2 Problem formulation and governing equations

We consider an incompressible infinitely electrically conducting plasma. In Cartesian coordinates ( $x,y,z$ ) the unperturbed state is an MHD tangential discontinuity. The unperturbed magnetic field and velocity are $\boldsymbol{B}=B\boldsymbol{e}_{x}$ and $\boldsymbol{U}=U_{0}\boldsymbol{e}_{x}$ , where $\boldsymbol{e}_{x}$ is the unit vector in the $x$ -direction. The equilibrium density $\unicode[STIX]{x1D70C}$ and the quantities $B$ and $U_{0}$ are given by

(2.1a-c ) $$\begin{eqnarray}\unicode[STIX]{x1D70C}=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D70C}_{1}, & z<0,\\ \unicode[STIX]{x1D70C}_{2}, & z>0,\end{array}\right.\quad B=\left\{\begin{array}{@{}ll@{}}B_{1}, & z<0,\\ B_{2}, & z>0,\end{array}\right.\quad U_{0}=\left\{\begin{array}{@{}ll@{}}0, & z<0,\\ U, & z>0.\end{array}\right.\end{eqnarray}$$

The plasma motion is described by the ideal MHD equations in the incompressible plasma approximation,

(2.2a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{v}=0,\quad \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{b}=0,\end{eqnarray}$$

(2.2c ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\boldsymbol{v}}{\unicode[STIX]{x2202}t}+U\frac{\unicode[STIX]{x2202}\boldsymbol{v}}{\unicode[STIX]{x2202}x}=-\frac{1}{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D735}P+\frac{B}{\unicode[STIX]{x1D707}_{0}\unicode[STIX]{x1D70C}}\frac{\unicode[STIX]{x2202}\boldsymbol{b}}{\unicode[STIX]{x2202}x}+\unicode[STIX]{x1D708}_{0}\unicode[STIX]{x1D6FB}^{2}\boldsymbol{v},\end{eqnarray}$$

(2.2d ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\boldsymbol{b}}{\unicode[STIX]{x2202}t}+U\frac{\unicode[STIX]{x2202}\boldsymbol{b}}{\unicode[STIX]{x2202}x}=B\frac{\unicode[STIX]{x2202}\boldsymbol{v}}{\unicode[STIX]{x2202}x},\end{eqnarray}$$

where $\boldsymbol{v}=(u,v,w)$ is the velocity perturbation, $\boldsymbol{b}=(b_{x},b_{y},b_{z})$ the magnetic field perturbation, $P$ the total pressure (plasma plus magnetic) perturbation, $\unicode[STIX]{x1D707}_{0}$ the magnetic permeability of free space and $\unicode[STIX]{x1D708}_{0}$ the kinematic viscosity equal to $\unicode[STIX]{x1D708}$ at $z<0$ and 0 at $z>0$ .

We write the equation of the perturbed tangential discontinuity as $z=\unicode[STIX]{x1D701}(t,x,y)$ . The perturbations must satisfy the kinematic boundary conditions and the conditions of the stress continuity at $z=0$ :

(2.3a ) $$\begin{eqnarray}\displaystyle & \displaystyle w_{1}=\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t}, & \displaystyle\end{eqnarray}$$

(2.3b ) $$\begin{eqnarray}\displaystyle & \displaystyle w_{2}=\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t}+U\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}x}, & \displaystyle\end{eqnarray}$$

(2.3c ) $$\begin{eqnarray}\displaystyle & \displaystyle P_{2}=P_{1}-2\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D708}\frac{\unicode[STIX]{x2202}w_{1}}{\unicode[STIX]{x2202}z}, & \displaystyle\end{eqnarray}$$

