# Thread: Normal Mode Oscillation in the Ocean

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## Normal Mode Oscillation in the Ocean

An Estimation of Normal Mode Oscillation in the Ocean.

The basic property of the Ocean or Atmosphere is that they are highly stratified. Simply speaking, in the case of Oceans, the high density water always stays deeper, because it is heavy. The lighter ones stays one over the other according to its density decreases with height. For instance, if the water in the upper level gets heavier than the lower layer, the system is unstable. The heavier water have to sink down and the lighter water has to come up. This is accomplished by a process called 'vertical convection'. This process is quite common in the surface layer of the ocean, where the water gets heavier (saltier) by evaporation. This stratification in the Ocean (as well as in the atmosphere), and the tendency of it to restore the stability with a stable stratification causes an important property of the system. Any parcel of water displaced vertically from its stably stratified form, the restoring tendency gives a fundamental mod of oscillation within the system. The fundamental quantity, we usually refer to such a system, is so called the 'buoyancy frequency' or Brunt Vaisala frequency. For a layer structure of the Ocean, this can be defined as squar(N) = (g/rho0)(drho/dz), where rho is the density of water in each layer, and rho0 is the mean density and 'g' is the gravity.

Now the real question is, in a stratified ocean, for instance, a small perturbation is produced in any of the interface, what adjustment will be happen in the Ocean. The answer of this question leads to the important concept in the Ocean dynamics, called the 'Normal Modes'.

What are normal modes ?

In order to understand the normal mode theory of the ocean, let us follow a simple experiment in the laboratory. For a while, let us forget about the Oceans, and consider a simple oscillations of a 'spring and mass system' in a laboratory. Follow the figure.1 shown below. It is a combination of three masses and two springs.

Figure.1

The system is composed of three masses M1, m2 and M3 and are connected by two springs K1 and K2, where K1 and K2 are their 'spring constants'. Consider the system is resting on a 'smooth friction less table'. Now, let us squeeze this 'system' and release (or stretch a little and release). It is EASY to imagine that the system will undergo some kind of Oscillation. Although, it is NOT EASY to
imagine, What is the type of Oscillation it executes.

In order to understand the type of Oscillation of the system, we have to consider the Basic Balance Equations of the system. The simple logic is to assign a set of equations which contains the balance, force = mass x acceleration Thus for the three above masses M1, m2 and M3 let us write three equations as given above (Figure.1). Please note that, here the Force means the restoring force of the spring, which is by rule, F = -k.x.

The next thing is to find out the solutions of the above set of three equations. Let us follow the analatical method to find the solution of the above system of equations. i.e. an x = f(t) form for each masses. This will represent the instantaneous displacement of each masses at any time.

How to find these solutions? .

As it is clear that, the system will exicute some kind of oscillation, the necessary solutions will also be in the oscillatory form. With out much explanation, we can attribute the solutions to a set of oscillating sine or cose curves or a combination of it. For analatycal convinience, let us assume each mass is vibrating with a common frequency, say 'w' (omega) Let us find this common frequency. If 'w' is the frequency of oscillation of the system, at any instant, the displacement of each massess can be written in the most general form xi = x0.exp(iwt), i=1,2,3. represents mass 1, 2 and 3 respectively. So this is the general solution of the above system of mass vibrations. Thus the solution contains both the real part and imaginary part. Let us concentrate on the real part of the solution.

Analatycal method to find the solutions .

Since the above system is composed only of limited numbr of mass (three in our mase), it is convinient to find the analatycal solution of the system. If xi = x0. exp(iwt) , is the solution of the system (instantanious displacement of each masses), we can find the second order derivative of this, and substitute substitute back to the equations 1,2 and 3, we will get a set of three equations as follows. A common factor exp(iwt) is divided out from each term in the following equation.

Figure.2

With a simple rearranging we shall get a system of matrix as given below

Figure.3

A close examination of this analatical form of the vibrating system, we can see that this is in the form A.m=(lamda).m, a matrix eigen value problem, that yields eigen values (characteristic roots) 'lamda' and eigen function m . Each Eigen value will represent a 'type of (kind of or a manner of) ' oscillation of the system. Also each 'eigen value' corresponds to an eigen function', which is the solution of the system representing that 'particular manner' of osccilation. and they are called as the Normal Mod oscillation of the system. Here in this experiment, the eigen value 'lamda' is w^2, or the square of frequency of the oscillation. The above set of equation gives 'three' eigen values and each eigen value represents eigenfunction or a mod of oscillation. So, inorder to find the solution of the system of vibration, (i.e. x=f(t) form) we need to find find the Eigen values and the corresponding Eigen functions.

