Generalization: damping, nonlinear spring, and external excitation

We shall now generalize the simple model problem from the section Finite difference discretization to include a possibly nonlinear damping term \(f(u')\), a possibly nonlinear spring (or restoring) force \(s(u)\), and some external excitation \(F(t)\):

(1)\[ mu'' + f(u') + s(u) = F(t),\quad u(0)=I,\ u'(0)=V,\ t\in (0,T] {\thinspace .}\]

We have also included a possibly nonzero initial value of \(u'(0)\). The parameters \(m\), \(f(u')\), \(s(u)\), \(F(t)\), \(I\), \(V\), and \(T\) are input data.

There are two main types of damping (friction) forces: linear \(f(u')=bu\), or quadratic \(f(u')=bu'|u'|\). Spring systems often feature linear damping, while air resistance usually gives rise to quadratic damping. Spring forces are often linear: \(s(u)=cu\), but nonlinear versions are also common, the most famous is the gravity force on a pendulum that acts as a spring with \(s(u)\sim \sin(u)\).

A centered scheme for linear damping

Sampling (1) at a mesh point \(t_n\), replacing \(u''(t_n)\) by \([D_tD_tu]^n\), and \(u'(t_n)\) by \([D_{2t}u]^n\) results in the discretization

\[[mD_tD_t u + f(D_{2t}u) + s(u) = F]^n,\]

which written out means

(2)\[ m\frac{u^{n+1}-2u^n + u^{n-1}}{\Delta t^2} + f(\frac{u^{n+1}-u^{n-1}}{2\Delta t}) + s(u^n) = F^n,\]

where \(F^n\) as usual means \(F(t)\) evaluated at \(t=t_n\). Solving (2) with respect to the unknown \(u^{n+1}\) gives a problem: the \(u^{n+1}\) inside the \(f\) function makes the equation nonlinear unless \(f(u')\) is a linear function, \(f(u')=bu'\). For now we shall assume that \(f\) is linear in \(u'\). Then

(3)\[ m\frac{u^{n+1}-2u^n + u^{n-1}}{\Delta t^2} + b\frac{u^{n+1}-u^{n-1}}{2\Delta t} + s(u^n) = F^n,\]

which gives an explicit formula for \(u\) at each new time level:

(4)\[ u^{n+1} = (2mu^n + (\frac{b}{2}\Delta t - m)u^{n-1} + \Delta t^2(F^n - s(u^n)))(m + \frac{b}{2}\Delta t)^{-1}\]\[ {\thinspace .}\]

For the first time step we need to discretize \(u'(0)=V\) as \([D_{2t}u = V]^0\) and combine with (4) for \(n=0\). The discretized initial condition leads to

(5)\[ u^{-1} = u^{1} - 2\Delta t V,\]

which inserted in (4) for \(n=0\) gives an equation that can be solved for \(u^1\):

(6)\[ u^1 = u^0 + \Delta t\, V + \frac{\Delta t^2}{2m}(-bV - s(u^0) + F^0) {\thinspace .}\]

A centered scheme for quadratic damping

When \(f(u')=bu'|u'|\), we get a quadratic equation for \(u^{n+1}\) in (2). This equation can straightforwardly be solved, but we can also avoid the nonlinearity by performing an approximation that is within other numerical errors that we have already committed by replacing derivatives with finite differences.

The idea is to reconsider (1) and only replace \(u''\) by \(D_tD_tu\), giving

(7)\[ [mD_tD_t u + bu'|u'| + s(u) = F]^n,\]

Here, \(u'|u'|\) is to be computed at time \(t_n\). We can introduce a geometric mean, defined by

\[(w^2)^n \approx w^{n-\frac{1}{2}}w^{n+\frac{1}{2}},\]

for some quantity \(w\) depending on time. The error in the geometric mean approximation is \({\mathcal{O}(\Delta t^2)}\), the same as in the approximation \(u''\approx D_tD_tu\). With \(w=u'\) it follows that

\[[u'|u'|]^n \approx u'(t_n+\frac{1}{2})|u'(t_n-\frac{1}{2})|{\thinspace .}\]

The next step is to approximate \(u'\) at \(t_{n\pm 1/2}\), but here a centered difference can be used:

