# Dirichlet-Neumann Waveform Relaxation Algorithm for Time Fractional
Diffusion Equation in Heterogeneous Media

Soura Sana111School of Basic Sciences, IIT Bhubaneswar, India
(ss87@iitbbs.ac.in). Bankim C Mandal222School of Basic Sciences, IIT
Bhubaneswar, India (bmandal@iitbbs.ac.in).

###### Abstract

In this article, we have studied the convergence behavior of the Dirichlet-
Neumann waveform relaxation algorithms for time-fractional sub-diffusion and
diffusion wave equations in 1D & 2D for regular domains, where the
dimensionless diffusion coefficient takes different constant values in
different subdomains. From numerical experiments, we first capture the optimal
relaxation parameters. Using these optimal relaxation parameters, our analysis
estimates the rate of change of the convergence behavior against the
fractional order and time. We have performed our analysis in multiple
subdomain cases for both in 1D & 2D.

###### keywords:

Dirichlet-Neumann, Waveform Relaxation, Domain Decomposition, Sub-diffusion,
Diffusion-wave

###### AMS:

65M55, 34K37

##  1 Model problem

We aim to extend the DNWR algorithm for time-fractional diffusion equations.
The model problem we have taken is linear, but the time derivative is of
fractional order $2\nu$, where $\nu$ can take any values from zero to one.
Therefore, our model incorporates three fundamental basic PDEs: elliptic,
parabolic, and hyperbolic. More details about the model and its application
can be found in [1, 3]. To define the sub-diffusion and diffusion-wave, first,
we consider the domain $\Omega$ in $\mathbb{R}^{d}$, and then we have:

(1) $\begin{cases}\frac{\partial^{2\nu}}{\partial
t^{2\nu}}u=\nabla\cdot\left(\kappa(\boldsymbol{x},t)\nabla
u\right)+f(\boldsymbol{x},t),&\textrm{in}\;\Omega\times(0,T),\\\
u(\boldsymbol{x},t)=g(\boldsymbol{x},t),&\textrm{on}\;\partial\Omega\times(0,T),\\\
u(\boldsymbol{x},0)=u_{0}(\boldsymbol{x}),&\textrm{in}\;\Omega.\end{cases}$

where $\kappa(\boldsymbol{x},t)>0$ be the dimensionless diffusion coefficient
and $\frac{\partial^{\alpha}}{\partial t^{\alpha}}$ is the Caputo fractional
derivative defined for order $\alpha$, $n-1<\alpha<n$ and $n\in\mathbb{N}$ as
follows:

$\frac{\partial^{\alpha}}{\partial
t^{\alpha}}x(t):=\frac{1}{\Gamma(n-\alpha)}\int_{0}^{t}(t-\tau)^{n-\alpha-1}x^{(n)}(\tau)d\tau.$

For further analysis, we assume that the solution $u(\boldsymbol{x},t)$ and
the flux is continuous even for finite jump discontinuous
$\kappa(\boldsymbol{x},t)$ in space. In support of the existence and
uniqueness of the weak solution of (1) we refer to [4]. With this assumption,
we next introduce the DNWR algorithm:

##  2 The Dirichlet-Neumann Waveform Relaxation algorithm

Many forms of DNWR algorithm depend on the placement of Dirichlet-Neumann,
Neumann-Neumann, and Dirichlet-Dirichlet boundary conditions already exist in
literature; for more details, see [2]. Here we have chosen the particular one,
which is given in Figure 1. The reason behind this combination is that it can
handle long-time processes much better than other ones. So to use this
combination in model problem (1), we subdivide the domain $\Omega$ into $N$
subdomains $\Omega_{i}$, $1\leq i\leq N$ with no overlaps and no cross points.

Fig. 1: Particular form of DNWR algorithm.

Set original boundary $\Gamma_{i}:=\partial\Omega\cap\partial\Omega_{i},1\leq
i\leq N$ and artificial one as
$\Lambda_{i}:=\partial\Omega_{i}\cup\partial\Omega_{i+1}$ for $1\leq i\leq
N-1$. We choose $\psi_{i}^{(0)}(\boldsymbol{x},t)$ as an initial guess in some
appropriate space along the interfaces $\Lambda_{i}\times(0,T)$. Then DNWR
algorithm reads:

(2) $\begin{array}[]{rcll}\begin{cases}\frac{\partial^{2\nu}}{\partial
t^{2\nu}}u_{i}^{(k)}=\nabla\cdot\left(\kappa(\boldsymbol{x},t)\nabla
u_{i}^{(k)}\right)+f,&\mbox{in $\Omega_{i}$},\\\
u_{i}^{(k)}(\boldsymbol{x},0)=u_{0}(\boldsymbol{x}),&\mbox{in
$\Omega_{i}$},\\\ u_{i}^{(k)}=g,&\mbox{on
$\Gamma_{i}$},\end{cases}\end{array}$

with the boundary conditions along the artificial interfaces: for $i<\lceil
N/2\rceil$

(3) $\begin{cases*}u_{i-1}^{(k)}=\psi_{i-1}^{(k-1)},&\mbox{on
$\Lambda_{i-1},i\neq 1$},\\\
\partial_{\boldsymbol{n}_{i,i+1}}u_{i}^{(k)}=-\partial_{\boldsymbol{n}_{i+1,i}}u_{i+1}^{(k)},&\mbox{on
$\Lambda_{i}$}\end{cases*}$

for $i=\lceil N/2\rceil$

(4) $\begin{cases*}u_{i-1}^{(k)}=\psi_{i-1}^{(k-1)},&\mbox{on
$\Lambda_{i-1}$},\\\ u_{i}^{(k)}=\psi_{i}^{(k-1)},&\mbox{on
$\Lambda_{i}$},\end{cases*}$

for $i>\lceil N/2\rceil$

(5)
$\begin{cases*}\partial_{\boldsymbol{n}_{i,i-1}}u_{i}^{(k)}=-\partial_{\boldsymbol{n}_{i-1,i}}u_{i-1}^{(k)},&\mbox{on
$\Lambda_{i-1}$}\\\ u_{i}^{(k)}=\psi_{i}^{(k-1)},&\mbox{on $\Lambda_{i},i\neq
N$},\end{cases*}$

and then with the relaxation parameter $\theta\in(0,1]$ update the interface
data using

(6) $\begin{cases*}h_{i}^{(k)}(\boldsymbol{x},t)=\theta
u_{i}^{(k)}\left|{}_{\Lambda_{i}\times(0,T)}\right.+(1-\theta)\psi_{i}^{(k-1)}(\boldsymbol{x},t),\quad
1\leq i<\lceil N/2\rceil\\\ h_{i}^{(k)}(\boldsymbol{x},t)=\theta
u_{i+1}^{(k)}\left|{}_{\Lambda_{i}\times(0,T)}\right.+(1-\theta)\psi_{i}^{(k-1)}(\boldsymbol{x},t),\quad\lceil
N/2\rceil\leq i\leq N-1.\end{cases*}$

Our goal is to study the convergence of
$w_{i}^{(k-1)}(\boldsymbol{x},t):=u_{i}|_{\Gamma_{i}\times(0,T)}-\psi_{i}^{(k-1)}(\boldsymbol{x},t)$
as $k\to\infty$. However, before going to the main theorems on convergence, we
need some estimates, which we will discuss in the next section.

##  3 Auxiliary Results

In this section, we first discuss a few norm inequalities and then some kernel
estimates, which are necessary for subsequent sections.

######  Lemma 1.

Let $g$ and $w$ be two real-valued functions in $(0,\infty)$ with
$\hat{w}(s)=\mathcal{L}\left\\{w(t)\right\\}$ the Laplace transform of $w$.
Then for $t\in(0,T)$, we have the following properties:

  1. (i)

$\|g*w\|_{L^{1}(0,t)}\leq\|g\|_{L^{1}(0,t)}\|w\|_{L^{1}(0,t)}.$

  2. (ii)

$\|g*w\|_{L^{\infty}(0,t)}\leq\|g\|_{L^{\infty}(0,t)}\|w\|_{L^{1}(0,t)}.$

  3. (iii)

If $w(\tau)\geq 0$ is $L^{1}$-integrable in $(0,t)$, then
$\|w(.)\|_{L^{1}(0,t)}\leq\lim_{s\to 0+}\hat{w}(Re(s))$.

######  Lemma 2.

For $\alpha\in(0,1)$ and $l,t>0$, the following results hold:

  1. (i)

The inverse Laplace transform of $e^{-ls^{\alpha}}$ is:

$\mathcal{L}^{-1}\left\\{e^{-ls^{\alpha}}\right\\}=l\alpha
t^{-(\alpha+1)}M_{\alpha}\left(lt^{-\alpha}\right),$

where $M_{\alpha}(x),x\in(0,\infty)$ be the M-Wright function.

  2. (ii)

$\left\|\mathcal{L}^{-1}\left\\{e^{-ls^{\alpha}}\right\\}\right\|_{L^{1}(0,t)}\leq\exp\left(-(1-\alpha)\left(\frac{\alpha}{t}\right)^{\alpha/(1-\alpha)}l^{1/(1-\alpha)}\right).$

######  Lemma 3.

Let $0\leq l_{1}<l_{2}$ and $s$ be a complex variable. Then, for
$t\in(0,\infty)$ and

  1. (i)

for $0<\alpha<1$

$\Phi(t):=\mathcal{L}^{-1}\left\\{\frac{\sinh(l_{1}s^{\alpha})}{\sinh(l_{2}s^{\alpha})}\right\\}\quad\mbox{and}\quad\Psi(t):=\mathcal{L}^{-1}\left\\{\frac{\cosh(l_{1}s^{\alpha})}{\cosh(l_{2}s^{\alpha})}\right\\}$

exist and $L^{1}$ integrable.

