$$ \newcommand{\half}{\frac{1}{2}} \newcommand{\tp}{\thinspace .} \newcommand{\uex}{{u_{\small\mbox{e}}}} \newcommand{\vex}{{v_{\small\mbox{e}}}} \newcommand{\Oof}[1]{\mathcal{O}(#1)} $$

Study guide: Truncation error analysis

Hans Petter Langtangen [1, 2]

[1] Center for Biomedical Computing, Simula Research Laboratory
[2] Department of Informatics, University of Oslo

Oct 17, 2015

Table of contents

Overview of what truncation errors are
      Abstract problem setting
      Various error measures
Truncation errors in finite difference formulas
      Example: The backward difference for \( u'(t) \)
      Taylor series
      Taylor series inserted in the backward difference approximation
      The forward difference for \( u'(t) \)
      The central difference for \( u'(t) \) (1)
      The central difference for \( u'(t) \) (1)
      Leading-order error terms in finite differences (1)
      Leading-order error terms in finite differences (2)
      Leading-order error terms in mean values (1)
      Leading-order error terms in mean values (2)
      Software for computing truncation errors
      Symbolic computing with difference operators
Truncation errors in exponential decay ODE
      Truncation error of the Forward Euler scheme
      Truncation error of the Crank-Nicolson scheme
      Test the understanding!
      Truncation error of the \( \theta \)-rule
      Using symbolic software
      Empirical verification of the truncation error (1)
      Empirical verification of the truncation error (2)
      Empirical verification of the truncation error in the Forward Euler scheme
      Empirical verification of the truncation error in the Forward Euler scheme
      Increasing the accuracy by adding correction terms
      Lowering the order of the derivative in the correction term
      With a correction term Forward Euler becomes Crank-Nicolson
      Correction terms in the Crank-Nicolson scheme (1)
      Correction terms in the Crank-Nicolson scheme (2)
      Extension to variable coefficients
      Exact solutions of the finite difference equations
      Computing truncation errors in nonlinear problems (1)
      Computing truncation errors in nonlinear problems (2)
Truncation errors in vibration ODEs
      Linear model without damping
      Truncation errors in the initial condition
      Computing correction terms
      Model with damping and nonlinearity
      Carrying out the truncation error analysis
      Extension to quadratic damping
      The truncation error for quadratic damping (1)
      The truncation error for quadratic damping (2)
      The general model formulated as first-order ODEs
      The forward-backward scheme
      Truncation error analysis
      A centered scheme on a staggered mesh
      Truncation error analysis (1)
      Truncation error analysis (2)

Overview of what truncation errors are

Abstract problem setting

Consider an abstract differential equation

 
$$ \mathcal{L}(u)=0$$

 
Example: \( \mathcal{L}(u)=u'(t)+a(t)u(t)-b(t) \).

The corresponding discrete equation:

 
$$ \mathcal{L}_{\Delta}(u) =0$$

 
Let now

 
$$ \begin{align*} \mathcal{L}(\uex)&=0\\ \mathcal{L}_\Delta(u)&=0 \end{align*} $$

 
\( u \) is computed at mesh points

Various error measures

Truncation errors in finite difference formulas

Example: The backward difference for \( u'(t) \)

Backward difference approximation to \( u' \):

 
$$ \lbrack D_t^- u\rbrack^n = \frac{u^{n} - u^{n-1}}{\Delta t} \approx u'(t_n) $$

 

Define the truncation error of this approximation as

 
$$ R^n = [D^-_tu]^n - u'(t_n) $$

 

The common way of calculating \( R^n \) is to

  1. expand \( u(t) \) in a Taylor series around the point where the derivative is evaluated, here \( t_n \),
  2. insert this Taylor series in \eqref{trunc:fd:bw3}, and
  3. collect terms that cancel and simplify the expression.

