The CF fractional operator is defined as follows [19, 20]:
$$ {}_{0}^{\mathrm{CF}}\mathcal{D}_{t}^{\alpha}f (t )= \frac{1}{1-\alpha } \int_{0}^{t}f ' (\tau )\exp \biggl(- \frac{\alpha (t-\tau )}{1-\alpha} \biggr)\, d\tau,\quad 0< \alpha\leq1, $$
(21)
where \({}_{0}^{\mathrm{CF}}\mathcal{D}_{t}^{\alpha}f (t )\) is the CF fractional operator with respect to t, \(M (\alpha )\) is a normalization function, such that \(M(0)=M(1)=1\); in this definition, the derivative of a constant is equal to zero, but unlike the usual Caputo definition [14], the kernel does not have a singularity at \(t=\tau\).
The Laplace transform (\(\mathscr{L}\)) of this novel definition (21) is defined as follows [19, 20]:
$$\begin{aligned} \mathscr{L} \bigl[{}_{0}^{\mathrm{CF}}\mathcal{D}_{t}^{ (\alpha+n )}f (t ) \bigr] & =\frac{1}{1-\alpha}\mathscr{L} \bigl[f^{ (\alpha +n )} (t ) \bigr] \mathscr{L} \biggl[\exp \biggl(-\frac{\alpha }{\alpha-1} t \biggr) \biggr] \\ & =\frac{s^{n+1}\mathscr{L} [f (t ) ]-s^{n}f (0 )-s^{n-1}f' (0 )-\cdots-f^{ (n )} (0 )}{s+\alpha (1-s )}. \end{aligned}$$
(22)
From this expression we have
$$ \begin{aligned} &\mathscr{L} \bigl[{}_{0}^{\mathrm{CF}} \mathcal{D}_{t}^{\alpha}f (t ) \bigr] =\frac{s\mathscr{L} [f (t ) ]-f (0 )}{s+\alpha (1-s )}, \quad n=0, \\ &\mathscr{L} \bigl[{}_{0}^{\mathrm{CF}}\mathcal{D}_{t}^{ (\alpha+1 )}f (t ) \bigr] =\frac{s^{2}\mathscr{L} [f (t ) ]-sf (0 )-f' (0 )}{s+\alpha (1-s )},\quad n=1. \end{aligned} $$
(23)
Description of the LHAM
An alternative procedure for constructing fractional differential equation was reported in [47], and successfully applied in [6–10]. In this context, to keep the dimensionality of the fractional differential equation a new parameter σ is introduced in the following way:
$$ \frac{d }{d t} \to\frac{1}{\sigma^{1 - \alpha}}\cdot{{}^{\mathrm{CF}}_{0}} { \mathcal{D}}_{t}^{\alpha}, \quad m - 1 < \alpha \leq m, m \in M=1,2,3,\ldots , $$
(24)
and
$$ \frac{d^{2} }{d t^{2}} \to\frac{1}{\sigma^{2(1 - \alpha)}}\cdot{{}^{\mathrm{CF}}_{0}} { \mathcal{D}}_{t}^{2\alpha},\quad m - 1 < \alpha \leq m, m \in M=1,2,3,\ldots , $$
(25)
where α represents the order of the fractional temporal operator and σ has the dimension of seconds, this auxiliary parameter is associated with the temporal components in the system (these components change the time constant of the system) [47]. In this context, the authors of [48] used the Planck time, \(t_{p}=5.39106\times10^{-44}\) seconds, with the finality to preserve the dimensional compatibility. Following [48] the σ parameter corresponds to the \(t_{p}\) in our calculations. For the case \(\alpha= 1\) the expressions (24) and (25) become ordinary temporal operators. Following this idea, we consider the following coupled linear fractional partial differential equations:
