南京工業(yè)大學(xué)研究生工程數(shù)學(xué)應(yīng)用B第一次作業(yè)

1、工程應(yīng)用數(shù)學(xué)作業(yè)（一）專業(yè)：安全科學(xué)與工程 姓名：* 學(xué)號(hào)：652083700*一、第三章練習(xí)題說(shuō)明：解題所需函數(shù)均提前保存為m文件，方便直接調(diào)用，以下解題過(guò)程不包括保存m文件過(guò)程。練習(xí)題1：設(shè)兩點(diǎn)邊值問(wèn)題的精確解為現(xiàn)以h為步長(zhǎng)劃分區(qū)間為100等份，用差分近似代替微分，將微分方程離散化為線性方程組，代入初始條件后，得到如下的方程組問(wèn)題其中，。(1) 分別用J迭代法，G-S迭代法和SOR迭代法求解，并與精確解進(jìn)行比較；(2) 如果，再求解該問(wèn)題解：（1）1）輸入：t=1;a=1/2;h=1/100;A=(-h-2*t)*eye(99)+diag(t+h)*ones(98,1),1)+diag(t

2、)*ones(98,1),-1);C(99,1)=zeros;C(99,1)=t+h;b=(a*h2)*ones(99,1)-C;x0=zeros(99,1);e=2e-10;y=jacobi(A,b,x0,e)進(jìn)行jacobi迭代。2）輸入：t=1;a=1/2;h=1/100;A=(-h-2*t)*eye(99)+diag(t+h)*ones(98,1),1)+diag(t)*ones(98,1),-1);C(99,1)=zeros;C(99,1)=t+h;b=(a*h2)*ones(99,1)-C;x0=zeros(99,1);e=2e-10;for i=1:15 w=0.95+i/20;

3、N=8; y,r,n=sor(a,b,x0,w,1e-15,N); h(i)=r;ends,j=min(h);w=0.95+j/20得松弛因子w=1，再輸入：x,r,n=sor(A,b,x0,w,e,6900)完成sor算法。3）輸入：t=1;a=1/2;h=0.001;A=(-h-2*t)*eye(99)+diag(t+h)*ones(98,1),1)+diag(t)*ones(98,1),-1);C(99,1)=zeros;C(99,1)=t+h;b=(a*h2)*ones(99,1)-C;x0=zeros(99,1);e=2e-10; y,r,n=seidel(A,b,x0,e)完成se

4、diel迭代。輸出結(jié)果如下表所示：Jacobi迭代Sediel迭代Sor迭代0.01290.01050.01280.0256 0.02090.02560.0383 0.03140.03820.0510 0.04190.05080.0635 0.05230.06340.0760 0.06270.07580.0884 0.07310.08820.1007 0.08350.10050.1130 0.09390.11270.1251 0.10430.12490.1372 0.11470.13700.1493 0.12510.14900.1613 0.13540.16090.1732 0.14580.1

5、7280.1850 0.15610.18460.1968 0.16650.19630.2085 0.17680.20800.2201 0.18710.21960.2317 0.19740.23120.2432 0.20770.24270.2546 0.21800.25410.2660 0.22820.26540.2773 0.23850.27670.2885 0.24870.28800.2997 0.25900.29910.3108 0.26920.31020.3219 0.27940.32130.3329 0.28960.33230.3438 0.29980.34320.3547 0.310

6、00.35410.3656 0.32020.36490.3763 0.33040.37570.3870 0.34060.38640.3977 0.35070.39700.4083 0.36090.40760.4188 0.37100.41820.4293 0.38110.42860.4398 0.39120.43910.4501 0.40130.44940.4605 0.41140.45980.4707 0.42150.47000.4810 0.43160.48030.4911 0.44170.49040.5013 0.45170.50060.5113 0.46180.51060.5214 0

7、.47180.52060.5313 0.48180.53060.5412 0.49190.54050.5511 0.50190.55040.5609 0.51190.56020.5707 0.52190.57000.5804 0.53180.57980.5901 0.54180.58940.5998 0.55180.59910.6093 0.56170.60870.6189 0.57170.61820.6284 0.58160.62770.6378 0.59150.63720.6473 0.60140.64660.6566 0.61140.65600.6659 0.62130.66530.67

