Numerical Methods for the Solution of Elliptic Partial Differential Equations
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1 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 Numeral Methods or the Soluton o Ellpt Partal Derental Equatons Davd Keer Department o Chemal Engneerng nverst o ennessee Knovlle September 999 able o Contents Introduton. Sngle Lnear Ellpt PDE A. Method o ransormaton to a sstem o parabol PDEs B. Method o the Laplaan Derene Equaton 3 C. Lebmann s method 4. Sngle Nonlnear Ellpt PDE 3. Sstems o Lnear Ellpt PDEs 3 4. Sstems o Nonlnear Ellpt PDEs 9
2 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 Introduton A lnear ellpt PDE has the general orm : ( (z (z b(z (z a(z(z (z (. whh an be rewrtten as: (z (z [ (z b(z ] (z a(z(z (z (. and whh when the oeents are onstants an be wrtten as: (z b (z a(z (z (.3 whh we abandon matr notaton beomes z b b bz a z z (.4 we have varaton n n three spatal dmensons or b b a (.5 we have varaton n n two spatal dmensons. I we have varaton n n onl one spatal dmenson then we have b a (.6 whh s an ordnar derental equaton whh we alread know how to solve numerall and analtall. Some smple and amous eamples o ellpt equatons are the two-dmensonal Laplae equaton: (.7 the three-dmensonal Laplae equaton: (.8 the two-dmensonal Posson equaton: ( (.9
3 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999. Sngle Lnear Ellpt PDEs A. Method o ransormaton to a sstem o parabol PDEs An ellpt PDE does not have tme as an ndepent varable. It s a PDE beause t has at least two ndepent spatal dmensons. Beause t does not have a tme depene we don t naturall thnk to solve t usng a numeral ntegraton method lke the Runge-Kutta method that we used or ODEs as well as or parabol and hperbol PDEs. O ourse we ould use that tehnque. he mathemats o our soluton methodolog doesn t are whether our ndepent varable s labeled t or tme or or spae. hereore we ould take the two-dmensonal verson o equaton. [ ] a b (. whh an be rewrtten as a b b (. where I have deded not to wrte the eplt (z depene o a b and and onsder as t and as and solve t as we dd a hperbol PDE. Frst we break the sngle seond order PDE nto two rst order PDEs usng the substtuton: and (.3 hen we an wrte the rst equaton as (.4 and the seond equaton as a b b (.5 a b b (.6 Equaton (.4 and (.6 orm a sstem o two lnear parabol PDEs whh are entrel onsstent wth the lnear ellpt PDE o equaton (.. Sne we alread know how to solve a sstem o parabol PDEs we are done. We use our ode or solvng a sstem o parabol PDEs to solve the ellpt PDE.
4 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 Note to ourselves: No one solves Ellpt PDEs usng ths tehnque. Wh not? People are raz and the lke to make thngs more omplated than the need to be. We have a tehnque or solvng sstems o parabol PDEs that s guaranteed to work. he onl plae where t wouldn t work s where equals n equaton (.6. A seond reason people don t solve ellpt PDEs as sstems o parabol PDEs s beause n dong so ou make a dstnton between and. In the above ormulaton we treat as a spatal dmenson. I we ollow our prevous work then we use a entered nte derene method or the rst and seond partal dervatves o wth respet to. he entered nte derene ormulae are aurate to order h. Now when we solved parabol PDEs we dd not use a entered nte derene ormulae or the tme varable sne that would have added an addtonal unknown to the problem. Instead we used a seond-order Runge-Kutta method. he thrd and bggest reason that people don t solve ellpt PDEs as sstems o parabol PDEs s beause a properl posed PDE has boundar ondtons at ( ( and ( ( (. However n order to solve the parabol PDE we need d d o o ( and ( (. Sne we don t have the ntal partal d d dervatve normaton we have to use a tehnque lke the shootng method (see ChE 3 notes whh we used to solve an ODE BVP as an ODE IVP n order to solve the ellpt PDE as a sstem o parabol PDEs. he bg drawbak wth ths s that we now have to onverge to the orret nal boundar ondton. Convergene s never a sure thng we an avod an sheme that reles on onvergene we would lke to. B. Method o the Laplaan Derene Equaton For small sstems we use the entered nte derene ormulae or both and : o o k k k k k (.7 k k k k k k k k k k k k (.8 (.9 k k k k k k (. 3
