To obtain the solution of the considered problem, the Laplace transformation technique was employed.
3.1 Exact solution of heat profile
Applying the Laplace transformation technique to write the solution of Eq. (11) with conditions as Eqs. (14), (15), (16), we have
$$\begin{aligned} \frac{d^{2}{\bar{ \theta }(\psi ,s)}}{d{\psi }^{2}}-P_{r} s{ \bar{\theta }(\psi ,s)}=0. \end{aligned}$$
(17)
with
$$\begin{aligned} {\bar{\theta }(0,s)}=\frac{1-e^{-s}}{s^{2}} \quad \text{and} \quad { \bar{\theta }(\psi ,s)}\rightarrow 0 \quad \text{as }\psi \rightarrow \infty . \end{aligned}$$
(18)
and its solution is given by
$$\begin{aligned} \bar{\theta }(\psi ,s)= c_{1} e^{\psi \sqrt{P_{r}s}} + c_{2} e^{- \psi \sqrt{P_{r}s}}. \end{aligned}$$
(19)
we applied boundary conditions for temperature given by Eq. (18) to determine the unknown constants and obtain:
$$\begin{aligned} \bar{\theta }(\psi ,s)= \frac{1}{s^{2}} \bigl(1-e^{-s} \bigr) e^{- \psi \sqrt{P_{r}s}}. \end{aligned}$$
(20)
which can be written as
$$\begin{aligned} \bar{\theta }(\psi ,s)&= \biggl(\frac{e^{-\psi \sqrt{P_{r}s}}}{s^{2}} \biggr)-{e^{-s}} \biggl(\frac{e^{-\psi \sqrt{P_{r}s}}}{s^{2}} \biggr), \\ &= \bar{\theta }_{r}(\psi ,s)-e^{-s}\bar{\theta }_{r}(\psi ,s). \end{aligned}$$
(21)
To obtain the required solution of Eq. (21), the Laplace inverse transformation was used, which is written as:
$$\begin{aligned} {\theta }(\psi ,\tau )= {\theta }_{r}(\psi ,\tau )-{\theta }_{r}(\psi , \Phi )g(\Phi ). \end{aligned}$$
(22)
with
$$\begin{aligned} {\theta }_{r}(\psi ,\tau )= \biggl( \frac{P_{r}}{2}{\psi }^{2}+\tau \biggr)\operatorname{erfc} \biggl( \sqrt{\frac{P_{r}}{4\tau }} {\psi }\biggr)-\biggl(\sqrt{ \frac{P_{r}\tau }{\pi }} {\psi }\biggr)e^{ \frac{-P_{r}{\psi }^{2}}{4\tau }}. \end{aligned}$$
(23)
where \(g(\Phi )\) represents a standard Heaviside function and \(\Phi =\tau -1\).
3.1.1 Nusselt number
An expression for the Nusselt number to efficiently forecast the generalized rate of heat transfer corresponding to ramped conditions is evaluated as:
$$\begin{aligned} N_{u}&=-\frac{\partial \theta (\psi ,\tau )}{\partial \psi }\biggm|_{\psi =0} \\ &=-\frac{\partial }{\partial \psi }\mathscr{L}^{-1} \bigl\lbrace \bar{\theta }( \psi ,s) \bigr\rbrace \biggm|_{\psi =0} \\ &=-\mathscr{L}^{-1} \biggl\lbrace \frac{\partial \bar{\theta }(\psi ,s)}{\partial \psi } \biggm|_{\psi =0} \biggr\rbrace \\ &=\mathscr{L}^{-1} \biggl\lbrace \frac{(1-e^{-s})\sqrt{Prs}}{s^{2}} \biggr\rbrace \\ &= \sqrt{\Pr } \biggl[\frac{2\sqrt{\tau }}{\sqrt{\pi }}- \frac{2\sqrt{\tau -1}}{\sqrt{\pi }}H(\tau -1) \biggr], \end{aligned}$$
(24)
where \(H(\tau -1)\) represents a standard Heaviside function.
