Post on 08-Dec-2016
transcript
Chapter 8Modeling: Simulation Examplesfor Distributed Parameters Processes
8.1 The Performance Indicator of Numerical Integration
Let us consider equations, or systems of equations with partial derivatives, usedespecially in engineering applications.
From the three types of solutions, general, particular or singular, the particularsolutions, respectively, usually the exponential or polynomial ones can approxi-mate quite well the numerous processes, from the control systems of electro-energetic, thermo-energetic, chemical and electromechanical processes.
For the performance indicator of numerical integration, we will also consider‘‘the cumulative relative error in percent’’ (crep) of the form:
crep y ¼ crep x0...0 ¼ 100 �Pkf
k0Dxkj j
Pkf
k0yANKj j
ð8:1Þ
where yAN; k
� �is the particular analytical solution (or in the optimum case, gen-
eral) and Dxk ¼ x0...k � yAN; k is the error of the solution numerically approximated(x0…k) compared with the analytical solution (yAN, k). Both components at thenominator and denominator of (8.1) are considered as having absolute values. The
notation ðPkf
k0Þ symbolizes the iterative sum of all sequences of calculus, from
k0 ¼ t0=Dt to kf ¼ tf =Dt, where (t0) and (tf) correspond to the initial and finalmoment, and (Dt) is the integration step, considered to be small enough.
Example PDE I.2 is of the form:
a00 � yþ a10oy
otþ a01
oy
op¼ uðt; pÞ ð8:2Þ
or
a00x00 þ a10x10 þ a01x01 ¼ u00: ð8:3Þ
T. Colos�i et al., Numerical Simulation of Distributed Parameter Processes,DOI: 10.1007/978-3-319-00014-5_8, � Springer International Publishing Switzerland 2013
83
We obtain
x10 ¼1
a10½u00 � ða00x00 þ a01x01Þ� ð8:4Þ
from which the partial derivatives are calculated:
x1þT ;P ¼1
a10½uTP � ða00xTP þ a01xT ;1þPÞ� ð8:5Þ
for P ¼ 1; 2; . . .30 and T ¼ 1; 2; . . .6. It results in the ‘‘matrix with partialderivatives of the state vector’’:
= =Ppdx
T TP
x xM
x x
30,6626160
30,2222120
30,1121110
30,0020100
xxxx
xxxx
xxxx
xxxx
ð8:6Þ
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC = x (t0, p), from which, by means of analytical derivative with respectto (p), we obtain all the elements of the line vector xP (t0, p). The transfer from thesequence (k - 1) to the sequence (k) is obtained:
xk ffi xk�1 þX6
T¼1
DtT
T !xT; k�1 ð8:7Þ
and
xP;k ffi xP;k�1 þX6
T¼1
DtT
T!xTP;k�1: ð8:8Þ
It was operated with the particular solution:
yANðt; pÞ ¼ y00 þ ðJ0T þ J1T � e�t=T1 þ J2T � e�t=T2ÞðJ0P þ J1P � e�p=P1 þ J2P
� e�p=P2Þ � Ku � u ð8:9Þ
where : y00 ¼ 1; t0 ¼ 0; tf ¼ 10; p0 ¼ 0; pf ¼ 10; Ku ¼ 100; u ¼ 10; kt ¼ 1:5;
lt ¼ 3; kp ¼ 2; lp ¼ 3; T1 ¼ tflt �ð1þktÞ ; T2 ¼ kt � T1; P1 ¼ pf
lp�ð1þkpÞ ; P2 ¼ kp � P1;
J0T ¼ 1; J1T ¼ � T1T1�T2
; J2T ¼ � T2T2�T1
; J0P ¼ 1; J1P ¼ � P1P1�P2
; J2P ¼ � P2P2�P1
:
Using the method of calculus presented in (4.3), for the particular solution (8.9),a00 = 1, 0.1 B a10 B 10 and 0.1 B a01 B 10 with the integration step Dt = 0.1,according to the program EDPTL 32(33), we obtained crep x00 B 3 9 10-5.
84 8 Modeling: Simulation Examples
Example PDE I.3 is of the form:
a000 � yþ a100 �oy
otþ a010 �
oy
opþ a001 �
oy
oq¼ u t; p; qð Þ ð8:10Þ
or
a000 � x000 þ a100 � x100 þ a010 � x010 þ a001 � x001 ¼ u000: ð8:11Þ
We obtain
x100 ¼1
a100u000 � a000 � x000 þ a010 � x010 þ a001 � x001ð Þ½ � ð8:12Þ
from which the partial derivatives are calculated:
x1þT ;P;Q ¼1
a100uTPQ � a000xTPQ þ a010xT ; 1þP;Q þ a001xT ;P; 1þQ
� �� �ð8:13Þ
for P = 0, 1,…5, Q = 0, 1, …5 and T = 0, 1, …6. It results in the ‘‘matrix withpartial derivatives of the state vector’’:
Mpdx TPQT
PQ
xx
xx
655
650
625
620
615
610
605
601
600
255
250
225
220
215
210
205
201
200
155
150
125
120
115
110
105
101
100
055
050
025
020
015
010
005
001
000
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
ð8:14Þ
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC = x (t0, p, q), from which, by analytical derivatives, we obtain all the
8.1 The Performance Indicator of Numerical Integration 85
elements of the line vector xPQ (t0, p, q). The transfer from the sequence (k - 1) tothe sequence (k) results from
xk ffi xk�1 þX6
T¼1
DtT
T !xT;k�1 ð8:15Þ
and
xPQ; k ffi xPQ;k�1 þX6
T¼1
DtT
T!xTPQ;k�1: ð8:16Þ
It was operated with the particular solution:
yANðt; p; qÞ ¼ y000 þ J � e� tT � e�
pP � e�
qQ ð8:17Þ
where: y000 ¼ 1; t0 ¼ 0; tf ¼ 10; p0 ¼ 0; pf ¼ 10; q0 ¼ 0; qf ¼ 10; J ¼ Ku � u;Ku ¼ 100; u ¼ 10; T ¼ 10; P ¼ 100; Q ¼ 20:
Using the method of calculus presented in (4.3), for the particular solution(8.17), a000 ¼ 1; a100 ¼ 1; a010 ¼ 1, with the integration step Dt = 0.5, accordingto the program EDPTL 35(36), we obtained the results shown in Table 8.1 and inFig. 8.1.
Table 8.1
p = pf tk 0.5 2.5 4.5 6.5 8.5 10.5x000k 532 437 360 296.5 244.6 202
q = qf crep x 2 � 10-5 1.6 � 10-5 1.5 � 10-5 1.3 � 10-5 1.3 � 10-5 1.2 � 10-5
p = 0 tk 0.5 2.5 4.5 6.5 8.5 10.5x000k 961 789 647 532 437.4 360
q = 0 crep x 6.3 � 10-6 4 � 10-6 2.5 � 10-6 4.5 � 10-6 6.2 � 10-6 6.6 � 10-6
0 1 2 3 4 5 6 7 8 9 100
100200300400500600700800900
10001100
t
Xoo
ok
p=pf ; q=qf
p=0 ; q=0
Fig. 8.1 The graphicalrepresentation of theTable 8.1 [x000k (t) for(p = pf; q = qf) and (p = 0;q = 0)]
86 8 Modeling: Simulation Examples
The integration domain is framed between the limits x000 tk; 0; 0ð Þ andx000 tk; pf ; qf
� �resulting crep x totally negligible. The range orders for crep x remain
comparable with intermediary values of the variables (p) and (q).Example PDE I.4 is of the form:
a0000yþ a1000 �oy
otþ a0100 �
oy
opþ a0010 �
oy
oqþ a0001 �
oy
or¼ uðt; p; q; rÞ ð8:18Þ
or
a0000x0000 þ a1000x1000 þ a0100x0100 þ a0010x0010 þ a0001x0001 ¼ u0000 ð8:19Þ
We obtain
x1000 ¼1
a1000½u0000 � ða0000x0000 þ a0100x0100 þ a0010x0010 þ a0001x0001Þ� ð8:20Þ
from which the partial derivatives are calculated:
x1þT1;PQR ¼1
a1000½uTPQR � ða0000xTPQR þ a0100xT ; 1þP;QR þ a0010xTP; 1þQ;R
þ a0001xTPQ; 1þRÞ� ð8:21Þ
for P ¼ 0; 1; . . .6; Q ¼ 0; 1; . . .6; R ¼ 0; 1; . . .6 and T ¼ 0; 1; . . .6:It results in the ‘‘matrix with partial derivatives of the state vector’’:
TPQRT
PQRpdx xx
xxM
PQR66000
PQR22000
PQR11000
PQR00000
xx
xx
xx
xx
ð8:22Þ
which has a more laborious form. The components (xPQR) and (xTPQR) are linevectors that contain all the possible combinations of the partial derivatives withrespect to the variables (p), (q) and (r), from the order (0) to the order (6)beginning with (r), then (q) and then (p), for instance
xTPQR ¼ xT; ð0�6Þ;ð0�6Þ;ð0�6Þ: ð8:23Þ
The elements of the line vector xPQR = x0PQR result from (8.23) for T = 0. Theelements of the matrix (xTPQR) also result from (8.23), for T = 1, then T = 2 andso on, until T = 6, fulfilling all the possible combinations of the above-mentionedpartial derivatives.
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC (t0, p, q, r), from which, by analytical derivatives, we obtain all theelements of the line vector xPQR (t0, p, q, r). The transfer from the sequence (k - 1)to the sequence (k) results from
8.1 The Performance Indicator of Numerical Integration 87
xk ffi xk�1 þX6
T¼1
DtT
T!xT;k�1 ð8:24Þ
and
xPQR;k ffi xPQR;k�1 þX6
T¼1
DtT
T !xTPQR;k�1: ð8:25Þ
It was operated with the particular solution:
yANðt; p; q; rÞ ¼ y0000 þ e� t
T1þ p
P1þ q
Q1þ r
R1
� �
� Ku � u ð8:26Þ
where: y0000 ¼ 1; t0 ¼ 0; p0 ¼ 0; q0 ¼ 0; r0 ¼ 0; tf ¼ 1; pf ¼ 1; qf ¼ 1; rf ¼1; Ku ¼ 1; u ¼ 1; a0000 ¼ 1; a1000 ¼ T1 ¼ tf =4; a0100 ¼ P1 ¼ pf =4; a0010 ¼ Q1 ¼qf =4; a0001 ¼ R1 ¼ rf =4:
Using the method of calculus presented in (4.3), for the particular solution(8.26), and the integration step Dt = 0.01, according to the program RBPD 1(2),we obtained the data presented in Table 8.2 and in Fig. 8.2.
It is noticeable that the entire domain of existence of the solution (analytical ornumerical) for y0000 = 0 frames in a hypercube, with the four unitary sides(respectively, tf, pf, qf, rf). By properly changing these final values, as well as thevalues (Ku) and (u), we can of course modify the existence space of thesesolutions.