(8)\[ u'(t_{n+1/2})\approx [D_t u]^{n+\frac{1}{2}},\quad u'(t_{n-1/2})\approx [D_t u]^{n-\frac{1}{2}} {\thinspace .}\]

We then get

\[[u'|u'|]^n \approx [D_tu]^{n+\frac{1}{2}}|[D_tu]^{n-\frac{1}{2}}| = \frac{u^{n+1}-u^n}{\Delta t} \frac{|u^n-u^{n-1}|}{\Delta t} {\thinspace .}\]

The counterpart to (2) is then

(9)\[ m\frac{u^{n+1}-2u^n + u^{n-1}}{\Delta t^2} + b\frac{u^{n+1}-u^n}{\Delta t}\frac{|u^n-u^{n-1}|}{\Delta t} + s(u^n) = F^n,\]

which is linear in \(u^{n+1}\). Therefore, we can easily solve with respect to \(u^{n+1}\) and achieve the explicit updating formula

\[u^{n+1} = \left( m + b|u^n-u^{n-1}|\right)^{-1}\times \nonumber\]

(10)\[ \qquad \left(2m u^n - mu^{n-1} + bu^n|u^n-u^{n-1}| + \Delta t^2 (F^n - s(u^n)) \right) {\thinspace .}\]

In the derivation of a special equation for the first time step we run into some trouble: inserting (5) in (10) for \(n=0\) results in a complicated nonlinear equation for \(u^1\). By thinking differently about the problem we can easily get away with the nonlinearity again. We have for \(n=0\) that \(b[u'|u'|]^0 = bV|V|\). Using this value in (7) gives

\[[mD_tD_t u + bV|V| + s(u) = F]^0 {\thinspace .}\]

Writing this equation out and using (5) results in the special equation for the first time step:

(11)\[ u^1 = u^0 + \Delta t V + \frac{\Delta t^2}{2m}\left(-bV|V| - s(u^0) + F^0\right) {\thinspace .}\]

A forward-backward discretization of the quadratic damping term

The previous section first proposed to discretize the quadratic damping term \(|u'|u'\) using centered differences: \([|D_{2t}|D_{2t}u]^n\). As this gives rise to a nonlinearity in \(u^{n+1}\), it was instead proposed to use a geometric mean combined with centered differences. But there are other alternatives. To get rid of the nonlinearity in \([|D_{2t}|D_{2t}u]^n\), one can think differently: apply a backward difference to \(|u'|\), such that the term involves known values, and apply a forward difference to \(u'\) to make the term linear in the unknown \(u^{n+1}\). With mathematics,

\[[\beta |u'|u']^n \approx \beta |[D_t^-u]^n|[D_t^+ u]^n = \beta\left\vert\frac{u^-u^{n-1}}{\Delta t}\right\vert \frac{u^{n+1}-u^n}{\Delta t}{\thinspace .}\]

The forward and backward differences have both an error proportional to \(\Delta t\) so one may think the discretization above leads to a first-order scheme. However, by looking at the formulas, we realize that the forward-backward differences result in exactly the same scheme as when we used a geometric mean and centered differences. Therefore, the forward-backward differences act in a symmetric way and actually produce a second-order accurate discretization of the quadratic damping term.

Implementation (2)

The algorithm arising from the methods in the sections A centered scheme for linear damping and A centered scheme for quadratic damping is very similar to the undamped case in the section A centered finite difference scheme. The difference is basically a question of different formulas for \(u^1\) and \(u^{n+1}\). This is actually quite remarkable. The equation (1) is normally impossible to solve by pen and paper, but possible for some special choices of \(F\), \(s\), and \(f\). On the contrary, the complexity of the nonlinear generalized model (1) versus the simple undamped model is not a big deal when we solve the problem numerically!

