Vibration problems lead to differential equations with solutions that oscillate in time, typically in a damped or undamped sinusoidal fashion. Such solutions put certain demands on the numerical methods compared to other phenomena whose solutions are monotone or very smooth. Both the frequency and amplitude of the oscillations need to be accurately handled by the numerical schemes. The forthcoming text presents a range of different methods, from classical ones (Runge-Kutta and midpoint/Crank-Nicolson methods), to more modern and popular symplectic (geometric) integration schemes (Leapfrog, Euler-Cromer, and Stoermer-Verlet methods), but with a clear emphasis on the latter. Vibration problems occur throughout mechanics and physics, but the methods discussed in this text are also fundamental for constructing successful algorithms for partial differential equations of wave nature in multiple spatial dimensions.

# Finite difference discretization¶

Many of the numerical challenges faced when computing oscillatory solutions to ODEs and PDEs can be captured by the very simple ODE \(u^{\prime\prime} + u =0\). This ODE is thus chosen as our starting point for method development, implementation, and analysis.

## A basic model for vibrations¶

The simplest model of a vibrating mechanical system has the following form:

Here, \(\omega\) and \(I\) are given constants. The section Applications of vibration models derives (1) from physical principles and explains what the constants mean.

The exact solution of (1) is

That is, \(u\) oscillates with constant amplitude \(I\) and
angular frequency \(\omega\).
The corresponding period of oscillations (i.e., the time between two
neighboring peaks in the cosine function) is \(P=2\pi/\omega\).
The number of periods per second
is \(f=\omega/(2\pi)\) and measured in the unit Hz.
Both \(f\) and \(\omega\) are referred to as frequency, but \(\omega\)
is more precisely named *angular frequency*, measured in rad/s.

In vibrating mechanical systems modeled by (1), \(u(t)\) very often represents a position or a displacement of a particular point in the system. The derivative \(u^{\prime}(t)\) then has the interpretation of velocity, and \(u^{\prime\prime}(t)\) is the associated acceleration. The model (1) is not only applicable to vibrating mechanical systems, but also to oscillations in electrical circuits.

## A centered finite difference scheme¶

To formulate a finite difference method for the model problem (1), we follow the four steps explained in Section 1.1.2 in [Langtangen_decay].

### Step 1: Discretizing the domain¶

The domain is discretized by introducing a uniformly partitioned time mesh. The points in the mesh are \(t_n=n\Delta t\), \(n=0,1,\ldots,N_t\), where \(\Delta t = T/N_t\) is the constant length of the time steps. We introduce a mesh function \(u^n\) for \(n=0,1,\ldots,N_t\), which approximates the exact solution at the mesh points. (Note that \(n=0\) is the known initial condition, so \(u^n\) is identical to the mathematical \(u\) at this point.) The mesh function \(u^n\) will be computed from algebraic equations derived from the differential equation problem.

### Step 2: Fulfilling the equation at discrete time points¶

The ODE is to be satisfied at each mesh point where the solution must be found:

### Step 3: Replacing derivatives by finite differences¶

The derivative \(u^{\prime\prime}(t_n)\) is to be replaced by a finite difference approximation. A common second-order accurate approximation to the second-order derivative is

We also need to replace the derivative in the initial condition by a finite difference. Here we choose a centered difference, whose accuracy is similar to the centered difference we used for \(u^{\prime\prime}\):

### Step 4: Formulating a recursive algorithm¶

To formulate the computational algorithm, we assume that we have already computed \(u^{n-1}\) and \(u^n\), such that \(u^{n+1}\) is the unknown value to be solved for:

The computational algorithm is simply to apply (7) successively for \(n=1,2,\ldots,N_t-1\). This numerical scheme sometimes goes under the name Stoermer’s method, Verlet integration, or the Leapfrog method (one should note that Leapfrog is used for many quite different methods for quite different differential equations!).

### Computing the first step¶

We observe that (7) cannot be used for \(n=0\) since the computation of \(u^1\) then involves the undefined value \(u^{-1}\) at \(t=-\Delta t\). The discretization of the initial condition then comes to our rescue: (6) implies \(u^{-1} = u^1\) and this relation can be combined with (7) for \(n=0\) to yield a value for \(u^1\):

which reduces to

Exercise 5: Use a Taylor polynomial to compute \(u^1\) asks you to perform an alternative derivation and also to generalize the initial condition to \(u^{\prime}(0)=V\neq 0\).

### The computational algorithm¶

The steps for solving (1) become

The algorithm is more precisely expressed directly in Python:

```
t = linspace(0, T, Nt+1) # mesh points in time
dt = t[1] - t[0] # constant time step
u = zeros(Nt+1) # solution
u[0] = I
u[1] = u[0] - 0.5*dt**2*w**2*u[0]
for n in range(1, Nt):
u[n+1] = 2*u[n] - u[n-1] - dt**2*w**2*u[n]
```

**Remark on using w for \(\omega\) in computer code.**

In the code, we use `w`

as the symbol for \(\omega\).
The reason is that the authors prefer `w`

for readability
and comparison with the mathematical \(\omega\) instead of
the full word `omega`

as variable name.

### Operator notation¶

We may write the scheme using a compact difference notation listed in the Finite difference operator notation section (see also Section 1.1.8 in [Langtangen_decay]). The difference (4) has the operator notation \([D_tD_t u]^n\) such that we can write:

Note that \([D_tD_t u]^n\) means applying a central difference with step \(\Delta t/2\) twice:

which is written out as

The discretization of initial conditions can in the operator notation be expressed as

where the operator \([D_{2t} u]^n\) is defined as

# Implementation¶

## Making a solver function¶

The algorithm from the previous section is readily translated to a complete Python function for computing and returning \(u^0,u^1,\ldots,u^{N_t}\) and \(t_0,t_1,\ldots,t_{N_t}\), given the input \(I\), \(\omega\), \(\Delta t\), and \(T\):

```
# NBVAL_IGNORE_OUTPUT
import numpy as np
import matplotlib.pyplot as plt
from devito import Dimension, Constant, TimeFunction, Eq, solve, Operator
```

```
# %load -s solver, src-vib/vib_undamped.py
def solver(I, w, dt, T):
"""
Solve u'' + w**2*u = 0 for t in (0,T], u(0)=I and u'(0)=0,
by a central finite difference method with time step dt.
"""
dt = float(dt)
Nt = int(round(T/dt))
t = Dimension('t', spacing=Constant('h_t'))
u = TimeFunction(name='u', dimensions=(t,),
shape=(Nt+1,), space_order=2)
u.data[:] = I
eqn = u.dt2 + (w**2)*u
stencil = Eq(u.forward, solve(eqn, u.forward))
op = Operator(stencil)
op.apply(h_t=dt, t_M=Nt-1)
return u.data, np.linspace(0, Nt*dt, Nt+1)
```

We have imported `numpy`

and `matplotlib`

under the names `np`

and `plt`

, as this is very common in the Python scientific computing community and a good programming habit (since we explicitly
see where the different functions come from).

A function for plotting the numerical and the exact solution is also convenient to have:

```
# %load -s u_exact, src-vib/vib_undamped.py
def u_exact(t, I, w):
return I*np.cos(w*t)
```

```
# %load -s visualize, src-vib/vib_undamped.py
def visualize(u, t, I, w):
plt.plot(t, u, 'r--o')
t_fine = np.linspace(0, t[-1], 1001) # very fine mesh for u_e
u_e = u_exact(t_fine, I, w)
plt.plot(t_fine, u_e, 'b-')
plt.legend(['numerical', 'exact'], loc='upper left')
plt.xlabel('t')
plt.ylabel('u')
dt = t[1] - t[0]
plt.title('dt=%g' % dt)
umin = 1.2*u.min(); umax = -umin
plt.axis([t[0], t[-1], umin, umax])
plt.savefig('tmp1.png'); plt.savefig('tmp1.pdf')
```

A corresponding main program calling these functions to simulate
a given number of periods (`num_periods`

) may take the form

```
I = 1
w = 2*np.pi
dt = 0.05
num_periods = 5
P = 2*np.pi/w # one period
T = P*num_periods
u, t = solver(I, w, dt, T)
visualize(u, t, I, w)
```

```
Operator `Kernel` run in 0.01 s
```

Adjusting some of the input parameters via the command line can be
handy. Here is a code segment using the `ArgumentParser`

tool in
the `argparse`

module to define option value (`--option value`

)
pairs on the command line:

```
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('--I', type=float, default=1.0)
parser.add_argument('--w', type=float, default=2*np.pi)
parser.add_argument('--dt', type=float, default=0.05)
parser.add_argument('--num_periods', type=int, default=5)
a = parser.parse_args()
I, w, dt, num_periods = a.I, a.w, a.dt, a.num_periods
```

Such parsing of the command line is explained in more detail in Section 5.2.3 in [Langtangen_decay].

A typical execution goes like

```
Terminal> python vib_undamped.py --num_periods 20 --dt 0.1
```

## Verification¶

### Manual calculation¶

The simplest type of verification, which is also instructive for understanding
the algorithm, is to compute \(u^1\), \(u^2\), and \(u^3\)
with the aid of a calculator
and make a function for comparing these results with those from the `solver`

function. The `test_three_steps`

function in
the file `vib_undamped.py`

shows the details of how we use the hand calculations to test the code:

```
# %load -s test_three_steps, src-vib/vib_undamped.py
def test_three_steps():
from math import pi
I = 1; w = 2*pi; dt = 0.1; T = 1
u_by_hand = np.array([1.000000000000000,
0.802607911978213,
0.288358920740053])
u, t = solver(I, w, dt, T)
diff = np.abs(u_by_hand - u[:3]).max()
tol = 1E-14
assert diff < tol
```

This function is a proper *test function*,
compliant with the pytest and nose testing
framework for Python code, because

the function name begins with

`test_`

the function takes no arguments

the test is formulated as a boolean condition and executed by

`assert`

We shall in this book implement all software verification via such proper test functions, also known as unit testing.

See Section 5.3.2 in [Langtangen_decay]
for more details on how to construct test functions and utilize nose
or pytest for automatic execution of tests. Our recommendation is to
use pytest. With this choice, you can
run all test functions in `vib_undamped.py`

by

```
Terminal> py.test -s -v vib_undamped.py
============================= test session starts ======...
platform linux2 -- Python 2.7.9 -- ...
collected 2 items
vib_undamped.py::test_three_steps PASSED
vib_undamped.py::test_convergence_rates PASSED
=========================== 2 passed in 0.19 seconds ===...
```

### Testing very simple polynomial solutions¶

Constructing test problems where the exact solution is constant or linear helps initial debugging and verification as one expects any reasonable numerical method to reproduce such solutions to machine precision. Second-order accurate methods will often also reproduce a quadratic solution. Here \([D_tD_tt^2]^n=2\), which is the exact result. A solution \(u=t^2\) leads to \(u^{\prime\prime}+\omega^2 u=2 + (\omega t)^2\neq 0\). We must therefore add a source in the equation: \(u^{\prime\prime} + \omega^2 u = f\) to allow a solution \(u=t^2\) for \(f=2 + (\omega t)^2\). By simple insertion we can show that the mesh function \(u^n = t_n^2\) is also a solution of the discrete equations. Problem 1: Use linear/quadratic functions for verification asks you to carry out all details to show that linear and quadratic solutions are solutions of the discrete equations. Such results are very useful for debugging and verification. You are strongly encouraged to do this problem now!

### Checking convergence rates¶

Empirical computation of convergence rates yields a good method for verification. The method and its computational details are explained in detail in Section 3.1.6 in [Langtangen_decay]. Readers not familiar with the concept should look up this reference before proceeding.

In the present problem, computing convergence rates means that we must

perform \(m\) simulations, halving the time steps as: \(\Delta t_i=2^{-i}\Delta t_0\), \(i=1,\ldots,m-1\), and \(\Delta t_i\) is the time step used in simulation \(i\);

compute the \(L^2\) norm of the error, \(E_i=\sqrt{\Delta t_i\sum_{n=0}^{N_t-1}(u^n-u(t_n))^2}\) in each case;

estimate the convergence rates \(r_i\) based on two consecutive experiments \((\Delta t_{i-1}, E_{i-1})\) and \((\Delta t_{i}, E_{i})\), assuming \(E_i=C(\Delta t_i)^{r}\) and \(E_{i-1}=C(\Delta t_{i-1})^{r}\), where \(C\) is a constant. From these equations it follows that \(r = \ln (E_{i-1}/E_i)/\ln (\Delta t_{i-1}/\Delta t_i)\). Since this \(r\) will vary with \(i\), we equip it with an index and call it \(r_{i-1}\), where \(i\) runs from \(1\) to \(m-1\).

The computed rates \(r_0,r_1,\ldots,r_{m-2}\) hopefully converge to the number 2 in the present problem, because theory (from the section Analysis of the numerical scheme) shows that the error of the numerical method we use behaves like \(\Delta t^2\). The convergence of the sequence \(r_0,r_1,\ldots,r_{m-2}\) demands that the time steps \(\Delta t_i\) are sufficiently small for the error model \(E_i=C(\Delta t_i)^r\) to be valid.

All the implementational details of computing the sequence \(r_0,r_1,\ldots,r_{m-2}\) appear below.

```
# %load -s convergence_rates, src-vib/vib_undamped.py
def convergence_rates(m, solver_function, num_periods=8):
"""
Return m-1 empirical estimates of the convergence rate
based on m simulations, where the time step is halved
for each simulation.
solver_function(I, w, dt, T) solves each problem, where T
is based on simulation for num_periods periods.
"""
from math import pi
w = 0.35; I = 0.3 # just chosen values
P = 2*pi/w # period
dt = P/30 # 30 time step per period 2*pi/w
T = P*num_periods
dt_values = []
E_values = []
for i in range(m):
u, t = solver_function(I, w, dt, T)
u_e = u_exact(t, I, w)
E = np.sqrt(dt*np.sum((u_e-u)**2))
dt_values.append(dt)
E_values.append(E)
dt = dt/2
r = [np.log(E_values[i-1]/E_values[i])/
np.log(dt_values[i-1]/dt_values[i])
for i in range(1, m, 1)]
return r, E_values, dt_values
```

The error analysis in the section Analysis of the numerical scheme is quite detailed and suggests that \(r=2\). mathcal{I}_t is also a intuitively reasonable result, since we used a second-order accurate finite difference approximation \([D_tD_tu]^n\) to the ODE and a second-order accurate finite difference formula for the initial condition for \(u^{\prime}\).

In the present problem, when \(\Delta t_0\) corresponds to 30 time steps
per period, the returned `r`

list has all its values equal to 2.00
(if rounded to two decimals). This amazingly accurate result means that all
\(\Delta t_i\) values are well into the asymptotic regime where the
error model \(E_i = C(\Delta t_i)^r\) is valid.

We can now construct a proper test function that computes convergence rates and checks that the final (and usually the best) estimate is sufficiently close to 2. Here, a rough tolerance of 0.1 is enough. Later, we will argue for an improvement by adjusting omega and include also that case in our test function here. The unit test goes like

```
# %load -s test_convergence_rates, src-vib/vib_undamped.py
def test_convergence_rates():
r, E, dt = convergence_rates(
m=5, solver_function=solver, num_periods=8)
# Accept rate to 1 decimal place
tol = 0.1
assert abs(r[-1] - 2.0) < tol
# Test that adjusted w obtains 4th order convergence
r, E, dt = convergence_rates(
m=5, solver_function=solver_adjust_w, num_periods=8)
print("adjust w rates:")
print(r)
assert abs(r[-1] - 4.0) < tol
```

where `solver_adjust_w`

is a slight variation on the original `solver`

function, as follows:

```
# %load -s solver_adjust_w, src-vib/vib_undamped.py
def solver_adjust_w(I, w, dt, T, adjust_w=True):
"""
Solve u'' + w**2*u = 0 for t in (0,T], u(0)=I and u'(0)=0,
by a central finite difference method with time step dt.
"""
dt = float(dt)
Nt = int(round(T/dt))
t = Dimension('t', spacing=Constant('h_t'))
u = TimeFunction(name='u', dimensions=(t,),
shape=(Nt+1,), space_order=2)
w_adj = w*(1 - w**2*dt**2/24.) if adjust_w else w
u.data[:] = I
eqn = u.dt2 + (w**2)*u
stencil = Eq(u.forward, solve(eqn, u.forward))
op = Operator(stencil)
op.apply(h_t=dt, t_M=Nt-1)
return u.data, np.linspace(0, Nt*dt, Nt+1)
```

The complete code appears in the file `vib_undamped.py`

.

### Visualizing convergence rates with slope markers¶

Tony S. Yu has written a script `plotslopes.py`

that is very useful to indicate the slope of a graph, especially
a graph like \(\ln E = r\ln \Delta t + \ln C\) arising from the model
\(E=C\Delta t^r\). A copy of the script resides in the `src-vib`

directory. Let us use it to compare the original method for \(u'' + \omega^2u =0\)
with the same method applied to the equation with a modified
\(\omega\). We make log-log plots of the error versus \(\Delta t\).
For each curve we attach a slope marker using the `slope_marker((x,y), r)`

function from `plotslopes.py`

, where `(x,y)`

is the position of the
marker and `r`

and the slope (\((r,1)\)), here (2,1) and (4,1).

```
# %load -s plot_convergence_rates, src-vib/vib_undamped.py
def plot_convergence_rates():
r2, E2, dt2 = convergence_rates(
m=5, solver_function=solver, num_periods=8)
plt.loglog(dt2, E2)
r4, E4, dt4 = convergence_rates(
m=5, solver_function=solver_adjust_w, num_periods=8)
plt.loglog(dt4, E4)
plt.legend(['original scheme', r'adjusted $\omega$'],
loc='upper left')
plt.title('Convergence of finite difference methods')
# from plotslopes import slope_marker
# slope_marker((dt2[1], E2[1]), (2,1))
# slope_marker((dt4[1], E4[1]), (4,1))
plt.savefig('tmp_convrate.png'); plt.savefig('tmp_convrate.pdf')
plt.show()
```

Figure displays the two curves with the markers. The match of the curve slope and the marker slope is excellent.

Empirical convergence rate curves with special slope marker.

## Scaled model¶

It is advantageous to use dimensionless variables in simulations, because fewer parameters need to be set. The present problem is made dimensionless by introducing dimensionless variables \(\bar t = t/t_c\) and \(\bar u = u/u_c\), where \(t_c\) and \(u_c\) are characteristic scales for \(t\) and \(u\), respectively. We refer to Section 2.2.1 in [Langtangen_scaling] for all details about this scaling.