Example PDE II.2 is of the form:
a00yþ a10oy
otþ a01
oy
opþ a20
o2y
ot2þ a11
o2y
otopþ a02
o2y
op2¼ uðt; pÞ ð8:27Þ
Table 8.2
tk 0.01 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x0000k 1.928 1.919 1.907 1.888 1.859 1.817 1.754 1.661 1.521 1.337 1.037crep 5 � 10-6 0.002 0.006 0.01 0.013 0.016 0.019 0.020 0.021 0.020 0.0194
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.20.40.60.8
11.21.41.61.8
2
t
Xoo
oo
Fig. 8.2 The graphicalrepresentation of theTable 8.2: x0000k (t)
88 8 Modeling: Simulation Examples
or
a00x00 þ a01x10 þ a01x01 þ a20x20 þ a11x11 þ a02x02 ¼ u00: ð8:28Þ
We obtain
x20 ¼1
a20½u00 � ða00x00 þ a10x10 þ a01x01 þ a11x11 þ a02x02Þ� ð8:29Þ
from which the partial derivatives are calculated:
x2þT ;P ¼1
a20½uTP � ða00xTP þ a10x1þT ;P þ a01xT ; 1þP þ a11x1þT ; 1þP þ a02xT ; 2þPÞ�
ð8:30Þ
for P = 0, 1,…30 and T = 0, 1, …6.It results in the ‘‘matrix with partial derivatives of the state vector’’:
TPT
Ppdx xx
xxM
30.8828180
00
30.3323130
30.2222120
30.1121110
30.0020100
xxxx
x
xxxx
xxxx
xxxx
xxxx
ð8:31Þ
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC (t0, p), from which, by analytical derivative, we obtain all the elementsof the two lines of the matrix xP (t0, p). The transfer from the sequence (k - 1) tothe sequence (k) results from
xk ffi xk�1 þX6
T¼1
DtT
T !xT;k�1 ð8:32Þ
and
xP;k ffi xP;k�1 þX6
T¼1
DtT
T !xTP;k�1: ð8:33Þ
It was operated with the particular solution:
yANðt; pÞ ¼ y00 þ ðJ0T þ J1Te�t=T1 þ J2Te�t=T2ÞðJ0P þ J1Pe�p=P1
þ J2Pe�p=P2ÞKuðu0 þ uA sin xtÞ ð8:34Þ
8.1 The Performance Indicator of Numerical Integration 89
where: y00 ¼ 1; t0 ¼ 0; tf ¼ 10; p0 ¼ 0; pf ¼ 10; Ku ¼ 100; u0 ¼ 10; uA ¼ 3;
kt ¼ 1:5; lt ¼ 2; kp ¼ 2; lp ¼ 3; T1 ¼ tfltð1þktÞ ; T2 ¼ ktT1; P1 ¼ pf
lpð1þkpÞ ; P2 ¼ kp
P1; x ¼ 2pT1þT2
; J0T ¼ 1; J1T ¼ � T1T2�T1
; J2T ¼ � T2T2�T1
; J0P ¼ 1; J1P ¼ � P1P1�P2
;
J2P ¼ � P2P2�P1
:
Using the method of calculus presented in (4.3), for the particular solution(8.34), a00 = 1, and the coefficients (a10, a01, a20, a11 and a02) with values betweenthe limits (0.1; 10), with the integration step Dt = 0.1, according to the programEDPTL 90(91), we have obtained the results shown in Table 8.3 and in Fig. 8.3.We can notice that the crep x00 within the limits (10-3 7 10-5), that is, com-pletely negligible even for Dt = 0.1, considered to be quite high. The particularsolution (8.34) contains a periodical component overlapped on the exponentialcomponents. For each integration step, we have operated with a number ofn.(1 ? M) = 2 (1 ? 30) = 62 Taylor series, and the final execution time of theprogram (for tf = 10.1) has not exceeded 3 s (computer IP4-2GHz-512MB).
Example PDE II.3 is of the form:
a000 � yþ a200 �o2y
ot2þ a020 �
o2y
op2þ a002 �
o2y
oq2¼ uðt; p; qÞ ð8:35Þ
or
a000 � x000 þ a200 � x200 þ a020 � x020 þ a002 � x002 ¼ u000: ð8:36Þ
We obtain
x200 ¼1
a200� ½u000 � ða000 � x000 þ a020 � x020 þ a002 � x002Þ� ð8:37Þ
from which the partial derivatives are calculated:
x2þT ; P; Q ¼1
a200½uTPQ � ða000 � xTPQ þ a020 � xT ; 2þP; Q þ a002 � xTP; 2þQÞ� ð8:38Þ
for P ¼ 0; 1; . . .7; Q ¼ 0; 1; . . .7 and T ¼ 0; 1; . . .6:
90 8 Modeling: Simulation Examples
Tab
le8.
3
t0.
11.
12.
13.
24.
25.
26.
27.
28.
29
10.1
p kp0
=0
x 00
1cr
epx 0
010
-4
10-
3
p kp0
=p f
/2x 0
02.
514
822
842
774
858
354
493
082
154
583
0cr
epx 0
05�1
0-3
2�1
0-4
7�1
0-5
4�1
0-5
3�1
0-5
3�1
0-5
5�1
0-5
4�1
0-5
4�1
0-5
4�1
0-5
6�1
0-5
p kp0
=p f
x 00
2.85
180
279
522
913
712
665
1137
1004
666
1.01
5cr
epx 0
06�1
0-3
10-
47�1
0-5
5�1
0-5
5�1
0-5
5�1
0-5
5�1
0-5
4�1
0-5
4�1
0-5
3�1
0-5
6�1
0-5
8.1 The Performance Indicator of Numerical Integration 91
It results in the partial derivatives matrix of the state vector:
ð8:39Þ
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC (t0, p, q), from which, by analytical derivative, we obtain all theelements of the two lines of the matrix xPQ (t0, p, q). The transfer from thesequence (k - 1) to the sequence (k) results from
xk ffi xk�1 þX7
T¼1
DtT
T !xT;k�1 ð8:40Þ
xPQ; k ffi xPQ;k�1 þX7
T¼1
DtT
T !xTPQ;k�1 ð8:41Þ
It was operated with the particular solution:
yANðt; p; qÞ ¼ y000 þ J � e� tTþ
pPþ
qQð Þ ð8:42Þ
0 1 2 3 4 5 6 7 8 9 100
200
400
600
800
1000
1200
tX
oo
pkpo=0pkpo=Pf/2pkpo=Pf
Fig. 8.3 The graphicalrepresentation of theTable 8.3: x00 (t) forpkp0 = 0, pkp0 = pf/2 andpkp0 = pf
92 8 Modeling: Simulation Examples
where: y000 ¼ 1; t0 ¼ 0; tf ¼ 10; p0 ¼ 0; pf ¼ 10; q0 ¼ 0; qf ¼ 10; J ¼ Ku�u; Ku ¼ 100; u ¼ 1; T ¼ 10; P ¼ 10; Q ¼ 10:
Using the method of calculus presented in (4.3), for the particular solution(8.42), with the coefficients (0.1 B a… B 10), with the integration step Dt = 0.1,according to the program EDPTL 48(49), we obtained the results exemplified inTable 8.4 and in Fig. 8.4.
The integration domain is framed between the limits x000(tk, 0, 0) and x000(tk, pf,0), resulting (crep x000) totally negligible, even if the integration step (Dt) is quitehigh.
Example PDE II.4 is of the form:
a0000 � yþ a1000 �oy
otþ a0100 �
oy
opþ a0010 �
oy
oqþ a0001 �
oy
orþ a2000 �
o2y
ot2
þ a1100 �o2y
otopþ a0200 �
o2y
op2þ a0110 �
o2y
opoqþ a0020 �
o2y
oq2þ a0011 �
o2y
oqor
þ a0002 �o2y
or2þ a1001 �
o2y
otorþ a1010 �
o2y
otoqþ a0101 �
o2y
opor¼ uðt; p; q; rÞ
ð8:43Þ
respectively,
0 2 4 6 8 10 120
20
40
60
80
100
120
t
Xoo
ok
p=pf ; q=qfp=0 ; q=0p=pf ; q=0
Fig. 8.4 The graphicalrepresentation of theTable 8.4 [x000k (t) forp = pf; q = qf, p = 0; q = 0and p = pf; q = 0]
Table 8.4
tk 0.1 2.1 4.2 6.2 8.2 10.1
p ¼ pf
q ¼ qf
x000k 14.4 11.9 9.9 8.3 7 5.9crep x000 6 � 10�6 1:2 � 10�5 2 � 10�5 5 � 10�5 2 � 10�7 8 � 10�7
p ¼ 0q ¼ 0
x000k 100 82 66.7 54.8 45 37.4crep x000 0 2 � 10�5 6 � 10�5 10�4 3 � 10�4 9 � 10�4
p ¼ pf
q ¼ 0x000k 37.4 30.8 25.1 20.8 17.2 14.4crep x000 0 2 � 10�5 7 � 10�5 10�4 2 � 10�4 9 � 10�4
8.1 The Performance Indicator of Numerical Integration 93
a0000 � x0000 þ a1000 � x1000 þ a0100 � x0100 þ a0010 � x0010 þ a0001 � x0001
þ a2000 � x2000 þ a1100 � x1100 þ a0200 � x0200 þ a0110 � x0110
þ a0020 � x0020 þ a0011 � x0011 þ a0002 � x0002 þ a1001 � x1001
þ a1010 � x1010 þ a0101 � x0101 ¼ u0000
: ð8:44Þ
We obtain
x2000 ¼1
a2000½u0000 � ða10000 � x0000 þ a1000 � x1000 þ a0100 � x0100
þ a0010 � x0010 þ a0001 � x0001 þ a1100 � x1100
þ a0200 � x0200 þ a0110 � x0110 þ a0020 � x0020
þ a0011 � x0011 þ a0002 � x0002 þ a1001 � x1001
þ a1010 � x1010 þ a0101 � x0101Þ�
ð8:45Þ
from which the partial derivatives are calculated:
x2þT ;PQR ¼1
a2000½uTPQR � ða1000 � x1þT ;PQR þ a0100 � xT ;Pþ1;QR
þ a0010 � xTP;Qþ1;R þ a0001 � xTPQ;1þR
þ a1100 � x1þT ;1þP;QR þ a0200 � xT ;2þP;QR
þ a0110 � xT ;1þP;1þQ;R þ a0020 � xTP;2þQ;R
þ a0011 � xTP;1þQ;1þR þ a0002 � xTPQ;2þR
þ a1001 � x1þT ;PQ;1þR þ a1010 � x1þT ;P;1þQ;R
þ a0101 � xT ;1þP;Q;1þRÞ�
ð8:46Þ
for P ¼ 0; 1; . . .6; Q ¼ 0; 1; . . .6; R ¼ 0; 1; . . .6 and T ¼ 0; 1; . . .6:It results in the ‘‘matrix with partial derivatives of the state vector’’:
MpdxPQR
T TPQR
x x
x x
PQR6
PQR3
PQR2
6000
3000
2000
PQR1
PQR0
1000
0000
x
x
x
x
x
xx
x
x
x
ð8:47Þ
which has a more complex form. The components of the (xPQR) and (xTPQR) areline vectors that contain all the possible combinations of the partial derivativeswith respect to the variables (p) (q) and (r), from the order (0) to the order (6)
94 8 Modeling: Simulation Examples
beginning with (r), then (q) and then (p). The details of calculations and theinterpretation of the results are formally identical with those presented at (8.23)(8.24) and (8.25), with respect to (8.47).
We have computed the analytical solution:
yAN t; p; q; rð Þ ¼ y0000 þ ðJ0T þ J1T � e�t=T1 þ J2T � e�t=T2Þ� ðJ0P þ J1P � e�p=P1 þ J2P � e�p=P2Þ� ðJ0Q þ J1Q � e�q=Q1 þ J2Q � e�q=Q2Þ� ðJ0R þ J1R � e�r=R1 þ J2R � e�r=R2Þ � Ku � u
ð8:48Þ
where: y0000 ¼ 1; t0 ¼ 0; p0 ¼ 0; q0 ¼ 0; r0 ¼ 0; tf ¼ 1; pf ¼ 1; qf ¼ 1;
rf ¼ 1; T1 ¼ 0:15 � tf ; T2 ¼ 0:1 � tf ; P1 ¼ 0:15 � pf ; P2 ¼ 0:1 � pf ; Q1 ¼ 0:15�qf ; Q2 ¼ 0:1 � qf ; R1 ¼ 0:15 � rf ; R2 ¼ 0:1 � rf ; Ku ¼ 1; u ¼ 1; J0T ¼ J0P ¼J0Q ¼ J0R ¼ 1; p ¼ pf ; q ¼ qf ; r ¼ rf ; J1T ¼ � T1
T1�T2; J2T ¼ � T2
T2�T1; J1P ¼
� P1P1�P2
; J2P ¼ � P2P2�P1
; J1Q ¼ � Q1Q1�Q2
; J2Q ¼ � Q2Q2�Q1
; J1R ¼ � R1R1�R2
;
J2R ¼ � R2R2�R1
:
Using the method of calculus presented in (4.3), for the particular solution(8.48), all a… = 1 with the integration step Dt = 0.01, according to the programRBPD 3(4), we obtained the data presented in Table 8.5 and in Fig. 8.5.
In this case, the entire domain of existence of the solution (analytical ornumerical) for y0000 = 0 frames in a hypercube is also noticeable, with the fourunitary sides (respectively, tf, pf, qf, rf). By properly changing these final values, aswell as the values (Ku) and (u), we can of course modify the space of existence ofthese solutions. Even if the order of the partial derivatives with respect to (t), (p),(q) and (r) has been limited to six, the values crep x0000 are maintained quite low.
Example PDE III.2 is of the form:
a00 � y þ a30 �o3y
ot3þ a03 �
o3y
op3¼ uðt; pÞ ð8:49Þ
or
a00 � x00 þ a30 � x30 þ a03 � x03 ¼ u00 ð8:50Þ
We obtain
x30 ¼1
a30½u00 � ða00 � x00 þ a03 � x03Þ� ð8:51Þ
from which the partial derivatives are calculated:
x3þT ;P ¼1
a30½uTP � ða00 � xTP þ a03 � xT ;3þPÞ� ð8:52Þ
for P = 0, 1,…20 and T = 6.
8.1 The Performance Indicator of Numerical Integration 95
Tab
le8.
5
t k0.
010.
10.
20.
30.
40.
50.
60.
70.
80.
91
x 0000
K1.
0015
1.12
41.
297
1.47
81.
622
1.72
61.
794
1.84
51.
876
1.89
71.