  2. (ii)

for $0<\alpha\leq 1/2$

$\Phi(t)\geq 0\textit{ and }\Psi(t)\geq 0.$

##  4 Convergence of DNWR Algorithm

We are now in a position to present the main convergence result for the DNWR
algorithm (2)-(6). To have a clear picture of theoretical convergence, we
choose the 1D form of the sub-diffusion and diffusion-wave model on
$2m+1,m\in\mathbb{N}$ subdomains, but for generality consider the
heterogeneous problem with $\kappa(\boldsymbol{x},t)=\kappa_{\mu}$, on
$\Omega_{\mu}$ where $1\leq\mu\leq m$ and
$\kappa(\boldsymbol{x},t)=\kappa_{\iota}$, on $\Omega_{\iota}$ where
$m+2\leq\iota\leq 2m+1$ and $\kappa(\boldsymbol{x},t)=\kappa_{m+1}$, on
$\Omega_{m+1}$. Considering $w_{i}^{(0)},1\leq i\leq 2m$ be the initial error
term on artificial interfaces, the Laplace transform in time converts the
error equations of (2)-(6) into the following sets of ODEs:

(7)
$\displaystyle\begin{cases}(s^{2\nu}-\kappa_{m+1}\frac{d^{2}}{dx^{2}})\hat{u}_{m+1}^{(k)}=0,&\textrm{in
$\Omega_{m+1}$},\\\ \hat{u}_{m+1}^{(k)}=\hat{w}_{m}^{(k-1)},&\textrm{at
$x_{m}$}\\\ \hat{u}_{m+1}^{(k)}=\hat{w}_{m+1}^{(k)},&\textrm{at
$x_{m+1}$}\end{cases}$ (8)
$\begin{array}[]{rcll}\begin{cases}(s^{2\nu}-\kappa_{\mu}\frac{d^{2}}{dx^{2}})\hat{u}_{\mu}^{(k)}=0,&\textrm{in
$\Omega_{\mu}$}\\\ \hat{u}_{\mu}^{(k)}=\hat{w}_{\mu-1}^{(k-1)},&\textrm{at
$x_{\mu-1}$}\\\
\kappa_{\mu}\frac{d}{dx}\hat{u}_{\mu}^{(k)}=\kappa_{\mu+1}\frac{d}{dx}\hat{u}_{\mu+1}^{(k)},&\textrm{at
$x_{\mu}$}\end{cases}\end{array}\begin{array}[]{rcll}\begin{cases}(s^{2\nu}-\kappa_{\iota}\frac{d^{2}}{dx^{2}})\hat{u}_{\iota}^{(k)}=0,&\textrm{in
$\Omega_{\iota}$}\\\
\kappa_{\iota}\frac{d}{dx}\hat{u}_{\iota}^{(k)}=\kappa_{\iota-1}\frac{d}{dx}\hat{u}_{\iota-1}^{(k)},&\textrm{at
$x_{\iota-1}$}\\\ \hat{u}_{\iota}^{(k)}=\hat{w}_{\iota}^{(k-1)},&\textrm{at
$x_{\iota}$}\end{cases}\end{array}$

then update the interface error by

(9) $\displaystyle\hat{w}_{\mu}^{(k)}(s)$
$\displaystyle=\theta_{\mu}\hat{u}_{\mu}^{(k)}(x_{\mu},s)+(1-\theta_{\mu})\hat{w}_{\mu}^{(k-1)}(s),\textit{
for }1\leq\mu\leq m,$ (10) $\displaystyle\hat{w}_{\iota}^{(k)}(s)$
$\displaystyle=\theta_{\iota}\hat{u}_{\iota}^{(k)}(x_{\iota},s)+(1-\theta_{\iota})\hat{w}_{\iota}^{(k-1)}(s),\textit{
for }m+1\leq\iota\leq 2m.$

Now we can solve for each $\hat{u}_{i}^{(k)}$ from (7) & (8), and substitute
those values in (9) & (10) at artificial interfaces give us the following
relation:

(11)
$\hat{w}_{i}^{(k)}(s)=\sum_{j=1}^{2m}\hat{\rho}_{i,j}\hat{w}_{j}^{(k-1)}(s)$

where $\hat{\rho}_{i,j}$, whose calculation is quite cumbersome so given in
Appendix A, takes the form for $1\leq i\leq m$ and $1\leq j\leq 2m$:

$\hat{\rho}_{i,j}=\begin{cases}1-\theta_{i}\left(1+\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}\right)+\theta_{i}\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}\left(1-\frac{\sigma_{i}\sigma_{i+1}}{\gamma_{i}\gamma_{i+1}}\right),&j=i,i<m\\\
1-\theta_{m}\left(1+\sqrt{\frac{\kappa_{m+1}}{\kappa_{m}}}\right)+\theta_{m}\sqrt{\frac{\kappa_{m+1}}{\kappa_{m}}}\left(1-\frac{\sigma_{m}\gamma_{m+1}}{\gamma_{m}\sigma{m+1}}\right),&j=i=m,\\\
\frac{\theta_{i}}{\gamma_{i}},&j=i-1,i>1,\\\
\theta_{i}\sqrt{\frac{\kappa_{j+1}}{\kappa_{i}}}\frac{-\sigma_{i}\sigma_{j+1}}{\gamma_{i}\gamma_{i+1}\cdots\gamma_{j}},&i+1\leq
j\leq m-1,\\\
\theta_{i}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\frac{-\gamma_{m+1}\sigma_{i}}{\gamma_{i}\gamma_{i+1}\cdots\gamma_{m}\sigma_{m+1}},&j=m,i<m,\\\
\theta_{i}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\frac{\sigma_{i}}{\gamma_{i}\gamma_{i+1}\cdots\gamma_{m}\sigma_{m+1}},&j=m+1,\\\
0,&\textit{ otherwise. }\end{cases}$

Due to the particular symmetric structure of the $\hat{\rho}$ matrix, the
lower half of it, that is, $\hat{\rho}_{i,j}$ such that $m+1\leq i\leq 2m$ for
all $j$ can be easily deduced from the upper half by just changing all indices
$i_{1}$ by $2m+2-i_{1}$; except $\hat{\rho}_{i,j},\theta_{i}$, which will be
replaced by $2m+1-i_{1}$, where $1\leq i_{1}\leq m$. For having super-linear
convergence, we choose
$\theta_{i}=1/(1+\sqrt{\kappa_{i+1}/\kappa_{i}})=:\theta^{*}_{i}$ for $1\leq
i\leq m$ and get:

(12)
$\hat{\rho}_{i,j}=\begin{cases}\theta^{*}_{i}\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}\left(1-\frac{\sigma_{i}\sigma_{i+1}}{\gamma_{i}\gamma_{i+1}}\right),&j=i,i<m\\\
\theta^{*}_{m}\sqrt{\frac{\kappa_{m+1}}{\kappa_{m}}}\left(1-\frac{\sigma_{m}\gamma_{m+1}}{\gamma_{m}\sigma{m+1}}\right),&j=i=m,\\\
\frac{\theta^{*}_{i}}{\gamma_{i}},&j=i-1,i>1,\\\
\theta^{*}_{i}\sqrt{\frac{\kappa_{j+1}}{\kappa_{i}}}\frac{-\sigma_{i}\sigma_{j+1}}{\gamma_{i}\gamma_{i+1}\cdots\gamma_{j}},&i+1\leq
j\leq m-1,\\\
\theta^{*}_{i}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\frac{-\gamma_{m+1}\sigma_{i}}{\gamma_{i}\gamma_{i+1}\cdots\gamma_{m}\sigma_{m+1}},&j=m,i<m,\\\
\theta^{*}_{i}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\frac{\sigma_{i}}{\gamma_{i}\gamma_{i+1}\cdots\gamma_{m}\sigma_{m+1}},&j=m+1,\\\
0,&\textit{ otherwise. }\end{cases}$

######  Lemma 4.

Suppose $l_{1},l_{2},\cdots,l_{n}$ are positive real numbers with
$l=\min_{i}l_{i}$ and $0<\alpha\leq 1/2$. Let $\hat{P}_{i}(s)$ be the Laplace
transform of $P_{i}(t)$. Then we have the followings bounds:

  1. (i)

$\|\mathcal{L}^{-1}\left\\{\hat{P}_{1}(s)\right\\}\|_{L^{1}(0,t)}=\|\mathcal{L}^{-1}\left\\{\frac{\cosh(ls^{\alpha})\cosh((l_{1}-l_{2})s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}\leq
1,$

  2. (ii)

$\|\mathcal{L}^{-1}\\{\hat{P}_{2}(s)\\}\|_{L^{1}(0,t)}=\|\mathcal{L}^{-1}\left\\{\frac{\cosh(ls^{\alpha})\sinh(l_{1}s^{\alpha})\sinh(l_{n}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}\leq
2,$

  3. (iii)

$\|\mathcal{L}^{-1}\\{\hat{P}_{3}(s)\\}\|_{L^{1}(0,t)}=\|\mathcal{L}^{-1}\left\\{\frac{\cosh(ls^{\alpha})\sinh(l_{1}s^{\alpha})\cosh(l_{n}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}\leq
1+\left|\frac{l_{1}-l_{n}}{l_{n}}\right|,$

  4. (iv)

$\|\mathcal{L}^{-1}\\{\hat{P}_{4}(s)\\}\|_{L^{1}(0,t)}=\|\mathcal{L}^{-1}\left\\{\frac{\cosh(ls^{\alpha})\sinh(l_{1}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}\leq\frac{l_{1}}{l_{n}}.$

###### Proof.

  1. (i)

If $l_{2}\geq l_{1}$, from part (i) and (ii) of Lemma 3, guarantee that the
inverse Laplace transform of
$\hat{P}_{11}(s):=\frac{\cosh(ls^{\alpha})}{\cosh(l_{1}s^{\alpha})},\hat{P}_{12}(s):=\frac{\cosh((l_{1}-l_{2})s^{\alpha})}{\cosh(l_{2}s^{\alpha})}$
exist, $L^{1}$ integrable and non-negative. Therefore using part (i) of Lemma
1 we have the following inequality:

$\displaystyle\|P_{1}(.)\|_{L^{1}(0,t)}\leq\left\|\mathcal{L}^{-1}\left\\{\hat{P}_{11}(s)\right\\}\right\|_{L^{1}(0,t)}\left\|\mathcal{L}^{-1}\left\\{\hat{P}_{12}(s)\right\\}\right\|_{L^{1}(0,t)}.$

Now using part (iii) of Lemma 1, we have:

$\displaystyle\|P_{1}(.)\|_{L^{1}(0,t)}$
$\displaystyle\leq\left\|\mathcal{L}^{-1}\left\\{\hat{P}_{11}(s)\right\\}\right\|_{L^{1}(0,t)}\left\|\mathcal{L}^{-1}\left\\{\hat{P}_{12}(s)\right\\}\right\|_{L^{1}(0,t)}$
$\displaystyle\leq\lim_{s\to 0}\hat{P}_{11}(Re(s))\hat{P}_{12}(Re(s))=1.$

If $l_{2}<l_{1}$, then we choose
$\hat{P}_{11}(s):=\frac{\cosh(ls^{\alpha})}{\cosh(l_{2}s^{\alpha})},\hat{P}_{12}(s):=\frac{\cosh((l_{1}-l_{2})s^{\alpha})}{\cosh(l_{1}s^{\alpha})}$
and follow the same procedure as earlier. Hence combining two cases give us
the result:

$\|\mathcal{L}^{-1}\left\\{\hat{P}_{1}(s)\right\\}\|_{L^{1}(0,t)}\leq 1.$

  2. (ii)

To have the $L^{1}$ estimate on $P_{2}(t)$, we reduce
$\hat{P}_{2}(s)=\hat{P}_{21}(s)-\hat{P}_{22}(s)$, where
$\hat{P}_{21}(s):=\frac{\cosh(ls^{\alpha})}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})}$
and
$\hat{P}_{22}(s):=\frac{\cosh(ls^{\alpha})\cosh((l_{1}-l_{n})s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n}s^{\alpha})}$.
Now we use the part(i) of Lemma 1 and have:

$\displaystyle\|\mathcal{L}^{-1}\left\\{\hat{P}_{21}(s)\right\\}\|_{L^{1}(0,t)}\leq\left\|\mathcal{L}^{-1}\left\\{\frac{\cosh(ls^{\alpha})}{\cosh(l_{2}s^{\alpha})}\right\\}\right\|_{L^{1}(0,t)}\cdots\left\|\mathcal{L}^{-1}\left\\{\frac{1}{\cosh(l_{n-1}s^{\alpha})}\right\\}\right\|_{L^{1}(0,t)}$

Again using the same kind of reasoning as part (i), we have:

(13)
$\displaystyle\|\mathcal{L}^{-1}\left\\{\hat{P}_{21}(s)\right\\}\|_{L^{1}(0,t)}\leq
1.$

Proof of the estimates for $\hat{P}_{22}(s)$ for the cases $l_{1}\leq l_{n}$
and $l_{1}>l_{n}$ can be done separately and are obvious. Therefore

(14) $\|\mathcal{L}^{-1}\left\\{\hat{P}_{22}(s)\right\\}\|_{L^{1}(0,t)}\leq
1.$

Hence, Adding (13) & (14), we obtain:

$\|\mathcal{L}^{-1}\\{\hat{P}_{2}(s)\\}\|_{L^{1}(0,t)}\leq 2.$

  3. (iii)

We split $\hat{P}_{3}(s)=\hat{P}_{31}(s)+\hat{P}_{32}(s)$ where,
$\hat{P}_{31}(s):=\frac{\cosh(ls^{\alpha})}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})}$,
so by using same technique as part (ii) , we have
$\|\mathcal{L}^{-1}\\{\hat{P}_{31}(s)\\}\|_{L^{1}(0,t)}\leq 1.$ Now
$\hat{P}_{32}(s):=\hat{P}_{31}(s)\hat{D}_{1}(s)$, where
$\hat{D}_{1}(s):=\frac{\sinh((l_{1}-l_{n})s^{\alpha})}{\cosh(l_{1}s^{\alpha})\sinh(l_{n}s^{\alpha})}$.
From Theorem 2 in LABEL:mandal2014time, we have
$\|\mathcal{L}^{-1}\\{\hat{D}_{1}(s)\\}\|_{L^{1}(0,t)}\leq\frac{|l_{1}-l_{n}|}{l_{n}}$.
Therefore adding both cases give:

$\|\mathcal{L}^{-1}\\{\hat{P}_{3}(s)\\}\|_{L^{1}(0,t)}\leq
1+\left|\frac{l_{1}-l_{n}}{l_{n}}\right|.$

  4. (iv)

Rearrange $\hat{P}_{4}(s)=\hat{P}_{41}(s)+\hat{P}_{42}(s)$ where,
$\hat{P}_{41}(s):=\frac{\sinh(ls^{\alpha})}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})}$.
Therefore from part (i) of Lemma 3 we have

$\displaystyle\|\mathcal{L}^{-1}\\{\hat{P}_{41}(s)\\}\|_{L^{1}(0,t)}$
$\displaystyle\leq\left\|\mathcal{L}^{-1}\left\\{\frac{1}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})}\right\\}\right\|_{L^{1}(0,t)}\left\|\mathcal{L}^{-1}\left\\{\frac{\sinh(ls^{\alpha})}{\sinh(l_{n}s^{\alpha})}\right\\}\right\|_{L^{1}(0,t)}\leq\frac{l}{l_{n}}$

Define
$\hat{D}_{2}(s):=\frac{\sinh((l_{1}-l)s^{\alpha})}{\cosh(l_{1}s^{\alpha})\sinh(l_{n}s^{\alpha})}$,
then $\hat{P}_{42}(s):=\hat{P}_{41}(s)\hat{D}_{2}(s)$. The proof of
$\|\mathcal{L}^{-1}\\{\hat{P}_{42}(s)\\}\|_{L^{1}(0,t)}\leq\|\mathcal{L}^{-1}\\{\hat{P}_{41}(s)\hat{D}_{2}(s)\\}\|_{L^{1}(0,t)}\leq\frac{l_{1}-l}{l_{n}}$
is same as part (iii). Hence adding both parts follows
$\|\mathcal{L}^{-1}\\{\hat{P}_{4}(s)\\}\|_{L^{1}(0,t)}\leq\frac{l_{1}}{l_{n}}.$

∎

######  Theorem 5 (Convergence of DNWR when $0<\nu\leq 1/2$).

For fixed $T>0$, $0<\nu\leq 1/2$ and
$\theta_{i}=1/(1+\sqrt{\kappa_{i+1}/\kappa_{i}}),1\leq i\leq m$ and
$\theta_{i}=1/(1+\sqrt{\kappa_{i}/\kappa_{i+1}}),m+1\leq i\leq 2m$ , the DNWR
algorithm (7)-(10) for $2m+1(m\in\mathbb{N})$ subdomains converge super-
linearly with the estimate:

$\displaystyle\max_{1\leq i\leq 2m}\|w_{i}^{(k)}\|_{L^{\infty}(0,T)}$
$\displaystyle\leq(2c)^{k}\exp(-Ak^{1/(1-\nu)})\max_{1\leq j\leq
2m}\|w_{j}^{(0)}\|_{L^{\infty}(0,T)},$

where, $h_{j}=a_{j}/\sqrt{\kappa_{j}},\,h=\min_{1\leq j\leq
2m+1}a_{j},\,A=(1-\nu)\nu^{\nu/(1-\nu)}(h/T^{\nu})^{1/(1-\nu)}$, and
$c=\max_{i}c_{i}$, where $c_{i}$ are ginen in (18) & (20).

###### Proof.

Solving the ODEs (7)-(10) in terms of errors on artificial boundaries give us:

(15)
$\hat{w}_{i}^{(k)}(s)=\sum_{j=1}^{2m}\hat{\rho}_{i,j}\hat{w}_{j}^{(k-1)}(s),$

where $\hat{\rho}_{i,j}$ takes the form from (12) for $1\leq i\leq m$.
Cumulative estimate of $\hat{\rho}_{i,j}^{k}$ at $k$-th iteration level will
be cumbersome, so we estimate some form of $\hat{\rho}_{i,j}$ at each
iteration. But before that define $\gamma=\cosh(hs^{\nu})$, where
$h=\min_{j}h_{j}$. setting
$\hat{v}^{(k)}_{i}(s):=\gamma^{k}\hat{w}^{(k)}_{i}(s)$. Which transform (15)
into:

$\hat{v}_{i}^{(k)}=\sum_{j=1}^{2m}\gamma\hat{\rho}_{i,j}\hat{v}_{j}^{(k-1)}.$

To have our final estimate result in time domain we take inverse Laplace
transform on both side:

$v_{i}^{(k)}=\sum_{j=1}^{2m}\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\ast
v_{j}^{(k-1)}.$

Before going to the sup-norm estimate we first assume that
$\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$ exist. Our assumption is
justified because each $\left\|w_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$
perfectly well-defined. Therefore taking $L^{\infty}$ norm on both side reduce
the above equation to:

$\displaystyle\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$
$\displaystyle\leq\sum_{j=1}^{2m}\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\ast
v_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)}$

Applying part (ii) of Lemma 1 give:

$\displaystyle\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$
$\displaystyle\leq\sum_{j=1}^{2m}\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}\left\|v_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)}$
(16)
$\displaystyle\leq\max_{j}\left\|v_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)}\sum_{j=1}^{2m}\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}.$

Now we use the Lemma 4 to estimate the summation part of the above inequality.
For $i=j,j<m$

$\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}=\theta_{i}^{*}\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}\mathcal{L}^{-1}\left\\{\gamma\left(1-\frac{\sigma_{i}\sigma_{i+1}}{\gamma_{i}\gamma_{i+1}}\right)\right\\},$

so from part(i) of Lemma 4 give us

$\left\|\mathcal{L}^{-1}\left\\{\gamma\left(1-\frac{\sigma_{i}\sigma_{i+1}}{\gamma_{i}\gamma_{i+1}}\right)\right\\}\right\|_{L^{1}(0,T)}\leq
1.$

Hence
$\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}\leq\theta_{i}^{*}\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}.$
In a similar manner we show
$\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{m,m}\\}\right\|_{L^{1}(0,T)}\leq\theta_{m}^{*}\sqrt{\frac{\kappa_{m+1}}{\kappa_{m}}}.$
and
$\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}\leq\theta_{i}^{*}$
for $j=i-1,i>1$. From part(ii) of Lemma 4, we get the estimate
$\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}\leq
2\theta_{i}^{*}\sqrt{\frac{\kappa_{j+1}}{\kappa_{i}}}$ for $i+1\leq j\leq
m-1$. And for $j=m,i<m$ from part (iii) give:

$\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}\leq\theta_{i}^{*}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\left(1+\left|\frac{h_{i}-h_{m+1}}{h_{m+1}}\right|\right).$

Finally for $j=m+1,i<m$ using part(iv) obtain
$\left\|\mathcal{L}^{-1}\\{\gamma\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}\leq\theta_{i}^{*}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\frac{h_{i}}{h_{m+1}}.$
Now putting each estimated value in (4) provide:

(17) $\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}\leq c_{i}\max_{1\leq j\leq
2m}\left\|v_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)},\hskip 20.0pt1\leq i\leq m,$

where

(18)
$\begin{cases*}c_{1}=\theta_{1}^{*}\left(\sqrt{\frac{\kappa_{2}}{\kappa_{1}}}+2\sum_{j=2}^{m-1}\sqrt{\frac{\kappa_{j+1}}{\kappa_{1}}}+\sqrt{\frac{\kappa_{m+1}}{\kappa_{1}}}\frac{2h}{h_{m+1}}\right)\\\
c_{i}=\theta_{i}^{*}\left(1+\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}+2\sum_{j=i+1}^{m-1}\sqrt{\frac{\kappa_{j+1}}{\kappa_{i}}}+\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\frac{2h}{h_{m+1}}\right),\hskip
20.0pt2\leq i\leq m.\end{cases*}$