The scaled ODE problem reads

A common choice is to take \(t_c\) as one period of the oscillations, \(t_c = 2\pi/w\), and \(u_c=I\). This gives the dimensionless model

Observe that there are no physical parameters in (13)! We can therefore perform a single numerical simulation \(\bar u(\bar t)\) and afterwards recover any \(u(t; \omega, I)\) by

We can easily check this assertion: the solution of the scaled problem is \(\bar u(\bar t) = \cos(2\pi\bar t)\). The formula for \(u\) in terms of \(\bar u\) gives \(u = I\cos(\omega t)\), which is nothing but the solution of the original problem with dimensions.

The scaled model can be run by calling `solver(I=1, w=2*pi, dt, T)`

.
Each period is now 1 and `T`

simply counts the number of periods.
Choosing `dt`

as `1./M`

gives `M`

time steps per period.

# Visualization of long time simulations¶

Figure shows a comparison of the exact and numerical solution for the scaled model (13) with \(\Delta t=0.1, 0.05\). From the plot we make the following observations:

The numerical solution seems to have correct amplitude.

There is an angular frequency error which is reduced by decreasing the time step.

The total angular frequency error grows with time.

By angular frequency error we mean that the numerical angular frequency differs from the exact \(\omega\). This is evident by looking at the peaks of the numerical solution: these have incorrect positions compared with the peaks of the exact cosine solution. The effect can be mathematically expressed by writing the numerical solution as \(I\cos\tilde\omega t\), where \(\tilde\omega\) is not exactly equal to \(\omega\). Later, we shall mathematically quantify this numerical angular frequency \(\tilde\omega\).

Here, we show the effect of halving the time step on the error.

```
I = 1
w = 2*np.pi
num_periods = 5
P = 2*np.pi/w # one period
T = P*num_periods
```

```
dt = 0.1
u_1, t = solver(I, w, dt, T)
visualize(u_1, t, I, w)
```

```
Operator `Kernel` run in 0.01 s
```

```
dt = 0.05
u_2, t = solver(I, w, dt, T)
visualize(u_2, t, I, w)
```

```
Operator `Kernel` run in 0.01 s
```

## Using a moving plot window¶

In vibration problems it is often of interest to investigate the system’s
behavior over long time intervals. Errors in the angular frequency accumulate
and become more visible as time grows. We can investigate long
time series by introducing a moving plot window that can move along with
the \(p\) most recently computed periods of the solution. The
SciTools package contains
a convenient tool for this: `MovingPlotWindow`

. Typing
`pydoc scitools.MovingPlotWindow`

shows a demo and a description of its use.
The function below utilizes the moving plot window and is in fact
called by the `main`

function in the `vib_undamped`

module
if the number of periods in the simulation exceeds 10.

```
# %load -s visualize_front, src-vib/vib_undamped.py
def visualize_front(u, t, I, w, savefig=False, skip_frames=1):
"""
Visualize u and the exact solution vs t, using a
moving plot window and continuous drawing of the
curves as they evolve in time.
Makes it easy to plot very long time series.
Plots are saved to files if savefig is True.
Only each skip_frames-th plot is saved (e.g., if
skip_frame=10, only each 10th plot is saved to file;
this is convenient if plot files corresponding to
different time steps are to be compared).
"""
import scitools.std as st
from scitools.MovingPlotWindow import MovingPlotWindow
from math import pi
# Remove all old plot files tmp_*.png
import glob, os
for filename in glob.glob('tmp_*.png'):
os.remove(filename)
P = 2*pi/w # one period
umin = 1.2*u.min(); umax = -umin
dt = t[1] - t[0]
plot_manager = MovingPlotWindow(
window_width=8*P,
dt=dt,
yaxis=[umin, umax],
mode='continuous drawing')
frame_counter = 0
for n in range(1,len(u)):
if plot_manager.plot(n):
s = plot_manager.first_index_in_plot
st.plot(t[s:n+1], u[s:n+1], 'r-1',
t[s:n+1], I*cos(w*t)[s:n+1], 'b-1',
title='t=%6.3f' % t[n],
axis=plot_manager.axis(),
show=not savefig) # drop window if savefig
if savefig and n % skip_frames == 0:
filename = 'tmp_%04d.png' % frame_counter
st.savefig(filename)
print('making plot file', filename, 'at t=%g' % t[n])
frame_counter += 1
plot_manager.update(n)
```

We run the scaled problem (the default values for the command-line arguments
`--I`

and `--w`

correspond to the scaled problem) for 40 periods with 20
time steps per period:

```
Terminal> python vib_undamped.py --dt 0.05 --num_periods 40
```

The moving plot window is invoked, and we can follow the numerical and exact solutions as time progresses. From this demo we see that the angular frequency error is small in the beginning, and that it becomes more prominent with time. A new run with \(\Delta t=0.1\) (i.e., only 10 time steps per period) clearly shows that the phase errors become significant even earlier in the time series, deteriorating the solution further.

## Making animations¶

### Producing standard video formats¶

The `visualize_front`

function stores all the plots in
files whose names are numbered:
`tmp_0000.png`

, `tmp_0001.png`

, `tmp_0002.png`

,
and so on. From these files we may make a movie. The Flash
format is popular,

```
Terminal> ffmpeg -r 25 -i tmp_%04d.png -c:v flv movie.flv
```

The `-r`

option should come first and
describes the number of frames per second in the movie (even if we
would like to have slow movies, keep this number as large as 25,
otherwise files are skipped from the movie). The
`-i`

option describes the name of the plot files.
Other formats can be generated by changing the video codec
and equipping the video file with the right extension:

Format | Codec and filename |
---|---|

Flash | `-c:v flv movie.flv` |

MP4 | `-c:v libx264 movie.mp4` |

WebM | `-c:v libvpx movie.webm` |

Ogg | `-c:v libtheora movie.ogg` |

The video file can be played by some video player like `vlc`

, `mplayer`

,
`gxine`

, or `totem`

, e.g.,

```
Terminal> vlc movie.webm
```

A web page can also be used to play the movie. Today’s standard is
to use the HTML5 `video`

tag:

```
<video autoplay loop controls
width='640' height='365' preload='none'>
<source src='movie.webm' type='video/webm; codecs="vp8, vorbis"'>
</video>
```

Modern browsers do not support all of the video formats. MP4 is needed to successfully play the videos on Apple devices that use the Safari browser. WebM is the preferred format for Chrome, Opera, Firefox, and Internet Explorer v9+. Flash was a popular format, but older browsers that required Flash can play MP4. All browsers that work with Ogg can also work with WebM. This means that to have a video work in all browsers, the video should be available in the MP4 and WebM formats. The proper HTML code reads

```
<video autoplay loop controls
width='640' height='365' preload='none'>
<source src='movie.mp4' type='video/mp4;
codecs="avc1.42E01E, mp4a.40.2"'>
<source src='movie.webm' type='video/webm;
codecs="vp8, vorbis"'>
</video>
```

The MP4 format should appear first to ensure that Apple devices will load the video correctly.

**Caution: number the plot files correctly.**

To ensure that the individual plot frames are shown in correct order,
it is important to number the files with zero-padded numbers
(0000, 0001, 0002, etc.). The printf format `%04d`

specifies an
integer in a field of width 4, padded with zeros from the left.
A simple Unix wildcard file specification like `tmp_*.png`

will then list the frames in the right order. If the numbers in the
filenames were not zero-padded, the frame `tmp_11.png`

would appear
before `tmp_2.png`

in the movie.

### Playing PNG files in a web browser¶

The `scitools movie`

command can create a movie player for a set
of PNG files such that a web browser can be used to watch the movie.
This interface has the advantage that the speed of the movie can
easily be controlled, a feature that scientists often appreciate.
The command for creating an HTML with a player for a set of
PNG files `tmp_*.png`

goes like

```
Terminal> scitools movie output_file=vib.html fps=4 tmp_*.png
```

The `fps`

argument controls the speed of the movie (“frames per second”).

To watch the movie, load the video file `vib.html`

into some browser, e.g.,

```
Terminal> google-chrome vib.html # invoke web page
```

Click on `Start movie`

to see the result. Moving this movie to
some other place requires moving `vib.html`

*and all the PNG files*
`tmp_*.png`

:

```
Terminal> mkdir vib_dt0.1
Terminal> mv tmp_*.png vib_dt0.1
Terminal> mv vib.html vib_dt0.1/index.html
```

### Making animated GIF files¶

The `convert`

program from the ImageMagick software suite can be
used to produce animated GIF files from a set of PNG files:

```
Terminal> convert -delay 25 tmp_vib*.png tmp_vib.gif
```

The `-delay`

option needs an argument of the delay between each frame,
measured in 1/100 s, so 4 frames/s here gives 25/100 s delay.
Note, however, that in this particular example
with \(\Delta t=0.05\) and 40 periods,
making an animated GIF file out of
the large number of PNG files is a very heavy process and not
considered feasible. Animated GIFs are best suited for animations with
not so many frames and where you want to see each frame and play them
slowly.

[hpl 2: Combine two simulations side by side!]

## Using Bokeh to compare graphs¶

Instead of a moving plot frame, one can use tools that allow panning
by the mouse. For example, we can show four periods of several signals in
several plots and then scroll with the mouse through the rest of the
simulation *simultaneously* in all the plot windows.
The Bokeh plotting library offers such tools, but the plots must be displayed in
a web browser. The documentation of Bokeh is excellent, so here we just
show how the library can be used to compare a set of \(u\) curves corresponding
to long time simulations. (By the way, the guidance to correct
pronunciation of Bokeh in
the documentation and on Wikipedia is not directly compatible with a YouTube video…).

Imagine we have performed experiments for a set of \(\Delta t\) values. We want each curve, together with the exact solution, to appear in a plot, and then arrange all plots in a grid-like fashion:

Furthermore, we want the axes to couple such that if we move into the future in one plot, all the other plots follows (note the displaced \(t\) axes!):

A function for creating a Bokeh plot, given a list of `u`

arrays
and corresponding `t`

arrays, is implemented below.
The code combines data from different simulations, described
compactly in a list of strings `legends`

.

A particular example using the `bokeh_plot`

function appears below.

```
# %load -s bokeh_plot, src-vib/vib_undamped.py
def bokeh_plot(u, t, legends, I, w, t_range, filename):
"""
Make plots for u vs t using the Bokeh library.
u and t are lists (several experiments can be compared).
legens contain legend strings for the various u,t pairs.
"""
if not isinstance(u, (list,tuple)):
u = [u] # wrap in list
if not isinstance(t, (list,tuple)):
t = [t] # wrap in list
if not isinstance(legends, (list,tuple)):
legends = [legends] # wrap in list
import bokeh.plotting as plt
plt.output_file(filename, mode='cdn', title='Comparison')
# Assume that all t arrays have the same range
t_fine = np.linspace(0, t[0][-1], 1001) # fine mesh for u_e
tools = 'pan,wheel_zoom,box_zoom,reset,'\
'save,box_select,lasso_select'
u_range = [-1.2*I, 1.2*I]
font_size = '8pt'
p = [] # list of plot objects
# Make the first figure
p_ = plt.figure(
width=300, plot_height=250, title=legends[0],
x_axis_label='t', y_axis_label='u',
x_range=t_range, y_range=u_range, tools=tools)
p_.xaxis.axis_label_text_font_size=font_size
p_.yaxis.axis_label_text_font_size=font_size
p_.line(t[0], u[0], line_color='blue')
# Add exact solution
u_e = u_exact(t_fine, I, w)
p_.line(t_fine, u_e, line_color='red', line_dash='4 4')
p.append(p_)
# Make the rest of the figures and attach their axes to
# the first figure's axes
for i in range(1, len(t)):
p_ = plt.figure(
width=300, plot_height=250, title=legends[i],
x_axis_label='t', y_axis_label='u',
x_range=p[0].x_range, y_range=p[0].y_range, tools=tools)
p_.xaxis.axis_label_text_font_size = font_size
p_.yaxis.axis_label_text_font_size = font_size
p_.line(t[i], u[i], line_color='blue')
p_.line(t_fine, u_e, line_color='red', line_dash='4 4')
p.append(p_)
# Arrange all plots in a grid with 3 plots per row
grid = [[]]
for i, p_ in enumerate(p):
grid[-1].append(p_)
if (i+1) % 3 == 0:
# New row
grid.append([])
plot = plt.gridplot(grid, toolbar_location='left')
plt.save(plot)
plt.show(plot)
```

```
# %load -s demo_bokeh, src-vib/vib_undamped.py
def demo_bokeh():
"""Solve a scaled ODE u'' + u = 0."""
from math import pi
w = 1.0 # Scaled problem (frequency)
P = 2*np.pi/w # Period
num_steps_per_period = [5, 10, 20, 40, 80]
T = 40*P # Simulation time: 40 periods
u = [] # List of numerical solutions
t = [] # List of corresponding meshes
legends = []
for n in num_steps_per_period:
dt = P/n
u_, t_ = solver(I=1, w=w, dt=dt, T=T)
u.append(u_)
t.append(t_)
legends.append('# time steps per period: %d' % n)
bokeh_plot(u, t, legends, I=1, w=w, t_range=[0, 4*P],
filename='tmp.html')
```

We can run this below, which should open a window with the Bokeh plots where you can experiment with the graphs yourself:

```
# NBVAL_IGNORE_OUTPUT
demo_bokeh()
```

```
Operator `Kernel` run in 0.01 s
```

```
Operator `Kernel` run in 0.01 s
```

```
Operator `Kernel` run in 0.01 s
```

```
Operator `Kernel` run in 0.01 s
```

```
Operator `Kernel` run in 0.01 s
```

## Using a line-by-line ascii plotter¶

Plotting functions vertically, line by line, in the terminal window
using ascii characters only is a simple, fast, and convenient
visualization technique for long time series. Note that the time
axis then is positive downwards on the screen, so we can let the
solution be visualized “forever”.
The tool
`scitools.avplotter.Plotter`

makes it easy to create such plots:

```
# %load -s visualize_front_ascii, src-vib/vib_undamped.py
def visualize_front_ascii(u, t, I, w, fps=10):
"""
Plot u and the exact solution vs t line by line in a
terminal window (only using ascii characters).
Makes it easy to plot very long time series.
"""
from scitools.avplotter import Plotter
import time
from math import pi
P = 2*pi/w
umin = 1.2*u.min(); umax = -umin
p = Plotter(ymin=umin, ymax=umax, width=60, symbols='+o')
for n in range(len(u)):
print(p.plot(t[n], u[n], I*np.cos(w*t[n])), '%.1f' % (t[n]/P))
time.sleep(1/float(fps))
```

The call `p.plot`

returns a line of text, with the \(t\) axis marked and
a symbol `+`

for the first function (`u`

) and `o`

for the second
function (the exact solution). Here we append to this text
a time counter reflecting how many periods the current time point
corresponds to.

The function can be run as follows:

```
visualize_front_ascii(u, t, I, w)
```

```
| o 0.0
| o+ 0.1
```

```
| o + 0.1
| o + 0.2
| o +
```

```
0.2
| + 0.2
```

```
o + | 0.3
o + | 0.4
```

```
o + | 0.4
o + | 0.5
```

```
o | 0.5
+o | 0.6
```

```
+ o | 0.6
+ o | 0.7
```

```
+ o | 0.7
+ | 0.8
```

```
| + o 0.8
| + o 0.9
```

```
| + o 0.9
| +o 1.0
```

```
| o 1.0
| o+ 1.1
| o + 1.1
```

```
| o + 1.2
| o + 1.2
```

```
| + 1.2
o + | 1.3
```

```
o + | 1.4
o + | 1.4
```

```
o+ | 1.5
o | 1.5
```

```
+o | 1.6
+ o | 1.6
+ o | 1.7
```

```
+ o | 1.7
+ | 1.8
```

```
| + o 1.8
| + o 1.9
| +o
```

```
1.9
| +o 2.0
```

```
| o 2.0
| o 2.1
| o + 2.1
```

```
| o + 2.1
| o + 2.2
| + 2.2
```

```
o + | 2.3
o + | 2.4
o+ | 2.4
```

```
o+ | 2.5
o | 2.5
```

```
o | 2.6
+o | 2.6
```

```
+o | 2.7
+ o | 2.7
+ | 2.8
```

```
| + o 2.8
| + o 2.9
| +o 2.9
```

```
| +o 3.0
| o 3.0
```

```
| o 3.1
| o+ 3.1
| o+ 3.2
```

```
| o+ 3.2
| + 3.2
```

```
o + | 3.3
o + | 3.4
```

```
o+ | 3.4
o+ | 3.5
```

```
o | 3.5
o | 3.6
+o | 3.6
```

```
+o | 3.7
+o | 3.7
+| 3.8
```

```
| + o 3.8
| +o 3.9
| +o 3.9
```

```
| +o 4.0
| o 4.0
```

```
| o 4.0
| o+ 4.1
```

```
| o+ 4.2
| o+ 4.2
```

```
|+ 4.2
o+ | 4.3
```

```
o+ | 4.4
o | 4.4
```

```
o+ | 4.5
o | 4.5
```

```
o | 4.5
+o | 4.6
o |
```

```
4.7
+o | 4.7
```

```
+| 4.8
| +o 4.8
```

```
| +o 4.9
| o 4.9
| o 5.0
| o 5.0
```

## Empirical analysis of the solution¶

For oscillating functions like those in Figure we may compute the amplitude and frequency (or period) empirically. That is, we run through the discrete solution points \((t_n, u_n)\) and find all maxima and minima points. The distance between two consecutive maxima (or minima) points can be used as estimate of the local period, while half the difference between the \(u\) value at a maximum and a nearby minimum gives an estimate of the local amplitude.

The local maxima are the points where

and the local minima are recognized by

In computer code this becomes

```
# %load -s minmax, src-vib/vib_empirical_analysis.py
def minmax(t, u):
"""
Compute all local minima and maxima of the function u(t),
represented by discrete points in the arrays u and t.
Return lists minima and maxima of (t[i],u[i]) extreme points.
"""
minima = []; maxima = []
for n in range(1, len(u)-1, 1):
if u[n-1] > u[n] < u[n+1]:
minima.append((t[n], u[n]))
if u[n-1] < u[n] > u[n+1]:
maxima.append((t[n], u[n]))
return minima, maxima
```

Note that the two returned objects are lists of tuples.