912
crep
x 0000
01.
6�1
0-5
10-
42.
6�1
0-4
3�1
0-4
1.1�1
0-4
3.5�1
0-3
8�1
0-3
1.4�1
0-2
1.8�1
0-2
1.9�1
0-2
96 8 Modeling: Simulation Examples
It results in the ‘‘matrix with partial derivatives of the state vector’’:
TPT
Ppdx xx
xxM
20.88281
20.44241
20.33231
80
40
30
20.22221
20.11211
20.00201
20
10
00
xxx
xxx
xxx
x
x
xxxx
xxx
xxx
x
xx
ð8:53Þ
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC = x (t0, p), from which, by analytical derivatives, we obtain all theelements of the three lines of the matrix xP (t0, p). The transfer from the sequence(k - 1) to the sequence (k) results from expressions formally identical to (8.32)and (8.33). It was operated with the particular solution (8.9) with the same calculusdetails for the parameters y00; t0; tf ; p0; pf ; Ku; u; lt; kp; . . .:J1P; J2P).
Using the method of calculus presented in (4.3), for the particular solution (8.9)and the integration step Dt = 0.1, according to the program EDPTL 55(56), wehave obtained crep x00, as exemplified in Table 8.6, for different combinations ofthe coefficients (a00), (a30) and (a03). For this example, the values for crep x00 arenegligible.
Example PDE III.3 is of the form:
a000 � y þ a300 �o3y
ot3þ a030 �
o3y
op3þ a003 �
o3y
oq3¼ uðt; p; qÞ ð8:54Þ
or
a000 � x000 þ a300 � x300 þ a030 � x030 þ a003 � x003 ¼ u000 ð8:55Þ
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.20.40.60.8
11.21.41.61.8
2
tX
oooo
k
Fig. 8.5 The graphicalrepresentation of theTable 8.5 [x0000k (t)]
8.1 The Performance Indicator of Numerical Integration 97
We obtain
x300 ¼1
a300½u000 � ða000 � x000 þ a030 � x030 þ a003 � x003Þ� ð8:56Þ
from which the partial derivatives (8.58) are calculated:
ð8:57Þ
x3þT ;PQ ¼1
a300½uTPQ � ða000 � xTPQ þ a030 � xT ;3þP;Q þ a003 � xTP;3þQÞ� ð8:58Þ
for P = 0, 1,…7, Q = 0, 1, …7 and T = 0, 1, …7.It results in the ‘‘matrix with partial derivatives of the state vector,’’ presented
in (8.57).
Table 8.6
a30 a03 a00 crep x00
p = pf p = 0.5 pf p = 0
0.1 0.1 0 1.7 � 10-4 1.8 � 10-4 5.3 � 10-3
1 7.3 � 10-3 1.1 � 10-2 3.6 � 10-3
1 0.1 0 1.7 � 10-4 1.6 � 10-4 7.5 � 10-5
1 6 � 10-5 2.5 � 10-4 01 1 0 1.7 � 10-4 1.8 � 10-4 5.3 � 10-3
1 4.7 � 10-5 2.6 � 10-4 3.4 � 10-3
10 1 0 1.7 � 10-4 1.6 � 10-4 7.5 � 10-5
1 7.8 � 10-5 8 � 10-5 6 � 10-5
10 10 0 1.7 � 10-4 1.7 � 10-4 5.3 � 10-3
1 8 � 10-5 6.4 � 10-5 4.4 � 10-3
98 8 Modeling: Simulation Examples
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC = x (t0, p, q), from which, by analytical derivatives, we obtain all theelements of the two lines of the matrix xPQ (t0, p, q).
The transfer from the sequence (k - 1) to the sequence (k) results from
xk ffi xk�1 þX7
T¼1
DtT
T !xT;k�1 ð8:59Þ
xPQ;k ffi xPQ;k�1 þX7
T¼1
DtT
T!xTPQ;k�1: ð8:60Þ
It was operated with the particular solution (8.42), with the same values of theparameters and the integration limits.
Using the method of calculus presented in (4.3), for the particular solution(8.42), with the coefficients (0.1 B a… B 10), with the integration step Dt = 0.1,according to the program EDPTL 62(63), we have obtained crep x000 = (10-6
7 10-2), with values that are practically identical to those presented in Table 8.4.Example PDE IV.2 is of the form:
a00 � yþ a40 �o4y
ot4þ a04 �
o4y
op4¼ uðt; pÞ ð8:61Þ
or
a00 � x00 þ a40 � x40 þ a04 � x04 ¼ u00 ð8:62Þ
We obtain
x40 ¼1
a40� ½u00 � ða00x00 þ a04x04Þ� ð8:63Þ
from which the partial derivatives are calculated:
x4þT ;P ¼1
a40� ½uTP � ða00xTP þ a04xT ;4þPÞ� ð8:64Þ
for P = 0, 1,…, 20 and T = 0, 1,…, 9.
8.1 The Performance Indicator of Numerical Integration 99
It results in the ‘‘matrix with partial derivatives of the state vector’’:
ð8:65Þ
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC (t0, p), from which, by means of analytical derivatives, we obtain allthe elements of the matrix xP (t0, p). The transfer from the sequence (k - 1) to thesequence (k) results from
xk ffi xk�1 þX9
T¼1
DtT
T !xT;k�1 ð8:66Þ
Table 8.7
a40 a04 a00 crep x00
p ¼ pf p ¼ 0:5 � pf p ¼ 0
1 1 0 1:7 � 10�4 2 � 10�4 6:5 � 10�4
1 8:3 � 10�4 6:6 � 10�4 9 � 10�4
10 1 0 1:3 � 10�4 1:6 � 10�4 9 � 10�5
1 1:8 � 10�4 1:7 � 10�4 1:2 � 10�4
1 10 0 1:8 � 10�4 1:1 � 10�3 6:6 � 10�2
1 8:1 � 10�4 3:3 � 10�4 8 � 10�2
10 10 0 1:7 � 10�4 1:8 � 10�4 6:1 � 10�4
1 1:7 � 10�4 1:5 � 10�4 5:6 � 10�4
0.1 0.1 0 1:8 � 10�4 2:6 � 10�4 3:8 � 10�4
1 2:8 � 10�2 2:6 � 10�2 2:8 � 10�3
1 0.1 0 1:7 � 10�4 1:6 � 10�4 1:6 � 10�4
1 8:3 � 10�4 7 � 10�4 9:3 � 10�5
0.1 1 0 1:8 � 10�4 1:3 � 10�3 5:3 � 10�2
1 3 � 10�2 2:4 � 10�2 10�1
100 8 Modeling: Simulation Examples
xP;k ffi xP;k�1 þX9
T¼1
DtT
T !xTP;k�1 ð8:67Þ
It was operated with the particular solution (8.9), with the same calculus detailsfor the parameters y00; t0; tf ; p0; pf ; Ku; u; lt; kp; . . .J1P; J2P).
Using the method of calculus presented in (4.3), for the particular solution (8.9)and the integration step Dt = 0.1, according to the program EDPTL 60(61), wehave obtained crep x00 presented in Table 8.7, for different combinations of thecoefficients (a00), (a40) and (a04).
Example PDE IV.3 is of the form:
a000 � yþ a400 �o4y
ot4þ a040 �
o4y
op4þ a004 �
o4y
oq4¼ uðt; p; qÞ ð8:68Þ
or
a000 � x000 þ a400 � x400 þ a040 � x040 þ a004 � x004 ¼ u000: ð8:69Þ
We obtain
x400 ¼1
a400� ½u000 � ða000x000 þ a040x040 þ a004x004Þ� ð8:70Þ
from which the partial derivatives are calculated:
x4þT ;P:Q ¼1
a400� ½uTPQ � ða000xTPQ þ a040xT ;4þP;Q þ a004xTP;4þQÞ� ð8:71Þ
for P = 0, 1,…7, Q = 0, 1, …7 and T = 0, 1, …7.It results in the ‘‘matrix with partial derivatives of the state vector’’:
ð8:72Þ
The initial conditions (IC) at t = t0 = 0 and k = k0 = 0 are considered to beknown, xIC (t0, p, q), from which, by analytical derivatives, we obtain all the
8.1 The Performance Indicator of Numerical Integration 101
elements of the matrix xPQ (t0, p, q). The transfer from the sequence (k - 1) to thesequence (k) results from expressions formally identical to (8.59) and (8.60). It wasoperated with the particular solution (8.17), with the same values of the parametersand the integration limits, except for T = 10, P = 10 and Q = 10.
Using the method of calculus presented in (4.3), for the particular solution(8.17) with the coefficients (0.1 B a … B 10), with the integration step Dt = 0.1,according to the program EDPTL 64(65), we have obtained crep x000 = (10-5
7 10-2), with values that are practically identical to those presented in Table 8.7.Example nonlinear PDE II.2 for the approximation of the modeling of an
isotopic separation column for N15 is of the form:
ða00 þ a01yÞ oy
opþ a10
oy
otþ a02;
o2y
op2¼ uðt; pÞ ð8:73Þ
or
ða00 þ a01 � x00Þ x01 þ a10 � x10 þ a02 � x02 ¼ u00 ð8:74Þ
where: (t), (p) and y = y (t, p) represent the time, the height of the column and theconcentration of the isotope N15, respectively.
In the hypothesis of the null productivity, we denote
a00 ¼ ða� 1Þ � L; a01 ¼ 2ða� 1Þ � L; a10 ¼ hþ H and a02 ¼�L2
K
where: a = 1.055; h = 2.8 atom�s N/m3; H = 430 atom�s N/m3; L = 1382.4atom�s N/day; K = 4060.8 atom�s/day�m.
The domains of integration are between the limits: t0 ¼ 0 days; tf ¼ 14 days;p0 ¼ 0 m; pf ¼ 7 m; the initial concentration of 15N is y00 ¼ y t0; p0ð Þ¼ 0:365 mol=m3; the final concentration is yff ¼ y tf ; pf
� �¼ 8:2 mol=m3:
From (8.74), we obtain
x10 ¼1
a10½u00 � ða00 � x01 þ a02 � x02 þ a01 � x00 � x01Þ� ð8:75Þ
from which we compute the partial derivatives (x1+T,P). Due to the nonlinear term(x00 � x01) from (8.75), the calculus of these partial derivatives becomes morecomplex (then, it was in the previous examples for PDE linear), without creatingspecial problems.
As a result, the Eqs. (8.6), (8.7) and (8.8) remain formally valid, equations thatare adapted for the particular solution:
yANðt; pÞ ¼ y00 þ ðJ0T þ JIT � e�t
T1 þ J2T � e�t
T2Þ
ðJop þ J1p � e�p
P1 þ J2p � e�p
P2Þðyff � y00Þð8:76Þ
102 8 Modeling: Simulation Examples
which proves to be very close to the index usual answers of the isotopic separationcolumn which is studied. We have also chosen: J0T ¼ 1; J0P ¼ 1; T1 ¼ 1:3
0 2 4 6 8 10 12 140
1
2
3
4
5
6
7
8
9
t[days]
8.27.956p=7 [m]
p=5 [m]
p=1 [m]
p=2 [m]
p=3 [m]
7.550
6.525
4.343
1.363
y 00[
t, p]
Fig. 8.6 The family of the analytical solutions yAN (t, p)
Table 8.8
t (days) p (m) 2 4 6 8 10 12 14 crep x00
7 yAN 3.367 6.181 7.443 7.918 8.083 8.138 8.156 0.085x00 3.385 6.190 7.447 7.920 8.084 8.138 8.156
6 yAN 3.347 6.142 7.396 7.868 8.032 8.086 8.104 0.085x00 3.365 6.151 7.400 7.869 8.032 8.086 8.104
5 yAN 3.290 6.031 7.261 7.724 7.885 7.939 7.956 0.085x00 3.307 6.041 7.265 7.726 7.886 7.939 7.956
4 yAN 3.133 5.729 6.893 7.331 7.484 7.534 7.550 0.085x00 3.150 5.738 6.897 7.333 7.484 7.534 7.551
3 yAN 2.738 4.963 5.961 6.337 6.467 6.511 6.525 0.084x00 2.753 4.971 5.964 6.338 6.468 6.511 6.525
2 yAN 1.898 3.335 3.979 4.222 4.306 4.334 4.343 0.081x00 1.907 3.339 3.981 4.223 4.306 4.334 4.343
1 yAN 0.749 1.110 1.272 1.333 1.354 1.361 1.363 0.062x00 0.752 1.111 1.272 1.333 1.354 1.361 1.363
0 yAN 0.365 0.365 0.365 0.365 0.365 0.365 0.365 0x00 0.365 0.365 0.365 0.365 0.365 0.365 0.365
Table 8.9
Dt 0.1 0.01 0.005 0.002 0.001 0.0001
crep x00 max 8.5 � 10-1 8.5 � 10-2 4.2 � 10-2 1.7 � 10-2 8 � 10-3 8 � 10-4
8.1 The Performance Indicator of Numerical Integration 103
days; T2 ¼ 1:7 days; P1 ¼ 0:7 m; P2 ¼ 0:9 m; J0T ¼ 1; J1T ¼ � T1T1�T2
; J2T
¼ � T2T2�T1
; J0P ¼ 1; J1P ¼ � P1P1�P2
and J2P ¼ � P2P2�P1
:
The two time constants (T1, T2) and the two ‘‘height constants’’ (P1, P2) havebeen chosen after multiple testing, so that the family of analytical solutions yAN (t,p), qualitatively presented in Fig. 8.6, are getting close to the index of experi-mental answers, appreciated by the specialists in technology.
