I'm trying to reproduce a result from a paper (T. Thatcher, Boundary Conditions for Grad's 13 moment equations, equation (32), page 6), however, I haven't been able to do so. Hopefully someone can provide some help.
The general problem concerns the Couette flow: A flow passes through two infinite parallel plates at distance $L$ move with velocity $v_{w}^{0}$ and $v_{w}^{L}$ ($w$ refers to the wall) relative to each other in their respective planes, each with temperature $\theta_{w}^{0}$ and $\theta_{w}^{L}$. The velocity $v$ of the flow only depends on the $y$ coordinate, that is the coordinate perpendicular to the direction of the flow. After solving the Navier-Stokes-Fourier equations for this problem, you can find the following equations
$$\frac{dv}{dy}=a=\text{constant}, \qquad \frac{d^{2}\theta}{dy^{2}}=-\frac{2\text{Pr}}{5}a^{2} \qquad \qquad (1)$$
which are linearized. Using as boundary conditions
\begin{array}
vv(L)-v_{w}^{L} = -\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\; a, \qquad \
\
v(0)-v_{w}^{0} = \frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\; a \qquad (2)
\end{array}
and
\begin{array}
\theta \theta(L)-\theta_{w}^{L} = -\frac{2-\chi}{\chi}\frac{5\beta_{1}}{4 Pr}\sqrt{\frac{\pi}{2}}\text{Kn}\frac{d\theta}{dy}, \quad \
\
\theta(0)-\theta_{w}^{0} = \frac{2-\chi}{\chi}\frac{5\beta_{1}}{4 Pr}\sqrt{\frac{\pi}{2}}\text{Kn}\frac{d\theta}{dy} \qquad (3)
\end{array}
the solution for the velocity should be
$$v = \frac{v_{w}^{L}}{2}+a\left(\frac{y}{L}-\frac{1}{2}\right) \qquad \text{with} \quad a = \frac{v_{w}^{L}-v_{w}^{0}}{1+\frac{2-\chi}{\chi}\sqrt{2\pi} \alpha_{1}\text{Kn}} \qquad \qquad (4)$$
Don't be concerned about $\alpha_{1}$ or $\beta_{1}$; they are correction terms. $\text{Kn}$ and $\text{Pr}$ are, as usual, the Knudsen number and the Prandtl number, respectively.
For now, I'm interested only in the solution for the velocity (equation $(4)$)
Note: First and second attempts are wrong.
First attempt:
Use equation $(2)$ as $v-v_{w}^{L}=-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\;a L$ and $v-v_{w}^{0}=\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\;a L$. Subtract the first from the second to obtain $a = \displaystyle \frac{v_{w}^{L}-v_{w}^{0}}{\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$, and from equation $(1)$ for the velocity, $v = ay+\text{constant}$, evaluating at the plate localized in $L$ we have $v_{w}^{L}=aL+\text{constant}$, which gives us the constant, and a solution but it's not the correct result.
Second attempt:
I think this can be useful:
Take equation $(2)$ and solve for $a$:
$$a=\frac{-v_{w}^{L}+v_{w}^{0}}{-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$$
Then we use the first part of equation $(1)$, so we have
$$\frac{dv}{dy}=\frac{-v_{w}^{L}+v_{w}^{0}}{-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$$
After integrating this equation, we evalute the velocity at the boundary $L$, which allows us to find the constant of integration. Substituting this constant, we obtain the following result
$$v-v_{w}^{L}=\frac{v_{w}^{L}-v_{w}^{0}}{\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}}\left(\frac{y}{L}-1\right)$$
which is
$$v-v_{w}^{L}=a\left(\frac{y}{L}-1\right)$$
Not quite the result I'm looking for, but maybe this is the right path.
Third attempt:
Taking $\displaystyle \frac{dv}{dy}=b$, it follows $v(L)=a L +b$ and $v(0)=b$. Using the boundary conditions to evalute $v(L)$ and $v(0)$ we obtain
$$v_{w}^{0}+\displaystyle\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}a = b$$
and
$$v_{w}^{L}-\displaystyle \frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}a = aL+b$$
After solving this system of equations:
$$a = \frac{v_{w}^{L}-v_{w}^{0}}{L+\text{Kn}\sqrt{2\pi}\alpha_{1}(\frac{2-\chi}{\chi})}$$
$$b = v_{w}^{0}-\frac{\left(v_{w}^{0}-v_{w}^{L}\right)\text{Kn}\sqrt{\frac{\pi}{2}}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)}{L+\text{Kn}\sqrt{2\pi}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)}$$
Substituting back in the original equation
$$v(y)-v_{w}^{0}=a\left(y-\text{Kn}\sqrt{\frac{\pi}{2}}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)\right)$$
Best Answer
Your mistake was to misinterpret the equation for the boundary conditions (2). Since those two equations (one at each wall) are boundary conditions, they involve two different quantities v(y=0) and v(y=L). Unfortunately these were both written as v in the paper, which may have given you the impression that you could subtract the equations and thereby cancel v. However, that is not the case. The boundary conditions apply only at the boundaries, so they do not hold for v in general! The correct approach is to take the solution, which you know is v = ay + const. and to solve for the constants (a and const) by applying the boundary conditions to v(0) and v(L), that is to a(0) + const and to a(L) + const. That will give you two linear equations and two unknowns (const and a). Presumably this gives the answer in the paper. But first you should fix the rest of your typos in equation (4), since it does not match that of the paper!
Update: There is a typo in the paper. This is clear since the solution given in the paper does not obey $dv/dy = a$ but rather $dv/dy=a/L$. I believe the correct solution has (y-L/2) rather than (y/L - 1/2).