數(shù)值分析2015-第六章3

O(hp1),(p O(hp1),(pTn1y(xn1)y(xn)h(xn,y(xn);y(xnh)y(xn)h(xn,y(xn);若按變量h在h0處作TaylorTn1[y(xn)hy'(xn)?]y(xn)h[(xn,y(xn);0)h[y'(xn)(xn,y(xn);0)] O(hp1p這相當(dāng)于Tn1含hy'(xn)(xn,y(xn);y'(xn滿足微分方程y'(xnf(xny(xn(xn,y(xn);0)f(xn,y(xn定義：?jiǎn)尾椒▂n1ynh(xnynh(x,y;0)f(x,y'(x)f(x,y(x0) h[y'(x)(x,y(x);0)]O(h2n O(h2p Tn10(當(dāng)h0)h相容單步法，若(xyhy滿足Lipschitz條件 yny(xn),hyn1ynh(xn,yn;yn1(x,y; h令xx0nh當(dāng)yn1(x,y; h令xx0nh當(dāng)h0y'(xn)(xn,y(xn);f(xn,y(xnyn1ynh(xn,yn;?nyn是單步法精確計(jì)算而得的準(zhǔn)確yn1ynh(xn,yn;?n?n?n?n?n n 1h(x,,n由于f有關(guān)，所以穩(wěn)定性與方程（微分）yy（Re( 解y(x)ex，yn先考慮Eulery先考慮Euleryyyn1ynhf(xn,ynynhyn(1h)yn1(1h)?nen1(1yn1(1h)yn1E(h)E(h依賴于方法選取，EulerE(h)(1hE(h令yn有誤差E(h)[yn]E(h)ynyn1yn有誤差yn1E(h)E(h)[yn1E(h)]yn2yn2?ynk有誤差[E(h)]k如果[E(h)]k定義：?jiǎn)尾椒?yn1ynh(xn,yn;h)解模型問題y'y，若得到的yn1E(hy如果[E(h)]k定義：?jiǎn)尾椒?yn1ynh(xn,yn;h)解模型問題y'y，若得到的yn1E(hyn在復(fù)hE(h對(duì)于每個(gè)方法是不同的，對(duì)于EulerE(h)11為復(fù)數(shù)，令hh1位圓的內(nèi)部，此為Euler11與實(shí)軸交的區(qū)間為(20對(duì)于改進(jìn)EulerE(h)1h21h2211h22 )即2考慮區(qū)間(220),[0，當(dāng)hh(20進(jìn)Euler方法的絕對(duì)穩(wěn)定性區(qū)間為(204階經(jīng)典R-K yh(K2K2KK 12 6K1f(xn,ynKf(x1h,y1hK2nn122Kf(x1h,y1hK3nn222K4f(xnh,ynhK3y'K1K(y1hK)y1hK(1h)2n1n1n2Kf(x1h,y1hK2nn122Kf(x1h,y1hK3nn222K4f(xnh,ynhK3y'K1K(y1hK)y1hK(1h)2n1n1n222K(y1hK)yh(y1hy3n2nnn222[1h(h)2]n K(yhK)yh[1h(h)2]4n3nn yhy2(1h)y2[1h(h)2]6 nnn2 (h)[1h24ynhynynyn2[1h2E(h)1h21，絕對(duì)穩(wěn)定絕對(duì)穩(wěn)定區(qū)間為0h向后Euleryn1ynhf(xn1,yn1用于yyn1yn1 ynh11等價(jià)1yn1yn1 ynh11等價(jià)1 圓外部，故絕對(duì)穩(wěn)定性區(qū)間為h0A-穩(wěn)定的。向后Euler方法是A穩(wěn)定的?！?yn1ynh(xn,yn;xnx0xnynxn1yn1xn1yn1xnnlyn,yn1,?,ynlyn1y'f(x,y(x) xnx0nhxn1x0(n1)h,?,xnlx0(nxxy'(x)dxf(x,右邊采用Simpsony(xn2)y(xn)2hf(x,y(x))4f, , ))f nn611fnf(xn,yn yh] 3yn2yn,yn1ffnf(xn,yn yh] 3yn2yn,yn1f(xn,yn),f(xn1,yn1fnikkjynjhjfnjj其中jj常數(shù)，k0;0,0kk1.ynkk1ynk1…1yn10nk1yn1…k1ynknkynkyn,yn1,?,ynk1nk1若k0k0khkf(xnk,ynk)g g(hjfnjjynjhf(xy() s,y(s 迭代收斂條件hL1，Lf(xyy的Lipschitz線性單步法是線性k步方法的特例，例如k1yn10ynh0fnh1不同0,0,（局部截?cái)嗾`差）y'f(x,y(x)y(x) kkjynjhjfnjjjkkjy(xnj)hjf(xnj,y(xnjj稱為線性k步法（*）xnkTnkhy'f(x,y(x)y(x) kkjynjhjfnjjjkkjy(xnj)hjf(xnj,y(xnjj稱為線性k步法（*）xnkTnkh hp1y(p1))O(hp2 np1( (xn為主局部截?cái)嗾`差，相應(yīng)的多步法稱為PSimpson h] 3y(x代入，并Taylor ) )h[y )4y'(x)y n3y(xn)hy'(xn) y(xn) y(xn)24 (xn)120 (xn)?