Using the method of calculus presented in (4.3), for the particular solution(8.76), with n = 1, N = 5, M = 9, and the integration step Dt = 0.01, accordingto the program RBPD 5(6), we obtained the results shown in Table 8.8, for theconcentration numerically approximated (x00) which is compared with the ana-lytical concentration (yAN).
It is also noticeable, in Table 8.9, that the decrease in (Dt) from (10-1) to (10-4)progressively decreases the maximum value (crep), from 0.85 to 8 9 10-4 %, forthe same dimensions of the operator matrix, respectively, Mpdx½ðnþ NÞ �ð1þMÞ� ¼Mpdx½ð1þ 5Þ � ð1þ 9Þ� ¼Mpdx 6� 10ð Þ. Thus, the remarkable
performances of this method are being underlined, method which is applied to theequation with partial derivatives, nonlinear (8.73), even for M = 9, of quite smallvalues.
It needs to be mentioned that if in (8.73) we would consider the right side asbeing null, u (t, p) = 0, then (8.73) would correspond to the Cohen equation, wellknown in the theory and practice of isotopic separations. This equation has been‘‘altered,’’ by u (t, p) = 0, for the particular solution (8.76), close enough to theindex experimental answers.
8.2 Control System of a Process, Modeled by PDE II.2
Let us consider the process with distributed parameters PDE II.2 from (8.27),which is included in the control system, with a controller PID, in Fig. 8.7: where(w), (a), (c), (u), (y) and (yM) correspond to the signals of reference, error, com-mand, action, output (controlled) and reaction, respectively. The coefficients(KPR), (KIR) and (KDR) underline the proportional effects (usually amplification),integrated with respect to time and, respectively, derivatives with respect to time,
cdt
daKadtKaK DRIRPR =∫ ++⋅
MK
VK 2IIEDP
y00
y
yM
aw c u
Fig. 8.7 The block diagram
104 8 Modeling: Simulation Examples
of the controller PID. The actuator and the reaction transducer are considered non-inertial, with the coefficients of proportionality (KV) and (KM).
The process with distributed parameters, designed for control, is defined byPDE II.2 of the form (8.27), respectively,
a00yþ a10oy
otþ a01
oy
opþ a20
o2y
ot2þ a11
o2y
otopþ a02
o2y
op2¼ uðt; pÞ ð8:27Þ
which admits the particular solution:
yðt; pÞ ¼ y ¼ y00 þ ðJOT þ J1T � e�t
T1 þ J2T � e�t
T2Þ
ðJOP þ J1P � e�p
p1 þ J2P � e�p
p2Þ � Ku � uðtÞð8:77Þ
where: y00 = 1, J0T = 1, J0P = 1 and Ku = 1. According to those presented for
(8.34) J1T ¼ �T1
T1 � T2; J2T ¼ �
T2
T2 � T1; J1P ¼ �
P1
P1 � P2and J2P ¼
� P2P2�P1
; kt ¼ 1:5; lt ¼ 2; kp ¼ 2; lp ¼ 3; tf ¼ 10; pf ¼ 10; T1 ¼ tfltð1þktÞ
¼ 2; T2 ¼ kt � T1 ¼ 3; P1 ¼ pf
lp�ð1þkpÞ ¼109 and P2 ¼ kp � P1 ¼ 20
9 :
Besides the two time constants (T1 and T2), the two coefficients (P1 and P2) wecould interpret, for instance, ‘‘length constants’’ for some of the spatial–temporalprocesses (thermo-energetic, chemistry, etc.)
Using the method of calculus presented in (4.3) and avoiding the presentation ofsome details of numerical routine calculus [for the integration component of thecontroller from Fig. (8.2)], in the program EDPM 1(2), there have been developedlots of control regimes, out of which we exemplify the following:
1. Checking for particular solution of the control in a proportional regime(KIR = 0; KDR = 0)
Due to the fact that PDE II.2 from (8.27), with the particular solution (8.34), hasbeen verified according to the results in Table 8.3 [and in (8.77) we can chooseu = u0 ? uA � sin xt], we will use this particular solution in the control loop inFig. 8.2. For the simplification of the presentation, this particular solution notes
yðt; pÞ ¼ y ¼ y00 þ Fðt; pÞ � Ku � uðtÞ ð8:78Þ
where
Fðt; pÞ ¼ ðJ0T þ J1T � e�tT1þJ2T � e
�tT2ÞðJ0P þ J1P � e
�PP1 þJ2P � e
�PP2 Þ ð8:79Þ
After including (8.78) in the control loop, results:
y ¼ y00 þ Fðt; pÞ � KPR � Ku � Kvðw� KM � yðt; pÞÞ ð8:80Þ
where
a ¼ w � KM � y ð8:81Þ
8.2 Control System of a Process, Modeled by PDE II.2 105
is the control error.Thus, it results
y ¼ y00 þ Fðt; pÞ � KPR � Ku � Kv � w1þ Fðt; pÞ � KPR � Ku � Kw � KM
ð8:82Þ
0 5 10 15 20 250
20
40
60
80
100
120
t
Xoo
k
WA=30 ; p=10WA=30 ; p=5WA=30 ; p=1
Fig. 8.9 The graphicalrepresentation of theTable 8.10 (x00k (t) forwA = 30; p = 10, wA = 30;p = 5 and wA = 30; p = 1)
Table 8.10
tk 0.05 1.05 3.1 5.05 7.05 9.05 11 13.05 15.05 20.1 25
wA = 0 p = 10 x00K 1 7.8 36.2 60.4 77.2 83.3 93 96.1 97.8 99.3 99.6
crep x00K 10-4 10-4 10-5 10-5 10-5 10-5 10-5 10-5 10-5 10-5 10-5
p = 5 x00K 1 6.8 31 51.8 66.1 74.7 79.6 82.2 83.7 84.9 85.2
crep x00K 10-4 10-4 10-5 10-5 10-5 10-5 10-5 10-5 10-5 10-5 10-5
p = 1 x00K 1 2.06 6.5 10.3 12.9 14.5 15.4 15.9 16.2 16.4 16.5
crep x00K 10-5 10-4 10-5 10-5 10-5 10-5 10-4 10-4 10-4 10-4 10-4
wA = 30 p = 10 x00K 1.02 9.8 29 61.6 89.6 63.2 119.7 77.9 99.6 103 103.3
crep x00K 10-4 10-4 10-5 10-5 10-4 10-4 10-4 10-4 10-4 10-4 10-4
p = 5 x00K 1.02 8.5 24.9 52.7 76.6 54.1 102.4 66.7 85.2 88.1 88.3
crep x00K 10-4 10-4 10-5 10-5 10-4 10-4 10-4 10-4 10-4 10-4 10-4
p = 1 x00K 1 2.4 5.4 10.5 15 10.7 19.6 13.1 16.5 17 17
crep x00K 10-5 10-4 10-5 10-5 10-4 10-4 10-4 10-4 10-4 10-4 10-3
0 5 10 15 20 250
102030405060708090
100
t
Xoo
k WA=0 ; p=10WA=0 ; p=5WA=0 ; p=1
Fig. 8.8 The graphicalrepresentation of theTable 8.10 (x00k (t) forwA = 0; p = 10, wA = 0;p = 5 and wA = 0; p = 1)
106 8 Modeling: Simulation Examples
Table 8.10 and Figs. 8.8 and 8.9 present a few results in open loop, according tothe program EDPM 1(2).
For which the following has been considered: t0 ¼ 0; tf ¼ 25; t ¼ t0; p0 ¼ 0;
pf ¼ 10; Kv ¼ 1; KPR ¼ 1; KM ¼ 0; Ku ¼ 1; KIR ¼ 0; KDR ¼ 0; w ¼ w0
þwA � sin xt; w0 ¼ 100; T1 ¼ 2; T2 ¼ 3; P1 ¼ 1; P2 ¼ 2; x ¼ 2p=s; s ¼T1 þ T2; y00 ¼ 1; J0T ¼ 1; J1T ¼ � T1
T1�T2; J2T ¼ � T2
T2�T1; J0P ¼ 1; J1P ¼
� P1P1�P2
; J2P ¼ � P2P2�P1
; a00 ¼ 1.
All the coefficients (a…) have been chosen between the limits (1 7 10).The integration step is big enough, Dt = 0.05.For wA = 0, the asymptotic evolution can be noticed toward w0 = 100, and for
wA = 30, for this behavior, it overlaps a periodical evolution, with the period (s).
2. Control for constant reference, w = w0 = 100, and the steady-state yST = 100.
For the steady-state (t ? ? and p ? ?), F (t, p) results from (8.79).
Fðt !1; p!1Þ ¼ J0T � J0P � Ku ¼ Ku ð8:83Þ
and
yST ¼ y00 þ Ku � uST ð8:84Þ
Table 8.11
tk 0.05 1.05 3.05 5.05 7.05 9.05 11.05 13.05 15.05 30
KPR = 1 p = 10 x00K 1.02 7.8 36.1 60.1 76.6 86.5 92.1 95.2 96.8 98.6
crep x00 10-4 10-4 10-5 10-5 10-5 10-5 10-5 10-5 10-5 10-5
KIR = 0 p = 8 x00K 1.02 7.6 35.3 58.7 74.9 84.6 90.06 92.93 94.6 96.4
crep x00 10-4 10-4 10-5 10-5 10-5 10-5 10-5 10-5 10-5 10-5
KDR = 0 p = 6 x00K 1.01 7.26 33.1 53.1 70.3 79.4 84.5 87.3 88.8 90.46
crep x00 10-4 10-4 10-5 10-5 10-5 10-5 10-5 10-5 10-5 10-5
KPR = 1 p = 10 x00K 1.02 7.86 36.7 61.15 77.84 87.8 93.4 96.5 98.2 100.013
KIR = 0.0135 p = 8 x00K 1.01 7.7 35.8 59.7 76.1 85.8 91.3 94.3 95.97 97.7
KDR = 0.1 p = 6 x00K 1.01 7.28 33.7 56 71.4 80.5 85.7 88.5 90.06 91.7
0 5 10 15 20 25 300
102030405060708090
100
t
Xoo
k
KPR=1 ; KIR=0; KDR=0; p=10
KPR=1 ; KIR=0; KDR=0; p=8
KPR=1 ; KIR=0; KDR=0; p=6
Fig. 8.10 The graphicalrepresentation of Table 8.11[x00k (t) for (KPR = 1;KIR = 0; KDR = 0; p = 10),(KPR = 1; KIR = 0;KDR = 0; p = 8) and(KPR = 1; KIR = 0;KDR = 0; p = 6)]
8.2 Control System of a Process, Modeled by PDE II.2 107
where
uST ¼ Kv � KPR � a ¼ Kv � KPR � w� KM � ySTð Þ: ð8:85Þ
So
yST ¼ y00 þ Ku � Kv � KPRðw� KM � ySTÞ ð8:86Þ
respectively,
yST ¼y00 þ Ku � Kv � KPR � w1þ Ku � Kv � KPR � KM
ð8:87Þ
where: y00 ¼ 1; Ku ¼ Kv ¼ 1; KM ¼ 0:01; KPR ¼ 1; KIR ¼ 0; KDR ¼ 0; w ¼w0 ¼ 100 and wA = 0. The other structural parameters T1; T2; P1; P2; x;ðs; tf ; J0T ; J1T ; J0P; J1P; J2P and a. . .Þ correspond to the previously presentedcase a).
Table 8.11 and Figs. 8.10 and 8.11 present a few results in closed loop, with thenegative feedback KM = 0.01, according to the program EDPM 3(4).
For the proportional controller, KPR = 1, KIR = 0 and KDR = 0, it results thecontrol error in steady-state regime, equal to 100�98:6
100 � 100 ¼ 1:4 %. For the pro-portional integrative derivative controller, KPR = 1, KIR = 0.0135, andKDR = 0.1, the annihilation of this error practically resulted.
The cumulated relative error in percentages (crep x00) is operated with respectto (8.87), which is valid for the proportional controller.
