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A semi-infinite layer of incompressible inviscid fluid with density ρ1 occupies the half-space $y < 0$ below a similar layer of fluid with density $\rho_2 < \rho_1$ occupying $y > 0$. The lower fluid flows at uniform speed U in the x-direction while the upper fluid flows at speed V in the x-direction. The surface tension on the interface between the two fluids is negligible. The flow is disturbed by small-amplitude irrotational waves such that the free surface initially at $y = 0$ is displaced to $y = \eta(x, t)$. You may assume that the velocities in the lower and upper fluids take the forms

$$\mathbf u_1 =U \mathbf i+ \nabla \varphi_1 , \quad \mathbf u_2 =V\mathbf i+∇φ_2$$

Show that harmonic travelling waves, with $ η(x, t) = A cos(kx − ωt)$ and k > 0, are possible provided ω and k satisfy the dispersion relation $$\rho_1(\omega − U k)^2 + \rho_2(ω−Vk)^2 =(\rho_1 − \rho_2)gk$$ Deduce that, whenever U $\neq$ V , the free surface is unstable to waves whose wavelength is smaller than the critical value

$$λ_c=\frac{2πρ_1ρ_2(U-V)^2}{(\rho_1^2−\rho_2^2)g}$$

I got the dispersion relation but don't know how to get to the stable free surface part.I know w can be negative for some value of k. For these wave numbers, w is pure imaginary so the disturbance grows exponentially.Which means no stable. And $\lambda =\frac{2\pi c}{w}$But how to get the critical $\lambda_c$ ?

Thank you so much!

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  • $\begingroup$ "For these wave numbers, w is pure imaginary so the disturbance grows exponentially". Can you please specify what your normal-mode decomposition looks like? Is it something of the form $\eta \propto e^{i (kx - \omega t)}$? $\endgroup$ – Dmoreno May 23 '17 at 21:36
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Given the dispersion relation, we can solve for the frequency and find:

$$\omega \,=\, \frac{(\rho_{1}U+\rho_{2}V)k \pm \sqrt{gk(\rho_{1}^{2}-\rho_{2}^{2}) - k^{2}(U-V)^{2}\rho_{1}\rho_{2}\;}}{\rho_{1}+\rho_{2}} $$

(I'll leave you to fill in the details!). As you note, the instability occurs when the frequency becomes imaginary; this occurs when the square root above goes negative or $k<k_{c}$ where this critical $k$ is found by solving:

$$\begin{align} gk_{c}(\rho_{1}^{2}- \rho_{2}^{2}) - k_{c}^{2}(U-V)^{2}\rho_{1}\rho_{2} &\,=\, 0 \end{align}$$

yielding $k_{c}=0$ or

$$\begin{align} k_{c} &\,=\, \frac{g(\rho_{1}^{2}-\rho_{2}^{2})}{\rho_{1}\rho_{2}(U-V)^{2}} \end{align}$$

(again, I will leave you to complete the algebra!). The relation $\lambda_{c}=\frac{2\pi}{k_{c}}$ then yields the relation you are after.

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  • $\begingroup$ Thank you to be stable $w$ is determined by the $\delta$ of the quadratic equation? $\endgroup$ – stedmoaoa May 23 '17 at 22:20
  • $\begingroup$ If by $\delta$ you mean the discriminant, then yes. It's when the term under the square root (the $b^2-4ac$ term in the quadratic formula for the roots of $ax^{2}+bx+c$) goes negative and hence $\omega$ becomes complex. $\endgroup$ – AloneAndConfused May 23 '17 at 23:22

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