В.Б. Андреев - Численные методы (1113834), страница 20
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Ìåòîä (16.7), (16.8) íàçûâàåòñÿ s-ýòàïíûì ìåòîäîì Ðóíãå-Êóòòû.Ýòîò ìåòîä ïðèíÿòî çàïèñûâàòü òàáëèöåé åãî êîýôôèöèåíòîâ, êîòîðàÿ íàçûâàåòñÿòàáëèöåé Áóò÷åðàc1 a11 a12 . . . a1sc2 a21 a22 . . . a2s.................ci =sXaij .(16.9)j=1cs as1 as2 . . . assb1 b2 . . . bsÇàìå÷àíèå 16.1.
Ïîñêîëüêó bi ñóòü âåñîâûå êîýôôèöèåíòû êâàäðàòóðíîé ôîðìóëûäëÿ èíòåãðàëà ïî åäèíè÷íîìó îòðåçêó, òîci =sPj=1sPi=1bi = 1. Èç àíàëîãè÷íûõ ñîîáðàæåíèéaij = θi .Îïðåäåëåíèå 16.2. Åñëè â òàáëèöå Áóò÷åðà (16.9) êîýôôèöèåíòû aij = 0 ïðè j > i,òî ìåòîä (16.7), (16.8) íàçûâàåòñÿ ÿâíûì s-ýòàïíûì ìåòîäîì Ðóíãå-Êóòòû.Îïðåäåëåíèå 16.3. Åñëè aij = 0 ïðè i > j è õîòÿ áû îäèí aii 6= 0, òî ìåòîä (16.7),(16.8) íàçûâàåòñÿ äèàãîíàëüíî íåÿâíûì.Âî âñåõ îñòàëüíûõ ñëó÷àÿõ ìû ãîâîðèì î íåÿâíûõ ìåòîäàõ Ðóíãå-Êóòòû.Êîýôôèöèåíòû â òàáëèöå Áóò÷åðà (16.9) ïðè çàäàííûõ îãðàíè÷åíèÿõ âûáèðàþòñÿèç ñîîáðàæåíèé ìàêñèìàëüíîé òî÷íîñòè ÷èñëåííîãî ìåòîäà.16.2 Îäíîýòàïíûå ìåòîäû Ðóíãå-ÊóòòûÈññëåäóåì îäíîýòàïíûå (s = 1) ìåòîäû Ðóíãå-Êóòòû.
Ïðè s = 1 ñîîòíîøåíèÿ (16.8),(16.7) ïðèíèìàþò âèäY1 = un + τ a11 f (Y1 ),(16.10)un+1 = un + τ b1 f (Y1 )(16.11)158 16. ÌÅÒÎÄÛ ÐÓÍÃÅ-ÊÓÒÒÛÈç ñîîáðàæåíèé àïïðîêñèìàöèè (êâàäðàòóðíàÿ ôîðìóëà äîëæíà áûòü òî÷íîé ïîêðàéíåé ìåðå íà const) íàõîäèì, ÷òî b1 = 1. Åñëè òåïåðü ïîëîæèòü a11 = 0, òî ìåòîäáóäåò ÿâíûì, ïðè÷åì Y1 = un , à (16.11) ìîæíî ïåðåïèñàòü â âèäåun+1 − un= f (un ).τÌû ïîëó÷èëè ìåòîä Ýéëåðà. Òåì ñàìûì, ìåòîä Ýéëåðà åñòü ÿâíûé îäíîýòàïíûéìåòîä Ðóíãå-Êóòòû .Åñëè âçÿòü a11 = 1, òî ìåòîä (16.10), (16.11) áóäåò íåÿâíûì. Ïðè ýòîì ïðàâûå÷àñòè (16.10) è (16.11) ñîâïàäàþò è ïðèâîäÿò ê ñîîòíîøåíèþ Y1 = un+1 .  ýòîì ñëó÷àåñèñòåìà (16.10), (16.11) ïðåîáðàçóåòñÿ ê âèäóun+1 − un= f (un+1 ).τÝòî íåÿâíûé ìåòîä Ýéëåðà (15.8).
