Preface

This section discusses

Heat Transfer in Boundary Layer

The algorithm for solving boundary value problems for differential equations on infinite or semi-infinite intervals include four steps. The first step is to compute high-order Taylor series expansions using an algebraic manipulation language such as Maple or Mathematica. These expansions will contain one or more unknown parameters z which will be determined by the boundary condition at infinity. The second step is to convert the Taylor expansions into diagonal Padé approximants. The boundary condition that u(x) decays to zero at infinity becomes the condition that the coefficient of the highest power of x in the numerator polynomial must be zero. The third step is to solve this equation for the free parameter z. The final step is to evaluate each of the multiple solutions of this equation for physical plausibility and convergence (as N increases). Methods for nonlinear problems are almost always iterative and need a first guess to initialize the iteration. The Padé algorithm is unusual in that it is a direct method that requires no a priori information about the solution.

Heat Transfer in Boundary Layer

We consider the heat transfer problem in steady two-dimensional boundary layer stagnation-point flow of viscous incompressible fluid toawrds a stretching/shrinking sheet in the presence of nonuniform heat flux. The governing equations of motion and the nergy equations may be written in usual notations as follows (see Krishnendu Bhattacharyya, Heat transfer in boundary layer stagnation-point flow towards a shrinking sheet with non-uniform heat flux, Chinese Physics B, 22, 2013, 074705):

$\begin{split} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} &= 0, \\ u\, \frac{\partial u}{\partial x} + v\, \frac{\partial u}{\partial y} &= U_s \, \frac{{\text d} U_s}{{\text d} x} + \nu\,\frac{\partial^2 u}{\partial y^2} , \\ u\, \frac{\partial T}{\partial x} + v\, \frac{\partial T}{\partial y} &= \frac{\kappa}{\rho\,c_p}\,\frac{\partial^2 T}{\partial y^2} , \end{split}$
where u and v are velocity components in the x direction and y direction, respectively, ν (= μ/ρ) is the kinematic viscosity of fluid, ρ is the density of fluid, μ is the coefficient of fluid viscosity, Us is the straining velocity, T is the temperature, κ is the thermal conductivity, cp is the specific heat, x is the distance along the sheet, and y is the distance perpendicular to the sheet.

The appropriate boundary conditions for the velocity components and temperature are as follows:

$\begin{split} u&= U_w = cx, \quad v=0 \quad\mbox{ at } \ y=0, \\ u&\to U_s = ax \quad \mbox{as } \ y \to \infty , \\ \frac{\partial T}{\partial y} &= - \frac{q_w}{\kappa} = -\frac{\lambda}{\kappa}\,x^m \quad\mbox{ at } \ y=0, \\ T&\to T_{\infty} \quad \mbox{as } \ y \to \infty , \end{split}$
where Uw is the shrinking velocity of the sheet, c(<0) and a(>0) are the shrinking constant and straining constant, respectively, qw = λxm is the nonuniform heat flux, $$T_{\infty}$$ is the free stream temperature assumed to be constant, λ is a given constant, and m is the heat flux parameter.

Upon introducing teh similarity transformations

$\begin{split} \psi &= \sqrt{a\nu}\,x\,f(\eta ) , \quad T- T_{\infty} = \frac{q_w}{\kappa} \left( \frac{\nu}{a} \right)^{1/2} \theta (\eta ) , \\ \eta &= y\,\left( \frac{a}{\nu} \right)^{1/2} , \end{split}$
where ψ is the stream function defined through its partial derivatives as $$u = \partial \psi / \partial y \quad\mbox{and}\quad v = - \partial \psi / \partial x,$$ η is the similarity variable, f is a dimensionless stream function, and θ is a dimensionless temperature, the governing equations become
$\begin{split} f''' + f\, f'' - (f')^2 +1 &=0 , \\ \theta'' + \mbox{Pr} \left( f\,\theta' - m\,f' \,\theta \right) &=0 . \end{split}$
Here primes denote differentiation with respect to η and Pr = cpμ/κ is the Prandtl number, and f' is the dimensionless velocity. The boundary conditions also reduce to
$\begin{split} f(0) &= 0, \quad f' (0) =B , \\ f' &\to 1 \quad \mbox{as } \ \eta \to \infty , \\ \theta'(0) &= -1, \\ \theta (\theta) &\to 0 \quad \mbox{as } \ \eta \to \infty , \end{split}$
where B = c/a is the velocity of ratio parameters.

