1、第八章常微分方程數值解姓名學號班級習題主要考察點：歐拉方法的構造，單步法的收斂性和穩定性的討論，線性多步法中亞當姆斯方法的構造和討論。1用改進的歐拉公式，求以下微分方程2xy=y_yx0,1y（0）=1的數值解（取步長h=0.2）,并與精確解作比較。（改進的尤拉公式的應用）2y2dz解：原萬程可轉化為yy=y-2乂，令2="，有2z=2x2dx解此一階線性微分方程，可得y=J2x+1。利用以下公式2xiyp=yi+0.2(yi)yi2xa=y+0.2<yp-)(i=0,1,2,3,4)Yp1，、Yt=2(YpYc)求在節點xj=0.2，i(i=1,2,3,4,5)處的數值解y,

2、其中，初值為x0=0,y0=1。MATLAB程序如下：x(1)=0;%初值節點y(1)=1;涮值fprintf('x(%d)=%f,y(%d)=%f,yy(%d)=%fn',1,x(1),1,y(1),1,y(1);fori=1:5yp=y(i)+0.2*(y(i)-2*x(i)/y(i);%預報值yc=y(i)+0.2*(yp-2*x(i)/yp);%正值y(i+1)=(yp+yc)/2;%進值x(i+1)=x(i)+0.2;初點值yy(i+1)=sqrt(2*x(i+1)+1);研確解fprintf('x(%d)=%f,y(%d)=%f,yy(%d)=%fn'

3、;,i+1,x(i+1),i+1,y(i+1),i+1,yy(i+1);end程序運行的結果如下：x(1)=0.000000,y(1)=1.000000,yy(1)=1.000000x(2)=0.200000,y(2)=1.220000,yy(2)=1.183216x(3)=0.400000,y(3)=1.420452,yy(3)=1.341641x(4)=0.600000,y(4)=1.615113,yy(4)=1.483240x(5)=0.800000,y(5)=1.814224,yy(5)=1.612452x(6)=1.000000,y(6)=2.027550,yy(6)=1.73205

4、1'yy12用四階龍格庫塔法求解初值問題3yy，取h=0.2，求x=0.2,0.4時的數值解y(0)=0.要求寫出由h,Xn,yn直接計算yn書的迭代公式，計算過程保留3位小數。(龍格一庫塔方法的應用)解：四階龍格-庫塔經典公式為hyn1.=yn-(kl2k22k3k)k1=f(Xn,yn),一1.1、k2=f(Xn2h,yn2hk1),一1.1,、k3=f(Xnh,ynhk2)22k4=f(Xnh,ynhk3)由于f(X,y)=1-y,在各點的斜率預報值分別為:k1=1-ynhhhk2=1-(yn2k1)=1-yn一2(1一yn)=(1一yn)(1-2)k3=1-(yn二k2)=1-

5、yn-二(1一yn)(1一二)=(1一yn)1一二(1一二)22222k4=1-(ynhk3)=1-yn-h(1-yn)1-1(1-h)=(1-yn)1-h(1-1(1-)2222四階經典公式可改寫成以下直接的形式：h2h3yn1=yn(1-yn)(6-3hh)64在x=X1=0.2處，有0.22(0.2)3y1=0(1-0)(6-30.2(0.2)-)-0.181364在x=x2=0.4處，有0.22(0.2)3y2=0.1813(1-0.1813)(6-30.2(0.2)-一-)=0.329764注：這兩個近似值與精確解y=1-e"在這兩點的精確值十分接近。3用梯形方法解初值問題

6、y+y=0:y(0)=1證明其近似解為2h丫yn=并證明當hT0時，它收斂于原初值問題的準確解y=e«。解：顯然，y=e”是原初值問題的準確解。求解一般微分方程初值問題的梯形公式的形式為h_一yn1=ynf(Xn,yn)f(Xn1,yn1)2對于該初值問題，其梯形公式的具體形式為hhh2-hyn4=yn1(一ynynd0，(1二)yn書=(1一二)yn，yn1=()yn2222h)/2-hyn(2_%)yn2y。-2-h)/2-h)yn<2+hJyn12+hJ亦即：yn=注意到：xn=0+nh=nh,n=&，令t=速匚，1=11有h2hht2yn=1Xn2h不i2h色=

