In this section we are interested in getting an approriate starting (or initial) solution for the iteration scheme (5). For this purpose, and under the assumptions
$$\begin{array}{}{\displaystyle {v}_{2}\equiv 0,\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{\mathrm{\partial}{v}_{1}}{\mathrm{\partial}{x}_{1}}\equiv 0,\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{\mathrm{\partial}\theta}{\mathrm{\partial}{x}_{1}}\equiv 0,}\end{array}$$(9)

we calculate for given values *F*_{1}, *F*_{2}, *R*, *θ*_{0} = 0, *θ*_{1}, *θ*_{2} and associated rheological parameters the flow fields and values *v*(*x*), *p*(*x*), *θ* (*x*), *p*_{0}, *h*_{∞}, *θ*_{∞}. The value *θ*_{∞} which has not been defined before describes the (asymptotic) value of the temperature *θ* at the free interface when *x*_{1} goes to +∞. Let us emphasize that the assumptions guarantee solution fields that are uniform and unidirectional (not depending on main-stream direction *x*_{1}).

Under the assumptions (9) the governing Eqs. (1) take the subsequent reduced form:
$$\begin{array}{}{\displaystyle -\nu \phantom{\rule{thinmathspace}{0ex}}{\mathbf{\nabla}}^{2}\phantom{\rule{thinmathspace}{0ex}}{v}_{1}+\frac{1}{\eta}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{\partial}p}{\mathrm{\partial}{x}_{1}}=0,\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{\mathrm{\partial}p}{\mathrm{\partial}{x}_{2}}=-\eta \phantom{\rule{thinmathspace}{0ex}}g,\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}-\lambda \phantom{\rule{thinmathspace}{0ex}}{\mathbf{\nabla}}^{2}\phantom{\rule{thinmathspace}{0ex}}\theta =0,}\end{array}$$

where the second equation replaces the continuity Eq. (1)_{2} Now it is possible to divide the original problem into three independent problems for the flow fields. Let us start with the problem for velocities **v**:
$$\begin{array}{}\left\{\begin{array}{ll}{\nu}_{1}{\eta}_{1}\phantom{\rule{thinmathspace}{0ex}}\frac{{\mathrm{d}}^{2}{v}_{1}^{(1)}}{\mathrm{d}{x}_{2}^{2}}=\frac{\mathrm{\partial}{p}^{(1)}}{\mathrm{\partial}{x}_{1}},& {\nu}_{2}{\eta}_{2}\phantom{\rule{thinmathspace}{0ex}}\frac{{\mathrm{d}}^{2}{v}_{1}^{(2)}}{\mathrm{d}{x}_{2}^{2}}=\frac{\mathrm{\partial}{p}^{(2)}}{\mathrm{\partial}{x}_{1}},\\ {v}_{1}^{(1)}(0)=R,& {v}_{1}^{(2)}(1)=0,\\ {v}_{1}^{(1)}{|}_{{x}_{2}={h}_{\mathrm{\infty}}}={v}_{1}^{(2)}{|}_{{x}_{2}={h}_{\mathrm{\infty}}},& {\nu}_{1}{\eta}_{1}{\left.\frac{\mathrm{d}{v}_{1}^{(1)}}{\mathrm{d}{x}_{2}}\right|}_{{x}_{2}={h}_{\mathrm{\infty}}}={\nu}_{2}{\eta}_{2}{\left.\frac{\mathrm{d}{v}_{1}^{(2)}}{\mathrm{d}{x}_{2}}\right|}_{{x}_{2}={h}_{\mathrm{\infty}}},\\ {\int}_{0}^{{h}_{\mathrm{\infty}}}{v}_{1}({x}_{2})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}{x}_{2}={F}_{1},& {\int}_{{h}_{\mathrm{\infty}}}^{1}{v}_{1}({x}_{2})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}{x}_{2}={F}_{2}.\end{array}\right.\end{array}$$(10)

For the pressure *p* one obtains the following equations
$$\begin{array}{}\left\{\begin{array}{lll}\frac{\mathrm{\partial}{p}^{(1)}}{\mathrm{\partial}{x}_{2}}=-{\eta}_{1}\phantom{\rule{thinmathspace}{0ex}}g,& & \frac{\mathrm{\partial}{p}^{(2)}}{\mathrm{\partial}{x}_{2}}=-{\eta}_{2}\phantom{\rule{thinmathspace}{0ex}}g,\\ & {p}^{(1)}{|}_{{x}_{2}={h}_{\mathrm{\infty}}}={p}^{(2)}{|}_{{x}_{2}={h}_{\mathrm{\infty}}}.& \end{array}\right.\end{array}$$(11)

