Small parameter expansion to obtain convective heat transfer solution of the heat equation I'm trying to follow along with this paper on perturbative transport analysis. The section I'm stuck on analyzes the heat equation in 1D slab geometry. To get a diffusion-only solution is relatively simple, but the paper then continues by doing a small parameter expansion to include the effect of convection and skips a couple of steps so that I get lost in the derivation and I don't know how to arrive at the same result. Can someone help me to fill in the blanks?
The starting equation is as follows (there is a reactive/damping term as well but for now I'd be happy just figuring out the convection so I've omitted it here):
$$\frac{\partial}{\partial t}u = D\frac{\partial^2}{\partial x^2}u + V \frac{\partial}{\partial x}u,$$
with boundary conditions $u(x=0,t) = u_0\rm{e}^{i\omega t}$ and $u(\infty,t) = 0.$
For the simple case with diffusion only ($V = 0$), the solution is given by:
$$u(x,t) = u_0\rm{e}^{-x/\lambda_d}\rm{e}^{i\left(\omega t - kx\right)}.$$
From this, you can see/derive that the amplitude decays with characteristic length $\lambda_d = \sqrt{\frac{2D}{\omega}}$ while the phase velocity $v_{\phi,d} = \omega/k$ is given by $v_{\phi,d} = \sqrt{2\omega D}$.
Now, we can quantify the relative importance of diffusion and convection by introducing normalized time $\theta = t/(\lambda_d/v_{\phi,d})$ and normalized distance $\xi=x/\lambda_d$:
$$\frac{\partial}{\partial \theta}u = 2\frac{\partial^2}{\partial \xi^2}u + \varepsilon_V \frac{\partial}{\partial \xi},$$
with the dimensionless parameter $\varepsilon_V = V/\sqrt{2D\omega}$.
Now, up to here, all is fine and dandy, and I can follow along just fine. However, what comes next is too big of a jump for me.
The paper goes on to say "[...] the effect of convection can be appreciated from an expansion in the small parameter $\varepsilon_V$", and gives the result as follows:
$$\lambda = \lambda_d\left(1-\varepsilon_V + O(\varepsilon_V^2)\right),$$
$$v_{\phi} = v_{\phi,d}\left(1-\frac{1}{4}\varepsilon_V^2 + O(\varepsilon_V^4)\right).$$
While I understand the principle of small parameter expansion, I can't figure out how to get to this result in this case. Can anyone show me how to get there?
 A: First, let me note that the dimensionless equation in the OP has an incorrect factor, it should read:
$$
\frac{\partial}{\partial \theta}u=\frac{1}{2}\frac{\partial^2}{\partial \xi^2}u+\varepsilon_V\frac{\partial}{\partial \xi}u
$$
The dimensionless solution at $V=0$ is
$$
u_0=e^\xi e^{i(\theta-\xi)}.
$$
The full equation can be solved exactly by first factoring out the time dependence,
$$
u(\xi, theta)=u(\xi)e^{i\theta}\Rightarrow iu(\xi)=\frac{1}{2}u''(\xi)+\varepsilon_V u'(\xi)
$$
One can then look for the solutions in the form $u\propto e^{\lambda \xi}$, which results in characteristic equation
$$
\lambda^2+2\varepsilon_V\lambda - 2i=0.
$$
If $\varepsilon_V=0$, teh roots of this equation are $\lambda=\pm(1+i)$, of which only the second gives us the solutions decaying for $\xi\rightarrow +\infty$.
The roots of the full characteristic equation can be approximated as
$$
\lambda_{1,2}=-\varepsilon_V\pm\sqrt{\varepsilon_V^2+2i}=
-\varepsilon_V \pm\sqrt{2i}\sqrt{1+\frac{\varepsilon^2}{2i}}\approx
-\varepsilon_V \pm (1+i)(1-\frac{\varepsilon^2}{4i}),
$$
where I used the Taylor expansion for the square root, $\sqrt{1+x}\approx 1 + x/2$.
Since $\varepsilon_V\ll 1$, the decaying solutions will be again given by the root with minus sign, i.e.,
$$
\lambda = -1-\varepsilon_V-\frac{\varepsilon_V^2}{4}-i(1-\frac{\varepsilon_V^2}{4})
$$
(The $\frac{\varepsilon_V^2}{4}$ in the real part can be omitted, since it is much smaller than the linear term in $\varepsilon_V$.)
The solution is then
$$
e^{-(1+\varepsilon_V)\xi-i(1-\frac{\varepsilon_V^2}{4})\xi}=
e^{-(1+\varepsilon_V)\frac{x}{\lambda_d}-i(1-\frac{\varepsilon_V^2}{4})\frac{\omega x}{v{\phi,d}}}=
e^{-\frac{x}{\lambda}-i\frac{\omega x}{v{\phi}}},
$$
from which we identify
$$
\lambda = \frac{\lambda_d}{1+\varepsilon_V}\approx\lambda_d(1-\varepsilon_V),\\
v_{\phi}=\frac{v_{\phi,d}}{1-\frac{\varepsilon_V^2}{4}}\approx
v_{\phi,d}(1+\frac{\varepsilon_V^2}{4})
$$

Let's write the equation as
$$
\frac{\partial}{\partial \theta}u-2\frac{\partial^2}{\partial \xi^2}u=\epsilon_V\frac{\partial}{\partial \xi}u = f(\xi,\theta)
$$
If $f(\xi,\theta)$ were a given function independent on $u$, this could be easily solved (a sum of the general solution of the homogeneous equation plus the particular solution of the inhomogeneous one).
In our case it depends on $u$, which can be represented as a Taylor expansion in the small parameter $\epsilon_V$:
$$
u(\xi,\theta)=\sum_{n=0}^{+\infty}\epsilon_V^n u_n(\xi,\theta),
$$
where $u_0(\xi,\theta)$ is the solution of the equation without the convection term, that we already know, and $u_n$ do not contain the small parameter $\epsilon_V$.
If we plug this into the expression for $f(\xi,\theta)$ we get
$$
f(\xi,\theta) = \epsilon_V\frac{\partial}{\partial \xi}u = 
\epsilon_V\frac{\partial}{\partial \xi}\sum_{n=0}^{+\infty}\epsilon_V^n
u_n(\xi,\theta)
\approx
\epsilon_V\frac{\partial}{\partial \xi}u_0(\xi,\theta),
$$
where in the last approximate equality I kept only the term of the first order in $\epsilon_V$. Thus, one needs to solve
$$
\frac{\partial}{\partial \theta}u-2\frac{\partial^2}{\partial \xi^2}u=\epsilon_V\frac{\partial}{\partial \xi}u_0.
$$
