Rayleigh–Taylor instability

The perturbation introduced to the system is described by a velocity field of infinitesimally small amplitude, ${\displaystyle (u'(x,z,t),w'(x,z,t)).\,}$ Because the fluid is assumed incompressible, this velocity field has the streamfunction representation

$${\displaystyle {\textbf {u}}'=(u'(x,z,t),w'(x,z,t))=(\psi _{z},-\psi _{x}),\,}$$

where the subscripts indicate partial derivatives. Moreover, in an initially stationary incompressible fluid, there is no vorticity, and the fluid stays irrotational, hence ${\displaystyle \nabla \times {\textbf {u}}'=0\,}$. In the streamfunction representation, ${\displaystyle \nabla ^{2}\psi =0.\,}$ Next, because of the translational invariance of the system in the x-direction, it is possible to make the ansatz

$${\displaystyle \psi \left(x,z,t\right)=e^{i\alpha \left(x-ct\right)}\Psi \left(z\right),\,}$$

where ${\displaystyle \alpha \,}$ is a spatial wavenumber. Thus, the problem reduces to solving the equation

$${\displaystyle \left(D^{2}-\alpha ^{2}\right)\Psi _{j}=0,\,\,\,\ D={\frac {d}{dz}},\,\,\,\ j=L,G.\,}$$

The domain of the problem is the following: the fluid with label 'L' lives in the region ${\displaystyle -\infty <z\leq 0\,}$, while the fluid with the label 'G' lives in the upper half-plane ${\displaystyle 0\leq z<\infty \,}$. To specify the solution fully, it is necessary to fix conditions at the boundaries and interface. This determines the wave speed c, which in turn determines the stability properties of the system.

The first of these conditions is provided by details at the boundary. The perturbation velocities ${\displaystyle w'_{i}\,}$ should satisfy a no-flux condition, so that fluid does not leak out at the boundaries ${\displaystyle z=\pm \infty .\,}$ Thus, ${\displaystyle w_{L}'=0\,}$ on ${\displaystyle z=-\infty \,}$, and ${\displaystyle w_{G}'=0\,}$ on ${\displaystyle z=\infty \,}$. In terms of the streamfunction, this is

$${\displaystyle \Psi _{L}\left(-\infty \right)=0,\qquad \Psi _{G}\left(\infty \right)=0.\,}$$

The other three conditions are provided by details at the interface ${\displaystyle z=\eta \left(x,t\right)\,}$.

Continuity of vertical velocity: At ${\displaystyle z=\eta }$, the vertical velocities match, ${\displaystyle w'_{L}=w'_{G}\,}$. Using the stream function representation, this gives

$${\displaystyle \Psi _{L}\left(\eta \right)=\Psi _{G}\left(\eta \right).\,}$$

Expanding about ${\displaystyle z=0\,}$ gives

$${\displaystyle \Psi _{L}\left(0\right)=\Psi _{G}\left(0\right)+{\text{H.O.T.}},\,}$$

where H.O.T. means 'higher-order terms'. This equation is the required interfacial condition.

The free-surface condition: At the free surface ${\displaystyle z=\eta \left(x,t\right)\,}$, the kinematic condition holds:

$${\displaystyle {\frac {\partial \eta }{\partial t}}+u'{\frac {\partial \eta }{\partial x}}=w'\left(\eta \right).\,}$$

Linearizing, this is simply

$${\displaystyle {\frac {\partial \eta }{\partial t}}=w'\left(0\right),\,}$$

where the velocity ${\displaystyle w'\left(\eta \right)\,}$ is linearized on to the surface ${\displaystyle z=0\,}$. Using the normal-mode and streamfunction representations, this condition is ${\displaystyle c\eta =\Psi \,}$, the second interfacial condition.

Pressure relation across the interface: For the case with surface tension, the pressure difference over the interface at ${\displaystyle z=\eta }$ is given by the Young–Laplace equation:

$${\displaystyle p_{G}\left(z=\eta \right)-p_{L}\left(z=\eta \right)=\sigma \kappa ,\,}$$

where σ is the surface tension and κ is the curvature of the interface, which in a linear approximation is

$${\displaystyle \kappa =\nabla ^{2}\eta =\eta _{xx}.\,}$$

Thus,

$${\displaystyle p_{G}\left(z=\eta \right)-p_{L}\left(z=\eta \right)=\sigma \eta _{xx}.\,}$$

However, this condition refers to the total pressure (base+perturbed), thus

$${\displaystyle \left[P_{G}\left(\eta \right)+p'_{G}\left(0\right)\right]-\left[P_{L}\left(\eta \right)+p'_{L}\left(0\right)\right]=\sigma \eta _{xx}.\,}$$