$$ \begin{aligned} &{{}^{\mathrm{CF}}_{0}} { \mathcal{D}}_{t}^{\alpha}x_{1} (z,t )-t_{p}^{1-\alpha}x_{2} (z,t )+t_{p}^{1-\alpha} \biggl(\frac {\mathcal{RC}+\mathcal{GL}}{\mathcal {LC}} \biggr)x_{1} (z,t )=0, \\ &{{}^{\mathrm{CF}}_{0}} {\mathcal{D}}_{t}^{\alpha}x_{2} (z,t )+t_{p}^{1-\alpha} \biggl(\frac{\mathcal{RG }}{\mathcal{LC}} \biggr)x_{1} (z,t )-t_{p}^{1-\alpha} \biggl( \frac{1}{\mathcal{LC}} \biggr)\frac {\partial^{2}x_{1} (z,t )}{\partial z^{2}}=0, \end{aligned} $$
(26)
with the initial conditions
$$ \frac{\partial^{k}x_{i} (z,0 )}{\partial t^{k}}=x_{i,k} (z,0 ) , \quad k=0,1,\ldots,n-1 , $$
(27)
and the boundary conditions
$$ x_{i} (0,t )=x_{i,0} (t ) ,\quad t\geq0. $$
(28)
The Laplace transform satisfies
$$ \mathscr{L} \bigl[{{}^{\mathrm{CF}}_{0}} {\mathcal{D}}_{t}^{\alpha}f(t) \bigr] (s )=\frac{1}{ (s+\alpha (1-s ) )} \bigl(s\mathscr {L} \bigl[f(t) \bigr] (s )-f (0 ) \bigr), \quad s>0, $$
(29)
we can define \(\Phi (x,s )=L [f(x,t) ] (s )\), for equation (26), we can write
$$\begin{aligned}& X_{1} (z,s ) =\mathscr{L} \bigl[x_{1}(z,t) \bigr] (s ) \\& \hphantom{X_{1} (z,s )}=\frac{x_{1} (z,0 )}{s} + (t_{p} )^{1-\alpha} \biggl(\frac{ (s+\alpha (1-s ) )}{s} \biggr) \biggl[X_{2} (z,s )- \biggl(\frac {\mathcal{RC}+\mathcal{GL}}{\mathcal{LC}} \biggr)X_{1} (z,s ) \biggr], \end{aligned}$$
(30)
$$\begin{aligned}& X_{2} (z,s )=\mathscr{L} \bigl[x_{2}(z,t) \bigr] (s ) \\& \hphantom{X_{2} (z,s )}=\frac{x_{2} (z,0 )}{s} + (t_{p} )^{1-\alpha} \biggl(\frac{ (s+\alpha (1-s ) )}{s} \biggr) \biggl[ \biggl(\frac{\mathcal{RG}}{\mathcal{LC}} \biggr)+ \biggl(\frac{1}{\mathcal {LC}} \biggr)\frac{\partial^{2}}{\partial z^{2}} \biggr]X_{1} (z,s ). \end{aligned}$$
(31)
According to LHAM, we construct the homotopy for Eq. (30) as follows:
$$ \begin{aligned} &X_{1} (z,s )=\frac{x_{1} (z,0 )}{s}+p (t_{p} )^{1-\alpha} \biggl(\frac{ (s+\alpha (1-s ) )}{s} \biggr) \biggl[X_{2} (z,s )- \biggl(\frac{\mathcal {RC}+\mathcal{GL}}{\mathcal{LC}} \biggr)X_{1} (z,s ) \biggr], \\ &X_{2} (z,s )=\frac{x_{2} (z,0 )}{s}+p (t_{p} )^{1-\alpha} \biggl(\frac{ (s+\alpha (1-s ) )}{s} \biggr) \biggl[ \biggl(\frac{\mathcal{RG}}{\mathcal{LC}} \biggr)+ \biggl(\frac{1}{\mathcal{LC}} \biggr)\frac{\partial^{2}}{\partial z^{2}} \biggr]X_{1} (z,s ), \end{aligned} $$
(32)
where \(X_{i} (z,s )\) denote the Laplace transform of \(x_{i} (z,t )\).