8、52 0.63110.67460.6845 0.64100.68390.6937 0.65090.69310.7028 0.66070.70220.7119 0.67060.71140.7210 0.68040.72050.7300 0.69030.72950.7390 0.70010.73850.7480 0.70990.74750.7569 0.71970.75640.7658 0.72950.76530.7746 0.73930.77410.7834 0.74910.78300.7922 0.75880.79170.8009 0.76860.80050.8096 0.77830.8092

9、0.8182 0.78810.81780.8268 0.79780.82650.8354 0.80750.83510.8440 0.81720.84360.8525 0.82690.85220.8609 0.83660.86060.8694 0.84630.86910.8778 0.85600.87750.8862 0.86560.88590.8945 0.87530.89430.9028 0.88490.90260.9111 0.89460.91090.9193 0.90420.91910.9275 0.91380.92740.9357 0.92340.93560.9438 0.93300.

10、94370.9520 0.94260.95190.9600 0.95220.96000.9681 0.96180.96800.9761 0.97140.97610.9841 0.98090.98410.9921 0.99050.9921于是得到以下結(jié)論：達(dá)到相同精度Jacobi迭代的迭代次數(shù)為n=30006達(dá)到相同精度sediel迭代的迭代次數(shù)n=15202達(dá)到相同精度sor迭代的迭代次數(shù)n=6900sor迭代最佳松弛因子w=1由結(jié)果可見(jiàn)對(duì)于此題達(dá)到相同精度迭代次數(shù)sor迭代<G-S迭代<J迭代。（2）t=0.1,輸出結(jié)果如下表所示：精確值Jacobi迭代Sediel迭代Sor迭代

11、0.00560.05050.01540.05050.10060.09680.03070.09680.14460.13940.04580.13940.18480.17850.06080.17850.22170.21460.07570.21460.25560.24780.09040.24780.28670.27840.10490.27840.31530.30680.11930.30680.34170.33300.13360.33300.36610.35730.14780.35730.38860.37980.16180.37980.40940.40070.17560.40070.42880.4202

12、0.18940.42020.44670.43840.20300.43840.46350.45530.21650.45530.47910.47120.22980.47120.49370.48610.24310.48610.50740.50010.25620.50010.52020.51330.26910.51330.53240.52570.28200.52570.54380.53750.29470.53750.55460.54860.30730.54860.56490.55920.31980.55920.57470.56930.33220.56930.58400.57890.34440.5789

13、0.59290.58810.35660.58810.60140.59690.36860.59690.60960.60540.38050.60540.61750.61350.39230.61350.62510.62140.40400.62140.63250.62900.41560.62900.63960.63640.42700.63640.64660.64350.43840.64350.65330.65050.44960.65050.65990.65720.46080.65720.66640.66390.47180.66390.67270.67030.48270.67030.67880.6767

14、0.49360.67670.68490.68290.50430.68290.69090.68900.51490.68900.69670.69500.52540.69500.70250.70090.53580.70090.70820.70670.54620.70670.71390.71250.55640.71250.71950.71820.56650.71820.72500.72380.57660.72380.73050.72940.58650.72940.73590.73490.59640.73490.74130.74040.60610.74040.74670.74580.61580.7458

15、0.75200.75120.62540.75120.75730.75650.63480.75650.76250.76180.64420.76180.76780.76710.65350.76710.77300.77240.66280.77240.77820.77760.67190.77760.78330.78290.68090.78290.78850.78800.68990.78800.79370.79320.69880.79320.79880.79840.70760.79840.80390.80350.71630.80350.80900.80870.72490.80870.81410.8138

16、0.73350.81380.81920.81890.74200.81890.82430.82400.75030.82400.82930.82910.75870.82910.83440.83420.76690.83420.83950.83930.77510.83930.84450.84430.78320.84430.84960.84940.79120.84940.85460.85450.79910.85450.85960.85950.80700.85950.86470.86460.81480.86460.86970.86960.82250.86960.87470.87460.83010.8746