Let \((t_i, e_i)\), \(i=0,\ldots,M-1\), be the sequence of all the \(M\) maxima points, where \(t_i\) is the time value and \(e_i\) the corresponding \(u\) value. The local period can be defined as \(p_i=t_{i+1}-t_i\). With Python syntax this reads

```
# %load -s periods, src-vib/vib_empirical_analysis.py
def periods(extrema):
"""
Given a list of (t,u) points of the maxima or minima,
return an array of the corresponding local periods.
"""
p = [extrema[n][0] - extrema[n-1][0]
for n in range(1, len(extrema))]
return np.array(p)
```

The list `p`

created by a list comprehension is converted to an array
since we probably want to compute with it, e.g., find the corresponding
frequencies `2*pi/p`

.

Having the minima and the maxima, the local amplitude can be calculated as the difference between two neighboring minimum and maximum points:

```
# %load -s amplitudes, src-vib/vib_empirical_analysis.py
def amplitudes(minima, maxima):
"""
Given a list of (t,u) points of the minima and maxima of
u, return an array of the corresponding local amplitudes.
"""
# Compare first maxima with first minima and so on
a = [(abs(maxima[n][1] - minima[n][1]))/2.0
for n in range(min(len(minima),len(maxima)))]
return np.array(a)
```

The code segments are found in the file `vib_empirical_analysis.py`

.

Since `a[i]`

and `p[i]`

correspond to
the \(i\)-th amplitude estimate and the \(i\)-th period estimate, respectively,
it is most convenient to visualize the `a`

and `p`

values with the
index `i`

on the horizontal axis.
(There is no unique time point associated with either of these estimate
since values at two different time points were used in the
computations.)

In the analysis of very long time series, it is advantageous to
compute and plot `p`

and `a`

instead of \(u\) to get an impression of
the development of the oscillations. Let us do this for the scaled
problem and \(\Delta t=0.1, 0.05, 0.01\).
A ready-made function

```
plot_empirical_freq_and_amplitude(u, t, I, w)
```

computes the empirical amplitudes and periods, and creates a plot
where the amplitudes and angular frequencies
are visualized together with the exact amplitude `I`

and the exact angular frequency `w`

.

```
# %load -s plot_empirical_freq_and_amplitude, src-vib/vib_undamped.py
def plot_empirical_freq_and_amplitude(u, t, I, w):
"""
Find the empirical angular frequency and amplitude of
simulations in u and t. u and t can be arrays or (in
the case of multiple simulations) multiple arrays.
One plot is made for the amplitude and one for the angular
frequency (just called frequency in the legends).
"""
from math import pi
if not isinstance(u, (list,tuple)):
u = [u]
t = [t]
legends1 = []
legends2 = []
for i in range(len(u)):
minima, maxima = minmax(t[i], u[i])
p = periods(maxima)
a = amplitudes(minima, maxima)
plt.figure(1)
plt.plot(range(len(p)), 2*pi/p)
legends1.append('frequency, case%d' % (i+1))
plt.figure(2)
plt.plot(range(len(a)), a)
legends2.append('amplitude, case%d' % (i+1))
plt.figure(1)
plt.plot(range(len(p)), [w]*len(p), 'k--')
legends1.append('exact frequency')
plt.legend(legends1, loc='lower left')
plt.axis([0, len(a)-1, 0.8*w, 1.2*w])
plt.savefig('tmp1.png'); plt.savefig('tmp1.pdf')
plt.figure(2)
plt.plot(range(len(a)), [I]*len(a), 'k--')
legends2.append('exact amplitude')
plt.legend(legends2, loc='lower left')
plt.axis([0, len(a)-1, 0.8*I, 1.2*I])
plt.savefig('tmp2.png'); plt.savefig('tmp2.pdf')
plt.show()
```

We can make a little program for creating the plot:

```
# NBVAL_IGNORE_OUTPUT
from math import pi
dt_values = [0.1, 0.05, 0.01]
u_cases = []
t_cases = []
for dt in dt_values:
# Simulate scaled problem for 40 periods
u, t = solver(I=1, w=2*pi, dt=dt, T=40)
u_cases.append(u)
t_cases.append(t)
plot_empirical_freq_and_amplitude(u_cases, t_cases, I=1, w=2*pi)
```

```
Operator `Kernel` run in 0.01 s
```

```
Operator `Kernel` run in 0.01 s
```

```
Operator `Kernel` run in 0.01 s
```

Empirical angular frequency (top) and amplitude (bottom) for three different time steps.

We can clearly see that lowering \(\Delta t\) improves the angular frequency significantly, while the amplitude seems to be more accurate. The lines with \(\Delta t=0.01\), corresponding to 100 steps per period, can hardly be distinguished from the exact values. The next section shows how we can get mathematical insight into why amplitudes are good while frequencies are more inaccurate.

# Analysis of the numerical scheme¶

## Deriving a solution of the numerical scheme¶

After having seen the phase error grow with time in the previous section, we shall now quantify this error through mathematical analysis. The key tool in the analysis will be to establish an exact solution of the discrete equations. The difference equation (7) has constant coefficients and is homogeneous. Such equations are known to have solutions on the form \(u^n=CA^n\), where \(A\) is some number to be determined from the difference equation and \(C\) is found as the initial condition (\(C=I\)). Recall that \(n\) in \(u^n\) is a superscript labeling the time level, while \(n\) in \(A^n\) is an exponent.

With oscillating functions as solutions, the algebra will be considerably simplified if we seek an \(A\) on the form

and solve for the numerical frequency \(\tilde\omega\) rather than \(A\). Note that \(i=\sqrt{-1}\) is the imaginary unit. (Using a complex exponential function gives simpler arithmetics than working with a sine or cosine function.) We have

The physically relevant numerical solution can be taken as the real part of this complex expression.

The calculations go as

The last line follows from the relation
\(\cos x - 1 = -2\sin^2(x/2)\) (try `cos(x)-1`

on
wolframalpha.com to see the formula).

The scheme (7) with \(u^n=Ie^{i\tilde\omega\Delta t\, n}\) inserted now gives

which after dividing by \(Ie^{i\tilde\omega t_n}\) results in

The first step in solving for the unknown \(\tilde\omega\) is

Then, taking the square root, applying the inverse sine function, and multiplying by \(2/\Delta t\), results in

## The error in the numerical frequency¶

The first observation following (18) tells that there is a phase error since the numerical frequency \(\tilde\omega\) never equals the exact frequency \(\omega\). But how good is the approximation (18)? That is, what is the error \(\omega - \tilde\omega\) or \(\tilde\omega/\omega\)? Taylor series expansion for small \(\Delta t\) may give an expression that is easier to understand than the complicated function in (18):

```
from sympy import *
dt, w = symbols('dt w')
w_tilde_e = 2/dt*asin(w*dt/2)
w_tilde_series = w_tilde_e.series(dt, 0, 4)
print(w_tilde_series)
```

```
w + dt**2*w**3/24 + O(dt**4)
```

This means that

The error in the numerical frequency is of second-order in \(\Delta t\), and the error vanishes as \(\Delta t\rightarrow 0\). We see that \(\tilde\omega > \omega\) since the term \(\omega^3\Delta t^2/24 >0\) and this is by far the biggest term in the series expansion for small \(\omega\Delta t\). A numerical frequency that is too large gives an oscillating curve that oscillates too fast and therefore “lags behind” the exact oscillations, a feature that can be seen in the left plot in Figure.

Figure plots the discrete frequency
(18) and its approximation
(19) for \(\omega =1\) (based on the
program `vib_plot_freq.py`

).
Although \(\tilde\omega\) is a function of \(\Delta t\) in
(19), it is misleading to think of
\(\Delta t\) as the important discretization parameter. mathcal{I}_t is the
product \(\omega\Delta t\) that is the key discretization
parameter. This quantity reflects the *number of time steps per
period* of the oscillations. To see this, we set \(P=N_P\Delta t\),
where \(P\) is the length of a period, and \(N_P\) is the number of time
steps during a period. Since \(P\) and \(\omega\) are related by
\(P=2\pi/\omega\), we get that \(\omega\Delta t = 2\pi/N_P\), which shows
that \(\omega\Delta t\) is directly related to \(N_P\).

The plot shows that at least \(N_P\sim 25-30\) points per period are necessary for reasonable accuracy, but this depends on the length of the simulation (\(T\)) as the total phase error due to the frequency error grows linearly with time (see Exercise 2: Show linear growth of the phase with time).

Exact discrete frequency and its second-order series expansion.

## Empirical convergence rates and adjusted \(\omega\)¶

The expression (19) suggests that adjusting omega to

could have effect on the *convergence rate* of the global error in \(u\)
(cf. the section Verification). With the `convergence_rates`

function
in `vib_undamped.py`

we can easily check this. A special solver, with
adjusted \(w\), is available as the function `solver_adjust_w`

. A
call to `convergence_rates`

with this solver reveals that the rate is
4.0! With the original, physical \(\omega\) the rate is 2.0 - as expected
from using second-order finite difference approximations,
as expected from the forthcoming derivation of the global error,
and as expected from truncation error analysis as explained in the truncation error analysis section.

Adjusting \(\omega\) is an ideal trick for this simple problem, but when adding damping and nonlinear terms, we have no simple formula for the impact on \(\omega\), and therefore we cannot use the trick.

## Exact discrete solution¶

Perhaps more important than the \(\tilde\omega = \omega + {\cal O}(\Delta t^2)\) result found above is the fact that we have an exact discrete solution of the problem:

We can then compute the error mesh function

From the formula \(\cos 2x - \cos 2y = -2\sin(x-y)\sin(x+y)\) we can rewrite \(e^n\) so the expression is easier to interpret:

The error mesh function is ideal for verification purposes and you are strongly encouraged to make a test based on (20) by doing Exercise 11: Use an exact discrete solution for verification.

## Convergence¶

We can use (19) and (21), or
(22), to show *convergence* of the numerical scheme,
i.e., \(e^n\rightarrow 0\) as \(\Delta t\rightarrow 0\), which implies
that the numerical solution approaches the exact solution as \(\Delta
t\) approaches to zero. We have that

by L’Hopital’s rule. This result could also been computed WolframAlpha, or
we could use the limit functionality in `sympy`

:

```
import sympy as sym
dt, w = sym.symbols('x w')
sym.limit((2/dt)*sym.asin(w*dt/2), dt, 0, dir='+')
```

Also (19) can be used to establish that \(\tilde\omega\rightarrow\omega\) when \(\Delta t\rightarrow 0\). mathcal{I}_t then follows from the expression(s) for \(e^n\) that \(e^n\rightarrow 0\).

## The global error¶

To achieve more analytical insight into the nature of the global
error, we can Taylor expand the error mesh function
(21). Since \(\tilde\omega\) in
(18) contains \(\Delta t\) in the denominator we
use the series expansion for \(\tilde\omega\) inside the cosine
function. A relevant `sympy`

session is

```
from sympy import *
dt, w, t = symbols('dt w t')
w_tilde_e = 2/dt*asin(w*dt/2)
w_tilde_series = w_tilde_e.series(dt, 0, 4)
w_tilde_series
```

Series expansions in `sympy`

have the inconvenient `O()`

term that
prevents further calculations with the series. We can use the
`removeO()`

command to get rid of the `O()`

term:

```
w_tilde_series = w_tilde_series.removeO()
w_tilde_series
```

Using this `w_tilde_series`

expression for \(\tilde w\) in
(21), dropping \(I\) (which is a common factor), and
performing a series expansion of the error yields

```
error = cos(w*t) - cos(w_tilde_series*t)
error.series(dt, 0, 6)
```

Since we are mainly interested in the leading-order term in
such expansions (the term with lowest power in \(\Delta t\), which
goes most slowly to zero), we use the `.as_leading_term(dt)`

construction to pick out this term:

```
error.series(dt, 0, 6).as_leading_term(dt)
```

The last result means that the leading order global (true) error at a point \(t\) is proportional to \(\omega^3t\Delta t^2\). Considering only the discrete \(t_n\) values for \(t\), \(t_n\) is related to \(\Delta t\) through \(t_n=n\Delta t\). The factor \(\sin(\omega t)\) can at most be 1, so we use this value to bound the leading-order expression to its maximum value

This is the dominating term of the error *at a point*.

We are interested in the accumulated global error, which can be taken as the \(\ell^2\) norm of \(e^n\). The norm is simply computed by summing contributions from all mesh points:

The sum \(\sum_{n=0}^{N_t} n^2\) is approximately equal to \(\frac{1}{3}N_t^3\). Replacing \(N_t\) by \(T/\Delta t\) and taking the square root gives the expression

This is our expression for the global (or integrated) error. A primary result from this expression is that the global error is proportional to \(\Delta t^2\).

## Stability¶

Looking at (20), it appears that the numerical
solution has constant and correct amplitude, but an error in the
angular frequency. A constant amplitude is not necessarily the case,
however! To see this, note that if only \(\Delta t\) is large enough,
the magnitude of the argument to \(\sin^{-1}\) in
(18) may be larger than 1, i.e., \(\omega\Delta
t/2 > 1\). In this case, \(\sin^{-1}(\omega\Delta t/2)\) has a complex
value and therefore \(\tilde\omega\) becomes complex. Type, for
example, `asin(x)`

in wolframalpha.com to see basic properties of \(\sin^{-1}
(x)\)).

A complex \(\tilde\omega\) can be written $\tilde\omega = \tilde\omega_r

i\tilde\omega_i\(. Since \)\sin^{-1}(x)\( has a *negative* imaginary part for \)x>1\(, \)\tilde\omega_i < 0\(, which means that \)e^{i\tilde\omega t}=e^{-\tilde\omega_i t}e^{i\tilde\omega_r t}\( will lead to exponential growth in time because \)e^{-\tilde\omega_i t}\( with \)\tilde\omega_i <0$ has a positive exponent.

**Stability criterion.**

We do not tolerate growth in the amplitude since such growth is not
present in the exact solution. Therefore, we
must impose a *stability criterion* so that
the argument in the inverse sine function leads
to real and not complex values of \(\tilde\omega\). The stability
criterion reads

With \(\omega =2\pi\), \(\Delta t > \pi^{-1} = 0.3183098861837907\) will give growing solutions. Figure displays what happens when \(\Delta t =0.3184\), which is slightly above the critical value: \(\Delta t =\pi^{-1} + 9.01\cdot 10^{-5}\).

Growing, unstable solution because of a time step slightly beyond the stability limit.

## About the accuracy at the stability limit¶

An interesting question is whether the stability condition \(\Delta t < 2/\omega\) is unfortunate, or more precisely: would it be meaningful to take larger time steps to speed up computations? The answer is a clear no. At the stability limit, we have that \(\sin^{-1}\omega\Delta t/2 = \sin^{-1} 1 = \pi/2\), and therefore \(\tilde\omega = \pi/\Delta t\). (Note that the approximate formula (19) is very inaccurate for this value of \(\Delta t\) as it predicts \(\tilde\omega = 2.34/pi\), which is a 25 percent reduction.) The corresponding period of the numerical solution is \(\tilde P=2\pi/\tilde\omega = 2\Delta t\), which means that there is just one time step \(\Delta t\) between a peak (maximum) and a through (minimum) in the numerical solution. This is the shortest possible wave that can be represented in the mesh! In other words, it is not meaningful to use a larger time step than the stability limit.

Also, the error in angular frequency when \(\Delta t = 2/\omega\) is severe: Figure shows a comparison of the numerical and analytical solution with \(\omega = 2\pi\) and \(\Delta t = 2/\omega = \pi^{-1}\). Already after one period, the numerical solution has a through while the exact solution has a peak (!). The error in frequency when \(\Delta t\) is at the stability limit becomes \(\omega - \tilde\omega = \omega(1-\pi/2)\approx -0.57\omega\). The corresponding error in the period is \(P - \tilde P \approx 0.36P\). The error after \(m\) periods is then \(0.36mP\). This error has reached half a period when \(m=1/(2\cdot 0.36)\approx 1.38\), which theoretically confirms the observations in Figure that the numerical solution is a through ahead of a peak already after one and a half period. Consequently, \(\Delta t\) should be chosen much less than the stability limit to achieve meaningful numerical computations.

Numerical solution with $\Delta t$ exactly at the stability limit.

**Summary.**

From the accuracy and stability analysis we can draw three important conclusions:

The key parameter in the formulas is \(p=\omega\Delta t\). The period of oscillations is \(P=2\pi/\omega\), and the number of time steps per period is \(N_P=P/\Delta t\). Therefore, \(p=\omega\Delta t = 2\pi/N_P\), showing that the critical parameter is the number of time steps per period. The smallest possible \(N_P\) is 2, showing that \(p\in (0,\pi]\).

Provided \(p\leq 2\), the amplitude of the numerical solution is constant.

The ratio of the numerical angular frequency and the exact one is \(\tilde\omega/\omega \approx 1 + \frac{1}{24}p^2\). The error \(\frac{1}{24}p^2\) leads to wrongly displaced peaks of the numerical solution, and the error in peak location grows linearly with time (see Exercise 2: Show linear growth of the phase with time).

# Alternative schemes based on 1st-order equations¶

A standard technique for solving second-order ODEs is to rewrite them as a system of first-order ODEs and then choose a solution strategy from the vast collection of methods for first-order ODE systems. Given the second-order ODE problem

we introduce the auxiliary variable \(v=u^{\prime}\) and express the ODE problem in terms of first-order derivatives of \(u\) and \(v\):

The initial conditions become \(u(0)=I\) and \(v(0)=0\).

## The Forward Euler scheme¶

A Forward Euler approximation to our \(2\times 2\) system of ODEs (24)-(25) becomes

or written out,

Let us briefly compare this Forward Euler method with the centered difference scheme for the second-order differential equation. We have from (28) and (29) applied at levels \(n\) and \(n-1\) that

Since from (28)

it follows that

which is very close to the centered difference scheme, but the last term is evaluated at \(t_{n-1}\) instead of \(t_n\). Rewriting, so that \(\Delta t^2\omega^2u^{n-1}\) appears alone on the right-hand side, and then dividing by \(\Delta t^2\), the new left-hand side is an approximation to \(u^{\prime\prime}\) at \(t_n\), while the right-hand side is sampled at \(t_{n-1}\). All terms should be sampled at the same mesh point, so using \(\omega^2 u^{n-1}\) instead of \(\omega^2 u^n\) points to a kind of mathematical error in the derivation of the scheme. This error turns out to be rather crucial for the accuracy of the Forward Euler method applied to vibration problems (the section Comparison of schemes has examples).

