# Gravity free-surface waves with two fluids

This is basically the same thing than free_surface_gravity.m but here there are two fluids on top of each other, so this is a little more complicated, but we do not have to neglect the density of the top fluid so this is more general. Technically it works like that, we build two dynamic systems for the two fluid layers with different physical properties, then we need to connect this two fluid layers through the moving interface.

We take a very low viscosity because we will compare with a theory for a perfect fluid.

clear all; clf;

% for domain 1 (top)
N1=50;      % number of gridpoints
L1=1;        % Fluid height in y
rho1=0.5;      % fluid density
mu1=0.00001;    % fuid viscosity

% for domain 2 (bottom)
N2=50;      % number of gridpoints
L2=1;        % Fluid height in y
rho2=1;      % fluid density
mu2=0.00001;    % fuid viscosity

alpha=2;    % wavenumber in x
g=1;        % gravity

% domain 1
% differentiation matrices
scale=-2/L1;
[y,DM] = chebdif(N1,2);
D=DM(:,:,1)*scale;
DD=DM(:,:,2)*scale^2;
y1=(y-1)/scale+L2;
I1=eye(N1); Z1=zeros(N1,N1);
% renaming the matrices
dy1=D; dyy1=DD;
dx1=i*alpha*I1; dxx1=-alpha^2*I1;
Delta1=dxx1+dyy1;

% domain 2
% differentiation matrices
scale=-2/L2;
[y,DM] = chebdif(N2,2);
D=DM(:,:,1)*scale;
DD=DM(:,:,2)*scale^2;
y2=(y-1)/scale;
I2=eye(N2); Z2=zeros(N2,N2);
% renaming the matrices
dy2=D; dyy2=DD;
dx2=i*alpha*I2; dxx2=-alpha^2*I2;
Delta2=dxx2+dyy2;


# System matrices

The motion of each fluid is described by the Navier-Stokes equations and continuity, linearized about a zero base flow, here written in a matrix form \displaystyle e_1q_{1t}=a_1q_1 for the top fluid and
\displaystyle e_2q_{2t}=e_2q_2 for the bottom fluid. We also have \displaystyle \eta_t=v_1(0) for the advection of the interface by the velocity next to the interface. Note that we could have as well written \eta_t=v_2(0) since anyway we will see for the boundary conditions that v is continuous across the interface so v_1(0)=v_2(0). The total state of the system is now called q \displaystyle q=\left(\begin{array}{c} q_1 \\ q_2 \\ \eta \end{array}\right) with all the information: both of the top and bottom fluids and the interface position \eta. Thus the system is Eq_t=Aq with \displaystyle E=\left(\begin{array}{ccc} \rho&0&0\\ 0&\rho&0\\ 0&0&1 \end{array}\right) , \quad A=\left(\begin{array}{ccc} e1 & 0 & 0\\ 0 & e2 & 0\\ I_{v_1(0)} & 0 & 0\\ \end{array}\right) where I_{v_1(0)} is the vector that multiplied by q_1 gives the value of v_1 at the interface (at y=0). This way we have in the dynamics the motion fo fluids 1 and 2, and the motion of the interface.

Here a little discussion on the different ways to build a matrix. What we have used here is a way to build the matrices by stacking blocks together. The function blkdiag means “block diagonal”, it build a larger matrices by stacking the input matrix arguments together to build a block diagonal matrix as output. An other way is like is done below for a1 and a2 is by concatenation of the differentiation matrices, the matrix full of zeros Z and the identity matrix I.

We can also access the elements of the big matrix using the indices of submatrices. In most of the codes we use also this, especially for the boundary conditions. So we build here the vectors of indices that correspond in the big matrix to the different components of the state: u1=1:N1 means that the u velocity of the fluid 1 is stored on the elements from 1 to N1 of the state q, and so on. Then we see also that the position \eta of the interface is the last element of the state q.

So this is why when we want to insert the connection between \eta and the vertical velocity at the bottom of fluid 1, we write

A(eta,v1(1))=1;

We will make use even more of these vectors of indices when implementing the boundary conditions below.

% location vectors
u1=1:N1; v1=u1+N1; p1=v1+N1;
u2=p1(end)+(1:N2); v2=u2+N2; p2=v2+N2;
eta=p2(end)+1;

% System matrices
a1=[mu1*Delta1, Z1, -dx1; ...
Z1, mu1*Delta1, -dy1; ...
dx1, dy1, Z1];

a2=[mu2*Delta2, Z2, -dx2; ...
Z2, mu2*Delta2, -dy2; ...
dx2, dy2, Z2];

e1=blkdiag(rho1*I1,rho1*I1,Z1);
e2=blkdiag(rho2*I2,rho2*I2,Z2);

A=blkdiag(a1,a2,0);
E=blkdiag(e1,e2,1);

% we add the equation for eta
A(eta,v1(1))=1;


# Boundary conditions

At the wall we have the classical no-slip boundary conditions. This means that u1 and v1 must be zero at the top of domain 1. here the top correspond the the first index, so u_1(1)=0, v_1(1)=0. For the fluid 2, the wall is at the bottom, so u_2(N2)=0, v_2(N2)=0. These are our 4 homogeneous Dirichlet boundary conditions.