17、0.87980.87970.83770.87970.88480.88470.84520.88470.88980.88970.85260.88970.89480.89480.86000.89480.89990.89980.86730.89980.90490.90480.87450.90480.90990.90980.88170.90980.91490.91490.88880.91490.91990.91990.89580.91990.92490.92490.90280.92490.92990.92990.90970.92990.93490.93490.91650.93490.93990.9399

18、0.92330.93990.94500.94490.93000.94490.95000.94990.93670.94990.95500.95500.94330.95500.96000.96000.94980.96000.96500.96500.95630.96500.97000.97000.96270.97000.97500.97500.96910.97500.98000.98000.97540.98000.98500.98500.98160.98500.99000.99000.98780.99000.99500.99500.99390.9950達(dá)到相同精度Jacobi迭代的迭代次數(shù)為n=11

19、156達(dá)到相同精度sediel迭代的迭代次數(shù)n=15064達(dá)到相同精度sor迭代的迭代次數(shù)n=5596sor迭代最佳松弛因子w=1（3）t=0.001,輸出結(jié)果如下表所示：Jacobi迭代Sediel迭代Sor迭代0.45950.47550.4595 0.5059 0.7135 0.5059 0.5146 0.8327 0.5146 0.5200 0.8926 0.5200 0.5250 0.9228 0.5250 0.5300 0.9382 0.5300 0.5350 0.9461 0.5350 0.5400 0.9503 0.5400 0.5450 0.9526 0.5450 0.5500

20、 0.9541 0.5500 0.5550 0.9550 0.5550 0.5600 0.9558 0.5600 0.5650 0.9564 0.5650 0.5700 0.9569 0.5700 0.5750 0.9575 0.5750 0.5800 0.9580 0.5800 0.5850 0.9585 0.5850 0.5900 0.9590 0.5900 0.5950 0.9595 0.5950 0.6000 0.9600 0.6000 0.6050 0.9605 0.6050 0.6100 0.9610 0.6100 0.6150 0.9615 0.6150 0.6200 0.962

21、0 0.6200 0.6250 0.9625 0.6250 0.6300 0.9630 0.6300 0.6350 0.9635 0.6350 0.6400 0.9640 0.6400 0.6450 0.9645 0.6450 0.6500 0.9650 0.6500 0.6550 0.9655 0.6550 0.6600 0.9660 0.6600 0.6650 0.9665 0.6650 0.6700 0.9670 0.6700 0.6750 0.9675 0.6750 0.6800 0.9680 0.6800 0.6850 0.9685 0.6850 0.6900 0.9690 0.69

22、00 0.6950 0.9695 0.6950 0.7000 0.9700 0.7000 0.7050 0.9705 0.7050 0.7100 0.9710 0.7100 0.7150 0.9715 0.7150 0.7200 0.9720 0.7200 0.7250 0.9725 0.7250 0.7300 0.9730 0.7300 0.7350 0.9735 0.7350 0.7400 0.9740 0.7400 0.7450 0.9745 0.7450 0.7500 0.9750 0.7500 0.7550 0.9755 0.7550 0.7600 0.9760 0.7600 0.7

23、650 0.9765 0.7650 0.7700 0.9770 0.7700 0.7750 0.9775 0.7750 0.7800 0.9780 0.7800 0.7850 0.9785 0.7850 0.7900 0.9790 0.7900 0.7950 0.9795 0.7950 0.8000 0.9800 0.8000 0.8050 0.9805 0.8050 0.8100 0.9810 0.8100 0.8150 0.9815 0.8150 0.8200 0.9820 0.8200 0.8250 0.9825 0.8250 0.8300 0.9830 0.8300 0.8350 0.