In a similar manner we obtain the estimate:

(19) $\left\|v_{m+i}^{(k)}\right\|_{L^{\infty}(0,T)}\leq c_{m+i}\max_{1\leq
j\leq 2m}\left\|v_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)},\hskip 20.0pt1\leq
i\leq m,$

where

(20)
$\begin{cases*}c_{2m+1-i}=\theta_{2m+1-i}^{*}\left(1+\sqrt{\frac{\kappa_{2m+1-i}}{\kappa_{2m+2-i}}}+2\sum_{j=i+1}^{m-1}\sqrt{\frac{\kappa_{2m+1-j}}{\kappa_{2m+2-i}}}+\sqrt{\frac{\kappa_{m+1}}{\kappa_{2m+2-i}}}\frac{2h}{h_{m+1}}\right),\hskip
5.0pt2\leq i\leq m,\\\
c_{2m}=\theta_{2m}^{*}\left(\sqrt{\frac{\kappa_{2m}}{\kappa_{2m+1}}}+2\sum_{j=2}^{m-1}\sqrt{\frac{\kappa_{2m+1-j}}{\kappa_{2m+1}}}+\sqrt{\frac{\kappa_{m+1}}{\kappa_{2m+1}}}\frac{2h}{h_{m+1}}\right).\end{cases*}$

Therefore combining (17) & (19) and define $c:=\max_{1\leq i\leq 2m}c_{i}$
deduce

$\displaystyle\max_{1\leq i\leq
2m}\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$ $\displaystyle\leq
c\max_{1\leq j\leq 2m}\left\|v_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)}.$

Further reduction in terms of the error on initial guesses give:

$\displaystyle\max_{1\leq i\leq
2m}\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$ $\displaystyle\leq
c^{k}\max_{1\leq j\leq 2m}\left\|v_{j}^{(0)}\right\|_{L^{\infty}(0,T)}$ (21)
$\displaystyle=c^{k}\max_{1\leq j\leq
2m}\left\|w_{j}^{(0)}\right\|_{L^{\infty}(0,T)}.$

Now the estimate of the error terms in $k^{th}$ iteration is obtained by

$\left\|w_{i}^{(k)}\right\|_{L^{\infty}(0,T)}\leq\left\|\mathcal{L}^{-1}\left\\{\frac{1}{\cosh^{k}(hs^{\nu})}\right\\}\right\|_{L^{1}(0,T)}\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}.$

Which using Lemma further reduce to

(22) $\left\|w_{i}^{(k)}\right\|_{L^{\infty}(0,T)}\leq
2^{k}\left\|\mathcal{L}^{-1}\left\\{\exp(-khs^{\nu})\right\\}\right\|_{L^{1}(0,T)}\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}.$

Finally taking the maximum over $i$ in (22) and plugging the values of
$\max_{i}\left\|v_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$ from (4) in (22) we
have:

$\displaystyle\max_{1\leq i\leq 2m}\|w_{i}^{(k)}\|_{L^{\infty}(0,T)}$
$\displaystyle\leq(2c)^{k}\exp(-Ak^{1/(1-\nu)})\max_{1\leq j\leq
2m}\|w_{j}^{(0)}\|_{L^{\infty}(0,T)}.$

∎

######  Lemma 6.

For $0<l_{1},l_{2},\alpha\in(0,1)$, the following inequality holds:

$\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-l_{1}s^{\alpha})}{1-\exp(-l_{2}s^{\alpha})}\right\\}\right\|_{L^{1}(0,t)}\leq\left[1+\frac{t^{\alpha}\Gamma(2-\alpha)}{l_{2}\Lambda^{(1-\alpha)}}\right]\exp\left(-\Lambda\left(\frac{l_{1}}{t^{\alpha}}\right)^{1/(1-\alpha)}\right),$

when $\Lambda=(1-\alpha)\alpha^{\alpha/(1-\alpha)}$.

######  Lemma 7.

Suppose $l_{1},l_{2},\cdots,l_{n}$ are positive real numbers with
$l=\min_{i}l_{i}$ and $1/2<\alpha<1$. Let $\mathcal{L}^{-1}\\{\hat{f}(s)\\}$
represents the inverse Laplace transform of function $f(t)$. Then we have the
following inequalities:

  1. (i)

$\displaystyle\|\mathcal{L}^{-1}\left\\{\frac{\exp(2ls^{\alpha})\cosh((l_{1}-l_{2})s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}$
$\displaystyle\leq
2\left[1+\frac{L(t)}{2l_{1}}\right]\left[1+\frac{L(t)}{2l_{2}}\right]\left[\exp\left(-2\Lambda\left(\frac{l_{1}-l}{t^{\nu}}\right)^{\beta}\right)+\exp\left(-2\Lambda\left(\frac{l_{2}-l}{t^{\nu}}\right)^{\beta}\right)\right],$

  2. (ii)

$\displaystyle\|\mathcal{L}^{-1}\left\\{\frac{\exp(2ls^{\alpha})}{\cosh(l_{1}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}\leq
2\left[1+\frac{L(t)}{2l_{1}}\right]\exp\left(-2\Lambda\left(\frac{l_{1}-2l}{t^{\nu}}\right)^{\beta}\right),$

  3. (iii)

$\displaystyle\|\mathcal{L}^{-1}\left\\{\frac{\exp(2ls^{\alpha})\sinh(l_{1}s^{\alpha})\sinh(l_{n}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}$
$\displaystyle\leq
2^{n-2}\left[1+\frac{L(t)}{2l_{2}}\right]\cdots\left[1+\frac{L(t)}{2l_{n-1}}\right]\exp\left(-(n-2)\Lambda\left(\frac{l_{2}+\cdots+l_{n-1}-2l}{(n-2)t^{\nu}}\right)^{\beta}\right)$
$\displaystyle\hskip
5.0pt\left[1+2\left[1+\frac{L(t)}{2l_{1}}\right]\left[1+\frac{L(t)}{2l_{n}}\right]\left[\exp\left(-2\Lambda\left(\frac{l_{1}-l}{t^{\nu}}\right)^{\beta}\right)+\exp\left(-2\Lambda\left(\frac{l_{n}-l}{t^{\nu}}\right)^{\beta}\right)\right]\right],$

  4. (iv)

$\displaystyle\|\mathcal{L}^{-1}\left\\{\frac{\exp(2ls^{\alpha})\sinh(l_{1}s^{\alpha})\cosh(l_{n}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}$
$\displaystyle\leq
2^{n-2}\left[1+\frac{L(t)}{2l_{2}}\right]\cdots\left[1+\frac{L(t)}{2l_{n-1}}\right]\exp\left(-(n-2)\Lambda\left(\frac{l_{2}+\cdots+l_{n-1}-2l}{(n-2)t^{\nu}}\right)^{\beta}\right)$
$\displaystyle\left[1+2\left[1+\frac{L(t)}{2l_{1}}\right]\left[1+\frac{L(t)}{2l_{n}}\right]\left[\exp\left(-2\Lambda\left(\frac{l_{1}-l}{t^{\nu}}\right)^{\beta}\right)+\exp\left(-2\Lambda\left(\frac{l_{n}-l}{t^{\nu}}\right)^{\beta}\right)\right]\right],$

  5. (v)

$\displaystyle\|\mathcal{L}^{-1}\left\\{\frac{\exp(2ls^{\alpha})\sinh(l_{1}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})}\right\\}\|_{L^{1}(0,t)}$
$\displaystyle\leq
2^{n}\left[1+\frac{L(t)}{2l_{1}}\right]\cdots\left[1+\frac{L(t)}{2l_{n}}\right]\exp\left(-n\Lambda\left(\frac{l_{2}+\cdots+l_{n}-2l}{nt^{\nu}}\right)^{\beta}\right).$

Where
$L(t)=t^{\alpha}\Gamma(2-\alpha)/\Lambda^{(1-\alpha)},\beta=1/(1-\alpha)$.

###### Proof.

All five estimates follow the same procedure. But according to the
combinations of the hyperbolic functions the results may vary, so we
particularly show each decomposition upto some known level. To avoid
inconvenience we only use $\|.\|$ instead of $\|.\|_{L^{1}(0,t)}$.

  1. (i)

Let

$\displaystyle\hat{Q}_{1}(s):=\frac{\exp(2ls^{\alpha})\cosh((l_{1}-l_{2})s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})}=\frac{\exp(-2(l_{1}-l)s^{\alpha})+\exp(-2(l_{2}-l)s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))(1+\exp(-2l_{2}s^{\alpha}))}.$

Therefore taking $L^{1}$ norm and using the triangular inequality give:

$\displaystyle\|\mathcal{L}^{-1}\\{\hat{Q}_{1}(s)\\}\|$ (23)
$\displaystyle\leq
2\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-2(l_{1}-l)s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))(1+\exp(-2l_{2}s^{\alpha}))}\right\\}\right\|$
$\displaystyle\hskip
50.0pt+2\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-2(l_{2}-l)s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))(1+\exp(-2l_{2}s^{\alpha}))}\right\\}\right\|.$

Now using the part (i) of Lemma 1 on each term of ((i)) we have:

$\displaystyle\|\mathcal{L}^{-1}\\{\hat{Q}_{1}(s)\\}\|$ $\displaystyle\leq
2\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-(l_{1}-l)s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))}\right\\}\right\|\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-(l_{1}-l)s^{\alpha})}{(1+\exp(-2l_{2}s^{\alpha}))}\right\\}\right\|$
$\displaystyle\hskip
50.0pt+2\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-(l_{2}-l)s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))}\right\\}\right\|\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-(l_{2}-l)s^{\alpha})}{(1+\exp(-2l_{2}s^{\alpha}))}\right\\}\right\|.$

Finally applying Lemma 6 we have our result.

  2. (ii)

The proof is obvious using part (i), hence omitted.