We then have 4 additional constraints comming from the connection of the two fluids through the interface. We have the continuity of vertical velocity v1(N1)=v2(1), the continuity of u: u1(N1)=u2(1), and then we have the continuity of the stress, that is the continuity of the pressure
\displaystyle p_1(0)-\rho_1 g\eta=p_2(0)-\rho_2 g\eta where we have done Taylor expantions to express the pressure at y=\eta using the pressure at y=0 like we did in the simpler case of free_surface_gravity.m#boundary-conditions. Now, instead of having the left hand side of this equation be zero (the atmospheric pressure), we have that the left hand side equals the right-hand side from the pressure of the other fluid at y=0.

The last constraint is the continuity of tangential stress \displaystyle \mu_1(u_{1y}(0)+v_{1x}(0))=\mu_2(u_{2y}(0)+v_{2x}(0).

Since we have many variables, we should find a nice and simple way to write all these constraints with the least chance of making mistakes. For this we make an extensive use of identity matrices and location indices.

The difficult thing with matrix representation of the boundary conditions is that we do not manipulate the state itself q, but we manipulate matrices that multiply this state vector. A way to work arround this difficulty is to use an identity matrix II of the same size than the state q, and say that for instance here for the u velocity of fluid 1

u_1=q(u1)=II(u1,:)*q

and an other example for \eta

\eta = q(eta)= II(eta,:)*q

where u1 and eta are the index vectors forthe location of U_1 and \eta in the state vector q. here a little trickier example for the contiuity of the viscous stress where we need to compute the y derivative of the velocity components

u_{1y}(0)=dy(N,:)II(u1,:)q

where from right to left of the multiplications we first extract the u_1 velocity from the state q using the identity matrix, then we multiply by the line number N of the y differentiation matrix to extract the derivative at position N, that is the y derivative at y=0. Now all these can be combined to get in compact form the constraint for viscous stress which is coded below

mu1(dy1(1,:)II(u1,:)+dx1(1,:)II(v1,:))-mu2(dy2(end,:)II(u2,:)+dx2(end,:)II(v2,:))

this line produces an array that multiplying q from the left must give the result 0.

% boundary conditions
dir=[u1(end),v1(end),u2(1),v2(1)]; % no-slip at both walls
loc=[dir,u1(1),v1(1),u2(end),v2(end)];
II=eye(3*N1+3*N2+1);

C=[II(dir,:); ... % no-slip at walls
II(v1(1),:)-II(v2(end),:); ... % continuity of v across interface
II(u1(1),:)-II(u2(end),:); ... % continuity of u across interface
(II(p1(1),:)-rho1*g*II(eta,:))-(II(p2(end),:)-rho2*g*II(eta,:)); ... % Continuity of pressure
mu1*(dy1(1,:)*II(u1,:)+dx1(1,:)*II(v1,:))-mu2*(dy2(end,:)*II(u2,:)+dx2(end,:)*II(v2,:))]; % continuity of shear stress

E(loc,:)=0;
A(loc,:)=C;

% compute eigenmodes
[U,S]=eig(A,E);
s=diag(S);  [t,o]=sort(-real(s)); s=s(o); U=U(:,o);
rem=abs(s)>1000; s(rem)=[]; U(:,rem)=[];


# validation

For a theory we will de the same derivation as for free_surface_gravity.m#validation, with the difference that we have an other fluid at the top. We call \phi_1 and \phi_2 the stream functions in domain 1and 2. They satisfy the Lalace equations \displaystyle \Delta \phi_1=0, \quad \Delta \phi_2=0 thus we have \displaystyle \hat{\phi}_1=A_1 \cosh (\alpha (y-L_1))+B_1 \sinh (\alpha (y-L_1)) and the no-penetration at the top wall at height y=L_1 gives A_1=0 thus \displaystyle \hat{\phi}_1=B_1 \sinh (\alpha (y-L_1)) we have chosen to parameterize the hyperbolic functions with y-L_1 just to simplify the imposition of the boundary conditions. For domain 2 we have \displaystyle \hat{\phi}_2=A_2 \cosh (\alpha (y+L_2))+B_2 \sinh (\alpha (y+L_2)) and the no-penetration at the bottom wall at height y=-L_2 gives A_2=0 thus \displaystyle \hat{\phi}_2=B_2 \sinh (\alpha (y+L_2)) The advection of the interface gives \displaystyle s\hat{\eta}=v_1(0)=\phi_{1x}(0)=i\alpha \phi_1(0)=I\alpha A_1\sinh(-\alpha L_1)