5 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 k k k k k k k k (. k k k k k k k k (. I we substtute these ormulae nto equaton (. we have a sstem o lnear algebra equatons whh we know how to solve. We have m m unknowns where m s the number o nodes where the temperature s unknown n the -dmenson and m s the number o nodes where the temperature s unknown n the -dmenson. he mappng o the unknowns rom ( oordnates n the spatal dmenson to the one-dmensonal orderng o the matr whh appears n our sstem o lnear algebra equatons s eatl analogous to the mappng that was lad out or the same mappng when solvng a sngle parabol PDE wth varaton n spatal dmensons so see those notes on the 55 webste. he pont s that ths s trval and works so long as we don t have too man unknowns. Note: ths tehnque onl works the problem s lnear but the lnear problem an be o the general orm o equaton (.. C. Lebmann s method For larger sstems where we don t have the ablt to solve the sstem o lnear equatons we an appl a Gauss-Sedel appromaton whh when appled to PDEs s known as Lebmann s method. Here we ust wrte: k k k k 4 k (.3 Beause the matr s dagonall domnant repeated applatons o ths appromaton wll onverge to the true soluton. hs alone would work. It s slow. hereore people tr to speed t up b usng the ollowng enhanement ater eah applaton o equaton (. new k λ k ( λ λ k (.4 where λ s a relaaton parameter wth a value between and and k k are the values rom the present and prevous teraton. hs proess o (. and (. s led through or eah node at whh the soluton unknown and then repeated untl the soluton prole no longer hanges. 4
6 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 Now onvergene even guaranteed onvergene s an busness. It requres ntal guesses. I wll do everthng n m power to avod havng to make ntal guesses. hereore at all possble I wll use method B above. Note ths method wll work or Laplae s equaton onl. An Eample o solvng Laplae s Equaton usng Lebmann s method. Consder a retangle plate wth the ollowng Drhlet boundar ondtons: ( ( ( 5 ( 75 We dsretze ths grd or n and n ntervals and solve or a varet o values o λ. We thereore n n n n unknown nodes. (he remanng nodes have have ( ( total nodes and ( ( values gven b the boundar ondtons. Our ntal guess s that all nteror nodes are. Below we show a table o results or the derent values o λ. Our error s alulated as the root mean square error. err n n new old ( ( n ( n 6 We stop the teratve soluton proedure when the error s less than. or we eeed teratons. Below are the errors or n n 4. (So we have 3 9 unknown nodes. We see that a value o λ. gave the qukest (4 teratons onvergene. lamdba. teraton 7 error. lamdba. teraton error. lamdba. teraton 4 error. lamdba.3 teraton 7 error. lamdba.4 teraton error. lamdba.5 teraton 7 error. lamdba.6 teraton 36 error. lamdba.7 teraton 5 error. lamdba.8 teraton 8 error. lamdba.9 teraton error.554 Below are the errors or n n. (So we have 9 8 unknown nodes. We see that a value o λ. 6 gave the qukest (4 teratons onvergene. 5
7 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 lamdba. teraton error.3 lamdba. teraton error.7 lamdba. teraton error. lamdba.3 teraton 8 error. lamdba.4 teraton 63 error. lamdba.5 teraton 43 error. lamdba.6 teraton 4 error. lamdba.7 teraton 5 error. lamdba.8 teraton 8 error. lamdba.9 teraton error.77 Below are the errors or n n 5. (So we have 49 4 unknown nodes. We see that none o the values o λ onverged wthn the aeptable tolerane n less than teratons. But we see that we had the smallest error ater teratons wth a value o λ. 9. lamdba. teraton error.5 lamdba. teraton error.5787 lamdba. teraton error.6469 lamdba.3 teraton error.779 lamdba.4 teraton error.755 lamdba.5 teraton error lamdba.6 teraton error lamdba.7 teraton error.75 lamdba.8 teraton error.4887 lamdba.9 teraton error.483 What we hange our ntal guess o the nteror nodes to be the average o the boundar ondtons values ( 5 75/ Below are the errors or n n. (So we have 9 8 unknown nodes. We see that a value o λ. 6 gave the qukest (39 teratons onvergene. It took one teraton less than the ntal guess wth all nteror nodes equal to zero. lamdba. teraton error.6 lamdba. teraton error. lamdba. teraton 89 error. lamdba.3 teraton 73 error. lamdba.4 teraton 57 error. lamdba.5 teraton 4 error. lamdba.6 teraton 39 error. lamdba.7 teraton 5 error. lamdba.8 teraton 78 error. lamdba.9 teraton error.593 What dd we learn rom ths? he value o λ s not onl depent on the boundar ondtons but on the sze o dsretzaton and the ntal guesses as well. 6
8 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 A two-dmensonal olor ontour plot o the onverged soluton s gven below A 3-dmensonal olor ontour plot o the onverged soluton s gven below