Eigen values and Eigen vector (function)

The eigen values can be found from the coefficient matrix, which is obtained by writing the above system in a secular form .ie. (A-lamda.I) = 0, where I is an unit matrix. The value of lamda is called [/b]'eigen values'[/b]. For any N x N matrix there are 'N' eigen values and for each eigen value there is an
Eigen vector. In the above experiment, the eigen values are (1) w^2 = 0, (2)w^2 = k/M and (3) w^2 = (k/M + 2k/m) . So, there will an associated for Eigen function for these three eigen values and that will give the solution (instantanious displacement of each mass ) of the system.

The following figure will help us to see the (tentative) oscillation of the system. For example, 1) corresponding the to the first eigen value, w^2 = 0.0, the eigenfunction is x1=x2=x3.. This means that there is no relative displacement between the masess or in other way we can say all the masses moves with the same displacement. Thus the motion (oscillation) corresponding to this eigen value is such that all the masses moves in same direction with same displacement. The other way of interpretation is that mod (type or kind) of oscillation of the system curresponding to the eigen value w^2 = 0.0 is, x1=x2=x3.

The second eigen value is w^2 = k/M. The eigenfunction of this mod is x1=-x3 and x2=0. The outer massesmoves oppositely whereas the central mass is at rest. Also the third eigen value is w^2 = (k/M + 2k/m). The eigen function of this mod is x1=x3, x2 = -(2M/m)x1, means the outer masses moves in same direction and the central mass moves opposite to the outer mass. Thus, the system composed of three masses (three equations), gives three eigen value, and represents three normal mod oscillation of the system. The word normal stands for the type of oscillation, it is perpendicular (normal) to the system.

The figure shows the oscillation of three modes separately. The displacement of the each masses is given as a graphical plot (please note that the values of the graphs are arbitrary, not produced from true solutions, but it gives the sketch of the real oscillations). The eigenfunction corresponding to the first mod is a straight line. In this case, all the masses moves in the same direction with same displacement. Thus the entire system behaves unique. The eigenfunction of the second mod shows that the outer massess moves opposite and the central mass is rest. The eigen function for this mod crosses the zero-line in the graph at one point. This is called the nodal point, or the point where amplitude of that mod is zero. Also, the third mod has two nodal points. The important thing should bear in mind that, at any instant the state of the system is a linear combination of this three mod of oscillation. Thus it turns out following coclussive points.

1. At any instant, the state of the system is a linear combination of the above three type of oscillations or mod of oscillations.

2. Thus the state of the system can be represented as the sum of these Normal Modes and it is an important tool to find the evolution of the system.

3. By observing the resultant oscillation of the system, we can not say exactly, which type of osccilation or Mod of osccilation is prominent in the system.

Figure.4

Why do we follow the above experiment ?

The above experiment gives us a simple explanation about the Normal Mod Osccilations. Now let us see, how this experiment help us to understand about the ocean. Let us apply the same principle to the Ocean. We can interpret an analogy in the Ocean very similar to this mass-spring oscillation. Consider, we are tilting the mass-spring system vertically, we can easily understand What is mean by the normal Mode Oscillation in the Ocean. see figure for an analogically view of spring-mass system and the ocean.

Figure.5

In the Ocean, each layer stands for M1, M2 and M3 of the above experiment and restoring forces due to the density difference between the interface of the two layers (or the 'reduced gravity') stands for the springs. Thus, for instance, any small perturbation produced in the ocean can excite oscillations vertically. Certainly for n-layer of ocean, there will be n-vertical (normal) modes.

Why these Oscillations are important ?

In the above experiment, we can see that, the entire oscillation in the mass-spring system can be represented by a 'combination of three mod of oscillation'. This is an effective tool because all the motion governing by the system can be represented by this three modes. Likewise in the case of ocean, with a finite vertical stratification, the solutions can be expressed as a sum of normal modes, each of which has a fixed vertical structure and behaves in the horizontal and time (x,y,t) dimensions in the same way as does a homogeneous fluid with a free surface (Gill, 1982). That means, the vertical oscillaion is accompanied with a fixed horizontal motion. To an extension to the above simple mass-spring system, the Ocean is a fluid body and it state is determined by various parameters such as composition, pressure, velocity etc... More over, the great utility of this Normal Mod construction in the ocean is that, with a sufficient approximation (The propogating waves are very long in horizontal scale compared to the vertical scale or the Hydrostatic approximation (pressure varies only with depth) ), the sysptem possess a seperable solutions into Horizontal and Vertical. And in that case, the each Mod of oscillation behaves indepndently and this also a great utility. Simply saying, at any instant, the state of the ocean, say u = u(x,y,z,t) can be found from the Normal Mod structure. In addition to that, if we apply the above approximations, the state of the system can be further simplified into u=u(x,y,t).u(z) .i.e., we seperated the Normal Mod and Horizontal motions into two.

