$F\left(k\right)=\frac{1}{k!}{\left[\frac{d^{k}f\left(\eta \right)}{d{\eta }^k}\right]}_{\eta ={\eta }_0}$

(3.107)

$f(\eta )=\sum _{k=0}^{\infty }F(k){\left(\eta -{\eta }_0\right)}^k.$

(3.108)

$f(\eta )={\sum _{k=0}^{\infty }\left[\frac{d^{k}f(\eta )}{d{(\eta )}^k}\right]}_{\eta ={\eta }_0}\frac{{\left(\eta -{\eta }_0\right)}^k}{k!}$

(3.109)

$f(\eta )=\sum _{k=0}^{N}F(k){\left(\eta -{\eta }_0\right)}^k$

(3.110)

Eq. (3.110) implies that $f(\eta )={\sum }_{k=N+1}^{\infty }\left(F(k){\left(\eta -{\eta }_0\right)}^k\right)$ is negligibly small, where N is series size. Theorems to be used in the transformation procedure, which can be evaluated from Eqs. (3.107) and (3.108), are given in Table 3.8.

Table 3.8

Some of the basic operations of differential transformation method (DTM)

Original function	Transformed function
f(η) = αg(η) ± βh(η)	F[k] = αG[k] ± βH[k]
$f(\eta )=\frac{d^{n}g(\eta )}{d{\eta }^n}$	$F[k]=\frac{(k+n)!}{k!}G[k+n]$
f(η) = g(η)h(η)	$F[k]={\sum }_{m-0}^{k}F(m)H[k-m]$
f(τ) = sin(ϖη + α)	$F[k]=\frac{{\varpi }^k}{k!}\sin \left(\frac{\pi k}{2}+\alpha \right)$
f(τ) = cos(ϖη + α)	$F[k]=\frac{{\varpi }^k}{k!}\cos \left(\frac{\pi k}{2}+\alpha \right)$
f(η) = eλη	$F[k]=\frac{{\lambda }^k}{k!}$
F(η) = (1 + η)m	$F(k)=\frac{m(m-1)\dots (m-k+1)}{k!}$
f(η) = ηm	$F(k)=\delta (k-m)=\left\{\begin{array}{c}\hfill 1,k=m\hfill \\ \hfill 0,k\ne m\hfill \end{array}\right.$

3.5.2.2. Application of DTM

$\begin{array}{l}(k+1)(k+2)(k+3)(k+4)F[k+4]+S\left(A_1/A_4\right)\sum _{m=0}^k\left(∆[k-m-1](m+1)(m+2)(m+3)F[m+3]\right)\\ -3S\left(A_1/A_4\right)(k+1)(k+2)F[k+2]-S(A_1/A_4)\sum _{m=0}^k\left(\left(k-m+1\right)F[k-m+1](m+1)(m+2)F[m+2]\right)\\ +S\left(A_1/A_4\right)\sum _{m=0}^k\left(F[k-m](m+1)(m+2)(m+3)F[m+3]\right)\\ -H{a}^2\left(A_5/A_4\right)(k+1)(k+2)F[k+2]=0,\\ ∆[m]=\left\{\begin{array}{l}1m=1\\ 0m\ne 1\end{array}\right.\end{array}$

(3.111)

$F[0]=0,F[1]=a_1,F[2]=0,F[3]=a_2$

(3.112)

$\begin{array}{l}(k+1)(k+2)\Theta [k+2]+Pr.S.\left(\frac{A_2}{A_3}\right)\sum _{m=0}^k\left(F[k-m](m+1)\Theta [m+1]\right)\\ -Pr.S.\left(\frac{A_2}{A_3}\right)\sum _{m=0}^k\left(∆[k-m](m+1)\Theta [m+1]\right)\\ +\frac{Hs}{A_3}\Theta [k]=0,\\ ∆[m]=\left\{\begin{array}{l}1m=1\\ 0m\ne 1\end{array}\right.\end{array}$

(3.113)

$\Theta [0]=a_3,\Theta [1]=0$

(3.114)

$\begin{array}{l}F[0]=0,F[1]=a_1,F[2]=0,F[3]=a{}_2,F[4]=0\hfill \\ F[5]=\frac{3}{20}S{a}_2\left(\frac{A_1}{A_2}\right)+\frac{1}{20}S{a}_1{a}_2\left(\frac{A_1}{A_2}\right)+\frac{1}{20}{a}_1{a}_2+\frac{1}{20}H{a}^2{a}_2,\mathrm{...}\hfill \end{array}$

(3.115)

$\begin{array}{l}\Theta [0]=a_3,\Theta [1]=0,\Theta [2]=-\frac{1}{2}\frac{Hs}{A_3}{a}_3,\Theta [3]=0.0,\hfill \\ \Theta [4]=\frac{1}{12}PrS\left(\frac{A_2}{A_3}\right)\frac{Hs}{A_3}{a}_3{a}_1,\Theta [5]=\mathrm{0,...}\hfill \end{array}$

(3.116)