Taylor series

General Taylor series expansion from calculus:

 
$$ f(x+h) = \sum_{i=0}^\infty \frac{1}{i!}\frac{d^if}{dx^i}(x)h^i$$

 

Here: expand \( u^{n-1} \) around \( t_n \):

 
$$ \begin{align*} u(t_{n-1}) = u(t-\Delta t) &= \sum_{i=0}^\infty \frac{1}{i!}\frac{d^iu}{dt^i}(t_n)(-\Delta t)^i\\ & = u(t_n) - u'(t_n)\Delta t + {\half}u''(t_n)\Delta t^2 + \Oof{\Delta t^3} \end{align*} $$

 

Taylor series inserted in the backward difference approximation

 
$$ \begin{align*} [D_t^-u]^n - u'(t_n) &= \frac{u(t_n) - u(t_{n-1})}{\Delta t} - u'(t_n)\\ &= \frac{u(t_n) - (u(t_n) - u'(t_n)\Delta t + {\half}u''(t_n)\Delta t^2 + \Oof{\Delta t^3} )}{\Delta t}\\ &\quad -u'(t_n)\\ &= -{\half}u''(t_n)\Delta t + \Oof{\Delta t^2} ) \end{align*} $$

 

Result:

 
$$ R^n = -{\half}u''(t_n)\Delta t + \Oof{\Delta t^2} ) $$

 
The difference approximation is of first order in \( \Delta t \). It is exact for linear \( \uex \).

The forward difference for \( u'(t) \)

Now consider a forward difference:

 
$$ u'(t_n) \approx [D_t^+ u]^n = \frac{u^{n+1}-u^n}{\Delta t}$$

 
Define the truncation error:

 
$$ R^n = [D_t^+ u]^n - u'(t_n)$$

 
Expand \( u^{n+1} \) in a Taylor series around \( t_n \),

 
$$ u(t_{n+1}) = u(t_n) + u'(t_n)\Delta t + {\half}u''(t_n)\Delta t^2 + \Oof{\Delta t^3} $$

 
We get

 
$$ R = {\half}u''(t_n)\Delta t + \Oof{\Delta t^2}$$

 

The central difference for \( u'(t) \) (1)

For the central difference approximation,

 
$$ u'(t_n)\approx [ D_tu]^n, \quad [D_tu]^n = \frac{u^{n+\half} - u^{n-\half}}{\Delta t} $$

 
the truncation error is

 
$$ R^n = [ D_tu]^n - u'(t_n)$$

 
Expand \( u(t_{n+\half}) \) and \( u(t_{n-1/2}) \) in Taylor series around the point \( t_n \) where the derivative is evaluated:

 
$$ \begin{align*} u(t_{n+\half}) = &u(t_n) + u'(t_n)\half\Delta t + {\half}u''(t_n)(\half\Delta t)^2 + \\ & \frac{1}{6}u'''(t_n) (\half\Delta t)^3 + \frac{1}{24}u''''(t_n) (\half\Delta t)^4 + \Oof{\Delta t^5}\\ u(t_{n-1/2}) = &u(t_n) - u'(t_n)\half\Delta t + {\half}u''(t_n)(\half\Delta t)^2 - \\ & \frac{1}{6}u'''(t_n) (\half\Delta t)^3 + \frac{1}{24}u''''(t_n) (\half\Delta t)^4 + \Oof{\Delta t^5} \end{align*} $$

 

The central difference for \( u'(t) \) (1)

 
$$ u(t_{n+\half}) - u(t_{n-1/2}) = u'(t_n)\Delta t + \frac{1}{24}u'''(t_n) \Delta t^3 + \Oof{\Delta t^5} \tp $$

 
By collecting terms in \( [D_t u]^n - u(t_n) \) we find \( R^n \) to be

 
$$ R^n = \frac{1}{24}u'''(t_n)\Delta t^2 + \Oof{\Delta t^4} $$

 
Note:

Leading-order error terms in finite differences (1)

 
$$ \begin{align*} \lbrack D_tu \rbrack^n &= \frac{u^{n+\half} - u^{n-\half}}{\Delta t} = u'(t_n) + R^n\\ R^n &= \frac{1}{24}u'''(t_n)\Delta t^2 + \Oof{\Delta t^4}\\ \lbrack D_{2t}u \rbrack^n &= \frac{u^{n+1} - u^{n-1}}{2\Delta t} = u'(t_n) + R^n\\ R^n &= \frac{1}{6}u'''(t_n)\Delta t^2 + \Oof{\Delta t^4}\\ \lbrack D_t^-u \rbrack^n &= \frac{u^{n} - u^{n-1}}{\Delta t} = u'(t_n) + R^n\\ R^n &= -{\half}u''(t_n)\Delta t + \Oof{\Delta t^2}\\ \lbrack D_t^+u \rbrack^n &= \frac{u^{n+1} - u^{n}}{\Delta t} = u'(t_n) + R^n\\ R^n &= {\half}u''(t_n)\Delta t + \Oof{\Delta t^2} \end{align*} $$

