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