3.2 Exact solution of mass profile
Solving Eq. (13) using Eqs. (14), (15) and (16), and employing the Laplace transformation technique, the resulting equations are written as:
$$\begin{aligned} \frac{d^{2}{\bar{C}(\psi ,s)}}{d{\psi }^{2}}- ({S_{c}}s+\lambda _{0} ) {\bar{C}(\psi ,s)}=0. \end{aligned}$$
(25)
with
$$\begin{aligned} {\bar{C}(0,s)}=\frac{1}{s} \quad \text{and} \quad {\bar{C}(\psi ,s)} \rightarrow 0 \quad \text{as }\psi \rightarrow \infty . \end{aligned}$$
(26)
the solution in general form is
$$\begin{aligned} \bar{C}(\psi ,s)= c_{1} e^{{\psi }\sqrt{{S_{c}}s+\lambda _{0}}} + c_{2} e^{-{ \psi }\sqrt{{S_{c}}s+\lambda _{0}}}. \end{aligned}$$
(27)
the values of constants \(c_{1}\) and \(c_{2}\), conditions are implemented as given in Eq. (26) for concentration, so
$$\begin{aligned} \bar{C}(\psi ,s)= \frac{1}{s}e^{-{\psi }\sqrt{{S_{c}}s+\lambda _{0}}}. \end{aligned}$$
(28)
To obtain the solution, taking the inverse Laplace transformation, we have:
$$\begin{aligned} C(\psi ,\tau )={}&\mathscr{L}^{-1} \biggl\lbrace \frac{1}{s}e^{-{\psi } \sqrt{Sc}\sqrt{s+\frac{\lambda _{0}}{Sc}}} \biggr\rbrace \\ ={}&\frac{1}{2} \biggl(e^{-{\psi }\sqrt{Sc}\sqrt{\frac{\lambda _{0}}{Sc}}}\operatorname{erfc} \biggl( \frac{\psi \sqrt{Sc}}{2\sqrt{\tau }}-\sqrt{ \frac{\lambda _{0}}{Sc}\tau } \biggr) \biggr) \\ &{}+ \frac{1}{2} \biggl(e^{{ \psi }\sqrt{Sc}\sqrt{\frac{\lambda _{0}}{Sc}}}\operatorname{erfc} \biggl( \frac{\psi \sqrt{Sc}}{2\sqrt{\tau }}+\sqrt{\frac{\lambda _{0}}{Sc}\tau } \biggr) \biggr). \end{aligned}$$
(29)
The following result, which exists in the literature, is used:
$$\begin{aligned} \mathscr{L}^{-1} \biggl\lbrace \frac{e^{-{\psi }\sqrt{c_{1}}\sqrt{s+b_{1}}}}{s-a_{1}} \biggr\rbrace ={}& \frac{e^{a_{1}\tau }}{2} \biggl(e^{-{\psi }\sqrt{c_{1}}\sqrt{a_{1}+b_{1}}} \operatorname{erfc} \biggl(\frac{\psi \sqrt{c_{1}}}{2\sqrt{\tau }}-\sqrt{(a_{1}+b_{1}) \tau } \biggr) \biggr) \\ &{}+\frac{e^{a_{1}\tau }}{2} \biggl(e^{{\psi }\sqrt{c_{1}}\sqrt{a_{1}+b_{1}}}\operatorname{erfc} \biggl( \frac{\psi \sqrt{c_{1}}}{2\sqrt{\tau }}+\sqrt{(a_{1}+b_{1})\tau } \biggr) \biggr). \end{aligned}$$
(30)
3.3 Exact solution of velocity profile
The solution of Eq. (10) by using a Laplace transform, is:
$$\begin{aligned}& \begin{aligned}[b] (1+\lambda s )\bar{u}(\psi ,s)= {}&\frac{d^{2}\bar{u}(\psi ,s)}{d{\psi }^{2}} + (1+ \lambda s )G_{r}{ \bar{\theta } (\psi ,s )}\\ &{}+ (1+\lambda s )G_{m}{ \bar{C} (\psi ,s )} - (1+\lambda s )M\bar{u}( \psi ,s), \end{aligned} \end{aligned}$$
(31)
$$\begin{aligned}& \frac{d^{2}\bar{u}(\psi ,s)}{d{\psi }^{2}}- \bigl( a + \lambda s^{2} + bs \bigr) \bar{u}(\psi ,s)= - (1+\lambda s ) ( G_{r} \bar{\theta }+G_{m}{\bar{C}} ). \end{aligned}$$
(32)