The reasoning above does not imply that the Forward Euler scheme is not correct, but more that it is almost equivalent to a second-order accurate scheme for the second-order ODE formulation, and that the error committed has to do with a wrong sampling point.

## The Backward Euler scheme¶

A Backward Euler approximation to the ODE system is equally easy to write up in the operator notation:

This becomes a coupled system for \(u^{n+1}\) and \(v^{n+1}\):

We can compare (32)-(33) with the centered scheme (7) for the second-order differential equation. To this end, we eliminate \(v^{n+1}\) in (32) using (33) solved with respect to \(v^{n+1}\). Thereafter, we eliminate \(v^n\) using (32) solved with respect to \(v^{n+1}\) and also replacing \(n+1\) by \(n\) and \(n\) by \(n-1\). The resulting equation involving only \(u^{n+1}\), \(u^n\), and \(u^{n-1}\) can be ordered as

which has almost the same form as the centered scheme for the second-order differential equation, but the right-hand side is evaluated at \(u^{n+1}\) and not \(u^n\). This inconsistent sampling of terms has a dramatic effect on the numerical solution, as we demonstrate in the section Comparison of schemes.

## The Crank-Nicolson scheme¶

The Crank-Nicolson scheme takes this form in the operator notation:

Writing the equations out and rearranging terms, shows that this is also a coupled system of two linear equations at each time level:

We may compare also this scheme to the centered discretization of the second-order ODE. It turns out that the Crank-Nicolson scheme is equivalent to the discretization

That is, the Crank-Nicolson is equivalent to (7) for the second-order ODE, apart from an extra term of size \(\Delta t^2\), but this is an error of the same order as in the finite difference approximation on the left-hand side of the equation anyway. The fact that the Crank-Nicolson scheme is so close to (7) makes it a much better method than the Forward or Backward Euler methods for vibration problems, as will be illustrated in the section Comparison of schemes.

Deriving (38) is a bit tricky. We start with rewriting the Crank-Nicolson equations as follows

and add the latter at the previous time level as well:

We can also rewrite (39) at the previous time level as

Inserting (40) for \(v^{n+1}\) in (39) and (41) for \(v^{n}\) in (39) yields after some reordering:

Now, \(v^n + v^{n-1}\) can be eliminated by means of (42). The result becomes

It can be shown that

meaning that (43) is an approximation to the centered scheme (7) for the second-order ODE where the sampling error in the term \(\Delta t^2\omega^2 u^n\) is of the same order as the approximation errors in the finite differences, i.e., \(\mathcal{O}{\Delta t^2}\). The Crank-Nicolson scheme written as (43) therefore has consistent sampling of all terms at the same time point \(t_n\).

## Comparison of schemes¶

We can easily compare methods like the ones above (and many more!) with the aid of the Odespy package. Below is a sketch of the code.

```
import odespy
import numpy as np
def f(u, t, w=1):
# v, u numbering for EulerCromer to work well
v, u = u # u is array of length 2 holding our [v, u]
return [-w**2*u, v]
def run_solvers_and_plot(solvers, timesteps_per_period=20,
num_periods=1, I=1, w=2*np.pi):
P = 2*np.pi/w # duration of one period
dt = P/timesteps_per_period
Nt = num_periods*timesteps_per_period
T = Nt*dt
t_mesh = np.linspace(0, T, Nt+1)
legends = []
for solver in solvers:
solver.set(f_kwargs={'w': w})
solver.set_initial_condition([0, I])
u, t = solver.solve(t_mesh)
```

There is quite some more code dealing with plots also, and we refer
to the source file `vib_undamped_odespy.py`

for details. Observe that keyword arguments in `f(u,t,w=1)`

can
be supplied through a solver parameter `f_kwargs`

(dictionary of
additional keyword arguments to `f`

).

Specification of the Forward Euler, Backward Euler, and Crank-Nicolson schemes is done like this:

```
solvers = [
odespy.ForwardEuler(f),
# Implicit methods must use Newton solver to converge
odespy.BackwardEuler(f, nonlinear_solver='Newton'),
odespy.CrankNicolson(f, nonlinear_solver='Newton'),
]
```

The `vib_undamped_odespy.py`

program makes two plots of the computed
solutions with the various methods in the `solvers`

list: one plot
with \(u(t)\) versus \(t\), and one *phase plane plot* where \(v\) is
plotted against \(u\). That is, the phase plane plot is the curve
\((u(t),v(t))\) parameterized by \(t\). Analytically, \(u=I\cos(\omega t)\)
and \(v=u^{\prime}=-\omega I\sin(\omega t)\). The exact curve
\((u(t),v(t))\) is therefore an ellipse, which often looks like a circle
in a plot if the axes are automatically scaled. The important feature,
however, is that the exact curve \((u(t),v(t))\) is closed and repeats
itself for every period. Not all numerical schemes are capable of
doing that, meaning that the amplitude instead shrinks or grows with
time.

Figure show the
results. Note that Odespy applies the label MidpointImplicit for what
we have specified as `CrankNicolson`

in the code (`CrankNicolson`

is
just a synonym for class `MidpointImplicit`

in the Odespy code). The
Forward Euler scheme in Figure has a pronounced spiral
curve, pointing to the fact that the amplitude steadily grows, which
is also evident in Figure. The
Backward Euler scheme has a similar feature, except that the spriral
goes inward and the amplitude is significantly damped. The changing
amplitude and the spiral form decreases with decreasing time step.
The Crank-Nicolson scheme looks much more accurate. In fact, these
plots tell that the Forward and Backward Euler schemes are not
suitable for solving our ODEs with oscillating solutions.

Comparison of classical schemes in the phase plane for two time step values.

Comparison of solution curves for classical schemes.

## Runge-Kutta methods¶

We may run two other popular standard methods for first-order ODEs, the 2nd- and 4th-order Runge-Kutta methods, to see how they perform. The figures below show the solutions with larger \(\Delta t\) values than what was used in the previous two plots.

Comparison of Runge-Kutta schemes in the phase plane.

Comparison of Runge-Kutta schemes.

The visual impression is that the 4th-order Runge-Kutta method is very accurate, under all circumstances in these tests, while the 2nd-order scheme suffers from amplitude errors unless the time step is very small.

The corresponding results for the Crank-Nicolson scheme are shown in Figure. It is clear that the Crank-Nicolson scheme outhinspace .erforms the 2nd-order Runge-Kutta method. Both schemes have the same order of accuracy \(\mathcal{O}{\Delta t^2}\), but their differences in the accuracy that matters in a real physical application is very clearly pronounced in this example. Exercise 13: Investigate the amplitude errors of many solvers invites you to investigate how the amplitude is computed by a series of famous methods for first-order ODEs.

Long-time behavior of the Crank-Nicolson scheme in the phase plane.

## Analysis of the Forward Euler scheme¶

We may try to find exact solutions of the discrete equations (28)-(29) in the Forward Euler method to better understand why this otherwise useful method has so bad performance for vibration ODEs. An “ansatz” for the solution of the discrete equations is

where \(q\) and \(A\) are scalars to be determined. We could have used a complex exponential form \(e^{i\tilde\omega n\Delta t}\) since we get oscillatory solutions, but the oscillations grow in the Forward Euler method, so the numerical frequency \(\tilde\omega\) will be complex anyway (producing an exponentially growing amplitude). Therefore, it is easier to just work with potentially complex \(A\) and \(q\) as introduced above.

The Forward Euler scheme leads to

We can easily eliminate \(A\), get \(q^2 + \omega^2=0\), and solve for

which gives

We shall take the real part of \(A^n\) as the solution. The two values of \(A\) are complex conjugates, and the real part of \(A^n\) will be the same for both roots. This is easy to realize if we rewrite the complex numbers in polar form, which is also convenient for further analysis and understanding. The polar form \(re^{i\theta}\) of a complex number \(x+iy\) has \(r=\sqrt{x^2+y^2}\) and \(\theta = \tan^{-1}(y/x)\). Hence, the polar form of the two values for \(A\) becomes

Now it is very easy to compute \(A^n\):

Since \(\cos (\theta n) = \cos (-\theta n)\), the real parts of the two numbers become the same. We therefore continue with the solution that has the plus sign.

The general solution is \(u^n = CA^n\), where \(C\) is a constant determined from the initial condition: \(u^0=C=I\). We have \(u^n=IA^n\) and \(v^n=qIA^n\). The final solutions are just the real part of the expressions in polar form:

The expression \((1+\omega^2\Delta t^2)^{n/2}\) causes growth of
the amplitude, since a number greater than one is raised to a positive
exponent \(n/2\). We can develop a series expression to better understand
the formula for the amplitude. Introducing \(p=\omega\Delta t\) as the
key variable and using `sympy`

gives

```
from sympy import *
p = symbols('p', real=True)
n = symbols('n', integer=True, positive=True)
amplitude = (1 + p**2)**(n/2)
amplitude.series(p, 0, 4)
```

The amplitude goes like \(1 + \frac{1}{2} n\omega^2\Delta t^2\), clearly growing linearly in time (with \(n\)).

We can also investigate the error in the angular frequency by a series expansion:

```
n*atan(p).series(p, 0, 4)
```

This means that the solution for \(u^n\) can be written as

The error in the angular frequency is of the same order as in the scheme (7) for the second-order ODE, but the error in the amplitude is severe.

# Energy considerations¶

The observations of various methods in the previous section can be
better interpreted if we compute a quantity reflecting
the total *energy of the system*. mathcal{I}_t turns out that this quantity,

is *constant* for all \(t\). Checking that \(E(t)\) really remains constant
brings evidence that the numerical computations are sound.
mathcal{I}_t turns out that \(E\) is proportional to the mechanical energy
in the system. Conservation of energy is
much used to check numerical simulations, so it is well invested time to
dive into this subject.

## Derivation of the energy expression¶

We start out with multiplying

by \(u^{\prime}\) and integrating from \(0\) to \(T\):

Observing that

we get

where we have introduced

The important result from this derivation is that the total energy is constant:

**\(E(t)\) is closely related to the system’s energy.**

The quantity \(E(t)\) derived above is physically not the mechanical energy of a vibrating mechanical system, but the energy per unit mass. To see this, we start with Newton’s second law \(F=ma\) (\(F\) is the sum of forces, \(m\) is the mass of the system, and \(a\) is the acceleration). The displacement \(u\) is related to \(a\) through \(a=u^{\prime\prime}\). With a spring force as the only force we have \(F=-ku\), where \(k\) is a spring constant measuring the stiffness of the spring. Newton’s second law then implies the differential equation

This equation of motion can be turned into an energy balance equation by finding the work done by each term during a time interval \([0,T]\). To this end, we multiply the equation by \(du=u^{\prime}dt\) and integrate:

The result is

where

is the *kinetic energy* of the system, and

is the *potential energy*. The sum \(\tilde E(t)\) is the total mechanical energy.
The derivation demonstrates the famous energy principle that, under
the right physical circumstances, any
change in the kinetic energy is due to a change in potential energy
and vice versa. (This principle breaks down when we introduce damping
in the system, as we do in the section Generalization: damping, nonlinearities, and
excitation.)

The equation \(mu^{\prime\prime}+ku=0\) can be divided by \(m\) and written as \(u^{\prime\prime} + \omega^2u=0\) for \(\omega=\sqrt{k/m}\). The energy expression \(E(t)=\frac{1}{2}(u^{\prime})^2 + \frac{1}{2}\omega^2u^2\) derived earlier is then \(\tilde E(t)/m\), i.e., mechanical energy per unit mass.

### Energy of the exact solution¶

Analytically, we have \(u(t)=I\cos\omega t\), if \(u(0)=I\) and \(u^{\prime}(0)=0\), so we can easily check the energy evolution and confirm that \(E(t)\) is constant:

### Growth of energy in the Forward Euler scheme¶

mathcal{I}_t is easy to show that the energy in the Forward Euler scheme increases when stepping from time level \(n\) to \(n+1\).

## An error measure based on energy¶

The constant energy is well expressed by its initial value \(E(0)\), so that the error in mechanical energy can be computed as a mesh function by

where

if \(u(0)=I\) and \(u^{\prime}(0)=V\). Note that we have used a centered approximation to \(u^{\prime}\): \(u^{\prime}(t_n)\approx [D_{2t}u]^n\).

A useful norm of the mesh function \(e_E^n\) for the discrete mechanical energy can be the maximum absolute value of \(e_E^n\):

Alternatively, we can compute other norms involving integration over all mesh points, but we are often interested in worst case deviation of the energy, and then the maximum value is of particular relevance.

A vectorized Python implementation of \(e_E^n\) takes the form

```
# import numpy as np and compute u, t
dt = t[1]-t[0]
E = 0.5*((u[2:] - u[:-2])/(2*dt))**2 + 0.5*w**2*u[1:-1]**2
E0 = 0.5*V**2 + 0.5**w**2*I**2
e_E = E - E0
e_E_norm = np.abs(e_E).max()
```

The convergence rates of the quantity `e_E_norm`

can be used for
verification. The value of `e_E_norm`

is also useful for comparing
schemes through their ability to preserve energy. Below is a table
demonstrating the relative error in total energy for various schemes
(computed by the `vib_undamped_odespy.py`

program). The test problem is
\(u^{\prime\prime} + 4\pi^2 u =0\) with \(u(0)=1\) and \(u'(0)=0\), so the
period is 1 and \(E(t)\approx 4.93\). We clearly see that the
Crank-Nicolson and the Runge-Kutta schemes are superior to the Forward
and Backward Euler schemes already after one period.

Method | $T$ | $\Delta t$ | $\max \left\vert e_E^n\right\vert/e_E^0$ |
---|---|---|---|

Forward Euler | $1$ | $0.025$ | $1.678\cdot 10^{0}$ |

Backward Euler | $1$ | $0.025$ | $6.235\cdot 10^{-1}$ |

Crank-Nicolson | $1$ | $0.025$ | $1.221\cdot 10^{-2}$ |

Runge-Kutta 2nd-order | $1$ | $0.025$ | $6.076\cdot 10^{-3}$ |

Runge-Kutta 4th-order | $1$ | $0.025$ | $8.214\cdot 10^{-3}$ |

Method | $T$ | $\Delta t$ | $\max \left\vert e_E^n\right\vert/e_E^0$ |
---|---|---|---|

Forward Euler | $10$ | $0.025$ | $1.788\cdot 10^{4}$ |

Backward Euler | $10$ | $0.025$ | $1.000\cdot 10^{0}$ |

Crank-Nicolson | $10$ | $0.025$ | $1.221\cdot 10^{-2}$ |

Runge-Kutta 2nd-order | $10$ | $0.025$ | $6.250\cdot 10^{-2}$ |

Runge-Kutta 4th-order | $10$ | $0.025$ | $8.288\cdot 10^{-3}$ |

Running multiple values of \(\Delta t\), we can get some insight into the convergence of the energy error:

Method | $T$ | $\Delta t$ | $\max \left\vert e_E^n\right\vert/e_E^0$ |
---|---|---|---|

Forward Euler | $10$ | $0.05$ | $1.120\cdot 10^{8}$ |

Forward Euler | $10$ | $0.025$ | $1.788\cdot 10^{4}$ |

Forward Euler | $10$ | $0.0125$ | $1.374\cdot 10^{2}$ |

Backward Euler | $10$ | $0.05$ | $1.000\cdot 10^{0}$ |

Backward Euler | $10$ | $0.025$ | $1.000\cdot 10^{0}$ |

Backward Euler | $10$ | $0.0125$ | $9.928\cdot 10^{-1}$ |

Crank-Nicolson | $10$ | $0.05$ | $4.756\cdot 10^{-2}$ |

Crank-Nicolson | $10$ | $0.025$ | $1.221\cdot 10^{-2}$ |

Crank-Nicolson | $10$ | $0.0125$ | $3.125\cdot 10^{-3}$ |

Runge-Kutta 2nd-order | $10$ | $0.05$ | $6.152\cdot 10^{-1}$ |

Runge-Kutta 2nd-order | $10$ | $0.025$ | $6.250\cdot 10^{-2}$ |

Runge-Kutta 2nd-order | $10$ | $0.0125$ | $7.631\cdot 10^{-3}$ |

Runge-Kutta 4th-order | $10$ | $0.05$ | $3.510\cdot 10^{-2}$ |

Runge-Kutta 4th-order | $10$ | $0.025$ | $8.288\cdot 10^{-3}$ |

Runge-Kutta 4th-order | $10$ | $0.0125$ | $2.058\cdot 10^{-3}$ |

# The Euler-Cromer method¶

While the Runge-Kutta methods and the Crank-Nicolson scheme work well for the vibration equation modeled as a first-order ODE system, both were inferior to the straightforward centered difference scheme for the second-order equation \(u^{\prime\prime}+\omega^2u=0\). However, there is a similarly successful scheme available for the first-order system \(u^{\prime}=v\), \(v^{\prime}=-\omega^2u\), to be presented below. The ideas of the scheme and their further developments have become very popular in particle and rigid body dynamics and hence are widely used by physicists.

## Forward-backward discretization¶

The idea is to apply a Forward Euler discretization to the first equation and a Backward Euler discretization to the second. In operator notation this is stated as

We can write out the formulas and collect the unknowns on the left-hand side:

We realize that after \(u^{n+1}\) has been computed from (52), it may be used directly in (53) to compute \(v^{n+1}\).

In physics, it is more common to update the \(v\) equation first, with a forward difference, and thereafter the \(u\) equation, with a backward difference that applies the most recently computed \(v\) value:

The advantage of ordering the ODEs as in (54)-(55) becomes evident when considering complicated models. Such models are included if we write our vibration ODE more generally as

We can rewrite this second-order ODE as two first-order ODEs,

This rewrite allows the following scheme to be used:

We realize that the first update works well with any \(g\) since old values \(u^n\) and \(v^n\) are used. Switching the equations would demand \(u^{n+1}\) and \(v^{n+1}\) values in \(g\) and result in nonlinear algebraic equations to be solved at each time level.

The scheme (54)-(55) goes under several names: forward-backward scheme, semi-implicit Euler method, semi-explicit Euler, symplectic Euler, Newton-Stoermer-Verlet, and Euler-Cromer. We shall stick to the latter name.

