Phase-ordering dynamics: numerical solution of the Mazenko equation in $D=2$ I'm considering the Mazenko equation as it's written in https://doi.org/10.1103/PhysRevB.46.10594 (eq. 7)
\begin{equation}
\label{a}
f''+\left(\frac{1}{x}+\frac x 4 \right)f'+\frac \lambda \pi \,\tan\left(\frac \pi 2 f \right)=0\tag{7}\end{equation}
with initial conditions
$f(0) = 1, f'(0) = -\underbrace{\frac {\sqrt{2\lambda}} \pi}_{\alpha(\lambda)}$.
While it is known that for $x\to 0$ the function behaves as
\begin{equation}f(x) \sim 1-\alpha(\lambda)x+O(x^3)\qquad x\to 0. \end{equation} I don't understand the way they authors numerically find a value for $\lambda$. They basically state that since $f(x)$ is small for $x\gg 1$ the equation can be linearised because it is
\begin{equation}
tan\left(\frac \pi 2 f \right) \sim \frac \pi 2 f
\end{equation}
which leads to a general expression 
\begin{equation}
f(x) \sim A(\lambda) \, x^{-(2-2\lambda)}\,\,e^{-x^2/8}+B(\lambda) \,x^{-2\lambda} \qquad x \gg 1.
\end{equation}
By matching the small-$x$ and large-$x$ behaviours it should be possible to numerically find a particular value ($\lambda_M\approx0.711$) of $\lambda$ such that it would be $B(\lambda)=0$ but I don't understand how. The original reference (https://doi.org/10.1103/PhysRevB.42.4487) doesn't seem to be particularly helpful.
 A: This is found using numerical integration of (eq. 7) and ensuring that only the exponentially decaying solution remains at long times. You can see it in the original paper at page 4499, part B. 4., the paragraph just after Eq 5.21
"But as explained above, for the band of p values of interest, the term proportional to $F_1$ is nonintegrable and
unphysical. Therefore, the branch of F determined by
$F(0)$ and $F'(0)$ must match onto $F_A$ with $F_1 =0$. Clearly
this reduces to a nonlinear eigenvalue problem where p
must be selected to give $F_1 =0$. Thus problem can be
solved relatively simply numerically and one obtains the
selected values..."
Also note that the parametrization is different, and in your case, $\lambda$ will be obtained from the $\mu$ parameter in the original paper by $\lambda = \pi /(4 \mu)$. For $\mu = 1.104$, this indeed yields $\lambda \simeq 0.711$.

By numerically integrating (eq.7) from $x=0$ to $x=50$ with steps of $dx = 0.005$ using a $RK4$ scheme, I find $\lambda \simeq 0.71129...$ (not sure about the precision, though, I could probably improve it by looking more finely at $\lambda$ values around $0.711$) - EDIT $\lambda \simeq 0.711277...$ after further investigation. This can be visualized quite nicely by the following graph, where I plot $log(|f(50)|)$ for different values of $\lambda$ ($100$ values between $0.6$ and $0.8$).

You can clearly see that just around the value of $\lambda$ for which the power-law term cancels, the value of $f(50)$ drops dramatically, as the only term remaining is exponentially small.
