1 The Euler Forward scheme (schéma d Euler explicite)

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1 TP : Fiite differece metod for a Europea optio M2 Modélisatio aléatoire - Uiversité Deis-Diderot Cours EDP e Fiace et Métodes Numériques December 8, Te Euler Forward sceme (scéma d Euler explicite) We look for a umerical approximatio of te Europea put fuctio v = v(t, s), t [0, T ], s [0, S max ] It satisfies i first approximatio te Black ad Scoles backward PDE o te trucated domai Ω = [S mi, S max ]: s2 2 s 2 v rs s v + rv = 0, t (0, T ), s (S mi, S max ) t v σ2 2 v(t, S mi ) = v l (t) Ke rt S mi, t (0, T ) v(t, S max ) = v r (t) 0, t (0, T ) v(0, s) = ϕ(s) := (K s) +, s (S mi, S max ) We will cosider te followig parameters: K = 100, S mi = 0, S max = 200, T = 1, σ = 02, r = 01 I particular, we aim at computig v(t, s) at fial time t = T We first itroduce a discrete mes as follows Let := Smax S mi I+1 be spatial mes step, ad τ := T N te time step Te s := S mi +, = 0,, I + 1 are te mes poits, ad t = τ, = 0,, N te time mes We are lookig for U, a approximatio of v(t, s ) For ay fuctio v C 2 (or v C 3 for (4)), we recall te followig approximatios, as 0, v (s ) = v(s ) v(s 1 ) v (s ) = v(s +1) v(s ) v (s ) = v(s +1) v(s 1 ) (1) + O() (2) + O() (3) + O( 2 ) (4) We terefore obtai several possible approximatios by fiite differeces for te first order derivative: s v(t, s ) U U 1 (backward differece approximatio) (5) s v(t, s ) U +1 U (forward differece approximatio) (6) s v(t, s ) U +1 U 1 (cetered approximatio) (7) Te first two approximatios are said to be cosistet of order 1 (i space), wile te secod oe is cosistet of order 2 1

2 We also recall te approximatio wic is secod-order cosistet i space 2 s 2 v(t, s ) U 1 + 2U U +1 2, (8) Hece we obtai te so-called "Euler Forward sceme" (or Explicit Euler sceme), abreviated "EE" usig te cetered approximatio, as follows: U +1 τ U + σ2 2 s2 U 1 + 2U U +1 2 rs U +1 U 1 + ru = 0 = 0,, N 1, = 1,, I U 0 = v l (t ) Ke rt S mi, = 0,, N (9) U I+1 = v r (t ) 0, U 0 = ϕ(s ) (K s ) +, = 0,, N = 1,, I Let us remark tat we ave take = 1 ad = I as extremal idices i For = 1, te sceme utilizes te kow value U0 := v l(t ) (left boudary value) For = I, te sceme utilizes te kow value UI+1 := v r(t ) (rigt boudary value) 2 Programmig Euler Forward 21 Prelimiaries We coose to work wit te ukow te vector correspodig to (v(t, s )) =1,,I : U = U 1 U I We would like to write (9) uder te vector form as follows: U +1 U + AU + q(t ) = 0, (10) τ were A is a square matrix of dimesio I ad q(t) is a colum vector of size I Let us deote α := σ2 s 2 2 2, β := r s We look for A ad q(t) suc tat α i ( U i 1 + 2U i U i+1) β i (U i+1 U i 1) + ru i = ( α i + β i )Ui 1 + (2α i + r)ui + ( α i β i )Ui+1 (AU + q(t )) i 2