How does the Euler-Cromer method preserve the total energy? We may run the example from the section An error measure based on energy:

Method | $T$ | $\Delta t$ | $\max \left\vert e_E^n\right\vert/e_E^0$ |
---|---|---|---|

Euler-Cromer | $10$ | $0.05$ | $2.530\cdot 10^{-2}$ |

Euler-Cromer | $10$ | $0.025$ | $6.206\cdot 10^{-3}$ |

Euler-Cromer | $10$ | $0.0125$ | $1.544\cdot 10^{-3}$ |

## Equivalence with the scheme for the second-order ODE¶

We shall now show that the Euler-Cromer scheme for the system of first-order equations is equivalent to the centered finite difference method for the second-order vibration ODE (!).

We may eliminate the \(v^n\) variable from (52)-(53) or (54)-(55). The \(v^{n+1}\) term in (54) can be eliminated from (55):

The \(v^{n}\) quantity can be expressed by \(u^n\) and \(u^{n-1}\) using (55):

and when this is inserted in (56) we get

which is nothing but the centered scheme (7)! The two seemingly different numerical methods are mathematically equivalent. Consequently, the previous analysis of (7) also applies to the Euler-Cromer method. In particular, the amplitude is constant, given that the stability criterion is fulfilled, but there is always an angular frequency error (19). Exercise 18: Analysis of the Euler-Cromer scheme gives guidance on how to derive the exact discrete solution of the two equations in the Euler-Cromer method.

Although the Euler-Cromer scheme and the method (7) are equivalent, there could be differences in the way they handle the initial conditions. Let us look into this topic. The initial condition \(u^{\prime}=0\) means \(u^{\prime}=v=0\). From (54) we get

and from (55) it follows that

When we previously used a centered approximation of \(u^{\prime}(0)=0\) combined with the discretization (7) of the second-order ODE, we got a slightly different result: \(u^1=u^0 - \frac{1}{2}\omega^2\Delta t^2 u^0\). The difference is \(\frac{1}{2}\omega^2\Delta t^2 u^0\), which is of second order in \(\Delta t\), seemingly consistent with the overall error in the scheme for the differential equation model.

A different view can also be taken. If we approximate \(u^{\prime}(0)=0\) by a backward difference, \((u^0-u^{-1})/\Delta t =0\), we get \(u^{-1}=u^0\), and when combined with (7), it results in \(u^1=u^0 - \omega^2\Delta t^2 u^0\). This means that the Euler-Cromer method based on (55)-(54) corresponds to using only a first-order approximation to the initial condition in the method from the section A centered finite difference scheme.

Correspondingly, using the formulation (52)-(53) with \(v^n=0\) leads to \(u^1=u^0\), which can be interpreted as using a forward difference approximation for the initial condition \(u^{\prime}(0)=0\). Both Euler-Cromer formulations lead to slightly different values for \(u^1\) compared to the method in the section A centered finite difference scheme. The error is \(\frac{1}{2}\omega^2\Delta t^2 u^0\).

## Implementation¶

### Solver function¶

The function below, found in `vib_undamped_EulerCromer.py`

, implements the Euler-Cromer scheme
(54)-(55):

```
import numpy as np
from devito import Dimension, TimeFunction, Eq, solve, Operator, Constant
```

```
# %load -s solver, src-vib/vib_undamped_EulerCromer.py
def solver(I, w, dt, T):
"""
Solve u'=v, v' = - w**2*u for t in (0,T], u(0)=I and v(0)=0,
by an Euler-Cromer method.
"""
dt = float(dt)
Nt = int(round(T/dt))
t = Dimension('t', spacing=Constant('h_t'))
v = TimeFunction(name='v', dimensions=(t,), shape=(Nt+1,), space_order=2)
u = TimeFunction(name='u', dimensions=(t,), shape=(Nt+1,), space_order=2)
v.data[:] = 0
u.data[:] = I
eq_v = Eq(v.dt, -(w**2)*u)
eq_u = Eq(u.dt, v.forward)
stencil_v = solve(eq_v, v.forward)
stencil_u = solve(eq_u, u.forward)
update_v = Eq(v.forward, stencil_v)
update_u = Eq(u.forward, stencil_u)
op = Operator([update_v, update_u])
op.apply(h_t=dt, t_M=Nt-1)
return u.data, v.data, np.linspace(0, Nt*dt, Nt+1)
```

We can now compare the output from this function with the exact solution to the original second-order vibration ODE using our `visualize`

function from before:

```
I = 1
w = 2*np.pi
dt = 0.05
num_periods = 5
P = 2*np.pi/w # one period
T = P*num_periods
u, v, t = solver(I, w, dt, T)
visualize(u, t, I, w)
```

```
Operator `Kernel` run in 0.01 s
```

### Verification¶

Since the Euler-Cromer scheme is equivalent to the finite difference
method for the second-order ODE \(u^{\prime\prime}+\omega^2u=0\) (see the section Equivalence with the scheme for the second-order ODE), the performance of the above
`solver`

function is the same as for the `solver`

function in the section Implementation. The only difference is the formula for the first time
step, as discussed above. This deviation in the Euler-Cromer scheme
means that the discrete solution listed in the section Exact discrete solution is not a solution of the Euler-Cromer
scheme!

To verify the implementation of the Euler-Cromer method we can adjust
`v[1]`

so that the computer-generated values can be compared with the
formula (20) from in the section Exact discrete solution. This adjustment is done in an alternative
solver function, `solver_ic_fix`

in `vib_undamped_EulerCromer.py`

. Since we now
have an exact solution of the discrete equations available, we can
write a test function `test_solver`

for checking the equality of
computed values with the formula (20):

```
# %load -s test_solver, src-vib/vib_undamped_EulerCromer.py
def test_solver():
"""
Test solver with fixed initial condition against
equivalent scheme for the 2nd-order ODE u'' + u = 0.
"""
I = 1.2; w = 2.0; T = 5
dt = 2/w # longest possible time step
u, v, t = solver_ic_fix(I, w, dt, T)
from vib_undamped import solver as solver2 # 2nd-order ODE
u2, t2 = solver2(I, w, dt, T)
error = np.abs(u - u2).max()
tol = 1E-14
assert error < tol
```

Another function, `demo`

, visualizes the difference between the
Euler-Cromer scheme and the scheme (7) for the
second-oder ODE, arising from the mismatch in the first time level.

### Convergence rates¶

We may use the `convergence_rates`

function in the file
`vib_undamped.py`

to investigate the convergence rate of the
Euler-Cromer method, see the `convergence_rate`

function in the file
`vib_undamped_EulerCromer.py`

. Since we could eliminate \(v\) to get a
scheme for \(u\) that is equivalent to the finite difference method for
the second-order equation in \(u\), we would expect the convergence
rates to be the same, i.e., \(r = 2\). However,
measuring the convergence rate of \(u\) in the Euler-Cromer scheme shows
that \(r = 1\) only! Adjusting the initial condition
does not change the rate. Adjusting \(\omega\), as outlined in the section The error in the numerical frequency, gives a 4th-order method there, while
there is no increase in the measured rate in the Euler-Cromer
scheme. It is obvious that the Euler-Cromer scheme is dramatically
much better than the two other first-order methods, Forward Euler and
Backward Euler, but this is not reflected in the convergence rate of
\(u\).

## The Stoermer-Verlet algorithm¶

Another very popular algorithm for vibration problems, especially
for long time simulations, is the
Stoermer-Verlet
algorithm. It has become *the* method among physicists
for molecular simulations as well as particle and rigid body dynamics.

The method can be derived by applying the Euler-Cromer idea twice, in a symmetric fashion, during the interval \([t_n,t_{n+1}]\):

solve \(v^{\prime}=-\omega u\) by a Forward Euler step in \([t_n,t_{n+\frac{1}{2}}]\)

solve \(u^{\prime}=v\) by a Backward Euler step in \([t_n,t_{n+\frac{1}{2}}]\)

solve \(u^{\prime}=v\) by a Forward Euler step in \([t_{n+\frac{1}{2}}, t_{n+1}]\)

solve \(v^{\prime}=-\omega u\) by a Backward Euler step in \([t_{n+\frac{1}{2}}, t_{n+1}]\)

With mathematics,

The two steps in the middle can be combined to

and consequently

Writing the last equation as \(v^n = v^{n-\frac{1}{2}} - \frac{1}{2}\Delta t\omega^2 u^n\) and using this \(v^n\) in the first equation gives \(v^{n+\frac{1}{2}} = v^{n-\frac{1}{2}} - \Delta t\omega^2 u^n\), and the scheme can be written as two steps:

which is nothing but straightforward centered differences for the
\(2\times 2\) ODE system on a *staggered mesh*, see
the section The Euler-Cromer scheme on a staggered mesh. We have thus seen that
four different reasonings (discretizing \(u^{\prime\prime}+\omega^2 u\)
directly, using Euler-Cromer, using
Stoermer-Verlet,
and using centered differences for the \(2\times 2\) system on a staggered
mesh) all end up with the same equations! The main difference is that
the traditional Euler-Cromer displays first-order convergence in \(\Delta t\)
(due to less symmetry in the way \(u\) and \(v\) are treated)
while the others are \(\mathcal{O}{\Delta t^2}\) schemes.

The most numerically stable scheme, with respect to accumulation of rounding errors, is (61)-(62). It has, according to [Hairer_Wanner_Norsett_bookI], better properties in this regard than the direct scheme for the second-order ODE.

# Staggered mesh¶

A more intuitive discretization than the Euler-Cromer method, yet
equivalent, employs solely centered differences in a natural way
for the \(2\times 2\) first-order ODE system.
The scheme is in fact fully equivalent to the second-order scheme for
\(u''+\omega u=0\), also for the first time step.
Such a scheme needs to operate on a *staggered mesh* in time.
Staggered meshes are very popular in many physical application, maybe foremost
fluid dynamics and electromagnetics, so the topic is important to learn.

## The Euler-Cromer scheme on a staggered mesh¶

In a staggered mesh, the unknowns are sought at different points in
the mesh. Specifically, \(u\) is sought at
integer time points \(t_n\) and \(v\) is sought at \(t_{n+1/2}\)
*between* two \(u\) points.
The unknowns are then \(u^1, v^{3/2}, u^2, v^{5/2}\), and so on.
We typically use the notation
\(u^n\) and \(v^{n+\frac{1}{2}}\) for the two unknown mesh functions.
Figure presents a graphical sketch of two
mesh functions \(u\) and \(v\) on a staggered mesh.

Examples on mesh functions on a staggered mesh in time.

On a staggered mesh it is natural to use centered difference approximations, expressed in operator notation as

or if we switch the sequence of the equations:

Writing out the formulas gives

We can eliminate the \(v\) values and get back the centered scheme based on the second-order differential equation \(u^{\prime\prime} +\omega^2 u = 0\), so all these three schemes are equivalent. However, they differ somewhat in the treatment of the initial conditions.

Suppose we have \(u(0)=I\) and \(u'(0)=v(0)=0\) as mathematical initial conditions. This means \(u^0=I\) and

Using the discretized equation (67) for \(n=0\) yields

and eliminating \(v^{-\frac{1}{2}} =- v^{\frac{1}{2}}\) results in

and

which is exactly the same equation for \(u^1\) as we had in the centered scheme based on the second-order differential equation (and hence corresponds to a centered difference approximation of the initial condition for \(u'(0)\)). The conclusion is that a staggered mesh is fully equivalent with that scheme, while the forward-backward version gives a slight deviation in the computation of \(u^1\).

We can redo the derivation of the initial conditions when \(u'(0)=V\):

Using this \(v^{-\frac{1}{2}}\) in

then gives \(v^\frac{1}{2} = V - \frac{1}{2}\Delta t\omega^2 I\). The general initial conditions are therefore

## Implementation of the scheme on a staggered mesh¶

The algorithm goes like this:

a. Compute \(u^{n}\) from (68).

b. Compute \(v^{n+\frac{1}{2}}\) from (67).

### Implementation with integer indices¶

Translating the schemes (68)
and (67) to computer code
faces the problem of how to store and access \(v^{n+\frac{1}{2}}\),
since arrays only allow integer indices with base 0.
We must then introduce a convention: \(v^{1+\frac{1}{2}}\) is stored
in `v[n]`

while \(v^{1-\frac{1}{2}}\) is stored in `v[n-1]`

.
We can then write the algorithm in Python as

```
# %load -s solver_v1, src-vib/vib_undamped_staggered.py
def solver_v1(I, w, dt, T):
"""
Solve u'=v, v' = - w**2*u for t in (0,T], u(0)=I and v(0)=0,
by a central finite difference method with time step dt.
"""
dt = float(dt)
Nt = int(round(T/dt))
t = Dimension('t', spacing=Constant('h_t'))
u = TimeFunction(name='u', dimensions=(t,), shape=(Nt+1,), space_order=2)
v = TimeFunction(name='v', dimensions=(t,), shape=(Nt+1,), space_order=2)
u.data[:] = I
v.data[:] = 0 - 0.5*dt*w**2*u.data[:]
eq_u = Eq(u.dt, v)
eq_v = Eq(v.dt, -(w**2)*u.forward)
stencil_u = solve(eq_u, u.forward)
stencil_v = solve(eq_v, v.forward)
update_u = Eq(u.forward, stencil_u)
update_v = Eq(v.forward, stencil_v)
op = Operator([update_u, update_v])
op.apply(h_t=dt, t_M=Nt-1)
t_mesh = np.linspace(0, Nt*dt, Nt+1) # mesh for u
t_v_mesh = (t_mesh + dt/2)[:-1] # mesh for v
return u.data, t_mesh, v.data, t_v_mesh
```

Note that \(u\) and \(v\) are returned together with the mesh points such
that the complete mesh function for \(u\) is described by `u`

and `t_mesh`

,
while `v`

and `t_v_mesh`

represent the mesh function for \(v\).

Once again, we can compare with the exact solution using the `visualize`

function:

```
I = 1
w = 2*np.pi
dt = 0.05
num_periods = 5
P = 2*np.pi/w # one period
T = P*num_periods
u, t, v, t_v = solver_v1(I, w, dt, T)
visualize(u, t, I, w)
```

```
Operator `Kernel` run in 0.01 s
```

Verification of this code is easy as we can just compare the computed
`u`

with the `u`

produced by the `solver`

function in
`vib_undamped.py`

(which solves \(u''+\omega^2u=0\) directly). The
values should coincide to machine precision since the two numerical
methods are mathematically equivalent. We refer to the file
`vib_undamped_staggered.py`

for the details of a unit test (`test_staggered`

) that checks this property.

# Exercises and Problems¶

## Problem 1: Use linear/quadratic functions for verification¶

Consider the ODE problem

**a)**
Discretize this equation according to \([D_tD_t u + \omega^2 u = f]^n\) and
derive the equation for the first time step (\(u^1\)).

**Solution.**
For the requested discretization, we get

To derive the equation for \(u^1\), we first find the expression for \(u^{n+1}\) from the discretized form of the equation. Isolating \(u^{n+1}\), we get

With \(n = 0\), this expression gives

Here, however, we get a problem with \(u^{-1}\), which appears on the right hand side. To get around that problem, we realize that the initial condition \(u^{\prime} = V\) might be approximated by use of a centered difference approximation as

which means that

Inserting this expression for \(u^{-1}\) into the expression for \(u^1\), we get

Finally, after isolating \(u^1\) on the left hand side, we arrive at [ u^1 = \left(1 - \frac{1}{2}(\Delta t\omega)^2\right)u^0 + \Delta t V + \frac{1}{2}\Delta t^2 f^n\thinspace .]

**b)**
For verification purposes, we use the method of manufactured solutions (MMS) with the
choice of \(\uex(t)= ct+d\). Find restrictions on \(c\) and \(d\) from
the initial conditions. Compute the corresponding source term \(f\).
Show that \([D_tD_t t]^n=0\) and use the fact
that the \(D_tD_t\) operator is linear,
\([D_tD_t (ct+d)]^n = c[D_tD_t t]^n + [D_tD_t d]^n = 0\), to show that
\(\uex\) is also a perfect solution of the discrete equations.

**Solution.**
The initial conditions \(u(0)=I\) and \(u^{\prime}(0)=V\) give demands
\(\uex(0)=I\) and \(\uex^{\prime}(0)=V\), which imply that
\(d = I\) and {c = V}.

To compute the source term \(f\), we insert the chosen solution \(\uex\) into the ODE. This gives

which implies that

To show that \([D_tD_t t]^n=0\), we proceed as

Finally, we show that the chosen \(\uex\) is also a perfect solution of the discrete equations. If we start by inserting \(\uex\) into

as well as the expression found for \(f\). We get

which can be rewritten as

Now, since the first term here is zero, we see that the discrete equation is fulfilled exactly for the chosen \(\uex\) function.

**c)**
Use `sympy`

to do the symbolic calculations above. Here is a
sketch of the program `vib_undamped_verify_mms.py`

:

```
import sympy as sym
V, t, I, w, dt = sym.symbols('V t I w dt') # global symbols
f = None # global variable for the source term in the ODE
def ode_source_term(u):
"""Return the terms in the ODE that the source term
must balance, here u'' + w**2*u.
u is symbolic Python function of t."""
return sym.diff(u(t), t, t) + w**2*u(t)
def residual_discrete_eq(u):
"""Return the residual of the discrete eq. with u inserted."""
R = ...
return sym.simplify(R)
def residual_discrete_eq_step1(u):
"""Return the residual of the discrete eq. at the first
step with u inserted."""
R = ...
return sym.simplify(R)
def DtDt(u, dt):
"""Return 2nd-order finite difference for u_tt.
u is a symbolic Python function of t.
"""
return ...
def main(u):
"""
Given some chosen solution u (as a function of t, implemented
as a Python function), use the method of manufactured solutions
to compute the source term f, and check if u also solves
the discrete equations.
"""
print '=== Testing exact solution: %s ===' % u
print "Initial conditions u(0)=%s, u'(0)=%s:" % \
(u(t).subs(t, 0), sym.diff(u(t), t).subs(t, 0))
# Method of manufactured solution requires fitting f
global f # source term in the ODE
f = sym.simplify(ode_lhs(u))
# Residual in discrete equations (should be 0)
print 'residual step1:', residual_discrete_eq_step1(u)
print 'residual:', residual_discrete_eq(u)
def linear():
main(lambda t: V*t + I)
if __name__ == '__main__':
linear()
```

Fill in the various functions such that the calls in the `main`

function works.

