Curs 12 13 Cmaie Micu

7/30/2019 Curs 12 13 Cmaie Micu

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Curs 12-13

Special Numerical Methods for solving Partial Differential Equations

- Electrical Engineering Applications -

Course: Numerical Methods forEngineers, Southern IllinoisUniversity Carbondale College ofEngineering, Dr. L.R. Chevalier, Dr.

B.A. DeVantier

Course: Applied NumericalComputation Methods, Shang-Xu, Hu

of ZJU

Course: Numerical Integration of PartialDifferential Equations, Th. Wiegelmann,International Max Planck Research Schoolon Physical Processes Universities of

Braunschweig and Gttingen

Course: Numerical Methods for electricalengineering, Dan D. Micu, ElectricalEngineering Department, UTCN

Complemente de matematiciMaster an I

2011-2012


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Modelul matematic cel mai des ntlnit al fenomenelor care

stau la baza multor aplicaii electrotehnice este ecuaia cuderivate partiale.

Rezolvarea exact a ecuaiilor cu derivate partiale este

posibil pentru o clas foarte restrns de aplicatii!!! Comportarea dinamic a sistemelor fizice conduce la modele

matematice formate din ecuaii cu derivate partiale care nu

pot fi rezolvate pe cale analitic (funcii complicate ca formsau funcii cunoscute doar pe baza unor valori n puncte date

tabelari obinute pe cale experimental). Din acest motiv se

recurge la rezolvarea numeric a acestora. Metodele numerice de aproximare a soluiilor conduc la tabele

de valori ale funciei necunoscute.


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Differential Equations

A differential equation is an equation for

an unknown function of one or severalvariables that relates the values of thefunction itself and of its derivatives ofvarious orders.

Ordinary Differential Equations:

Function has 1 independent variable. Partial Differential Equations (PDEs):

At least 2 independent variables.


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An equation involving partial derivatives of an unknownfunction of two or more independent variables

General (implicit) form for one function u(x,y):

Highest derivative defines order of PDE Explicit PDE => We can resolve the equation

to the highest derivative of u. Linear PDE => PDE is linear in u(x,y) and

for all derivatives of u(x,y) Semi-linear PDEs are nonlinear PDEs, which

are linear in the highest order derivative.

Partial Differential Equations

)(u)(ax

u)(a

xx

u)(a f

n

i i

i

n

i

n

j ji

ji xxxx 011 1

2

=++ == =


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Aceste ecuaii sunt de fapt modele matematice carerezult din studiul unor fenomene i aplicaii diningineria electric ca de exemplu:

propagarea undelor electromagnetice funcionarea liniilor electrice lungi n regim armonic

ecuaia de tip Poisson care definete potenialul

electric ntr-un punct probleme de echilibru sau stri staionare

ecuaia Laplace etc.


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Multe fenomene studiate n cadrul electrotehnicii sunt rezultatul aciunii combinate a maimultor factori i prin urmare sunt descrise din punct de vedere matematic de ecuaii demai multe variabile. Cele mai des ntlnite sunt fenomenele dependente de doi factorii care satisfac ecuaiile cu derivate pariale de forma (unde a, b, c, d, e, f, g sunt funciicare pot depinde de cele dou variabile independente precum i de soluia nsiu=u(x,y)):

0gufy

u

ex

u

dy

u

cyx

u

bx

u

a 2

22

2

2

=++

+

+

+

+

Observations: We still speak of linear PDEs, even if the coefficients a(x,y) ... e(x,y)might be nonlinear in x and y. Linearity is required only in the unknown function uand all derivatives of u.Further simplification are: constant coefficients a-e; vanishing mixed derivatives (b=0);no lower order derivates (d=e=0)

ecuaia lui Poisson:

0gy

u

x

u2

2

2

2

=+

+

ecuaia lui Laplace: 0y

u

x

u2

2

2

2

=

+

0y

v

x

v2

2

2

2

=

+

+

- fluxul total prin orice suprafa

nchis Gauss

2

2

22

2

xuc

tu

=

- propagarea undelorelectromagnetice

0t

H

x

H2

z2

2z

2

0t

E

x

E2

y2

2y

2ecuaia undelor electromagnetice:


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How to solve PDEs?

PDEs are solved together with appropriateBoundary Conditions (B.C) and/orInitial Conditions.

B.C.

-Dirichlet B.C.: Specify u(x,y,...) on boundaries (at x=0, x=Lx, y=0, y=Ly in arectangle)-von Neumann B.C.: Specify normal gradient of u(x,y,...) on boundaries.

In principle boundary can be arbitrary shaped (but difficult to implement in

computer codes)


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Initial value problem

Boundary values are usually specified on

all boundaries of the computational domain. Initial conditions are specified in the entire

computational (spatial) domain, but only

for the initial time t=0. Initial conditions as a Cauchy problem:-Specify u and du/dt for t=0[for hyperbolic problems like electromagnetic wave

equation.]


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Cauchy Boundary conditions

Cauchy B.C. impose both Dirichlet and Von Neumann B.C. onpart of the boundary (for PDEs of 2. order).

More general: For PDEs of order n the Cauchy problemspecifies u and all derivatives of u, up to the ordern-1 on partsof the boundary.

In electrical engineering applications the Cauchy problem isoften related to temporal evolution (initial conditions specifiedfor t=0)

Fenomenele care apar in aplicatii din IE au semnificaii fizice clare, condiiile la limit (pefrontiera domeniului n care are loc procesul fizic) i condiiile iniiale (care descriu stareainiial a sistemului analizat) definindu-se alturi de problema principal.

Avem urmtoarele categorii de probleme:

1

Cauchy

initialeconditiiledause limitlaconditiiledausenu ;

2

la limit

initialeconditiiledausenu

limitlaconditiiledause;

3 mixt

initialeconditiiledauselimitlaconditiiledause .


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B2-4AC Category Example

< 0 El lipt ic Laplace equation (steady state with2 spatial dimensions)

= 0 Parabolic Heat conduct ion equat ion (time variable

with one spatial dimension)

>0 Hyperbolic Wave equat ion (t ime-variable with onespatial dimension

kT

x

T

t

2

2=

2

2 2

2

2

1y

x c

y

t=

0y

u

x

u2

2

2

2

=

+

Tipul de metode numerice preferat n rezolvarea ecuaiilor cu derivate pariale suntmetodele de diferene finite. Vom insista pe ecuaiile utilizate mai mult n electrotehnicadic hiperbolice i eliptice!

