I bid you good day.
I have a question about this paper which can be found here: https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1957.0157
I am working through the paper as an exercise. The paper studies the boundary layer flow over a rotating cylinder. The paper states the boundary layer equations and continuity equations respectively,
$ u \frac{\partial u} { \partial x} + v \frac{\partial u} { \partial y}= U\frac{dU}{dx}+\nu \frac{\partial^2 y}{\partial y^2 } $
$ \frac{\partial u} { \partial x} + \frac{\partial v}{\partial y}=0 $
where $ U=Q+2U_0\cos (\frac {2x}{d})$
and boundary conditions
$ u=q , v=0$ at $y=0$
$u \rightarrow U $ as $y\rightarrow \infty $
Here $q$ is the rotational velocity of the cylinder, $U_0$ is the free stream velocity, $d$ is the diameter of the cylinder,$\nu$ is the kinematic viscosity, and $Q$ is the circulatory component of the velocity at the edge of the boundary layer.
By the continuity equations there is a stream function that satisfies the following relationship
$u= \frac{\partial \psi}{\partial y} $ and $v= -\frac{\partial \psi}{\partial x} $
Glauert assumes the solution has the form
$\frac1Q\psi =y+\alpha f_1(y)e^{i\theta}+\alpha^2[f_2(y)e^{2i\theta}+g_2(y)]+O(\alpha^3)$
Then
$\frac1Q u=1+\alpha f_1'e^{i\theta}+\alpha^2[f_2'e^{2i\theta}+g_2']$
$\frac{1}{Q} v = -i\alpha f_1e^{i\theta}-2 i \alpha^2 f_2e^{2i\theta} $
Also, rewrite $U=Q(1+\alpha e^{1\theta })$.
The paper carries out the calculation up to the fourth order of $\alpha$ however, my question comes from the second order ODEs.
The derivatives that are to be inserted into boundary layer equations are as followed
$\frac{1}{Q}\frac{dU}{dx}=i\alpha e^{i\theta }$
$\frac{1}{Q}\frac {\partial u}{\partial x}= i\alpha f_1' e^{i\theta}+2i\alpha^2 f_2' e^{2i\theta}$
$\frac1Q \frac{\partial u}{\partial y}=\alpha f_1''e^{i\theta}+\alpha^2[f_2''e^{2i\theta}+g_2'']$
$\frac1Q \frac{\partial^2 u}{\partial y^2}=\alpha f_1'''e^{i\theta}+\alpha^2[f_2'''e^{2i\theta}+g_2''']$
After inserting these derivatives and the velocities into the boundary layer equations, multiplying both sides by $\frac{d}{2i}$ for the chain rule due to the change of coordinates from $x$ to $\theta$, collecting the terms of order $\alpha$ and simplifying you obtain
$f_1'=1+\frac{1}{\gamma^2}f_1''' \space \rightarrow f_1'''=\gamma^2 (f_1'-1)$
where $\gamma^2=\frac{2Qi}{\nu d}$. This parameter is like a Reynolds number. Solving for $\gamma=(1+i)\beta$ where $\beta=\sqrt\frac{Q}{\nu d}$
The solution to $f_1=y+\frac1\gamma (e^{-\gamma y}-1)$ that was solved using the boundary conditions $f_1(0)=f_1'(0)=0$ and $f_1'(\infty)=1$
When collecting terms of $\alpha^2$ you obtain the following ODE
$2f_2' e^{2i\theta}+f_1' e^{i\theta} f_1' e^{i\theta}-f_1 e^{i\theta} f_1'' e^{i\theta}=e^{i\theta} e^{i\theta}+\frac{1}{\gamma^2}(f_2'''e^{2i\theta}+g_2''')$
Reorganizing the terms $2f_2' e^{2i\theta}+f_1' e^{i\theta} f_1' e^{i\theta}-f_1 e^{i\theta} f_1'' e^{i\theta}=e^{i\theta} e^{i\theta}+\frac{1}{\gamma^2}(f_2'''e^{2i\theta}+g_2''')$
$\frac{1}{\gamma^2}f_2'''e^{2i\theta} - 2f_2' e^{2i\theta} +\frac{1}{\gamma^2}g_2''' = f_1' e^{i\theta} f_1' e^{i\theta}-f_1 e^{i\theta} f_1'' e^{i\theta}-e^{i\theta} e^{i\theta}$
The paper then states that care must be taken into consideration when calculating terms quadratic in $\alpha$ because if $A$ and $B$ are complex numbers,
$\Re(A)\Re(B)=\Re[\frac12A(B+\bar B)]$ where $\bar B$ denotes the complex conjugate therefore terms independent of $\theta$ arise as well as terms proportional to $e^{2i\theta}$. Applying this rule to each $f_1$ term
$\Re(f_1' e^{i\theta})\Re(f_1' e^{i\theta})= \Re[\frac12 f_1' e^{i\theta}(f_1' e^{i\theta}+\bar f_1' e^{-i\theta})]=\frac12 f_1'f_1' e^{2i\theta}+ \frac{1}{2} f_1' \bar f_1'$
$\Re(f_1 e^{i\theta})\Re(f_1'' e^{i\theta})= \Re[\frac12 f_1'' e^{i\theta}(f_1 e^{i\theta}+\bar f_1 e^{-i\theta})]=\frac12 f_1f_1'' e^{2i\theta}+ \frac{1}{2} \bar f_1 \ f_1''$
$\Re(e^{i\theta})\Re(e^{i\theta})= \Re[\frac12 e^{i\theta}(f_1' e^{i\theta}+e^{-i\theta})]=\frac12 e^{2i\theta}+ \frac12$
Inserting these back in you get
$\frac{1}{\gamma^2}f_2'''e^{2i\theta} - 2f_2' e^{2i\theta} +\frac{1}{\gamma^2}g_2''' = \frac12 f_1'f_1' e^{2i\theta}+ \frac{1}{2} \bar f_1' f_1'-(\frac12 f_1f_1'' e^{2i\theta}+ \frac{1}{2} \bar f_1 \ f_1'')-( \frac12 e^{2i\theta}+ \frac12)$
Separating terms proportional to $e^{21\theta}$ and not proportional to $\theta$ you get
$\frac{1}{\gamma^2}f_2''' - 2f_2' = \frac12 f_1'f_1' -\frac12 f_1f_1'' -\frac12$
$\frac{1}{\gamma^2}g_2''' = \frac{1}{2} \bar f_1' f_1'-\frac{1}{2} \bar f_1 \ f_1''-\frac12$
Here are my questions:
1) Why did the author choose to conjugate $f_1$ instead of $f_1''$ on the $ f_1 f_1''$ term to get $ \bar f_1 f_1''$ rather than $ f_1 \bar f_1''$ ?
2) With that same term, as you can see from my calculations that term has a negative sign whereas in the paper (equation 18) the $\bar f_1 f_1''$ is positive. How did it jump from a negative to a positive? Am I missing something when doing that rule that is stated for the quadratic term. Below is what the paper has.
$\frac{1}{\gamma^2}g_2''' = \frac{1}{2} \bar f_1' f_1'+\frac{1}{2} \bar f_1 \ f_1''-\frac12$
3) Last question unrelated to the math of the problem; is there a name for the method used to solve this PDE? I know its a perturbative method but does it have a particular name?
Thank you for any input.