**Solution.**
This part of the code goes as follows:

```
import sympy as sym
import numpy as np
V, t, I, w, dt = sym.symbols('V t I w dt') # global symbols
f = None # global variable for the source term in the ODE
def ode_source_term(u):
"""Return the terms in the ODE that the source term
must balance, here u'' + w**2*u.
u is symbolic Python function of t."""
return sym.diff(u(t), t, t) + w**2*u(t)
def residual_discrete_eq(u):
"""Return the residual of the discrete eq. with u inserted."""
R = DtDt(u, dt) + w**2*u(t) - f
return sym.simplify(R)
def residual_discrete_eq_step1(u):
"""Return the residual of the discrete eq. at the first
step with u inserted."""
half = sym.Rational(1,2)
R = u(t+dt) - I - dt*V - \
half*dt**2*f.subs(t, 0) + half*dt**2*w**2*I
R = R.subs(t, 0) # t=0 in the rhs of the first step eq.
return sym.simplify(R)
def DtDt(u, dt):
"""Return 2nd-order finite difference for u_tt.
u is a symbolic Python function of t.
"""
return (u(t+dt) - 2*u(t) + u(t-dt))/dt**2
def main(u):
"""
Given some chosen solution u (as a function of t, implemented
as a Python function), use the method of manufactured solutions
to compute the source term f, and check if u also solves
the discrete equations.
"""
print '=== Testing exact solution: %s ===' % u(t)
print "Initial conditions u(0)=%s, u'(0)=%s:" % \
(u(t).subs(t, 0), sym.diff(u(t), t).subs(t, 0))
# Method of manufactured solution requires fitting f
global f # source term in the ODE
f = sym.simplify(ode_source_term(u))
# Residual in discrete equations (should be 0)
print 'residual step1:', residual_discrete_eq_step1(u)
print 'residual:', residual_discrete_eq(u)
```

**d)**
The purpose now is to choose a quadratic function
\(\uex = bt^2 + ct + d\) as exact solution. Extend the `sympy`

code above with a function `quadratic`

for fitting `f`

and checking
if the discrete equations are fulfilled. (The function is very similar
to `linear`

.)

**Solution.**
Yes, a quadratic function will fulfill the discrete equations exactly.
The implementation becomes

```
def quadratic():
"""Test quadratic function q*t**2 + V*t + I."""
q = sym.Symbol('q') # arbitrary constant in t**2 term
u_e = lambda t: q*t**2 + V*t + I
main(u_e)
```

Calling `quadratic()`

shows that the residual vanishes, and the quadratic
function is an exact solution of the discrete equations.

**e)**
Will a polynomial of degree three fulfill the discrete equations?

**Solution.**
We can easily make a test:

```
def cubic():
r, q = sym.symbols('r q')
main(lambda t: r*t**3 + q*t**2 + V*t + I)
```

When running the final code presented below, the printout shows that the step1 residual for the cubic function is not zero.

**f)**
Implement a `solver`

function for computing the numerical
solution of this problem.

**Solution.**
The `solver`

function may take the form

```
# NBVAL_SKIP
def solver(I, V, f, w, dt, T):
"""
Solve u'' + w**2*u = f for t in (0,T], u(0)=I and u'(0)=V,
by a central finite difference method with time step dt.
f(t) is a callable Python function.
"""
dt = float(dt)
Nt = int(round(T/dt))
u = np.zeros(Nt+1)
t = np.linspace(0, Nt*dt, Nt+1)
u[0] = I
u[1] = u[0] - 0.5*dt**2*w**2*u[0] + 0.5*dt**2*f(t[0]) + dt*V
for n in range(1, Nt):
u[n+1] = 2*u[n] - u[n-1] - dt**2*w**2*u[n] + dt**2*f(t[n])
return u, t
```

We can verify the implementation by the following test function:

```
def test_quadratic_exact_solution():
"""Verify solver function via quadratic solution."""
# Transform global symbolic variables to functions and numbers
# for numerical computations
global p, V, I, w
p, V, I, w = 2.3, 0.9, 1.2, 1.5
global f, t
u_e = lambda t: p*t**2 + V*t + I # use p, V, I, w as numbers
f = ode_source_term(u_e) # fit source term
f = sym.lambdify(t, f) # make function numerical
dt = 2./w
u, t = solver(I=I, V=V, f=f, w=w, dt=dt, T=3)
u_e = u_e(t)
error = np.abs(u - u_e).max()
tol = 1E-12
assert error < tol
print 'Error in computing a quadratic solution:', error
```

**g)**
Write a test function for checking that the quadratic solution
is computed correctly (to machine precision, but the
round-off errors accumulate and increase with \(T\)) by the `solver`

function.

**Solution.**
Here is the complete code for this exercise:

```
import sympy as sym
import numpy as np
V, t, I, w, dt = sym.symbols('V t I w dt') # global symbols
f = None # global variable for the source term in the ODE
def ode_source_term(u):
"""Return the terms in the ODE that the source term
must balance, here u'' + w**2*u.
u is symbolic Python function of t."""
return sym.diff(u(t), t, t) + w**2*u(t)
def residual_discrete_eq(u):
"""Return the residual of the discrete eq. with u inserted."""
R = DtDt(u, dt) + w**2*u(t) - f
return sym.simplify(R)
def residual_discrete_eq_step1(u):
"""Return the residual of the discrete eq. at the first
step with u inserted."""
half = sym.Rational(1,2)
R = u(t+dt) - I - dt*V - \
half*dt**2*f.subs(t, 0) + half*dt**2*w**2*I
R = R.subs(t, 0) # t=0 in the rhs of the first step eq.
return sym.simplify(R)
def DtDt(u, dt):
"""Return 2nd-order finite difference for u_tt.
u is a symbolic Python function of t.
"""
return (u(t+dt) - 2*u(t) + u(t-dt))/dt**2
def main(u):
"""
Given some chosen solution u (as a function of t, implemented
as a Python function), use the method of manufactured solutions
to compute the source term f, and check if u also solves
the discrete equations.
"""
print '=== Testing exact solution: %s ===' % u(t)
print "Initial conditions u(0)=%s, u'(0)=%s:" % \
(u(t).subs(t, 0), sym.diff(u(t), t).subs(t, 0))
# Method of manufactured solution requires fitting f
global f # source term in the ODE
f = sym.simplify(ode_source_term(u))
# Residual in discrete equations (should be 0)
print 'residual step1:', residual_discrete_eq_step1(u)
print 'residual:', residual_discrete_eq(u)
def linear():
"""Test linear function V*t+I: u(0)=I, u'(0)=V."""
main(lambda t: V*t + I)
def quadratic():
"""Test quadratic function q*t**2 + V*t + I."""
q = sym.Symbol('q') # arbitrary constant in t**2 term
u_e = lambda t: q*t**2 + V*t + I
main(u_e)
def cubic():
r, q = sym.symbols('r q')
main(lambda t: r*t**3 + q*t**2 + V*t + I)
def solver(I, V, f, w, dt, T):
"""
Solve u'' + w**2*u = f for t in (0,T], u(0)=I and u'(0)=V,
by a central finite difference method with time step dt.
f(t) is a callable Python function.
"""
dt = float(dt)
Nt = int(round(T/dt))
u = np.zeros(Nt+1)
t = np.linspace(0, Nt*dt, Nt+1)
u[0] = I
u[1] = u[0] - 0.5*dt**2*w**2*u[0] + 0.5*dt**2*f(t[0]) + dt*V
for n in range(1, Nt):
u[n+1] = 2*u[n] - u[n-1] - dt**2*w**2*u[n] + dt**2*f(t[n])
return u, t
def test_quadratic_exact_solution():
"""Verify solver function via quadratic solution."""
# Transform global symbolic variables to functions and numbers
# for numerical computations
global p, V, I, w
p, V, I, w = 2.3, 0.9, 1.2, 1.5
global f, t
u_e = lambda t: p*t**2 + V*t + I # use p, V, I, w as numbers
f = ode_source_term(u_e) # fit source term
f = sym.lambdify(t, f) # make function numerical
dt = 2./w
u, t = solver(I=I, V=V, f=f, w=w, dt=dt, T=3)
u_e = u_e(t)
error = np.abs(u - u_e).max()
tol = 1E-12
assert error < tol
print 'Error in computing a quadratic solution:', error
if __name__ == '__main__':
linear()
quadratic()
cubic()
test_quadratic_exact_solution()
```

Filename: `vib_undamped_verify_mms`

.

## Exercise 2: Show linear growth of the phase with time¶

Consider an exact solution \(I\cos (\omega t)\) and an approximation \(I\cos(\tilde\omega t)\). Define the phase error as the time lag between the peak \(I\) in the exact solution and the corresponding peak in the approximation after \(m\) periods of oscillations. Show that this phase error is linear in \(m\).

**Solution.**
From (19) we have that

Dropping the \(\mathcal{O}{\Delta t^4}\) term, and since \(\omega=\frac{2\pi}{P}\) and \(\tilde\omega=\frac{2\pi}{\tilde P}\), we have that

Now, \(2\pi\) cancels and the remaining equation may be rewritten as

This implies that the periods differ by a constant. Since the exact and the numerical solution start out identically, the phase error \(P - \tilde P\) will become \(m\frac{1}{24}\omega^2\Delta t^2\) after \(m\) periods, i.e. the phase error is linear in \(m\).

Filename: `vib_phase_error_growth`

.

## Exercise 3: Improve the accuracy by adjusting the frequency¶

According to (19), the numerical
frequency deviates from the exact frequency by a (dominating) amount
\(\omega^3\Delta t^2/24 >0\). Replace the `w`

parameter in the algorithm
in the `solver`

function in `vib_undamped.py`

by `w*(1 - (1./24)*w**2*dt**2`

and test how this adjustment in the numerical
algorithm improves the accuracy (use \(\Delta t =0.1\) and simulate
for 80 periods, with and without adjustment of \(\omega\)).

**Solution.**
We may take a copy of the `vib_undamped.py`

file and edit the `solver`

function to

```
# NBVAL_SKIP
from numpy import *
from matplotlib.pyplot import *
def solver(I, w, dt, T, adjust_w=True):
"""
Solve u'' + w**2*u = 0 for t in (0,T], u(0)=I and u'(0)=0,
by a central finite difference method with time step dt.
"""
dt = float(dt)
Nt = int(round(T/dt))
u = zeros(Nt+1)
t = linspace(0, Nt*dt, Nt+1)
if adjust_w:
w = w*(1 - 1./24*w**2*dt**2)
u[0] = I
u[1] = u[0] - 0.5*dt**2*w**2*u[0]
for n in range(1, Nt):
u[n+1] = 2*u[n] - u[n-1] - dt**2*w**2*u[n]
return u, t
```

The modified code was run for 80 periods with, and without, the given adjustment of \(\omega\). A substantial difference in accuracy was observed between the two, showing that the frequency adjustment improves the situation.

Filename: `vib_adjust_w`

.

## Exercise 4: See if adaptive methods improve the phase error¶

Adaptive methods for solving ODEs aim at adjusting \(\Delta t\) such
that the error is within a user-prescribed tolerance. Implement the
equation \(u^{\prime\prime}+u=0\) in the Odespy
software. Use the example from Section 3.2.11 in [Langtangen_decay].

Run the scheme with a very low
tolerance (say \(10^{-14}\)) and for a long time, check the number of
time points in the solver’s mesh (`len(solver.t_all)`

), and compare
the phase error with that produced by the simple finite difference
method from the section A centered finite difference scheme with the same number of (equally
spaced) mesh points. The question is whether it pays off to use an
adaptive solver or if equally many points with a simple method gives
about the same accuracy.

**Solution.**
Here is a code where we define the test problem, solve it by the
Dormand-Prince adaptive method from Odespy, and then call `solver`

```
import odespy
import numpy as np
import sys
#import matplotlib.pyplot as plt
import scitools.std as plt
def f(s, t):
u, v = s
return np.array([v, -u])
def u_exact(t):
return I*np.cos(w*t)
I = 1; V = 0; u0 = np.array([I, V])
w = 1; T = 50
tol = float(sys.argv[1])
solver = odespy.DormandPrince(f, atol=tol, rtol=0.1*tol)
Nt = 1 # just one step - let scheme find its intermediate points
t_mesh = np.linspace(0, T, Nt+1)
t_fine = np.linspace(0, T, 10001)
solver.set_initial_condition(u0)
u, t = solver.solve(t_mesh)
# u and t will only consist of [I, u^Nt] and [0,T], i.e. 2 values
# each, while solver.u_all and solver.t_all contain all computed
# points. solver.u_all is a list with arrays, one array (with 2
# values) for each point in time.
u_adaptive = np.array(solver.u_all)
# For comparison, we solve also with simple FDM method
import sys, os
sys.path.insert(0, os.path.join(os.pardir, 'src-vib'))
from vib_undamped import solver as simple_solver
Nt_simple = len(solver.t_all)
dt = float(T)/Nt_simple
u_simple, t_simple = simple_solver(I, w, dt, T)
# Compare in plot: adaptive, constant dt, exact
plt.plot(solver.t_all, u_adaptive[:,0], 'k-')
plt.hold('on')
plt.plot(t_simple, u_simple, 'r--')
plt.plot(t_fine, u_exact(t_fine), 'b-')
plt.legend(['tol=%.0E' % tol, 'u simple', 'exact'])
plt.savefig('tmp_odespy_adaptive.png')
plt.savefig('tmp_odespy_adaptive.pdf')
plt.show()
raw_input()
```

The program may produce the plots seen in the figure below, which shows how the adaptive solution clearly outhinspace .erforms the simpler method, regardless of the accuracy level.

Filename: `vib_undamped_adaptive`

.

## Exercise 5: Use a Taylor polynomial to compute \(u^1\)¶

As an alternative to computing \(u^1\) by (8), one can use a Taylor polynomial with three terms:

With \(u^{\prime\prime}=-\omega^2 u\) and \(u^{\prime}(0)=0\), show that this method also leads to (8). Generalize the condition on \(u^{\prime}(0)\) to be \(u^{\prime}(0)=V\) and compute \(u^1\) in this case with both methods.

**Solution.**
With \(u^{\prime\prime}(0)=-\omega^2 u(0)\) and \(u^{\prime}(0)=0\), the given Taylor series
becomes

which may be written as

but this is nothing but (8).

Now, consider \(u^{\prime}(0)=V\). With a centered difference approximation, this initial condition becomes

which implies that

When \(n=0\), (7) reads

Inserting the expression for \(u^{-1}\), we get

which implies that

With the Taylor series approach, we now get

which also gives

Filename: `vib_first_step`

.

## Problem 6: Derive and investigate the velocity Verlet method¶

The velocity Verlet method for \(u^{\prime\prime} + \omega^2u=0\) is based on the following ideas:

step \(u\) forward from \(t_n\) to \(t_{n+1}\) using a three-term Taylor series,

replace \(u^{\prime\prime}\) by \(-\omega^2u\)

discretize \(v^{\prime}=-\omega^2u\) by a Crank-Nicolson method.

Derive the scheme, implement it, and determine empirically the convergence rate.

**Solution.**
Stepping \(u\) forward from \(t_n\) to \(t_{n+1}\) using a three-term Taylor
series gives

Using \(u^{\prime}=v\) and \(u^{\prime\prime}=-\omega^2u\), we get the updating formula

Second, the first-order equation for \(v\),

is discretized by a centered difference in a Crank-Nicolson fashion at \(t_{n+\frac{1}{2}}\):

To summarize, we have the scheme

known as the velocity Verlet algorithm. Observe that this scheme is explicit since \(u^{n+1}\) in the second equation is already computed by the first equation.

The algorithm can be straightforwardly implemented as shown below:

```
from vib_undamped import convergence_rates, main
def solver(I, w, dt, T, return_v=False):
"""
Solve u'=v, v'=-w**2*u for t in (0,T], u(0)=I and v(0)=0,
by the velocity Verlet method with time step dt.
"""
dt = float(dt)
Nt = int(round(T/dt))
u = np.zeros(Nt+1)
v = np.zeros(Nt+1)
t = np.linspace(0, Nt*dt, Nt+1)
u[0] = I
v[0] = 0
for n in range(Nt):
u[n+1] = u[n] + v[n]*dt - 0.5*dt**2*w**2*u[n]
v[n+1] = v[n] - 0.5*dt*w**2*(u[n] + u[n+1])
if return_v:
return u, v, t
else:
# Return just u and t as in the vib_undamped.py's solver
return u, t
```

We provide the option that this `solver`

function returns the same data
as the `solver`

function from the section Making a solver function (if `return_v`

is `False`

), but alternatively, it may return `v`

along with `u`

and `t`

.

The error in the Taylor series expansion behind the first equation
is \(\mathcal{O}{\Delta t^3}\), while the error
in the central difference for \(v\) is \(\mathcal{O}{\Delta t^2}\). The overall
error is then no better than \(\mathcal{O}{\Delta t^2}\), which can be verified
empirically using the `convergence_rates`

function from
the section Verification:

```
import vib_undamped_velocity_Verlet as m
m.convergence_rates(4, solver_function=m.solver)
```

The output confirms that the overall convergence rate is 2.

## Problem 7: Find the minimal resolution of an oscillatory function¶

Sketch the function on a given mesh which has the highest possible frequency. That is, this oscillatory “cos-like” function has its maxima and minima at every two grid points. Find an expression for the frequency of this function, and use the result to find the largest relevant value of \(\omega\Delta t\) when \(\omega\) is the frequency of an oscillating function and \(\Delta t\) is the mesh spacing.

**Solution.**
The smallest period must be \(2\Delta t\). Since the period \(P\) is related
to the angular frequency \(\omega\) by \(P=2\pi/\omega\), it means that
\(\omega = \frac{2\pi}{2\Delta t} = \frac{\pi}{\Delta t}\) is the smallest
meaningful angular frequency.
This further means that the largest value for \(\omega\Delta t\) is \(\pi\).

Filename: `vib_largest_wdt`

.

## Exercise 8: Visualize the accuracy of finite differences for a cosine function¶

We introduce the error fraction

to measure the error in the finite difference approximation \(D_tD_tu\) to \(u^{\prime\prime}\). Compute \(E\) for the specific choice of a cosine/sine function of the form \(u=\exp{(i\omega t)}\) and show that

Plot \(E\) as a function of \(p=\omega\Delta t\). The relevant
values of \(p\) are \([0,\pi]\) (see Problem 7: Find the minimal resolution of an oscillatory function
for why \(p>\pi\) does not make sense).
The deviation of the curve from unity visualizes the error in the
approximation. Also expand \(E\) as a Taylor polynomial in \(p\) up to
fourth degree (use, e.g., `sympy`

).