 

Leading-order error terms in finite differences (2)

 
$$ \begin{align*} [\bar D_tu]^{n+\theta} &= \frac{u^{n+1} - u^{n}}{\Delta t} = u'(t_{n+\theta}) + R^{n+\theta}\\ R^{n+\theta} &= \half(1-2\theta)u''(t_{n+\theta})\Delta t - \frac{1}{6}((1 - \theta)^3 - \theta^3)u'''(t_{n+\theta})\Delta t^2 + \Oof{\Delta t^3}\\ \lbrack D_t^{2-}u \rbrack^n &= \frac{3u^{n} - 4u^{n-1} + u^{n-2}}{2\Delta t} = u'(t_n) + R^n\\ R^n &= -\frac{1}{3}u'''(t_n)\Delta t^2 + \Oof{\Delta t^3}\\ \lbrack D_tD_t u \rbrack^n &= \frac{u^{n+1} - 2u^{n} + u^{n-1}}{\Delta t^2} = u''(t_n) + R^n\\ R^n &= \frac{1}{12}u''''(t_n)\Delta t^2 + \Oof{\Delta t^4}\\ \end{align*} $$

 

Leading-order error terms in mean values (1)

Weighted arithmetic mean:

 
$$ \begin{align*} [\overline{u}^{t,\theta}]^{n+\theta} & = \theta u^{n+1} + (1-\theta)u^n = u(t_{n+\theta}) + R^{n+\theta}\\ R^{n+\theta} &= {\half}u''(t_{n+\theta})\Delta t^2\theta (1-\theta) + \Oof{\Delta t^3} \end{align*} $$

 
Standard arithmetic mean:

 
$$ \begin{align*} [\overline{u}^{t}]^{n} &= \half(u^{n-\half} + u^{n+\half}) = u(t_n) + R^{n}\\ R^{n} &= \frac{1}{8}u''(t_{n})\Delta t^2 + \frac{1}{384}u''''(t_n)\Delta t^4 + \Oof{\Delta t^6} \end{align*} $$

 

Leading-order error terms in mean values (2)

Geometric mean:

 
$$ \begin{align*} [\overline{u^2}^{t,g}]^{n} &= u^{n-\half}u^{n+\half} = (u^n)^2 + R^n\\ R^n &= - \frac{1}{4}u'(t_n)^2\Delta t^2 + \frac{1}{4}u(t_n)u''(t_n)\Delta t^2 + \Oof{\Delta t^4} \end{align*} $$

 

Harmonic mean:

 
$$ \begin{align*} [\overline{u}^{t,h}]^{n} &= u^n = \frac{2}{\frac{1}{u^{n-\half}} + \frac{1}{u^{n+\half}}} + R^{n+\half}\\ R^n &= - \frac{u'(t_n)^2}{4u(t_n)}\Delta t^2 + \frac{1}{8}u''(t_n)\Delta t^2 \end{align*} $$

 

Software for computing truncation errors

>>> from truncation_errors import TaylorSeries
>>> from sympy import *
>>> u, dt = symbols('u dt')
>>> u_Taylor = TaylorSeries(u, 4)
>>> u_Taylor(dt)
D1u*dt + D2u*dt**2/2 + D3u*dt**3/6 + D4u*dt**4/24 + u
>>> FE = (u_Taylor(dt) - u)/dt
>>> FE
(D1u*dt + D2u*dt**2/2 + D3u*dt**3/6 + D4u*dt**4/24)/dt
>>> simplify(FE)
D1u + D2u*dt/2 + D3u*dt**2/6 + D4u*dt**3/24

Notation: D1u for \( u' \), D2u for \( u'' \), etc.

See trunc/truncation_errors.py.