(2.3d,e ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}u_{1}}{\unicode[STIX]{x2202}z}+\frac{\unicode[STIX]{x2202}w_{1}}{\unicode[STIX]{x2202}x}=0,\quad \frac{\unicode[STIX]{x2202}v_{1}}{\unicode[STIX]{x2202}z}+\frac{\unicode[STIX]{x2202}w_{1}}{\unicode[STIX]{x2202}x}=0,\end{eqnarray}$$

where the indices 1 and 2 refer to quantities below $(z<0)$ and above ( $z>0$ ) the discontinuity. In addition, we assume that all perturbations tend to zero as $|z|\rightarrow \infty$ . Finally, we assume that the magnetic field lines are frozen in the immovable plasma at two planes orthogonal to the $x$ -axis at $x=0$ and $x=L$ . In terms of the velocity and magnetic field perturbation this condition is written as $U\boldsymbol{b}_{\bot }-B\boldsymbol{v}_{\bot }=0$ at $x=0,\,L$ , where $\boldsymbol{b}_{\bot }$ and $\boldsymbol{v}_{\bot }$ are the components of the magnetic field and velocity perturbation orthogonal to the $x$ -axis (see e.g. Ruderman Reference Ruderman2010). In particular, it follows that

(2.4) $$\begin{eqnarray}\unicode[STIX]{x1D701}=0\quad \text{at }x=0,L.\end{eqnarray}$$

3 Instability of standing waves

Ruderman & Goossens (Reference Ruderman and Goossens1995) used the system of (2.2) and boundary conditions (2.3) to derive the equation describing the propagation of surface waves on the tangential discontinuity. This equation reads

(3.1) $$\begin{eqnarray}(\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t^{2}}+2\unicode[STIX]{x1D70C}_{2}U\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t\unicode[STIX]{x2202}x}+\unicode[STIX]{x1D70C}_{2}(U^{2}-U_{c}^{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}x^{2}}=4\unicode[STIX]{x1D708}\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D6FB}^{2}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t},\end{eqnarray}$$

where

(3.2a,b ) $$\begin{eqnarray}U_{c}^{2}=\frac{\unicode[STIX]{x1D70C}_{1}V_{1}^{2}+\unicode[STIX]{x1D70C}_{2}V_{2}^{2}}{\unicode[STIX]{x1D70C}_{1}}=\frac{\unicode[STIX]{x1D70C}_{1}V_{KH}^{2}}{\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2}},\quad V_{1,2}^{2}=\frac{B_{1,2}^{2}}{\unicode[STIX]{x1D707}_{0}\unicode[STIX]{x1D70C}_{1,2}}.\end{eqnarray}$$

This equation is only valid in the long wavelength approximation, that is when $\text{Re}=LV_{0}/\unicode[STIX]{x1D708}\gg 1$ where $V_{0}=(V_{1}+V_{2})/2$ . When $U>V_{KH}$ the discontinuity is subject to the Kelvin–Helmholtz instability. Equation (3.1) describes two wave modes propagating in the opposite directions in the absence of flow. They are called the forward wave mode and the backward wave mode, respectively. Ruderman & Goossens (Reference Ruderman and Goossens1995) showed that the backward mode becomes a negative energy wave when $U>U_{c}$ . This mode is unstable due to the presence of viscosity. The forward mode is always a positive energy wave that decays due to the presence of viscosity. When $U_{c}<U<V_{KH}$ the increment of the negative energy wave is

(3.3) $$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{-}=\frac{2\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D708}k^{2}C_{-}}{\sqrt{\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D70C}_{2}(V_{KH}^{2}-U^{2})}},\end{eqnarray}$$

and the decrement of the positive energy wave is

(3.4) $$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{+}=\frac{2\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D708}k^{2}C_{+}}{\sqrt{\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D70C}_{2}(V_{KH}^{2}-U^{2})}}.\end{eqnarray}$$

In these expressions $k^{2}=k_{x}^{2}+k_{y}^{2}$ , $k_{x}$ and $k_{y}$ are the $x$ and $y$ -component of the wave vector, $\boldsymbol{C}_{\pm }=\unicode[STIX]{x1D714}/k_{x}$ and $\unicode[STIX]{x1D714}$ is the wave frequency. The quantity $\boldsymbol{C}_{\pm }$ is given by

(3.5) $$\begin{eqnarray}\boldsymbol{C}_{\pm }=\frac{\unicode[STIX]{x1D70C}_{2}U\pm \sqrt{\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D70C}_{2}(V_{KH}^{2}-U^{2})}}{\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2}}.\end{eqnarray}$$

