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Discuss the graphics for the choose of grid for the linearised shallow water equations without friction!  

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The linearized shallow water equations or the gravity-inertia wave equations are:

\(\frac{\partial u}{\partial t}-fv=-g\frac{\partial h}{\partial x}\\ \frac{\partial v}{\partial t}+fu=-g\frac{\partial h}{\partial y}\\ \frac{\partial h}{\partial t}+H(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y})=0\)


If we consider solutions in wave form than we obtain:


For each of the three grids we use the simplest centred approximations for the space derivative and the Coriolis terms.

If we substitute the wave solution into the numerical schemes we get the frequency equations. This are basically dependent of two paramters: \(k\Delta x\) and \(gH/f^2\) (Rossby Radius to the square)

The equations are not shown but discussed with the following graphic:

Grid A:

  1. The frequency reaches a maximum at  \(k\Delta x=\pi/2\) , i.e. a wave length of 4 grids. The group velocity for this wave length is zero, which means the wave energy stays near that point. 
  2. For \(\pi/2 < k \Delta x < \pi\)the frequency decreases as the wave number increases. Thus, the group velocity vector has the wrong sign. 
  3. For the two-grid-interval wave (\(k\Delta x=\pi\)) the group velocity vector is again zero. 

Grid B:

  1. The frequency increases monotonically. 
  2. Reaches a maximum for \(k\Delta x=\pi\) . There the group velocity is again zero. 

Grid C: 

  1. The frequency increases monotonically in a similiar way to the one for the B-grid, when the Rossby Radius is longer than a half grid (\((\sqrt{(gH)/f}> \frac{\Delta x}{2})\) .
  2. If the Rossby-Radius resolution fall beneath that, then the frequency will decrease for increasing wavenumber throughout \(0< k\Delta x <\pi \)
  3. The big advantage of the C-grid: Velocities are perpendicular on the walls of the grid-box, which makes differentiating straightforward. 
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Taylor Series. 

Taylor series of f(x) at the point a:


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Which method is typically used for the numerical solution of differential equations?

The grid point method

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Consistency: What is it? 

  • A numerical scheme needs to be above all consistent.
  • This means that the approximation should approach the derivative when the grid interval approaches zero. 
  • For a consistent scheme the truncation error needs to be at least of order one. 
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What is the truncation error?

The truncation error gives a measure of how accurately the difference quotient approximates the derivative for small values of \(\Delta x\).

The usual measure of this is the order of accuracy. This is the lowest power of \(\Delta x\) that appears in the truncation error. 

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How is the algebric equation, obtained by replacing derivatives with finite difference approximations, called?

finite difference approximation or finite difference scheme. 


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What is the error of a numerical solution?

The difference between the numerical and the true solution. 

\(u_j^n-u(j\Delta x,n\Delta t)\)

Most of the time we cannot find the error of the solution. However, we can always find a meassure of accuracy. 

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What is Convergence? A consistent scheme does not need to be convergent. Why?

The truncation error of a consitent scheme can be made abitrarily small by a suffiecient reduction of the increments \(\Delta x\;and\;\Delta t\).

Although, this cannot be expected for the error of the numerical solution: \(u^n_j-u(j\Delta x, n \Delta t)\)

Definition: If the error of the numerical solution approaches zero as the grid is refined, the solution is called convergent. If this holds for any initial conditions, than the scheme is called convergent too. 

Consistency od a scheme does not guarantee convergence!

For convergence of a scheme, the initial conditions needs to lie within the domain of dependency.

For the advection equations  (Euler forward and upstream) this is the case if the CFL criterion holds. That is:

\(c\Delta t \leq\Delta x \)

(If a scheme is convergent it is also stable!)

consistency plus stability gives convergence!