How to find Normal Mod oscillation of the Ocean ?

In the above spring-mass experiment, we sought a solution of the form, x=x0.exp(iwt). In the case of stratified Ocean, the similar solutions of the perturbation can be expressed as w = w0.exp[i(kx+ly-wt)], where k and l are horizontal wavenumber associated with the vertical perturbationw.. Also, in the case of Oceans, there exists boundaries at the bottom and surface, which confines the energy to a region of finite vertical extent, though the horizontal propagations are permitted. Thus Ocean can be considered as a waveguide that causes energy to propagate horizontally (Gill, 1982). This wave guide property of the Ocean provides only a possible number of eigen values and in the long-wave limit (i.e. with very small values of 'k' and 'l' ) the corresponding eigenfunction are independent of the horizontal wave numbers. Here, more theoretical aspects are not discussed, because this post is mainly discussing 'How to calculate normal mod oscillation in Ocean'. The readers are refereed to (Gill, 1982) for the theoretical discussion of the Normal modes.

( Continues to the next post..... )
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## .... continuation

Numerical way to find Normal Mod oscillations of the Ocean

With sufficient approximations, the above theory lead us to a equation of normal mode of oscillation as d2w/dz2 + (N/Ce)2.w = 0, where 'w' is the vertical perturbation, and N is the Buoyancy frequency. (See the figure below for the more aligned view ). Here the quantiy 'Ce' is called the seperation constant, which is come from the seperation of vertical mition form the horizontal motion. This quanity has the dimension of the 'velocity', and can be considered as the 'Horizontal Speed of the Mod'. Certainly, it is clear that, for n-layer of ocean, there will be 'n' Normal mod oscillations. Thus there curresponds to a 'Horizontal phase speed, Ce, for each of these Mode. i.e. Cn, n=0,1,2, etc... These are called the 'Phase speed of each Mod. The first mod (n=0. this is the convention we followed in oceanography to represent the first mod as 'zero', instead of 'one'.) is called the 'Barotropic Mod'. Let us remeber back, in our mass-spring experiment, the mod with w^2=0.0 gives a uniform motion of the entire system. Likewise, in the case of ocean, the first mod, n=0, gives a uniform oscillation (in vertical) or motion (in horizontal) of the ocean curresponding to the first mod (n=0). This is called the Barotropic Mod. Thus for a Barotrpic Mode, the entire ocean will have a unique motion (either vertical or Horizontal). The phase speed Ce for this mod, in the case of ocean is, around 250m/second. This is the speed of surface gravity wave in shallow water. (for a depth of approximately 4 km). The another modes, i.e., n=1,2,3, etc ... are called the Baroclinic Modes. They are arranged in order from 1 to n, the first Baroclinic Mode (n=1), generally has a phase speed of Ce ~ 3 m/sec for the ocean. If we remember back, a corresponding analogy for this mod can be attributed to the mass-spring system with Eigen value w^2 = -k/M. The motion (eigen function) curresponding to this eigen values was, the outer masses moves opposite and the central mass is at rest. Likewise, a First Baroclinic Mode in the ocean approximate to a two-layer system, with each layer has opposite motion. This is approximation is rather justifiable in the real ocean, by assuming an active thermocline and a deep quiscent layer. The discussion about a model of this kind is given in this post. The contributions from the higher modes (second Baroclinic, third Baroclinic and so on) are rather small.

Figure.6

Let us consider the way to find the normal mod oscillation of the ocean. All we need is, to solve this equation for the Ocean (Figure.6) for each layer and apply the boundary condition. Let us compare this equation with the equations of the mass-spring system. In the mass-spring system, we represent three equation corresponding to the three mass. Likewise, in the case of n-layer ocean (say), we have to express the above equation for each layer. Thus for
each layer, we will get an equation with n-terms. So a resulting coefficient matrix will be a symmetric matrix with dimension n x n. For example, for a 34-layer ocean, we get a matrix of size 34 x 34. However NOT all the elements in a row are non-zero elements. This can be easily understand from the mass-spring equations. Look at the coefficient matrix of the mass-spring equation. (figure.2). In the first row, the last element is zero. And in the 3rd row the first element is zero.

