1 Correction to: J. Math. Fluid Mech. (2019) 21:6 https://doi.org/10.1007/s00021-019-0406-9

This note provides a list of errata and their correction for Reference [1].

  • There is a typo in Eq. (5.12): the term in \((\tilde{\lambda }-1)\) should come with a negative sign. This expression becomes:

    $$\begin{aligned}&\left[ i\rho _0\left( -\alpha +V_0\cdot k \right) +\gamma |k|^2\rho _0 -\frac{(\tilde{\lambda }-1)\tilde{b}}{2}\frac{ \rho ^2_0k_0^2}{|k|^2}\right] \bar{\Omega }\nonumber \\&\quad = \left[ \frac{1}{2}\frac{\rho _0}{|k|^2}\left( 2\tilde{\lambda }k_0 \bar{p}+\tilde{b} (\tilde{\lambda }+1) ( \rho _0 \bar{k} + \bar{\rho }k_0)\right) -\frac{i}{\kappa }\bar{\rho }\right] k^\perp . \end{aligned}$$
    (0.1)

  • In the statement of Th. 5.1 (Linear Stability Analysis), we have that Eq. (5.5) is not true. As a consequence equation (5.6) is slightly modified into

    $$\begin{aligned} D(\alpha , k)= & {} \Bigg \{\frac{\tilde{b} \rho _0}{2|k|^2}\Bigg [ \left( -4\tilde{\lambda }\frac{k_0^2}{|k|^2}+\tilde{\lambda }+1 \right) \left( -\alpha +U_0\cdot k \right) \nonumber \\&-c_1 k_0\left( -2\tilde{\lambda }\frac{k_0^2}{|k|^2}+\tilde{\lambda }+1 \right) \Bigg ]+\frac{i}{\kappa } c_1 \Bigg \}(|k|^2-k_0^2) \nonumber \\&- (-\alpha +U_0\cdot k) \left[ i(-\alpha +V_0\cdot k) - \frac{(\tilde{\lambda }-1)\tilde{b} \rho _0}{2}\frac{k_0^2}{|k|^2}+\gamma |k|^2\right] \!,\qquad \end{aligned}$$
    (0.2)

    when \(\bar{k}\ne 0\). And Eq. (5.7) is valid when \(\bar{k}\ne 0\). When \(\bar{k}=0\) we have a different scenario where \(\bar{\rho }=0\) and \(\bar{\Omega }\ne 0\) arbitrary and

    $$\begin{aligned} \alpha = V_0\cdot k+ i \left( \frac{(\tilde{\lambda }-1)\tilde{b}}{2}\frac{\rho _0 k_0^2}{|k|^2}-\gamma |k|^2 \right) , \end{aligned}$$
    (0.3)

    which gives only stable modes since \(\tilde{\lambda }\in [-1,1]\) and so \(\text{ Im }(\alpha )\le 0\). (Finally, notice that, the parameter \(\eta \) in the original article does not play a role any more in this corrected version.)

To obtain these results we modify the proof of Th. 5.1 Case B, which we rewrite fully next (notice that the number of the equations that do not start by 0 refer to the number of the equations as they appear in the original article).

Proof

Case (B) Suppose that \(k^\perp \ne 0\). Doing the inner product of Eq. (0.1) with k and using that \(k^\perp \cdot k= |k|^2-k_0^2\), the dispersion relation is given by (using Eq. (5.11)):

$$\begin{aligned}&\left\{ \tilde{b} \frac{\rho _0}{2|k|^2}\left[ \rho _0 \bar{k} \left( -4\tilde{\lambda }\frac{k_0^2}{|k|^2}+\tilde{\lambda }+1\right) + \bar{\rho }k_0 \left( -\frac{2\tilde{\lambda }k_0^2}{|k|^2}+\tilde{\lambda }+1\right) \right] - \frac{i}{\kappa }\bar{\rho }\right\} (|k|^2-k_0^2)\qquad \nonumber \\&\quad = \left[ i\rho _0 (-\alpha + V_0 \cdot k) - (\tilde{\lambda }-1) \frac{\tilde{b} \rho ^2_0}{2}\frac{k_0^2}{|k|^2}+\gamma |k|^2 \rho _0 \right] \, \bar{k}, \end{aligned}$$
(0.4)

and from Eq. (5.10b) we have the relation

$$\begin{aligned} (-\alpha + U_0\cdot k) \bar{\rho }+ \rho _0 c_1 \bar{k}=0. \end{aligned}$$
(0.5)

Next we distinguish between the cases \(\bar{k}\ne 0\) and \(\bar{k}=0\). Suppose that \(\bar{k}\ne 0\), then from Eq. (0.5) we have that \(-\alpha + U_0 \cdot k\ne 0\) and multiplying Eq. (0.4) by \(-\alpha + U_0 \cdot k\ne 0\) and using Eq. (0.5), we get the dispersion relation in Eq. (0.2), after simplifying \(\bar{k}\).

If, on the contrary, \(\bar{k}=0\), then one can check that \(k^\perp \cdot \bar{\Omega }=0\) (thanks to Eq. (5.10a)) and, therefore, they are normal. Since by assumption \(k^\perp \ne 0\) from Eq. (0.1) we conclude that the coefficient in front of \(k^\perp \) on the right hand side must be zero. Rewriting this term using that \(\bar{k}=0\) and Eq. (5.11) we have that

$$\begin{aligned} \bar{\rho }\, I =0, \end{aligned}$$
(0.6)

with

$$\begin{aligned} I:= \tilde{b} \frac{\rho _0}{2|k|^2} k_0 \left( -\frac{2\tilde{\lambda }k_0^2}{|k|^2}+\tilde{\lambda }+1\right) - \frac{i}{\kappa }. \end{aligned}$$

From here we deduce that \(\bar{\rho }=0\) because otherwise we should have that \(I=0\) but the imaginary part of I is non-zero, so \(\bar{\rho }=0\). In particular this implies that Eq. (0.5) is fulfilled and that \(\bar{\Omega }\ne 0\) (otherwise we would have null perturbation). Since \(\bar{\Omega }\ne 0\), from Eq. (0.1) again (remembering that \(k^\perp \perp \bar{\Omega }\)) we must have that the coefficient in front of \(\bar{\Omega }\) is equal to zero. This gives the dispersion relation (0.3).

Now, we go back to the case \(\bar{k}\ne 0\). To simplify the analysis we will restrict ourselves to the case where \(k^\perp =k\), i.e. \(k_0=k\cdot \Omega _0=0\) and \(\bar{k}\ne 0\). This implies, in particular, that \(U_0\cdot k=V_0\cdot k= v_0\cdot k\). With these considerations one can simplify the dispersion relation (0.2) into

$$\begin{aligned} \tilde{D}(\alpha ,k)=0, \end{aligned}$$

where \(\tilde{D}(\alpha ,k)\) is given in Eq. (5.7). \(\square \)

  • There are two typos at the end of page 21: in the third line of Case (A) part b) should read \(\text{ Im }(\alpha )\) instead of \(\text{ Im }(\omega )\); and in the fifth line should read “\(\bar{\Omega }\) is arbitrary with \(\bar{\Omega }\cdot \Omega _0 = 0\)” rather than “\(\bar{\Omega }\) is arbitrary with \(\bar{\Omega }, \Omega _0 \ne 0\)”.