We now consider a solution to (3.1) in the form of standing wave satisfying the boundary conditions (2.4). We assume that $U_{c}<U<V_{KH}$ . We only consider waves standing in the $y$ -direction and take $\unicode[STIX]{x1D701}$ proportional to $\cos (k_{y}y)$ . The characteristic time of variation of the wave amplitude is equal to the characteristic period of oscillations times $\text{Re}$ . In accordance with this we introduce the ‘slow’ time $T=\unicode[STIX]{x1D716}t$ , where $\unicode[STIX]{x1D716}=\text{Re}^{-1}$ . We also introduce the scaled shear viscosity $\bar{\unicode[STIX]{x1D708}}=\unicode[STIX]{x1D716}^{-1}\unicode[STIX]{x1D708}$ . Then (3.1) is transformed to

(3.6) $$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t^{2}}+2\unicode[STIX]{x1D70C}_{2}U\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t\unicode[STIX]{x2202}x}+\unicode[STIX]{x1D70C}_{2}(U^{2}-U_{c}^{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}x^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad +\,2\unicode[STIX]{x1D716}\left((\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t\unicode[STIX]{x2202}T}+\unicode[STIX]{x1D70C}_{2}U\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}x\unicode[STIX]{x2202}T}\right)-4\unicode[STIX]{x1D716}\bar{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D70C}_{1}\left(\frac{\unicode[STIX]{x2202}^{2}}{\unicode[STIX]{x2202}x^{2}}-k_{y}^{2}\right)\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}{\unicode[STIX]{x2202}t}=O(\unicode[STIX]{x1D716}^{2}).\end{eqnarray}$$

We look for the solution to this equation in the form of series

(3.7) $$\begin{eqnarray}\unicode[STIX]{x1D701}=\unicode[STIX]{x1D701}_{1}+\unicode[STIX]{x1D716}\unicode[STIX]{x1D701}_{2}+O(\unicode[STIX]{x1D716}^{2}).\end{eqnarray}$$

Substituting this expression in (3.6) and collecting terms of the order of unity yields

(3.8) $$\begin{eqnarray}(\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}_{1}}{\unicode[STIX]{x2202}t^{2}}+2\unicode[STIX]{x1D70C}_{2}U\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}_{1}}{\unicode[STIX]{x2202}t\unicode[STIX]{x2202}x}+\unicode[STIX]{x1D70C}_{2}(U^{2}-U_{c}^{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}_{1}}{\unicode[STIX]{x2202}x^{2}}=0.\end{eqnarray}$$

It follows from (2.4) that $\unicode[STIX]{x1D701}_{1}$ must satisfy the boundary conditions

(3.9) $$\begin{eqnarray}\unicode[STIX]{x1D701}_{1}=0\quad \text{at }x=0,L.\end{eqnarray}$$

We look for the solution to the boundary value problem constituted by (3.8) and the boundary conditions (3.9) in the form

(3.10) $$\begin{eqnarray}\unicode[STIX]{x1D701}_{1}=A(T)[\cos (\unicode[STIX]{x1D714}t-k_{+}x)-\cos (\unicode[STIX]{x1D714}t-k_{-}x)],\end{eqnarray}$$

where $k_{\pm }=\unicode[STIX]{x1D714}/\boldsymbol{C}_{\pm }$ . It is straightforward to verify that this expression satisfies (3.8). It also satisfies the condition $\unicode[STIX]{x1D701}_{1}=0$ at $x=0$ . The condition $\unicode[STIX]{x1D701}_{1}=0$ at $x=L$ can be written as

(3.11) $$\begin{eqnarray}\sin \left(\unicode[STIX]{x1D714}t-\frac{L(k_{+}+k_{-})}{2}\right)\sin \frac{L(k_{+}-k_{-})}{2}=0.\end{eqnarray}$$

It follows from this equation that $L(k_{-}-k_{+})=2\unicode[STIX]{x03C0}n$ , which leads to

(3.12) $$\begin{eqnarray}\unicode[STIX]{x1D714}=\frac{2\unicode[STIX]{x03C0}nC_{+}C_{-}}{L(C_{+}-C_{-})}=\frac{\unicode[STIX]{x03C0}n(U^{2}-U_{c}^{2})}{LU},\quad n=1,2,\ldots .\end{eqnarray}$$