How does it appear?.

Because, when we write equation for the mass M1, we considered M1 and the mass adjacent to it (i.e mass M2), and contribution from mass M3 is zero. Likewise, whatever mass we add after M3, the resulting elements in the first row of the
coefficient matrix will be zero. This picture will be more clear, when we look at the second row of the Matrix. All elements are non-zero, because second row represents the coefficient of the second mass. Thus the M2 and adjacent masses (M1 and M3) are considered. In the third row (coefficients for third mass M3), the first element is zero, because M1 is not the adjacent mass. (spend some time to understand the situation by looking at the figure.2)

In the case of n-layer ocean, an Analytical solution for such a system will be a tedious task (as we have done for the mass-spring example). Instead, we can find the solutions in numerical methods, by a finite-differencing scheme. All we need is a coefficient matrix to find the eigen values. So our first task is to find the coefficient matrix for n-layer of ocean for the equation (Figure.6). This is in the form [(1/N2)d2/dz2].w = (-1/Ce2).w, which is equivalent to A.w = lamda.w, where the A=[(1/N2)d2/dz2] is the coefficient matrix and lamda=(-1/Ce2) is the eigen value. Just like the above discussion, not all the elements in the coefficient matrix of 'A', are non-zero elements. Instead, for a particular layer-m (say), the m-1, m and m+1 elements Will be non-zero and others are zero.

What else we require to find the matrix A ?.

The A=[(1/N2)d2/dz2]. So we need layer density, to find N2 (N-squar) (see above for What is N), and vertical difference 'dz'. N- is the buoyancy frequency and it comes from the density structure. Thus we need the density observations of the ocean to find the matrix-A. The following figure gives a systematic sampling of density in the real ocean. Let us say, the density sampling are coming from '7' depth levels of the ocean. Thus we can find the value of 'N', at midle level of each locations ZZ(k), following the finite difference form (see Figure.6). Once we find the 'N', at the locations ZZ(k), we can express the matrix A in the finite difference form as follows. 1/(N2k.dZk.dZk-1) - [ (1/dZk.dZZk-1) + (1/dZk.dZZk) ].1/N2k + 1/(dZk.dZZk.N2k) ; where dZk = Z(k) - Z(k+1) is the layer thickness from observation points and dZZk = ZZ(k) - ZZ(k+1) is the thickness of the mid-points between observational levels (see Figure.7 for detailed understanding). Thus we can find the coefficient of each layer at the location ZZ(k).

Figure.7

Boundary Conditions .

As mentioned earlier, the ocean has fixed upper (surface) and lower (bottom) boundary conditions. All the motion is confined to these boundaries so that the wave are reflected at this boundaries back into the ocean and propogates horizontally. For the case of vertical petrubation 'w', the boundary conditions would be w = 0 at z=0 and z= -H. For the horizontal velocity (u), the boundary conditions would be 'no normal flow' across the boundary, i.e. dU/dn = 0 at z=0 and z=-H. The seperable solutions for w, u and p can be represented as, In the following example, let us upply the boundary conditions for 'U', (the horizontal velocity) and let us find the normal mod structure of the 'U'.

How does this Coefficient Matrix (A) look like ?

As we seen earlier, the coeffiecient matrix will be a symmatric matrix with size 7 x 7 for n-layer ocean. Als NOT all elements are non-zero. More over applying the boundary condition, i.e. no-normal flow across the boundary for the 'w' components, the coefficient matrix will reduce to a 'Tri-diagonal system', with non-zero elements along the leading diagonal, upper diagonal and lower diagonal, and all other elements as zero. See the following figure and spend some time to understand the situation. The color coding of the elemetns will help to compare with the above finite difference form which is also given in color.

Figure.8

Thus our 7-layer ocean gives a coefficient matrix of size (7-1) x (7-1) or 6 x 6. Thus for n layer ocean, a finite difference scheme will yeild a (n-1) x (n-1) matrix as above.

What is next step ?

So now we have a coefficient matrix. The immediate next step is to find the eigen values for this matrix. There are built-in packages to find the Numerical Eigen Values for FORTRAN or any other softwere. For example, in FORTRAN, the libraries called (LAPACK or BLAS) (see this post for how to invoke library linking in a FORTRAN programming) are available. If we call the sub-routines from these libraries, it will calculate the Eigen values and Eigen vectors (functions).