3. Control for the variable reference, respectively,
w ¼ w0 þ wA sin xt ¼ 100þ 30 sin2p
T1 þ T2
�
ð8:88Þ
The previous example is repeated, for the same data, but the reference corre-sponds to (8.88), w0 = 100 and wA = 30. The results are presented in Table 8.12and in Figs. 8.12 and 8.13.
0 5 10 15 20 25 300
20
40
60
80
100
120
tX
ook
KPR=1 ; KIR=0.0135 ; KDR=0.1; p=10KPR=1 ; KIR=0.0135 ; KDR=0.1; p=8KPR=1 ; KIR=0.0135 ; KDR=0.1; p=6
Fig. 8.11 The graphicalrepresentation of theTable 8.11 [x00k (t) for(KPR = 1; KIR = 0.0135;KDR = 0.1; p = 10),(KPR = 1; KIR = 0.0135;KDR = 0.1; p = 8) and(KPR = 1; KIR = 0.0135;KDR = 0.1; p = 6)]
108 8 Modeling: Simulation Examples
If control is not strictly proportional, that is, KIR = 0 and KDR = 0, then theanalytical solution of the equivalent system does not correspond to (8.82), and thecalculus (crep x00) does not have any sense.
For the program EDPM 5(6) that was used in this example, in the analogicalcontroller–actuator model from Fig. 8.2, respectively,
u ¼ KvðKPR � aþ KIR �Z
a dt þ KDRda
dtÞ ð8:89Þ
0 5 10 15 20 25 300
20
40
60
80
100
120
140
t
KPR=1 ; KIR=0.0135 ; KDR=0.1 ; p=10
KPR=1 ; KIR=0.0135 ; KDR=0.1 ; p=8
KPR=1 ; KIR=0.0135 ; KDR=0.1 ; p=6
Xoo
k
Fig. 8.13 The graphicalrepresentation of theTable 8.12 [x00k (t) forKPR = 1; KIR = 0.0135;KDR = 0.1; p = 10,KPR = 1; KIR = 0.0135;KDR = 0.1; p = 8 andKPR = 1; KIR = 0.0135;KDR = 0.1; p = 6]
0 5 10 15 20 25 300
20
40
60
80
100
120
t
Xoo
k
KPR=1 ; KIR=0 ; KDR=0; p=10
KPR=1 ; KIR=0 ; KDR=0; p=8
KPR=1 ; KIR=0 ; KDR=0; p=6
Fig. 8.12 The graphicalrepresentation of theTable 8.12 (x00k (t) forKPR = 1; KIR = 0; KDR = 0;p = 10, KPR = 1; KIR = 0;KDR = 0; p = 8 andKPR = 1; KIR = 0; KDR = 0;p = 6)
Table 8.12tk 0.05 1.05 3.05 5.05 7.05 9.05 11.05 13.05 15.05 30
KPR = 1 p = 10 x00K 1.02 9.8 28.9 61.2 88.8 62.7 118.6 77.2 98.7 102.3
crep x00 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4
KIR = 0 p = 8 x00K 1.02 9.58 28.2 59.8 86.8 61.3 115.9 75.4 96.4 100.01
crep x00 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4
KDR = 0 p = 6 x00K 1.02 9.04 26.5 56.1 81.4 57.5 108.7 70.8 90.5 93.8
crep x00 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4 10-4
KPR = 1 p = 10 x00K 1.02 9.92 28.56 64.3 87.9 64.8 120.8 76 103.2 106.9
KIR = 0.0135 p = 8 x00K 1.02 9.7 27.9 62.8 85.9 63.4 118 74.3 100.8 104.5
KDR = 0.1 p = 6 x00K 1.02 9.1 26.2 58.9 80.6 59.5 110.8 69.7 94.6 98.06
8.2 Control System of a Process, Modeled by PDE II.2 109
the numerical approximation for the integration component
uI ¼ Kv � KIRZ
a dt ¼ Kv � KIRZ ðw� KM � x00Þdt ð8:90Þ
it has been operated through
uIK ffi uI; K�1 þ Kv � KIR �X4
n¼1
Dtn
n!� dn
dtnðw� KM � x00Þk�1 ð8:91Þ
where: w = w (t) corresponds to (8.88), and (xn0)k-1 frames within the usual stagesof calculus for the matrix (Mpdx).
The results in Table 8.12 have been obtained from the program EDPM 5(6), fory00 = 1, Ku = Kv = 1, KM = 0.01, w0 = 100 and wA = 30. The other structuralparameters (T1, T2, P1, P2, x, s, tf, J0T, J1T, J2T, J0P, J1P, J2P and a…), correspondto the previously presented cases (a) and (b).
Due to the fact that wA = 30, it can be noticed that (x00K) presents a periodicalevolution (s = T1 ? T2), overlapped on the asymptotic evolution obtained inEDPM 3(4) with wA = 0. In this case, the relative error cumulated in percentages(crep x00) is operated with respect to (8.87), which is valid for the proportionalcontroller.
8.3 Control System of a Process, Modeled by PDE II.3
It is considered that the complete form defined by PDE II.3 of a process withdistributed parameters, that is
a000 � yþ a100oyot þ a010 � oy
opþ a001 � oyoqþ a200 � d
2y
dt2þ a110 � o2y
otopþ a020 � d2y
op2
þa011 � o2yopoqþ a002 � o
2yoq2 þ a101 � d2
yotoq ¼ uðt; p; qÞ
ð8:92Þ
or
a000 � x000 þ a100 � x100 þ a010 � x010 þ a001 � x001 þ a200 � x200
þ a110 � x110 þ a020 � x020 þ a011 � x011 þ a002 � x002 þ a101 � x101 ¼ u000
ð8:93Þ
The particular solution, associated for (8.92), presents the form:
yðt; p; qÞ ¼ y ¼y000 þ ðJ0T þ J1T � t þ J2T � t2 þ J3T � t3ÞðJ0P þ J1P � pþ J2P � p2 þ J3P � p3Þ� ðJ0Q þ J1Q � qþ J2Q � q2 þ J3Q � q3Þ � Ku � u
ð8:94Þ
110 8 Modeling: Simulation Examples
or
y t; p; qð Þ ¼ y ¼ y000 þ Fðt; p; qÞ � Ku � u ð8:95Þ
where: F (t, p, q) represents the product of the three polynomials of third order,with respect to (t, p, q) from (8.94).
This process with distributed parameters (8.92) is included in the control sys-tem, with a PID controller from Fig. 8.3, formally identical to Fig. 8.2, and havingthe same interpretations of the components (Fig. 8.14).
It is noticeable that the particular solution (8.94) is made out of products ofpolynomials of third order, representing a quite usual form in the technical field.
The same as in the stages of calculations (8.78), …, (8.82), results:
y ¼ y000 þ Fðt; p; qÞ � KPR � Ku � Kv � w1þ Fðt; p; qÞ � KPR � Ku � Kv � KM
ð8:96Þ
which represents the particular solution for (8.92), included in the control loopfrom Fig. 8.3, for a behavior that is strictly proportional to the one of the controller(PID), KPR = 0, and KIR = 0 and KDR = 0.
The partial derivatives matrix of the state vector for PDE II.3 has the form:
ð8:97Þ
with the remark that the order of the partial derivatives with respect to (p) and(q) from the constitution of the matrices (xPQ) and (xTPQ) is limited to three,because the degree of the polynomials from (8.94), with respect to (p) and (q), isnot higher than three.
ct
aKadtKaK DRIRPR =
∂∂
∫ ++⋅
MK
VK 3IIEDP
y00
y
yM
aw c u
Fig. 8.14 EDP II
8.3 Control System of a Process, Modeled by PDE II.3 111
The first element (x200) of the vector (xT) results from (8.93), respectively,
x200 ¼1
a200½u000 � ða000 � x000 þ a100 � x100 þ a010 � x010 þ a001 � x001
þ a110 � x110 þ a020 � x020 þ a011 � x011 þ a002 � x002 þ a101 � x101Þ�ð8:98Þ
from which the partial derivatives are calculated:
x2þT ; P; Q ¼1
a200½uTPQ � ða000 � xTPQ þ a100 � x1þT ; PQ þ a010 � xT ; 1þP;Q
þ a001 � xTP;1þQ þ a110 � x1þT ; 1þP;Q þ a020 � xT ; 2þP; Q
þ a011 � xT ; 1þP;1þQ þ a002 � xTP; 2þQ þ a101 � x1þT ; P; 1þQÞ�
ð8:99Þ
for P = 0, 1, 2, 3; Q = 0, 1, 2, 3 and T = 0, 1, 2, 3.For choosing the coefficients (J…) from (8.94), we have factored the function F
(t, p, q) from (8.95) of the form:
Fðt; p; qÞ ¼ FtðtÞ � FpðpÞ � FqðqÞ ð8:100Þ
Fig. 8.15 Inflexion points
112 8 Modeling: Simulation Examples
where:
Ft tð Þ ¼ J0T þ J1T � t þ J2T � t2 þ J3T � t3 ð8:101Þ
FpðpÞ ¼ J0P þ J1p � pþ J2P � p2 þ J3P � p3 ð8:102Þ
FqðqÞ ¼ J0Q þ J1Q � qþ J2Q � q2 þ J3Q � q3 ð8:103Þ
We consider for these functions that t0 = 0, p0 = 0, q0 = 0, tf = 10, pf = 10and qf = 10 as it is shown in Fig. 8.4 and 8.15.
The following conditions are imposed for the function Ft (t): Ft (t0) = 0 and Ft
(tf) = 1, and the slopes in the points (t0) and (tf) are null. The following conditionsare imposed to the function Fp (p) and Fq (q): Fp (p0) = 0.05, Fq (q0) = 0.05, Fp
(pf) = 1 and Fq (qf) = 1. The slope of the function Fp (p) in the points (p0) and (pf)will correspond to 1
2 � 0:95pf
, and the slope of the function Fq (q) in the points (q0) and
(qf) will correspond to 12 � 0:95
qf. In the inflexion points (i), the slopes of the functions
Ft (ti), Fp (pi) and Fq (qi) are bigger than 1tf
� �, 0:95
pf
� �, respectively, 0:95
qf
� �.
Out of the above conditions, the expressions of the coefficients (J…) result from(8.101), (8.102) and (8.103).
The other parameters of structure belonging to the control loop from Fig. 8.3are identical to those in the previous example, respectively, Ku = 1, kv = 1,KM = 1, KPR = 1, w0 = 10, wA = 5, a… = 1, s = tf and x ¼ 2p
s .In accordance with the method of calculus presented in (4.3) and avoiding the
presentation of some details of numerical routine calculus [due to the integrationcomponent of the controller from Fig. (8.2)], in the program EDPM 7(8), therehave been developed lots of control regimes, out of which we exemplify thefollowing:
1. Open loop, for KM = 0 and Dt = 10-2, with the results presented in Table 8.13and in Figs. 8.16 and 8.17 [EDPM 7(8)].
It can be noticed that the results for (p = 10; q = 2) and (p = 2; q = 10), forwA = 0 and wA = 5 are identical, which underlines the symmetry and correctnessof the results. In this case, the coefficients (a…) frame between the values(1 7 10). For the two cases of the reference signal, that is, w = w0 andw = w0 ? wA sin xt, the values (crep x000) are completely negligible, of the order(10-5 7 10-4), for the integration step (0.005 B Dt B 0.05), which proves thecorrectness and stability of the used method.
2. Control loop, for KM = 0.5 and Dt = 10-2, with the results presented inTable 8.14 and in Figs. 8.18, 8.19, 8.20 and 8.21 [EDPM 9(10)]:
The first two cases correspond to the proportional control (KPR = 1, KIR = 0and KDR = 0) for which (x000) is numerically approximated and (y) from (8.96)allows the calculus (crep x000). The third variant includes the control PID(KPR = 1, KIR = 0.056 and KDR = 0.2) for which (y) does not correspond to
8.3 Control System of a Process, Modeled by PDE II.3 113
Tab
le8.
13
t K0
12
34
56
78
910
wA
=0
p=
10x 0
00K
11.
292.
063.
184.
546.
027.
498.
859.
9710
.72
10.9
9q
=10
crep
x10
-5
10-
410
-4
10-
410
-5
10-
410
-4
10-
410
-4
10-
410
-4
p=
10x 0
00K
11.
051.
201.
421.
691.
972.
262.
522.
742.
892.
94q
=2
crep
x10
-5
10-
510
-5
10-
510
-5
10-
510
-4
10-
410
-4
10-
410
-4
p=
2x 0
00K
11.
051.
201.
421.
691.
972.
262.
522.
742.
892.
94q
=10
crep
x0
10-
510
-5
10-
510
-5
10-
510
-4
10-
410
-4
10-
410
-4
p=
2x 0
00K
11.
011.
041.
081.
131.
191.
241.
291.
341.
371.
