NAP4
NUMERICAL ANALYSIS OF
PROCESSES
NONLINEAR MODELS of systems described by
ordinary differential equations
Initial problem (Runge Kutta, Adams)
Time delay
Strange attractors
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
NAP4
Models
ODE initial problem
Systems which are described by a system of ordinary differential equations and
their solutions are fully described by the initial state, are for example integral
models of mass and enthalpy balances elementary units ("lumped parameter" or
"compartment" models). The aim is to determine the evolution of state variables
(concentration, temperature) at the time, so it is an evolutionary problems.
These elementary units may be specific devices (reactors, absorbers, columns, tanks, filters, ...) in
"spread sheets" or zones in apparatuses in which the state can be characterized by mean temperature,
and the concentration of components (ideal mixers, plug flow zone). An example is the description of a
stirred vessel system of ideal mixers below and above a turbine stirrer or a replacement shell and tube
heat exchanger by a systém of elementary exchangers between the baffles.
Each elementary unit correspond to ordinary differential equations describing the mass balance (one for
each component of the differential equation describing the rate of change of concentration)
dc n
V
  Qi ci
dt i 1
volumetric flowrate from
neigbouring compartments
and a enthalpy balance (or total energy balance) describing temperature evolution in the department
n
dT
mc p
  mi hi  H ( g )
dt i 1
wnthalpic flows from
neighbouring compartments
volumetric
hear source
NAP4
Models
ODE initial problem
A system need not be described by ordinary differential equations with only first
derivatives. Eg. the equations of vibration include also the second time derivative
(acceleration)
2
m
d y
dy

d
 ky  f (t )
2
dt
dt
But every such equation can be replaced by a system of equations with only
first derivatives, e.g.
dv
 dv  ky  f (t )
dt
dy
v
dt
m
The initial problem is characterized by the fact that each computed variable
corresponds to one differential equation of the first order and to one initial
condition (value) corresponding to the initial state.
…or how many equations so many initial
conditions corresponding to one value of
independent variable.
NAP4
Models
ODE initial problem
backmixing model
(dispersion,
diffusiion)
Example of continuous system:
(1+r+b)Q
Q,x(t)
(1+r)Q
2
1
5
4
6
bQ
Ci, Ti
sQ
rQ
3
7
model of stagnant zone
recirkulation with
transport delay
Differential equations for concentrations
dc1
 Q( x  c1 )  sQ (c3  c1 )
dt
dc
V2 2  Q(c1  c2 )
dt
dc
V3 3  sQ (c1  c3 )
dt
V1
dc4
 Q(c2  rc7  bc5  (1  r  b)c4 )
dt
dc
V5 5  Q((1  r  b)c4  bc6  (1  r  2b)c5 )
dt
dc
V6 5  Q(1  r  b)(c5  c6 )
dt
c7 (t )  c6 (t  )
V4
Parameters determining distribution of flowrates (s-stagnant zone, r-recirculation, b-backmixing) must be
specified, r-from flowmeter measurement, b-from correlations. Also the volume of individual compartments
must be known. These parameters are most frequently identiofied from fit of predicted characteristics (for
example RTD - residence time distribution) with experiment. As a result information about the size of dead
zones, recirculation zones, shortcuts, intensity of dispersion etc. are obtained.
NAP4
Models
ODE Fourier transform
The models can be solved by the Fourier transformation method (see previous
lecture NAZP3).
It is not enough to apply only a convolution, because the model includes compartments with back
mixing, stagnant zone ... However, you can come out of these differential equations, transformed by
FFT. For the Fourier transform of derivation applies (assuming that the concentration at time zero is
infinity)