$F(\eta )=a_1\eta +a_2{\eta }^3+\left(\frac{3}{20}S{a}_2\left(\frac{A_1}{A_2}\right)+\frac{1}{20}S{a}_1{a}_2\left(\frac{A_1}{A_2}\right)+\frac{1}{20}{a}_1{a}_2+\frac{1}{20}H{a}^2{a}_2\right){\eta }^4+\cdots$

(3.117)

$\theta \left(\eta \right)=a_3+\left(-\frac{1}{2}\frac{Hs}{A_3}{a}_3\right){\eta }^2+\left(\frac{1}{12}\mathrm{Pr}S\left(\frac{A_2}{A_3}\right)\frac{Hs}{A_3}{a}_3{a}_1\right){\eta }^4+\mathrm{..}\text{.}$

(3.118)

by substituting the boundary condition from Eq. (3.103) into Eqs. (3.117) and (3.118) in point η = 1, the values of a₁, a₂, and a₃ can be obtained. By substituting obtained a₁, a₂, and a₃ into Eqs. (3.117) and (3.118), the expression of F(η) and Θ(η) can be obtained.

Consider the steady, two-dimensional flow of a nanofluid near the stagnation point on a stretching sheet saturated in the presence of transverse magnetic field as shown in Fig. 3.42 [12]. The stretching velocity $U_{\text{w}}(x)$ and the free stream velocity U_∞(x) are assumed to vary proportionally to the distance x from the stagnation point, that is, $U_{\text{w}}(x)=ax$ and U_∞(x) = bx, where a and b are constants with a > 0 and b ≥ 0. It is also assumed that the surface of the sheet is subjected to a prescribed temperature $T_{\text{w}}\left(x\right)=T_{\infty }+c{x}^n$, where T_∞ is the ambient fluid temperature and c and n are constants with c > 0 (heated surface). Further, a uniform magnetic field of strength B₀ is assumed to be applied in the positive y-direction normal to the stretching sheet. The magnetic Reynolds number is assumed to be small, and thus the induced magnetic field is negligible. Also the thermal radiation effect is considered in this problem.

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

(3.119)

${\rho }_{\text{nf}}\left(u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}-U_{\infty }\frac{d{U}_{\infty }}{dx}\right)={\mu }_{\text{nf}}\frac{{\partial }^{2}u}{\partial y^2}+{\sigma }_{\text{nf}}{B}_0^2\left(U_{\infty }-u\right)$

(3.120)

${\left(\rho C_{\text{p}}\right)}_{\text{nf}}\left(u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}\right)=k_{\text{nf}}\frac{{\partial }^{2}T}{\partial y^2}-\frac{\partial q_{\text{r}}}{\partial y}$

(3.121)

$\begin{array}{l}y=0:u=U_{\text{w}}(x),v=0,T=T_{\text{w}}(x)\hfill \\ y\to \infty :u\to U_{\infty }(x),T\to T_{\infty }\hfill \end{array}$

(3.122)

where u and $v$ are the velocity components along the x and y axes, respectively and T is fluid temperature. The radiation heat flux q_r is considered according to Rosseland approximation, such that $q_{\text{r}}=-\frac{4{\sigma }_{\text{e}}}{3{\beta }_{\text{R}}}\frac{\partial T^4}{\partial y}$ where σ_e, β_R are the Stefan–Boltzmann constant and the mean absorption coefficient, respectively. The fluid-phase temperature differences within the flow are assumed to be sufficiently small, such that T ⁴ may be expressed as a linear function of temperature. This is done by expanding T ⁴ in a Taylor series about the temperature T_c and neglecting the higher-order terms to yield, $T^4\cong 4T_{\text{c}}^{3}T-3T_{\text{c}}^4$.

The effective density ρ_nf, the heat capacitance ${\left(\rho C_{\text{p}}\right)}_{\text{nf}}$, and electrical conductivity (σ_nf) of the nanofluid are given as:

${\rho }_{\text{nf}}={\rho }_{\text{f}}\left(1-\phi \right)+{\rho }_{\text{s}}\phi$

(3.123)

${\left(\rho C_{\text{p}}\right)}_{\text{nf}}={\left(\rho C_{\text{p}}\right)}_{\text{f}}\left(1-\phi \right)+{\left(\rho C_{\text{p}}\right)}_{\text{s}}\phi$

(3.124)

$\frac{{\sigma }_{\text{nf}}}{{\sigma }_{\text{f}}}=1+\frac{3\left({\sigma }_{\text{s}}/{\sigma }_{\text{f}}-1\right)\phi }{\left({\sigma }_{\text{s}}/{\sigma }_{\text{f}}+1\right)-\left({\sigma }_{\text{s}}/{\sigma }_{\text{f}}-1\right)\phi }$

(3.125)