  3. (iii)

We define

$\displaystyle\hat{Q}_{2}(s):=\frac{\exp(2ls^{\alpha})\sinh(l_{1}s^{\alpha})\sinh(l_{n}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n}s^{\alpha})}$
$\displaystyle=-\frac{\exp(2ls^{\alpha})}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})}\left(1-\frac{\sinh(l_{1}s^{\alpha})\sinh(l_{n}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{n}s^{\alpha})}-1\right).$

Hence

$\displaystyle\left\|\mathcal{L}^{-1}\\{\hat{Q}_{2}(s)\\}\right\|$
$\displaystyle\leq\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(2ls^{\alpha})}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})}\right\\}\right\|$
$\displaystyle\hskip
30.0pt+\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(2ls^{\alpha})\cosh((l_{1}-l_{n})s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n}s^{\alpha})}\right\\}\right\|=:I_{1}+I_{2}.$

Now we consider $I_{1}$ and $I_{2}$ separately to obtain the estimates using
the part (i) of Lemma 1:

$\displaystyle I_{1}$
$\displaystyle=2^{n-2}\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-(l_{2}+\cdots+l_{n-1}-2l)s^{\alpha})}{(1+\exp(-2l_{2}s^{\alpha}))\cdots(1+\exp(-2l_{n-1}s^{\alpha}))}\right\\}\right\|$
$\displaystyle\leq
2^{n-2}\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n-1}-2l}{n-2}s^{\alpha})}{(1+\exp(-2l_{2}s^{\alpha}))}\right\\}\right\|\cdots\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n-1}-2l}{n-2}s^{\alpha})}{(1+\exp(-2l_{n-1}s^{\alpha}))}\right\\}\right\|:=J_{1}.$

And we have

$\displaystyle I_{2}$ $\displaystyle\leq
2J_{1}\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-2(l_{1}-l)s^{\alpha})+\exp(-2(l_{n}-l)s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))(1+\exp(-2l_{n}s^{\alpha}))}\right\\}\right\|.$

Finally applying Lemma 6 on both $I_{1}$ and $I_{2}$, we have our result.

  4. (iv)

The procedure of the proof is similar to part (iii), hence omitted.

  5. (v)

Define

$\displaystyle\hat{Q}_{3}(s)$
$\displaystyle:=\frac{\exp(2ls^{\alpha})\sinh(l_{1}s^{\alpha})}{\cosh(l_{1}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})}$
$\displaystyle=\frac{\exp(2ls^{\alpha})}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})(1+\exp(-2l_{1}s^{\alpha}))}$
$\displaystyle\hskip
50.0pt-\frac{\exp(-2(l_{1}-l)s^{\alpha})}{\cosh(l_{2}s^{\alpha})\cdots\cosh(l_{n-1}s^{\alpha})\sinh(l_{n}s^{\alpha})(1+\exp(-2l_{1}s^{\alpha}))}$
$\displaystyle=:\hat{I}_{3}-\hat{I}_{4},$

Separately considering $\hat{I}_{3}$ and $\hat{I}_{4}$ we have:

$\displaystyle\|\mathcal{L}^{-1}\\{\hat{I}_{3}\\}\|$ $\displaystyle\leq
2^{n-1}\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-(l_{2}+\cdots+l_{n}-2l)s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))\cdots(1+\exp(-2l_{n-1}s^{\alpha}))(1-\exp(-2l_{n}s^{\alpha}))}\right\\}\right\|$
$\displaystyle\leq
2^{n-1}\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n}-2l}{n}s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))}\right\\}\right\|\cdots\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n}-2l}{n}s^{\alpha})}{(1+\exp(-2l_{n-1}s^{\alpha}))}\right\\}\right\|$
$\displaystyle\hskip
205.0pt\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n}-2l}{n}s^{\alpha})}{(1-\exp(-2l_{n}s^{\alpha}))}\right\\}\right\|,$

and

$\displaystyle\|\mathcal{L}^{-1}\\{\hat{I}_{4}\\}\|$ $\displaystyle\leq
2^{n-1}\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n}+2l_{1}-2l}{n}s^{\alpha})}{(1+\exp(-2l_{1}s^{\alpha}))}\right\\}\right\|\cdots\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n}+2l_{1}-2l}{n}s^{\alpha})}{(1+\exp(-2l_{n-1}s^{\alpha}))}\right\\}\right\|$
$\displaystyle\hskip
190.0pt\left\|\mathcal{L}^{-1}\left\\{\frac{\exp(-\frac{l_{2}+\cdots+l_{n}+2l_{1}-2l}{n}s^{\alpha})}{(1-\exp(-2l_{n}s^{\alpha}))}\right\\}\right\|.$

Finally applying Lemma 6 on both $\hat{I}_{3}$ and $\hat{I}_{4}$, we have our
result.

∎

######  Theorem 8 (Convergence of DNWR for $1/2<\nu<1$).

For fixed $T>0$, $0<\nu\leq 1/2$ and
$\theta_{i}=1/(1+\sqrt{\kappa_{i+1}/\kappa_{i}}),1\leq i\leq m$ and
$\theta_{i}=1/(1+\sqrt{\kappa_{i}/\kappa_{i+1}}),m+1\leq i\leq 2m$ , the DNWR
algorithm (7)-(10) for $2m+1(m\in\mathbb{N})$ subdomains converge super-
linearly with the estimate:

$\displaystyle\max_{1\leq i\leq 2m}\|w_{i}^{(k)}\|_{L^{\infty}(0,T)}\leq
d^{k}\exp(-Ak^{1/(1-\nu)})\max_{1\leq j\leq
2m}\|w_{j}^{(0)}\|_{L^{\infty}(0,T)},$

where, $h_{j}=a_{j}/\sqrt{\kappa_{j}}$, $h=\min_{1\leq j\leq 2m+1}h_{j}/2$,
$A=(1-\nu)\nu^{\nu/(1-\nu)}(2h/T^{\nu})^{1/(1-\nu)}$, $d=\|D\|_{\infty}$, and
$\|.\|_{\infty}$ imply the infinity norm of matrix $D$, given in (25).

###### Proof.

The idea to proceed the proof for diffusion-wave case is similar to the sub-
diffusion one but the estimation technique is different due to the lack of
positivity results, which is one of the key ingredients in sub-diffusion. The
recursive equation in terms of $k^{th}$ and $(k-1)^{th}$ iteration on the
error of artificial boundary takes the form from (11):

$\hat{w}_{i}^{(k)}(s)=\sum_{j=1}^{2m}\hat{\rho}_{i,j}\hat{w}_{j}^{(k-1)}(s),\quad
1\leq i,j\leq 2m.$

Define $\zeta:=\exp(2hs^{\nu})$ and
$\hat{\chi}^{(k)}(s):=\zeta^{k}\hat{w}^{(k)}(s)$. Therefore, the error
equation converts to:

$\hat{\chi}_{i}^{(k)}(s)=\sum_{j=1}^{2m}\zeta\hat{\rho}_{i,j}\hat{\chi}_{j}^{(k-1)}(s),\quad
1\leq i,j\leq 2m.$

Taking the inverse Laplace transform on both side and then sup-norm on time
reduce the equation:

$\displaystyle\left\|\chi_{i}^{(k)}\right\|_{L^{\infty}(0,T)}$
$\displaystyle\leq\sum_{j=1}^{2m}\left\|\mathcal{L}^{-1}\\{\zeta\hat{\rho}_{i,j}\\}\ast\chi_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)}.$

Using part (ii) of Lemma 1 and taking maximum over $j$ give us:

(24)
$\displaystyle\left\|\chi_{i}^{(k)}\right\|_{L^{\infty}(0,T)}\leq\max_{1\leq
j\leq
2m}\left\|\chi_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)}\sum_{j=1}^{2m}\left\|\mathcal{L}^{-1}\\{\zeta\hat{\rho}_{i,j}\\}\right\|_{L^{1}(0,T)}.$

Now taking the $L^{1}$ norm on each term on
$\mathcal{L}^{-1}\\{\zeta\hat{\rho}_{i,j}\\}$ give us the estimate matrix
$D=d_{i,j}$, which follow directly from Lemma 7. Therefore for $1\leq i\leq
m,1\leq j\leq 2m$

(25) $\displaystyle\bar{d}_{i,i}$
$\displaystyle=2\theta^{*}_{i}\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}\left[1+\frac{L_{T}}{2h_{i}}\right]\left[1+\frac{L_{T}}{2h_{i+1}}\right]\left[\exp\left(-2B\left(\frac{h_{i}-h}{T^{\nu}}\right)^{\nu_{1}}\right)\right.$
$\displaystyle\hskip
225.0pt\left.\kern-1.2pt+\exp\left(-2B\left(\frac{h_{i+1}-h}{T^{\nu}}\right)^{\nu_{1}}\right)\right]$
$\displaystyle\leq
4\theta^{*}_{i}\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}}\left[1+\frac{L_{T}}{4h}\right]^{2}\exp\left(-2B\left(\frac{h}{T^{\nu}}\right)^{\nu_{1}}\right):=d_{i,j},$
$\displaystyle\bar{d}_{i,j}$
$\displaystyle=2\theta^{*}_{i}\left[1+\frac{L_{T}}{2h_{i}}\right]\exp\left(-2B\left(\frac{h_{i}-2h}{T^{\nu}}\right)^{\nu_{1}}\right)\leq
2\theta^{*}_{i}\left[1+\frac{L_{T}}{4h}\right]:=d_{i,j},\hskip 10.0pt\mbox{for
$j=i-1,i>1$}$

for $i+1\leq j\leq m$

$\displaystyle\bar{d}_{i,j}$
$\displaystyle=2^{j-i}\theta^{*}_{i}\sqrt{\frac{\kappa_{j+1}}{\kappa_{i}}}\left[1+\frac{L_{T}}{2h_{i+1}}\right]\cdots\left[1+\frac{L_{T}}{2h_{j}}\right]\exp\left(-(j-i)B\left(\frac{h_{i+1}+\cdots+h_{j}-2h}{(j-i)T^{\nu}}\right)^{\nu_{1}}\right)$
$\displaystyle\left[1+2\left[1+\frac{L_{T}}{2h_{i}}\right]\left[1+\frac{L_{T}}{2h_{j+1}}\right]\left[\exp\left(-2B\left(\frac{h_{i}-h}{T^{\nu}}\right)^{\nu_{1}}\right)+\exp\left(-2B\left(\frac{h_{j+1}-h}{T^{\nu}}\right)^{\nu_{1}}\right)\right]\right]$
$\displaystyle\leq
2^{j-i}\theta^{*}_{i}\sqrt{\frac{\kappa_{j+1}}{\kappa_{i}}}\left[1+\frac{L_{T}}{4h}\right]^{j-i}\exp\left(-(j-i)B\left(\frac{2(j-i-1)h}{(j-i)T^{\nu}}\right)^{\nu_{1}}\right)$
$\displaystyle\hskip
160.0pt\left[1+4\left[1+\frac{L_{T}}{4h}\right]^{2}\exp\left(-2B\left(\frac{h}{T^{\nu}}\right)^{\nu_{1}}\right)\right]:=d_{i,j},$
$\displaystyle\bar{d}_{i,m+1}$
$\displaystyle=2^{m-i+2}\theta^{*}_{i}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\left[1+\frac{L_{T}}{2h_{i}}\right]\cdots\left[1+\frac{L_{T}}{2h_{m+1}}\right]$
$\displaystyle\hskip
130.0pt\exp\left(-(m-i+2)B\left(\frac{h_{i+1}+\cdots+h_{m+1}-2h}{(m-i+2)T^{\nu}}\right)^{\nu_{1}}\right)$
$\displaystyle\leq
2^{m-i+2}\theta^{*}_{i}\sqrt{\frac{\kappa_{m+1}}{\kappa_{i}}}\left[1+\frac{L_{T}}{4h}\right]^{m-i+2}\exp\left(-(m-i+2)B\left(\frac{2(m-i+1)h}{(m-i+2)T^{\nu}}\right)^{\nu_{1}}\right):=d_{i,j},$
$\displaystyle d_{i,j}=0\hskip 290.0pt\mbox{otherwise},$