Because of their widespread application in engineering, our study of PDE will focuson linear, second-order equations (specific for electrical engineering applications)

The following general form will be evaluated forB2- 4AC

Au

xB

u

x yC

u

yD

2

2

2 2

2 0+ + + =


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Exemple de aplicaii care au ca model ecuaiidiferentiale sau ecuatii cu derivate partiale


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Analiza comportrii descarctoarelor de supratensiuni, datorate comutrii liniilor

electrice cu sarcin capacitiv, presupune modelarea liniei ca i circuit, innd cont deprezena sursei de energie, de amplasarea descrctoarelor (surge-arresters) i de naturasarcinii electrice (capacitiv), precum n figura de mai jos:

Modelul de cricuit electric

Soluionarea numeric a ecuaiei difereniale

corespunztoare circuitului, cu variabil necunoscut tensiunea la bornele descrctorului, indicvariaiile care apar pentru diferite sarcini capacitive:

Variaa tensiunii pentru diferite sarcini capacitive

Analiza comportrii descarctoarelor de supratensiuni


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Analiza stabilitii la mari perturbaii a unui generator electric racordat la SE

Se consider

un sistem electroenergetic (SEE) de configuraie simpl

:generator sincron (GS), legat la un sistem de putere infinit printr-o reearadial (transformator bloc i linie dublu circuit de nalt tensiune).

Cunoscnd parametrii elementelor de sistem i datele referitoare la un anumitregim de funcionare, se cere s se elaboreze un program de calcul pentruanaliza stabilitii la mari perturbaii a GS prin rezolvarea numeric a

ecuaiilor difereniale care descriu funcionarea n regim tranzitoriu a SEE.


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Modelul matematicAnaliza stabilitii la mari perturbaii se face prin integrarea ecuaiei

difereniale de micare a ansamblului rotoarelor generatorului iturbinei:d

d tM

P P Hm e

2

2

1= ( )

rezultnd curba de variaie n timp a unghiului intern al generatorului(curba de oscilaie) i cea a vitezei unghiulare (reprezentnd, de fapt,abaterea vitezei unghiulare fa de turaia sincron

s

= 314 rad/s)

Analiza formei acestor curbe ofer informaii n privina stabilitii saua instabilitii generatorului la perturbaia considerat.

M - constanta mecanic a ansamblului turbin-generator, Pm - putereamecanic a GS, Pe - puterea electric a GS, H - constanta de amortizare(nglobnd efectele tuturor surselor de amortizare a oscilaiilor).


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d

d td

d t MP P Hm e

=

=

1( )

t

t

t

0

0 0

0 0

0=

=

=

( )

( )

Ecuaia diferenial de ordin II se poate scrie sub forma unui sistem de dou ecuaiidifereniale de ordinul nti:

unde variabila independent este timpul t, iar puterea electric are expresia:

P P Pe e e= + 1 2 sin( )

unde Pe1, Pe2 i sunt constante, avnd valorile dependente de parametrii elementelor de sistemi de regimul de funcionare a sistemului!!!

Condiiile iniiale cunoscute sunt definite de relaia

n concluzie, analiza stabilitii tranzitorii se face prin rezolvareanumeric a sistemului de ecuaii difereniale cu condiiile iniialepentru t [ 0;tfinal ] i interpretarea rezultatelor obinute.


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0.30

0.34

0.38

0 0.5 1 1.5 2 2.5t [s]

[rad]

-0.20

-0.10

0.00

0.10

0.20

0.30

0 0.5 1 1.5 2 2.5

t [s]

[rad/s]

Curba de variaie n timp a unghiului intern

Curba de variaie n timp a vitezei unghiulare


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Studiul performanelor dinamice ale motoarelor liniare de inducie, atunci cnd serealizeaz compensarea serie este o problem studiat n domeniul proiectriimainilor electrice, acolo unde este necesar s s obin cupluri de pornire iacceleraii ridicate. Aplicaii: utilaje industriale, traciune electric. Figura de mai josindic modelul unei maini liniare de inducie si transpunerea acestui model fizic ntr-

un circuit electric:

Studiul performanelor dinamice ale unui motor de inductie


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Modelul matematic al acestui circuit este constituit dintr-o ecuaiediferenial, de unde rezult variaia curentului n condiii dinamice.

Pe baza expresiei numerice a curentului, se deduce variaia cuplului, nraport cu reglajul vitezei mainii liniare:

Fig. 13.3. Variaia cuplului n raport cu reglajul vitezei

S-a prezentat astfel un model de problem a crei soluionare estecondiionat de integrarea numeric a unor ecuaii difereniale.


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How to obtain uncoupled 2. orderPDEs from physical laws?

Example: From Maxwell equations to

wave equations. Maxwell equations are a coupled system of

first order vector PDEs. Can we reformulate this equations

to a more simple form?

Here we use the electromagnetic potentials,vector potential and scalar potential.

Analiza ecuatiilor de camp electromagnetic


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What do we win with wave equations?

Inhomogenous coupled system of

Maxwell reduces to wave equations. We get 2. order scalar PDEs

for components of electric andmagnetic potentials.

Equation are not coupled and have

same form. Well known methods exist to solvethese wave equations.


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Wave equation Electric charges and currents on right side of

wave-equation can be computed from othersources

Moments of electron and ion-distribution inspace-plasma.

The particle-distributions can be derived from

numerical simulations Here we study the wave equation in vacuum for

simplicity.


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Wave equation in vacuum


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Ecuaia liniilor de camp create de o sarcin n micare n planul 21oxx sub aciuneaunui cmp de fore, este o ecuaie diferenial total exact;

Pentru rezolvarea numeric a ecuaiilor difereniale ca model matematic rezultat prinstudiul (analiza) numeric a circuitelor electrice n regim tranzitoriu avem de construit mai

muli pai. Finalitatea este determinarea variaiei n timp a curenilor prin laturile unui

circuit RLC alimentate de la surse de tensiune sau curent variabile n timp. Intervalul de

timp n care se face analiza [ ]maxt,t0 este mprit ntr-o reea de discretizare uniform cupasul h si n noduri kt ;

Micarea unui electron supus unui cmp electric Er

i a unui cmp magnetic Hr

satisface ecuaia diferenial vectorial: )HvE(m

|e|

dt

vd rrrr

0+= ;

Rezolvarea unei ecuaii difereniale asociate unui circuit electric de ordin I sau IIexcitat cu un impuls regim tranzitoriuCondensator de capacitate C care se ncarc de la o surs de tensiune continu E,

printr-un rezistor de rezisten R.

Descrcare unui condensator de capacitate C, ncrcat iniial la tensiunea E, pe unrezistor de rezisten R.