by using Eqs. (21), (28) to determine the values \({\bar{\theta } (\psi ,s )}\) and \({\bar{C} (\psi ,s )}\), then Eq. (32) has the general form of the solution, which is represented as:
$$\begin{aligned} \bar{u} (\psi ,s )={}&c_{3}e^{\psi \sqrt{ ( a + \lambda s^{2} + bs )}}+c_{4}e^{-\psi \sqrt{ ( a + \lambda s^{2} + bs )}}- \biggl( \frac{G_{r} (1+\lambda s ) (1-e^{-s} )e^{-\psi \sqrt{P_{r}s}}}{s^{2} ( P_{r}s- ( a + \lambda s^{2} + bs ) ) } \biggr) \\ &{} - \biggl( \frac{G_{m} (1+\lambda s )e^{-\psi \sqrt{S_{c}s+{\lambda _{0}}}}}{s ( S_{c}s+{\lambda _{0}}- ( a + \lambda s^{2} + bs ) ) } \biggr) \end{aligned}$$
(33)
Applying \({\bar{u}(\psi ,s)}\rightarrow 0\), as \(\psi \rightarrow \infty \) and \({\bar{u}(0,s)}=\frac{1-e^{-s}}{s^{2}}\), to determine the values of unknowns \(c_{3}\) and \(c_{4}\), we obtain:
$$\begin{aligned} \bar{u} (\psi ,s )={}& \biggl( \frac{1-e^{-s}}{s^{2}} \biggr)e^{- \psi \sqrt{ ( a + \lambda s^{2} + bs )}}+ \biggl( \frac{1-e^{-s}}{s^{2}} \biggr) \biggl( \frac{G_{r} (1+\lambda s ) (e^{-\psi \sqrt{ ( a + \lambda s^{2} + bs )}}-{e^{-\psi \sqrt{P_{r}s}}} ) }{P_{r}s- ( a + \lambda s^{2} + bs )} \biggr) \\ &{}+ \biggl( \frac{G_{m} (1+\lambda s ) (e^{-\psi \sqrt{ ( a + \lambda s^{2} + bs )}}-e^{-\psi \sqrt{S_{c}s+{\lambda _{0}}}} ) }{s ( S_{c}s+{\lambda _{0}}- ( a + \lambda s^{2} + bs ) ) } \biggr). \end{aligned}$$
(34)
To find the Laplace inverse of Eq. (34), we rearrange the above equation as:
$$\begin{aligned} \bar{u} (\psi ,s )=\bar{F} (\psi ,s )+ Gr \bar{\chi } (\psi ,s ) \bigl[\bar{\theta } (\psi ,s ) - \bar{F} (\psi ,s ) \bigr] +Gm \bar{\xi } ( \psi ,s ) \bigl[\bar{C} (\psi ,s )-\bar{H} ( \psi ,s ) \bigr]. \end{aligned}$$
(35)
Employing an inverse Laplace transformation on Eq. (35), the obtained solution is written as:
$$\begin{aligned} {u} (\psi ,\tau )={F} (\psi ,\tau )+ Gr \bigl[(\chi \ast \theta ) (\tau ) - (\chi \ast F) (\tau ) \bigr] +Gm \bigl[(\xi \ast C) (\tau )-(\xi \ast H) (\tau ) \bigr]. \end{aligned}$$
(36)
with
$$\begin{aligned} {F} (\psi ,\tau )={F_{1}} (\psi ,\tau ) - {F_{1}} (\psi ,\Phi )g(\Phi ). \end{aligned}$$
(37)
where \(g(\Phi )\) represents a standard Heaviside function and \(\Phi =\tau -1\), also
$$\begin{aligned} \bar{F_{1}} (\psi ,s )&=\frac{1}{s^{2}}e^{-\psi \sqrt{a + \lambda s^{2} + bs}} \end{aligned}$$
(38)
$$\begin{aligned} &=\sum_{\alpha =0}^{\infty }\sum _{\beta =0}^{\infty }\sum_{\gamma =0}^{ \infty } \frac{(-\psi )^{\alpha }(a)^{\frac{\alpha }{2}-\beta }(b)^{\beta -\gamma }(\lambda )^{\gamma }\Gamma (\frac{\alpha }{2}+1)}{\alpha ! \gamma ! \Gamma (\beta - \gamma +1)} \frac{1}{s^{2- \beta -\gamma }}, \end{aligned}$$
(39)
