FINITE DIFFERENCE AND PDE. Haryo Tomo

Transkripsi

1 FINIE DIFFERENCE AND PDE Haryo omo

2 Numercal methods: propertes Fnte derences - tme-dependent PDEs -> robust, smple concept, easy to parallelze, regular grds, eplct method Fnte elements - statc and tme-dependent PDEs -> mplct approach, matr nverson, well ounded, rregular grds, more comple algorthms, engneerng problems Fnte volumes - tme-dependent PDEs -> robust, smple concept, rregular grds, eplct method

3 Other numercal methods Partcle-based methods - lattce gas methods - molecular dynamcs - granular problems - lud low - earthquake smulatons -> very heterogeneous problems, nonlnear problems Boundary element methods - problems wth boundares (rupture) - based on analytcal solutons - only dscretzaton o planes -> good or problems wth specal boundary condtons (rupture, cracks, etc) Pseudospectral methods - orthogonal bass unctons, specal case o FD - spectral accuracy o space dervatves - wave propagaton, ground penetratng radar -> regular grds, eplct method, problems wth strongly heterogeneous meda

4 What s a nte derence? Common dentons o the dervatve o (): lm d0 ( d) d ( ) lm d0 ( ) ( d d) lm d0 ( d) d ( d) hese are all correct dentons n the lmt d->0. But we want d to reman FINIE

5 What s a nte derence? he equvalent appromatons o the dervatves are: ( d) d ( ) orward derence ( ) ( d d) backward derence ( d) ( d d) centered derence

6 he bg queston: How good are the FD appromatons? hs leads us to aylor seres...

7 aylor Seres aylor seres are epansons o a uncton () or some nte dstance d to (+d) ( d) ( ) d ' ( ) d! '' ( ) d 3! 3 ''' ( ) d 4! 4 '''' ( )... What happens, we use ths epresson or ( d) d ( )?

8 aylor Seres... that leads to : ( d) d ( ) 1 d ' d ( ) ( ) O( d) ' d! '' ( ) d 3! 3 ''' ( )... he error o the rst dervatve usng the orward ormulaton s o order d. Is ths the case or other ormulatons o the dervatve? Let s check!

9 aylor Seres... wth the centered ormulaton we get: ( d / ) d ( d / ) 1 d ' d ( ) ( ) O( d ' d 3! ) 3 ''' ( )... he error o the rst dervatve usng the centered appromaton s o order d. hs s an mportant results: t DOES matter whch ormulaton we use. he centered scheme s more accurate!

10 Hgher order operators *a ( d) (d) ' (d)! '' (d) 3! 3 ''' *b ( d) ( d) ' ( d)! '' ( d) 3! 3 ''' *c ( d) ( d) ' ( d)! '' ( d) 3! 3 ''' *d ( d) (d) ' (d)! '' (d) 3! 3 '''... agan we are lookng or the coecents a,b,c,d wth whch the uncton values at ±()d have to be multpled n order to obtan the nterpolated value or the rst (or second) dervatve!... Let us add up all these equatons lke n the prevous case...

11 Problems: Stablty p( t dt) c dt d p( d) p( ) p( d) p( t) p( t dt) sdt Stablty: Careul analyss usng harmonc unctons shows that a stable numercal calculaton s subject to specal condtons (condtonal stablty). hs holds or many numercal problems. (Dervaton on the board). c dt d 1

12 Problems: Dsperson p( t dt) c dt d p( d) p( ) p( d) p( t) p( t dt) sdt rue velocty Dsperson: he numercal appromaton has artcal dsperson, n other words, the wave speed becomes requency dependent (Dervaton n the board). You have to nd a requency bandwdth where ths eect s small. he soluton s to use a sucent number o grd ponts per wavelength.

