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stability analysis with the Neumann method: advection equation with leap-frog scheme.

\(\frac{u^{n+1}_j-u^{n-1}_j}{2\Delta t}+c\frac{u^n_{j+1}-u^n_{j-1}}{2\Delta x}=0\)

\(|\lambda|^2= \left(\frac{c\Delta t}{\Delta x}*sin(k\Delta x) \right)^2+1-\frac{c\Delta t}{\Delta x}*sin(k\Delta x) \)

The scheme is conditionally stable with the Courant-Fredrichs-Lewy or CFL criterion! 

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What is the Computational Mode? Show it for the leap-frog scheme and the oscillation equation!

A spurious solution to a finite-difference approximation to a differential equation that is not related to the physical solutions of the differential equation.

For instance, the leapfrog differencing scheme can introduce a computational mode.

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three level schemes like the leap-frog-scheme requires more than one initial condition (the physical initial condition \(u^{n=0}\) and the computational initial condition \(u^{n=1}\) .

taken the oscillation equation: \(\frac{du}{dt}=i\omega u\;\;where\;\;u=u(t)\)

with the solution:\(u=u_0e^{i\omega t}\)

the leap-frog scheme can be written as: \(u^{n+1}=u^{n-1}+2i\omega \Delta tu^n\)

Then there exists two amplification roots: \(\lambda_{1,2}=i\omega \Delta t\pm \sqrt{1-(\omega \Delta t)^2} \)

Thus there are two solutions of the form: \(u^{n+1}=\lambda u^n \)

Since we are solving a linear equation its solution will be a linear combination of the two solutions:

\(u^n=a\lambda_1^nu^0_1+b\lambda_2^nu^0_2\)

The physical mode is the one where \(\lambda \rightarrow 1\) when \(\Delta t \rightarrow 0\) . The other is the computational mode, which is introduced by the numerical scheme.

 

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Stability analysis; Neumann method; advection equation; The upstream or upwind scheme

The upstream or upwind scheme: 

\(\frac{u_j^{n+1}-u_j^n}{\Delta t}+c\frac{u^n_j-u^n_{j-1}}{\Delta x}\;\;\;if\;c>0\\ \frac{u_j^{n+1}-u_j^n}{\Delta t}+c\frac{u^n_{j+1}-u^n_{j}}{\Delta x}\;\;\;if\;c<0\)

The von Neumann stability analysis gives us:

\(\lambda=1-\frac{c\Delta t}{\Delta x}[1-cos(k\Delta x)+i*sin(k\Delta x)]\)

it follows:

\(0\leq\frac{c\Delta t}{\Delta x}\leq 1\)

which implies that c has to be positive to have stability. The scheme is hence conditionally stable. 

advantage: A disturbance cannot propagate in the direction opposite to the physical advection. Thus, no parasitic waves will contaminate the numerical solution. 

disadvantage: The upstream scheme is highly diffusive!

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Alles über die Welle!

\(u(x,t)=u_0*exp(i(kx-\omega t))\)

with:

  • angular frequency... \(\omega=2\pi f\)  mit  \(f=1/T\)
  • wave number... \(k=2\pi/\lambda\)
  • Dispersion-relation: \(\omega =kc\)

\(u(x,t)=u_0*exp(ik(x-ct))\)

group velocity: \(c_g=\frac{\partial \omega }{\partial k}\)

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How to integrate the Rayleigh friction numerically?

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Overview: 

  1. The leap-frog scheme is not a good idea since the stability analysis of Neumann gives unconditionally instability
  2. Instead one should use the Crank- Nicholson scheme
    1. This is unconditional stable but gives meaningful solutions only for \(\gamma \Delta t<1\)
    2. The scheme is strictly seen implicit, but since the discretization doesn't vary in space, one can rearrange the solution to an "explicit scheme".

 

\(\frac{\partial u}{\partial t}=-\gamma u\) where \(\gamma >0\)

the analytical solution is: \(u(t)=u_0*e^{-\gamma t}\)

leap-frog discretization brings:

\(\frac{u^{n+1}_j-u^{n-1}_j}{2\Delta t}=-\gamma u_j\)

The stability analysis by Neumann shows:

\(\lambda_{1,2}=-\Delta t*\gamma \pm \sqrt{(\gamma \Delta t)^2+1}\)

\(|\lambda |>1 \), which shows unconditional instability!

Typically used for the Rayleigh friction is the Crank-Nickolson scheme: here the right hand side is taken as an average of n-1 and n+1 such that:

\(\frac{u_j^{n+1}-u_j^{n-1}}{2\Delta t}=-\frac{\gamma }{2}(u_j^{n+1}+u_j^{n-1})\)

The stability analysis by Neumann gives:

\(\lambda^2=\frac{1-\gamma\Delta t}{1+\gamma \Delta t}<1\)

Hence, the scheme is absolutely stable. Although, one needs to choose \(\gamma \Delta t<1\) to get a realisitc solution. 

Furthermore, the scheme seems to be implicit, but since the discretization doesn't vary in space one can just rearrange the equation to an explicit scheme! 

 

Arakawa:

  • Euler scheme: conditionally stable
  • backward scheme: stable if \(\gamma \Delta t>0\) (unconditional stable)
  • trapezoidal scheme: unconditionally stable

Fazit: I would use the Crank-Nicolson scheme, since this is unconditional stable and has an order of accuracy of 2. The popular leap-frog scheme is unconditional unstable and can't be used. 