9 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 he MALAB ode we used to solve ths problem s gven below. unton ell_lebmann lear all % % Solve Laplae's equaton usng Lebmann's method % % Assume Drhlet Boundar Condtons % % Author: Davd Keer % Department o Chemal Engneerng % nverst o ennessee Knovlle % Last pdated: Otober 4 % %solve or several values o lambda lambdav [.:.:.9]; or k ::length(lambdav lambda lambdav(k; % number o ntervals n_nt ; n_nt ; % number o nodes n n_nt ; n n_nt ; % grd ponts n and o.;.; o.;.; d ( - o/n_nt; d ( - o/n_nt; grd [o:d:]; grd [o:d:]; % ntalze and old zeros(nn; old zeros(nn; % Fll n old wth our drhlet BCs or ::n % BC or o old(.; % BC or old(n.; % BC or o old( 5.; % BC or old(n 75.; % ll n the nteror nodes o old wth ntal guess or ::n- or ::n- %old(.; old( 56.5; old; % % teratvel solve % mat ; err ; tol.e-6; ount ; whle (ount < mat & err > tol ount ount ; old ; or ::n- or ::n- % appl Lebmann's method ( (( (- ( (- *.5; ( lambda*( (-lambda*old(; 8
10 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 % alulate error err.; or ::n- or ::n- err err (( - old(^; err sqrt(err/((n-*(n-; prnt(' lamdba % teraton % error % \n'lambda ount err; % plot mn grd(; ma grd(n; mn grd(; ma grd(n; nontourlnes 5; mn mn(mn(; ma ma(ma(; (abs(mn-ma <.e-8; prnt('soluton s a lat plane wth value % \n'mn else plot_dmensons ; (plot_dmensons 3 ontour3(grd grdnontourlnes; as([mn ma mn ma mn ma] else ontour(grd grdnontourlnes; as([mn ma mn ma] label(''; label(''; olorbar 9
11 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999. Sngle Nonlnear Ellpt PDE Now the trouble starts. he Method o the Laplaan Derene Equaton onl works or lnear ellpt PDEs. he Method o Lebmann onl works or Laplae s equaton. So we have to use the method o transormaton to a sstem o parabol PDEs. he Method o ransormaton to a sstem o parabol PDEs s our best bet or at least two reasons. Frst we alread have a ode that does 9% o (A. It won t take muh eort to add an eteror loop whh onverges on the nal boundar ondton. Seond n (A we have to d d make ntal guesses or the partal dervatve ( ( at eah o m nodes. d d In ths method we begn b rewrtng the PDE n the same wa that we dd or the lnear ase namel b onvertng t to a sstem o two parabol PDEs. ( ( (. ( ( b ( b ( a ( (. Now our PDE s nonlnear so the orm ma not be as we have wrtten t n equaton (.. Let s be a lttle more general and rewrte equatons (. and (. as general nonlnear untons: ( ( K K ( ( ( ( ( ( ( ( ( ( ( ( ( ( (.3 (.4 (l A omment on notaton: we wll wrte ( as ( l so that supersrpts desgnate temporal ( nrements subsrpts desgnate spatal ( nrements l and k supersrpts nsde parentheses desgnate derent untons o reap what we dd n the parabol PDE ase we rst dsretzed tme and spae. Seond we used the seond order Classal Runge-Kutta method to solve the tme omponent o the PDE lke an ODE. ( K ( l l ( l ( l K (.5 where ( K l ( s the partal dervatve o l wth respet to
12 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 K ( l K ( l { } (.6 he braes n equaton (.6 now stand or the omplete set over both poston and unton (l. he seond unton n equaton (.5 s gven b K( K K ( l K ( l { } (.7 where the value o s gven b K (.8 Note: ths temperature s used not onl or the eplt temperatures but s also used n the nte derene ormulae to obtan the rst and seond spatal partal dervatves. From ths pont on we proeed n generall same manner as we dd or a sstem o parabol PDEs. Hopeull we see that n treatng an ellpt equaton as a set o parabol PDEs we are perormng the same transormaton that we dd n solvng the hperbol PDE. In ths transormaton we wll that we are mssng ertan boundar ondtons. Remember the ellpt PDE on a retangular doman gves boundar ondtons (n ths ase Drhlet o the orm: o ( ( ( (. o 3 ( ( ( (. 4 In order to solve ths as a sstem o parabol PDEs we need boundar ondtons on and on (. We have the boundar ondtons on an obtan the boundar ondtons on ( o ( b realzng that ( ( rom the orgnal set o BCs. We ( ( so ( ( ( ( (. ( ( o ( ( (. (
13 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 As or the ntal ondtons we have one and we have to guess the other along the lnes o the shootng method to obtan the ourth boundar ondton ( 4(. hus our ntal ondtons beome: ( o ( ( o guess(. 3 We are gong to have to make guesses or the value o the slope at all o the nodes. (hat stnks but that s le.