When there is no viscosity equation (3.10) with $A=\text{const.}$ gives an exact solution to (3.1) with $\unicode[STIX]{x1D708}=0$ . The presence of viscosity causes the variation of the wave amplitude $A$ with time. To determine this variation we proceed to the next-order approximation. Collecting terms of the order of $\unicode[STIX]{x1D716}$ in (3.8) and using (3.10) and (3.12) yields

(3.13) $$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}_{2}}{\unicode[STIX]{x2202}t^{2}}+2\unicode[STIX]{x1D70C}_{2}U\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}_{2}}{\unicode[STIX]{x2202}t\unicode[STIX]{x2202}x}+\unicode[STIX]{x1D70C}_{2}(U^{2}-U_{c}^{2})\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D701}_{2}}{\unicode[STIX]{x2202}x^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\text{d}A}{\text{d}T}\sqrt{\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D70C}_{2}(V_{KH}^{2}-U^{2})}[k_{+}\sin (\unicode[STIX]{x1D714}t-k_{+}x)+k_{-}\sin (\unicode[STIX]{x1D714}t-k_{-}x)]\nonumber\\ \displaystyle & & \displaystyle \qquad +\,4\bar{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70C}_{1}A[(k_{y}^{2}+k_{+}^{2})\sin (\unicode[STIX]{x1D714}t-k_{+}x)-(k_{y}^{2}+k_{-}^{2})\sin (\unicode[STIX]{x1D714}t-k_{-}x)].\end{eqnarray}$$

It follows from (2.4) that $\unicode[STIX]{x1D701}_{2}$ must satisfy the boundary conditions

(3.14) $$\begin{eqnarray}\unicode[STIX]{x1D701}_{2}=0\quad \text{at }x=0,L.\end{eqnarray}$$

The left-hand side of (3.13) coincides with (3.8). This implies that $\unicode[STIX]{x1D701}_{1}$ is a solution to the homogeneous counterpart of the boundary value problem constituted by (3.13) and the boundary conditions (3.14). Then the inhomogeneous boundary value problem has solutions only if the right-hand side of (3.13) satisfies the compatibility condition, which is the condition that it is orthogonal to $\unicode[STIX]{x1D701}_{1}$ . To obtain the compatibility condition we multiply (3.13) by $\unicode[STIX]{x1D701}_{1}$ , integrate with respect to $x$ and use (3.11) and the boundary conditions (3.14). Then, after long but straightforward calculation we obtain

(3.15) $$\begin{eqnarray}\displaystyle & & \displaystyle \sin \left(2\unicode[STIX]{x1D714}t-\frac{L(k_{+}+k_{-})}{2}\right)\sin \frac{L(k_{+}+k_{-})}{2}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\left(\frac{\text{d}A}{\text{d}T}\sqrt{\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D70C}_{2}(V_{KH}^{2}-U^{2})}-\frac{2\bar{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70C}_{1}A(k_{-}-k_{+})(k_{-}k_{+}-k_{y}^{2})}{k_{-}k_{+}}\right)=0.\end{eqnarray}$$

Obviously the first multiplier on the right-hand side of this equation cannot be equal to zero for all values of $t$ . If we assume that the second multiplier is zero, then it follows from this condition and (3.13) that $C_{+}/C_{-}$ is a rational number, which is only possible for a countable set of parameters. Eliminating this set from the consideration we obtain that the second multiplier is not equal to zero. Hence, the only possibility left is that the third multiplier is zero, which gives

(3.16) $$\begin{eqnarray}\frac{1}{A}\frac{\text{d}A}{\text{d}T}=\frac{2\bar{\unicode[STIX]{x1D708}}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70C}_{1}(\boldsymbol{k}_{-}-\boldsymbol{k}_{+})(\boldsymbol{k}_{-}\boldsymbol{k}_{+}-k_{y}^{2})}{\boldsymbol{k}_{-}\boldsymbol{k}_{+}\sqrt{\unicode[STIX]{x1D70C}_{1}\unicode[STIX]{x1D70C}_{2}(V_{KH}^{2}-U^{2})}}.\end{eqnarray}$$

