=k and rn=rr_{n}=r are constant for all nn, so that the formulae of Example 12 apply. To get a sense of how the matrix entries Hij¯H^{\bar{\ell}}_{ij} behave, we computed

𝑯0=[1.088070.155440.070650.042390.155440.494580.093260.048340.070650.093260.338390.068930.042390.048340.068930.26319]\boldsymbol{H}^{0}=\left[\begin{array}[]{rrrr}1.08807&0.15544&0.07065&0.04239\\ -0.15544&0.49458&0.09326&0.04834\\ 0.07065&-0.09326&0.33839&0.06893\\ -0.04239&0.04834&-0.06893&0.26319\end{array}\right]

and

𝑯1=[0.346230.134280.068840.042190.134280.084140.054050.036900.068840.054050.040500.030480.042190.036900.030480.02472],\boldsymbol{H}^{1}=\left[\begin{array}[]{rrrr}-0.34623&-0.13428&-0.06884&-0.04219\\ 0.13428&0.08414&0.05405&0.03690\\ -0.06884&-0.05405&-0.04050&-0.03048\\ 0.04219&0.03690&0.03048&0.02472\end{array}\right],

which illustrate the property (22). The factor (1+Δ¯(τσ))α2\bigl{(}1+\Delta_{\bar{\ell}}(\tau-\sigma)\bigr{)}^{\alpha-2} in (21) becomes very smooth as ¯\bar{\ell} increases, with the result that Hij¯H^{\bar{\ell}}_{ij} decays rapidly to zero as i+ji+j increases. Even for ¯=2\bar{\ell}=2, we have

𝑯2=101×[0.914830.102200.012610.001640.102200.020270.003550.000590.012610.003550.000800.000160.001640.000590.000160.00004],\boldsymbol{H}^{2}=10^{-1}\times\left[\begin{array}[]{rrrr}-0.91483&-0.10220&-0.01261&-0.00164\\ 0.10220&0.02027&0.00355&0.00059\\ -0.01261&-0.00355&-0.00080&-0.00016\\ 0.00164&0.00059&0.00016&0.00004\end{array}\right],

and Figure 2 shows this behaviour for larger values of ¯\bar{\ell}, with entries in the lower right corner of the matrix reaching the order of the machine epsilon (2522.22×10162^{-52}\approx 2.22\times 10^{-16}) once ¯\bar{\ell} is of order 100100.

Figure 2: Decay of maxi+j=m|Hij¯|\max_{i+j=m}|H^{\bar{\ell}}_{ij}| for increasing mm and ¯\bar{\ell}, when α=3/4\alpha=3/4.
Refer to caption

The value of Hij0H^{0}_{ij} can be computed to machine precision using Gauss quadrature with Mσ=j/2M_{\sigma}=\lceil j/2\rceil and My=(i+j)/21M_{y}=\lceil(i+j)/2\rceil-1 points for the integrals with respect to σ\sigma and yy in (19), and using Mz=(i+j)/21M_{z}=\lceil(i+j)/2\rceil-1 points for the integral with respect to zz in (20). When 1\ell\geq 1, let Hij¯(M)H^{\bar{\ell}}_{ij}(M) denote the value of Hij¯H^{\bar{\ell}}_{ij} computed by applying MM-point Gauss rules to (21), that is, M2M^{2} points for the double integral. For a given absolute tolerance 𝚊𝚝𝚘𝚕\mathtt{atol}, let Mr¯(𝚊𝚝𝚘𝚕)M^{\bar{\ell}}_{r}(\mathtt{atol}) denote the smallest MM for which

|Hij¯(M)Hij¯(12)|<𝚊𝚝𝚘𝚕for all ij{1,2,,r}.\bigl{|}H^{\bar{\ell}}_{ij}(M)-H^{\bar{\ell}}_{ij}(12)\bigr{|}<\mathtt{atol}\quad\text{for all $i$, $j\in\{1,2,\ldots,r\}$.}

Table 1 lists some values of Mr¯(𝚊𝚝𝚘𝚕)M^{\bar{\ell}}_{r}(\mathtt{atol}) for 𝚊𝚝𝚘𝚕=1014\mathtt{atol}=10^{-14}. Unsurprisingly, fewer quadrature points are needed as ¯\bar{\ell} increases.