Îí òàêæå ÿâëÿåòñÿ îäíîýòàïíûì ìåòîäîì ÐóíãåÊóòòû.Èññëåäóåì òåïåðü íàèáîëåå öåëåñîîáðàçíûé âûáîð ïàðàìåòðîâ b1 è a11 ñ òî÷êèçðåíèÿ ìèíèìèçàöèè ïîãðåøíîñòè àïïðîêñèìàöèè. ×òîáû íàéòè ïîãðåøíîñòü àïïðîêñèìàöèè, ïåðåïèøåì óðàâíåíèå (16.11) â âèäåun+1 − un= b1 f (Y1 )τ(16.12)(ñð. ñ (15.7), (15.8), (15.10) è (15.12)), à ðåøåíèå óðàâíåíèÿ (16.10) îáîçíà÷èì ÷åðåçY1 (un ). Åñëè, êàê è âûøå, zn = un − u(tn ), òîzn+1 − znu(tn+1 ) − u(tn )= b1 f (Y1 (u(tn ) + zn )) −.ττÈ ñíîâà, ðàñêëàäûâàÿ ïåðâîå ñëàãàåìîå ïðàâîé ÷àñòè ïî ôîðìóëå Òåéëîðà, íàõîäèì,÷òî·¸zn+1 − zn∂fu(tn+1 ) − u(tn )= b1 f (Y1 (u(tn ))) +(ũ)zn −=τ∂uτ∂f ∂Y1= b1(ũ)zn + ψn ,∂Y1 ∂uãäåu(tn+1 ) − u(tn )(16.13)ψn = b1 f (Y1 (u(tn ))) −τ ïîãðåøíîñòü àïïðîêñèìàöèè, à Y1 (u(tn )) ðåøåíèå óðàâíåíèÿ (16.10) ñ u(tn ) âìåñòîun , ò.å.Y1 (u(tn )) = u(tn ) + τ a11 f (Y1 (u(tn ))).(16.14)Çàìå÷àíèå 16.2.
Ïîãðåøíîñòü àïïðîêñèìàöèè (16.13) ïðåäñòàâëÿåò ñîáîé ðàçíîñòüìåæäó ïðàâîé è ëåâîé ÷àñòÿìè óðàâíåíèÿ (16.12), åñëè òóäà âìåñòî ïðèáëèæåííîãîðåøåíèÿ ïîäñòàâèòü òî÷íîå (ñð. ñ çàìå÷àíèåì 15.3).16.2. ÎÄÍÎÝÒÀÏÍÛÅ ÌÅÒÎÄÛ ÐÓÍÃÅ-ÊÓÒÒÛ159Ðàçëîæèì ïîãðåøíîñòü àïïðîêñèìàöèè (16.13) ïî ñòåïåíÿì τ . Èìååì"#2fd f (Y1 ) ¯¯τ 2 dgψn = b1+τ+−τ =0τ =0dτ2 dτ 2·¸τ 00τ 2 ˜0000− u (tn ) + u (tn ) + ũ .26¯f (Y1 ) ¯Èç (16.14) íàõîäèì, ÷òî Y1 |τ =0 = u(tn ) è, ñëåäîâàòåëüíî,¯f (Y1 ) ¯τ =0 = f (u(tn )).Ñíîâà ñ èñïîëüçîâàíèåì (16.14)df (Y1 ) ¯¯df dY1 ¯¯df==(u(tn ))a11 f (u(tn )),τ=0τ=0dτdY1 dτduà èç óðàâíåíèÿ (16.10 )u0 (tn ) = f (u(tn )),u00 (tn ) =dfdfdudf(u(tn )) =(u(tn )) (tn ) =f.dtdudtduÏîýòîìó·¸1dfψn = (b1 − 1)f (u(tn )) + τ b1 a11 −f (u(tn )) (u(tn )) + O(τ 2 ).2duÒåì ñàìûì, äëÿ òîãî, ÷òîáû ïîãðåøíîñòü àïïðîêñèìàöèè áûëà O(τ 2 ), íåîáõîäèìî èäîñòàòî÷íî, ÷òîáû âûïîëíÿëèñü óñëîâèÿb1 = 1,a11 b1 = 1/2.(16.15)Îòñþäà íàõîäèìb1 = 1,a11 = 1/2è, ñëåäîâàòåëüíî, íåÿâíûé îäíîýòàïíûé ìåòîä Ðóíãå-ÊóòòûτY1 = un + f (Y1 ),2un+1 = un + τ f (Y1 )(16.16)èìååò âòîðîé ïîðÿäîê àïïðîêñèìàöèè.Çàìå÷àíèå 16.3.