Finally, the problem for temperature *θ* reads
$$\begin{array}{}\left\{\begin{array}{ll}\frac{{\mathrm{d}}^{2}{\theta}^{(1)}}{\mathrm{d}{x}_{2}^{2}}=0,& \frac{{\mathrm{d}}^{2}{\theta}^{(2)}}{\mathrm{d}{x}_{2}^{2}}=0,\\ {\theta}^{(1)}{|}_{{x}_{2}=0}={\theta}_{0}=0,& {\theta}^{(2)}{|}_{{x}_{2}=1}={\theta}_{2},\\ {\theta}^{(1)}{|}_{{x}_{2}={h}_{\mathrm{\infty}}}={\theta}^{(2)}{|}_{{x}_{2}={h}_{\mathrm{\infty}}},& {\lambda}_{1}{\left.\frac{\mathrm{d}{\theta}^{(1)}}{\mathrm{d}{x}_{2}}\right|}_{{x}_{2}={h}_{\mathrm{\infty}}}={\lambda}_{2}{\left.\frac{\mathrm{d}{\theta}^{(2)}}{\mathrm{d}{x}_{2}}\right|}_{{x}_{2}={h}_{\mathrm{\infty}}},\end{array}\right.\end{array}$$(12)

In Eqs. (10), (11), (12) the superscripts (*k*), (*k* = 1, 2) or (+) denote the corresponding fluid layer and the subregion *x*_{1} ⩾ 1. The solutions of these three (independent) problems are of NUSSELT type (cf. also [19]) and allow the representation
$$\begin{array}{}\left\{\begin{array}{l}{v}_{1}^{(+)}({x}_{2})=\left\{\begin{array}{ll}\phantom{\rule{thinmathspace}{0ex}}0.5{a}_{1}{x}_{2}^{2}+{b}_{1}{x}_{2}+R,& 0\u2a7d{x}_{2}\u2a7d{h}_{\mathrm{\infty}}\\ \phantom{\rule{thinmathspace}{0ex}}0.5{a}_{2}({x}_{2}^{2}-1)+{b}_{2}({x}_{2}-1),& {h}_{\mathrm{\infty}}\u2a7d{x}_{2}\u2a7d1\end{array},\right.\\ {v}_{2}^{(+)}({x}_{2})\equiv 0,\phantom{\rule{2em}{0ex}}{p}_{0}={a}_{1}{\nu}_{1}{\eta}_{1}={a}_{2}{\nu}_{2}{\eta}_{2},\phantom{\rule{2em}{0ex}}r=({\nu}_{1}{\eta}_{1})/({\nu}_{2}{\eta}_{2}),\end{array}\right.\end{array}$$(13)

The coefficients in (13) are given by
$$\begin{array}{c}{\displaystyle {a}_{1}=\left[-3\frac{{F}_{1}-R{h}_{\mathrm{\infty}}}{{h}_{\mathrm{\infty}}^{2}}-3\frac{{F}_{2}}{r(1-{h}_{\mathrm{\infty}}^{2})}\right],\phantom{\rule{1em}{0ex}}{a}_{2}=r\phantom{\rule{thinmathspace}{0ex}}{a}_{1},}\\ \\ {\displaystyle {b}_{1}=\left[(2+{h}_{\mathrm{\infty}})\frac{{F}_{1}-R{h}_{\mathrm{\infty}}}{{h}_{\mathrm{\infty}}^{2}}+{h}_{\mathrm{\infty}}\frac{{F}_{2}}{r(1-{h}_{\mathrm{\infty}}^{2})}\right],\phantom{\rule{1em}{0ex}}{b}_{2}=r\phantom{\rule{thinmathspace}{0ex}}{b}_{1}.}\end{array}$$

Note, that the values *h*_{∞} and *p*_{0} are already known for these expressions (see Eq. (16) below). That is why it follows *θ*_{∞} = (*λ*_{2} *θ*_{2} *h*_{∞})/[*λ*_{1} (1−*h*_{∞}) + *λ*_{2} *h*_{∞}] and for the complete temperature and pressure fields one obtains
$$\begin{array}{}{\theta}^{(+)}({x}_{2})=\left\{\begin{array}{ll}\phantom{\rule{thinmathspace}{0ex}}\frac{{\theta}_{\mathrm{\infty}}}{{h}_{\mathrm{\infty}}}\phantom{\rule{thinmathspace}{0ex}}{x}_{2},& 0\u2a7d{x}_{2}\u2a7d{h}_{\mathrm{\infty}}\\ \phantom{\rule{thinmathspace}{0ex}}{\theta}_{\mathrm{\infty}}+\frac{{\theta}_{2}-{\theta}_{\mathrm{\infty}}}{1-{h}_{\mathrm{\infty}}}\phantom{\rule{thinmathspace}{0ex}}{x}_{2}-\frac{{\theta}_{2}-{\theta}_{\mathrm{\infty}}}{1-{h}_{\mathrm{\infty}}}\phantom{\rule{thinmathspace}{0ex}}{h}_{\mathrm{\infty}},& {h}_{\mathrm{\infty}}\u2a7d{x}_{2}\u2a7d1,\end{array}\right.\end{array}$$(14)
$$\begin{array}{}{p}^{(+)}(\mathit{x})=\left\{\begin{array}{l}{p}_{0}{x}_{1}-{\eta}_{1}g{x}_{2}+k,\\ {p}_{0}{x}_{1}-{\eta}_{2}g({x}_{2}-{h}_{\mathrm{\infty}})-{\eta}_{1}g{h}_{\mathrm{\infty}}+k,\end{array}\right.\end{array}$$(15)