3 By idetificatio we see tat A is a tridiagoal matrix 2α 1 + r α 1 β 1 0 α 2 + β 2 2α 2 + r α 2 β 2 A := α i + β i 2α i + r α i β i 0 α I + β I 2α I + r ad q(t ) cotais te boudary values U0 = v l(t ) ad UI+1 = v r(t ) as follows: ( α 1 + β 1 )U0 ( α 1 + β 1 )v l (t ) 0 0 q(t ) := 0 ( α I β I )U I+1 0 ( α I β I )v r (t ) 22 Gettig ito te program a) Dowload te workig program 1 (tp1m if usig Matlab, or tp1sci if usig Scilab) Tis program as some lies to be completed i order to work properly I te case of te matlab program, you sould also dowload te oter files (BSm, plootm, orderm ad te solutio file tp1solm) b) Ceck te program for te payoff fuctio u0 (for ϕ), ul (for v l ) ad ur (for v r ) ad program tem correctly (otice tat te istructio y=max(k-s,0) works for s vector, K scalar, ad retur a vector of same size as s) c) I te sectio MESH, complete te value of te mes step ad program te vector s cotaiig te (s ) 1 I values Typically it sould look like: dt=t/n; =(Smax-Smi)/(I+1); s=smi+ (1:I) *; // colum vector (s 1,, s I ) T d) Program te matrix A ad test te program Uder te matlab commad widow, type tp1 (Scilab: type exec tp1sci or exec( tp1sci,-1)) e) Program te fuctio q(t) For istace i matlab: q [(-alpa(1) + bet(1))* ul(t); zeros(i-2,1); (-alpa(ed) - bet(ed))* ur(t)]; I Scilab tis may look like fuctio y=ul(t); y=k*exp(-r*t); edfuctio fuctio y=ur(t); y=0; edfuctio; fuctio y=q(t) y=zeros(s); // vector of zeros wit same size as te s vector y(1)= (-a(1)+b(1))*ul(t); u($)= (-a($)-b($))*ur(t); edfuctio 1 See ttps://lllmatupmcfr/bokaowski/eseigemet/2017/m2mo/m2motml 3

4 22 Euler Forward sceme (or "Euler Explicit" sceme) a) Program te explicit form of te vector U +1, i terms of U, i te mai loop, usig te matrix A ad te fuctio q I te ed, it sould look like (i matlab) case EE P = (Id - dt*a)*p - dt*q(t); Note tat te grapic fuctio ploot also plots te exact Black ad Scoles formula (see te give fuctio BS i te file BSm) b) Correct te lig errli=0 i te mai loop, i order to compute correctly te maximum orm betwee te sceme values ad te Black ad Scoles values: U V BS = max 1 i I U i V BS (s i ) (use orm(x, if ) to get te l orm of a give vector x) 23 Solutio A solutio file is give i tp1solm 3 First umerical tests a) Test te Euler forward sceme (EE) First fix N = 10 ad take I = 10, 20, 50, Te take te followig N = I values : 10, 20, 50, 100 Observe tat: - te sceme is ot always umerically stable - it does ot always give a positive solutio (ie we do ot always ave U 0, ) b) Uderstad te origi of te oscillatios we tey occur For istace, fix N = 10 ad I = 50, ad look at te "amplificatio" matrix defied as B := I d τa (ceck tat te coefficiets of B are ot all positive ad tat tey may ave a modulus greater ta 1) c) Fill i te CFL umber defied ere as µ := τ 2 σ2 S 2 max ad prit it Ceck tat tere is o stability problem we µ is sufficietly small d) I te case we σ = 0002 ad for istace wit N = 10 ad I = 50: observe tat te sceme is stable but tat te solutio give by te sceme is ot positive e) Order of te sceme We first cosider te followig values of I ad of N: I = 10, 20, 40, 80, 160, ad N = I 2 /10 (tat is, N = 10, 40, ) Tis is i order to ave τ 2 Fill i te followig error tables ad compute te correspodig (spatial) order of te metod (you may use te file orderm) More precisely, if te error is e k for a give parameter I = I k, we will compute te order at step k by te formula α k := log(e k 1/e k ) log( k 1 / k ) were k is te spatial mes step at step k Te idea is to try to detect a beavior of te form e k = C α k, were C est a costat ad were k is te spatial mes size correspodig to I k 4