Applying the LHAM to obtain the solution of Eq. (32), we can start by the hypothesis that the solution \(\Phi (x,z )\) is expressed as
$$ X_{i} (z,s )=\sum_{j=0}^{\infty}p^{j}X_{i , j} (z,s ) , $$
(33)
where \(X_{i j} (z,s )\), \(j=0,1,2,\ldots\) , are the unknown functions. Substituting (33) into (32), we get
$$ \begin{aligned} &\sum_{j=0}^{\infty}p^{j}X_{1,j} (z,s ) \\ &\quad =\frac{x_{1} (z,0 )}{s}+p (t_{p} )^{1-\alpha} \biggl( \frac{ (s+\alpha (1-s ) )}{s} \biggr) \\ &\qquad {}\times\Biggl[\sum_{j=0}^{\infty}p^{j}X_{2,j} (z,s )- \biggl(\frac{\mathcal{RC} +\mathcal{GL}}{\mathcal{LC}} \biggr)\sum_{j=0}^{\infty}p^{j}X_{1,j} (z,s ) \Biggr], \\ &\sum_{j=0}^{\infty}p^{j}X_{2,j} (z,s ) \\ &\quad =\frac{x_{2} (z,0 )}{s}+p (t_{p} )^{1-\alpha } \biggl( \frac{ (s+\alpha (1-s ) )}{s} \biggr) \\ &\qquad {}\times\biggl[ \biggl(\frac{\mathcal{RG}}{\mathcal{LC}} \biggr)+ \biggl( \frac{1}{\mathcal{LC}} \biggr)\frac{\partial^{2}}{\partial z^{2}} \biggr]\sum _{j=0}^{\infty}p^{j}X_{1,j} (z,s ), \end{aligned} $$
(34)
which, on comparing the coefficients of powers of p, yields
$$\begin{aligned}& p^{0}: X_{1,0} (z,s ) =\frac{x_{1} (z,0 )}{s}, \\& \hphantom{p^{0}:{}} X_{2,0} (z,s ) =\frac{x_{2} (z,0 )}{s}, \\& p^{1}: X_{1,1} (z,s ) = (t_{p} )^{1-\alpha} \biggl(\frac{ (s+\alpha (1-s ) )}{s} \biggr) \biggl[X_{2,0} (z,s )- \biggl( \frac{\mathcal{RC}+\mathcal {GL}}{\mathcal{LC}} \biggr)X_{1,0} (z,s ) \biggr], \\& \hphantom{p^{1}:{}} X_{2,1} (z,s ) = (t_{p} )^{1-\alpha} \biggl(\frac { (s+\alpha (1-s ) )}{s} \biggr) \biggl[ \biggl(\frac {\mathcal{RG}}{\mathcal{LC}} \biggr)+ \biggl(\frac {1}{\mathcal{LC}} \biggr)\frac{\partial^{2}}{\partial z^{2}} \biggr]X_{1,0} (z,s ), \\& \ldots, \\& p^{n+1}: X_{1, n+1} (z,s ) = (t_{p} )^{1-\alpha } \biggl(\frac{ (s+\alpha (1-s ) )}{s} \biggr) \biggl[X_{2, n} (z,s )- \biggl( \frac{\mathcal{RC}+\mathcal {GL}}{\mathcal{LC}} \biggr)X_{1, n} (z,s ) \biggr], \\& \hphantom{p^{n+1}:{}} X_{2, n+1} (z,s ), = (t_{p} )^{1-\alpha} \biggl(\frac{ (s+\alpha (1-s ) )}{s} \biggr) \biggl[ \biggl(\frac {\mathcal{RG}}{\mathcal{LC}} \biggr)+ \biggl(\frac {1}{\mathcal{LC}} \biggr)\frac{\partial^{2}}{\partial z^{2}} \biggr]X_{1, n} (z,s ). \end{aligned}$$
(35)
In the limit \(p\rightarrow1\), we note that (35) becomes the approximate solution for the problem of (26)-(27) and is given by
$$ \begin{aligned} &H_{1, n} (z,s )= \sum_{i=0}^{n}X_{1, n} (z,s ), \\ &H_{2, n} (z,s )= \sum_{i=0}^{n}X_{2, n} (z,s ). \end{aligned} $$
(36)
Taking the inverse Laplace transform of (36), we obtain
$$ \begin{aligned} &x_{1} (z,t )\approx \mathscr{L}^{-1} \bigl[H_{1 ,n} (z,s ) \bigr] , \\ &x_{2} (z,t )\approx \mathscr{L}^{-1} \bigl[H_{2 ,n} (z,s ) \bigr]. \end{aligned} $$
(37)