$$\begin{aligned} {F_{1}} (\psi ,\tau )&=\sum_{\alpha =0}^{\infty } \sum_{ \beta =0}^{\infty }\sum _{\gamma =0}^{\infty } \frac{(-\psi )^{\alpha }(a)^{\frac{\alpha }{2}-\beta }(b)^{\beta -\gamma }(\lambda )^{\gamma }\Gamma (\frac{\alpha }{2}+1)}{\alpha ! \gamma ! \Gamma (\beta - \gamma +1)} \frac{t^{1- \beta -\gamma }}{\Gamma (2- \beta -\gamma )}, \end{aligned}$$
(40)
$$\begin{aligned} \bar{H} (\psi ,s )&=\frac{1}{s}e^{-\psi \sqrt{a + \lambda s^{2} + bs}}, \end{aligned}$$
(41)
$$\begin{aligned} &=\sum_{\alpha =0}^{\infty }\sum _{\beta =0}^{\infty }\sum_{\gamma =0}^{ \infty } \frac{(-\psi )^{\alpha }(a)^{\frac{\alpha }{2}-\beta }(b)^{\beta -\gamma }(\lambda )^{\gamma }\Gamma (\frac{\alpha }{2}+1)}{\alpha ! \gamma ! \Gamma (\beta - \gamma +1)} \frac{1}{s^{1- \beta -\gamma }}, \end{aligned}$$
(42)
$$\begin{aligned} {H} (\psi ,\tau )&=\sum_{\alpha =0}^{\infty }\sum _{ \beta =0}^{\infty }\sum _{\gamma =0}^{\infty } \frac{(-\psi )^{\alpha }(a)^{\frac{\alpha }{2}-\beta }(b)^{\beta -\gamma }(\lambda )^{\gamma }\Gamma (\frac{\alpha }{2}+1)}{\alpha ! \gamma ! \Gamma (\beta - \gamma +1)} \frac{t^{- \beta -\gamma }}{\Gamma (1- \beta -\gamma )}, \end{aligned}$$
(43)
$$\begin{aligned} \bar{\chi } (\psi ,s )&=(1 + \lambda s) \frac{1}{\lambda s^{2}- (\Pr -b)s+a}, \end{aligned}$$
(44)
$$\begin{aligned} &=\sum_{n=0}^{\infty }\frac{(-1)^{n}(a)^{n}}{(\lambda )^{n+1}} \frac{(s)^{-n-1}}{(s - \epsilon )^{n+1}} + \lambda \sum_{n=0}^{ \infty } \frac{(-1)^{n}(a)^{n}}{(\lambda )^{n+1}} \frac{(s)^{-n}}{(s - \epsilon )^{n+1}}, \end{aligned}$$
(45)
$$\begin{aligned} {\chi } (\psi ,\tau )&=\sum_{n=0}^{\infty } \frac{(-1)^{n}(a)^{n}}{(\lambda )^{n+1}} \bigl[G_{1,-n-1, n + 1}( \epsilon ,\tau ) + G_{1,-n, n + 1}(\epsilon ,\tau ) \bigr], \end{aligned}$$
(46)
$$\begin{aligned} \bar{\xi } (\psi ,s )&=(1 + \lambda s) \frac{1}{\lambda s^{2}- (Sc-b)s+(a-\lambda _{0})}, \end{aligned}$$
(47)
$$\begin{aligned} &=\sum_{m=0}^{\infty } \frac{(-1)^{m}(a-\lambda _{0})^{m}}{(\lambda )^{m+1}} \frac{(s)^{-m-1}}{(s - \delta )^{m+1}} + \lambda \sum_{m=0}^{\infty } \frac{(-1)^{m}(a-\lambda _{0})^{m}}{(\lambda )^{m+1}} \frac{(s)^{-m}}{(s - \delta )^{m+1}}, \end{aligned}$$
(48)
$$\begin{aligned} {\xi } (\psi ,\tau )&=\sum_{m=0}^{\infty } \frac{(-1)^{m}(a-\lambda _{0})^{m}}{(\lambda )^{m+1}} \bigl[G_{1,-m-1, m + 1}(\delta ,\tau ) + G_{1,-m, m + 1}(\delta ,\tau ) \bigr]. \end{aligned}$$
(49)
with \(\epsilon = \frac{\Pr - b}{\lambda }\) and \(\delta = \frac{Sc - b}{\lambda }\), the function \(G_{h,b,l}(.,\tau )\) used in the above expressions is known as the generalized Lorenzo–Hartley function and is defined as \(\mathscr{L}^{-1} \lbrace \frac{s^{b}}{(s^{h}-j)^{l}} \rbrace =G_{h,b,l}(j,\tau )\); \(\operatorname{Re}(hl-b)>0\), \(\operatorname{Re}(s)>0\), \(\vert \frac{j}{s^{h}} \vert <1 \).