Symbolic computing with difference operators

A class DiffOp represents many common difference operators:

>>> from truncation_errors import DiffOp
>>> from sympy import *
>>> u = Symbol('u')
>>> diffop = DiffOp(u, independent_variable='t')
>>> diffop['geometric_mean']
-D1u**2*dt**2/4 - D1u*D3u*dt**4/48 + D2u**2*dt**4/64 + ...
>>> diffop['Dtm']
D1u + D2u*dt/2 + D3u*dt**2/6 + D4u*dt**3/24
>>> diffop.operator_names()
['geometric_mean', 'harmonic_mean', 'Dtm', 'D2t', 'DtDt',
 'weighted_arithmetic_mean', 'Dtp', 'Dt']

Names in diffop: Dtp for \( D^+_t \), Dtm for \( D_t^- \), Dt for \( D_t \), D2t for \( D_{2t} \), DtDt for \( D_tD_t \).

Truncation errors in exponential decay ODE

 
$$ u'(t)=-au(t) $$

 

Truncation error of the Forward Euler scheme

The Forward Euler scheme:

 
$$ \lbrack D_t^+ u = -au \rbrack^n \tp $$

 
Definition of the truncation error \( R^n \):

 
$$ \lbrack D_t^+ \uex + a\uex = R \rbrack^n \tp $$

 
From \eqref{trunc:table:fd1:fw:eq}-\eqref{trunc:table:fd1:fw}:

 
$$ [D_t^+ \uex]^n = \uex'(t_n) + \half\uex''(t_n)\Delta t + \Oof{\Delta t^2}\tp$$

 
Inserted in \eqref{trunc:decay:FE:uex}:

 
$$ \uex'(t_n) + \half\uex''(t_n)\Delta t + \Oof{\Delta t^2} + a\uex(t_n) = R^n \tp $$

 
Note: \( \uex'(t_n) + a\uex^n = 0 \) since \( \uex \) solves the ODE. Then

 
$$ R^n = \half\uex''(t_n)\Delta t + \Oof{\Delta t^2} \tp $$

 

Truncation error of the Crank-Nicolson scheme

Crank-Nicolson:

 
$$ [D_t u = -au]^{n+\half} $$

 
Truncation error:

 
$$ [D_t \uex + a\overline{\uex}^{t} = R]^{n+\half} $$

 

From \eqref{trunc:table:fd1:center:eq}-\eqref{trunc:table:fd1:center} and \eqref{trunc:table:avg:arith:eq}-\eqref{trunc:table:avg:arith}:

 
$$ \begin{align*} \lbrack D_t\uex \rbrack^{n+\half} &= u'(t_{n+\half}) + \frac{1}{24}\uex'''(t_{n+\half})\Delta t^2 + \Oof{\Delta t^4}\\ [a\overline{\uex}^{t}]^{n+\half} &= u(t_{n+\half}) + \frac{1}{8}u''(t_{n})\Delta t^2 + + \Oof{\Delta t^4} \end{align*} $$

 
Inserted in the scheme we get

 
$$ R^{n+\half} = \left( \frac{1}{24}\uex'''(t_{n+\half}) + \frac{1}{8}u''(t_{n}) \right)\Delta t^2 + \Oof{\Delta t^4} $$

 
\( R^n = \Oof{\Delta t^2} \) (second-order scheme)

Test the understanding!

Analyze the the truncation error of the Backward Euler scheme and show that it is \( \Oof{\Delta t} \) (first order scheme).

Truncation error of the \( \theta \)-rule

The \( \theta \)-rule:

 
$$ [\bar D_t u = -a\overline{u}^{t,\theta}]^{n+\theta} \tp $$

 

Truncation error:

 
$$ [\bar D_t \uex + a\overline{\uex}^{t,\theta} = R]^{n+\theta} \tp $$

 
Use \eqref{trunc:table:fd1:theta:eq}-\eqref{trunc:table:fd1:theta} and \eqref{trunc:table:avg:theta:eq}-\eqref{trunc:table:avg:theta} along with \( \uex'(t_{n+\theta}) + a\uex(t_{n+\theta})=0 \) to show

 
$$ \begin{align} R^{n+\theta} = &({\half}-\theta)\uex''(t_{n+\theta})\Delta t + \half\theta (1-\theta)\uex''(t_{n+\theta})\Delta t^2 + \nonumber\\ & \half(\theta^2 -\theta + 3)\uex'''(t_{n+\theta})\Delta t^2 + \Oof{\Delta t^3} \tag{1} \end{align} $$

 
Note: 2nd-order scheme if and only if \( \theta =1/2 \).