**Solution.**

Since \(u(t)=\exp{(i\omega t)}\), we have that \(u^{\prime}(t)=i\omega\exp{(i\omega t)}\) and \(u^{\prime\prime}(t)=(i\omega)^2\exp{(i\omega t)}=-\omega^2\exp{(i\omega t)}\), so we may proceed with \(E\) as

Now, since \(\cos(\omega\Delta t)=1-2\sin^2\left(\frac{\omega\Delta t}{2}\right)\), we finally get

```
import matplotlib.pyplot as plt
import numpy as np
import sympy as sym
def E_fraction(p):
return (2./p)**2*(np.sin(p/2.))**2
a = 0; b = np.pi
p = np.linspace(a, b, 100)
E_values = np.zeros(len(p))
# create 4th degree Taylor polynomial (also plotted)
p_ = sym.symbols('p_')
E = (2./p_)**2*(sym.sin(p_/2.))**2
E_series = E.series(p_, 0, 4).removeO()
print(E_series)
E_pyfunc = sym.lambdify([p_], E_series, modules='numpy')
# To avoid division by zero when p is 0, we rather take the limit
E_values[0] = sym.limit(E, p_, 0, dir='+') # ...when p --> 0, E --> 1
E_values[1:] = E_fraction(p[1:])
plt.plot(p, E_values, 'k-', p, E_pyfunc(p), 'k--')
plt.xlabel('p'); plt.ylabel('Error fraction')
plt.legend(['E', 'E Taylor'])
plt.savefig('tmp_error_fraction.png')
plt.savefig('tmp_error_fraction.pdf')
plt.show()
```

```
1.0 - 0.0833333333333333*p_**2
```

From the plot seen below, we realize how the error fraction \(E\) deviates from unity as \(p\) grows.

Filename: `vib_plot_fd_exp_error`

.

## Exercise 9: Verify convergence rates of the error in energy¶

We consider the ODE problem \(u^{\prime\prime} + \omega^2u=0\), \(u(0)=I\), \(u^{\prime}(0)=V\), for \(t\in (0,T]\). The total energy of the solution \(E(t)=\frac{1}{2}(u^{\prime})^2 + \frac{1}{2}\omega^2 u^2\) should stay constant. The error in energy can be computed as explained in the section Energy considerations.

Make a test function in a separate file, where code from
`vib_undamped.py`

is imported, but the `convergence_rates`

and
`test_convergence_rates`

functions are copied and modified to also
incorporate computations of the error in energy and the convergence
rate of this error. The expected rate is 2, just as for the solution
itself.

**Solution.**
The complete code with test functions goes as follows.

```
import os, sys
sys.path.insert(0, os.path.join(os.pardir, 'src-vib'))
from vib_undamped import solver, u_exact, visualize
import numpy as np
def convergence_rates(m, solver_function, num_periods=8):
"""
Return m-1 empirical estimates of the convergence rate
based on m simulations, where the time step is halved
for each simulation.
solver_function(I, w, dt, T) solves each problem, where T
is based on simulation for num_periods periods.
"""
from math import pi
w = 0.35; I = 0.3 # just chosen values
P = 2*pi/w # period
dt = P/30 # 30 time step per period 2*pi/w
T = P*num_periods
energy_const = 0.5*I**2*w**2 # initial energy when V = 0
dt_values = []
E_u_values = [] # error in u
E_energy_values = [] # error in energy
for i in range(m):
u, t = solver_function(I, w, dt, T)
u_e = u_exact(t, I, w)
E_u = np.sqrt(dt*np.sum((u_e-u)**2))
E_u_values.append(E_u)
energy = 0.5*((u[2:] - u[:-2])/(2*dt))**2 + \
0.5*w**2*u[1:-1]**2
E_energy = energy - energy_const
E_energy_norm = np.abs(E_energy).max()
E_energy_values.append(E_energy_norm)
dt_values.append(dt)
dt = dt/2
r_u = [np.log(E_u_values[i-1]/E_u_values[i])/
np.log(dt_values[i-1]/dt_values[i])
for i in range(1, m, 1)]
r_E = [np.log(E_energy_values[i-1]/E_energy_values[i])/
np.log(dt_values[i-1]/dt_values[i])
for i in range(1, m, 1)]
return r_u, r_E
def test_convergence_rates():
r_u, r_E = convergence_rates(
m=5,
solver_function=solver,
num_periods=8)
# Accept rate to 1 decimal place
tol = 0.1
assert abs(r_u[-1] - 2.0) < tol
assert abs(r_E[-1] - 2.0) < tol
if __name__ == '__main__':
test_convergence_rates()
```

Filename: `test_error_conv`

.

## Exercise 10: Use linear/quadratic functions for verification¶

This exercise is a generalization of Problem 1: Use linear/quadratic functions for verification to the extended model problem
(vib:ode2) where the damping term is either linear or quadratic.
Solve the various subproblems and see how the results and problem
settings change with the generalized ODE in case of linear or
quadratic damping. By modifying the code from Problem 1: Use linear/quadratic functions for verification, `sympy`

will do most
of the work required to analyze the generalized problem.

**Solution.**
With a linear spring force, i.e. \(s(u)=cu\) (for constant \(c\)),
our model problem becomes

First we consider linear damping, i.e., when \(f(u^{\prime}) = bu^{\prime}\), and follow the text in the section vib:ode2:fdm:flin. Discretizing the equation according to

implies that

The explicit formula for \(u\) at each new time level then becomes

For the first time step, we use \(n=0\) and a centered difference approximation for the initial condition on the derivative. This gives

Next, we consider quadratic damping, i.e., when \(f(u^{\prime})=bu^{\prime}|u^{\prime}|\), and follow the text in the chapter vib:ode2:fdm:fquad. Discretizing the equation according to

gives us

We solve for \(u^{n+1}\) to get the explicit updating formula as

and the equation for the first time step as

Turning to verification with MMS and \(u_e(t)=ct+d\), we get \(d=I\) and \(c=V\) independent of the damping term, so these parameter values stay as for the undamped case.

Proceeding with linear damping, we get from the chapter vib:ode2:verify that

(Note that there are two different c parameters here, one from \(u_e=ct+d\) and one from the spring force \(cu\). The first one disappears, however, as it is switched with \(V\).)

To show that \(u_e\) is a perfect solution also to the discrete equations, we insert \(u_e\) and \(F\) into

This gives

which may be split up as

Simplifying, we note that the first term is zero and that \(c[(Vt+I)]^n\) appears with the same sign on each side of the equation. Thus, dropping these terms, and cancelling the common factor \(b\), we are left with

mathcal{I}_t therefore remains to show that \([D_{2t}(Vt+I)]^n\) is equal to \([V]^n = V\). We write out the left hand side as

which shows that the two sides of the equation are equal and that the discrete equations are fulfilled exactly for the given \(u_e\) function.

If the damping is rather quadratic, we find from the chapter vib:ode2:verify that

As with linear damping, we show that \(u_e\) is a perfect solution also to the discrete equations by inserting \(u_e\) and \(F\) into

We then get

which simplifies to

and further to

which simply states that

Thus, \(u_e\) fulfills the discrete equations exactly also when the damping term is quadratic.

When the exact solution is changed to become quadratic or cubic, the situation is more complicated.

For a quadratic solution \(u_e\) combined with (zero damping or) linear damping, the output from the program below shows that the discrete equations are fulfilled exactly. However, this is not the case with nonlinear damping, where only the first step gives zero residual.

For a cubic solution \(u_e\), we get a nonzero residual for (zero damping and) linear and nonlinear damping.

```
import sympy as sym
import numpy as np
# The code in vib_undamped_verify_mms.py is here generalized
# to treat the model m*u'' + f(u') + c*u = F(t), where the
# damping term f(u') = 0, b*u' or b*V*abs(V).
def ode_source_term(u, damping):
"""Return the terms in the ODE that the source term
must balance, here m*u'' + f(u') + c*u.
u is a symbolic Python function of t."""
if damping == 'zero':
return m*sym.diff(u(t), t, t) + c*u(t)
elif damping == 'linear':
return m*sym.diff(u(t), t, t) + \
b*sym.diff(u(t), t) + c*u(t)
else: # damping is nonlinear
return m*sym.diff(u(t), t, t) + \
b*sym.diff(u(t), t)*abs(sym.diff(u(t), t)) + c*u(t)
def residual_discrete_eq(u, damping):
"""Return the residual of the discrete eq. with u inserted."""
if damping == 'zero':
R = m*DtDt(u, dt) + c*u(t) - F
elif damping == 'linear':
R = m*DtDt(u, dt) + b*D2t(u, dt) + c*u(t) - F
else: # damping is nonlinear
R = m*DtDt(u, dt) + b*Dt_p_half(u, dt)*\
abs(Dt_m_half(u, dt)) + c*u(t) - F
return sym.simplify(R)
def residual_discrete_eq_step1(u, damping):
"""Return the residual of the discrete eq. at the first
step with u inserted."""
half = sym.Rational(1,2)
if damping == 'zero':
R = u(t+dt) - u(t) - dt*V - \
half*dt**2*(F.subs(t, 0)/m) + half*dt**2*(c/m)*I
elif damping == 'linear':
R = u(t+dt) - (I + dt*V + \
half*(dt**2/m)*(-b*V - c*I + F.subs(t, 0)))
else: # damping is nonlinear
R = u(t+dt) - (I + dt*V + \
half*(dt**2/m)*(-b*V*abs(V) - c*I + F.subs(t, 0)))
R = R.subs(t, 0) # t=0 in the rhs of the first step eq.
return sym.simplify(R)
def DtDt(u, dt):
"""Return 2nd-order finite difference for u_tt.
u is a symbolic Python function of t.
"""
return (u(t+dt) - 2*u(t) + u(t-dt))/dt**2
def D2t(u, dt):
"""Return 2nd-order finite difference for u_t.
u is a symbolic Python function of t.
"""
return (u(t+dt) - u(t-dt))/(2.0*dt)
def Dt_p_half(u, dt):
"""Return 2nd-order finite difference for u_t, sampled at n+1/2,
i.e, n pluss one half... u is a symbolic Python function of t.
"""
return (u(t+dt) - u(t))/dt
def Dt_m_half(u, dt):
"""Return 2nd-order finite difference for u_t, sampled at n-1/2,
i.e, n minus one half.... u is a symbolic Python function of t.
"""
return (u(t) - u(t-dt))/dt
def main(u, damping):
"""
Given some chosen solution u (as a function of t, implemented
as a Python function), use the method of manufactured solutions
to compute the source term f, and check if u also solves
the discrete equations.
"""
print '=== Testing exact solution: %s ===' % u(t)
print "Initial conditions u(0)=%s, u'(0)=%s:" % \
(u(t).subs(t, 0), sym.diff(u(t), t).subs(t, 0))
# Method of manufactured solution requires fitting F
global F # source term in the ODE
F = sym.simplify(ode_source_term(u, damping))
# Residual in discrete equations (should be 0)
print 'residual step1:', residual_discrete_eq_step1(u, damping)
print 'residual:', residual_discrete_eq(u, damping)
def linear(damping):
def u_e(t):
"""Return chosen linear exact solution."""
# General linear function u_e = c*t + d
# Initial conditions u(0)=I, u'(0)=V require c=V, d=I
return V*t + I
main(u_e, damping)
def quadratic(damping):
# Extend with quadratic functions
q = sym.Symbol('q') # arbitrary constant in quadratic term
def u_e(t):
return q*t**2 + V*t + I
main(u_e, damping)
def cubic(damping):
r, q = sym.symbols('r q')
main(lambda t: r*t**3 + q*t**2 + V*t + I, damping)
def solver(I, V, F, b, c, m, dt, T, damping):
"""
Solve m*u'' + f(u') + c*u = F for t in (0,T], u(0)=I and u'(0)=V,
by a central finite difference method with time step dt.
F(t) is a callable Python function.
"""
dt = float(dt)
Nt = int(round(T/dt))
u = np.zeros(Nt+1)
t = np.linspace(0, Nt*dt, Nt+1)
if damping == 'zero':
u[0] = I
u[1] = u[0] - 0.5*dt**2*(c/m)*u[0] + \
0.5*dt**2*F(t[0])/m + dt*V
for n in range(1, Nt):
u[n+1] = 2*u[n] - u[n-1] - \
dt**2*(c/m)*u[n] + dt**2*F(t[n])/m
elif damping == 'linear':
u[0] = I
u[1] = u[0] + dt*V + \
0.5*(dt**2/m)*(-b*V - c*u[0] + F(t[0]))
for n in range(1, Nt):
u[n+1] = (2*m*u[n] + (b*dt/2.-m)*u[n-1] + \
dt**2*(F(t[n])-c*u[n]))/(m+b*dt/2.)
else: # damping is quadratic
u[0] = I
u[1] = u[0] + dt*V + \
0.5*(dt**2/m)*(-b*V*abs(V) - c*u[0] + F(t[0]))
for n in range(1, Nt):
u[n+1] = 1./(m+b*abs(u[n]-u[n-1])) * \
(2*m*u[n] - m*u[n-1] + b*u[n]*\
abs(u[n]-u[n-1])+dt**2*(F(t[n])-c*u[n]))
return u, t
def test_quadratic_exact_solution(damping):
# Transform global symbolic variables to functions and numbers
# for numerical computations
global p, V, I, b, c, m
p, V, I, b, c, m = 2.3, 0.9, 1.2, 2.1, 1.6, 1.3 # i.e., as numbers
global F, t
u_e = lambda t: p*t**2 + V*t + I
F = ode_source_term(u_e, damping) # fit source term
F = sym.lambdify(t, F) # ...numerical Python function
from math import pi, sqrt
dt = 2*pi/sqrt(c/m)/10 # 10 steps per period 2*pi/w, w=sqrt(c/m)
u, t = solver(I=I, V=V, F=F, b=b, c=c, m=m, dt=dt,
T=(2*pi/sqrt(c/m))*2, damping=damping)
u_e = u_e(t)
error = np.abs(u - u_e).max()
tol = 1E-12
assert error < tol
print 'Error in computing a quadratic solution:', error
if __name__ == '__main__':
damping = ['zero', 'linear', 'quadratic']
for e in damping:
V, t, I, dt, m, b, c = sym.symbols('V t I dt m b c') # global
F = None # global variable for the source term in the ODE
print '---------------------------------------Damping:', e
linear(e) # linear solution used for MMS
quadratic(e) # quadratic solution for MMS
cubic(e) # ... and cubic
test_quadratic_exact_solution(e)
```

Filename: `vib_verify_mms`

.

## Exercise 11: Use an exact discrete solution for verification¶

Write a test function in a separate file
that employs the exact discrete solution
(20) to verify the implementation of the
`solver`

function in the file `vib_undamped.py`

.

**Solution.**
The code goes like this:

```
from vib_undamped import solver
from numpy import arcsin as asin, pi, cos, abs
def test_solver_exact_discrete_solution():
def tilde_w(w, dt):
return (2./dt)*asin(w*dt/2.)
def u_numerical_exact(t):
return I*cos(tilde_w(w, dt)*t)
w = 2.5
I = 1.5
# Estimate period and time step
P = 2*pi/w
num_periods = 4
T = num_periods*P
N = 5 # time steps per period
dt = P/N
u, t = solver(I, w, dt, T)
u_e = u_numerical_exact(t)
error= abs(u_e - u).max()
# Make a plot in a file, but not on the screen
from scitools.std import plot
plot(t, u, 'bo', t, u_e, 'r-',
legend=('numerical', 'exact'), show=False,
savefig='tmp.png')
assert error < 1E-14
if __name__ == '__main__':
test_solver_exact_discrete_solution()
```

Filename: `test_vib_undamped_exact_discrete_sol`

.

## Exercise 12: Use analytical solution for convergence rate tests¶

The purpose of this exercise is to perform convergence tests of the
problem (vib:ode2) when \(s(u)=cu\), \(F(t)=A\sin\phi t\) and there
is no damping. Find the complete analytical solution to the problem
in this case (most textbooks on mechanics or ordinary differential
equations list the various elements you need to write down the exact
solution, or you can use symbolic tools like `sympy`

or `wolframalpha.com`

).
Modify the `convergence_rate`

function from the
`vib_undamped.py`

program to perform experiments with the extended
model. Verify that the error is of order \(\Delta t^2\).

**Solution.**
The code:

```
import numpy as np
import matplotlib.pyplot as plt
from vib_verify_mms import solver
def u_exact(t, I, V, A, f, c, m):
"""Found by solving mu'' + cu = F in Wolfram alpha."""
k_1 = I
k_2 = (V - A*2*np.pi*f/(c - 4*np.pi**2*f**2*m))*\
np.sqrt(m/float(c))
return A*np.sin(2*np.pi*f*t)/(c - 4*np.pi**2*f**2*m) + \
k_2*np.sin(np.sqrt(c/float(m))*t) + \
k_1*np.cos(np.sqrt(c/float(m))*t)
def convergence_rates(N, solver_function, num_periods=8):
"""
Returns N-1 empirical estimates of the convergence rate
based on N simulations, where the time step is halved
for each simulation.
solver_function(I, V, F, c, m, dt, T, damping) solves
each problem, where T is based on simulation for
num_periods periods.
"""
def F(t):
"""External driving force"""
return A*np.sin(2*np.pi*f*t)
b, c, m = 0, 1.6, 1.3 # just some chosen values
I = 0 # init. cond. u(0)
V = 0 # init. cond. u'(0)
A = 1.0 # amplitude of driving force
f = 1.0 # chosen frequency of driving force
damping = 'zero'
P = 1/f
dt = P/30 # 30 time step per period 2*pi/w
T = P*num_periods
dt_values = []
E_values = []
for i in range(N):
u, t = solver_function(I, V, F, b, c, m, dt, T, damping)
u_e = u_exact(t, I, V, A, f, c, m)
E = np.sqrt(dt*np.sum((u_e-u)**2))
dt_values.append(dt)
E_values.append(E)
dt = dt/2
#plt.plot(t, u, 'b--', t, u_e, 'r-'); plt.grid(); plt.show()
r = [np.log(E_values[i-1]/E_values[i])/
np.log(dt_values[i-1]/dt_values[i])
for i in range(1, N, 1)]
print(r)
return r
def test_convergence_rates():
r = convergence_rates(
N=5,
solver_function=solver,
num_periods=8)
# Accept rate to 1 decimal place
tol = 0.1
assert abs(r[-1] - 2.0) < tol
if __name__ == '__main__':
test_convergence_rates()
```

The output from the program shows that \(r\) approaches 2.