The computational algorithm takes the form

1. \(u^0=I\)
2. compute \(u^1\) from (6) if linear damping or (11) if quadratic damping
3. for \(n=1,2,\ldots,N_t-1\):

1. compute \(u^{n+1}\) from (4) if linear damping or (10) if quadratic damping

Modifying the solver function for the undamped case is fairly easy, the big difference being many more terms and if tests on the type of damping:

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 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.
    """
    dt = float(dt); b = float(b); m = float(m) # avoid integer div.
    Nt = int(round(T/dt))
    u = zeros(Nt+1)
    t = linspace(0, Nt*dt, Nt+1)

    u[0] = I
    if damping == 'linear':
        u[1] = u[0] + dt*V + dt**2/(2*m)*(-b*V - s(u[0]) + F(t[0]))
    elif damping == 'quadratic':
        u[1] = u[0] + dt*V + \
               dt**2/(2*m)*(-b*V*abs(V) - s(u[0]) + F(t[0]))

    for n in range(1, Nt):
        if damping == 'linear':
            u[n+1] = (2*m*u[n] + (b*dt/2 - m)*u[n-1] +
                      dt**2*(F(t[n]) - s(u[n])))/(m + b*dt/2)
        elif damping == 'quadratic':
            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]) - s(u[n])))/\
                      (m + b*abs(u[n] - u[n-1]))
    return u, t

The complete code resides in the file vib.py.

Verification (2)

Constant solution

For debugging and initial verification, a constant solution is often very useful. We choose \({u_{\small\mbox{e}}}(t)=I\), which implies \(V=0\). Inserted in the ODE, we get \(F(t)=s(I)\) for any choice of \(f\). Since the discrete derivative of a constant vanishes (in particular, \([D_{2t}I]^n=0\), \([D_tI]^n=0\), and \([D_tD_t I]^n=0\)), the constant solution also fulfills the discrete equations. The constant should therefore be reproduced to machine precision.

Linear solution

Now we choose a linear solution: \({u_{\small\mbox{e}}} = ct + d\). The initial condition \(u(0)=I\) implies \(d=I\), and \(u'(0)=V\) forces \(c\) to be \(V\). Inserting \({u_{\small\mbox{e}}}=Vt+I\) in the ODE with linear damping results in

\[0 + bV + s(Vt+I) = F(t),\]

while quadratic damping requires the source term

\[0 + b|V|V + s(Vt+I) = F(t){\thinspace .}\]

Since the finite difference approximations used to compute \(u'\) all are exact for a linear function, it turns out that the linear \({u_{\small\mbox{e}}}\) is also a solution of the discrete equations. Exercise 9: Use linear/quadratic functions for verification asks you to carry out all the details.

Quadratic solution

Choosing \({u_{\small\mbox{e}}} = bt^2 + Vt + I\), with \(b\) arbitrary, fulfills the initial conditions and fits the ODE if \(F\) is adjusted properly. The solution also solves the discrete equations with linear damping. However, this quadratic polynomial in \(t\) does not fulfill the discrete equations in case of quadratic damping, because the geometric mean used in the approximation of this term introduces an error. Doing Exercise 9: Use linear/quadratic functions for verification will reveal the details. One can fit \(F^n\) in the discrete equations such that the quadratic polynomial is reproduced by the numerical method (to machine precision).

Visualization

The functions for visualizations differ significantly from those in the undamped case in the vib_undamped.py program because we in the present general case do not have an exact solution to include in the plots. Moreover, we have no good estimate of the periods of the oscillations as there will be one period determined by the system parameters, essentially the approximate frequency \(\sqrt{s'(0)/m}\) for linear \(s\) and small damping, and one period dictated by \(F(t)\) in case the excitation is periodic. This is, however, nothing that the program can depend on or make use of. Therefore, the user has to specify \(T\) and the window width in case of a plot that moves with the graph and shows the most recent parts of it in long time simulations.

The vib.py code contains several functions for analyzing the time series signal and for visualizing the solutions.

User interface

The main function has substantial changes from the vib_undamped.py code since we need to specify the new data \(c\), \(s(u)\), and \(F(t)\). In addition, we must set \(T\) and the plot window width (instead of the number of periods we want to simulate as in vib_undamped.py). To figure out whether we can use one plot for the whole time series or if we should follow the most recent part of \(u\), we can use the plot_empricial_freq_and_amplitude function’s estimate of the number of local maxima. This number is now returned from the function and used in main to decide on the visualization technique.