Alte aplicatii

T t i t l id d i l d i d t t


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Testarea varistoarelor cu oxid de zinc, la descrcri de trsnet;Calculul regimului tranzitoriu al unui motor electric asincron; sistem de ecuaiidifereniale;

Studiul efectului de stimulare magnetic a esuturilor nervoase;Determinarea caracteristicilor magnetice neliniare ale unor dispozitiveelectromagnetice, prin testarea experimental cu semnale alternative sinusoidale, sau ntrepte;

Caracterizarea comportrii n regim dinamic a motoarelor cu reluctan variabil(SRM), n vederea mbuntirii parametrilor constructivi pentru reducerea variaiilorrapide de cuplu;Proiectarea cuplajelor motor main de lucru, care utilizeaz fluide magneto-rheologice, cu proprieti de orientare sub aciunea unui cmp magnetic;Proiectarea estimatoarelor de vitez la motoarele electrice cu flux transversal ireluctan variabil (Transverse Flux Reluctance Machines, TFRM); se realizeazstudiul comportrii n condiiile reglajului de vitez;Reprezentarea ca i circuit i simularea funcional a unei celule nervoase;

Studiul comportrii unui galvanometru cu cadru mobilSimularea comportrii n regim tranzitoriu a miezurilor magnetice din diferitemateriale, utilizate pentru fabricarea transformatoarelor de putere;Studiul declanrii accidentale a proteciilor difereniale, datorit componentei

continue i a armonicilor care apar n curentul de magnetizare la cuplareatransformatoarelor de putere.

1. S se arate care este soluia problemei Neumann (dac exist) pentru sfera de raza cuAlt li tii


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. S se e c e es e so u p ob e e eu (d c e s ) pe u s e de a cu

condiia 3cos4cos2 =

n

upe suprafaa sferei.

2. S se afle soluia ecuaiei 22

2

2

2

=

+

=

y

u

x

uu , n dreptunghiul ax


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; y

limit 0),0( =yu ; 0),( =yau ;

=a

xAxu 1)0,( ; 0),( =xu .

2. S se afle repartiia potenialului n interiorul unui cilindru drept nchis la baze tiindc:

a) pe ambele baze potenialul are valoarea zerob) pe suprafaa lateral potenialul este o funcie dat de z.

3. S se afle repartiia potenialului n interiorul unui cilindru circular drept nchis la bazainferioar, prezentat n figura, tiind c:

a. Pe suprafaa laterali pe baza inferioar potenialul este nul.

b. La baza superioar aplicm un capac circular plan pe care potenialul este constant i

egal cu0

u .

4. Se considera o linie de transmisie fr distorsiuni CGLR = de lungime finit l, estencrcat la un potenial E. Captul x=l este izolat, iar la momentul t=0 captul x=0

este pus la pmnt. S se demonstreze c potenialul n punctul x este:

=

+

+

+=

0 2)12(sin

2)12(cos

1214

n

t

lxn

ltan

neEv

; unde LR= ; LCa 12 = .

h

z

y

xO

A(a,0,0) u=0

u=0

u=u0

I Rezolvarea numerica a ecuatiilor


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Ecuaiile difereniale cu derivate pariale de tip eliptic sunt asociate problemelor de stristaionare sau de echilibru. Exemplele cele mai cunoscute de ecuaii eliptice sunt cele de tipPoisson, i cele de tip Laplace

Typically used to characterize steady-state boundary value problems

Au

xB

u

x yC

u

yD

2

2

2 2

20+ + + =

I. Rezolvarea numerica a ecuatiilor

eliptice

)y,x(fy

u

x

uu

2

2

2

22 =

+

= 0

y

u

x

uu

2

2

2

22 =

+

=

Poisson i Laplace:

Observaie: Ne vom mrgini la cazul bidimensional. Extinderea procedeelor de

rezolvare pentru cazul tridimensional nu prezint dificulti teoretice. Domeniul deintegrare a ecuaiilor eliptice n dou dimensiuni este o suprafa S mrginit de conturulnchis C. Condiiile pe frontier specific de obicei fie valoarea soluiei fie valoareaderivatei acesteia pe normala n orice punct al lui C. Aceste dou tipuri de condiiidefinesc problemele Dirichlet respectiv Neumann. Alteori condiiile pe frontier sunt datesub forma unei combinaii liniare a celor dou tipuri.

P i E i


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2

2

2

2

u

x

u

y

f x y+ = ( , )

Now lets consider solution techniques.

Poisson Equation

Ecuaia care definete potenialul electric ntr-un punct de coordonate (x,y,z) dininteriorul unei suprafee nchise pe care este distribuit o sarcin electric de densitatesuperficial este de tip Poisson, fiind permitivitatea dielectric a mediului

2 V =2

2

x

V

+

2

2

y

V

+

2

2

z

V

= -

unde variabilele x [0,a]; y [0,b].

Condiiile pe frontier reprezentnd valorile lui u(x,y) pentru x = 0 i x = a ,respectiv pentru y = 0 i y = b , unde s(y) , d(y) , f(x) i g(x) sunt funcii cunoscute.

u y s y( , ) ( )0 = u a y d y( , ) ( )=

u x f x( , ) ( )0 = u x b g x( , ) ( )=

We will consider cases involving electromagnetic field computation

Domeniul de definiie limitat i n partea superioar, se discretizeaz cu pasul hx dup


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o e u de de e a pa ea supe oa , se d sc e ea cu pasu x dupvariabila x , respectiv cu pasul hy dup variabila y . Se consider c exist n intervale hxpe axa x , respectiv m intervale hy pe axa y . Valorile funciei u n nodurile reelei de

discretizare se noteaz cu ui,j, semnificnd u(xi,yj), pentru xi = ihx i yj = jhy .Rezolvarea ecuaiei difereniale cu derivate pariale de tip eliptic nseamn, de fapt, calcululvalorilor u(xi,y ) pentru i = 1, 2, ..., n-1 ,j = 1, 2, ..., m-1

Metode de soluionare cu diferene finite


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Metode de soluionare cu diferene finiteElementul specific al metodei - valorile necunoscute ale funciei u(x,y) se calculeaz n modindirect n funcie de cele cunoscute, aproximnd operatorii difereniali prin operatori de diferene

finite i soluionnd sistemele de ecuaii liniare care rezult n aceste condiii.Metoda foloseste pentru operatorii difereniali de ordinul al doilea - relaiile (deduse in cursulprecedent) care conin numai valori ale funciei u(x,y) corespunztoare lui xi , respectiv lui yj :

2

2

1 1

2

2u

x

u u u

hi j

i j i j i j

x,

, , ,

=

++

2

21 1

2

2u

y

u u u

hi j

i j i j i j

y,

, , ,= ++

nlocuind expresiile n relaia care definete ecuaia diferenial de tip eliptic, rezult:

j,i

y

j,ij,ij,i

x

j,ij,ij,i

fh

uuu

h

uuu

=

+

+

+ ++2

11

2

11 22

2

2

2

2

u

x

u

yf x y+ = ( , )

Pentru cazul particular cnd se adopt acelai pas de discretizare h dup cele dou axe ( hx = hy= h ), relaia ajunge la forma relativ mai simpla:

j,ij,ij,ij,ij,ij,i fhuuuuu =+++ ++ 21111 4

Relaiile de forma scrise pentru i = 1, 2, ..., n-1 , j = 1, 2, ..., m-1 , conduc la un sistem


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Dirichlet boundary conditions: u is specified at theboundary

Neumann boundary condition: the derivative of u isspecified

Combination of both u and its derivative (mixedboundary condition) Algoritmul este formula n 5 puncte i trebuie scris acest

algoritm sub form matricial.