Table 1: Numbers of Gauss points Mr¯(𝚊𝚝𝚘𝚕)M^{\bar{\ell}}_{r}(\mathtt{atol}) required for atol=1014\texttt{atol}=10^{-14}, when α=3/4\alpha=3/4.
rr ¯=1\bar{\ell}=1 ¯=2\bar{\ell}=2 ¯=10\bar{\ell}=10 ¯=100\bar{\ell}=100 ¯=1000\bar{\ell}=1000
1 9 9 5 3 2
2 9 9 5 3 2
3 9 9 5 4 3
4 10 10 6 4 3
5 10 10 6 5 4
6 11 11 7 5 4
Figure 3: The exact solution uu of (3) in the case (34), together with the piecewise-quadratic (r=3r=3) dG solution with N=3N=3 subintervals.
Refer to caption

7.2 A fractional ODE

We consider the initial-value problem (3) in the case

α=1/2,λ=1/2,f(t)=cosπt,u0=1,T=2,\alpha=1/2,\quad\lambda=1/2,\quad f(t)=\cos\pi t,\quad u_{0}=1,\quad T=2, (34)

for which the solution is

u(t)=u0E1/2(λt)+0tE1/2(λts)f(s)𝑑s,u(t)=u_{0}E_{1/2}(-\lambda\sqrt{t})+\int_{0}^{t}E_{1/2}(-\lambda\sqrt{t-s})f(s)\,ds,

where Eα(x)=n=0tn/Γ(1+nα)E_{\alpha}(x)=\sum_{n=0}^{\infty}t^{n}/\Gamma(1+n\alpha) denotes the Mittag–Leffler function. The substitution s=(1y2)ts=(1-y^{2})t yields a smooth integrand, allowing u(t)u(t) to be computed accurately via Gauss quadrature on the unit interval [0,1][0,1]. Note that E1/2(x)=erfcx(x)=ex2erfc(x)E_{1/2}(-x)=\operatorname{erfcx}(x)=e^{x^{2}}\operatorname{erfc(x)} is just the scaled complementary error function.

Figure 3 shows uu, together with the dG solution UU using piecewise quadratics (r=3r=3) and only N=3N=3 subintervals. In Figure 4 we plot the absolute errors,

E^(t)=|U^(t)u(t)|andEjn={|U(tn0+0)u(tn0)|,j=0,|U(tnj)u(tnj)|,1jr1,|U(tnr0)u(tnr)|,j=r,\widehat{E}(t)=|\widehat{U}(t)-u(t)|\quad\text{and}\quad E^{n}_{j}=\begin{cases}|U(t^{*}_{n0}+0)-u(t^{*}_{n0})|,&j=0,\\ |U(t^{*}_{nj})-u(t^{*}_{nj})|,&1\leq j\leq r-1,\\ |U(t^{*}_{nr}-0)-u(t^{*}_{nr})|,&j=r,\end{cases} (35)

again using piecewise quadratics but now with N=5N=5 subintervals of uniform size kn=k=T/Nk_{n}=k=T/N. Two features are immediately apparent. First, the accuracy is poor near t=0t=0, reflecting the singular behaviour of the solution: for m1m\geq 1, the mmth derivative u(m)(t)u^{(m)}(t) blows up like t(m1/2)t^{-(m-1/2)} as t0t\to 0. Second, on intervals InI_{n} away from 0, the error is notably smaller at the right-Radau points (tnjt^{*}_{nj} for 1j31\leq j\leq 3) than at the left endpoint (tn0=tn1t^{*}_{n0}=t_{n-1}).

Figure 4: Absolute errors in the reconstruction U^(t)\widehat{U}(t) for 0tT=20\leq t\leq T=2, and in the dG solution U(t)U(t) for t=tnjt=t^{*}_{nj}, using piecewise quadratics (r=3r=3) and N=5N=5 uniform subintervals; see (31).
Refer to caption
Table 2: Maximum weighted errors (36) at the points tnjt^{*}_{nj} using piecewise quadratics (r=3r=3) on a uniform grid.
NN E0maxE^{\max}_{0} E1maxE^{\max}_{1} E2maxE^{\max}_{2} E3maxE^{\max}_{3}
8 8.0e-03 8.8e-05 1.3e-04 1.0e-04
16 1.2e-03 2.69 1.4e-05 2.62 1.4e-05 3.15 9.3e-06 3.46
32 1.7e-04 2.87 1.5e-06 3.25 1.3e-06 3.40 8.2e-07 3.50
64 2.2e-05 2.94 1.4e-07 3.42 1.2e-07 3.47 7.2e-08 3.51
128 2.8e-06 2.97 1.3e-08 3.47 1.1e-08 3.49 6.3e-09 3.51
256 3.6e-07 2.98 1.1e-09 3.49 9.5e-10 3.50 5.5e-10 3.51