38q
=2
crep
x0
10-
510
-5
10-
510
-5
10-
510
-5
10-
510
-5
10-
510
-5
wA
=5
p=
10x 0
00K
11.
382.
564.
225.
576
5.57
5.11
5.71
7.89
11.0
3q
=10
crep
x10
-5
10-
510
-5
10-
510
-5
10-
510
-5
10-
510
-4
10-
410
-4
p=
10x 0
00K
11.
073
1.3
1.62
51.
891.
971.
891.
81.
912.
342.
95q
=2
crep
x0
10-
510
-5
10-
510
-5
10-
510
-5
10-
410
-4
10-
410
-4
p=
2x 0
00K
11.
073
1.3
1.62
51.
891.
971.
891.
81.
912.
342.
95q
=10
crep
x0
10-
510
-5
10-
510
-5
10-
510
-5
10-
410
-4
10-
410
-4
p=
2x 0
00K
11.
014
1.05
91.
121
1.17
31.
188
1.17
21.
155
1.17
81.
261.
38q
=2
crep
x0
10-
510
-5
10-
510
-5
10-
510
-5
10-
410
-4
10-
410
-4
114 8 Modeling: Simulation Examples
(8.96) and the calculus of crep x000 does not have any sense. In this case also, theapproximation method of the integration component of the control law (PID)operates according to (8.90) and (8.91). Sometimes, the particular solutions ofpolynomial form (8.94) could become advantageous, because they use a limitednumber (for instance 3) of the order of partial derivatives with respect to (p),(q) and (r), thus lowering the dimensions of the matrix (Mpdx). The approximationmethod for the coefficients (J…) from Fig. 8.4 could belong to some technologicalapplications, on which we do not insist at this time.
3. General remark for the examples 8.13 and 8.14.
Within this frame, we present some details of numerical routine calculus,associated with the integrating component of the controllers, from Figs. 8.2,respectively 8.3,
c ¼ KPR � aþ KDR �da
dtþ KIR �
Z
a dt ð8:104Þ
and
u ¼ KV � KPR � aþ KDR �da
dtþ KIR �
Z
a dt
�
ð8:105Þ
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
12
tX
ook
WA=0 ; p=10 ; q=10WA=0 ; p=10 ; q=2WA=0 ; p=2 ; q=10WA=0 ; p=2 ; q=2
Fig. 8.16 The graphicalrepresentation of theTable 8.13 (x00k (t) forwA = 0; p = 10; q = 10,wA = 0; p = 10; q = 2,wA = 0; p = 2; q = 10 andwA = 0; p = 2; q = 2)
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
12
t
Xoo
k
WA=5 ; p=10 ; q=10
WA=5 ; p=10 ; q=2WA=5 ; p=2 ; q=10
WA=5 ; p=2 ; q=2
Fig. 8.17 The graphicalrepresentation of theTable 8.13: x00k (t) forwA = 5; p = 10; q = 10,wA = 5; p = 10; q = 2,wA = 5; p = 2; q = 10 andwA = 5; p = 2; q = 2
8.3 Control System of a Process, Modeled by PDE II.3 115
Tab
le8.
14t K
12
34
56
78
910
wA
=0
KP
R=
1p
=10
x 00
0K
1.27
1.95
2.87
3.86
4.80
5.65
6.35
6.88
7.21
7.33
q=
10cr
ep10
-5
10-
410
-4
10-
310
-3
10-
310
-3
10-
310
-3
10-
3
KIR
=0
p=
10x 0
00K
1.05
1.19
1.39
1.63
1.88
2.13
2.35
2.52
2.64
2.68
q=
2cr
ep10
-5
10-
510
-5
10-
510
-4
10-
410
-4
10-
410
-4
10-
4
KD
R=
0p
=2
x 00
0K
1.05
1.19
1.39
1.63
1.88
2.13
2.35
2.52
2.64
2.68
q=
10cr
ep10
-5
10-
510
-5
10-
510
-4
10-
410
-4
10-
410
-4
10-
4
KP
R=
1p
=2
x 00
0K
1.01
1.03
71.
078
1.12
1.18
1.23
1.28
1.31
1.34
1.35
KIR
=0.
056
q=
2cr
ep10
-5
10-
510
-5
10-
510
-5
10-
510
-5
10-
510
-4
10-
4
KD
R=
0.2
p=
10x 0
00K
1.28
62.
052
3.15
24.
426
5.72
46.
978.
078.
989.
6410
.01
q=
10p
=10
x 00
0K
1.05
61.
215
1.46
01.
772.
122.
493
2.85
23.
172
3.42
3.56
7q
=2
p=
2x 0
00K
1.05
61.
215
1.46
01.
772.
122.
493
2.85
23.
172
3.42
3.56
7q
=10
p=
2x 0
00K
1.01
1.04
21.
091
1.15
51.
228
1.30
71.
386
1.45
71.
514
1.54
7q
=2
wA
=5
p=
10x 0
00K
1.35
82.
436
3.80
4.73
4.79
4.20
3.67
3.94
5.31
7.35
q=
10cr
ep10
-5
10-
410
-3
10-
310
-3
10-
310
-3
10-
210
-2
10-
2
KP
R=
1p
=10
x 00
0K
1.07
01.
291.
592
1.82
61.
881.
776
1.67
1.76
22.
137
2.68
8q
=2
crep
10-
510
-5
10-
410
-4
10-
410
-4
10-
410
-3
10-
310
-3
KIR
=0
p=
2x 0
00K
1.07
01.
291.
592
1.82
61.
881.
776
1.67
1.76
22.
137
2.68
8q
=10
crep
10-
510
-5
10-
410
-4
10-
410
-4
10-
410
-3
10-
310
-3
KD
R=
0p
=2
x 00
0K
1.01
1.05
71.
117
1.16
51.
177
1.15
81.
138
1.15
81.
237
1.35
3q
=2
crep
10-
510
-5
10-
510
-5
10-
510
-5
10-
510
-5
10-
510
-4
KP
R=
1p
=10
x 00
0K
1.38
82.
577
4.15
5.38
25.
836
5.70
35.
646
6.36
58.
102
10.4
1q
=10
KIR
=0.
056
p=
10x 0
00K
1.07
61.
322
1.67
31.
984
2.14
52.
179
2.23
42.
492
3.02
3.68
1q
=2
KD
R=
0.2
p=
2x 0
00K
1.07
61.
322
1.67
31.
984
2.14
52.
179
2.23
42.
492
3.02
3.68
1q
=10
p=
2x 0
00K
1.01
1.06
31.
133
1.19
71.
232
1.24
31.
258
1.31
61.
430
1.57
2q
=2
116 8 Modeling: Simulation Examples
For a more general hypothesis of the control of the errors, for PDE II.4
a ¼ w� KM � x0000 ð8:106Þ
0 1 2 3 4 5 6 7 8 9 100
1
2
3
4
5
6
7
8
t
Xoo
k
WA=0 ; KPR=1; KIR=0 ; KDR=0 ; p=10 ; q=10WA=0 ; KPR=1; KIR=0 ; KDR=0 ; p=10 ; q=2WA=0 ; KPR=1; KIR=0 ; KDR=0 ; p=2 ; q=10WA=0 ; KPR=1; KIR=0 ; KDR=0 ; p=2 ; q=2
Fig. 8.18 The graphicalrepresentation of theTable 8.14 (x00k (t) forwA = 0; KPR = 1; KIR = 0;KDR = 0; p = 10; q = 10,wA = 0; KPR = 1; KIR = 0;KDR = 0; p = 10; q = 2,wA = 0; KPR = 1; KIR = 0;KDR = 0; p = 2; q = 10 andwA = 0; KPR = 1; KIR = 0;KDR = 0; p = 2; q = 2)
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
12
t
WA=0 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; p=10 ; q=10
WA=0 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; p=10 ; q=2WA=0 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; p=2 ; q=10WA=0 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; p=2 ; q=2
Xoo
k
Fig. 8.19 The graphicalrepresentation of heTable 8.14 (x00k (t) forwA = 0; KPR = 1;KIR = 0.056; KDR = 0.2;p = 10; q = 10, wA = 0;KPR = 1; KIR = 0.056;KDR = 0.2; p = 10; q = 2,wA = 0; KPR = 1;KIR = 0.056; KDR = 0.2;p = 2; q = 10 and wA = 0;KPR = 1; KIR = 0.056;KDR = 0.2; p = 2; q = 2)
0 1 2 3 4 5 6 7 8 9 100
1
2
3
4
5
6
7
8
t
Xoo
k
WA=5 ; KPR=1 ; KIR=0 ; KDR=0 ; p=10 ; q=10
WA=5 ; KPR=1 ; KIR=0 ; KDR=0 ; p=10 ; q=2WA=5 ; KPR=1 ; KIR=0 ; KDR=0 ;p=2 ; q=10
WA=5 ; KPR=1 ; KIR=0 ; KDR=0 ; p=2 ; q=2
Fig. 8.20 The graphicalrepresentation of theTable 8.14 [x00k (t) for(wA = 5; KPR = 1; KIR = 0;KDR = 0; p = 10; q = 10),(wA = 5; KPR = 1; KIR = 0;KDR = 0; p = 10; q = 2),(wA = 5; KPR = 1; KIR = 0;KDR = 0; p = 2; q = 10) and(wA = 5; KPR = 1; KIR = 0;KDR = 0; p = 2; q = 2)]
8.3 Control System of a Process, Modeled by PDE II.3 117
result:
uk ¼ Kv � KPR � w� KM � x0000ð ÞkþKDR _w� KM � x1000ð ÞkþKIR �RtK
tK �Dt
w� KM � x0000ð Þ � dt
" #
¼ uPK þ uDK þ uIK
ð8:107Þ
As a result, the three components of the command signal (uk) are
• the proportional component:
uPK ¼ KV � KPR � w� KM � x0000ð ÞK ð8:108Þ
• the derivative component with respect to time:
uDK ¼ KV � KDR � _w� KM � x1000ð ÞK ð8:109Þ
• the integrating component with respect to time:
uIK ffi uI; K�1 þ DuI; K�1 ð8:110Þ
where
DuI; K�1 ffi KV � KIR �X4
n¼1
Dtn
n!� dnw
dtn� KM � xn000
�
K�1
ð8:111Þ
Thus, (8.111) corresponds to an iterative sum, at each integration step (Dt),specific to the numerical integration.
If the process with distributed parameters included in the control loops from theFigs. (8.2) or (8.3) is defined by PDE II.4, then the first element (x2000) of thevector (xT) results from
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
12
t
WA=5 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; P=10 ; Q=10
WA=5 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; P=10 ; Q=2WA=5 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; P=2 ; Q=10
WA=5 ; KPR=1 ; KIR=0.056 ; KDR=0.2 ; P=2 ; Q=2
Xoo
k
Fig. 8.21 The graphicalrepresentation of Table 8.14[x00k (t) for (wA = 5;KPR = 1; KIR = 0.056;KDR = 0.2; p = 10; q = 10),(wA = 5; KPR = 1;KIR = 0.056; KDR = 0.2;p = 10; q = 2), (wA = 5;KPR = 1; KIR = 0.056;KDR = 0.2; p = 2; q = 10)and (wA = 5; KPR = 1;KIR = 0.056; KDR = 0.2;p = 2; q = 2)]
118 8 Modeling: Simulation Examples
x2000 ¼1
a0000u0000 � . . .ð Þ½ �: ð8:112Þ
The notation (…) represents a sum of monomials (or polynomials) in the elements(x…) from the composition of (x) and (xPQR).
Because
u0000 ¼ u0000 . . .; u; _uð Þ ð8:113Þ
from (8.108) and (8.109) results
_uPK ¼ KV � KPR � _w� KM � x1000ð ÞK ð8:114Þ
_uDK ¼ KV � KDR � €w� KM � x2000ð ÞK ð8:115Þ
and for the integrating component from (8.107), by derivatives in accordance withtime, we obtain
_uIK ¼ KV � KIR � w� KM � x0000ð ÞK ð8:116Þ
Further on,
€uPK ¼ KV � KPR � €w� KM � x2000ð ÞK ð8:117Þ
€uDK ¼ KV � KDR � vw� KM � x3000� �
K ð8:118Þ
€uIK ¼ KV � KIR � _w� KM � x1000ð ÞK : ð8:119Þ
The derivatives forms ðoT u...=otTÞK ¼ u...T ; K and oT w=otTKwT ; K lead to
uP; T ; K ¼ KV � KPR � wT ; K � KM � xT ; 000� �
K; T ¼ 0; 1; 2; . . .ð Þ ð8:120Þ
uD; T ; K ¼ KV � KDR � w1þT ; K � KM � x1þT ; 000� �
K ; T ¼ 0; 1; 2; . . .ð Þ ð8:121Þ
uI; T ; K ¼ KV � KIR � w�1þT ; K � KM � x�1þT ; 000
� �K; T ¼ 1; 2; . . .ð Þ ð8:122Þ
with the remark that the reference signal is w = w (t).