Using symbolic software

Can use sympy and the tools in truncation_errors.py:

def decay():
    u, a = sm.symbols('u a')
    diffop = DiffOp(u, independent_variable='t',
                    num_terms_Taylor_series=3)
    D1u = diffop.D(1)   # symbol for du/dt
    ODE = D1u + a*u     # define ODE

    # Define schemes
    FE = diffop['Dtp'] + a*u
    CN = diffop['Dt' ] + a*u
    BE = diffop['Dtm'] + a*u
    # Residuals (truncation errors)
    R = {'FE': FE-ODE, 'BE': BE-ODE, 'CN': CN-ODE}
    return R

The returned dictionary becomes

decay: {
 'BE': D2u*dt/2 + D3u*dt**2/6,
 'FE': -D2u*dt/2 + D3u*dt**2/6,
 'CN': D3u*dt**2/24,
}

\( \theta \)-rule: see truncation_errors.py (long expression, very advantageous to automate the math!)

Empirical verification of the truncation error (1)

Ideas:

For the Forward Euler scheme:

 
$$ R^n = \lbrack D_t^+\uex + a\uex \rbrack^n \tp $$

 
Insert correct \( \uex(t)=Ie^{-at} \) (or use method of manufactured solution in more general cases).

Empirical verification of the truncation error (2)

See the text for more details and an implementation.

Empirical verification of the truncation error in the Forward Euler scheme


Figure 1: Estimated truncation error at mesh points for different meshes.

Empirical verification of the truncation error in the Forward Euler scheme


Figure 2: Difference between theoretical and estimated truncation error at mesh points for different meshes.

Increasing the accuracy by adding correction terms

Question.

Can we add terms in the differential equation that can help increase the order of the truncation error?

To be precise for the Forward Euler scheme, can we find \( C \) to make \( R \) $\Oof{\Delta t^2}$?

 
$$ \lbrack D_t^+ \uex + a\uex = C + R \rbrack^n\tp $$

 

 
$$ \half\uex''(t_n)\Delta t - \frac{1}{6}\uex'''(t_n)\Delta t^2 + \Oof{\Delta t^3} = C^n + R^n\tp$$

 
Choosing

 
$$ C^n = \half\uex''(t_n)\Delta t$$

 
makes

 
$$ R^n = \frac{1}{6}\uex'''(t_n)\Delta t^2 + \Oof{\Delta t^3}\tp$$

 

Lowering the order of the derivative in the correction term

 
$$ u' = -au,\quad\Rightarrow\quad u''=-au'=a^2u\tp$$

 
Result for \( u''=a^2u \): apply Forward Euler to a perturbed ODE,

 
$$ u' = -\hat au ,\quad \hat a = a(1 - {\half}a\Delta t) $$

 
to make a second-order scheme!

With a correction term Forward Euler becomes Crank-Nicolson

Use the other alternative \( u''=-au' \):

 
$$ u'=-au - {\half}a\Delta t u'\quad\Rightarrow\quad \left( 1 + {\half}a\Delta t\right) u' = -au\tp$$

 
Apply Forward Euler:

 
$$ \left( 1 + {\half}a\Delta t\right)\frac{u^{n+1}-u^n}{\Delta t} = -au^n$$

 
which after some algebra can be written as

 
$$ u^{n+1} = \frac{1 - {\half}a\Delta t}{1+{\half}a\Delta t}u^n\tp$$

 
This is a Crank-Nicolson scheme (of second order)!

Correction terms in the Crank-Nicolson scheme (1)

 
$$ [D_t u = -a\overline{u}^t]^{n+\half}$$

 
Definition of the truncation error \( R \) and correction terms \( C \):

 
$$ [D_t \uex + a\overline{\uex}^{t} = C + R]^{n+\half}\tp$$

 

Must Taylor expand

 
$$ C^{n+\half} + R^{n+\half} = \frac{1}{24}\uex'''(t_{n+\half})\Delta t^2 + \frac{a}{8}\uex''(t_{n+\half})\Delta t^2 + \Oof{\Delta t^4}\tp$$

 
Let \( C^{n+\half} \) cancel the \( \Delta t^2 \) terms:

 
$$ C^{n+\half} = \frac{1}{24}\uex'''(t_{n+\half})\Delta t^2 + \frac{a}{8}\uex''(t_{n})\Delta t^2\tp$$

 

Correction terms in the Crank-Nicolson scheme (2)

Result: solve the perturbed ODE by a Crank-Nicolson method,

 
$$ u' = -\hat a u,\quad \hat a = a(1 - \frac{1}{12}a^2\Delta t^2)\tp$$

 
and experience an error \( \Oof{\Delta t^4} \).