$\begin{array}{l}{k}_{\text{nf}}=1+\frac{3\left(\frac{k_{\text{p}}}{k_{\text{f}}}-1\right)\phi }{\left(\frac{k_{\text{p}}}{k_{\text{f}}}+2\right)-\left(\frac{k_{\text{p}}}{k_{\text{f}}}-1\right)\phi }+5\times {10}^4{g}^{\prime }\left(\phi ,T,d_{\text{p}}\right)\phi {\rho }_{\text{f}}{c}_{\text{p,f}}\sqrt{\frac{{\kappa }_{\text{b}}T}{{\rho }_{\text{p}}{d}_{\text{p}}}}\hfill \\ g^{\prime }\left(\phi ,T,d_{\text{p}}\right)=\left(a_6+a_7\ln \left(d_{\text{p}}\right)+a_8\ln (\phi )+a_9\ln (\phi )\ln \left(d_{\text{p}}\right)+a_{10}\ln {\left(d_{\text{p}}\right)}^2\right)\hfill \\ +\ln (T)\left(a_1+a_2\ln \left(d_{\text{p}}\right)+a_3\ln (\phi )+a_4\ln (\phi )\ln \left(d_{\text{p}}\right)+a_5\ln {\left(d_{\text{p}}\right)}^2\right)\hfill \\ R_{\text{f}}+d_{\text{p}}/k_{\text{p}}=d_{\text{p}}/k_{\text{p,eff}},R_{\text{f}}=4\times {10}^{-8}\text{K}{\text{m}}^2\text{/W}\hfill \end{array}$

(3.126)

${\mu }_{\text{nf}}=\frac{{\mu }_{\text{f}}}{{\left(1-\phi \right)}^{2.5}}+\frac{k_{\text{Brownian}}}{k_{\text{f}}}\times \frac{{\mu }_{\text{f}}}{Pr}$

(3.127)

$u=\frac{\partial \psi }{\partial y}\text{and}\upsilon =-\frac{\partial \psi }{\partial x}\text{.}$

(3.128)

$\eta ={\left(\frac{a}{{\upsilon }_{\text{f}}}\right)}^{1/2}y,f\left(\eta \right)=\frac{\psi }{{\left(a{\upsilon }_{\text{f}}\right)}^{1/2}x},\theta \left(\eta \right)=\frac{T-T_{\infty }}{T_{\text{w}}-T_{\infty }}\text{.}$

(3.129)

$f^{‴}+f{f}^{″}-{f^{\prime }}^2+{\lambda }^2+\frac{M}{A_1}{A}_5\left(\lambda -f^{\prime }\right)=0\text{,}$

(3.130)

$\left(1+\frac{4}{3}\frac{Rd}{A_3}\right){\theta }^{″}+\mathrm{Pr}\left(\frac{A_4{A}_2}{A_3{A}_1}\right)\left(f{\theta }^{\prime }-n{f}^{\prime }\theta \right)=0\text{,}$

(3.131)

$\begin{array}{l}f(0)=0,f^{\prime }(0)=1,\theta (0)=1,\hfill \\ f^{\prime }(\infty )\to \lambda ,\theta (\infty )\to 0.\hfill \end{array}$

(3.132)

Here prime denote differentiation with respect to η, λ = b/a is the velocity ratio parameter, $Pr={\mu }_{\text{f}}{\left(\rho C_{\text{p}}\right)}_{\text{f}}/\left({\rho }_{\text{f}}{k}_{\text{f}}\right)$, Pr is the Prandtl number, $M={\sigma }_{\text{f}}{B}_0^2/\left({\rho }_{\text{f}}a\right)$ is the magnetic parameter and $Rd=4{\sigma }_{\text{e}}{T}_{\text{c}}^3/\left({\beta }_{\text{R}}{k}_{\text{f}}\right)$ is Radiation parameter. Also A₁, A₂, A₃, A₄, and A₅ are parameters having the following form:

$A_1=\frac{{\rho }_{\text{nf}}}{{\rho }_{\text{f}}},A_2=\frac{{\left(\rho C_{\text{p}}\right)}_{\text{nf}}}{{\left(\rho C_{\text{p}}\right)}_{\text{f}}},A_3=\frac{k_{\text{nf}}}{k_{\text{f}}},A_4=\frac{{\mu }_{\text{nf}}}{{\mu }_{\text{f}}},A_5=\frac{{\sigma }_{\text{nf}}}{{\sigma }_{\text{f}}}$

(3.133)

$\begin{array}{l}{C}_{\text{f}}=A_4{f}^{″}(0)\\ Nu=-A_3{\theta }^{\prime }(0)\end{array}$

(3.134)

3.6.1.2. Numerical method

$\left(\begin{array}{l}{x^{\prime }}_1\\ {x^{\prime }}_2\\ {x^{\prime }}_3\\ {x^{\prime }}_4\\ {x^{\prime }}_5\\ {x^{\prime }}_6\end{array}\right)=\left(\begin{array}{c}1\\ x_3\\ x_4\\ -x_2{x}_4+x_3^2-{\lambda }^2-\frac{M}{A_1}{A}_5\left(\lambda -x_3\right)\\ x_6\\ x_7\\ -Pr\left(\frac{A_4{A}_2}{A_3{A}_1}\right)\left(x_2{x}_6-n{x}_4{x}_5\right)/\left(1+\frac{4}{3}\frac{Rd}{A_3}\right)\end{array}\right)$

(3.135)

$\left(\begin{array}{l}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 1\\ u_1\\ 1\\ u_2\end{array}\right)$

(3.136)