In Table 2, we show how the quantities

Ejmax=max1nN(tnj)rαEjnE^{\max}_{j}=\max_{1\leq n\leq N}(t^{*}_{nj})^{r-\alpha}E^{n}_{j} (36)

behave as NN grows. These results, together with similar computations using other choices of α\alpha and r2r\geq 2, lead us to conjecture that, in general, using a constant time step kk,

E0nC(tn0)αrkrfor 2nN,E^{n}_{0}\leq C(t^{*}_{n0})^{\alpha-r}k^{r}\quad\text{for $2\leq n\leq N$,}

whereas

EjnC(tnj)αrkr+αfor 1nN and 1jr,E^{n}_{j}\leq C(t^{*}_{nj})^{\alpha-r}k^{r+\alpha}\quad\text{for $1\leq n\leq N$ and $1\leq j\leq r$,}

and that, consequently,

|U^(t)u(t)|Ctαrkr+αfor t1tT.|\widehat{U}(t)-u(t)|\leq Ct^{\alpha-r}k^{r+\alpha}\quad\text{for~{}$t_{1}\leq t\leq T$.}

However, using piecewise-constants (r=1r=1) we do not observe any superconvergence, with both E0maxE^{\max}_{0} and E1maxE^{\max}_{1} behaving like Ctn1αkCt_{n}^{1-\alpha}k, albeit with a noticably smaller constant in the case of E1maxE^{\max}_{1}.

To suppress the growth in the error as tt approaches 0, we can use a graded mesh of the form

tn=(n/N)qTfor 0nN,t_{n}=(n/N)^{q}T\quad\text{for $0\leq n\leq N$,} (37)

with a suitable grading exponent q1q\geq 1. Table 3 shows the maximum error in the reconstruction, i.e., max0tT|U^(t)u(t)|\max_{0\leq t\leq T}|\widehat{U}(t)-u(t)|, together with the associated convergence rates, for four choices of qq and using T=1T=1 as the final time. These errors appear to be of order kmin(3.5,qα)k^{\min(3.5,q\alpha)} where k=max1nNknCN1k=\max_{1\leq n\leq N}k_{n}\leq CN^{-1}. We conjecture that, in general,

|U^(t)u(t)|Ckmin(r+α,qα)for 0tT, provided r2.|\widehat{U}(t)-u(t)|\leq Ck^{\min(r+\alpha,q\alpha)}\quad\text{for $0\leq t\leq T$, provided $r\geq 2$.} (38)
Table 3: Maximum error in the reconstruction U^(t)\widehat{U}(t) for 0tT=10\leq t\leq T=1, using piecewise quadratics (r=3r=3) for four choices of the mesh grading exponent qq; see (37).
NN q=1q=1 q=3q=3 q=5q=5 q=6q=6
8 1.1e-02 1.4e-03 4.1e-04 7.1e-04
16 6.0e-03 0.84 3.8e-04 1.89 5.2e-05 2.95 9.1e-05 2.97
32 4.3e-03 0.50 1.3e-04 1.50 7.9e-06 2.72 1.0e-05 3.19
64 3.0e-03 0.50 4.7e-05 1.50 1.4e-06 2.51 9.8e-07 3.35
128 2.1e-03 0.50 1.7e-05 1.50 2.5e-07 2.50 9.2e-08 3.41
256 1.5e-03 0.50 5.9e-06 1.50 4.3e-08 2.50 8.4e-09 3.45

7.3 A fractional PDE

Consider the elliptic operator A=2/x2A=-\partial^{2}/\partial x^{2} for the 1D spatial domain Ω=(0,L)\Omega=(0,L). To construct a reference solution, we exploit fact that the Laplace transform of uu,

u~(x,z)=0eztu(x,t)𝑑t,\tilde{u}(x,z)=\int_{0}^{\infty}e^{-zt}u(x,t)\,dt,

satisfies the two-point boundary-value problem

ω2u~u~xx=g(x,z)for 0<x<L,with u~(0,z)=0=u~(L,z),\omega^{2}\tilde{u}-\tilde{u}_{xx}=g(x,z)\quad\text{for $0<x<L$,}\quad\text{with $\tilde{u}(0,z)=0=\tilde{u}(L,z)$,}

where

ω=zα/2andg(x,z)=zα1[u0(x)+f~(x,z)].\omega=z^{\alpha/2}\quad\text{and}\quad g(x,z)=z^{\alpha-1}[u_{0}(x)+\tilde{f}(x,z)].