3.4 Solution of shear stress
In the industrial and mechanical fields, wall shear stress is of indispensable significance and increasing shear stress is considered as a disadvantage. To estimate the wall shear stress and skin-friction coefficient for a Maxwell fluid we employ a Laplace transform on Eq. (12), and we have:
$$\begin{aligned} (1+\lambda s )\bar{S} (\psi ,s ) = \frac{d\bar{u} (\psi ,s )}{d{\psi }}. \end{aligned}$$
(50)
To determine the value of \(\frac{d\bar{u} (\psi ,s )}{d{\psi }}\), differentiating Eq. (34) with respect to ψ yields:
$$\begin{aligned} \frac{d\bar{u} (\psi ,s )}{d{\psi }}={}& \biggl( \frac{1-e^{-s}}{s^{2}} \biggr) \bigl( -\sqrt{\S }e^{-\psi \sqrt{\S }} \bigr) \\ &{} + \biggl( \frac{1-e^{-s}}{s^{2}} \biggr) \biggl( \frac{G_{r} (1+\lambda s ) (-\sqrt{\S }e^{-\psi \sqrt{\S }}+{\sqrt{P_{r}s}e^{-\psi \sqrt{P_{r}s}}} ) }{P_{r}s-\S } \biggr) \\ &{}+ \biggl( \frac{G_{m} (1+\lambda s ) (-\sqrt{\S }e^{-\psi \sqrt{\S }}+\sqrt{S_{c}s+{\lambda _{0}}}e^{-\psi \sqrt{S_{c}s+{\lambda _{0}}}} ) }{s ( S_{c}s+{\lambda _{0}}-\S ) } \biggr), \end{aligned}$$
(51)
where \(\S = a + \lambda s^{2} + bs\).
Inserting Eq. (51) into Eq. (50), gives the expression for shear stress:
$$\begin{aligned} \bar{S} (\psi ,s )={}& \biggl( \frac{1-e^{-s}}{s^{2}} \biggr) \biggl(\frac{ -\sqrt{\S }e^{-\psi \sqrt{\S }}}{1+\lambda s } \biggr) \\ &{} + \biggl( \frac{1-e^{-s}}{s^{2}} \biggr) \biggl( \frac{G_{r} (1+\lambda s ) (-\sqrt{\S }e^{-\psi \sqrt{\S }}+{\sqrt{P_{r}s}e^{-\psi \sqrt{P_{r}s}}} ) }{ (1+\lambda s ) ( P_{r}s-\S ) } \biggr) \\ &{}+ \biggl( \frac{G_{m} (1+\lambda s ) (-\sqrt{\S }e^{-\psi \sqrt{\S }}+\sqrt{S_{c}s+{\lambda _{0}}}e^{-\psi \sqrt{S_{c}s+{\lambda _{0}}}} ) }{s (1+\lambda s ) ( S_{c}s+{\lambda _{0}}-\S ) } \biggr). \end{aligned}$$
(52)
The skin-friction coefficient is estimated as
$$\begin{aligned} \bar{C_{f}} ={}&\frac{1}{ (1+\lambda s )} \frac{d\bar{u} (\psi ,s )}{d{\psi }}\biggm|_{\psi =0}, \\ ={}& \biggl( \frac{1-e^{-s}}{s^{2}} \biggr) \biggl( \frac{ -\sqrt{\S }}{1+\lambda s } \biggr) + \biggl( \frac{1-e^{-s}}{s^{2}} \biggr) \biggl( \frac{G_{r} (1+\lambda s ) (-\sqrt{\S }+{\sqrt{P_{r}s}} ) }{ (1+\lambda s ) ( P_{r}s-\S ) } \biggr) \\ &{}+ \biggl( \frac{G_{m} (1+\lambda s ) (-\sqrt{\S }+\sqrt{S_{c}s+{\lambda _{0}}} ) }{s (1+\lambda s ) ( S_{c}s+{\lambda _{0}}-\S ) } \biggr). \end{aligned}$$
(53)
Since the solution given in Eq. (52) and Eq. (53) contains complex terms of the Laplace parameter s, to derive the solution in real time τ, we applied the numerical inversion method known as the Durbin Method [27].