dc 2ift
2ift
e
dt

c
(
t
)
e
 dt




 2if

2ift
~( f )
c
(
t
)
e
dt


2

if
c


Differential equations are transformed to algebraic equations, where instead of time t is frequency f.
2 iV1 fc1  Q( x  c1 )  sQ(c3  c1 )
2 iV4 fc4  Q(c2  rc7  bc5  (1  r  b)c4 )
2 iV2 fc2  Q(c1  c2 )
2 iV5 fc5  Q((1  r  b)c4  bc6  (1  r  2b)c5 )
2 iV3 fc3  sQ(c1  c3 )
2 iV6 fc6  Q(1  r  b)(c5  c6 )
c7  exp(2 if )c6
This system of algebraic equations (linear!) is already fairly easy to solve, and the result will be the Fourier
transform of concentration. Concentration courses in time are then get back by FFT. The solution is
accurate and fast (using FFT), but is limited to linear models. The procedure is preferred when the
coefficients of the system (r, b, s) are constant. Otherwise, it was necessary to compute the Fourier
transform of the product of two functions (and it is not equal to the product of the transformation). In the
general case with nonlinear terms is therefore better to stay in the time domain and seek solutions to the
numerical integration of the ODE.
NAP4
Models
ODE Laplace transform
Similar relationships hold also for Laplace transform: the same convolution,
correlation and transformation of derivatives


dc  st
 st 
 st


e
dt

c
(
t
)
e

s
c
(
t
)
e
dt  c(0)  sc (s)
0 dt
0

0
initial
Laplace transform
value
Inverse Laplace transform is usually searched in dictionaries, for example


L t m e  t 
sc~1  ~x  c~1
sc~2  c~1  c~2
...
sc~m  c~m 1  c~m
m!
( s   ) m 1
L  (t )  1
NAP4
ODE
dc
 f (t , c)
dt
initial problem numerical solutions
t  t k c  ck
One-step methods approximate solution at time tk+1=tk+ by estimating mean
value of derivative f in interval <tk,tk+1>
ck 1  ck  f
c
c
Euler metods are
easily programmable
even in Excel
ck
ck
t
tk
tk+1
ck 1  ck  f (tk , ck )
t
tk
tk+1
ck 1  ck  f (tk 1 , ck 1 )
Euler method explicit
Euler method implicit
order of accuracy  (first)
order of accuracy  (first)
stable only for small
integration steps 
stable, but iterations are
necessary (because ck+1 is
unknown)
first order of accuracy means
that the error is directly
proportional to the step 
when you write a Taylor expansion for
one time step, the error is O(2), a
second-order accuracy. Errors,
however, accumulate, so 1/ steps will
result in an error O().
NAP4
ODE
dc
 f (t , c)
dt
initial problem numerical solutions
t  t k c  ck
One-step methods approximate solution at time tk+1=tk+ by estimating mean
value of derivative f in interval <tk,tk+1>
ck 1  ck  f
Runge Kutta methods calculate mean value of
derivative as a weighted average of derivatives f(t,c).
The most often used version calculates four points
(endpoints tk, tk+1 and two midpoints at tk+1/2)
c
f1 f 2 f 3 f 4
  