7、(1t)7xn2XnXn二(1骨廣(13-2從而lhimQyn=limQ(1t)_Xn-t_Xn!im°(1t)一2=En即：當hT0時，yn收斂于原初值問題的準確解y(Xn)=e'n。4對于初值問題；y'=-10y:y(0)=1,證明當h<0.2時，歐拉公式絕對穩定。(顯式和隱式歐拉公式的穩定性討論)證明：顯式的歐拉公式為yn1=yn,hf(Xn,yn)=(1-10h)yn從而en+=(110h)en,由于0<h<0.2,-1<1-10h<1,en由<en因此，顯式歐拉公式絕對穩定。隱式的歐拉公式為yn1ynhf(xn1,yn1)

8、=yn-10hynd_yn_enyn1=r0h5en1=?d0h，一C-1/由于0<h,0<<1,en+<en1+10h因此，隱式的歐拉公式也是絕對穩定的。h.5證明：梯形公式yn由=yn+£儀口，丫口)+£儀口書3口中)無條件穩定。(梯形公式的穩定2性討論)解：對于微分方程初值問題|y=一九y:y(0)=1(0)其隱式的梯形公式的具體形式可表示為h5)y"=(1一myn1=(G)ynyn1=yn2'Yn一Wn1,(12-1h從而en1"()en2h,、,2+入h1由hA0,九A0可知，en書<(-一h)-en=en

9、,故隱式的梯形公式無條件穩定。2+Khy=f(x,y)6設有常微分方程的初值問題y,力，試用泰勒展開法，構造線性兩步法數值計算Y(x0)=Y0公式Yn書=a(Yn+Yn_L)+h(P°fn+$)，使其具有二階精度，并推導其局部截斷誤差主項。(局部截斷誤差和主項的計算)解：假設Yn=y(Xn),YnJ=Y(Xnj),利用泰勒展式，有Yn二Y(Xn)=Y(Xn)-Y(Xn)h以當h2一="h326fn=f(Xn，Yn)=f(Xn,y(Xn)=Y(Xn)fnl=f(Xni,Yn,)=f(Xnj,Y(Xn4)=Y(Xn二尸Y仰)-Y(Xn)hY'h?-2ot0tpYn1=2

10、-Y(Xn)(：0：1-:)Y(Xn)h(-：)y佟方(1)Y(Xn)h32621213又以一之)y(Xn)h2y(xn)h6yd)h欲使其具有盡可能高的局部截斷誤差，必須2ct=1,Po+Pi-Ct=1,；，1=122.1.7.1從而,：0=,:1=一2441 71、于無數值計算公式為yn書=-(yn+yn)+h(fnfn)。2 44該數值計算公式的局部截斷誤差的主項為y(xn1)7n1嗎6(4*3Bn)"7已知初值問題y=2xy(o)=oy(0.1)=0.01取步長h=0.1,利用阿當姆斯公式yn+=yn+h(3fn-fn)，求此微分方程在上的數值解，求此公式的局部截斷誤差的首項

11、。(阿當姆斯公式的應用)解：假設yn=y(Xn),yn=Y(Xnj),利用泰勒展開，有yn=y(xn),fn=y'(xn),fn/RGeXnigh+3h2yn1=y(xn)y(xn)h1y(xn)h2-':424_.1213而y(xn1)=y(xn)y(xn)hy(xn)hy(xn)h26y(xn1)yn1=(；1)y(xn)h3=-5y(xn)h364125該阿當姆斯兩步公式具有2階精度，其局部截斷誤差的主項為y(xn)h3o12取步長h=0.1,節點xn=0.1n(n=0,1,2，100),注意到f(x,y)=2x,式可改寫為0.1-yn1=yn£(6xn-2。=

12、丫00.02n0.01僅需取一個初值y0=0,可實現這一公式的實際計算。其MATLAB下的程序如下：x0=0;%初值節點0,10其計算公y0=0;班值forn=0:99y1=y0+0.02*n+0.01;x1=x0+0.1;,n+1,x1,n+1,y1);fprintf('x(%3d)=%10.8f,y(%3d)=%10.8fn'x0=x1;y0=y1;end運行結果如下：x(1)=0.10000000,y(1)=0.01000000x(2)=0.20000000,y(2)=0.04000000x(3)=0.30000000,y(3)=0.09000000x(4)=0.4000