An example

Thus if we have a density profile, we can find the Normal Mod structure of the ocean. The following figure gives the normal mod structure calculated by following above procedure for the Indian Ocean in a location at 83 degree East and 18 degree South. . The eigen functions (for the horizontal velocity) are shown. The first mode, n=0, is the Barotropic Mod, the eigen function is a straight line, showing that all the layer have unique velocity. The next mode, n=1, first Baroclinic mode has a eigen function (solution) with a zero-crossing at nearly 1500m. Thus the layer above 1500m will act as a single layer for this mode and layer below will act as a another layer (a two-layer approximation). The second Baroclinic Mode corsses the zero-line at two points, and so on the higher modes.

Figure.9

Thus the vertical structure of the Normal Mod oscillation help us to understand the location and amplitudes of each mode as well as its significance in the resultant ocean dynamics. Thus examining the Normal Mod sturcture will be a usefull research component. More over, it shed light into the concept of analysis of OGCM (Ocean General Circulation Model) resutls. The sophesticated OGCM will thus contains the dynamics associated with large number of modes. So one cannot state with confidence, which mode is prominent. Perhaps, by looking at the mod speed one can assess the curresponding Baroclinic Mod. Here is the real advantage of a simpel layered model compared to more sophesticated OGCMS. A layer model (for exaple 1&1/2 layer) with 'motion-less lower layer and active upper layer' has only one Baroclinic (first) mod. (There is no Barotropic mod because the lower layer is at rest). For 2&1/2 layer model, there are first and second Baroclinic Modes. Thus, these simple model gives us better confidence for the interpretations of results. However it comes with the penalty that, simpler model does not represent the detailed physics.

Reference

Chelton, D. B., R. A. de Szoeke, M. G. Schlax, K. El Naggar, and N. Siwertz, 1998: Geographical variability of the first-baroclinic Rossby radius of deformation. Journal of Physical Oceanography, 28, 433-460.

Kantha, L. H, C.A. Clayson (2000): Numerical Models of Oceans and Oceanic Processes, International Geophysics Series, Vol.66, 820p.

Gent P. R. and K. O'Neill, 1983. A model of the Semiannual Oscillation in the Equatorial Indian Ocean. Journal of Physical Oceanogrpahy, Vol. 13, 2148-2160.

Gill A. E., Atmospher-Ocean Dynamics, 1982, International Geophysics Series, Vol.30

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PS: If any of the above statments feels miss-leading or discussion is incomplete please feed back. Excusse is requested for spelling mistakes.

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## Hi

Wonderful Post
Cheers
Nuncio

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Dr.Vinu,
Very interesting and infomative thead. I wish to get few clarifications. i) the equation for ocean data as given in the thread is 1/(N2k.dZk.dZk-1) - [ (1/dZk.dZZk-1) + (1/dZk.dZZk) ].1/N2k + 1/(dZk.dZZk.N2k) ;. Are the 3 tems in this equation the
A1,A2,and A3 in trigonal matrix. ii) Is it possible to get eigen functions of the matrix in mathlab,using any command. Our attempts produced wrong values. iii) In the example presented, eigen functions are plotted up to 4000m. If data intervals are close, then the trigonal matrix would be very large, I suppose. What could be ideal sampling interval ?
regards, k k varma

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## Eigen values in Matlab

Numerical methods to solve for Eigen Values.

Thank you for your interest in this post. Here are the clarification for your questions.

i) Yes, the Coefficients A1, A2 and A3 are the three terms in the equation. It can identified by looking at the color of each term.
In the above post, the 'text' representation of the equations may cause a little confusion. A clear format of the equations are
given below. The colors are given seperately so that one can easily assess, what values are we need to find the coefficient
A1, A2 and A3 respectively. The W(k-1) , W(k) and W(k+1) are the variables here, just like x1, x3 and x3 (the 'displacements')
of the mass-spring experiment.

ii)Yes, Eigenvalues are possible to obtain by using Matlab. In Matlab, the function Eig (X) will give the Eigenvalues and
Eigenvector (Eigenfunction) of a Matrix 'X'. A more complete syntax is,

[V,D] = EIG(X);

which produces a diagonal matrix D of eigenvalues and a full matrix V whose columns are the corresponding eigenvectors
so that X*V = V*D

Please give a command "help eig" in the Matlab window and you can get the complete description of the Eig function there.

iii) The data intervals in vertical profiles are usually specified in standard depths, for example, as in the World Ocean Atlas
(Levitus et al, 1994). This contains about 40 'non-uniform' depth interval. This is a usual practice, only because the
availability of data is limited to that intervals. If one has observations in a more-finer vertical resolution,
that can be used to solve the above equations. As you pointed out, the size of the resulting Matrix will be as big as the
number of vertical data points. However, using the computer resources vailable now, it is not so computational-expensive
to find Eigen values of a 100 by 100 matrix (for example).