6 3 3 6
f1  f (t k , ck )
f 
ck
t
tk
tk+1
This variant is
implemented in MATLABu
as a function ode45
f 2  f (t k   / 2, ck  f1 / 2)
f 3  f (t k   / 2, ck  f 2  / 2)
f 4  f (t k  , ck  f 3 )
Weight coefficients (1/6,1/3,1/3,1/6) and increments c are designed so that the
order of accuracy is the highest (4th order of accuracy in this case). Generally, the
order of accuracy of RK methods is the same as the number of evaluated points.
ODE
NAP4
dc
 f (t , c )
dt
initial problem numerical solutions
t  t k c  ck
t  t k 1 c  ck 1
Multistep methods approximate solution at time tk+1 on the basis of several
previous values for example at times tk, tk-1 (two), or tk-2 (three-step method).
This is usually a combination of predictor (explicit) and
corrector (default formula with higher precision and
stable)
c
ck 1  ck 2 
ck
ck-1
t
tk-2
tk-1
tk
tk+1
ck 1  ck 
3
( f (tk , ck )  f (tk 1 , ck 1 ))
2
1
( f (tk , ck )  f (tk 1 , ck 1 ))
2
it is sufficient to
evaluate just one value
Both formulas are second order accuracy but corrector (implicit formula) has the
absolute value of the error about 9-fold lower (implicit formulas tend to have an
error less than explicit, even if the order errors is the same).
The disadvantage is that for the first two steps a single-step method, for example
RK, must be used. On the other hand, number of computed derivatives is smaller.
NAP4
ODE initial problem numerical solutions
dc
 f (t , c)
dt
Dynamic adjustment of integration step 
The simplest solution is that each step is performed twice, once with step  and then with
increments /2 (twice). If the difference is greater than the specified tolerance, the step  shrinks.
Using the two calculated values (for full and half step) the accuracy of the final value can be
improved by so called Aitken extrapolation. MATLAB uses a dynamic choice of the time step
automatically.
Stiff problem
When you solve a system of equations may be that the integration step is dictated by the
equation with an extremely short time constant, and the stability of the solution then requires
extremely short integration step (the calculation time is prohibitively increased). We say that the
system is stiff. Some methods of integrating ODE this unpleasant impact suppresses (an
example is Gearova method).
Transport delay (time shift)
Time delay was included in the previous example. In this case the solver should remember the
whole history of calculated results and not only the last step. Even the values for negative times
(before the initial time) have to be defined.
NAP4
ODE initial problem numerical solutions
dc
 f (t , c)
dt
Stability analysis
Consider a differential equation describing e.g. the rate of a chemical reaction of the first order
dc
  Ac
dt
stability condition restriction of
time step  (coef. of gain less
than 1)
Euler explicit integration formula of the first order
ck  ck 1  Ack 1
ck  c0 (1  A) k
| 1  A | 1   
2
A
Euler implicit integration formula of the first order
gain less than 1 for
arbitrary 
ck  ck 1  Ack
ck  c0
1
(1  A) k
|
1
| 1
1  A
NAP4
ODE examples in MATLAB
ode45, ode23 Runge-Kutta (one step method with different order of accuracy)
ode113 Adams (multistep method)
ode15s Stiff (Gear multistep method with variable accuracy order-suitable for stiff problems)
dde23 transport delay
NAP4
RTD
series of mixers 1/4
An example has been solved in the previous lectures using FFT. Series 4 ideally
mixed vessels (container volume, flow rate Q). The aim is to find the time courses
of concentrations c1,…,c4 for arbitrary initial conditions and for arbitrary course of
concentration at inlet x(t).
x(t)
y(t)
Mass balances
(t)
dc1
dt
dc2
dt
dc3
dt
dc4
dt
Q
( x  c1 )
V
Q
 (c1  c2 )
V
Q
 (c2  c3 )
V
Q
 (c3  c4 )
V