24、9835 0.8350 0.8400 0.9840 0.8400 0.8450 0.9845 0.8450 0.8500 0.9850 0.8500 0.8550 0.9855 0.8550 0.8600 0.9860 0.8600 0.8650 0.9865 0.8650 0.8700 0.9870 0.8700 0.8750 0.9875 0.8750 0.8800 0.9880 0.8800 0.8850 0.9885 0.8850 0.8900 0.9890 0.8900 0.8950 0.9895 0.8950 0.9000 0.9900 0.9000 0.9050 0.9905 0

25、.9050 0.9100 0.9910 0.9100 0.9150 0.9915 0.9150 0.9200 0.9920 0.9200 0.9250 0.9925 0.9250 0.9300 0.9930 0.9300 0.9350 0.9935 0.9350 0.9400 0.9940 0.9400 0.9450 0.9945 0.9450 0.9500 0.9950 0.9500 0.9550 0.9955 0.9550 0.9600 0.9960 0.9600 0.9650 0.9965 0.9650 0.9700 0.9970 0.9700 0.9750 0.9975 0.9750

26、0.9800 0.9980 0.9800 0.9850 0.9985 0.9850 0.9900 0.9990 0.9900 0.9950 0.9995 0.9950于是得到以下結(jié)論：達(dá)到相同精度Jacobi迭代的迭代次數(shù)為n=184達(dá)到相同精度sediel迭代的迭代次數(shù)n=417達(dá)到相同精度sor迭代的迭代次數(shù)n=142sor迭代最佳松弛因子w=1練習(xí)題2：設(shè)是一個(gè)對(duì)稱正定矩陣，分別是其最大和最小特征值，對(duì)于線性方程組建立迭代法求出的范圍使迭代法收斂，并求出最優(yōu)使迭代法的漸進(jìn)收斂速度最大。解：水平有限，暫未解出。練習(xí)題3：對(duì)某電路的分析，可以歸結(jié)為下面的線性方程組，其中R(1,1)=31；R

27、(1,2)=-13；R(1,6)=-10；R(2,1)=-13；R(2,2)=35；R(2,3)=-9；R(2,5)=-11；R(3,2)=-9；R(3,3)=31；R(3,4)=-10；R(4,3)=-10；R(4,4)=79；R(4,5)=-30；R(4,9)=-9；R(5,4)=-30；R(5,5)=57；R(5,6)=-7；R(5,8)=-5；R(6,5)=-7；R(6,6)=47；R(6,7)=-30；R(7,6)=-30；R(7,7)=41；R(8,5)=-5；R(8,8)=27；R(8,9)=-2；R(9,4)=-9；R(9,8)=-2；R(9,9)=29；V=(-15, 27,

28、 -23, 0, -20, 12, -7, 7, 10)T其余元素為零。要求：（1）用高斯列主元消去法求解該方程組；（2）用SOR方法迭代求解該方程組，誤差，近似最佳松弛因子由試算法確定，設(shè)解：（1）首先輸入：A=zeros(9,9);A(1,1)=31;A(1,2)=-13;A(1,6)=-10;A(2,1)=-13;A(2,2)=35;A(2,3)=-9;A(2,5)=-11;A(3,2)=-9;A(3,3)=31;A(3,4)=-10;A(4,3)=-10;A(4,4)=79;A(4,5)=-30;A(4,9)=-9;A(5,4)=-30;A(5,5)=57;A(5,6)=-7;A(5,

29、8)=-5;A(6,5)=-7;A(6,6)=47;A(6,7)=-30;A(7,6)=-30;A(7,7)=41;A(8,5)=-5;A(8,8)=27;A(8,9)=-2;A(9,4)=-9;A(9,8)=-2;A(9,9)=29;L,U,P=lu(A)得：L = 1.0000 0 0 0 0 0 0 0 0 -0.4194 1.0000 0 0 0 0 0 0 0 0 -0.3046 1.0000 0 0 0 0 0 0 0 0 -0.3539 1.0000 0 0 0 0 0 0 0 0 -0.3976 1.0000 0 0 0 0 0 0 0 0 -0.1569 1.0000 0 0