13 Fnte Derences - Summary Conceptually the most smple o the numercal methods and can be learned qute quckly Dependng on the physcal problem FD methods are condtonally stable (relaton between tme and space ncrement) FD methods have dcultes concernng the accurate mplementaton o boundary condtons (e.g. ree suraces, absorbng boundares) FD methods are usually eplct and thereore very easy to mplement and ecent on parallel computers FD methods work best on regular, rectangular grds

14 Partal Derental Equatons by Lale Yurttas, eas A&M Unversty Part 8 14

15 PERSAMAAN DIFERENSIAL PARSIAL Persamaan Umum Menyatakan bagamana varabel tak bebas Ø berubah terhadap varabel bebas,y. Dsn a,b,c,d,e,, dan g mungkn merupakan ungs dar Ø 0 g y e d y c y b a

16 Jens PDP Dtentukan oleh harga b -4ac < 0, elptc = 0, parabolc > 0, hyperbolc Adveks Dus. Gelombang

17 JENIS- JENIS PDP

18 by Lale Yurttas, eas A&M Unversty 18

19 he Laplacan Derence Equatons/ , 1, 1, 1, 1, 1,, 1, 1,, 1, 1,, 1, 1,, 1, j j j j j j j j j j j j j j j j j y y y y y by Lale Yurttas, eas A&M Unversty Chapter 9 Laplacan derence equaton. Holds or all nteror ponts Laplace Equaton O[() ] O[(y) ]

20 by Lale Yurttas, eas A&M Unversty Chapter 9 0

21 In addton, boundary condtons along the edges must be speced to obtan a unque soluton. he smplest case s where the temperature at the boundary s set at a ed value, Drchlet boundary condton. A balance or node (1,1) s: Smlar equatons can be developed or other nteror ponts to result a set o smultaneous equatons. by Lale Yurttas, eas A&M Unversty Chapter 9 1

22 he result s a set o nne smultaneous equatons wth nne unknowns:

23 h Duson h t D h Duson Equaton

24 u d h D h ud u d u d u h d t

25 u h d t t 0 u d h t u d h D h t D h Duson Equaton

26 Numercal Soluton o Duson Eq. h h ( 1) h () h ( 1) h h( ) h( 1) 1 h h( 1) h( ) h 1 h h 1

27 1 j-1 j j+ J Numercal Calculaton o Duson Equaton t t h( 1, j 1) h(, j) N

28 h h(, j 1) h(, j ) t t D h D h( 1, j) h(, j) h(, j) h( 1, j) D Unknown h( 1, j) h(, j) h( 1, j) h(, j 1) h(, j) Known Dt { h( 1, j) h(, j) h( 1, j)}

29 1 j-1 j j+ J Numercal Soluton o Duson Equaton t Boundary Condton (Gven) t N Intal Condton (Gven)

30 Dt h(, j 1) h(, j) { h( 1, j) h(, j) h( 1, j)} do,n-1 t h ( ) ( ) ( 1) ( ) ( 1) new hold hold hold hold D end do do,n-1 h ( ) h ( ) old end do new

31 Contoh d y d 3 dy d y 4, y(1) 1, y() 6 Car solusnya dengan step sze 0, Jumlah ttk solus n=((-1)/0,)-1=4 Kta dapatkan 4 persamaan, satu untuk tap ttk yang dcar.

32 Penyelesaan dengan Beda Hngga Persamaannya: Buat persamaan untuk semua ttk, mula dar =1, hngga =4 0=a =b y h y y h y y y

33 Doman Solus y5=6 y0=1 0=1 5=

34 Penyelesaan dg Fnte D. Dengan h = 0, 3,5 y 48y 17,5 y Buat persamaan untuk semua ttk, mula dar =1, hngga =4 Masukkan nla-nla 1=1, hngga 4 =1,8 dan konds batas y0 = 1 dan y5 = 6

35 Konds batas Drchlet atau ed boundary, msal C(0) = Co Neuman atau natural boundary, msal dc/d = 0 Robn/ Cauchy boundary condton, msal dc/d + C = 0 Penerapan dalam nte derence dengan menambahkan magnary node

36 penyelesaan persamaan parabolk dengan skema eksplst

37 SEHINGGA persamaan derensal, dtuls dalam bentuk metode beda hngga,

38 laplace.equaton penyelesaan persamaan elptk

39 Stabltas skema eksplst k k

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42 penyelesaan persamaan parabolk dengan skema mplst

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