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('--c', 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=140)
    parser.add_argument('--damping', type=str, default='linear')
    parser.add_argument('--window_width', type=float, default=30)
    parser.add_argument('--savefig', action='store_true')
    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, c, dt, T, window_width, savefig, damping = \
       a.I, a.V, a.m, a.c, a.dt, a.T, a.window_width, a.savefig, \
       a.damping

    u, t = solver(I, V, m, c, s, F, dt, T)
    num_periods = empirical_freq_and_amplitude(u, t)
    if num_periods <= 15:
        figure()
        visualize(u, t)
    else:
        visualize_front(u, t, window_width, savefig)
    show()

The program vib.py contains the above code snippets and can solve the model problem (1). As a demo of vib.py, we consider the case \(I=1\), \(V=0\), \(m=1\), \(c=0.03\), \(s(u)=\sin(u)\), \(F(t)=3\cos(4t)\), \(\Delta t = 0.05\), and \(T=140\). The relevant command to run is

Terminal> python vib.py --s 'sin(u)' --F '3*cos(4*t)' --c 0.03

This results in a moving window following the function on the screen. Figure Damped oscillator excited by a sinusoidal function shows a part of the time series.

Damped oscillator excited by a sinusoidal function

A staggered Euler-Cromer scheme for the generalized model

The model

\[mu'' + f(u') + s(u) = F(t),\quad u(0)=I,\ u'(0)=V,\ t\in (0,T],\]

can be rewritten as a first-order ODE system

(12)\[ u' = v,\]

(13)\[ v' = m^{-1}\left(F(t) - f(v) - s(u)\right){\thinspace .}\]

It is natural to introduce a staggered mesh (see the section The Euler-Cromer scheme on a staggered mesh) and seek \(u\) at mesh points \(t_n\) (the numerical value is denoted by \(u^n\)) and \(v\) between mesh points at \(t_{n+1/2}\) (the numerical value is denoted by \(v^{n+\frac{1}{2}}\)). A centered difference approximation to (12)-(13) can then be written in operator notation as

(14)\[ \lbrack D_t u = v\rbrack^{n-\frac{1}{2}},\]

(15)\[ \lbrack D_tv = m^{-1}\left(F(t) - f(v) - s(u)\right)\rbrack^n{\thinspace .}\]

Written out,

(16)\[ \frac{u^n - u^{n-1}}{\Delta t} = v^{n-\frac{1}{2}},\]

(17)\[ \frac{v^{n+\frac{1}{2}} - v^{n-\frac{1}{2}}}{\Delta t} = m^{-1}\left(F^n - f(v^n) - s(u^n)\right){\thinspace .}\]

With linear damping, \(f(v)=bv\), we can use an arithmetic mean for \(f(v^n)\): \(f(v^n)\approx = \frac{1}{2}(f(v^{n-\frac{1}{2}}) + f(v^{n+\frac{1}{2}}))\). The system (16)-(17) can then be solved with respect to the unknowns \(u^n\) and \(v^{n+\frac{1}{2}}\):

(18)\[ u^n = u^{n-1} + {\Delta t}v^{n-\frac{1}{2}},\]

(19)\[ v^{n+\frac{1}{2}} = \left(1 + \frac{b}{2m}\Delta t\right)^{-1}\left( v^{n-\frac{1}{2}} + {\Delta t} m^{-1}\left(F^n - {\frac{1}{2}}f(v^{n-\frac{1}{2}}) - s(u^n)\right)\right){\thinspace .}\]

In case of quadratic damping, \(f(v)=b|v|v\), we can use a geometric mean: \(f(v^n)\approx b|v^{n-\frac{1}{2}}|v^{n+\frac{1}{2}}\). Inserting this approximation in (16)-(17) and solving for the unknowns \(u^n\) and \(v^{n+\frac{1}{2}}\) results in

(20)\[ u^n = u^{n-1} + {\Delta t}v^{n-\frac{1}{2}},\]

(21)\[ v^{n+\frac{1}{2}} = (1 + \frac{b}{m}|v^{n-\frac{1}{2}}|\Delta t)^{-1}\left( v^{n-\frac{1}{2}} + {\Delta t} m^{-1}\left(F^n - s(u^n)\right)\right){\thinspace .}\]

The initial conditions are derived at the end of the section The Euler-Cromer scheme on a staggered mesh:

(22)\[ u^0 = I,\]

(23)\[ v^\frac{1}{2} = V - \frac{1}{2}\Delta t\omega^2I {\thinspace .}\]