Condiiile pe frontier ne dau valorile urmtoare: j,0u , (j = 0, 1, 2, , p); 0,iu , (i = 0, 1, 2, , p);

j,pu , (j = 0, 1, 2, , p); p,iu , (i = 0, 1, 2, , p).

Prin urmare avem de determinat (p-1) x (p-1) valori j,iu , (i,j= 0, 1, 2,,p-1)

p , , , , j , , , ,liniar de ecuaii de ordinul (n-1) (m-1) = n m - n - m + 1 , necunoscutele fiind ui , j = u(x i , y j) , i = 1, 2, ..., n-1 , j = 1, 2, ..., m-1 , care se rezolv cu metodele prezentate n

cursul 3, aplicate ntr-o manier ct mai eficient.Equation holds on inner points onlyon the boundary we specify

Boundary Conditions :

Avem trei cazuri:


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1. Dou din cele 5 valori ale lui u sunt cunoscute din condiiile pe frontier.Acest caz se ntmpl n cele 4 coluri:

1ji == ; 1pji == ; 1pi = ; 1j = ; 1i = ; 1pj = .n acest caz algoritmul are trei necunoscute n membrul stng.De exemplu pentru 1ji == avem:

1,11,22,11,1 buuu4 =++ 1,00,112

1,1 uufhb =

2. Una din cele 5 valori ale lui u este cunoscut din condiiile iniiale. Pentrutoate punctele orizontale i verticale adiacente frontierei, cu excepiacolurilor

1i = sau 1pi = i (j = 2, 3 , p-2);1j = sau 1pj = i (i = 2, 3 , p-2);

n acest caz algoritmul conine patru necunoscute.De exemplu pentru 1i = , 3j = avem:

3,13,24,13,12,1 buuu4u =++

3. Cazul punctelor interioare cnd nici una din valorile lui u care intervin n(2) nu este furnizat de condiiile pe frontier. Acesta este cazul cel mai

frecvent i are loc pentru orice pereche (i, j) i = 2, 3 , p-2 i j = 2, 3 , p-

iar algoritmul se scrie:

j,ij,1i1j,ij,i1j,ij,1i buuu4uu =+++ ++

3,03,12

3,1 ufhb =

j,i2

j,i fhb =

Avem toate elementele necesare pentru a scrie algoritmul n forma matricial:


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buA =

A matrice ptratic (p-1) x (p-1)u, b matrici coloan cu (p-1) x (p-1) elemente.

Matricea A are structura:

=

M

I

I

MI

......

IMIIM

A

=

41

14

141

141

14

.......M

=

1p,1p

2,1p

1,1p

1p,2

2,2

1,2

1p,1

2,1

1,1

u

uu

u

u

u

u

u

u

u

M

K

M

K

M

=

11

2111

12

22

12

11

21

11

p,p

,p

,p

p,

,

,

p,

,

,

b

bb

b

b

b

b

b

b

b

M

K

M

K

M

Deci rezolvarea numeric a ecuaiei Poisson conduce la rezolvarea unui sistem deecuaii liniare.

Aplicatia 1.


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Aplicatia 1.

Se consider ecuaia diferenial cu derivate pariale de tip eliptic definit de relaia cu

condiiile pe frontier, domeniul de variaie pentru cele dou variabile fiind precizat2

2

2

24

u

x

u

y+ =

u y s y( , ) ( ) = y20 = u a y d y( , ) ( ) = 1+ y2=

u x f x( , ) ( ) = x20 = u x b g x( , ) ( ) =1+ x2=

xy

0;10 ;1[ ]

[ ]

Soluionarea se efectueaz cu algoritmul prezentat (Metoda cu diferene finite), pentru paiide discretizare adoptndu-se valorile h x = 0,25 ( n = 4 ), respectiv h t = 0,25 ( m = 4 ),

ceea ce conduce la x0 = y 0 = 0 , x 1 = y 1 = 0,25 , x 2 = y 2 = 0,50 , x 3 = y 3 = 0,75 , x 4 =y 4 = 1 . Rezult un sistem liniar de ordinul 9 , avnd ecuaiile de forma

4 1 0 1 0 0 0 0 0

1 4 1 0 1 0 0 0 0

0 1 4 0 0 1 0 0 01 0 0 4 1 0 1 0 0

0 1 0 1 4 1 0 1 0

0 0 1 0 1 4 0 0 1

0 0 0 1 0 0 4 1 0

0 0 0 0 1 0 1 4 10 0 0 0 0 1 0 1 4

1 1

2 1

3 1

1 2

2 2

3 2

1 3

2 33 3

u

u

uu

u

u

u

uu

,

,

,

,

,

,

,

,,

=

0 0625 0 0625 0 0625

0 0625 0 2500

0 0625 0 5625 106250 0625 0 2500

0 0625

0 0625 12500

0 25 0 5625 10625

0 25 125000 25 15625 15625

44

44

4

44

44

, , ,

, ,

, , ,, ,

,

, ,

, , ,

, ,, , ,


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Rezultatele obinute, mpreun cu valorile corespunztoare condiiilor pe frontier, adictoate valorile funciei u(x,y) , sunt prezentate n tabel. Aa cum era de ateptat, pe baza

formei expresiilor care definesc condiiile pe frontiere, se remarc o simetrie a valorilorcalculate fat de diagonala principal.

j \ i 0 1 2 3 4 i / j0 0,0000 0,0625 0,2500 0,5625 1,0000 01 0,0625 0,1250 0,3125 0,6250 1,0625 12 0,0625 0,3125 0,5000 0,8125 1,2500 23 0,5625 0,6250 0,8125 1,1250 1,5625 3

4 1,0000 1,0625 1,2500 1,5625 2,0000 4j / i 0 1 2 3 4 i \ j

Aplicatia 2.