30、 0 0 0 0 0 0 -0.6540 1.0000 0 0 0 0 0 0 -0.1121 -0.0175 -0.0246 1.0000 0 0 0 0 -0.1193 -0.0834 -0.0142 -0.0200 -0.0923 1.0000U = 31.0000 -13.0000 0 0 0 -10.0000 0 0 0 0 29.5484 -9.0000 0 -11.0000 -4.1935 0 0 0 0 0 28.2587 -10.0000 -3.3504 -1.2773 0 0 0 0 0 0 75.4613 -31.1856 -0.4520 0 0 -9.0000 0 0

31、0 0 44.6020 -7.1797 0 -5.0000 -3.5780 0 0 0 0 0 45.8732 -30.0000 -0.7847 -0.5615 0 0 0 0 0 0 21.3807 -0.5132 -0.3672 0 0 0 0 0 0 0 26.4131 -2.4200 0 0 0 0 0 0 0 0 27.3895P = 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0

32、0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1之后，輸入：b=-15;27;-23;0;-20;12;-7;7;10;x=Ab得：x = -0.2892 0.3454 -0.7128 -0.2206 -0.4304 0.1543 -0.0578 0.20110.2902（2）輸入：A=zeros(9,9);A(1,1)=31;A(1,2)=-13;A(1,6)=-10;A(2,1)=-13;A(2,2)=35;A(2,3)=-9;A(2,5)=-11;A(3,2)=-9;A(3,3)=31;A(3,4)=-10;A(4,3)=-10;A(4,4)=79;A(4,5)=-

33、30;A(4,9)=-9;A(5,4)=-30;A(5,5)=57;A(5,6)=-7;A(5,8)=-5;A(6,5)=-7;A(6,6)=47;A(6,7)=-30;A(7,6)=-30;A(7,7)=41;A(8,5)=-5;A(8,8)=27;A(8,9)=-2;A(9,4)=-9;A(9,8)=-2;A(9,9)=29;b=-15;27;-23;0;-20;12;-7;7;10;x0=0;0;0;0;0;0;0;0;0;e=1e-6;>> for i=1:99 w=0.95+i/50;N=8; y,r,n=sor(A,b,x0,w,1e-15,N); h(i)=r;end

34、s,j=min(h);w=0.95+j/50得：w=1.2100，再輸入：x,r,n=sor(A,b,x0,w,e,100)得：x = -0.2892 0.3454 -0.7128 -0.2206 -0.4304 0.1543 -0.0578 0.2011 0.2902r = 4.5333e-007n =12再輸入：x,r,n=sor(A,b,x0,w,e,12)得：x = -0.2892 0.3454 -0.7128 -0.2206 -0.4304 0.1543 -0.0578 0.2011 0.2902r = 4.5333e-007n = 12二、第四章練習(xí)題說(shuō)明：解題所需函數(shù)均提前保存為

35、m文件，方便直接調(diào)用，以下解題過(guò)程可能不包括保存m文件過(guò)程。練習(xí)題1：分別用不動(dòng)點(diǎn)迭代法和牛頓迭代法求解方程其中初值，計(jì)算精度為。解：（1）使用進(jìn)行迭代f=inline('-0.9*x2+1.7*x+2.5','x');df=inline('-1.8*x+1.7','x');x0=5;e=1e-6;n1=0;x1=17/9+25/9/x0;n1=n1+1;while (norm(x1-x0)>=e)&(n1<=1000) x0=x1; disp(n1,x1); x1=17/9+25/9/x0; n1=n1+1;

36、endx_b=x1x0=5;e=1e-6;n=0;x1=x0-feval(f,x0)/feval(df,x0);n=n+1;while (norm(x1-x0)>=e)&(n<=1000) x0=x1;x1=x0-feval(f,x0)/feval(df,x0);n=n+1;end%輸出fprintf('不動(dòng)點(diǎn)迭代次數(shù)為n=%g',n1);fprintf('結(jié)果為，x=%gn',x_b);fprintf('牛頓迭代迭代次數(shù)為n=%g，結(jié)果為',n);fprintf('x=%gn',x1);得到：不動(dòng)點(diǎn)迭代次數(shù)為