Since the associated linear problem is completely decomposed, we got the same polynomial equation for the determination of the value *h*_{∞} as in the former paper [17].
$$\begin{array}{}0=r(r-1)R{h}_{\mathrm{\infty}}^{5}+\left[-4r(r-1)R-r(r-1){F}_{1}-(r-1){F}_{2}\right]{h}_{\mathrm{\infty}}^{4}\\ +\left[r(6r-5)R+2r(2r-3){F}_{1}-2r{F}_{2}\right]{h}_{\mathrm{\infty}}^{3}+[2r(-2r+1)R\\ +3r(-2r+3){F}_{1}+3r{F}_{2}]{h}_{\mathrm{\infty}}^{2}+\left[{r}^{2}R+4r(r-1){F}_{1}\right]{h}_{\mathrm{\infty}}-{r}^{2}{F}_{1}.\end{array}$$(16)

In [17] the subsequent two lemmas were proved.

#### Lemma 4.1

*If F*_{1} *F*_{2} > 0, *then Eq*. (16) *has at least one root h*_{∞} *within the open interval* ]0, 1[.

#### Lemma 4.2

*If F*_{1} *F*_{2} ⩾ 0, *then Eq*. (16) *has at most three different roots h*_{∞} ∈]0, 1[.

Note that in the subregion
$\begin{array}{}{\mathit{\Omega}}^{(-)}:={\mathit{\Omega}}_{1}^{(-)}\cup {\mathit{\Omega}}_{2}^{(-)},\end{array}$ i.e. for *x*_{1} ⩽ −1, the corresponding problems are even simpler due to the fact that there is no free boundary. In order not to repeat simple things we restrict the presentation to the basic solution *v*^{(−)}, p^{(−)}, *θ*^{(−)} in the double - channel which can also be determined very simple by straightforward calculations in the left part *Ω*^{−} of the (double) - channel. The corresponding velocities and temperatures do not depend on *x*_{1}. In
$\begin{array}{}{\mathit{\Omega}}_{1}^{-}\end{array}$ one obtains
$$\begin{array}{}\left\{\begin{array}{l}{v}_{1}^{(-)}({x}_{2})=\left(\frac{3R}{{h}_{1}^{2}}-\frac{6{F}_{1}}{{h}_{1}^{3}}\right){x}_{2}^{2}+\left(-\frac{4R}{{h}_{1}}+\frac{6{F}_{1}}{{h}_{1}^{2}}\right){x}_{2}+R,\\ {v}_{2}^{(-)}(\mathit{x})\equiv 0,\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{\theta}^{(-)}({x}_{2})=\frac{{x}_{2}{\theta}_{1}}{{h}_{1}},\\ {p}^{(-)}(\mathit{x})=2{\nu}_{1}{\eta}_{1}\left(\frac{3R}{{h}_{1}^{2}}-\frac{6{F}_{1}}{{h}_{1}^{3}}\right){x}_{1}-{\eta}_{1}g{x}_{2}+{k}_{1}.\end{array}\right.\end{array}$$(17)

In
$\begin{array}{}{\mathit{\Omega}}_{2}^{-}\end{array}$ one gets, respectively,
$$\begin{array}{}\left\{\begin{array}{l}{v}_{1}^{(-)}({x}_{2})=-\frac{6{F}_{2}}{(1-{h}_{1}{)}^{3}}{x}_{2}^{2}+\frac{6(1+{h}_{1}){F}_{2}}{(1-{h}_{1}{)}^{3}}{x}_{2}-\frac{6{h}_{1}{F}_{2}}{(1-{h}_{1}{)}^{3}},\\ {v}_{2}^{(-)}(\mathit{x})\equiv 0,\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{\theta}^{(-)}({x}_{2})={\theta}_{1}+\frac{({x}_{2}-{h}_{1})}{(1-{h}_{1})}({\theta}_{2}-{\theta}_{1}),\\ {p}^{(-)}(\mathit{x})=-\frac{12{\nu}_{2}{\eta}_{2}{F}_{2}}{(1-{h}_{1}{)}^{3}}{x}_{1}-{\eta}_{2}g{x}_{2}+{k}_{2}.\end{array}\right.\end{array}$$(18)

It is well-known that the pressure *p* can be determined only up to an additive constant in channel flows (cf. *k*_{1}, *k*_{2} in formulae (17), (18)).

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