Exercises and Problems

Problem 1: Use linear/quadratic functions for verification

Consider the ODE problem

\[u'' + \omega^2u=f(t), \quad u(0)=I,\ u'(0)=V,\ t\in(0,T]{\thinspace .}\]

Discretize this equation according to \([D_tD_t u + \omega^2 u = f]^n\).

a) Derive the equation for the first time step (\(u^1\)).

b) For verification purposes, we use the method of manufactured solutions (MMS) with the choice of \({u_{\small\mbox{e}}}(x,t)= ct+d\). Find restrictions on \(c\) and \(d\) from the initial conditions. Compute the corresponding source term \(f\) by term. 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 \({u_{\small\mbox{e}}}\) is also a perfect solution of the discrete equations.

c) Use sympy to do the symbolic calculations above. Here is a sketch of the program vib_undamped_verify_mms.py:

import sympy as sp
V, t, I, w, dt = sp.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 sp.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 sp.simplify(R)

def residual_discrete_eq_step1(u):
    """Return the residual of the discrete eq. at the first
    step with u inserted."""
    R = ...
    return sp.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), sp.diff(u(t), t).subs(t, 0))

    # Method of manufactured solution requires fitting f
    global f  # source term in the ODE
    f = sp.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.

d) The purpose now is to choose a quadratic function \({u_{\small\mbox{e}}} = bt^2 + ct + d\) 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.)

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

f) Implement a solver function for computing the numerical solution of this problem.

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

Filenames: vib_undamped_verify_mms.pdf, vib_undamped_verify_mms.py.

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 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 \(m\). Filename: vib_phase_error_growth.pdf.

Exercise 3: Improve the accuracy by adjusting the frequency

According to (1.11), 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 \(\omega\)).

Filename: vib_adjust_w.py.

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 \(u''+u=0\) in the Odespy software. Use the example on adaptive schemes in [Ref1]. 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. Filename: vib_undamped_adaptive.py.

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

As an alternative to the derivation of (1.8) for computing \(u^1\), one can use a Taylor polynomial with three terms for \(u^1\):

\[u(t_1) \approx u(0) + u'(0)\Delta t + {\frac{1}{2}}u''(0)\Delta t^2\]

With \(u''=-\omega^2 u\) and \(u'(0)=0\), show that this method also leads to (1.8). Generalize the condition on \(u'(0)\) to be \(u'(0)=V\) and compute \(u^1\) in this case with both methods. Filename: vib_first_step.pdf.

Exercise 6: 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. Filename: vib_largest_wdt.pdf.

Exercise 7: Visualize the accuracy of finite differences for a cosine function

We introduce the error fraction

\[E = \frac{[D_tD_t u]^n}{u''(t_n)}\]

to measure the error in the finite difference approximation \(D_tD_tu\) to \(u''\). Compute \(E\) for the specific choice of a cosine/sine function of the form \(u=\exp{(i\omega t)}\) and show that

\[E = \left(\frac{2}{\omega\Delta t}\right)^2 \sin^2(\frac{\omega\Delta t}{2}) {\thinspace .}\]

Plot \(E\) as a function of \(p=\omega\Delta t\). The relevant values of \(p\) are \([0,\pi]\) (see Exercise 6: 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). Filename: vib_plot_fd_exp_error.py.

Exercise 8: Verify convergence rates of the error in energy

We consider the ODE problem \(u'' + \omega^2u=0\), \(u(0)=I\), \(u'(0)=V\), for \(t\in (0,T]\). The total energy of the solution \(E(t)=\frac{1}{2}(u')^2 + \frac{1}{2}\omega^2 u^2\) should stay constant. The error in energy can be computed as explained in the section Enegy considerations.

Make a nose test in a file test_error_conv.py, 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. Filename: test_error_conv.py.

Exercise 9: 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 (1) 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. Filename: vib_verify_mms.py.

Exercise 10: Use an exact discrete solution for verification

Write a nose test function in a separate file that employs the exact discrete solution (1.12) to verify the implementation of the solver function in the file vib_undamped.py. Just import solver and make functions for the exact discrete solution and the nose test. Filename: vib_verify_discrete_omega.py.