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Se prezint structura sistemului liniar de ecuaii pentru situaia particular n = 5 i m =4 , care rezult de ordinul 12 , remarcndu-se forma matricei coeficienilor. Reeaua dediscretizare corespunztoare este prezentat n figura.

4 1 0 0 1 0 0 0 0 0 0 0

1 4 1 0 0 1 0 0 0 0 0 0

0 1 4 1 0 0 1 0 0 0 0 0

0 0 1 4 0 0 0 1 0 0 0 0

1 0 0 0 4 1 0 0 1 0 0 0

0 1 0 0 1 4 1 0 0 1 0 0

0 0 1 0 0 1 4 1 0 0 1 0

0 0 0 1 0 0 1 4 0 0 0 1

0 0 0 0 1 0 0 0 4 1 0 0

0 0 0 0 0 1 0 0 1 4 1 0

0 0 0 0 0 0 1 0 0 1 4 1

0 0 0 0 0 0 0 1 0 0 1 4

=

u

u

u

u

u

u

u

u

u

u

u

u

h s f

h f

h f

h d f

h s

h

h

1 1

2 1

3 1

4 1

1 2

2 2

3 2

4 2

1 3

2 3

3 3

4 3

21 1 1 1

21 2 2

21 3 3

21 4 1 4

22 1 2

22 2

22

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

3

22 4 2

23 1 3 1

23 2 2

23 3 3

2 3 4 3 4

h d

h s g

h g

h g

h d g

p

II Rezolvarea numerica a ecuaiilor


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II. Rezolvarea numerica a ecuaiilor

hiperboliceUn exemplu clasic n electrotehnic de ecuaie cu derivate de tip hiperbolic este ceacare descrie funcionarea liniilor electrice lungi n regim armonic, numit ecuaiatelegrafitilor:

Dac se dorete rezolvarea ecuaiei telegrafitilor n cazul general al regimuluitranzitoriu trebuiesc avute n vedere unde multiple care se propag pe linie, unde carese reflect, se refract i se atenueaz n funcie de conductoarele liniei, de mediul(dielectricul) nconjurtori de condiiile iniiale i la limit impuse de regimul tranzitoriu.n cazul regimului staionar rezolvarea ecuaiei pune n eviden propagarea pe linie a

unor unde asociate avnd frecvena sursei de alimentare.

( )2 2

0 0 0 0 0 0 0 02 2

u u uR G u R C G L L C

x t t

= + + +

unde u(x,t) reprezint valoarea instantanee a tensiunii de faz la momentul t > 0 ntr-un

anumit punct al liniei, de lungime total , definit prin distana x [ 0 ; ] fa de sfrit,iar R0,L0,G0 i C0 sunt parametrii unitari ai liniei (rezistena, inductana, conductana icapacitatea pe unitate de lungime).


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Consider one segment at position x along the line:

S d d t l h ' ti


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Second order telegrapher's equation

( )2 2

0 0 0 0 0 0 0 02 2

u u uR G u R C G L L C

x t t

= + + +

( )

2 2

0 0 0 0 0 0 0 02 2i i iR G i R C G L L Cx t t = + + + If the line is lossless, then R0 = G0 = 0, therefore:

2

2

22

2

002

2 1

t

i

vt

iCL

x

i

=

=

22

222 1

tu

vxu

=

00

1CL

v

=

The velocity of the propagating wave for a 2parallel wire TL with diameters muchsmaller than the distance between the axesand the length of the TL

The voltage and current traveling along the TL are governed by waveequations


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BMTL-Software Applications


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Branched Multi-conductor TL

A1. Signal propagation along

a 3 TL network

A3 Propagation of the signal along

A2. Propagation of a signal along a

coaxial cable


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A3. Propagation of the signal along

a TL with 2 parallel conductors

A4. Signal propagation along a TL

excited by a incident plane wave

coaxial cable

Dac se consider linii electrice lungi ideale, cu alte cuvinte se neglijeaz pierderilelongitudinale i transversale de putere activ (R=G=0 )


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longitudinale i transversale de putere activ (R G 0 )

2

2

x

u

= LC 2

2

t

u

; u(x,t) = u

Prezentarea metodelor de soluionare numeric a ecuaiilor hiperbolice va fi fcutpentru ecuaia diferenial de forma

22

22

u

x

u

t=

domeniul de variaie pentru cele dou variabile fiindxt

0,0,t

f

[ ]

[ ]

Condiiile iniiale reprezentnd valorile funciei u(x,t) i ale derivatei sale pariale nraport cu variabila t pentru t = 0 sunt

u x f x( , ) ( )0 = u x tt

g x

t

( , )( )

=

=

0

f(x) i g(x) - funcii cunoscute

Condiiile pe frontier reprezentnd valorile funciei u(x,t) pentru x = 0 si pentru x = 1

u t s t( , ) ( )0 = u t d t( , ) ( )1 = s(t) i d(t) - funcii cunoscute.

Domeniul de definiie se discretizeaz cu pasul hxdup variabila x , respectiv cu pasul ht dup


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dup variabila x , respectiv cu pasul ht dupvariabila t . Se consider c exist n intervale hx

pe axa x , respectiv m intervale ht pe axa t .Valorile funciei u n nodurile reelei de discretizarese noteaz cu ui,j , semnificnd u(xi,tj) ,pentru xi = ihx i tj = jht .

Rezolvarea ecuaiei cu derivate partiale se va face progresiv: pornind de la condiiileiniiale definite se determin succesiv valorile funciei u pentru

t1 = h t , t2 = 2ht , ... , tm = mht , adic ui,1, i = 1, 2, ..., n-1,ui,2, i = 1, 2, ..., n-1 , ... , ui,m , i = 0, 1, 2, ..., n-1 .

Algoritm explicit pentru soluionarea ecuatiilor hiperbolice


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Elementul specific al metodelor cu algoritm explicit const n aceea c valorilenecunoscute se calculeaz n mod direct n funcie de cele cunoscute, prin aproximarea

corespunztoare a operatorilor difereniali cu operatori de diferene finite.

Versiunea clasic a metodei cu algoritm explicit utilizeaz cea mai simpl modalitate deaproximare numeric a operatorilor difereniali, astfel nct s rezulte relaii directe decalcul al valorilor funciei u(x,t) n punctele definite de nodurile reelei de discretizare.