Now, returning to the original non-scaled time and shear viscosity we obtain after some algebra

(3.17) $$\begin{eqnarray}A=A_{0}\text{e}^{\unicode[STIX]{x1D6FE}t},\end{eqnarray}$$

where $A_{0}$ is the initial value of $A$ and

(3.18) $$\begin{eqnarray}\unicode[STIX]{x1D6FE}=4\unicode[STIX]{x1D708}\unicode[STIX]{x1D70C}_{1}\left(\frac{\unicode[STIX]{x03C0}^{2}n^{2}(U^{2}-U_{c}^{2})}{\unicode[STIX]{x1D70C}_{2}L^{2}U^{2}}-\frac{k_{y}^{2}}{\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2}}\right).\end{eqnarray}$$

We see that for sufficiently small $k_{y}$ satisfying

(3.19) $$\begin{eqnarray}k_{y}^{2}<\frac{\unicode[STIX]{x03C0}^{2}n^{2}(\unicode[STIX]{x1D70C}_{1}+\unicode[STIX]{x1D70C}_{2})(U^{2}-U_{c}^{2})}{\unicode[STIX]{x1D70C}_{2}L^{2}U^{2}},\end{eqnarray}$$

$\unicode[STIX]{x1D6FE}>0$ and the amplitude of the standing wave grows exponentially.

As we have already seen the standing wave is a superposition of the negative energy wave with the wave vector $\boldsymbol{k}_{-}$ and the positive energy wave with the wave vector $\boldsymbol{k}_{+}$ . Using (3.3)–(3.5) to calculate the increment $\unicode[STIX]{x1D6FE}_{-}$ for the first wave and the decrement $\unicode[STIX]{x1D6FE}_{+}$ for the second wave we easily obtain

(3.20) $$\begin{eqnarray}\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D6FE}_{-}-\unicode[STIX]{x1D6FE}_{+}.\end{eqnarray}$$

4 Summary and conclusions

In this article we studied the dissipative instability of standing waves caused by the flow. We considered a simple equilibrium where two incompressible plasmas are separated by a tangential discontinuity. There is the plasma flow with constant velocity parallel to the discontinuity at one of its sides while the plasma at the other side is at rest. There is also magnetic field parallel to the flow velocity. The plasma that is at rest is viscous, while the moving plasma is ideal. We only consider long waves satisfying the condition that the Reynolds number is large. In the linear approximation these waves are described by (3.1) derived by Ruderman & Goossens (Reference Ruderman and Goossens1995).

When the flow speed $U$ exceeds the Kelvin–Helmholtz threshold $V_{KH}$ the discontinuity is subject to the Kelvin–Helmholtz instability. When $U<V_{KH}$ there are two surface wave modes that can propagate on the surface of the discontinuity. In the absence of flow these modes propagate in the opposite directions. We call the mode propagating in the flow velocity direction forward, and the mode propagating in the opposite direction backward. When $U_{c}<U<V_{KH}$ both modes propagate in the same direction. In this case the forward mode is a positive energy wave, while the backward mode is a negative energy wave. In the absence of viscosity both modes are neutrally stable. However viscosity causes the forward mode to decay, and the backward mode to grow. This growth of the backward mode is called the negative energy wave instability.

We assumed that the magnetic field lines are frozen in a dense plasma at two planes perpendicular to the flow direction lines. These plane are a distance $L$ from each other. This, in particular, implies that the tangential discontinuity is fixed at two lines perpendicular to the flow direction that are a distance $L$ from each other. Then we consider surface waves that are standing both in the direction of the flow velocity and in direction orthogonal to the flow velocity. A standing wave is a linear superposition of a forward wave and a backward wave. We restricted our analysis to the case where $U_{c}<U<V_{KH}$ and showed that the standing wave can grow due to the presence of viscosity. However it is only possible when its wavelength in the direction orthogonal to the flow velocity is sufficiently large. The increment of the standing wave is equal to the difference between the increment of the backward wave and the decrement of the forward wave.

We emphasise that our analysis is only valid for slowly growing waves with the increment much smaller than the wave frequency. In the case when the increment is of the order of the wave frequency the stability analysis must be modified substantially.