The significance of the sampling interval in the resulting normal mod structure.

I think, this need further explanation. The mod structure depends on the density startification or the Brunt-Vaisala
frequency (N^2). The largest values of this N^2 is usually seen at the thermocline depth (100m-200m).
From the definition of the thermocline, it is obvious, Why this is so. Because the thermocline is the depth
range where maximum gradient in the temperature is found. Now, if the sampling intervel is very poor
at these surface depth, the N^2 profile will be affected. This may lead to incorrect structure of the Normal mod.
But, the standard depth interval for sampling the density is finer enough to resolve the thermocline gradient.
In short, a finer sampling intervel at the surface with a coarse interval at deeper level will be a reasonable choice.

Thanks

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Hi cheers Vinuetta
Gr8 thread just saw it. Just kooked superficially. Expect more of this kind.

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Dr.Vinu
Thanks for the clarifications. It would be useful,if you could give some more information i) when one uses closely spaced data, say 20 layers in upper 1000m, one gets 19 modes. Are all the modes important( we find that the amplitudes of all modes are nearly equal). ii) when number of levels are more the profile becomes very noisy unlike the one that was presented (fig 9 in your note). iii) zero crossings uccur in the upper ocean itself (100m) and below 200m all the modes are negligible.iv) are there some publications that look into the nomal modes in the Arabian Sea.
regards,K.K.varma

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## Vertical structure

Significance of vertical modes

Here are the possible ways to explain the behaviour of your plot.

1) It is to be taken care that what boundary condition one use. If you use
a limited depth at the middle of the ocean, say upto 1000 m, you can not
apply the boundary condition like; "vertical velocity is zero at boundaries".
Because for the vertical velocities to be zero (or if that is our boundary
condition) then our boundary must be either at the ocean surface or
at the ocean bottom. Because through these surface water cannot
cross the boundary, so the vertical velocity is zero.
So when we apply boundary conditions, it should follow a physical sense.

Now, if you want to use 1000m as your boundary, you can'not
treat your normal model equations in terms of 'w'. Instead, you have to
use the equation in terms of either 'u' or 'v', where they are horizontal
velocities. The basic equation will NOT change even if you choose your
state variable as u or v. Only the boundary condition changes. That means you
will have a definite velocity at the both the ends (top and bottom) and
one have to specify that into the matrix. It is a very uncertain quantiy
because we dont know the veolcoties a priori.
So, better to use the vertical velocity as the state variable so that
we are SURE about the boundary conditions. It can only have a zero-value
at the boundaries because water will not penetrate across the seas surface
and ocean floor.

Now, a comment on limiting your domain to 1000m, I am rather confused over
this. You may not get the real normal mode structure if you limit your bottom
at 1000m. Because, I think, the normal mod structure
depends on the N^2 profile of the ocean from surface to bottom.
The boundaries are so critical that, it is from where the oscillatons reflects
back into the ocean interior. Actually the boundaries acts as the oceanic
wave guide. If you assume 1000m as your boundary, the wave solutions
can still penetrate through the bottom. So I think
one should use the entire water colum depth to resolve the correct picture.
It is a very interestng question to ask, At what conditions one may able to limit the bottom level to a
specific depth? I dont know the answer exactly. I think, it depends on the
N^2 profile. If the N^2 is "dead" below a particual depth (means not varying
after a particual depth) one may possible to limit the boundary upto that
depth. I am not sure on this statement

2) I can not figure out about why the amplitude of different mode looks
similar. A nosiy profile may represent higher modes. So if you use more levels
it may resolve higher modes, which is very nosiy about the zero axis.

Can you post an example figure?

3) I am not awire about a publication on Arabian Sea about normal mode
structure. But, Shankar et al (1999, JGR, ?????) (NIO, Goa)
resolved the normal mode structure of Bay of Bangal.
I think, they have published two papers as part-I and 2. If you could chek
with them, it may be possible to find a publication on Arabian Sea.

However, it will be highly useful if you look at the Chelton et al. (1998)
paper which I given in reference in Post-2. In their appendix, it is given
with different discretization/boundary conditions they used and its
significance on the resulting structure.

hope some of this helps.

-Thanks

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