# Lernkarten

Karten 41 Karten 1 Lernende English Universität 21.07.2018 / 25.07.2018 Keine Angabe
0 Exakte Antworten 41 Text Antworten 0 Multiple Choice Antworten

What is the connection between consitency, stability and convergence?

consistency plus stability gives convergence!

Finite time differences: facts without formula (with order of accuracy, centred/uncentred, explicit/implicit)

To define time schemes, we consider the equation:

$$\frac{du}{dt}=f(u,t)\;\;\;where\;\;u=u(t)$$

1. Two-level-schemes: use timesteps n and n+1
1. $$u^{n+1}=u^n+\int^{(n+1)\Delta t}_{n\Delta t}f(u,t)dt$$
2. Euler or forward scheme:
• $$u^{n+1}=u^n+\Delta tf^n$$

• first order accurate scheme: $$O(\Delta t)$$

• uncentred

• explicit

3. Backward scheme:

• $$u^{n+1}=u^n+\Delta tf^{n+1}$$

• first order accurate scheme: $$O(\Delta t)$$

• uncentred
• implicit
4. Trapezoidal scheme
• $$u^{n+1}=u^n+\frac{1}{2}\Delta t(f^n+f^{n+1})$$
• second order accurate scheme: $$O((\Delta t)^2)$$

• implicit
5. Matsuno or Euler-backward scheme
• to increase the accuracy: first an Euler forward time step, than a backward step
• $$1.\;u^{n+1/2}=u^n+\Delta t f^n\\ 2.\;u^{n+1}=u^n+\Delta tf^{n+1/2}$$
• explicit
• first order accurate scheme: $$O(\Delta t)$$

6. Heun scheme
• first Euler forward, then Trapezoidal
• $$1.\;U^{n+1*}=U^{n}+\Delta t*f^n\\2.\;U^{n+1}=U^n+\frac{1}{2}\Delta t(f^n+f^{n+1})$$
• second order of accuracy
• explicit

1. Three-level-schemes: use timesteps n-1, n, n+1
1. $$u^{n+1}=u^{n-1}+\int^{(n+1)\Delta t}_{(n-1)\Delta t}f(u,t)dt$$
2. leapfrog scheme
• $$u^{n+1}=u^{n-1}+2\Delta tf^n$$
• second order accurate scheme: $$O((\Delta t)^2)$$

•

most widely used scheme in atmospheric and ocean models.

What is stability?

A solution $$u^n_{j}$$ is stable if the error $$u^n_j-u(j\Delta x,n\Delta t)$$remains bounded as n increases, for fixed values of $$\Delta x$$ and $$\Delta t$$.

Of course, we assume that also the true solution is bounded, which is usually the case.

A finite difference scheme is stable if this is true for any initial condition.

Explain the Von Neumann's method!

• only able to test the stability of linear equations or linearized versions
• most frequently used
• main concept: every solution to a linear equation can be expressed in a Fourier series, where each harmonic component is also a solution to the equation. Thus, it is enough to study the stability of one Fourier Component on it's own! That is: $$u_j^n=u_0e^{ik(j\Delta x-C_Dn\Delta t)}$$
• We analyze stability with the amplification factor $$\lambda$$. If the true solution doesn't grow over time it is sufficient fo stability that: $$|\lambda|\leq1$$.
• we define: $$U^{n+1}=\lambda U^{(n)}$$

Are in general centred or uncentred schemes with higher order of accuracy?

In general, centred scheme has an accuracy of order 2, whereas uncentred scheme are only of order 1 in accuracy!

The oscillation equation

equation: $$\frac{dU}{dt}=i\omega U$$

analytical solution: $$U(t)=U(0)e^{i\omega t}$$

start with the stability analysis: $$U^{(n+1)}=\lambda U^{n}$$

We define $$\lambda=|\lambda|e^{i\Theta}$$

Then the numerical solution can formally be written as:

$$U^{(n)}=|\lambda|^nU^{(0)}e^{in\Theta}$$

$$|\lambda|(<,>,=) 1$$ leads to: damping, unstable or neutral

The relative phase change leads to:

$$\frac{\Theta}{\omega\Delta t}(>=<)1$$ leads to: accelerating, no effect, decelerating

For accuracy we want to have the ampflification factor and the relative phase speed close to unity. Unless it is a computational mode. Then we want strong damping!

Euler scheme:

• $$\lambda =1+ip$$
• unconditionally unstable

Backward Scheme:

• $$\lambda=\frac{1}{1+\omega\Delta t}(1+i\omega\Delta t)$$
• unconditionally stable
• it is always damping and damping increases with frequency.

Trapezoidal scheme:

• $$\lambda=\frac{1}{1+\frac{1}{4}p^2}(1-\frac{1}{4}p^2+ip)$$
• $$|\lambda|=1$$
• the trapezoidal scheme is always neutral (and thus stable)

Matsuno scheme:

• stable if: $$|\omega\Delta t|\leq 1$$ (so time step must be small enough!)

Heun scheme:

• very weak unstable. With small time steps often the instability can be tolerated

Leap-frog scheme:

• stable and neutral if $$|\omega \Delta t|\leq1$$

Fazit: Ich would use leap frog because it has a high accuracy and under specific conditions it is stable and even neutral! Although it is implicit what is in general more complicated to deal with, so then trapezoidal would be an alternative: unconditionally neutral. Unconditionally unstable is the euler forward and thus not useable. The Heun Scheme is also unconditionally unstable but at some situation tolerable.

Is it possible to avoid the computational mode?

• It's through the "round of error" that one can not eliminate the computational mode completely, even when it was avoided in the initail conditions.
• Although " round of errors" are of little effect in atmospheric models.
• It can be shown that the Heun scheme is a good choice to compute the computional initial condition, since it will give a smaller amplitude of the computational mode.

with oscillation and friction term: A combination of schemes:

We might like to use the leapfrog scheme for the oscillation term but we cannot use it for the friction term. So we use two different schemes with Euler forward for the friction!

$$\frac{du}{dt}=i\omega u-\gamma u\\\frac{u_j^{n+1}-u_j^{n-1}}{2\Delta t}=i\omega u^n-\gamma u^{n-1}$$