Extension to variable coefficients

 
$$ u'(t) = -a(t)u(t) + b(t)$$

 

Forward Euler:

 
$$ [D_t^+ u = -au + b]^n \tp $$

 
The truncation error is found from

 
$$ [D_t^+ \uex + a\uex - b = R]^n \tp $$

 
Using \eqref{trunc:table:fd1:fw:eq}-\eqref{trunc:table:fd1:fw}:

 
$$ \uex'(t_n) - \half\uex''(t_n)\Delta t + \Oof{\Delta t^2} + a(t_n)\uex(t_n) - b(t_n) = R^n \tp $$

 
Because of the ODE, \( \uex'(t_n) + a(t_n)\uex(t_n) - b(t_n) =0 \), and

 
$$ R^n = -\half\uex''(t_n)\Delta t + \Oof{\Delta t^2} \tp $$

 
No problems with variable coefficients!

Exact solutions of the finite difference equations

How does the truncation error depend on \( \uex \) in finite differences?

Consequence:

Problem: harmonic and geometric mean (error depends on \( \uex' \) and \( \uex \))

Computing truncation errors in nonlinear problems (1)

 
$$ u'=f(u,t) $$

 
Crank-Nicolson scheme:

 
$$ [D_t u =\overline{f}^{t}]^{n+\half} $$

 
Truncation error:

 
$$ [D_t \uex - \overline{f}^{t}= R]^{n+\half}\tp $$

 
Using \eqref{trunc:table:avg:arith:eq}-\eqref{trunc:table:avg:arith} for the arithmetic mean:

 
$$ \begin{align*} \lbrack\overline{f}^{t}\rbrack^{n+\half} &= \half(f(\uex^n,t_n) + f(\uex^{n+1},t_{n+1}))\\ &= f(\uex^{n+\half},t_{n+\half}) + \frac{1}{8}\uex''(t_{n+\half})\Delta t^2 + \Oof{\Delta t^4}\tp \end{align*} $$

 

Computing truncation errors in nonlinear problems (2)

With \eqref{trunc:table:fd1:center:eq}-\eqref{trunc:table:fd1:center}, \eqref{trunc:decay:gen:ode:CN} leads to \( R^{n+\half} \) equal to

 
$$ \uex'(t_{n+\half}) + \frac{1}{24}\uex'''(t_{n+\half})\Delta t^2 - f(\uex^{n+\half},t_{n+\half}) - \frac{1}{8}\uex''(t_{n+\half})\Delta t^2 + \Oof{\Delta t^4} \tp $$

 
Since \( \uex'(t_{n+\half}) - f(\uex^{n+\half},t_{n+\half})=0 \), the truncation error becomes

 
$$ R^{n+\half} = (\frac{1}{24}\uex'''(t_{n+\half}) - \frac{1}{8}\uex''(t_{n+\half})) \Delta t^2\tp $$

 
The computational techniques worked well even for this nonlinear ODE!

Truncation errors in vibration ODEs

Linear model without damping

 
$$ u''(t) + \omega^2 u(t) = 0,\quad u(0)=I,\ u'(0)=0\tp $$

 
Centered difference approximation:

 
$$ [D_tD_t u + \omega^2u=0]^n $$

 
Truncation error:

 
$$ [D_tD_t \uex + \omega^2\uex =R]^n \tp $$

 
Use \eqref{trunc:table:fd2:center:eq}-\eqref{trunc:table:fd2:center} to expand \( [D_t D_t\uex]^n \):

 
$$ [D_tD_t \uex]^n = \uex''(t_n) + \frac{1}{12}\uex''''(t_n)\Delta t^2$$

 
Collect terms: \( \uex''(t) + \omega^2\uex(t)=0 \). Then,

 
$$ R^n = \frac{1}{12}\uex''''(t_n)\Delta t^2 + \Oof{\Delta t^4} \tp $$

 