The variation-of-parameters formula leads to the integral representation

u~(x,z)=sinhω(Lx)ωsinhωL0xg(x,z)sinhωξdξ+sinhωxωsinhωLxLg(x,z)sinhωξdξ,\tilde{u}(x,z)=\frac{\sinh\omega(L-x)}{\omega\sinh\omega L}\int_{0}^{x}g(x,z)\sinh\omega\xi\,d\xi\\ +\frac{\sinh\omega x}{\omega\sinh\omega L}\int_{x}^{L}g(x,z)\sinh\omega\xi\,d\xi,

and the Laplace inversion formula then gives

u(x,t)=12πiΓeztu~(x,z)𝑑z,u(x,t)=\frac{1}{2\pi i}\int_{\Gamma}e^{zt}\tilde{u}(x,z)\,dz, (39)

for a contour Γ\Gamma homotopic to the imaginary axis and passing to the right of all singularities of the integrand.

Figure 5: The reference solution for the 1D problem with data given by (40) and (41).
Refer to caption
Figure 6: Comparison of the jumps Un1\|\llbracket U\rrbracket^{n-1}\| with the dG error U(t)u(t)\|U(t)-u(t)\| and the reconstruction error U^(t)u(t)\|\widehat{U}(t)-u(t)\|. Top: a uniform mesh with N=12N=12 time steps. Bottom: a graded mesh with N=40N=40 time steps.
Refer to caption
Refer to caption

We choose as data the functions

u0(x)=C0x(Lx)andf(x,t)=Cftet,u_{0}(x)=C_{0}x(L-x)\quad\text{and}\quad f(x,t)=C_{f}te^{-t}, (40)

for constants C0C_{0} and CfC_{f}, and find that

u~(x,z)\displaystyle\tilde{u}(x,z) =C0zρ1(x)sinhω(Lx)+ρ1(Lx)sinhωxsinhωL\displaystyle=\frac{C_{0}}{z}\,\frac{\rho_{1}(x)\sinh\omega(L-x)+\rho_{1}(L-x)\sinh\omega x}{\sinh\omega L}
+Cfz(z+1)2ρ2(x)sinhω(Lx)+ρ2(Lx)sinhωxsinhωL,\displaystyle\qquad{}+\frac{C_{f}}{z(z+1)^{2}}\frac{\rho_{2}(x)\sinh\omega(L-x)+\rho_{2}(L-x)\sinh\omega x}{\sinh\omega L},

where

ρ1(x)=(ωx(Lx)2ω1)coshωx+(2xL)sinhωx+2ω1\rho_{1}(x)=\bigl{(}\omega x(L-x)-2\omega^{-1}\bigr{)}\cosh\omega x+(2x-L)\sinh\omega x+2\omega^{-1}

and

ρ2(x)=coshωx1.\rho_{2}(x)=\cosh\omega x-1.

To evaluate the contour integral (39) we apply an optimised equal-weight quadrature rule that arises after deforming Γ\Gamma into the left branch of an hyperbola [16]. Figure 5 shows the reference solution over the time interval [0,2][0,2] in the case

α=0.6,L=2,C0=1,Cf=2.\alpha=0.6,\qquad L=2,\qquad C_{0}=1,\qquad C_{f}=2. (41)

In Figure 6, we plot the L2L_{2}-norms of the jumps, Un1\|\llbracket U\rrbracket^{n-1}\|, together with the errors in U(t)U(t) and its reconstruction U^(t)\widehat{U}(t). The dG method used piecewise-quadratics (r=3r=3), first with a uniform mesh of N=12N=12 subintervals (top), and then with a non-uniform mesh of N=40N=40 subintervals (bottom). In both cases, the spatial discretisation used (continuous) piecewise cubics on a uniform grid with 2020 subintervals. Since u0u_{0} is a quadratic polynomial in this instance, we simply put U0=u0U_{0}=u_{0}. Consistent with our conjecture (33), we observe that

suptn1<t<tnU(t)u(t)Un1.\sup_{t_{n-1}<t<t_{n}}\|U(t)-u(t)\|\approx\bigl{\|}\llbracket U\rrbracket^{n-1}\bigr{\|}.