37、n=15，結(jié)果為x=2.8601牛頓迭代迭代次數(shù)為n=5，結(jié)果為x=2.8601練習(xí)題2：分別用不動(dòng)點(diǎn)迭代法、牛頓迭代法和逆Broyden秩1方法求解方程組并對(duì)結(jié)果進(jìn)行比較。其中逆Broyden秩1方法的初始矩陣分別取I和，取初值，計(jì)算精度。解：(1) 不動(dòng)點(diǎn)迭代法解方程組這個(gè)不會(huì)（2）首先，建立如下內(nèi)容的m文件：function f=F(x) f(1)=12*x(1)-x(2)2-4*x(3)-7;f(2)=x(1)2+10*x(2)-x(3)-11;f(3)=x(2)3+10*x(3)-8;f=f(1) f(2) f(3);接著，建立如下內(nèi)容的m文件：function df=DF(x) d

38、f=12,-2*x(2),4;2*x(1),10,-1;0,3*x(2)2,10;最后，建立如下內(nèi)容的m文件（文件名為NEWTON）：clear; clc x=1,1,1' f=F(x); df=DF(x); fprintf('%d %.7f %.7fn',0,x(1),x(2),x(3);N=4; for i=1:N y=dff' x=x-y; f=F(x); df=DF(x); fprintf('%d %.7f %.7fn',i,x(1),x(2),x(3); if norm(y)<0.00000001 break; else ende

39、nd此時(shí)，在命令框中輸入NEWTON，可得結(jié)果：0 1.0000000 1.0000000 1 1 1.1123077 1.0461538 6.861538e-001 2 0.9151577 1.0873339 6.719843e-001 3 0.9054088 1.0852514 6.721837e-001 4 0.9055333 1.0852203 6.721933e-001（3）首先，建立名為fun423的m文件：function y=fun423(x)y(1)=12*x(1)-x(2)2-4*x(3)-7;y(2)=x(1)2+10*x(2)-x(3)-11;y(3)=x(2)3+10

40、*x(3)-8;y=y(1);y(2);y(3);再建立名為fun423d的m文件：function dy=fun423d(x)dy(1)=12,-2*x(2),-4;dy(2)=2*x(1),10,-1;dy(3)=0,3*x(2)2,10;dy=dy(1) dy(2) dy(3);接著建立名為ex423的m文件：w='n_Broyden I' x0=1;1;1'e=0.00000001;b0=1,0,0;0,1,0;0,0,1;y,r=nbroyden('fun423',b0,x0,e); w='n_Broyden A0' x0=1;

41、1;1'e=0.00000001;b0=inv(feval('fun423d',x0)' y,r=nbroyden('fun423',b0,x0,e); w='Newton' x0=1;1;1;e=0.00000001;y,r=newtpro1('fun423','fun423d',x0,e);然后在命令窗口輸入：ex423，程序報(bào)錯(cuò)：? Error using => minusMatrix dimensions must agree.Error in => nbroyden at 7x

42、1=x0-b0*feval(f,x0);Error in => ex423 at 6y,r=nbroyden('fun423',b0,x0,e);暫無(wú)力調(diào)試。練習(xí)題3：理想與非理想氣體定律考慮固定容器中一定量的氣體的壓強(qiáng)、體積和溫度之間，有如下定律：理想氣體定律其中p為壓強(qiáng)，V為體積，n為摩爾數(shù)，R為普適氣體常數(shù)，T為絕對(duì)溫度；非理想氣體的van der Waals方程其中稱為摩爾體積，a, b是與具體氣體有關(guān)的經(jīng)驗(yàn)常數(shù)。在某化工設(shè)計(jì)中，需要精確估計(jì)二氧化碳和氧氣在不同溫度和壓強(qiáng)下的摩爾體積，以選擇適當(dāng)?shù)娜萜?，其中各參?shù)分別為R=0.082054 L atm/(mol K)；二氧化碳：a=3.592，b=0.04267；氧 氣：a=1.360，b=0.03183；設(shè)計(jì)壓強(qiáng)分別為1，10，50，100 atm；溫度分別為400，60