 

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numerical integration of the heat equation (diffusion equation)!

\(\frac{\partial u}{\partial t}=A\frac{\partial^2u}{\partial x^2}\;\;where\;\;A>0\)

has the analytical solution: \(u(x,t)=u_0e^{\pm ikx-Ak^2t}\)

Euler forward: conditionally stable

Leap-frog: unconditionally unstable

Crank-Nicholson scheme/trapezoidal scheme: unconditionally stable

heat equation with leap frog scheme:

\(\frac{u_j^{n+1}-u_j^{n-1}}{2\Delta t}=A\frac{u_{j+1}-2u_j+u_{j-1}}{(\Delta x)^2}\)

for Neumann method set: \(u^n_j=u_0\lambda^ne^{ikj\Delta x}\)

depends on the time step (n-1,n,n+1) for the spatial derivative on the right hand side: 

  • for n: (leap frog)
    • \(\lambda^2 +\frac{8A\Delta t}{(\Delta x)^2}sin^2(\frac{k\Delta x}{2})\lambda-1=0\)
    • the scheme is unconditionally unstable
    • (identity of use: \(sin^2(x)=\frac{1}{2}(1-cos(2x))\))
  • for n-1: (Euler forward)
    • the scheme is conditionally stable with \(\frac{A\Delta t}{(\Delta x)^2}<1/4\) (\(-1\leq\lambda\leq 1\))
    • to avoid imaginary roots and oscillating of the solution we recommend: \(\frac{A\Delta t}{(\Delta x)^2}<1/8\) (\(\lambda^2>0\))
    • \(\lambda^2=1-\frac{8A\Delta t}{(\Delta x)^2}sin^2(\frac{k\Delta x}{2})\)
  • for an average of n-1 and n+1 (Crank-Nicholson Scheme)
    • absolute stable but imaginary roots which are leading to oscillating solutions can occur
    • \(\lambda^2=\frac{1-\frac{4A\Delta t}{(\Delta x)^2}sin^2(\frac{k\Delta x}{2})}{1+\frac{4A\Delta t}{(\Delta x)^2}sin^2(\frac{k\Delta x}{2})}\)
    • to avoid oscillating imaginary roots on should use: \(\frac{A\Delta t}{(\Delta x)^2}<1/4\)

There leap-frog can be replaced by an Euler forward in time (replace all n-1 by n). The stability analysis doesn't change. 

Fazit: The leap-frog scheme is unconditionally unstable. Although the Crank-Nickolson scheme is unconditionally stable but one should use \(\frac{A\Delta t}{(\Delta x)^2}<1/4\) to avoid oscillating solutions. 

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Numerical integration of the Poisson and Laplace equations (elliptic)

Poisson's equation in two dimensions: 

\(\nabla^2 u=(\frac{\partial^2}{\partial x^2}+\frac{\partial ^2}{\partial y^2})u=f(x,y)\)

If \(f(x,y)=0\) then it is the Laplace equation. 

The equation can be discretised as 

\(\frac{u_{i-1,j}-2u_{i,j}+u_{i+1,j}}{(\Delta x)^2}+\frac{u_{i,j-1}-2u_{i,j}+u_{i,j+1}}{(\Delta y)^2 }=f_{i,j}\)

If we consider a square grid \((\Delta x=\Delta y)\) we get:

\(u_{i,j}=\frac{1}{4}[u_{i-1,j}+u_{i+1,j}+u_{i,j-1}+u_{i,j+1}-(\Delta x)^2f_{i,j}]\)

To solve this we need boundary values for the domain and initial values at the iteration level \(u_{i,j}^m\;\;(m=0)\)

 

most simple form to solve this by iteration is the Jacobi iteration. Quite inefficient and not used for practical purposes. 

\(u_{i,j}^{m+1}=\frac{1}{4}[u_{i-1,j}^m+u_{i+1,j}^m+u_{i,j-1}^m+u_{i,j+1}^m-(\Delta x)^2f_{i,j}]\)

 

An improvement in efficiency is the Gauss-Seidel iteration where newly computed values are partly used in the same iteration level: iteration level m+1 values are available for nodes (i-1,j) and (i,j-1).

\(u_{i,j}^{m+1}=\frac{1}{4}[u_{i-1,j}^{m+1}+u_{i+1,j}^m+u_{i,j-1}^{m+1}+u_{i,j+1}^m-(\Delta x)^2f_{i,j}]\)

 

Gauss-Seidel iteration can be further improved by increasing the convergence rate using the method of SOR (Successive Over Relaxation). The change between two successive Gauss-Seidel iterations is called the residual c, which is defined as: 

\(c=u_{i,j}^{m+1}-u_{i,j}^m\)

The Gauss-Seidel residual is multiplied by a relaxation factor \(\omega\) and new iteration value is obtained from: 

\(u_{i,j}^{m+1}=u_{i,j}^m+\omega c=(1-\omega )u^m_{i,j}+\omega\widehat{u}_{i,j}^{m+1}\)

where  \(\widehat{u}_{i,j}^{m+1}\) denotes the new iteration value obtained from Gauss-Seidel method. 

\(u_{i,j}^{m+1}=(1-\omega)u_{i,j}^m+\)

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Sketch the A,B and C-grid!

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