# Revisiting the Meandering Instability During Step-Flow Epitaxy
## 2 Stability Analysis
### The \(\Theta\) Dependence of the Stability Function \(\omega(k)\)  
Now we consider the general situation that the kinetic coefficients for the attachment-detachment of adatoms from the lower and upper terrace onto the step edge are finite, that is, \(d_{\pm}\) are finite. To find the solution \(u\) in the form of (13), we first need to obtain \(u_{0}(z)=A\sinh(z/x_{s})+B\cosh(z/x_{s})\) for the case of perfectly straight steps where the curvature \(\kappa=0\) and the normal vector \(n=(0,1)^{T}\). Notice that  
\[u_{0}^{+}=u_{0}|_{z=0}=B,\] \[u_{0}^{-}=u_{0}|_{z=l}=A\sinh(l/x_{s})+B\cosh(l/x_{s}),\] \[(\partial_{n}u_{0})^{+}=(\partial_{n}u_{0})|_{z=0}=\frac{A}{x_{s }},\] \[(\partial_{n}u_{0})^{-}=(\partial_{n}u_{0})|_{z=l}=\frac{A}{x_{s }}\cosh(l/x_{s})+\frac{B}{x_{s}}\sinh(l/x_{s}).\]  
Consequently, the boundary condition (9) for the straight step case becomes  
\[M\frac{A}{x_{s}} = C_{+}(B+\Phi-\Theta[B-A\sinh(l/x_{s})-B\cosh(l/x_{s})]),\] \[-M\left(\frac{A}{x_{s}}\cosh(l/x_{s})+\frac{B}{x_{s}}\sinh(l/x_{s })\right) = C_{-}\bigg{(}A\sinh(l/x_{s})+B\cosh(l/x_{s})+\Phi \tag{16}\] \[-\Theta\left(B-A\sinh(l/x_{s})-B\cosh(l/x_{s})\right)\bigg{)}.\]  
\(A\) and \(B\) in \(u_{0}\) can be solved from the linear system (16). Now we allow the accuracy up to the first order of \(\epsilon\). We have \(u^{+}=u|_{z=\epsilon\sin(kx)}\) and \(u^{-}=u|_{z=l+\epsilon\sin(kx)}\). Then the first-order Taylor expansion about \(\epsilon\) gives that  
\[ \begin{array}{ll}u^{+}&=B+\epsilon\sin(kx)(\frac{A}{x_{s}}+B_{k}),\\ u^{-}&=\frac{A}{x_{s}}\sinh(l/x_{s})+\frac{B}{x_{s}}\cosh(l/x_{s})+\epsilon \sin(kx)\bigg{[}\frac{A}{x_{s}}\cosh(l/x_{s})+\frac{B}{x_{s}}\sinh(l/x_{s})\\ &+A_{k}\sinh(\Lambda_{k}l)+B_{k}\cosh(\Lambda_{k}l)\bigg{]},\\ \partial_{n}u^{+}&=\frac{A}{x_{s}}+\epsilon\sin(kx)\left(\frac{B}{x_{s}^{2}}+ A_{k}\Lambda_{k}\right),\\ \partial_{n}u^{-}&=\frac{A}{x_{s}}\cosh(l/x_{s})+\frac{B}{x_{s}}\sinh(l/x_{s}) +\epsilon\sin(kx)\bigg{[}\frac{A\sinh(l/x_{s})}{x_{s}^{2}}+\frac{B\cosh(l/x_{ s})}{x_{s}^{2}}\\ &+A_{k}\Lambda_{k}\cosh(\Lambda_{k}l)+B_{k}\Lambda_{k}\sinh(\Lambda_{k}l) \bigg{]}.\end{array}  \tag{17}\]  
Substituting (17) into the boundary condition (9) and equating the like powers of \(\epsilon\) show that, for the first order in \(\epsilon\), we have  
\[ \begin{array}{ll}&M\left(\frac{B}{x_{s}^{2}}+A_{k}\Lambda_{k}\right)=C_{+} \bigg{(}\frac{A}{x_{s}}+B_{k}-\Theta\big{(}\frac{A}{x_{s}}+B_{k}-\frac{A}{x_{s }}\cosh(l/x_{s})\\ &-\frac{B}{x_{s}}\sinh(l/x_{s})-A_{k}\sinh(\Lambda_{k}l)-B_{k}\cosh(\Lambda_{k }l)\big{)}-\Gamma k^{2}\bigg{)},\\ &-M\bigg{(}\frac{A}{x_{s}^{2}}\sinh(l/x_{s})+\frac{B}{x_{s}^{2}}\cosh(l/x_{s}) +A_{k}\Lambda_{k}\cosh(\Lambda_{k}l)+B_{k}\Lambda_{k}\sinh(\Lambda_{k}l)\bigg{)} =\\ &C_{-}\bigg{(}\frac{A}{x_{s}}\cosh(l/x_{s})+\frac{B}{x_{s}}\sinh(l/x_{s})+A_{ k}\sinh(\Lambda_{k}l)+B_{k}\cosh(\Lambda_{k}l)\\ &-\Theta\big{(}\frac{A}{x_{s}}+B_{k}-\frac{A}{x_{s}}\cosh(l/x_{s})-\frac{B}{x_ {s}}\sinh(l/x_{s})-A_{k}\sinh(\Lambda_{k}l)-B_{k}\cosh(\Lambda_{k}l)\big{)}- \Gamma k^{2}\bigg{)}.\end{array}  \tag{18}\]  
The coefficients \(A_{k}\) and \(B_{k}\) can be solved from the linear system (18). Therefore, we obtain the solution \(u\) in the form of (13). Substituting \(u\) into (10), we can compute the velocity \(V\) of a single step in the form of  
\[V=V_{0}+\omega(k)\epsilon\sin(kx).\]  
Here  
\[V_{0}=MN\left(\frac{A}{x_{s}}-\frac{A}{x_{s}}\cosh(l/x_{s})-\frac{B}{x_{s}} \sinh(l/x_{s})\right) \tag{19}\]  
and  
\[\omega(k)=MN\left(\frac{B}{x_{s}^{2}}+A_{k}\Lambda_{k}-\frac{A\sinh(l/x_{s})}{ x_{s}^{2}}-\frac{B\cosh(l/x_{s})}{x_{s}^{2}}-A_{k}\Lambda_{k}\cosh(\Lambda_{k}l)-B_{ k}\Lambda_{k}\sinh(\Lambda_{k}l)\right) \tag{20}\]  
which is in the form of \(g(k)-k^{2}f(k)\).