13、0000,y(4)=0.16000000x(5)=0.50000000,y(5)=0.25000000x(6)=0.60000000,y(6)=0.36000000x(7)=0.70000000,y(7)=0.49000000x(8)=0.80000000,y(8)=0.64000000x(9)=0.90000000,y(9)=0.81000000x(10)=1.00000000,y(10)=1.00000000x(11)=1.10000000,y(11)=1.21000000x(12)=1.20000000,y(12)=1.44000000x(13)=1.30000000,y(13)=1.6

14、9000000x(14)=1.40000000,y(14)=1.96000000x(15)=1.50000000,y(15)=2.25000000x(16)=1.60000000,y(16)=2.56000000x(17)=1.70000000,y(17)=2.89000000x(18)=1.80000000,y(18)=3.24000000x(19)=1.90000000,y(19)=3.61000000x(20)=2.00000000,y(20)=4.00000000x(21)=2.10000000,y(21)=4.41000000x(22)=2.20000000,y(22)=4.8400

15、0000x(23)=2.30000000,y(23)=5.29000000x(24)=2.40000000,y(24)=5.76000000x(25)=2.50000000,y(25)=6.25000000x(26)=2.60000000,y(26)=6.76000000x(27)=2.70000000,y(27)=7.29000000x(28)=2.80000000,y(28)=7.84000000x(29)=2.90000000,y(29)=8.41000000x(30)=3.00000000,y(30)=9.00000000x(31)=3.10000000,y(31)=9.6100000

16、0x(32)=3.20000000,y(32)=10.24000000x(33)=3.30000000,y(33)=10.89000000x(34)=3.40000000,y(34)=11.56000000x(35)=3.50000000,y(35)=12.25000000x(36)=3.60000000,y(36)=12.96000000x(37)=3.70000000,y(37)=13.69000000x(38)=3.80000000,y(38)=14.44000000x(39)=3.90000000,y(39)=15.21000000x(40)=4.00000000,y(40)=16.0

17、0000000x(41)=4.10000000,y(41)=16.81000000x(42)=4.20000000,y(42)=17.64000000x(43)=4.30000000,y(43)=18.49000000x(44)=4.40000000,y(44)=19.36000000x(45)=4.50000000,y(45)=20.25000000x(46)=4.60000000,y(46)=21.16000000x(47)=4.70000000,y(47)=22.09000000x(48)=4.80000000,y(48)=23.04000000x(49)=4.90000000,y(49

18、)=24.01000000x(50)=5.00000000,y(50)=25.00000000x(51)=5.10000000,y(51)=26.01000000x(52)=5.20000000,y(52)=27.04000000x(53)=5.30000000,y(53)=28.09000000x(54)=5.40000000,y(54)=29.16000000x(55)=5.50000000,y(55)=30.25000000x(56)=5.60000000,y(56)=31.36000000x(57)=5.70000000,y(57)=32.49000000x(58)=5.8000000

19、0,y(58)=33.64000000x(59)=5.90000000,y(59)=34.81000000x(60)=6.00000000,y(60)=36.00000000x(61)=6.10000000,y(61)=37.21000000x(62)=6.20000000,y(62)=38.44000000x(63)=6.30000000,y(63)=39.69000000x(64)=6.40000000,y(64)=40.96000000x(65)=6.50000000,y(65)=42.25000000x(66)=6.60000000,y(66)=43.56000000x(67)=6.7

20、0000000,y(67)=44.89000000x(68)=6.80000000,y(68)=46.24000000x(69)=6.90000000,y(69)=47.61000000x(70)=7.00000000,y(70)=49.00000000x(71)=7.10000000,y(71)=50.41000000x(72)=7.20000000,y(72)=51.84000000x(73)=7.30000000,y(73)=53.29000000x(74)=7.40000000,y(74)=54.76000000x(75)=7.50000000,y(75)=56.25000000x(76)=7.60000000,y(76)=57.76000000x(77)=7.70000000,y(77)=59.29000000x(78)=7.80000000,y(78)=60.84000000x(79)=7.90000000,y(79)=62.41000000x(80)=8.00000000,y(80)=64.0000000