Taking into consideration the general case oTþPþQþRu...
otTopPoqQorR
� �
K¼ u...;TPQR; K ; we
obtain
uP; TPQR; K ¼ KV � KPR � wT � KM � xTPQRð ÞK ð8:123Þ
uD; TPQR; K ¼ KV � KDR � w1þT � KMx1þT ; PQR
� �K ð8:124Þ
uI; TPQR; K ¼ KV � KIR � w�1þT � KM � x�1þT ; PQR
� �K
ð8:125Þ
results which respect the validity from the order (T) of the partial derivatives, from(8.120), …, (8.122).
8.3 Control System of a Process, Modeled by PDE II.3 119
Thus,
uTPQR; K ¼ uP; TPQR; K þ uD; TPQR; K þ uI; TPQR; K ð8:126Þ
so that (8.113) becomes
u0000; K ¼ u0000 . . .; u0000; K ; u1000; K
� �ð8:127Þ
or in a general manner
uTPQR; K ¼ uTPQR . . .; uTPQR; K ; u1þT ; PQR; K
� �ð8:128Þ
As a conclusion, the details of numerical calculus presented at (8.104),…(8.125)allow a direct approach of the integral–differential equations from (8.106), thusensuring the maintenance of the component (u0000, k) from (8.127).
8.4 System of Two PDE II.2
The following system is considered:
a1 �oy1
otþ a2 �
o2y1
ot2þ a3 �
o2y2
otop¼ u1 t; pð Þ ð8:129Þ
a4 �o2y2
ot2þ a5 �
o2y1
otopþ a6 �
o2y2
op2¼ u2 t; pð Þ ð8:130Þ
which can be rewritten as
a1 � x1:10 þ a2 � x1:20 þ a3 � x2:11 ¼ u1:00 ð8:131Þ
a4 � x2:20 þ a5 � x1:11 þ a6 � x2:02 þ u2:00 ð8:132Þ
and for which the following particular solutions are considered:
y1 ¼ x1:00 ¼ t2p3 ð8:133Þ
y2 ¼ x2:00 ¼ t3p2 ð8:134Þ
As a result, the components of the matrix (Mpdx) become
x =
10.2
10.1
00.2
00.1
x
x
x
x
=
10.2
10.1
00.2
00.1
y
y
y
y
ð8:135Þ
120 8 Modeling: Simulation Examples
xP =
P1.213.212.211.2
P1.113.112.111.1
P0.203.202.201.2
P0.103.102.101.1
x......xxx
x......xxx
x.....xxx
x......xxx
ð8:136Þ
xT =
0T.2
0T.1
40.2
40.1
30.2
30.1
20.2
20.1
x
x
...
x
x
x
x
x
x
ð8:137Þ
xTP =
TP.23T.22T.21T.2
TP.13T.12T.11T.1
P4.243.242.241.2
P4.143.142.141.1
P3.233.232.231.2
P3.133.132.131.1
P2.223.222.221.2
P2.123.122.121.1
x...xxx
x...xxx
...............
x...xxx
x...xxx
x...xxx
x...xxx
x...xxx
x...xxx
ð8:138Þ
It can be noticed that, for the integration in accordance with time (t), the twoPDE II.2 from (8.129) and (8.130) allow two state variables from the constitutionof the state vector x (4 9 1), from (8.135), in accordance with which the devel-opments from (8.136),…(8.138) become obvious.
From (8.131) and (8.132), we obtain
x1:20 ¼1a2
u1:00 � a1 � x1:10 þ a3 � x2:11ð Þ½ � ð8:139Þ
8.4 System of Two PDE II.2 121
and
x2:20 ¼1a4
u2:00 � a5 � x1:11 þ a6 � x2:02ð Þ½ � ð8:140Þ
from which the partial derivatives are calculated:
x1:2þT ; P ¼1a2
u1:TP � a1 � x1:1þT ; P þ a3 � x2:1þT ; 1þP
� �� �ð8:141Þ
and
x2:2þT ; P ¼1a4
u2; TP � a5 � x1:1þT ; 1þP þ a6 � x2:T ; 2þP
� �� �ð8:142Þ
In accordance with the particular solution (8.133) and (8.134), from (8.131) and(8.132),
u1 t; pð Þ ¼ u1:00 ¼ 2a1tp3 þ 2a2p3 þ 6a3t2p ð8:143Þ
u2 t; pð Þ ¼ u2:00 ¼ 6 a4 þ a5ð Þtp2 þ 2a6t3 ð8:144Þ
The initial conditions for xIC = x (t0, p) are known, which is of the form:
ð8:145Þ
ð8:146Þ
with the observation that in (8.145) and (8.146), eventual boundary conditionscould be included, in accordance with the variable (p).
The stages of calculus reiterate, for the elements of the vector (xT) and thematrix (xTP), according to (8.139),…, (8.142), as follows:
1. x1.2P, for P = 1, 2, 3.2. x2.2P, for P = 1, 2.3. (x1.30) and (x2.30) are calculated.4. x1.3P, for P = 1, 2, 3.
122 8 Modeling: Simulation Examples
5. x2.3P, for P = 1, 2.6. (x1.40) and (x2.40) are calculated.7. x1.4P, for P = 1, 2, 3.8. x2.4P, for P = 1, 2.
These stages continue until the derivative order (T) in accordance with thepredetermined time, which was considered to be T = 5 for this example.
For each element obtained in the above stages (1), (2), (3), …, (x1.2+T,P) and(x2.2+T,P), we also calculate (u1.TP), (u2.TP), according to (8.143) and (8.144),respectively. In the program, these expressions (u1.TP) and (u2.TP) will be declaredbefore the correspondent expressions are calculated (x1.2+T, P) and (x2.2+T, P).
It is noticeable that for the two particular solutions (8.133) and (8.134), strictlynecessary in order to ensure the initial conditions (and eventually for the verifi-cation of the results), the maximum order of the derivatives with respect to (p) islimited to P = 3 for (x1…P), respectively, for P = 2 for (x2…P), from which thesepartial derivatives are cancelled.
As a result, the final form of the matrix (Mpdx) becomes
TPT
Ppdx xx
xxM
0xxx
xxxx
0xxx
xxxx
0xxx
xxxx
0xxx
xxxx
0xxx
xxxx
0xxx
xxxx
52.251.250.2
53.152.151.150.1
42.241.240.2
43.142.141.140.1
32.231.230.2
33.132.131.130.1
22.221.220.2
23.122.121.120.1
12.211.210.2
13.112.111.110.1
02.201.200.2
03.102.101.100.1
ð8:147Þ
with the dimensions x (4 9 1), xP (4 9 3), xT (8 9 1) and xTP (8 9 3), so thatMpdx (12 9 4). At the sequence (k - 1), we know (xk-1) and (xP, k-1), and eachelement from the constitution of (xT, k-1) and (xTP, k-1) results, according to(8.139),…(8.142) from the lines previously calculated of (xk-1), (xP, k-1), (xT, k-1)and (xTP, k-1).
At the sequence (k), in a formal presentation, we obtain
xk ffi xk�1 þX
T¼1
DtT
T !xT; k�1 ð8:148Þ
8.4 System of Two PDE II.2 123
and
xP; k ffi xP; k�1 þX
T¼1
DtT
T !xTP; k�1 ð8:149Þ
where we will operate separately for each one of the elements (x1.0…), (x1.1…),(x2.0…) and (x2.1…) from the constitution of (xk) and (xP, Q), the number ofderivatives from these Taylor series being five for (x1.0…) and (x2.0…), respectivelyfour for (x1.1…) and (x2.1…). So, for each sequence of calculus, with the integrationstep (Dt), we will operate with a number of (14) Taylor series, with the observanceof the method of calculus presented in (4.3).
Based on the elements presented above, we have made the programS2EDP22(P) for t0 = 0, p0 = 0, tf = 10 and pf = 10, the coefficients a… = 1 andDt = 10-3, with a few results exemplified in Table 8.15 and in Figs. 8.22 and8.23. It is noticeable that the values of crep x1.00 and crep x2.00 are very small, evenif the increased slopes of these state variables are extremely steep. By increasing
Table 8.15
t 0 2 4 6 8 10
p = 10 x1:00 0 4000 16,000 36,000 64,000 100,000crep x1.00 0 2:7 � 10�3 2:8 � 10�3 8 � 10�3 8:9 � 10�3 7:3 � 10�3
x2.00 0 800 6,400 21,600 51,200 100,000crep x2.00 0 3:6 � 10�3 4:3 � 10�3 8:4 � 10�3 1:1 � 10�2 1:0 � 10�2
p = 5 x1:00 0 500 2,000 4,500 8,000 12,500crep x1.00 0 2:7 � 10�3 2:8 � 10�3 7:5 � 10�3 8:1 � 10�3 6:7 � 10�3
x2:00 0 200 1,600 5,400 12,800 25,000crep x2.00 0 3:6 � 10�3 4:2 � 10�3 8:1 � 10�3 1 � 10�2 1 � 10�2
p = 1 x1:00 0 4 16 36 64 100crep x1.00 0 2:1 � 10�3 2:9 � 10�3 1:1 � 10�3 2 � 10�2 2 � 10�2
x2:00 0 8 64 216 512 1,000crep x2.00 0 4:4 � 10�3 6:2 � 10�3 6 � 10�3 2:3 � 10�2 5:7 � 10�2
0 1 2 3 4 5 6 7 8 9 100123456789
10 x 104
t
X1.
00
p=10p=5p=1
Fig. 8.22 The graphicalrepresentation of theTable 8.15 (x1.00 (t) forp = 10, p = 5 and p = 1)
124 8 Modeling: Simulation Examples
the number of lines of (xT) and implicitly (xTP) to values higher than (8), theseapproximation errors could decrease.
As a conclusion, the use of the operator matrix (Mpdx) could be easily extendedfor systems of partial derivative equations.
8.5 Comparison Between the Numerical Integrationof a ODE II Through Taylor Series and TaylorSeries-LIL
We consider the following ODE II:
a0yþ a1 �oy
otþ a2 �
o2y
ot2¼ K � uðtÞ ð8:150Þ
which allows the analytical solution
yAN ¼ J � e�t=T ð8:151Þ
for which
uðtÞ ¼ J
Kða0 �
a1
Tþ a2
T2Þ � e�t=T : ð8:152Þ
When we rewritten (8.150) in the usual matrix–vector form, it results
_x ¼ A � xþ B � u ð8:153Þ
2
1
x
xð8:154Þ
0 1 2 3 4 5 6 7 8 9 100123456789
10 x 104
tX
2.00
p=10p=5p=1
Fig. 8.23 The graphicalrepresentation of theTable 8.15 (x2.00 (t) forp = 10, p = 5 and p = 1)
8.4 System of Two PDE II.2 125
A2
1
2
0
aa
aa
10ð8:155Þ
B2a
K
0ð8:156Þ
where the state variables are x1 = y and x2 = dy/dt, and the initial conditions (IC)from t0 = 0 correspond to x1IC = J and x2IC = -J/T.
Numerical integration for (8.150) by Taylor series leads to
xk ffi xk�1 þXx
m¼1
Dtm
m!� xk�1ðmÞ
ð8:157Þ
respectively,
xk ffi xk�1 þXx
m¼0
Dtmþ1
ðmþ 1Þ! � ðA xk�1ðmÞþB uk�1
ðmÞÞ ð8:158Þ
where (x) corresponds to the maximum order of the derivatives with respect totime. It is noticeable that all the components from the right side for (8.157) and(8.158) correspond to the sequences (k - 1), thus including the derivatives
uðmÞ
k�1 ¼ d um
dtm
� �
k�1:
Numerical integration for (8.150) by Taylor series LIL in (8.158) introduces theobvious substitution:
uk�1 ffi uk �Xx
m¼1
Dtm
m!uðmÞ
k�1 ð8:159Þ
where the index (k) corresponds to the moment tk = k � Dt, and for (k - 1), itcorresponds to tk-1 = (k-1) � Dt. As a result, (8.158) becomes
xk ffiDt
1!� B � uk þ xk�1 þ
Dt
1!ðA � xk�1 � B �
Xx
m¼1
Dtm
m!uðmÞ
k�1Þ
þXx
m¼1
Dtmþ1
ðmþ 1Þ!ðA � xðmÞ
k�1 þ B � uðmÞ
k�1Þð8:160Þ
which can be rewritten
xk ffi g � uk þ hk ð8:161Þ
126 8 Modeling: Simulation Examples
Tab
le8.
16
t0.
11.
12.
13.
24.
25.
26.
27.
28.
29.