Dac se consider ecuaia hiperbolic, atunci trebuie s se efectueze discretizareapentru operatorul diferenial de ordinul al doilea 2/ x2 , 2/ t2 . n acest scop, seutilizeaz elementele prezentate n legtur cu diferenele finite, precum i cu metodele

i formulele de derivare numeric, n cursul 10:2

21 1

2

2u

x

u u u

hi j

i j i j i j

x,

, , ,= ++ 2

21 1

2

2u

t

u u u

hi j

i j i j i j

t,

, , ,= ++

2

2

2

2

u

x

u

t=

Inlocuind in ecuatia hiperbolica rezulta relaia caracteristic a metodei explicite

u u r u r u ui j i j i j i j i j, , , , ,( ) ( )+ + = + + +1 1 1 12 1

unde coeficientul r se determin cu relaiar

h

h

t

x=

2

2

Rezolvarea efectiv a ecuaiei hiperbolice cu condiiile iniiale i cu condiiile pefrontier date implic aplicarea relaiei explicite pentru determinarea succesiv a


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p p p pvalorilor

ui,2, i = 1, 2, 3, ..., n-1 , ui,3 , i = 1, 2, 3, ..., n-1 , ... , ui,m , i = 1, 2, ..., n-1 , adic pentru j = 1,2, ..., m-1 , i = 1, 2 , ..., n-1

Se observ c relaia dedusa nu se poate utiliza la calculul valorilor func iei u(x,t) pentrut1 = ht , deoarece apare i indicele j -1 (pentru j+1= 1 sunt disponibile numai valorile

corespunztoare lui j = 0, date chiar de condiiile iniiale). n consecin, pentru demarareaalgoritmului trebuie gsit o alt relaie de calcul, care s utilizeze numai condiiile iniiale. nacest scop, pentru operatorul diferenial / t se folosete cea mai simpl relaie de derivarenumeric care definesc diferen ele finite de ordinul 1 , rezultnd

ut

u u

hii i

t,, ,

001=

ceea ce conduce la expresia pentru valorile cutate u i,1 , i = 1, 2, ..., n-1

u h g f i t i i,1 = +

Algoritmul prezentat este stabil numeric pentru valori subunitare ale parametrului, exist i

posibilitatea adoptrii unor alte relaii de discretizare pentru operatorii difereniali, rezultnd metodedirecte mai sofisticate de rezolvare a ecuaiilor difereniale de tip hiperbolic

Aplicatie 1.Soluionarea numeric cu metoda explicita a unei ecuaii difereniale hiperbolice pentru tf= 1


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2

2

2

2uxu

t= u x f x( , ) ( ) = (x )0 = sin u x tt g xt

( , ) ( ) = 0=

=0

u t s t( , ) ( ) = 00 = u t d t( , ) ( ) = 01 =xt

0;10;t

f

[ ]

[ ]

Soluionarea se efectueaz cu algoritmul prezentat, pentru paii de discretizare adoptndu-se

valorile hx = 0,1 ( n = 10 ), respectiv ht = 0,1 ( m = 10 ), ceea ce pentru coeficientul rnseamn r = 0,12/0,1

2 = 1 .Forma condiiei iniiale i a condiiilor pe frontier indic simetria funciei u(x,t)

fa de x = 0,5 , ceea ce nseamn ui,j = u10-i,j , i = 1, 2,3,4. Rezultatele complete suntdate n tabelul 12.5.8. Pentru primele dou linii (corespunztoare lui j = 1, 2 ) se prezint

n detaliu calculul valorilor u1,j, u2,j , u3,j , u4,j i u5,j (valorile aferente ale variabilei xfiind x1= 0,1,x2 = 0,2,x3 = 0,3,x4 = 0,4, x5 = 0,5, iar x0 = 0 i x10 = 1 reprezentndfrontierele domeniului :

a) valorile pentru t1 = 0,1, innd cont de faptul c 1- r = 0 , respectiv g i = 0

u h g r f r f f t1 1 1 1 2 01 0 5 0 5 1 0 5878 0 0 2939, ( ) , ( ) , ( , ) ,= + + + = + =

u h g r f r f f t2 1 2 2 3 11 0 5 0 5 1 0 8090 0 3090 0 5590, ( ) , ( ) , ( , , ) ,= + + + = + =

u h g r f r f f t3 1 3 3 4 21 0 5 0 5 1 0 9511 0 5878 07694, ( ) , ( ) , ( , , ) ,= + + + = + =

u h g r f r f f t4 1 4 4 5 31 0 5 0 5 1 10000 0 8090 0 9045, ( ) , ( ) , ( , , ) ,= + + + = + =

u h g r f r f f t5 1 5 5 6 41 0 5 0 5 1 0 9511 0 9511 0 9511, ( ) , ( ) , ( , , ) ,= + + + = + =

b) valorile pentru t2 = 0,2 (innd cont c 1- r = 0 )

u u r u r u u2 1( ) ( )= + + + =


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u u r u r u u1 2 1 0 11 2 1 0 12 1

0 3090 1 0 5590 0 0000 0 2500, , , , ,( ) ( )

, ( , , ) ,

= + + + =

= + + =

u u r u r u u2 2 2 0 2 1 3 1 112 10 5878 1 0 7694 0 2939 0 4755

, , , , ,( ) ( ), ( , , ) ,

= + + + == + + =

u u r u r u u3 2 3 0 3 1 4 1 2 12 1

0 8090 1 0 9045 0 5590 0 6545, , , , ,( ) ( )

, ( , , ) ,

= + + + =

= + + =

u u r u r u u

4 2 4 0 4 1 5 1 3 1

2 1

0 9511 1 0 9511 07694 0 7694, , , , ,( ) ( )

, ( , , ) ,

= + + + =

= + + =

u u r u r u u5 2 5 0 5 1 6 1 4 12 1

10000 1 0 9045 0 9045 08090, , , , ,( ) ( )

, ( , , ) ,

= + + + == + + =

j tj u0, j u1, j u2, j u3, j u4, j u5, j0 0,0 0,0000 0,3090 0,5878 0,8090 0,9511 1,0000

1 0,1 0,0000 0,2939 0,5590 0,7694 0,9045 0,9511

2 0,2 0,0000 0,2500 0,4755 0,6545 0,7694 0,8090

3 0,3 0,0000 0,1816 0,3455 0,4755 0,5590 0,5878

4 0,4 0,0000 0,0955 0,1816 0,2500 0,2939 0,3090

5 0,5 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000

6 0,6 0,0000 -0,0955 -0,1816 -0,2500 -0,2939 -0,3090

7 0,7 0,0000 -0,1816 -0,3455 -0,4755 -0,5590 -0,5878

8 0,8 0,0000 -0,2500 -0,4755 -0,6545 -0,7694 -0,8090

9 0,9 0,0000 -0,2939 -0,5590 -0,7694 -0,9045 -0,9511

10 1,0 0,0000 -0,3090 -0,5878 -0,8090 -0,9511 -1,0000

Observat ion 1.