Truncation errors in the initial condition

Computing correction terms

 
$$ [D_tD_t \uex + \omega^2\uex =C + R]^n$$

 

 
$$ C^n = \frac{1}{12}\uex''''(t_n)\Delta t^2\tp$$

 

 
$$ [D_tD_t u + \omega^2(1 - \frac{1}{12}\omega^2\Delta t^2)u=0]^n$$

 

Model with damping and nonlinearity

Linear damping \( \beta u' \), nonlinear spring force \( s(u) \), and excitation \( F \):

 
$$ mu'' + \beta u' + s(u) =F(t) $$

 
Central difference discretization:

 
$$ [mD_tD_t u + \beta D_{2t} u + s(u)=F]^n \tp $$

 
Truncation error is defined by

 
$$ [mD_tD_t \uex + \beta D_{2t} \uex + s(\uex)=F + R]^n \tp $$

 

Carrying out the truncation error analysis

Using \eqref{trunc:table:fd2:center:eq}-\eqref{trunc:table:fd2:center} and \eqref{trunc:table:fd1:center2:eq}-\eqref{trunc:table:fd1:center2} we get

 
$$ \begin{align*} \lbrack mD_tD_t \uex + \beta D_{2t} \uex\rbrack^n &= m\uex''(t_n) + \beta\uex'(t_n) + \\ &\quad \left(\frac{m}{12}\uex''''(t_n) + \frac{\beta}{6}\uex'''(t_n)\right)\Delta t^2 + \Oof{\Delta t^4} \end{align*} $$

 
The terms

 
$$ m\uex''(t_n) + \beta\uex'(t_n) + \omega^2\uex(t_n) + s(\uex(t_n)) - F^n$$

 
correspond to the ODE (= zero).

Result: accuracy of \( \Oof{\Delta t^2} \) since

 
$$ R^n = \left(\frac{m}{12}\uex''''(t_n) + \frac{\beta}{6}\uex'''(t_n)\right)\Delta t^2 + \Oof{\Delta t^4} $$

 

Correction terms: complicated when the ODE has many terms...

Extension to quadratic damping

 
$$ mu'' + \beta |u'|u' + s(u) =F(t) $$

 
Centered scheme: \( |u'|u' \) gives rise to a nonlinearity.

Linearization trick: use a geometric mean,

 
$$ [|u'|u']^n \approx |[u']^{n-\half}|[u']^{n+\half}\tp$$

 

Scheme:

 
$$ [mD_t D_t u]^n + \beta |[D_{t} u]^{n-\half}|[D_t u]^{n+\half} + s(u^n)=F^n\tp $$

 

The truncation error for quadratic damping (1)

Definition of \( R^n \):

 
$$ [mD_t D_t \uex]^n + \beta |[D_{t} \uex]^{n-\half}|[D_t \uex]^{n+\half} + s(\uex^n)-F^n = R^n\tp $$

 

Truncation error of the geometric mean, see \eqref{trunc:table:avg:geom:eq}-\eqref{trunc:table:avg:geom},

 
$$ \begin{align*} |[D_{t} \uex]^{n-\half}|[D_t \uex]^{n+\half} &= [|D_t\uex|D_t\uex]^n - \frac{1}{4}u'(t_n)^2\Delta t^2 + \\ &\quad \frac{1}{4}u(t_n)u''(t_n)\Delta t^2 + \Oof{\Delta t^4}\tp \end{align*} $$

 
Using \eqref{trunc:table:fd1:center:eq}-\eqref{trunc:table:fd1:center} for the \( D_t\uex \) factors results in

 
$$ \begin{align*} [|D_t\uex|D_t\uex]^n &= |\uex' + \frac{1}{24}\uex'''(t_n)\Delta t^2 + \Oof{\Delta t^4}|\times\\ &\quad (\uex' + \frac{1}{24}\uex'''(t_n)\Delta t^2 + \Oof{\Delta t^4}) \end{align*} $$

 

The truncation error for quadratic damping (2)