where
$B=(1-\nu)\nu^{\nu/(1-\nu)},L_{T}=T^{\nu}\Gamma(2-\nu)/B^{(1-\nu)},\nu_{1}=1/(1-\nu)$.
Again the symmetric structure helps us to find $d_{i,j}$ for $m+1\leq i\leq
2m$ for all $j$ and we achieve it by replacing all indices $i_{1}$ by
$2m+2-i_{1}$, except $d_{i,j},\theta^{*}_{i}$, where $i_{1}$ is replaced by
$2m+1-i_{1}$. Now we define infinity norm on matrix $D$ by
$d:=\|D\|_{\infty}$. Therefore taking maximum over $i$ in (24) give:

(26) $\displaystyle\max_{1\leq i\leq
2m}\left\|\chi_{i}^{(k)}\right\|_{L^{\infty}(0,T)}\leq d\max_{1\leq j\leq
2m}\left\|\chi_{j}^{(k-1)}\right\|_{L^{\infty}(0,T)}.$

Now using the same reduction technique from $\chi_{i}$ to $w_{i}$ as sub-
diffusion on (26) we end up with the estimate:

$\displaystyle\max_{1\leq i\leq 2m}\|w_{i}^{(k)}\|_{L^{\infty}(0,T)}\leq
d^{k}\exp(-Ak^{1/(1-\nu)})\max_{1\leq j\leq
2m}\|w_{j}^{(0)}\|_{L^{\infty}(0,T)}.$

Hence the theorem. ∎

######  Remark 4.1.

The estimates for even number of subdomains that is $2m,m\in\mathbb{N}$ is
almost same as the odd case. Here first we construct recursive error equation
(11) for $1\leq i\leq m-1$ and then for $m\leq i\leq 2m-1$. Except $m=1$, one
can directly construct $\hat{\rho}_{i,j}$ matrix from general formula (12).

##  5 Convergence of DNWR Algorithm in 2D

In $2D$, we consider the homogeneous variant, that is, diffusion coefficient
$\kappa$ same throughout the domain, of our model problem (1) in rectangular
spatial domain $\Omega:=(0,L)\times(0,\pi)$. Zero Dirichlet boundary condition
on $\partial\Omega$ is imposed, and subdivide $\Omega$ into
$2m+1,m\in\mathbb{N}$ rectangular strips; size
$\Omega_{i}:=(x_{i-1},x_{i})\times(0,\pi),1\leq i\leq 2m+1$. If we take
$\psi_{i}(x,y,t)$ be the initial guess on artificial boundary $x=x_{i}$, then
the error update on artificial boundary takes the form:

(27)
$\hat{W}_{i}^{(k)}(s,n)=\sum_{j=1}^{2m}\hat{\rho}_{i,j}\left(\sqrt{s^{2\nu}+kn^{2}}\right)\hat{W}_{j}^{(k-1)}(s,n),\quad
1\leq i,j\leq 2m.$

Here, $\hat{W}_{i}^{(k)}(s,n)$ represent first Fourier sine transform on $y$
axis and then Laplace transform on time and $\hat{\rho}_{i,j}$ takes the same
form as in (12) but definitions of $\sigma_{i},\gamma_{i}$ change to
$\sigma_{i}:=\sinh(\frac{a_{i}}{\sqrt{\kappa_{i}}}\sqrt{s^{2\nu}+\kappa
n^{2}}),\gamma_{i}:=\cosh(\frac{a_{i}}{\sqrt{\kappa_{i}}}\sqrt{s^{2\nu}+\kappa
n^{2}})$. Now we need some lemmas on estimates for further progression.

######  Lemma 9.

Let $g\in L^{\infty}(0,\pi)$, and $\tau>0$. Then for any $y\in\mathbb{R}$,

$\frac{2}{\pi}\left|\int_{0}^{\pi}g(\eta)\sum_{n\geq
1}\exp(-n^{2}\eta)\sin(n\eta)\sin(ny)d\eta\right|\leq\|g\|_{L^{\infty}(0,\pi)}.$

######  Lemma 10.

Let $K(n)$ represents the Fourier coefficient of $k(y)$ and

$\hat{K_{1}}(s,n)=\frac{\exp(-l_{1}{\sqrt{s^{2\alpha}+\lambda
n^{2}}})}{1+\exp(-l_{2}{\sqrt{s^{2\alpha}+\lambda n^{2}}})}\hat{K_{2}}(s,n)$

then for $l_{1}<l_{2},\lambda>0$ and $0<\alpha\leq 1/2$

$\|k_{1}(.,.)\|_{L^{\infty}(0,t;L^{\infty}(0,\pi))}\leq\left[1+\frac{t^{\alpha}\Gamma(2-\alpha)}{l_{2}\Lambda^{(1-\alpha)}}\right]\exp\left(-\Lambda\left(\frac{l_{1}}{t^{\alpha}}\right)^{1/(1-\alpha)}\right)\|k_{2}(.,.)\|_{L^{\infty}(0,t;L^{\infty}(0,\pi))},$

where $\Lambda=(1-\alpha)\alpha^{\alpha/(1-\alpha)}$.

###### Proof.

Our procedure to prove this lemma is straightforward; we first take the
inverse Laplace transform and then add all the sine frequencies to achieve our
estimate. Therefore first take the binomial expansion and then inverse Laplace
transform we have:

$\displaystyle
K_{1}(t,n)=\mathcal{L}^{-1}\left\\{\sum_{j=0}^{\infty}(-1)^{j}\exp(-(l_{1}+jl_{2})\sqrt{s^{2\alpha}+\lambda
n^{2}})\right\\}\ast K_{2}(t,n),$

Using Efros theorem and part (i) of Lemma 2 on each exponential term give:

$\displaystyle
K_{1}(t,n)=\left[\sum_{j=0}^{\infty}(-1)^{n}\int_{0}^{\infty}\exp(-\lambda
n^{2}\tau)F(\tau,t)d\tau\right]\ast K_{2}(t,n),$

where $F(\tau,t):=2\alpha\tau t^{-(2\alpha+1)}M_{2\alpha}(\tau
t^{-2\alpha})\frac{l_{1}+jl_{2}}{\sqrt{4\pi
t^{3}}}\exp(\frac{-(l_{1}+jl_{2})^{2}}{4\tau})$. Now adding all the Fourier
sine terms with coefficients provide:

$\displaystyle k_{1}(t,y)$ $\displaystyle=\sum_{n\geq 1}K_{1}(t,n)\sin(ny)$
$\displaystyle=\sum_{n\geq
1}\int_{0}^{t}d\tau_{1}\sum_{j=0}^{\infty}(-1)^{n}\int_{0}^{\infty}d\tau\exp(-\lambda
n^{2}\tau)F(\tau,\tau_{1})K_{2}(t-\tau_{1},n)\sin(ny)$
$\displaystyle=\sum_{n\geq
1}\int_{0}^{t}d\tau_{1}\sum_{j=0}^{\infty}(-1)^{n}\int_{0}^{\infty}d\tau\exp(-\lambda
n^{2}\tau)F(\tau,\tau_{1})\left[\frac{2}{\pi}\int_{0}^{\pi}d\eta
k_{2}(t-\tau_{1},\eta)\sin(n\eta)\right]\sin(ny)$
$\displaystyle=\int_{0}^{t}d\tau_{1}\sum_{j=0}^{\infty}(-1)^{n}\int_{0}^{\infty}d\tau
F(\tau,\tau_{1})\left[\frac{2}{\pi}\int_{0}^{\pi}d\eta
k_{2}(t-\tau_{1},\eta)\sum_{n\geq 1}\exp(-\lambda
n^{2}\tau)\sin(n\eta)\sin(ny)\right].$

We have applied Fubini’s theorem to exchange integration and summation.
Further taking absolute value on both sides and apply Lemma 9 give:

$\displaystyle|k_{1}(t,y)|$
$\displaystyle\leq\int_{0}^{t}d\tau_{1}\sum_{j=0}^{\infty}\int_{0}^{\infty}d\tau
F(\tau,\tau_{1})\left|\frac{2}{\pi}\int_{0}^{\pi}d\eta
k_{2}(t-\tau_{1},\eta)\sum_{n\geq 1}\exp(-\lambda
n^{2}\tau)\sin(n\eta)\sin(ny)\right|$
$\displaystyle\leq\int_{0}^{t}d\tau_{1}\sum_{j=0}^{\infty}\int_{0}^{\infty}d\tau
F(\tau,\tau_{1})\left\|k_{2}(t-\tau_{1},.)\right\|_{L^{\infty}(0,\pi)}$
$\displaystyle\leq\left\|k_{2}(.,.)\right\|_{L^{\infty}(0,t;L^{\infty}(0,\pi))}\int_{0}^{t}d\tau_{1}\sum_{j=0}^{\infty}\int_{0}^{\infty}d\tau
F(\tau,\tau_{1}).$

Again using Efros theorem and Lemma 6, we have our proof:

$\displaystyle\|k_{1}(.,.)\|_{L^{\infty}(0,t;L^{\infty}(0,\pi))}\leq\left[1+\frac{t^{\alpha}\Gamma(2-\alpha)}{l_{2}\Lambda^{(1-\alpha)}}\right]\exp\left(-\Lambda\left(\frac{l_{1}}{t^{\alpha}}\right)^{1/(1-\alpha)}\right)\|k_{2}(.,.)\|_{L^{\infty}(0,t;L^{\infty}(0,\pi))}.$

∎

######  Theorem 11 (Convergence of DNWR for sub-diffusion in $2D$ ).

For fixed $T>0$, $0<\nu\leq 1/2$ and
$\theta_{i}=1/(1+\sqrt{\kappa_{i+1}/\kappa_{i}}),1\leq i\leq m$ and
$\theta_{i}=1/(1+\sqrt{\kappa_{i}/\kappa_{i+1}}),m+1\leq i\leq 2m$ , the DNWR
algorithm in $2D$ for the update error equation (27) for
$2m+1(m\in\mathbb{N})$ subdomains converge super-linearly with the estimate:

$\displaystyle\max_{1\leq i\leq
2m}\|w_{i}^{(k)}\|_{L^{\infty}(0,T;L^{\infty}(0,\pi))}\leq
d^{k}\exp(-Ak^{1/(1-\nu)})\max_{1\leq j\leq
2m}\|w_{j}^{(0)}\|_{L^{\infty}(0,T;L^{\infty}(0,\pi))},$

where, $h_{j}=a_{j}/\sqrt{\kappa_{j}}$, $h=\min_{1\leq j\leq 2m+1}h_{j}/2$,
$A=(1-\nu)\nu^{\nu/(1-\nu)}(2h/T)^{1/(1-\nu)}$, $d=\|D\|_{\infty}$, and
$\|.\|_{\infty}$ imply the infinity norm of matrix $D$, given in (25).