5 I N = I e k order α k I N I 2 e k order α k So if k 0 /2 k, as i te left table, te we get e k /e k 1 1/2 α ad te previous formula gives α k = log(e k 1/e k ) log(2) wic sould give a estimate of α Te umerical (spatial) order sould be close to two But ere settig N cost I 2 is costly i terms of umber of operatios More precisely, sice τ = cost 2 oe ca see tat error also beaves as O(τ), tat is, a first order error beavior wit respect to te time approximatio Tis motivates te use of implicit scemes i order to avoid te time-step coditio (or "CFL" coditio) Exercice: backward ad forward differece approximatios a) Program i te same way te oter approximatios (backward differeces / décetrage droit : DROIT ad te forward differeces - décetrage gauce GAUCHE ) Te parameter CENTRAGE is defied at te begiig of te program (sectio NUMERICAL DATA) ad determies te type of fiite differece approximatio tat is used b) Usig te forward differeces, fix temporarily σ = 0002 ad N = 10, I = 50, ad ceck tat ow te solutio keeps positive c) Usig te backward differeces, fix temporarily σ = 0002 ad N = 10, I = 50, ad ceck tat ow te umerical solutio is ustable (uderstad te problem by lookig at te coefficiets of te amplificatio matrix) 4 Implicit Euler sceme (scéma d Euler implicite) Te CFL costrait imposes some restrictio o te time step τ Implicit scemes may allow us to get rid of tis restrictio For istace, te implicit euler sceme, ereafter abrieviated "EI" (for Frec "Euler Implicit"), wit cetered differece approximatio for te first spatial derivatives reads U +1 τ U + σ2 2 s2 U U U+1 +1 U rs U ru +1 = 0 = 0,, N 1, = 1,, I U0 +1 = v l (t +1 ) Ke rt +1 S mi, = 0,, N 1 (11) U +1 I+1 = v r(t +1 ) 0, = 0,, N 1 U 0 = (K s ), = 1,, I 5

6 a) Sow tat te sceme, i vector form, i aalogy wit (10), is (U +1 U )/τ + AU +1 + q(t +1 ) = 0 b) Program EI: set te parameter SCHEMA= EI at te begiig of te program, ad complete accordigly, i te mai loop, te part case EI Note tat i order to solve a liear system of te form Ax = b oe may use te liear solver i matlab (resp Scilab) as follows x=a\b; c) Ceck tat wit te EI sceme tere is o more stability problems (wit for istace N = 10 ad I = 50) 5 Crak-Nicolso sceme a) Program te Crak-Nicolso sceme (CN) (θ-sceme wit θ = 1 2 ) I vector form te sceme reads (U +1 U )/τ (AU +1 + q(t +1 ) (AU + q(t )) = 0 b) Numerically study te covergece of te cetered CN sceme To do so, oe ca compute te umerical error obtaied wit te followig values N = I + 1 {10, 20, 40, 80, 160} Te do te same study as for te Euler implicit sceme ad compare by makig a error table (ad wit a order estimatio) Oe sould observe a secod order beavior for CN, wile it sould be first order for EI c) Ceck tat tese results are coeret wit te teory 6 Exercices Exercice 1 ("Call optio") a) Usig te put-call parity formula, do a fuctio tat also computes te black ad scoles formula for te call b) Propose adequate left ad rigt boudary coditios (tat is, at s = S mi ad s = S max ) for te call optio, i te form (v(t, S mi ) = v l (t) ad v(t, S max ) = v r (t), were v l, v r are fuctio to determie 2 c) Propose a apropriate PDE for te call optio, usig tese boudary coditios d) Write a Euler explicit sceme for te call optio, ad program it Exercice 2 (Improved efficiecy usig sparse matrices) Te matrix A as oly a few o zero elemets (about 3I o zero elemets) It is possible to code oly te o-zero elemets of A by usig te matrix type sparse Type elp sparse or doc sparse for documetatio i matlab A=spzeros(I,I) full(a) sparse(a) speye(i,i), speye(i) iitialize a sparse 0-matrix of size I I create correspodig full matrix of a sparse matrix create te correspodig sparse matrix of A create a sparse idetity matrix of size I Program te EI or CN sceme by usig oly sparse matrices, ad compare te speed of te ew code wit respect to te old oe for large I values Executio time is i geeral reduced It ca be evaluated by usig te commads tic; toc; (resp time() ad time() i Scilab) as follows: 2 v l (t, s) 0, v r(t, s) s Ke rt 6

7 tic; % INSTRUCTIONS; % t=toc; pritf( cpu time : t=%52f, t); (te te variable t cotais te time elapsed betwee te calls of tic ad toc) 7