c1 c2
c3
c4
tm=2 odpovídá poměru V/Q=0.5 (to
je střední doba prodlení jedné nádoby)
As an initial concentration x(t) is applied impulse response of a set of two
mixers (the same volume of tanks) and zero initial concentration in all tanks.
The reason for such an assignment is that there is an analytical solution that
can be compared with the results of numerical solution of the ODE. Stimulus
function x(t) is the impulse response of E (t) for N=2 and tm=1; outlet
concentration y(t)=c4(t) shou;ld be impulse response for N=6 and tm=3
N N t N 1  Nt / tm
E (t ) 
e
( N  1)!t mN
NAP4
RTD
series of mixers 2/4
MATLAB provides for the solution of ODE variety of methods (ode45, ode113,
ode15s, ...) which can automatically adjust the integration step so as to achieve the
required accuracy. Procedure is always the same, first define a function whose
output is the vector of first derivatives for a given value of integration variable (time)
and a solution vector.
In our case it is necessary to define a function whose result is a column vector (4x1) of right-hand sides of 4
differential equations. It will be assumed that the stimulus function x(t) is the impulse response of nx series
mixers, with a mean time tmx. The parameter tm is the mean time of the modeled system of 4 mixers.
function dy = serie4(t,y,tm,tmx,nx)
dy = zeros(4,1);
xx=efun(t,tmx,nx);
It would probably be smarter if the stimulus
tm1=tm/4;
function x(t) was directly the next formal
dy(1)=(xx-y(1))/tm1;
parameter of the function serie4. I have a
dy(2)=(y(1)-y(2))/tm1;
little problem with that in Matlab, I need
dy(3)=(y(2)-y(3))/tm1;
advice.
dy(4)=(y(3)-y(4))/tm1;
The Efun function can be defined as a sub-function in the m-file serie4.m, but maybe it's better to define it as a
separate m-file efun.m
function y=efun(t,tm,n)
y=n^n*t^(n-1)/(factorial(n-1)*tm^n)*exp(-t*n/tm);
N N t N 1  Nt / tm
E (t ) 
e
( N  1)!t mN
NAP4
RTD
series of mixers 3/4
Integration ODE by Runge Kutta method
Sol = ode45(function describing right side, time range, initial conditions)
in our case for example
sol=ode45(@(t,y)serie4(t,y,2,1,1), [0 10] , [0 0 0 0])
Vector of initial conditions
(zero concentrations)
solver: 'ode45'
extdata: [1x1 struct]
x: [1x29 double]
y: [4x29 double]
stats: [1x1 struct]
idata: [1x1 struct].
Result is the structure sol
containing vector of time steps
sol.x (in this specific case 29
steps was necessary for
prescribed accuracy) and a
corresponding matrix sol.y of
calculated concentrations (four
rows for four concentrations)
Integration from t=0, to t=10. It is
not possible to specify a constant
integration step in Matlab
Using a function as an actual parameter looks a bit
strange in Matlab. It is not possible to write simply the
name of this function (serie4) because ode45 expects
only two parameters (t, y) and the function serie4 has
5 parameters. Therefore, the so-called anonymous
function @(t,y)expression.must be used.
NAP4
RTD
series of mixers 4/4
Numerical solution by Runge Kutta (ode45) can be
compared with the analytical solution E(t) for N=6 and tm=3
N N t N 1  Nt / tm
E (t ) 
e
( N  1)!t mN
This is demonstration how to
define a one-line function
(without using a m-file)
e=@(t,tm,n) n^n.*t.^(n-1)./(factorial(n-1)*tm^n).*exp(-n.*t/tm)
t=linspace(0,10.23,1024);
Standard function linspace
y=e(t,3,6);
0.4
generates vector of 1024
0.7
equidistant time steps
0.35
0.6
these points are results of
ode45 with automatically
adjusted integration time
steps
0.3
0.25
Sy=deval(sol,t)
0.5
function deval interpolates the
nonequidistant results sol.y
according to the prescribed
time scale t
0.4
0.2
0.3
0.15
0.2
0.1
0.1
0.05
0
0
2
4
6
8
10
0
0
2
4
6
8
10
NAP4
RTD
time delay (transport delay)
In the case that there are time delays (long connecting pipelines,
recycles) it is not possible to use the standard integration functions
(ode23,ode45,…), but a different „family“ of MATLAB functions
which is capable to save previous results (with prescribed vector of
time delays 1 2 …)
NAP4
RTD
delay in recycle 1/2
Example: continuous system formed by two mixed tanks with time delay  in
recycle. It is quite common: connecting pipelines are usually modeled by a
time delay.
x(t)
(1+r)Q
Q
1
2
rQ
The aim is to find time courses of concentrations c1(t), c2(t), for given
flowrate (Q), recirculation ratio (r) and volume of tanks (V1,V2), for
prescribed time course of inlet concentration x(t). The system is described
by the mass balances:
dc1 Q
 ( x(t )  c1 (t ))
dt V1
dc2 Q
 (c1 (t )  rc2 (t   )  (1  r )c2 (t ))
dt V2
time delay

Vd
rQ
NAP4
RTD
delay in recycle 2/2
Special ODE solvers must be used in MATLAB (dde23 for a constant time delay,
and ddesd for variabl;e time delay)
sol=dde23(@derivatives, [1, 2…], history for t<tmin, [tmin,tmax])
function dcdt=dclag(t,c,z)
q=1;
q-flowrate,v1,v2 volumes, r-recycle ratio, impulse
v1=2;
stimulus function x(t) with duration 0.2 seconds
v2=1;
r=10;
if t<.2
x=1;
else
z is the matrix of solutions in
x=0;
times t-k. Column index (1)
is the index in vector of
end
delays
dcdt=zeros(2,1);
clag=z(:,1);
dcdt(1)=q/v1*(x-c(1));
dcdt(2)=q/v2*(c(1)+r*clag(2)-(1+r)*c(2));
zero history c1=c2=0 for
t<0
0.1
0.09
0.08
0.07
0.06
0.05
0.04
dc1 Q
 ( x(t )  c1 (t ))
dt V1
0.03
dc2 Q
 (c1 (t )  rc2 (t   )  (1  r )c2 (t ))
dt V2
0.02
0.01
0
sol=dde23(@dclag,1,[0;0],[0,20])
plot(sol.x,sol.y)
vector of delays 1=1 (there is
only one delay)
0
2
4
6
8
10
12
14
16
18
20
NAP4
Trajectory of particles
The initial problem is also
description of motion of particles
under action of forces. Examples
are trajectory of droplets in the
spray drying chamber, motion of
particles in a fluidized bed, burning
coal particles in the combustion
chamber, particles in the mixer, a
settling tank, etc. The equations of
motion are ODE of the type:

du
  
m
 F (v  u )  g
dt
acceleratio
n
drag for in fluid moving with
velocity v
buoyant, centrifugal,
electrical forces
NAP4
Trajectories of particles
In the last lecture of the course we will pay attention to the type of modern
methods of DEM (Discrete Element Method), which simulates the dynamic
behavior of particulate systems such as hoppers, silos, mills, mixers,
separators, solid and fluid bed. Calculations are based on the solution of
ordinary differential equations of motion and the consideration of interactions of
individual particles.
Verlet integration of dynamic equations
d 2x

A( x )
2
dt
forces acting upon a particle/
acceleration
mass of particle
xk 1  2 xk  xk 1
t 2
 A( xk )
second derivative substituted by second differences
xk 1  2 xk  xk 1  t 2 A( xk )  O (t 4 )
NAP4
Theory of chaos strange attractor
Trajectories of particles calculated also the meteorologist E.Lorenz (1963)
who tried to model the natural convection in atmosphere, the layer of air
that is heated from below and cooled from above.
After drastic simplifications he obtained the three ordinary differential
equations for the coordinates x, y, z of a particle in the atmosphere
dx
 Ra ( y  x)
dt
dy
  xz  Pr x  y
dt
dz
 xy  bz
dt
Rayleigh-Bénárdova
instability
The model has only 3 parameters, Rayleigh number Ra, Prandtl number Pr
and the parameter b is a slenderness ratio of rotating cells.
These 3 equations can be solve easily in MATLAB see next slide…
NAP4
Theory of chaos strange attractor
50
45
trajektory pf particle
function dy = chaos(t,y)
fopr Ra=10, Pr=28,
b=8/3
dy = zeros(3,1);
dy(1) = 10*(y(2)-y(1));
dy(2) = -y(1)*y(3)+28*y(1)-y(2);
dy(3) = y(1)*y(2)-8/3*y(3);
40
35
30
25
20
15
10
sol=ode45(@chaos,[0 30],[.10001 .1 .1])
5
0
-30
integration up to 30 s
initial
coordinates
x,y,z
plot(sol.y(2,:),sol.y(3,:))
-10
0
10
-20
-10
0
10
20
30
50
45
40
plot graph y-z fro automatically
calculated integration steps
tt=linspace(0,30,3000);
sy=deval(sol,tt);
plot(sy(2,:),sy(3,:))
-20
35
30
25
20
15
10
5
0
-30
20
30
NAP4
Theory of chaos strange attractor
Numerically calculated trajectories behave erratically for Pr> 1 (they do not
converge to a trivial steady steate solution x = y = z = 0), but they create
something that looks like a counter-rotating swirls of final dimensions, and
these seemingly random trajectories "jump" from the left to the right wing,
and never intersect. The trajectories are extremely sensitive to initial
conditions and to any disturbances (including approximation errors of
numerical integration).
This limiting state is called „strange attractor“
parametric trajectory of particle
z
Such behavior (deterministic chaos)
appears at nonlinear systems when a
stability limit is exceeded (eg Prantl,
Reynolds or Rayleigh numbers).
Deterministic chaos is typical for
turbulent flow.
50
45
40
35
30
25
20
15
10
5
0
-30
y
-20
-10
0
10
20
30
NAP 4
EXAM
ODE Initial problem
numerical solutions, stability
NAP4
Exam - remember

Fourier transform of derivatives
dc 2 ift
 dt e dt  2 ifc( f )
What is it initial problem
What is it order of accuracy O(n)
Explicit and implicit Euler method
ck 1  ck  f (tk , ck )
ck 1  ck  f (tk 1 , ck 1 )
Principles of RK and multistep methods
(predictor, corrector)
Principle of stability analysis
Principle of integration step optimisation