###### Proof.

As one can see that the estimate for diffusion-wave case for $1D$ and sub-
diffusion case for $2D$ is exactly same, so we only go through the proof
briefly and only mention those terms which are different from $1D$ case.

Define $\xi:=\exp(2h\sqrt{s^{2\nu}+kn^{2}})$, where $h=\min_{j}h_{j}/2$.
Setting $\hat{V}^{(k)}_{i}(s,n):=\xi^{k}\hat{W}^{(k)}_{i}(s,n)$ transform the
error update equation (27) to:

(28)
$\hat{V}_{i}^{(k)}(s,n)=\sum_{j=1}^{n}\xi\hat{\rho}_{i,j}(\sqrt{s^{2\nu}+kn^{2}})\hat{V}_{j}^{(k-1)}(s,n).$

Now first simplify the cosine and sine term to exponential as given in the
proof of Lemma 7 and then repeatedly applying Lemma 10 on the simplify terms,
we obtain:

$\displaystyle\left\|v_{i}^{(k)}(.,.)\right\|_{L^{\infty}(0,T;L^{\infty}(0,\pi))}$
$\displaystyle\leq d\max_{1\leq j\leq
2m}\left\|v_{j}^{(k-1)}(.,.)\right\|_{L^{\infty}(0,T;L^{\infty}(0,\pi))}$ (29)
$\displaystyle\leq d^{k}\max_{1\leq j\leq
2m}\left\|w_{j}^{(0)}(.,.)\right\|_{L^{\infty}(0,T;L^{\infty}(0,\pi))},$

where $d=\|D\|_{\infty}$. Now again applying the Lemma 10 letting
$l_{2}\to\infty$ on $\hat{W}^{(k)}_{i}(s,n):=\xi^{-k}\hat{V}^{(k)}_{i}(s,n)$
give:

(30)
$\|w_{i}(.,.)\|_{L^{\infty}(0,t;L^{\infty}(0,\pi))}\leq\exp\left(-Ak^{1/(1-\nu)}\right)\|v_{i}(.,.)\|_{L^{\infty}(0,t;L^{\infty}(0,\pi))}.$

Finally combining (5) & (30) and taking maximum over $i$ we have our estimate:

$\displaystyle\max_{1\leq i\leq
2m}\|w_{i}^{(k)}\|_{L^{\infty}(0,T;L^{\infty}(0,\pi))}\leq
d^{k}\exp(-Ak^{1/(1-\nu)})\max_{1\leq j\leq
2m}\|w_{j}^{(0)}\|_{L^{\infty}(0,T;L^{\infty}(0,\pi))}.$

∎

######  Remark 5.1.

We keep the $2D$ bounded multi-subdomain diffusion-wave proof for future work
due to some estimation issue. If we take $y$-axis unbounded then the two sub-
domain proof can be easily done. We leave that one because the procedure is
same as the two sub-domain NNWR diffusion-wave proof, given in.

##  Appendix A Necessary calculation for DNWR

To simplify the calculation we assign:

(31) $\displaystyle\hat{z}_{\mu}^{(k)}(s)$
$\displaystyle=\partial_{x}\hat{u}_{\mu+1}^{(k)}(x_{\mu},s)\textit{ for
}1\leq{\mu}\leq m,$ (32) $\displaystyle\hat{z}_{\iota}^{(k)}(s)$
$\displaystyle=-\partial_{x}\hat{u}_{\iota}^{(k)}(x_{\iota},s)\textit{ for
}m+1\leq{\iota}\leq 2m.$

Solving the BVP in Dirichlet and Neumann step in (7)-(8) we get

(33) $\displaystyle\hat{u}_{m+1}^{(k)}(x,s)$
$\displaystyle=\frac{1}{\sigma_{m+1}}\left(\hat{w}_{m+1}^{(k-1)}\sinh(x-x_{m})\sqrt{s^{2\nu}/\kappa_{m+1}}+\hat{w}_{m}^{(k-1)}\sinh(x_{m+1}-x)\sqrt{s^{2\nu}/\kappa_{m+1}}\right),$
for $1\leq\mu\leq m$ (34) $\displaystyle\hat{u}_{\mu}^{(k)}(x,s)$
$\displaystyle=\frac{1}{\gamma_{\mu}}\left(\frac{\kappa_{\mu+1}\hat{z}_{\mu}^{(k)}}{\sqrt{\kappa_{\mu}s^{2\nu}}}\sinh(x-x_{\mu-1})\sqrt{s^{2\nu}/\kappa_{\mu}}+\hat{w}_{\mu-1}^{(k-1)}\cosh(x_{\mu}-x)\sqrt{s^{2\nu}/\kappa_{\mu}}\right),$
for $m+2\leq{\iota}\leq 2m+1$ (35) $\displaystyle\hat{u}_{\iota}^{(k)}(x,s)$
$\displaystyle=\frac{1}{\gamma_{\iota}}\left(\hat{w}_{\iota}^{(k-1)}\cosh(x-x_{\iota-1})\sqrt{s^{2\nu}/\kappa_{\iota}}+\frac{\kappa_{\iota-1}\hat{z}_{\iota-1}^{(k)}}{\sqrt{\kappa_{\iota}s^{2\nu}}}\sinh(x_{\iota}-x)\sqrt{s^{2\nu}/\kappa_{\iota}}\right),$

Substituting (34) into (9) and (35) into (10) and using the notation
$\sigma_{\mu}:=\sinh(\frac{a_{\mu}}{\sqrt{\kappa_{\mu}}}s^{\nu})$,
$\gamma_{\mu}:=\cosh(\frac{a_{\mu}}{\sqrt{\kappa_{\mu}}}s^{\nu})$ and define
$\sigma_{\iota},\gamma_{\iota}$ same way, the recurrence relation for
$k\in\mathbb{N}$ becomes:

(36) $\displaystyle\hat{w}_{\mu}^{(k)}(s)$
$\displaystyle=\frac{\theta_{\mu}}{\gamma_{\mu}}\left(\frac{\kappa_{\mu+1}\hat{z}_{\mu}^{(k)}}{\sqrt{\kappa_{\mu}s^{2\nu}}}\sigma_{\mu}+\hat{w}_{\mu-1}^{(k-1)}\right)+(1-\theta_{\mu})\hat{w}_{\mu}^{(k-1)},$
(37) $\displaystyle\hat{w}_{\iota}^{(k)}(s)$
$\displaystyle=\frac{\theta_{\iota}}{\gamma_{{\iota}+1}}\left(\hat{w}_{{\iota}+1}^{(k-1)}+\frac{\kappa_{\iota}\hat{z}_{{\iota}+1}^{(k)}}{\sqrt{\kappa_{{\iota}+1}s^{2\nu}}}\sigma_{{\iota}+1}\right)+(1-\theta_{\iota})\hat{w}_{\iota}^{(k-1)},$

Where for $1\leq\mu\leq m-1$

(38) $\displaystyle\hat{z}_{\mu}^{(k)}$
$\displaystyle=-\sqrt{\frac{s^{2\nu}}{\kappa_{\mu+1}}}\frac{\sigma_{\mu+1}}{\gamma_{\mu+1}}\hat{w}_{\mu}^{(k-1)}+\frac{\kappa_{\mu+2}}{\kappa_{\mu+1}\gamma_{\mu+1}}\hat{z}_{\mu+1}^{(k)},$

for $m+2\leq{\iota}\leq 2m+1$

(39) $\displaystyle\hat{z}_{\iota}^{(k)}$
$\displaystyle=\frac{\kappa_{{\iota}-1}}{\kappa_{\iota}\gamma_{\iota}}\hat{z}_{{\iota}-1}^{(k)}-\sqrt{\frac{s^{2\nu}}{\kappa_{\iota}}}\frac{\sigma_{\iota}}{\gamma_{\iota}}\hat{w}_{\iota}^{(k-1)},$

and

(40) $\displaystyle\hat{z}_{m}^{(k)}$
$\displaystyle=-\sqrt{\frac{s^{2\nu}}{\kappa_{m+1}}}\frac{\gamma_{m+1}}{\sigma_{m+1}}\hat{w}_{m}^{(k-1)}+\sqrt{\frac{s^{2\nu}}{\kappa_{m+1}}}\frac{1}{\sigma_{m+1}}\hat{w}_{m+1}^{(k-1)},$
(41) $\displaystyle\hat{z}_{m+1}^{(k)}$
$\displaystyle=\sqrt{\frac{s^{2\nu}}{\kappa_{m+1}}}\frac{1}{\sigma_{m+1}}\hat{w}_{m}^{(k-1)}-\sqrt{\frac{s^{2\nu}}{\kappa_{m+1}}}\frac{\gamma_{m+1}}{\sigma_{m+1}}\hat{w}_{m+1}^{(k-1)}.$