(26[y(x)hy'(x) y(xn) y(xn)24 (xn)120 (xn)nn26h[y'(x)hy(x) y(x) (x) (x) 264y'(xny'(xn)hy(xn) y(xn) (xn)24 (xn)263h35h2hy'(xn(xn)?y(xnyh[6y'(x)hy(x) (x) 2[11]h5y(5)(x)?1h5y(5))?nn 1局部截?cái)嗾`差主項(xiàng)為h5y(5))n（II）Adams方1.Adams方為[xnk1xnkyf(x,y[xnk1xnk ))1局部截?cái)嗾`差主項(xiàng)為h5y(5))n（II）Adams方1.Adams方為[xnk1xnkyf(x,y[xnk1xnk ))f(x,nkxf(xnkf(xnk1,y(xnk1f(xn,y(xnf(xn1,y(xn1這樣可以求得k-1次Lagrange??Lk1(x)f(x,y(x))l(x)?f,nk nk nk (x)為對(duì)應(yīng)點(diǎn) ,j0,1,?,k1,上的k-1次插值多項(xiàng)式基函nnkl)(nkl )) nkkxf(x,y(x l?(x)dx?f,nk nknk nxxnknkhnf(xn,y(xn))?hnk1f(xnk1,y(xnk11(x)dx,j0,1,?,knnhxynkynk1hnfnhn1fn1?hnk1fnkf(xnj,ynj),j0,1,?,kfn這是顯式Adams方法，稱Adams-Bashforthf(xn,y(xnf(xn1,y(xn1L1(x)f(x,y(x))l(x)f(xn, ??n )n R1(x)fn1xxf(xn,y(xnf(xn1,y(xn1L1(x)f(x,y(x))l(x)f(xn, ??n )n R1(x)fn1xxy'(x)dxf(x, n2L(x)dx n2R11xxf(x,y(x))?(x)dx,y(x f(x n xxxfxxf(x,y(x x?f(x,y(xnnnxxfxn2l?(x)dxn2x (x)dx3x?2x )dx5xn26xh yn2yn12(3fn1fn)（二步法）) )h[3y )y'(x n223y(xn)2hy'(xny(xny(xn262h23h6[y(xn)hy'(xny(xn)y(xn2hh[3(y'(x)hy(x y23y(xn)2hy'(xny(xny(xn262h23h6[y(xn)hy'(xny(xn)y(xn2hh[3(y'(x)hy(x y(xn)?)y'(xn2hy'(x)3h2y(x)7h3y(x)nnn2634hy'(xn 2hy(xny(xn)25 h3y(x)O(h4n2.隱式Adams方在顯式Adamsnk1y(xn),y(xn1),?,y(xnk1ynkxnkf(xnk,y(xnknkf(xnk1,y(xnk1f(xn,y(xnf(xn1,y(xn1共有k+1個(gè)節(jié)點(diǎn) Lk(x)插值多項(xiàng) ))L(x)dxnkkxxnk取k1L(x)f(x,y(x1 R(x)fxn1y'(x)dxL(x)dx n1nL(x)dxf(x,y(x1h[f(x,y(x))f , 26n1)dxf[xn,R(x)dxfxxnn3dx12 f(x,12h[f] 23 y(Adams方法orAdams-Moulton方法6n1)dxf[xn,R(x)dxfxxnn3dx12 f(x,12h[f] 23 y(Adams方法orAdams-Moulton方法ynk1nk下面對(duì)常用顯式Adams和隱式Adamsp1 (xn) 顯式Adams（Adams-kpyn1ynhf(xn,yn112538 h(3f22) n2h 33] nh 944] n隱式AdamsAdams-Moulton方法kp方1 yh(f12 21h 23] nh f 34 n1yn1…k1ynknk設(shè)yxn處進(jìn)行Taylory(xnk)[0y(xn)1y(xn1)…k1y(xnk1h[0y'(xn)1y'(xnnk設(shè)yxn處進(jìn)行Taylory(xnk)[0y(xn)1y(xn1)…k1y(xnk1h[0y'(xn)1y'(xn1)?k1y'(xnk1)ky'(xnk)]2(y(x )y(xnjh)y(xn) y'(xny(xn)?n2(y )y'(xnjh)y'(xn) y(xny(xn)?nk [y(x hy Cy(xChy'(x?Chly(lx? n n nlnjC001?k,kC1122?kk(01?kC1(22?k2212k(122?kkC1(23?k3312k1(22?k212kC1(24?k4412k1(23?k312k?C1(2l?kll12kl12l1?kl1(l12k4步方法k401yn12yn23n234fn4C001231234() C214293C001231234() C214293162(1223344)827643(4916) 31234C411628132564(182273644)59取0120,403 2(12233)373(491231754(182273由此得到k4p4的Adams-Bashforthh 9f 如果在（*）中取1230,4848011323,33，可以有Milne公式 y4h)n 3 14h5y(5)(x)O(h6n)y(x)4h[2f , ))f, n32f(xn1,y(xn1)y(x)4h[2y)y)2y n3[y(xn)4hy'(xn)y(xn)y(xn) (xn) (xn)?y(xn26[2(y'(xn)3hy(xn)y(xn) (xn)( (xn)26(y'(xn)2hy(xn)y(xn[2(y'(xn)3hy(xn)y(xn) (xn)( (xn)26(y'(xn)2hy(xn)y(xn)(xn)(xn)yy262(y'(xn)hy(xn) (xn)24 (xn)614h5y(5)(x)O(h6n01yn12h(03fn3n2C00121C11223(0123)0C214292(12233)0C3182273(14293)C41162814(182273)7個(gè)未知數(shù)，5個(gè)方程，取10,001,02883,6,321888得到1(9