14 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August Sstems o Lnear Ellpt PDEs sng the Method o the Laplaan Derene Equaton a sngle lnear ellpt PDE an be onverted to a sstem o m m lnear algebra equatons. Analogousl a sstem o n oupled lnear ellpt PDEs an be onverted to a sstem o n m m lnear algebra equatons. t reall doesn t matter how bg the sstem o equatons s the methodolog needed to solve t remans unhanged. he onl lmtaton s omputatonal power and memor requrements whh we are assumng to be suent or our problems. Eample: Consder the problem ( ( ( a ( ( b ( ( (3. ( ( ( a ( ( b ( ( (3. on the retangle dened b and wth the boundar ondtons: ( o ( ( o h3 ( ( h ( ( h ( (3.3 ( ( h4 ( ( ( o ( ( o g3 ( ( g ( ( g ( (3.4 wth m 3 and m 3. ( ( g4 ( ( Ptorall we have our nodes wth unknowns eah: 3
15 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August 999 node numberng k k k k3 k k κ k κ3 k3 k k κ k κ4 k3 3 k 3 k 3 k 3 k3 nde: m 3 nde: k m 3 matr nde: κ 3m u 4 (numbers reer to nodes loated to the upper let sng the nte-entered derene ormulae n equatons (.6-(. or the partal dervatves n equatons (3. and (3. we have ( k ( k ( k ( k ( k ( k ( a ( ( b ( ( ( ( ( ( ( k k k k k k ( ( ( a( b( (3.6 ( ( We an wrte an equaton suh as (3.5 and (3.6 or eah node where and are unknown. In ths ase sne we have our unknown nodes we have 8 unknown varables and 8 equatons. We an easl solve the 8 b 8 matr. ( (3.5 R (3.7 where the vetor o unknowns s ( ( ( ( ( ( ( ( 4
16 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August he matr s lr lr lr lr ll ll ll ll ur ur ur ur ul ul ul ul (3.5 where ( a ul (3.6 ( a ll (3.7 ( b ur (3.8 ( b lr (3.9 he resdual has the orm:
17 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August ( g ( g ( ( g ( g ( ( g ( g ( ( g ( g ( ( h ( h ( ( h ( h ( ( h ( h ( ( h ( h ( R (3. We solve ths sstem o lnear algebra equatons or the soluton.
18 D. Keer ChE 55 nverst o ennessee Department o Chemal Engneerng August Sstems o Nonlnear Ellpt PDEs sng the Method o ransormaton to a sstem o parabol PDEs a sngle nonlnear ellpt PDE an be onverted to a sstem o parabol PDEs whh s solved as a sstem o m nonlnear ODEs. Analogousl a sstem o n oupled nonlnear ellpt PDEs an be onverted to a sstem o n m nonlnear ODEs. It reall doesn t matter how bg the sstem o equatons s the methodolog needed to solve t remans unhanged. he onl lmtaton s omputatonal power and memor requrements whh we are assumng to be suent or our problems. As beore we solve the sstem o n m nonlnear ODEs based on an ntal guess o the ntal ondton o the slope. We see how well that guess delvers us to the nal boundar ondtons. We then revse our guess o the ntal ondton o the slope and tr agan untl we obtan the nal boundar ondton. Methods or updatng the guess are gven n the ChE 3 notes on the shootng method. 7
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