Choose: for $1\leq\mu\leq m$,
$\bar{w}_{\mu}^{(k)}=\gamma_{\mu}\hat{w}_{\mu}^{k},\bar{z}_{\mu}^{(k)}=\frac{\sigma_{\mu}}{\sqrt{s^{2\nu}/\kappa_{\mu}}}\hat{z}_{\mu}^{k}$
and for $m+1\leq{\iota}\leq 2m+1$,
$\bar{w}_{\iota}^{(k)}=\gamma_{{\iota}+1}\hat{w}_{\iota}^{k},\bar{z}_{\iota}^{(k)}=\frac{\sigma_{{\iota}+1}}{\sqrt{s^{2\nu}/\kappa_{{\iota}+1}}}\hat{z}_{\iota}^{k}$.
Obtain:

for $1\leq\mu\leq m$ (43) $\displaystyle\bar{w}_{\mu}^{(k)}$
$\displaystyle=\frac{\theta_{\mu}}{\gamma_{\mu-1}}\bar{w}_{\mu-1}^{(k-1)}+(1-\theta_{\mu})\bar{w}_{\mu}^{(k-1)}+\theta_{\mu}\frac{\kappa_{\mu+1}}{\kappa_{\mu}}\bar{z}_{\mu}^{(k)},$
for $m+1\leq{\iota}\leq 2m$ (44) $\displaystyle\bar{w}_{\iota}^{(k)}$
$\displaystyle=\frac{\theta_{\iota}}{\gamma_{{\iota}+2}}\bar{w}_{{\iota}+1}^{(k-1)}+(1-\theta_{\iota})\bar{w}_{\iota}^{(k-1)}+\theta_{\iota}\frac{\kappa_{\iota}}{\kappa_{{\iota}+1}}\bar{z}_{\iota}^{(k)},$

and

for $1\leq\mu\leq m-1$ (46)
$\displaystyle\sqrt{\frac{\kappa_{\mu+1}}{\kappa_{\mu}}}\bar{z}_{\mu}^{(k)}$
$\displaystyle=-\frac{\sigma_{\mu}\sigma_{\mu+1}}{\gamma_{\mu}\gamma_{\mu+1}}\bar{w}_{\mu}^{(k-1)}+\frac{\kappa_{\mu+2}}{\kappa_{\mu+1}}\frac{\sigma_{\mu}}{\sigma_{\mu+1}\gamma_{\mu+1}}\bar{z}_{\mu+1}^{(k)},$
(47) $\displaystyle\sqrt{\frac{\kappa_{m+1}}{\kappa_{m}}}\bar{z}_{m}^{(k)}$
$\displaystyle=-\frac{\sigma_{m}\gamma_{m+1}}{\gamma_{m}\sigma_{m+1}}\bar{w}_{m}^{(k-1)}+\frac{\sigma_{m}}{\sigma_{m+1}\gamma_{m+1}}\bar{w}_{m+1}^{(k-1)},$
(48)
$\displaystyle\sqrt{\frac{\kappa_{m+1}}{\kappa_{m+2}}}\bar{z}_{m+1}^{(k)}$
$\displaystyle=\frac{\sigma_{m+2}}{\gamma_{m}\sigma_{m+1}}\bar{w}_{m}^{(k-1)}-\frac{\gamma_{m+1}\sigma_{m+2}}{\sigma_{m+1}\gamma_{m+2}}\bar{w}_{m+1}^{(k-1)},$
for $m+2\leq{\iota}\leq 2m$ (49)
$\displaystyle\sqrt{\frac{\kappa_{\iota}}{\kappa_{{\iota}+1}}}\bar{z}_{\iota}^{(k)}$
$\displaystyle=\frac{\kappa_{{\iota}-1}}{\kappa_{\iota}}\frac{\sigma_{{\iota}+1}}{\sigma_{\iota}\gamma_{\iota}}\bar{z}_{{\iota}-1}^{(k)}-\frac{\sigma_{\iota}\sigma_{{\iota}+1}}{\gamma_{\iota}\gamma_{{\iota}+1}}\bar{w}_{\iota}^{(k-1)},$

Define: for $1\leq\mu\leq m$ and
$\left(\bar{\mathbf{w}}_{L}^{(k)}\right)_{\mu}=\bar{w}_{\mu}^{(k)},\left(\bar{\mathbf{w}}_{R}^{(k)}\right)_{\mu}=\bar{w}_{m+\mu}^{(k)},\left(\bar{\mathbf{z}}_{L}^{(k)}\right)_{\mu}=\bar{z}_{\mu}^{(k)},\left(\bar{\mathbf{z}}_{R}^{(k)}\right)_{\mu}=\bar{z}_{\mu+m}^{(k)}$.
From (43) (44)

(50)
$\displaystyle\bar{\mathbf{w}}_{L}{(k)}=T_{L}\bar{\mathbf{w}}_{L}^{(k-1)}+P_{L}\bar{\mathbf{z}}_{L}^{(k)},\bar{\mathbf{w}}_{R}{(k)}=T_{R}\bar{\mathbf{w}}_{R}^{(k-1)}+P_{R}\bar{\mathbf{z}}_{R}^{(k)},$

From (46) (49)

(51) $\displaystyle
U_{L}\bar{\mathbf{z}}_{L}{(k)}=D_{L}\bar{\mathbf{w}}_{L}^{(k-1)}+E_{L}\bar{\mathbf{w}}_{R}^{(k-1)},U_{R}\bar{\mathbf{z}}_{R}{(k)}=D_{R}\bar{\mathbf{w}}_{R}^{(k-1)}+E_{R}\bar{\mathbf{w}}_{L}^{(k-1)},$
$T_{L}=\begin{bmatrix}1-\theta_{1}\\\
\frac{\theta_{2}}{\gamma_{1}}&1-\theta_{2}\\\ \quad\quad\ddots&\quad\ddots\\\
&\frac{\theta_{m}}{\gamma_{m-1}}&1-\theta_{m}\\\ \end{bmatrix}\quad
T_{R}=\begin{bmatrix}1-\theta_{m+1}&\frac{\theta_{m+1}}{\gamma_{m+3}}\\\
&1-\theta_{m+2}&\frac{\theta_{m+2}}{\gamma_{m+4}}\\\
&\quad\ddots&\quad\ddots\\\ &&\ddots&\frac{\theta_{2m-1}}{\gamma_{2m+1}}\\\
&&&1-\theta_{2m}\end{bmatrix}$
$U_{L}=\begin{bmatrix}\sqrt{\frac{\kappa_{2}}{\kappa_{1}}}&-\frac{\kappa_{3}}{\kappa_{2}}\frac{\sigma_{1}}{\sigma_{2}\gamma_{2}}\\\
\quad\ddots&\quad\ddots\\\
&\ddots&-\frac{\kappa_{m+1}}{\kappa_{m}}\frac{\sigma_{m-1}}{\sigma_{m}\gamma_{m}}\\\
&&\sqrt{\frac{\kappa_{m+1}}{\kappa_{m}}}\end{bmatrix}\quad
U_{R}=\begin{bmatrix}\sqrt{\frac{\kappa_{m+1}}{\kappa_{m+2}}}\\\
-\frac{\kappa_{m+1}}{\kappa_{m+2}}\frac{\sigma_{m+3}}{\sigma_{m+2}\gamma_{m+2}}&\ddots\\\
\quad\ddots&\quad\ddots\\\
&-\frac{\kappa_{2m-1}}{\kappa_{2m}}\frac{\sigma_{2m+1}}{\sigma_{2m}\gamma_{2m}}&\sqrt{\frac{\kappa_{2m}}{\kappa_{2m+1}}}\end{bmatrix}$

and

$\displaystyle D_{L}$
$\displaystyle=diag\left(-\frac{\sigma_{1}\sigma_{2}}{\gamma_{1}\gamma_{2}},\cdots,-\frac{\sigma_{m-1}\sigma_{m}}{\gamma_{m-1}\gamma_{m}},-\frac{\sigma_{m}\gamma_{m+1}}{\gamma_{m}\sigma_{m+1}}\right)$
$\displaystyle D_{R}$
$\displaystyle=diag\left(-\frac{\gamma_{m+1}\sigma_{m+2}}{\sigma_{m+1}\gamma_{m+2}}-\frac{\sigma_{m+2}\sigma_{m+3}}{\gamma_{m+2}\gamma_{m+3}},\cdots,-\frac{\sigma_{2m}\sigma_{2m+1}}{\gamma_{2m}\gamma_{2m+1}},\right)$
$\displaystyle P_{L}$
$\displaystyle=diag\left(\theta_{1}\frac{\kappa_{2}}{\kappa_{1}},\cdots,\theta_{m}\frac{\kappa_{m+1}}{\kappa_{m}}\right),$
$\displaystyle P_{R}$
$\displaystyle=diag\left(\theta_{m+1}\frac{\kappa_{m+1}}{\kappa_{m+2}},\cdots,\theta_{2m}\frac{\kappa_{2m}}{\kappa_{2m+1}}\right),$

and

$(E_{L})_{m,1}=\frac{\sigma_{m}}{\sigma_{m+1}\gamma_{m+2}},(E_{R})_{1,m}=\frac{\sigma_{m+2}}{\gamma_{m}\sigma_{m+1}},$

Using (50) & (51):

(52) $\displaystyle\bar{\mathbf{w}}_{L}^{(k)}$
$\displaystyle=(T_{L}+P_{L}U_{L}^{-1}D_{L})\bar{\mathbf{w}}_{L}^{(k-1)}+P_{L}U_{L}^{-1}E_{L}\bar{\mathbf{w}}_{R}^{(k-1)},$
(53) $\displaystyle\bar{\mathbf{w}}_{R}^{(k)}$
$\displaystyle=(T_{R}+P_{R}U_{R}^{-1}D_{R})\bar{\mathbf{w}}_{r}^{(k-1)}+P_{R}U_{R}^{-1}E_{R}\bar{\mathbf{w}}_{L}^{(k-1)},$

Now introduce $i,j$ for matrix indices, we have:

$(P_{L}U_{L}^{-1})_{i,j}=\begin{cases}\theta_{i}\sqrt{\frac{\kappa_{i+1}}{\kappa_{i}}},i=j,\\\
\theta_{i}\sqrt{\frac{\kappa_{j+1}}{\kappa_{i}}}\frac{\sigma_{i}}{\gamma_{i+1}\cdots\gamma_{j}\sigma_{j}},i<j,\\\
0,\textit{ otherwise
}\end{cases}\quad(P_{R}U_{R}^{-1})_{i,j}=\begin{cases}\theta_{m+i}\sqrt{\frac{\kappa_{m+i+1}}{\kappa_{m+i}}},i=j,\\\
\theta_{m+i}\sqrt{\frac{\kappa_{m+j+1}}{\kappa_{m+i}}}\frac{\sigma_{m+j}}{\sigma_{m+i}\gamma_{m+j+1}\cdots\gamma_{m+i}},i>j,\\\
0,\textit{ otherwise }\end{cases}$

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