# 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

classification of second-order linear and homogeneous PDE

second-order linear and homogeneous PDE are from the kind:

$$.\\a \frac{\partial^2u}{\partial x^2}+b \frac{\partial^2 u}{\partial x \partial y}+c \frac{\partial ^2 u}{\partial y^2} + d \frac{\partial u}{\partial x}+ e \frac{\partial u}{\partial y}+ fu+g=0$$

resembles the equation for a conic section

$$ax^2+bxy+cy^2+dx+ey+f=0$$

give the three main types of finite difference schemes!

1. forward difference: $$(\frac{du}{dx})_j\rightarrow \frac{u_{j+1}-u_j}{\Delta x}$$
2. centered difference: $$(\frac{du}{dx})_j \rightarrow \frac{u_{j+1}-u_{j-1}}{2\Delta x}$$
3. backward difference: $$(\frac{du}{dx})_j \rightarrow \frac{u_{j}-u_{j-1}}{\Delta x}$$

Why are time schemes that are used for the PDE's are relatively simple?

1. the error of the numerical solution is brought up by:
2. insufficient information about the initial consitions (only known at discrete time points)

-> Thus, an increase of accuracy of the scheme improves only one of these two components, and the result is not too impressive.

2.  for stability requirements, it is necessary to choose a time step significantly smaller than that required for adequate accuracy.

-> With timesteps usually chosen, other errors, for example in the space differencing, are much greater that those to the time differencing.

Show that the order of accuracy of the forward difference scheme is $$\varepsilon=O(\Delta x)$$ .

NMiMaO, S.14

What is the CFL criterion?

In mathematics, the Courant–Friedrichs–Lewy (CFL) condition is a necessary condition for convergence while solving certain partial differential equations (usually hyperbolic PDEs) numerically by the method of finite differences. It arises in the numerical analysis of explicit time integration schemes, when these are used for the numerical solution.

If a wave is moving across a discrete spatial grid and we want to compute its amplitude at discrete time steps of equal duration, then this duration must be less than the time for the wave to travel to adjacent grid points.

$$c\leq \frac{\Delta x}{\Delta t}$$

What is an implicit and what an explicit scheme?

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

if a value of f for integration is taken at time level n+1, than this scheme is implicit.

if this is not the case, than the scheme is implicit.

Give the three main types of finite differences of second order!

second order central:

$$(\frac{d^2u}{dx^2 })_j \rightarrow \frac{u_{j+1}-2u_j+u_{j-1}}{\Delta x^2}$$

second order forward:

$$(\frac{d^2u}{dx^2})_j \rightarrow \frac{u_{j+2}-2*u_{j+1}+ u_j}{\Delta x^2}$$

second order backward:

$$(\frac{d^2u}{dx^2 })_j \rightarrow \frac{u_j-2*u_{j-1}+u_{j-2}}{\Delta x^2 }$$

stability analysis Neumann method

advection equation: Euler (forward) scheme in time and centred scheme in space

$$\frac{\partial u}{\partial t}+c*\frac{\partial u}{\partial x}=0$$

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

A solution to a linear equation can be expressed as a Fourier series where each Fourier component is a solution. Thus, we can test the stability with one single Fourier component of the form:

$$.\\u^n_j=u_0*e^{ik(j\Delta x-C_Dn\Delta t)}$$

for the time derivatives:

$$u^{n+m}{j}=u^n_j*\lambda ^m$$

----------------------------

$$u^{n+1}_j=\lambda u^n_j\;\;\;where\;\;\;\lambda=e^{-ikC_D\Delta t}$$

for stability it's necessary that:

$$|\lambda|=|e^{-ikC_D\Delta t}|\leq1$$

-----------------------

we get:

$$\lambda =1-i\frac{c\Delta t}{\Delta x}*sin(k\Delta x)$$

The absolute value of $$\lambda$$ is:

$$|\lambda|^2=1+[\frac{c\Delta t}{\Delta x}*sin(k\Delta x)]^2$$

$$|\lambda|>1$$

The scheme is unconditionally unstable!

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!

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.

--------

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.

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!

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: $$c_g=\frac{\partial \omega }{\partial k}$$

How to integrate the Rayleigh friction numerically?

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.

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.

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.

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

Sketch the A,B and C-grid!

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

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:

$$.\\\omega^2=f^2+gH(k^2+l^2)$$

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.

Taylor Series.

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

$$Tf(x;a)=\sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n$$

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

The grid point method

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.

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.

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

finite difference approximation or finite difference scheme.

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.

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!

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.