Exercise 11: Use analytical solution for convergence rate tests

The purpose of this exercise is to perform convergence tests of the problem (1) when \(s(u)=\omega^2u\) and \(F(t)=A\sin\phi t\). Find the complete analytical solution to the problem in this case (most textbooks on mechanics list the various elements you need to write down the exact solution). 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 \(\Delta t^2\). Filename: vib_conv_rate.py.

Exercise 12: Investigate the amplitude errors of many solvers

Use the program vib_undamped_odespy.py from the section Standard methods for 1st-order ODE systems and the amplitude estimation from the amplitudes function in the vib_undamped.py file (see the section Empirical analysis of the solution) 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 section vib:model2x2:ueq. Filename: vib_amplitude_errors.py.

Exercise 13: Minimize memory usage of a vibration solver

The program vib.py store 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. 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). Filename: vib_memsave.py.

Exercise 14: Implement the solver via classes

Reimplement the vib.py program using a class Problem to hold all the physical parameters of the problem, a class Solver to hold the numerical parameters and compute the solution, and a class Visualizer to display the solution.

Hint. Use the ideas and examples for an ODE model. More specifically, make a superclass Problem for holding the scalar physical parameters of a problem and let subclasses implement the \(s(u)\) and \(F(t)\) functions as methods. Try to call up as much existing functionality in vib.py as possible.

Filename: vib_class.py.

Exercise 15: Show equivalence between schemes

Show that the schemes from the sections A centered finite difference scheme, The Euler-Cromer method, and The Euler-Cromer scheme on a staggered mesh are all equivalent. Filename: vib_scheme_equivalence.pdf.

Exercise 16: 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 forward and backward differences. Filename: vib_DtDt_fw_bw.pdf.

Exercise 17: Use the forward-backward scheme with quadratic damping

We consider the generalized model with quadratic damping, expressed as a system of two first-order equations as in the section A staggered Euler-Cromer scheme for the generalized model:

\[\begin{split}u' &= v,\\ v' &= \frac{1}{m}\left( F(t) - \beta |v|v - s(u)\right){\thinspace .}\end{split}\]

However, contrary to what is done in the section A staggered Euler-Cromer scheme for the generalized model, we want to apply the idea of the forward-backward discretization in the section The Euler-Cromer method. Express the idea in operator notation and write out the scheme. Unfortunately, the backward difference for the \(v\) equation creates a nonlinearity \(|v^{n+1}|v^{n}\). To linearize this nonlinearity, use the known value \(v^n\) inside the absolute value factor, i.e., \(|v^{n+1}|v^{n}\approx |v^n|v^{n+1}\). Show that the resulting scheme is equivalent to the one in the section A staggered Euler-Cromer scheme for the generalized model for some time level \(n\geq 1\).

What we learn from this exercise is that the first-order differences and the linearization trick play together in “the right way” such that the scheme is as good as when we (in the section A staggered Euler-Cromer scheme for the generalized model) carefully apply centered differences and a geometric mean on a staggered mesh to achieve second-order accuracy. There is a difference in the handling of the initial conditions, though, as explained at the end of the section The Euler-Cromer method. Filename: vib_gen_bwdamping.pdf.

Exercise 18: Use a backward difference for the damping term

As an alternative to discretizing the damping terms \(\beta u'\) and \(\beta |u'|u'\) by centered differences, we may apply backward differences:

\[\begin{split}[u']^n &\approx [D_t^-u]^n,\\ & [|u'|u']^n &\approx [|D_t^-u|D_t^-u]^n = |[D_t^-u]^n|[D_t^-u]^n{\thinspace .}\end{split}\]

The advantage of the backward difference is that the damping term is evaluated using known values \(u^n\) and \(u^{n-1}\) only. Extend the vib.py code with a scheme based on using backward differences in the damping terms. Add statements to compare the original approach with centered difference and the new idea launched in this exercise. Perform numerical experiments to investigate how much accuracy that is lost by using the backward differences.

Filename: vib_gen_bwdamping.pdf.

References

[Ref1]

H. P. Langtangen. Introduction to Computing With Finite Difference Methods, Simula Research Laboratory and University of Oslo, 2013, http://hplgit.github.com/INF5620/doc/notes/decay-sphinx/main_decay.html.