210
.1
x 1k
90.5
54.9
33.3
19.2
11.6
7.06
4.3
2.6
1.6
0.95
0.61
crep
x 1k
TA
YL
OR
1.2�1
0-5
8.2�1
0-6
7�1
0-6
10-
51.
7�1
0-5
1.9�1
0-5
1.9�1
0-5
1.9�1
0-5
2�1
0-5
2.2�1
0-5
2.3�1
0-5
crep
x 1k
TA
YL
OR
LIL
1.2�1
0-5
8.2�1
0-6
7�1
0-6
10-
51.
7�1
0-5
1.91�1
0-5
1.91�1
0-5
2�1
0-5
1.9�1
0-5
2�1
0-5
2.18�1
0-5
8.5 Comparison Between the Numerical Integration of a ODE II 127
where
g ffi Dt
1!� B ð8:162Þ
and
hk ¼ xk�1 þDt
1!ðA � xk�1 � B �
Xx
m¼1
Dtm
m!ðmÞ
k�1Þ þXx
m¼2
Dtmþ1
ðmþ 1Þ!ðA � xðmÞ
k�1 þ B
� uðmÞ
k�1Þð8:163Þ
It can be noticed that the form of the solution (8.161) is specific to the methodTaylor–LIL, detailed in the Sect. 1.5.
The two variants of solution, by (8.5) for Taylor series and by (8.160) forTaylor series–LIL, are exemplified in Table 8.16, where for (8.150), we haveconsidered: a01 = 1; a1 = 1; a2 = 1; K = 10; J = 100; T = 2; t0 = 0; tf = 5;T = 10; Dt = 0.1; and x = 5. The programs run on the computer correspond toEDO2TL 1(2), respectively, TLLLI1(2), and the errors cumulated in percentages,crep x1k–Taylor and crep x1k–Taylor–LIL, have been calculated with respect to theanalytical solution (8.151). Because the results from Table 8.16 and Fig. 8.24 andfrom the two variants of numerical integration are practically identical, the validityof the subsolution (8.159) is proved, with (x) big enough, that is, x = 5, in orderto obtain the method Taylor–LIL.
0 1 2 3 4 5 6 7 8 9 100
102030405060708090
100
tX
1k
Fig. 8.24 The graphicalrepresentation of theTable 8.16 [x1k (t)]
128 8 Modeling: Simulation Examples
8.6 Comparison Between the Numerical Integrationof a PDE II.2 Through Taylor Series and TaylorSeries-LIL
Let us consider PDEII.2 from (8.27), that is
a00 � yþ a10oy
otþ a01
oy
opþ a20
o2y
ot2þ a11
o2y
otopþ a02
o2y
op2¼ uðt; pÞ ð8:164Þ
or
a00 � x00 þ a10x10 þ a01x01 þ a20x20 þ a11x11 þ a02x02 ¼ u00 ð8:165Þ
where we denote
uðt; pÞ ¼ u00 ¼ uðt; pÞ ð8:166Þ
(x20) from (8.164) is
x20; K�1 ¼1
a20½u00; K�1 � a00x00 þ a10x10 þ a01x01 þ a11x11 þ a02x02ð Þk�1�
ð8:167Þ
where the index (k - 1) corresponds to the stage of the regressive sequence fromthe general form (8.32). In order to introduce, in the simplest form, the effect oflocal-iterative linearization (LIL), we introduce in (8.167) the approximationsubstitution:
u00; K�1 ffi u00; K �X6
m¼1
ðþDtÞm
m!� um0; K�1 ð8:168Þ
obtaining
x20; K�1 ¼1
a20½u00; k �
X6
m¼1
ðþDtÞm
m!� um0; K�1
� a00x00 þ a10x10 þ a01x01 þ a11x11 þ a02x02ð Þk�1�ð8:169Þ
where
um0; K¼1 ¼omu00
otm
�
k�1
ð8:170Þ
So, the form (8.169) in which we have introduced (u00k), specific to the methodLIL, represents the first element of the vector (xT,k-1) from (8.32). For the calculusof all the other elements from the constitution of (xP,k-1), then (xT,k-1) and finally(xTP,k-1), we operate with the partial derivatives (x2+T, P,k-1) from (8.167),respectively,
8.6 Comparison Between the Numerical Integration of a PDE II.2 129
x2þT ; P; K�1 ¼1
a20½uTP; K�1 � ða00xTP þ a10x1þT ; P þ a01xT ; 1þP þ a11x1þT ; 1þP
þ xT ; 2þPÞk�1�ð8:171Þ
where uTP,K-1 results from the partial derivations of (8.168).The stages presented above, which include (8.167),…, (8.171), represent the
simplest form of the Taylor series–LIL that only includes (u00, k) in (8.169). Themore complete form of the Taylor–LIL could also include the derivatives (yTP, k-1)from (8.171), operated on (8.168), respectively,
uTP; K�1 ffi uTP; K �X6
m¼1
ðþDtÞm
m!� umþT ; P; K�1 ð8:172Þ
where
uTP; K ¼ ðoTþPu00
otT � oppÞk ð8:173Þ
and
umþT ; P; K�1 ¼omþTþPu00
omþT t � oPp
�
k�1
ð8:174Þ
It needs to be noticed that at the sequence (k), all values (u00k) from (8.168) and(uTP,k) are known; thus, all the intermediary stages of calculus of the matrix (Mpdx)could easily be elaborated.
According to (8.164), (8.165),…, (8.171), changes in the program EDPTL90(91) were done, resulting inthe new program TLLLI 5(6). For both programshaving identical initial conditions, according to 8.5, we have presented inTable 8.17 and in Fig. 8.25 the results obtained for (x00) and (crep x00), at the samemoments (t). Due to the fact that the results of the two variants of numericalintegration are also practically identical in this example, the validity of the sub-stitution (8.168) is proved, for (x) big enough, x = 6, in order to obtain themethod Taylor–LIL
8.7 PDE of the IVth Order with Four Variables
The following PDE IV.2 is considered:
a4000 �o4y
ot4þ a1111 �
o4y
otopoqor¼ uðt; p; q; rÞ ð8:174Þ
130 8 Modeling: Simulation Examples
Tab
le8.
17
Pro
gram
t0.
11.
12.
13.
24.
25.
26.
27.
28.
29.
210
.1
ED
PT
L90
/91
x 00
2.75
143
357
564
703
800
865
907
934
951
961
crep
x 00
6�1
0-3
1.6�1
0-4
4�1
0-5
2�1
0-5
2�1
0-5
2�1
0-5
2�1
0-5
2�1
0-5
2�1
0-5
2�1
0-5
2�1
0-5
TL
LL
I5/
6x 0
02.
758
143.
435
7.1
564
703.
480
0.6
865.
490
7.5
934.
495
1.4
961.
1cr
epx 0
06�1
0-3
1.7�1
0-4
7.8�1
0-5
3.9�1
0-5
2.7�1
0-5
2.5�1
0-5
2.5�1
0-5
2�1
0-5
1.7�1
0-5
1.5�1
0-5
1.5�1
0-5
8.7 PDE of the IVth Order with Four Variables 131
respectively,
a4000 � x4000 þ a1111 � x1111 ¼ u0000 ð8:175Þ
which, if it verifies the particular solution:
yAN ¼ x0000 AN ¼ t4pqr ð8:176Þ
leads to
u0000 ¼ uðt; p; q; rÞ ¼ 24a4000 � pqr þ 4a1111 � t3: ð8:177Þ
As a result, the following initial conditions (IC) need to be known, consideredat t = t0: x0000IC = t0
4 pqr; x1000IC = 4t03 pqr; x2000IC = 12t0
2 pqr and x3000IC =
24t0 pqr. In the particular hypothesis of t0 = 0, these four initial conditionsbecome null.
From (8.175), we obtain
x4000 ¼1
a4000� ðu0000 � a1111 � x1111Þ ð8:178Þ
a result which is subjected to multiple partial derivatives, with respect to (t), (p),(q) and (r), respectively,
x4þT ; P; Q; R ¼1
a4000� ðuTPQR � a1111 � x1þT ; 1þP; 1þQ; 1þRÞ: ð8:179Þ
The state vector x (4 9 1) from the constitution of
MpdxTPQRT
PQR
xx
xxð8:180Þ
0 1 2 3 4 5 6 7 8 9 100
100200300400500600700800900
1000
tX
oo
EDPTL 90/91TLLLI 5/6
Fig. 8.25 The graphicalrepresentation of theTable 8.17 [x00(t)]
132 8 Modeling: Simulation Examples
is defined by four state variables, that is
x
3000
2000
1000
0000
x
x
x
x
ð8:181Þ
The elements of the matrix (xPQR) from (8.180) are calculated from the fourknown initial conditions of the above, that is
First line:
x0100 ¼ t04qr; x0010 ¼ t0
4pr; x0001 ¼ t04pq; x0110 ¼ t0
4r; x0011 ¼ t04p;
x0101 ¼ t04q; x0111 ¼ t0
4:
Second line:
x1100 ¼ 4t03qr; x1010 ¼ 4t0
3pr; x1001 ¼ 4t03pq; x1110 ¼ 4t0
3r;
x1011 ¼ 4t03p; x1101 ¼ 4t3
0q; x1111 ¼ 4t03:
Third line:
x2100 ¼ 12t02qr; x2010 ¼ 12t0
2pr; x2001 ¼ 12t02pq; x2110 ¼ 12t0
2r;
x2011 ¼ 12t02p; x2101 ¼ 12t2
0q; x2111 ¼ 12t02:
Fourth line:
x3100 ¼ 24t0qr; x3010 ¼ 24t0pr; x3001 ¼ 24t0pq; x3110 ¼ 24t0r;
x3011 ¼ 24t0p; x3101 ¼ 24t0q; x3111 ¼ 24t0:
If the first line from (Mpdx) is associated with Taylor series with seven deriv-atives, then this matrix has the form in (8.182).
x0000 x0100 x0010 x0001 x0110 x0011 x0101 x0111
x1000 x1100 x1010 x1001 x1110 x1011 x1101 x1111
x2000 x2100 x2010 x2001 x2110 x2011 x2101 x2111
x3000 x3100 x3010 x3001 x3110 x3011 x3101 x3111
x4000 x4100 x4010 x4001 x4110 x4011 x4101 x4111
x5000 x5100 x5010 x5001 x5110 x5011 x5101 x5111
x6000 x6100 x6010 x6001 x6110 x6011 x6101 x6111
x7000 x7100 x7010 x7001 x7110 x7011 x7101 x7111
Mpdx
ð8:182Þ
8.7 PDE of the IVth Order with Four Variables 133
The elements of the vector (xT) and the matrix (xTPQR) are obtained from theelements belonging to (x) and (xPQR), as well as from the lines disposed above theline, from which we calculate, based on the general Eq. (8.179). The function(uTPQR) from (8.179) is obtained by partial derivatives conveniently operated on(8.177).
It needs to be noticed that all the elements of the matrix (xPQR) are null, if theyexist in the columns 9, 10 etc., as we can also observe from the expressions 1, 2, 3and 4 presented above.
Using the method of calculus presented in (4.3), according to the programEDP44.1(2), for a1111 = 1, a4000 = 1 and Dt = 0.01, we have extracted a fewresults which are presented in Table 8.18 and in Fig. 8.26. Because crep x0000
maintains at negligible values, the validity of the method used is proved for thisexample too.
Due to the fact that in the lines 1, 2, 3 and 4 from the matrix (MPQR) included inthe matrix (Mpdx) from (8.182) we have wasted all the partial derivatives withrespect to (p, q and r), as well as with respect to time (t), the choice of theintegration step (Dt) becomes arbitrary.
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
12 x 106
t
Xoo
oo
p=1 ; q=1 ;r=1
p=1 ; q=5 ; r=10
p=10 ; q=10 ; r=10
Fig. 8.26 The graphicalrepresentation of Table 8.18(x0000 (t) for (p = 1; q = 1;r = 1), (p = 1; q = 5;r = 10) and (p = 10;q = 10; r = 10)
Table 8.18
t 0 2 4 6 8 10
p = 1 x0000 10-8 16.65 261.15 1,304.6 4,116.5 10,040.6q = 1 crep 10-5 2 � 10-4 4 � 10-4 1.8 � 10-3 3.4 � 10-3 2.5 � 10-3
r = 1p = 1 x0000 5 � 10-7 832.5 13,058 65,233 205,825 502,002q = 5 crep 0 3 � 10-4 3.8 � 10-4 1.9 � 10-3 3.7 � 10-3 4.7 � 10-3
r = 10p = 10 x0000 10-5 16.650 261.159 1304.661 4.116.507 1.004 � 107
q = 10 crep 0 2.8 � 10-4 4.4 � 10-4 1.6 � 10-3 3.4 � 10-3 4.7 � 10-3
r = 10
134 8 Modeling: Simulation Examples