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This simplest case is where the boundaries are specif ied as f ixedvalues. This case is known as the Dirichlet boundary condit ions.

u1

u2

u3

u4

Obse at oLet s consider how to model the Dir ichlet boundary condit ion

Consider how we can deal with the lower node shown, u1,1


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u1

u2

u3

u4

-4u1,1 +u1,2+u2,1+u1 +u4 = 0

1,2

1,1 2,1

Note:

This grid would resultin nine simult aneous

equat ions.

Observat ion 2


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Observat ion 2Let s consider how to model the Neumann boundary condit ion

u

x

u u

x

i j i j+ 1 1

2, ,

centered finite divided differenceapproximation

suppose we wanted to consider this endgrid point

(0,100)

(0,0) (200,0)

(200,100)u = 0.05x + 100

0=xu

0

22

22 =+

y

u

x

u

0=y

u

y

x

0=xu


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1,2

1,1 2,1

0=x

u

0=yu

The two boundaries areconsider to be symmetrylines due to the fact thatthe BC translates in the

f inite dif ference form to:

ui+1,j = ui-1,j

and

ui,j+1

= ui , j-1


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1,2

1,1 2,1

0=yu

u1,1 = (2u1,2 + 2 u2,1)/ 4

u1,2 = (u1,1 + u1,3+2u22)/ 42,2

0=x

u

III. Semi-analytic methods to solve

PDE


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PDEs.

From systems of coupled first order PDEs(which are difficult to solve) to uncoupledPDEs of second order.

Example: From Maxwell equations

to wave equation. (Semi) analytic methods to solve the

wave equation by separation of variables.

Exercise: Solve Diffusion equationby separation of variables.

Semi-analytic methods to solve PDEs.

Variable separation


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Example: Homogenous wave equation

Can be solved by any analytic function f(x-ct) andg(x+ct).

As the homogenous wave equation is a linearequation any linear combination of f and g is also asolution of the PDE.

This property can be used to specify boundary andinitial conditions.

The appropriate coefficients have to be found often

numerically.

Variable separation


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IV. Remember metode numerice de

rezolvare a sistemelor de ecuatii


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We can interpret u as a vector and write the equationformally as an algebraic matrix equation:

Theoretical one could solve this algebraicequation by well known algebraicequation solvers like Gauss-Jordan elimination.

This is very unpractical, however, because Ais very large and contains almost only zeros.

j,ij,ij,ij,ij,ij,i fhuuuuu =+++ ++2

1111 4

How large is A ?


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g

For a very moderate 2D-grid of 100x100-u has 100 x 100= 104 gridpoints, but-A has 104 x 104 =108 entries!

For 3D-grids typically used in scienceapplication of about 300 x 300 x 300

-u has 3003= 2.7 *107 gridpoints,-A has (2.7 *107 ) 2 =7.29*1014 entries!

=> Memory requirement for 300-cube to store

u ~100 MB, A~3Million GByte

Structure ofA ?


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0

0

0 0

How to proceed?


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We have reduced our original PDEto algebraic equations (systemof linear equations, because we started

from a linear PDE.) To do: Solve these equations

As exact Matrix solvers are of no much usewe solve the equations numerically byRelaxation methods.

Relaxation: Jacobi method


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Relaxation: Jacobi method

we derived the algebraic equations:

From

Assume any initial value, say u=0 on all grid points (exceptthe specified boundary values of course) and compute:

Use the new values ofu as input for the right side and

repeat the iteration until u converges. (n: iteration step)

Jacobi method converge for diagonal


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g g

dominant matrices A.(diagonal entries of A larger than the others)

This condition is usually fulfilled for Matrixequations derived from finite differencing.(Tridiagonal block matrix: Most entries in A arezeros!)

Jacobi method converges (but slowly) and can beused in principle, but maybe we can improve it?

For practice: Method should converge fast!

Gauss Seidel method


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Gauss Seidel method

Similar as Jacobi method.

Difference: Use on the right-handside already the new (assumed to be better)approximation un+1

How fast do the methods


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converge?To solve:

We split A as:LowerTriangle

DiagonalElements

UpperTriangle

For the rth iteration step of the Jacobi method we get:

How fast do the methods


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converge?We have to investigate the iteration matrix

Eigenvalues of iteration matrix define howfast residual are suppressed. Slowest decaying

Eigenmode (largest factor) defines convergencerate. => Spectral radius of relaxation operator.0 <


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the overall error by a factor of10p

?In general:

For a J x J grid and Dirichlet B.C. one gets:

Jacobi method Gauss Seidel method

Can we do better?


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Gauss Seideliteration:

Can be rewritten as:

Successive Overrelaxation


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(SOR)

Now we overcorrect the residual error by

overrelaxationparameter

Method is only convergent for 0


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For the optimal overrelaxation parameter w the number

of iteration steps to reduce the error by 10-p are:

Number of iteration steps increases only linear with

the number of mesh points J for SOR method.For Jacobi and Gauss Seidel it was ~J2

SOR method only more effective when


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SOR method only more effective whenoverrelaxation parameter w is close its optimum.

Some analytic methods exist to estimate optimum w,but often one has to find it empirically.

Unfortunately the optimum value w does not dependonly on the PDE, but also on the grid resolution.

The optimum asymptotic w is not necessarily a

good initial choice. Chebyshev acceleration changes w during iteration.

Generalization of SOR-method.


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Finite difference schemes from 2D-elliptic PDEs have the form:

for our example

We iterate for the solution by

and get:

Generalization to 3D is straight forward

Computing time for relaxation methods

For a J x J 2D-PDE the number of iteration


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For a J x J 2D PDE the number of iterationsteps is ~J2 (Jacobi GS) or ~J (SOR)

But: Each iteration step takes ~J2

Total computing time: ~J4

(Jacobi, GaussSeidel)

~J3 (SOR-method)

Computing time depends also on other factors:-required accuracy

-computational implementation-IDL is much slower as C or Fortran-Hardware and parallelization

How fast are errors smoothed out?


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We use Gauss-Seidel 40x40 box and investigatea high frequency (k=10) disturbance.

Converged (Error


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We use Gauss-Seidel 40x40 box and investigatea low frequency (k=1) disturbance.

Converged (Error


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We use Gauss-Seidel on JxJ boxes and investigatenumber of steps to converge for different frequencies

28722168800160

142667261580

11132474740

4020101k

J

Gauss-Seidel method is very good smoother!


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Same test with SOR method

189179173844160

15214614121380

1191121098140

4020101k

J

SOR is a faster solver, but NOT good smoother!


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High frequency errors are smoothed out fast.

Low frequency errors take very longto vanish.