Motivated by our conjecture (38), the second mesh was graded for 0tn10\leq t_{n}\leq 1 by taking q=(r+α)/αq=(r+\alpha)/\alpha, N=34N=34 and T=1T=1 in the formula (37), followed by a uniform mesh on the other half [1,2][1,2] of the time interval. We see that the mesh grading is effective at resolving the solution for tt near zero, albeit with a substantial increase in the overal computational cost.

Acknowledgements

This project was supported by a UNSW Faculty Research Grant (PS47152/IR001/MATH).

References

  • [1] Michael G. Duffy “Quadrature over a pyramid or cube of integrands with a singularity at a vertex” In SIAM J. Numer. Anal. 19, 1982, pp. 1260–1262 DOI: 10.1137/0719090
  • [2] Kenneth Eriksson, Claes Johnson and Vidar Thomée “Time discretization of parabolic problems by the discontinuous Galerkin method” In ESAIM: M2AN 19, 1985, pp. 611–643 DOI: 10.1051/m2an/1985190406111
  • [3] J. Klafter and I. M. Sokolov “First Steps in Random Walks” Oxford University Press, 2011
  • [4] Vladimir Ivanovich Krylov “Approximate Calculation of Integrals”, ACM Monographs New York: Macmillan, 1962
  • [5] Kim-Ngan Le, William McLean and Martin Stynes “Existence, uniqueness and regularity of the solution of the time-fractional Fokker–Planck equation with general forcing” In Commun. Pure Appl. Anal. 18, 2019, pp. 2765–2787 DOI: 10.3934/cpaa.2019124
  • [6] Charalambos Makridakis and Richardo H. Nochetto “A posteriori error analysis for higher order dissipative methods for evolution problems” In Numer. Math. 104, 2006, pp. 489–514 DOI: 10.1007/s00211-006-0013-6
  • [7] William McLean “FractionalTimeDG: Generate coefficient arrays needed for discontinuous Galerkin time-stepping of fractional diffusion problems” Github, https://github.com/billmclean/FractionalTimeDG.jl, 2020
  • [8] William McLean “Regularity of solutions to a time-fractional diffusion equation” In ANZIAM J. 52, 2010, pp. 123–138 DOI: 10.1017/S1446181111000617
  • [9] William McLean and Kassem Mustapha “Convergence analysis of a discontinuous Galerkin method for a sub-diffusion equation” In Numer. Algor. 52, 2009, pp. 69–88 DOI: 10.1007/s11075-008-9258-8
  • [10] William McLean, Kassem Mustapha, Raed Ali and Omar Knio “Well-posedness of time-fractional advection-diffusion-reaction equations” In Fract. Calc. Appl. Anal. 22, 2019, pp. 918–944 DOI: 10.1515/fca-2019-0050
  • [11] Ralf Metzler and Joseph Klafter “The random walk’s guide to anomalous diffusion: a fractional dynamics approach” In Physics Reports 339, 2000, pp. 1–77 DOI: 10.1016/S0370-1573(00)00070-3
  • [12] Kassem Mustapha “Time-stepping discontinuous Galerkin methods for fractional diffusion problems” In Numer. Math. 130, 2015, pp. 497–516 DOI: 10.1007/s00211-014-0669-2
  • [13] Lars Schmutz and Thomas P. Wihler “The variable-order discontinuous Galerkin time stepping scheme for parabolic evolution problems is uniformly LL^{\infty}-stable” In SIAM J. Numer. Anal. 57, 2019, pp. 293–319 DOI: 10.1137/17M1158835
  • [14] Dominik Schötzau and Christoph Schwab “Time discretization of parabolic problems by the hp-version of the discontinuous Galerkin finite element method” In SIAM J. Numer. Anal. 38, 2001, pp. 837–875 DOI: 10.1137/S0036142999352394
  • [15] Vidar Thomée “Galerkin Finite Element Methods for Parabolic Problems” Springer, 2006
  • [16] J. A. C. Weideman and L. N. Trefethen “Parabolic and hyperbolic contours for computing the Bromwich integral” In Math. Comp. 76, 2007, pp. 1341–1356 DOI: 10.1090/S0025-5718-07-01945-X

Author address