Èç ïåðâîãî óðàâíåíèÿ (16.16) ñëåäóåò, ÷òî ìîìåíò âðåìåíè, íàêîòîðûé Y1 ïðèáëèæàåò u(t), åñòü t + τ /2, èáî äëÿ çàäà÷è u0 = 1, u(0) = 0, èìåþùåéðåøåíèå u = t, Y1 = un + τ /2 = tn + τ /2.160 16. ÌÅÒÎÄÛ ÐÓÍÃÅ-ÊÓÒÒÛÑîîòíîøåíèÿ (16.16) ìîæíî ïðåîáðàçîâàòü. Èñêëþ÷èâ f (Y1 ), íàéäåì, ÷òî un+1 =2Y1 − un . Âûðàæàÿ îòñþäà Y1 è ïîäñòàâëÿÿ åãî âî âòîðîå óðàâíåíèå (16.16), ïîëó÷è쵶un+1 + unun+1 = un + τ f.2Çàìå÷àíèå 16.4.
Ìåòîä (16.16) î÷åíü ñèëüíî íàïîìèíàåò ìåòîä Ðóíãå (15.10), (15.11).Îòëè÷èå ìåæäó íèìè ñîñòîèò â òîì, ÷òî çäåñü ïðîìåæóòî÷íîå çíà÷åíèå íàõîäèòñÿ ïîíåÿâíîé ôîðìóëå, à â ìåòîäå Ðóíãå ïî ÿâíîé ôîðìóëå (15.11). Ìåòîä (16.16), êàêìû óæå ñêàçàëè, ÿâëÿåòñÿ îäíîýòàïíûì (íåÿâíûì) ìåòîäîì Ðóíãå-Êóòòû, à ìåòîä(15.10), (15.11) äâóõýòàïíûì (ÿâíûì) ìåòîäîì. Ïîä÷åðêíåì, ÷òî ñëîâó ýòàï çäåñüìû ïðèäàåì ÷åòêèé ìàòåìàòè÷åñêèé ñìûñë.16.3 Ìåòîäû òðåòüåãî ïîðÿäêà àïïðîêñèìàöèèÂûÿñíèì îãðàíè÷åíèÿ íà êîýôôèöèåíòû (16.9), îáåñïå÷èâàþùèå òðåòèé ïîðÿäîê àïïðîêñèìàöèè s-ýòàïíîãî ìåòîäà Ðóíãå-Êóòòû. Äëÿ ýòîãî íóæíî èññëåäîâàòü ïîãðåøíîñòü àïïðîêñèìàöèèψn := ψn (τ ) :=sXbi f (Yi (u(tn ))) −i=1ãäåYi (u(tn )) = u(tn ) + τsXu(tn+1 ) − u(tn ),τaij f (Yj (u(tn ))) =: Yi (u(tn ); τ ).(16.17)j=1Ðàñêëàäûâàÿ ψn (τ ) ïî τ äî òðåòüåãî ïîðÿäêà, áóäåì èìåòü¯¯¯·¸s2 2X¯¯¯df(Y)τdf(Y)ii3¯¯ψn (τ ) =bi f (Yi ) ¯¯++O(τ ) −+τ¯¯2dτ2dττ=0τ=0τ=0¸·i=1τ 2 000τ 0003− u (tn ) + u (tn ) + u (tn ) + O(τ ) .26(16.18)Ïîñêîëüêó f (Yi (u(tn ))) åñòü ñëîæíàÿ ôóíêöèÿ τ , òî âû÷èñëèì ñíà÷àëà ïðîèçâîäíûåïî τ ôóíêöèè Yi (u(tn ); τ ) ïðè τ = 0.