We convert the given problem for f to a regular boundary value problem by choosing a large number L so that get

$\begin{split} \texttt{D}^3 f + N \left[ f\right] +1 &=0 , \\ f(0) &= 0, \quad f' (0) =B , \\ f' (L)&= 1, \end{split}$
where $$\texttt{D} = {\text d} / {\text d}x$$ is the derivative operator (we change independent variable η to x for x for convenience) and the nonlinear operator is
$N \left[ f \right] = f\, f'' - (f')^2 .$
According to ADM method, the solution is approximated by the series
$f(x) = \sum_{n\ge 0} u_n (x),$
where coefficients in the above series are determined recursively.

To derive the recurrence, we consider generating functions for f and for nonlinear operator:

$Y(x,\lambda ) = \sum_{n\ge 0} u_n (x)\, \lambda^n \qquad \mbox{and} \qquad N \left[ Y(x,\lambda ) \right] = \sum_{n\ge 0} A_n \left( u_0 , u_1 , \ldots , u_n \right) \lambda^n .$
The first term, u0 incorporates the nonhomogeneous term and the given boundary conditions:
$u'''_0 +1 =0, \qquad u_0 (0) =0, \quad u'_0 (0) =B, \quad u'_0 (L) =1 .$
DSolve[{u'''[x] + 1 == 0, u[0] == 1, u'[0] == B, u'[L] == 1}, u[x], x]
Out[1]= {{u[x] -> (12 L + 12 B L x + 6 x^2 - 6 B x^2 + 3 L^2 x^2 - 2 L x^3)/( 12 L)}}
So
$u_0 (x) = 1 + B\, x + \frac{1}{4L} \left( 2 - 2B + L^2 \right) x^2 - \frac{1}{6}\, x^3 .$
Substituting the generating function into the differential equation and boundary conditions, we obtain the sequence of boundary value problems with respect to linear differential equation of the third order:
$u'''_1 + u_0 u''_0 - \left( u_0' \right)^2 =0 , \qquad u_1 (0) =0, \quad u'_1 (0) =0, \quad u'_1 (L) =0 .$
We solve this boundary problem with mathematica:
u[0][x_] = (12 L + 12 B L x + 6 x^2 - 6 B x^2 + 3 L^2 x^2 - 2 L x^3)/( 12 L)
DSolve[{u'''[x] + 1 u[0][x]*D[u[0][x], x, x] - (D[u[0][x], x])^2 == 0, u[0] == 0, u'[0] == B, u'[L] == 0}, u[x], x]
Out[3]= {{u[x] -> (1/( 20160 L^2))(20160 B L^2 x - 10080 B L x^2 + 5040 L^2 x^2 - 5040 B L^2 x^2 - 420 L^3 x^2 - 840 B L^3 x^2 - 3780 B^2 L^3 x^2 + 840 L^4 x^2 - 252 L^5 x^2 - 588 B L^5 x^2 - 49 L^7 x^2 - 3360 L x^3 + 3360 B L x^3 + 3360 B^2 L^2 x^3 - 1680 L^3 x^3 + 840 B L x^4 - 840 B^2 L x^4 + 840 L^2 x^4 + 420 B L^3 x^4 + 168 x^5 - 336 B x^5 + 168 B^2 x^5 + 168 L^2 x^5 - 168 B L^2 x^5 + 42 L^4 x^5 - 56 L x^6 + 56 B L x^6 - 28 L^3 x^6 + 8 L^2 x^7)}}
So we define the next Adomian approximation:
u[1][x_] = (1/( 20160 L^2))(20160 B L^2 x - 10080 B L x^2 + 5040 L^2 x^2 - 5040 B L^2 x^2 - 420 L^3 x^2 - 840 B L^3 x^2 -
3780 B^2 L^3 x^2 + 840 L^4 x^2 - 252 L^5 x^2 - 588 B L^5 x^2 - 49 L^7 x^2 - 3360 L x^3 + 3360 B L x^3 + 3360 B^2 L^2 x^3 -
1680 L^3 x^3 + 840 B L x^4 - 840 B^2 L x^4 + 840 L^2 x^4 + 420 B L^3 x^4 + 168 x^5 - 336 B x^5 + 168 B^2 x^5 +
168 L^2 x^5 - 168 B L^2 x^5 + 42 L^4 x^5 - 56 L x^6 + 56 B L x^6 - 28 L^3 x^6 + 8 L^2 x^7)

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