For simplicity, remove the absolute value. The product becomes

 
$$ [D_t\uex D_t\uex]^n = (\uex'(t_n))^2 + \frac{1}{12}\uex(t_n)\uex'''(t_n)\Delta t^2 + \Oof{\Delta t^4}\tp$$

 

With

 
$$ m[D_t D_t\uex]^n = m\uex''(t_n) + \frac{m}{12}\uex''''(t_n)\Delta t^2 +\Oof{\Delta t^4}$$

 
and using \( mu'' + \beta (u')^2 + s(u)=F \), we end up with

 
$$ R^n = (\frac{m}{12}\uex''''(t_n) + \frac{\beta}{12}\uex(t_n)\uex'''(t_n)) \Delta t^2 + \Oof{\Delta t^4}\tp$$

 
Second-order accuracy! Thanks to

The general model formulated as first-order ODEs

 
$$ mu'' + \beta |u'|u' + s(u) =F(t) $$

 
Rewritten as first-order system:

 
$$ \begin{align*} u' &= v\\ v' &= \frac{1}{m}\left( F(t) - \beta |v|v - s(u)\right) \end{align*} $$

 

To solution methods:

The forward-backward scheme

Forward step for \( u \), backward step for \( v \):

 
$$ \begin{align*} [D_t^+ u &= v]^n\\ [D_t^-v &= \frac{1}{m}( F(t) - \beta |v|v - s(u))]^{n+1} \end{align*} $$

 

 
$$ \begin{align*} [D_t^+ u &= v]^n\\ [D_t^-v]^{n+1} &= \frac{1}{m}( F(t_{n+1}) - \beta |v^n|v^{n+1} - s(u^{n+1})) \end{align*} $$

 

Truncation error analysis

A centered scheme on a staggered mesh

Staggered mesh:

Centered differences in \eqref{trunc:vib:gen:2x2model:ode:u}-\eqref{trunc:vib:gen:2x2model:ode:u}:

 
$$ \begin{align*} [D_t u &= v]^{n-\half}\\ [D_t v &= \frac{1}{m}( F(t) - \beta |v|v - s(u))]^{n} \end{align*} $$

 

 
$$ |v^n|v^n \approx |v^{n-\half}|v^{n+\half}\tp$$

 

Truncation error analysis (1)

Resulting scheme:

 
$$ \begin{align*} [D_t u]^{n-\half} &= v^{n-\half}\\ [D_t v]^n &= \frac{1}{m}( F(t_n) - \beta |v^{n-\half}|v^{n+\half} - s(u^n)) \end{align*} $$

 

The truncation error in each equation is found from

 
$$ \begin{align*} [D_t \uex]^{n-\half} &= \vex(t_{n-\half}) + R_u^{n-\half}\\ [D_t \vex]^n &= \frac{1}{m}( F(t_n) - \beta |\vex(t_{n-\half})|\vex(t_{n+\half}) - s(u^n)) + R_v^n\tp \end{align*} $$

 
Using \eqref{trunc:table:fd1:center:eq}-\eqref{trunc:table:fd1:center} for derivatives and \eqref{trunc:table:avg:geom:eq}-\eqref{trunc:table:avg:geom} for the geometric mean:

 
$$ \uex'(t_{n-\half}) + \frac{1}{24}\uex'''(t_{n-\half})\Delta t^2 + \Oof{\Delta t^4} = \vex(t_{n-\half}) + R_u^{n-\half}$$

 
and

 
$$ \vex'(t_n) = \frac{1}{m}( F(t_n) - \beta |\vex(t_n)|\vex(t_n) + \Oof{\Delta t^2} - s(u^n)) + R_v^n\tp $$

 

Truncation error analysis (2)

Resulting truncation error is \( \Oof{\Delta t^2} \):

 
$$ R_u^{n-\half}= \Oof{\Delta t^2}, \quad R_v^n = \Oof{\Delta t^2}\tp$$

 

Observation.

Comparing The schemes \eqref{trunc:vib:gen:2x2model:ode:u:staggered2}-\eqref{trunc:vib:gen:2x2model:ode:v:staggered2} and \eqref{trunc:vib:gen:2x2model:ode:u:fw2}-\eqref{trunc:vib:gen:2x2model:ode:v:bw2} are equivalent. Therefore, the forward/backward scheme with ad hoc linearization is also \( \Oof{\Delta t^2} \)!