(As usual, The perturbed quantities can be linearized onto the surface z=0.) Using hydrostatic balance, in the form

$${\displaystyle P_{L}=-\rho _{L}gz+p_{0},\qquad P_{G}=-\rho _{G}gz+p_{0},\,}$$

this becomes

$${\displaystyle p'_{G}-p'_{L}=g\eta \left(\rho _{G}-\rho _{L}\right)+\sigma \eta _{xx},\qquad {\text{on }}z=0.\,}$$

The perturbed pressures are evaluated in terms of streamfunctions, using the horizontal momentum equation of the linearised Euler equations for the perturbations,

$${\displaystyle {\frac {\partial u_{i}'}{\partial t}}=-{\frac {1}{\rho _{i}}}{\frac {\partial p_{i}'}{\partial x}}\,}$$

with ${\displaystyle i=L,G,\,}$ to yield

$${\displaystyle p_{i}'=\rho _{i}cD\Psi _{i},\qquad i=L,G.\,}$$

Putting this last equation and the jump condition on ${\displaystyle p'_{G}-p'_{L}}$ together,

$${\displaystyle c\left(\rho _{G}D\Psi _{G}-\rho _{L}D\Psi _{L}\right)=g\eta \left(\rho _{G}-\rho _{L}\right)+\sigma \eta _{xx}.\,}$$

Substituting the second interfacial condition ${\displaystyle c\eta =\Psi \,}$ and using the normal-mode representation, this relation becomes

$${\displaystyle c^{2}\left(\rho _{G}D\Psi _{G}-\rho _{L}D\Psi _{L}\right)=g\Psi \left(\rho _{G}-\rho _{L}\right)-\sigma \alpha ^{2}\Psi ,\,}$$

where there is no need to label ${\displaystyle \Psi \,}$ (only its derivatives) because ${\displaystyle \Psi _{L}=\Psi _{G}\,}$ at ${\displaystyle z=0.\,}$

Solution

Now that the model of stratified flow has been set up, the solution is at hand. The streamfunction equation ${\displaystyle \left(D^{2}-\alpha ^{2}\right)\Psi _{i}=0,\,}$ with the boundary conditions ${\displaystyle \Psi \left(\pm \infty \right)\,}$ has the solution

$${\displaystyle \Psi _{L}=A_{L}e^{\alpha z},\qquad \Psi _{G}=A_{G}e^{-\alpha z}.\,}$$

The first interfacial condition states that ${\displaystyle \Psi _{L}=\Psi _{G}\,}$ at ${\displaystyle z=0\,}$, which forces ${\displaystyle A_{L}=A_{G}=A.\,}$ The third interfacial condition states that

$${\displaystyle c^{2}\left(\rho _{G}D\Psi _{G}-\rho _{L}D\Psi _{L}\right)=g\Psi \left(\rho _{G}-\rho _{L}\right)-\sigma \alpha ^{2}\Psi .\,}$$

Plugging the solution into this equation gives the relation

$${\displaystyle Ac^{2}\alpha \left(-\rho _{G}-\rho _{L}\right)=Ag\left(\rho _{G}-\rho _{L}\right)-\sigma \alpha ^{2}A.\,}$$

The A cancels from both sides and we are left with

$${\displaystyle c^{2}={\frac {g}{\alpha }}{\frac {\rho _{L}-\rho _{G}}{\rho _{L}+\rho _{G}}}+{\frac {\sigma \alpha }{\rho _{L}+\rho _{G}}}.\,}$$

To understand the implications of this result in full, it is helpful to consider the case of zero surface tension. Then,

$${\displaystyle c^{2}={\frac {g}{\alpha }}{\frac {\rho _{L}-\rho _{G}}{\rho _{L}+\rho _{G}}},\qquad \sigma =0,\,}$$

and clearly

• If ${\displaystyle \rho _{G}<\rho _{L}\,}$, ${\displaystyle c^{2}>0\,}$ and c is real. This happens when the lighter fluid sits on top;
• If ${\displaystyle \rho _{G}>\rho _{L}\,}$, ${\displaystyle c^{2}<0\,}$ and c is purely imaginary. This happens when the heavier fluid sits on top.

Now, when the heavier fluid sits on top, ${\displaystyle c^{2}<0\,}$, and

$${\displaystyle c=\pm i{\sqrt {\frac {g{\mathcal {A}}}{\alpha }}},\qquad {\mathcal {A}}={\frac {\rho _{G}-\rho _{L}}{\rho _{G}+\rho _{L}}},\,}$$

where ${\displaystyle {\mathcal {A}}\,}$ is the Atwood number. By taking the positive solution, we see that the solution has the form

$${\displaystyle \Psi \left(x,z,t\right)=Ae^{-\alpha |z|}\exp \left[i\alpha \left(x-ct\right)\right]=A\exp \left(\alpha {\sqrt {\frac {g{\tilde {\mathcal {A}}}}{\alpha }}}t\right)\exp \left(i\alpha x-\alpha |z|\right)\,}$$

and this is associated to the interface position η by: ${\displaystyle c\eta =\Psi .\,}$ Now define ${\displaystyle B=A/c.\,}$