Èç (16.17) ñ ó÷åòîì (16.9), íàõîäèì, ÷òYi ¯¯ = u(tn ),τ =0" s#¯¯¯sXX¯¯¯dYdfi ¯Yi0 ¯¯ ==aij f (Yi ) + τaijYj0 ¯¯ = f (u(tn ))ci ,¯d τ τ =0d Yjτ =0τ =0j=1j=1"#¯¯ssss2XXX¯¯dfdf Xdfdf0 200 ¯aijYi00 ¯¯ = 2=2f(u(t))Yj0 +τaij(Y)+τaYaij cj .nijjj ¯2dYdYdYdujjjτ =0τ=0j=1j=1j=1j=116.3. ÌÅÒÎÄÛ ÒÐÅÒÜÅÃÎ ÏÎÐßÄÊÀ ÀÏÏÐÎÊÑÈÌÀÖÈÈ161Òåïåðü ìîæíî íàéòè ïðîèçâîäíûå f :¯¯f (Yi (u(tn ))) ¯¯ = f (u(tn )),τ =0¯¯df (Yi ) ¯¯df 0 ¯¯df=Y=f(u(t))ci ,nidτ ¯τ =0 dYi ¯τ =0du¯· 2¸¯µ ¶2 Xsd f 0 2 df 00 ¯¯d2 f 2d2 f (Yi ) ¯¯df2==f(u(t))Yc+2f(u(t))aij cj .(Y)+nndτ 2 ¯dY 2 idYi i ¯du2 iduiτ =0τ =0j=1Äàëåå, èç (16.10 )u0 = f,u00 =df 0u = f 0 f,duu000 = f 00 u0 f + (f 0 )2 u0 = f 00 f 2 + (f 0 )2 fè, ñëåäîâàòåëüíî,¢u(tn+1 ) − u(tn )ττ 2 ¡ 00 2= f + f 0f +f f + (f 0 )2 f + O(τ 3 ).τ26Ïîäñòàâëÿÿ òåïåðü íàéäåííûå ðàçëîæåíèÿ â (16.18), áóäåì èìåòüψn =·sXi=1"biτ2f + τ f f 0 ci +2Ãf 2 f 00 c2i + 2f f 02sX!#aij cjj=1¸¢ττ ¡ 2 00− f + ff0 +f f + f f 02 + O(τ 3 ).262−(16.19)Îòñþäà, ïðèðàâíèâàÿ êîýôôèöèåíòû ïðè îäèíàêîâûõ ñòåïåíÿõ τ , íàõîäèì, ÷òî óñëîâèÿ òðåòüåãî ïîðÿäêà àïïðîêñèìàöèè ñóòüsXi=1sXi=1sXi=1bi = 1 ,1bi ci =2(16.20),1bi c2i = ,3sX1bi aij cj = .6i,j=1Ïðè ýòîì (16.20) ñóòü óñëîâèÿ âòîðîãî ïîðÿäêà àïïðîêñèìàöèè.(16.21)162 16.
ÌÅÒÎÄÛ ÐÓÍÃÅ-ÊÓÒÒÛÇàìå÷àíèå 16.5. ×òîáû èìåòü óñëîâèÿ ÷åòâåðòîãî ïîðÿäêà àïïðîêñèìàöèè, ê óñëîâèÿì (16.20), (16.21) íóæíî äîáàâèòü ñëåäóþùèå óñëîâèÿ:sXi=11bi c3i = ,4sX1bi ci aij cj = ,8i,j=1sXbi aij c2ji,j=1sX(16.22)1= ,12bi aij ajk ck =i,j,k=11.24Çàìå÷àíèå 16.6. Óñëîâèÿ (16.20) ñ ó÷åòîì çàìå÷àíèÿ 16.1 ìîæíî òðàêòîâàòü êàêóñëîâèÿ òî÷íîñòè êâàäðàòóðíîé ôîðìóëû èç (16.6) íà ëèíåéíûõ ôóíêöèÿõ.1 Äîáàâëåíèå ê ýòèì óñëîâèÿì ïåðâîãî èç ñîîòíîøåíèé (16.21), à çàòåì è ïåðâîãî èç ñîîòíîøåíèé(16.22) íà óêàçàííóþ êâàäðàòóðíóþ ôîðìóëó íàêëàäûâàåò äîïîëíèòåëüíûå óñëîâèÿòî÷íîñòè íà êâàäðàòè÷íûõ è êóáè÷íûõ ôóíêöèÿõ.Óïðàæíåíèå 16.1. Ïîêàçàòü, ÷òî ìåòîä òðàïåöèé (15.12) ÿâëÿåòñÿ íåÿâíûì äâóõ-ýòàïíûì ìåòîäîì Ðóíãå-Êóòòû âòîðîãî ïîðÿäêà àïïðîêñèìàöèè.