Filename: `vib_conv_rate`

.

## Exercise 13: Investigate the amplitude errors of many solvers¶

Use the program `vib_undamped_odespy.py`

from the section Comparison of schemes (utilize the function `amplitudes`

) to investigate
how well famous methods for 1st-order ODEs can preserve the amplitude of \(u\) in undamped
oscillations. Test, for example, the 3rd- and 4th-order Runge-Kutta
methods (`RK3`

, `RK4`

), the Crank-Nicolson method (`CrankNicolson`

),
the 2nd- and 3rd-order Adams-Bashforth methods (`AdamsBashforth2`

,
`AdamsBashforth3`

), and a 2nd-order Backwards scheme
(`Backward2Step`

). The relevant governing equations are listed in
the beginning of the section Alternative schemes based on 1st-order equations.

Running the code, we get the plots seen in Figure,
vib:exer:fig:ampl_CNB2, and vib:exer:fig:ampl_AB. They
show that `RK4`

is superior to the others, but that also `CrankNicolson`

performs well. In fact, with `RK4`

the amplitude changes by less than \(0.1\) per cent over the interval.

The amplitude as it changes over 100 periods for RK3 and RK4.

The amplitude as it changes over 100 periods for Crank-Nicolson and Backward 2 step.

The amplitude as it changes over 100 periods for Adams-Bashforth 2 and 3.

**Solution.**
We modify the proposed code to the following:

```
import scitools.std as plt
#import matplotlib.pyplot as plt
from vib_empirical_analysis import minmax, amplitudes
import sys
import odespy
import numpy as np
def f(u, t, w=1):
# v, u numbering for EulerCromer to work well
v, u = u # u is array of length 2 holding our [v, u]
return [-w**2*u, v]
def run_solvers_and_check_amplitudes(solvers, timesteps_per_period=20,
num_periods=1, I=1, w=2*np.pi):
P = 2*np.pi/w # duration of one period
dt = P/timesteps_per_period
Nt = num_periods*timesteps_per_period
T = Nt*dt
t_mesh = np.linspace(0, T, Nt+1)
file_name = 'Amplitudes' # initialize filename for plot
for solver in solvers:
solver.set(f_kwargs={'w': w})
solver.set_initial_condition([0, I])
u, t = solver.solve(t_mesh)
solver_name = \
'CrankNicolson' if solver.__class__.__name__ == \
'MidpointImplicit' else solver.__class__.__name__
file_name = file_name + '_' + solver_name
minima, maxima = minmax(t, u[:,0])
a = amplitudes(minima, maxima)
plt.plot(range(len(a)), a, '-', label=solver_name)
plt.hold('on')
plt.xlabel('Number of periods')
plt.ylabel('Amplitude (absolute value)')
plt.legend(loc='upper left')
plt.savefig(file_name + '.png')
plt.savefig(file_name + '.pdf')
plt.show()
# Define different sets of experiments
solvers_CNB2 = [odespy.CrankNicolson(f, nonlinear_solver='Newton'),
odespy.Backward2Step(f)]
solvers_RK34 = [odespy.RK3(f),
odespy.RK4(f)]
solvers_AB = [odespy.AdamsBashforth2(f),
odespy.AdamsBashforth3(f)]
if __name__ == '__main__':
# Default values
timesteps_per_period = 30
solver_collection = 'CNB2'
num_periods = 100
# Override from command line
try:
# Example: python vib_undamped_odespy.py 30 RK34 50
timesteps_per_period = int(sys.argv[1])
solver_collection = sys.argv[2]
num_periods = int(sys.argv[3])
except IndexError:
pass # default values are ok
solvers = eval('solvers_' + solver_collection) # list of solvers
run_solvers_and_check_amplitudes(solvers,
timesteps_per_period,
num_periods)
```

Filename: `vib_amplitude_errors`

.

## Problem 14: Minimize memory usage of a simple vibration solver¶

We consider the model problem \(u''+\omega^2 u = 0\), \(u(0)=I\), \(u'(0)=V\),
solved by a second-order finite difference scheme. A standard implementation
typically employs an array `u`

for storing all the \(u^n\) values. However,
at some time level `n+1`

where we want to compute `u[n+1]`

, all we need
of previous `u`

values are from level `n`

and `n-1`

. We can therefore avoid
storing the entire array `u`

, and instead work with `u[n+1]`

, `u[n],`

and `u[n-1]`

, named as `u`

, `u_n`

, `u_nmp1`

, for instance. Another
possible naming convention is `u`

, `u_n[0]`

, `u_n[-1]`

.
Store the solution in a file
for later visualization. Make a test function that verifies the implementation
by comparing with the another code for the same problem.

**Solution.**
The modified solver function needs more manual steps initially, and it needs
shuffling of the `u_n`

and `u_nm1`

variables at each time level. Otherwise
it is very similar to the previous `solver`

function with an array `u`

for
the entire mesh function.

```
# NBVAL_SKIP
import numpy as np
import matplotlib.pyplot as plt
def solver_memsave(I, w, dt, T, filename='tmp.dat'):
"""
As vib_undamped.solver, but store only the last three
u values in the implementation. The solution is written to
file `tmp_memsave.dat`.
Solve u'' + w**2*u = 0 for t in (0,T], u(0)=I and u'(0)=0,
by a central finite difference method with time step dt.
"""
dt = float(dt)
Nt = int(round(T/dt))
t = np.linspace(0, Nt*dt, Nt+1)
outfile = open(filename, 'w')
u_n = I
outfile.write('%20.12f %20.12f\n' % (0, u_n))
u = u_n - 0.5*dt**2*w**2*u_n
outfile.write('%20.12f %20.12f\n' % (dt, u))
u_nm1 = u_n
u_n = u
for n in range(1, Nt):
u = 2*u_n - u_nm1 - dt**2*w**2*u_n
outfile.write('%20.12f %20.12f\n' % (t[n], u))
u_nm1 = u_n
u_n = u
return u, t
```

Verification can be done by comparing with the `solver`

function in
the `vib_undamped`

module. Note that to compare both time series, we need
to load the data written to file in `solver_memsave`

back in memory again.
For this purpose, we can use the `numpy.loadtxt`

function, which reads
tabular data and returns them as a table `data`

. Our interest is in the
second column of the data (the `u`

values).

```
# NBVAL_SKIP
def test_solver_memsave():
from vib_undamped import solver
_, _ = solver_memsave(I=1, dt=0.1, w=1, T=30)
u_expected, _ = solver (I=1, dt=0.1, w=1, T=30)
data = np.loadtxt('tmp.dat')
u_computed = data[:,1]
diff = np.abs(u_expected - u_computed).max()
assert diff < 5E-13, diff
```

Filename: `vib_memsave0`

.

## Problem 15: Minimize memory usage of a general vibration solver¶

The program `vib.py`

stores the complete
solution \(u^0,u^1,\ldots,u^{N_t}\) in memory, which is convenient for
later plotting. Make a memory minimizing version of this program
where only the last three \(u^{n+1}\), \(u^n\), and \(u^{n-1}\) values are
stored in memory under the names `u`

, `u_n`

, and `u_nm1`

(this is the
naming convention used in this book).
Write each computed \((t_{n+1}, u^{n+1})\) pair to
file. Visualize the data in the file (a cool solution is to read one
line at a time and plot the \(u\) value using the line-by-line plotter
in the `visualize_front_ascii`

function - this technique makes it
trivial to visualize very long time simulations).

**Solution.**
Here is the complete program:

```
import numpy as np
import scitools.std as plt
def solve_and_store(filename, I, V, m, b, s,
F, dt, T, damping='linear'):
"""
Solve m*u'' + f(u') + s(u) = F(t) for t in (0,T], u(0)=I and
u'(0)=V, by a central finite difference method with time step
dt. If damping is 'linear', f(u')=b*u, while if damping is
'quadratic', f(u')=b*u'*abs(u'). F(t) and s(u) are Python
functions. The solution is written to file (filename).
Naming convention: we use the name u for the new solution
to be computed, u_n for the solution one time step prior to
that and u_nm1 for the solution two time steps prior to that.
Returns min and max u values needed for subsequent plotting.
"""
dt = float(dt); b = float(b); m = float(m) # avoid integer div.
Nt = int(round(T/dt))
outfile = open(filename, 'w')
outfile.write('Time Position\n')
u_nm1 = I
u_min = u_max = u_nm1
outfile.write('%6.3f %7.5f\n' % (0*dt, u_nm1))
if damping == 'linear':
u_n = u_nm1 + dt*V + dt**2/(2*m)*(-b*V - s(u_nm1) + F(0*dt))
elif damping == 'quadratic':
u_n = u_nm1 + dt*V + \
dt**2/(2*m)*(-b*V*abs(V) - s(u_nm1) + F(0*dt))
if u_n < u_nm1:
u_min = u_n
else: # either equal or u_n > u_nm1
u_max = u_n
outfile.write('%6.3f %7.5f\n' % (1*dt, u_n))
for n in range(1, Nt):
# compute solution at next time step
if damping == 'linear':
u = (2*m*u_n + (b*dt/2 - m)*u_nm1 +
dt**2*(F(n*dt) - s(u_n)))/(m + b*dt/2)
elif damping == 'quadratic':
u = (2*m*u_n - m*u_nm1 + b*u_n*abs(u_n - u_nm1)
+ dt**2*(F(n*dt) - s(u_n)))/\
(m + b*abs(u_n - u_nm1))
if u < u_min:
u_min = u
elif u > u_max:
u_max = u
# write solution to file
outfile.write('%6.3f %7.5f\n' % ((n+1)*dt, u))
# switch references before next step
u_nm1, u_n, u = u_n, u, u_nm1
outfile.close()
return u_min, u_max
def main():
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('--I', type=float, default=1.0)
parser.add_argument('--V', type=float, default=0.0)
parser.add_argument('--m', type=float, default=1.0)
parser.add_argument('--b', type=float, default=0.0)
parser.add_argument('--s', type=str, default='u')
parser.add_argument('--F', type=str, default='0')
parser.add_argument('--dt', type=float, default=0.05)
parser.add_argument('--T', type=float, default=10)
parser.add_argument('--window_width', type=float, default=30.,
help='Number of periods in a window')
parser.add_argument('--damping', type=str, default='linear')
parser.add_argument('--savefig', action='store_true')
# Hack to allow --SCITOOLS options
# (scitools.std reads this argument at import)
parser.add_argument('--SCITOOLS_easyviz_backend',
default='matplotlib')
a = parser.parse_args()
from scitools.std import StringFunction
s = StringFunction(a.s, independent_variable='u')
F = StringFunction(a.F, independent_variable='t')
I, V, m, b, dt, T, window_width, savefig, damping = \
a.I, a.V, a.m, a.b, a.dt, a.T, a.window_width, a.savefig, \
a.damping
filename = 'vibration_sim.dat'
u_min, u_max = solve_and_store(filename, I, V, m, b, s,
F, dt, T, damping)
read_and_plot(filename, u_min, u_max)
def read_and_plot(filename, u_min, u_max):
"""
Read file and plot u vs t line by line in a
terminal window (only using ascii characters).
"""
from scitools.avplotter import Plotter
import time
umin = 1.2*u_min; umax = 1.2*u_max
p = Plotter(ymin=umin, ymax=umax, width=60, symbols='+o')
fps = 10
infile = open(filename, 'r')
# read and treat one line at a time
infile.readline() # skip header line
for line in infile:
time_and_pos = line.split() # gives list with 2 elements
t = float(time_and_pos[0])
u = float(time_and_pos[1])
#print 'time: %g position: %g' % (time, pos)
print p.plot(t, u), '%.2f' % (t)
time.sleep(1/float(fps))
if __name__ == '__main__':
main()
```

Filename: `vib_memsave`

.

## Exercise 16: Implement the Euler-Cromer scheme for the generalized model¶

We consider the generalized model problem

**a)**
Implement the Euler-Cromer method from the section vib:ode2:EulerCromer.

**Solution.**
A suitable function is

```
import numpy as np
from math import pi
def solver(I, V, m, b, s, F, dt, T, damping='linear'):
"""
Solve m*u'' + f(u') + s(u) = F(t) for t in (0,T], u(0)=I,
u'(0)=V by an Euler-Cromer method.
"""
f = lambda v: b*v if damping == 'linear' else b*abs(v)*v
dt = float(dt)
Nt = int(round(T/dt))
u = np.zeros(Nt+1)
v = np.zeros(Nt+1)
t = np.linspace(0, Nt*dt, Nt+1)
v[0] = V
u[0] = I
for n in range(0, Nt):
v[n+1] = v[n] + dt*(1./m)*(F(t[n]) - s(u[n]) - f(v[n]))
u[n+1] = u[n] + dt*v[n+1]
#print 'F=%g, s=%g, f=%g, v_prev=%g' % (F(t[n]), s(u[n]), f(v[n]), v[n])
#print 'v[%d]=%g u[%d]=%g' % (n+1,v[n+1],n+1,u[n+1])
return u, v, t
```

**b)**
We expect the Euler-Cromer method to have first-order convergence rate.
Make a unit test based on this expectation.

**Solution.**
We may use SymPy to derive a problem based on a manufactured solution
\(u=3\cos t\). Then we may run some \(\Delta t\) values, compute the error
divided by \(\Delta t\), and check that this ratio remains approximately
constant. (An alternative is to compute true convergence rates and check
that they are close to unity.)

```
def test_solver():
"""Check 1st order convergence rate."""
m = 4; b = 0.1
s = lambda u: 2*u
f = lambda v: b*v
import sympy as sym
def ode(u):
"""Return source F(t) in ODE for given manufactured u."""
print 'ode:', m*sym.diff(u, t, 2), f(sym.diff(u,t)), s(u)
return m*sym.diff(u, t, 2) + f(sym.diff(u,t)) + s(u)
t = sym.symbols('t')
u = 3*sym.cos(t)
F = ode(u)
F = sym.simplify(F)
print 'F:', F, 'u:', u
F = sym.lambdify([t], F, modules='numpy')
u_exact = sym.lambdify([t], u, modules='numpy')
I = u_exact(0)
V = sym.diff(u, t).subs(t, 0)
print 'V:', V, 'I:', I
# Numerical parameters
w = np.sqrt(0.5)
P = 2*pi/w
dt_values = [P/20, P/40, P/80, P/160, P/320]
T = 8*P
error_vs_dt = []
for n, dt in enumerate(dt_values):
u, v, t = solver(I, V, m, b, s, F, dt, T, damping='linear')
error = np.abs(u - u_exact(t)).max()
if n > 0:
error_vs_dt.append(error/dt)
for i in range(len(error_vs_dt)):
assert abs(error_vs_dt[i]-
error_vs_dt[0]) < 0.1
```

**c)**
Consider a system with \(m=4\), \(f(v)=b|v|v\), \(b=0.2\), \(s=2u\), \(F=0\).
Compute the solution using the centered difference scheme
from the section vib:ode2:fdm:flin and the Euler-Cromer scheme
for the longest possible time step \(\Delta t\). We can use the
result from the case without damping, i.e., the largest \(\Delta t= 2/\omega\),
\(\omega\approx
\sqrt{0.5}\) in this case, but since \(b\) will modify the frequency, we
take the longest possible time step as a safety factor 0.9 times \(2/\omega\).
Refine \(\Delta t\) three times by a factor of two and compare the
two curves.

**Solution.**
We rely on the module `vib`

for the implementation of the method
from the section vib:ode2:fdm:flin. A suitable function for making
the comparisons is then

```
# NBVAL_SKIP
def demo():
"""
Demonstrate difference between Euler-Cromer and the
scheme for the corresponding 2nd-order ODE.
"""
I = 1.2; V = 0.2; m = 4; b = 0.2
s = lambda u: 2*u
F = lambda t: 0
w = np.sqrt(2./4) # approx freq
dt = 0.9*2/w # longest possible time step
w = 0.5
P = 2*pi/w
T = 4*P
from vib import solver as solver2
import scitools.std as plt
for k in range(4):
u2, t2 = solver2(I, V, m, b, s, F, dt, T, 'quadratic')
u, v, t = solver(I, V, m, b, s, F, dt, T, 'quadratic')
plt.figure()
plt.plot(t, u, 'r-', t2, u2, 'b-')
plt.legend(['Euler-Cromer', 'centered scheme'])
plt.title('dt=%.3g' % dt)
raw_input()
plt.savefig('tmp_%d' % k + '.png')
plt.savefig('tmp_%d' % k + '.pdf')
dt /= 2
```

Filename: `vib_EulerCromer`

.

## Problem 17: Interpret \([D_tD_t u]^n\) as a forward-backward difference¶

Show that the difference \([D_t D_tu]^n\) is equal to \([D_t^+D_t^-u]^n\) and \(D_t^-D_t^+u]^n\). That is, instead of applying a centered difference twice one can alternatively apply a mixture of forward and backward differences.

**Solution.**

Similarly, we get that

Filename: `vib_DtDt_fw_bw`

.

## Exercise 18: Analysis of the Euler-Cromer scheme¶

The Euler-Cromer scheme for the model problem \(u^{\prime\prime} + \omega^2 u =0\), \(u(0)=I\), \(u^{\prime}(0)=0\), is given in (55)-(54). Find the exact discrete solutions of this scheme and show that the solution for \(u^n\) coincides with that found in the section Analysis of the numerical scheme.

**Hint.**
Use an “ansatz” \(u^n=I\exp{(i\tilde\omega\Delta t\,n)}\) and
\(v^n=qu^n\), where \(\tilde\omega\) and \(q\) are unknown parameters. The
following formula is handy:

**Solution.**
We follow the ideas in the section Analysis of the numerical scheme. Inserting
\(u^n=I\exp{(i\tilde\omega\Delta t\,n)}\) and
\(v^n=qu^n\) in
(55)-(54)
and dividing by \(I\exp{(i\tilde\omega\Delta t\,n)}\) gives

Solving (74) with respect to \(q\) gives

Inserting this expression for \(q\) in (73) results in

Using the relation \(\exp{(i\tilde\omega(\Delta t))} + \exp{(i\tilde\omega(-\Delta t))} - 2 = -4\sin^2(\frac{\tilde\omega\Delta t}{2})\) gives

or after dividing by 4,

which is the same equation for \(\tilde\omega\) as found in the section Analysis of the numerical scheme, such that \(\tilde\omega\) is the same. The accuracy, stability, and formula for the exact discrete solution are then all the same as derived in the section Analysis of the numerical scheme.