But the long frequency errors are

reduced faster on low resolution grids. Can we use this property to speed up

the relaxation? Yes! The answer is Multigrid

V. Remember - Finite Differences


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How to compute the differential quotient witha finite number of grid points?

First order and higher order approximations.

Central and one-sided finite differences.

Accuracy of methods for smooth and notsmooth functions.

Higher order derivatives.


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Most PDEs cannot be solved analytically. Variable separation works only for somesimple cases and in particular usually notfor inhomogenous and/or nonlinear PDEs.

Numerical methods require that the PDEbecome discretized on a grid.

Finite difference methods are popular/most commonly used in science. They replacedifferential equation by difference equations)

Engineers (and a growing number ofscientists too) often use Finite Elements.


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How to compute differential quotient

numerically? Just apply the formula above for a finite h.

For simplicity we use an equidistant grid in

x=[0,h,2h,3h,......(n-1) h] and evaluate f(x)on the corresponding grid points xi.

Grid resolution h must be sufficient high.

Depends strongly on function f(x)!

Remember the definition of differential quotient:

Accuracy of finite differences0


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We approximate the derivative of f(x)=sin(n x) on a grid x=0 ...360 with50 (and 500) grid points by f/dx=(f(x+h)-f(x))/h and comparewith the exact solution df/dx= n cos(n x)

Average error done bydiscretisation:50 grid points: 0.040500 grid points: 0.004


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Average error done bydiscretisation:50 grid points: 2.49

500 grid points: 0.256

Higher accuracy methods


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Can we use more points for higher accuracy?

Higher accuracy: Central differences


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df/dx=(f(x+h)-f(x))/h computes the derivativeat x+h/2 and not exactly at x.

The alternative formulae df/dx=(f(x)-f(x-h))/hhas the same shortcomings.

We introduce central differences:

df/dx=(f(x+h)-f(x-h))/(2 h) which providesthe derivative at x.

Central differences are of 2. order accuracy

instead of 1. order for the simpler methodsabove.

How to find higher order formulas?

F ffi i t th f ti d ib th f ti f( )


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For sufficient smooth functions we describe the function f(x)locally by polynomial of nth order. To do so n+1grid points are required; n defines the order of the scheme.

Make a Taylor expansion (Definition ):

And by linear combination we get the central difference:


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At boundary points central differences might not bepossible (because the point i-1 does not exist at theboundary i=0), but we can still find schemes of the

same order by one-sited (here right-sited) derivative:

Alternatives to one sited derivatives are periodic

boundary conditions or to introduce ghost-cells.

Higher derivatives

H t d i hi h d i ti ?


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How to derive higher derivatives?From the Taylor expansion

we derive by a linear combination:

Basic formulae used to discretise2 order Partial Differential Equations

By using more points (higher order polynomials) toapproximate f(x) locally we can get higher orders, e.g. by an

i t bi ti f


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appropriate combination of:

we get 4th order central differences:

Accuracy of finite differences

We approximate the derivative of f(x)=sin(nx) on a grid x=0 3600

with


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We approximate the derivative of f(x)=sin(nx) on a grid x=0..360 with50 (and 500) grid points with 1th, 2th and 4th order schemes:

0.00170.131.60n=20, 500 pixel8.19.913.5n=20, 50 pixel

4.5 10-50.00860.26n=8, 500 pixel0.150.822.49n=8, 50 pixel4.9 10-61.7 10-50.004n=1, 500 pixel5.4 10-60.00170.04n=1, 50 pixel4

th

order2th

order1th

order

What scheme to use?

Higher order schemes give significant better


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Higher order schemes give significant betterresults only for problems which are smoothwith respect to the used grid resolution.

Implementation of high order schemes makes moreeffort and take longer computing time,in particular for solving PDEs.

Popular and a kind of standard aresecond order methods.

If we want to feed our PDE-solver with(usually unsmooth) observed data higherorder schemes can cause additional problems.

VI. Remember - Finite Element

Method (FEM)


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Arbitrary shaped boundaries are difficult to implement infinite difference methods.Alternative: Finite Elements, popular in particular to solve

PDEs in engineering/structural mechanics.

Finite Elements


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FEM covers the space with finite elements (in 2D oftentriangles, in 3D tetrahedra). The elements do not needto have the same size and shape.

This allows to use a higher resolution where needed.

Variational formulation: 1D

example


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exampleIf u fulfills P1 and v(x) is an arbitrary function whichvanishes on the boundary:

Partial integration of right side

Weak formulation

of the PDE

Solution of weak problem and original PDE are identical.

Variational formulation: 2D example

Poisson equation


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Poisson equation

For an arbitrary function v the PDE can again

be formulated in weak form (using Greens theorem):

If we find a solution for the weak problem, we

solved our (strong form) original PDE.Order of derivatives is reduced in weak form,which is helpful to treat discontinuities.

Shape function v

How to choose the function v ?


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How to choose the function v ? v must be at least once differentiable.

For FEM-approach one takes polynomialsor in lowest order piecewise linearfunctions:

1D 2D

Basis of functions for v

We choose piecewise linear functions which are one


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We choose piecewise linear functions which are oneat a particular grid-point and zero at all other grid-points (triangle or tent-function)

Basic tent-function (blue)and superposition topiecewise linear function (red)

We get function value andderivative by interpolation.

For such base-functions almost allintegrals in the form:


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For such base functions almost allintegrals in the form:

1D 2D

are zero. Only integrals of elements sharinggrid points (edges of triangles in 2D) arenon-zero.

From FEM to matrix form

Lets try to describe the unknown function u(x) and thek f( ) ith th b i f ti


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Let s try to describe the unknown function u(x) and theknown f(x) with these basis functions:

Aim: Find the parameters uk!This will be the solution in FEM-approach.

How to find this solution?Insert this approaches for u and f into the weak formulation ofthe PDE.


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which leads to a system of equations which has to be solvedfor uk. We can write in matrix form:

Lkj Mkj

This sparse matrix system can be solved with numericalmethods studied in II year Numerical Methods!

Lets remember all steps:

Original PDE( f )


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g(strong form)

PDE inweak form

PDE indiscretizedform

Solve corresponding sparse Matrix system:=> Solution of PDE in FEM-approach.

Finite Element Method are an alternative toFinite Difference Method


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Finite Difference Method

Very good alternative for complicated

boundaries. PDE is solved in weak form.

More flexible as finite differences, but alsomore complicated to implement in code.

Setting up the optimal grid can be tricky.

(Some research groups only work on this.)

http://www.comsol.com/shared/downloads/comsol_v4_building_model.pdf


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Mathematically the relation between the magnetic and electric field at the boundary are


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