(Íàéòè âñå bi , aij èïîêàçàòü íåâûïîëíåíèå õîòÿ áû îäíî èç óñëîâèé (16.21))Îòâåò:00011/2 1/21/2 1/2µ1 2 1 20 + 122¶16= .3Óïðàæíåíèå 16.2. Ïîêàçàòü, ÷òî ìåòîä Ðóíãå (15.10), (15.11) ÿâëÿåòñÿ ÿâíûìäâóõýòàïíûì ìåòîäîì Ðóíãå-Êóòòû âòîðîãî ïîðÿäêà.Îòâåò:1/21/201·¸110·0+1·6=.4316.4 Äâóõýòàïíûå íåÿâíûå ìåòîäû òðåòüåãî ïîðÿäêàÏîëîæèì â (16.20), (16.21) ïàðàìåòð s = 2.  ðåçóëüòàòå ñèñòåìà ïðèìåò âèäb1 + b2 = 1,c1 b1 + c2 b2 = 1/2,c21 b1 + c22 b2 = 1/3,b1 (a11 c1 + a12 c2 ) + b2 (a21 c1 + a22 c2 ) = 1/6.1 Âåäüci = θi , ò.å. êîîðäèíàòà ïåðåìåííîé èíòåãðèðîâàíèÿ â i-îì óçëå.(16.23)16.4.
ÄÂÓÕÝÒÀÏÍÛÅ ÍÅßÂÍÛÅ ÌÅÒÎÄÛ ÒÐÅÒÜÅÃÎ ÏÎÐßÄÊÀ163Ýòà ñèñòåìà ñîäåðæèò ÷åòûðå óðàâíåíèÿ è øåñòü íåèçâåñòíûõ (Åñëè íå ñ÷èòàòü c1è c2 , çàäàâàåìûå (16.9)). Ïîýòîìó, âîîáùå ãîâîðÿ, äâà èç ýòèõ íåèçâåñòíûõ äîëæíûîñòàòüñÿ ñâîáîäíûìè, à îñòàëüíûå âûðàçèòüñÿ ÷åðåç íèõ. Ñèñòåìà (16.23) íåëèíåéíàÿ,è íåò ðåãóëÿðíûõ ñïîñîáîâ åå ðåøåíèÿ. Óêàæåì îäèí ïóòü, ïðèâîäÿùèé ê ðåøåíèþýòîé ñèñòåìû.Äëÿ îòûñêàíèÿ ðåøåíèÿ ñèñòåìû (16.23) ïðåäïîëîæèì ñíà÷àëà, ÷òî íåèçâåñòíûåc1 è c2 íàéäåíû, è ðàññìîòðèì ïåðâûå òðè óðàâíåíèÿ (16.23) êàê ñèñòåìó ëèíåéíûõóðàâíåíèé îòíîñèòåëüíî b1 è b2 . Ïîñêîëüêó ýòà ñèñòåìà ïåðåîïðåäåëåíà, òî äëÿ ååðàçðåøèìîñòè íåîáõîäèìî îáðàùåíèå â íóëü îïðåäåëèòåëÿ ðàñøèðåííîé ìàòðèöû¯¯¯1 1 1 ¯¯¯¯c1 c2 1/2¯ = 1 c2 + 1 c21 + c1 c22 − c21 c2 − 1 c22 − 1 c1 =¯¯ 2 2223¯c1 c2 1/3¯ 311= (c2 − c1 ) − (c2 + c1 )(c2 − c1 ) + c1 c2 (c2 − c1 ) =3· 2¸1 c1 + c2= (c2 − c1 )−+ c1 c2 = 0.32(16.24)Ïðîàíàëèçèðóåì ýòî ñîîòíîøåíèå.
Åñëè áû c1 = c2 , òî ïîñëåäíåå óðàâíåíèå (16.23)ïðèíÿëî áû âèäc21 b1 + c22 b2 = 1/6,÷òî ïðîòèâîðå÷èò òðåòüåìó óðàâíåíèþ (16.23), è ïîýòîìóc1 − c2 6= 0.(16.25)Òåì ñàìûì, èç (16.24) ñëåäóåò, ÷òî2 − 3(c1 + c2 ) + 6c1 c2 = 0èëè(3 − 6c1 )c2 = 2 − 3c1 .Ïîñêîëüêó c1 = 1/2 íå óäîâëåòâîðÿåò ýòîìó óðàâíåíèþ, òîc1 6= 1/2è ìîæíî íàéòèc2 =2 − 3c1.3 − 6c1(16.26)(16.27)Ðàçðåøèì òåïåðü ïåðâûå äâà óðàâíåíèÿ (16.23) îòíîñèòåëüíî b1 è b2 ïðè ïîìîùèïðàâèëà Êðàìåðà.
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