We now use the continuity of v across the interface \displaystyle \begin{array}{l} v_1(\eta)=v_2(\eta)\\ v_1(0)=v_2(0)\\ i\alpha \hat{\phi}_1(0)=i\alpha \hat{\phi}_2(0)\\ A_1\sinh(-\alpha L_1)=A_2\sinh(\alpha L_2)\\ A_2=A_1\sinh(-\alpha L_1)/\sinh(\alpha L_2)\\ \end{array}

The continuity of the pressure at the interface gives \displaystyle \begin{array}{l} p_1(\eta)=p_2(\eta) \\ p_1(0)+P_1(0)+\eta P_{1y}(0)=p_2(0)+P_2(0)+\eta P_{2y}(0) \\ p_1(0)-p_2(0)=(\rho_1-\rho_2)g\eta\\ \end{array} where we have used the fact that the base pressure P is continuous accross the interface, the base pressure gradient is P_y=-\rho g with \rho different in the two fluids, and that \eta p_y is too small to be kept.

we need now to express the pressure as a function of \phi, for this we use the u component of the Euler equation at y=0 \displaystyle \begin{array}{l} \rho u_t=-p_x\\ \rho s\hat{u}(0)=-i\alpha \hat{p}(0)\\ \end{array} thus \hat{p}(0)=\rho s\hat{\phi}_y/i\alpha. We can thus express the continuity of the pressure using the streamfunctions \displaystyle i(\hat{p}_1-\hat{p}_2)=\rho_1A_1\cosh(-\alpha L_1)-\rho_2 A_2 \cosh(\alpha L_2) we can now use the relation between A_1 and A_2 to get \displaystyle sA_1 \sinh(\alpha L_1)\left[ \frac{\rho_1}{\tanh(\alpha L_1)}+\frac{\rho_2}{\tanh(\alpha L_2)}\right]=i(\rho_1-\rho_2)g \eta if we now use the relation obatined from the advection of \eta we obtain \displaystyle s^2=\frac{\alpha(\rho_1-\rho_2)g}{\frac{\rho_1}{\tanh(\alpha L_1)} + \frac{\rho_2}{\tanh(\alpha L_2)}} and we have used the fact that \cosh is even and \sinh is uneven. This is the final relation telling the behaviour of the eigenmodes of the flow.

We see that the flow is unstable if the top fluid is heavier than the bottom fluid (Rayleigh-Taylor instability), since then there are two solutions for s, one positive and one negative. The positive one correspond to the unstable eigenmode.

If now \rho_1 tends to zero, we recover the solution from free_surface_gravity.m#validation \displaystyle s^2=\alpha \tanh(\alpha L_2) g

For the validation we will compare the wave velocity from teh code to this. The wave speed is minus the imaginary part of the eigenvalue divided by the wavenumber \alpha. The theory for this is \displaystyle c=\sqrt{\frac{-(\rho_1-\rho_2)g}{\alpha \left[\frac{\rho_1}{\tanh(\alpha L_1)} + \frac{\rho_2}{\tanh(\alpha L_2)}\right] }}

% validation
alphavec=linspace(0,5,100);
ctheo=sqrt(-(rho1-rho2)*g./(alphavec.*(rho1./tanh(alphavec*L1)+rho2./tanh(alphavec*L2))));

cnum=max(imag(s(1:50)))/alpha;
plot(alphavec,ctheo,'r-',alpha,cnum,'b.');
xlabel('alpha');ylabel('wave velocity');
title('validation')
grid on

set(gcf,'paperpositionmode','auto');
print('-dpng','-r100','free_surface_gravity_twofluids.png'); The wave velocity as a function of the wavenumber \alpha

# Exercices/Contributions

• Please check that the codes behaves well in the limit of low density of the top fluid (thus its results tend to that of free_surface_gravity.m)
• Please check what happens when the top fluid is heavier than the bottom fluid (Rayleigh_taylor instability)————-> rayleigh_taylor.m
• Please add the surface tension at the interface (the pressure is nolonger continuous at the interface, it is equal to the Laplace pressure jump \Delta p=\sigma/R where \sigma is the intensty of the surface tension and R is the radius of curvature of the interface)————-> rayleigh_taylor.m
• Please compare the results of this code to a simulation with Gerris by using the first eigenmode as an initial condition in Gerris.