南京工业大学研究生工程数学应用B第一次作业

1、工程应用数学作业（一）专业：安全科学与工程 姓名：* 学号：652083700*一、第三章练习题说明：解题所需函数均提前保存为m文件，方便直接调用，以下解题过程不包括保存m文件过程。练习题1：设两点边值问题的精确解为现以h为步长划分区间为100等份，用差分近似代替微分，将微分方程离散化为线性方程组，代入初始条件后，得到如下的方程组问题其中，。(1) 分别用J迭代法，G-S迭代法和SOR迭代法求解，并与精确解进行比较；(2) 如果，再求解该问题解：（1）1）输入：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)进行jacobi迭代。2）输入：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，再输入：x,r,n=sor(A,b,x0,w,e,6900)完成sor算法。3）输入：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。输出结果如下表所示：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于是得到以下结论：达到相同精度Jacobi迭代的迭代次数为n=30006达到相同精度sediel迭代的迭代次数n=15202达到相同精度sor迭代的迭代次数n=6900sor迭代最佳松弛因子w=1由结果可见对于此题达到相同精度迭代次数sor迭代<G-S迭代<J迭代。（2）t=0.1,输出结果如下表所示：精确值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达到相同精度Jacobi迭代的迭代次数为n=11

19、156达到相同精度sediel迭代的迭代次数n=15064达到相同精度sor迭代的迭代次数n=5596sor迭代最佳松弛因子w=1（3）t=0.001,输出结果如下表所示：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于是得到以下结论：达到相同精度Jacobi迭代的迭代次数为n=184达到相同精度sediel迭代的迭代次数n=417达到相同精度sor迭代的迭代次数n=142sor迭代最佳松弛因子w=1练习题2：设是一个对称正定矩阵，分别是其最大和最小特征值，对于线性方程组建立迭代法求出的范围使迭代法收敛，并求出最优使迭代法的渐进收敛速度最大。解：水平有限，暂未解出。练习题3：对某电路的分析，可以归结为下面的线性方程组，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其余元素为零。要求：（1）用高斯列主元消去法求解该方程组；（2）用SOR方法迭代求解该方程组，误差，近似最佳松弛因子由试算法确定，设解：（1）首先输入：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，输入：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）输入：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，再输入：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再输入：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、第四章练习题说明：解题所需函数均提前保存为

35、m文件，方便直接调用，以下解题过程可能不包括保存m文件过程。练习题1：分别用不动点迭代法和牛顿迭代法求解方程其中初值，计算精度为。解：（1）使用进行迭代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%输出fprintf('不动点迭代次数为n=%g',n1);fprintf('结果为，x=%gn',x_b);fprintf('牛顿迭代迭代次数为n=%g，结果为',n);fprintf('x=%gn',x1);得到：不动点迭代次数为

37、n=15，结果为x=2.8601牛顿迭代迭代次数为n=5，结果为x=2.8601练习题2：分别用不动点迭代法、牛顿迭代法和逆Broyden秩1方法求解方程组并对结果进行比较。其中逆Broyden秩1方法的初始矩阵分别取I和，取初值，计算精度。解：(1) 不动点迭代法解方程组这个不会（2）首先，建立如下内容的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);接着，建立如下内容的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，建立如下内容的m文件（文件名为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此时，在命令框中输入NEWTON，可得结果：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，建立名为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);再建立名为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);接着建立名为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);然后在命令窗口输入：ex423，程序报错：? 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);暂无力调试。练习题3：理想与非理想气体定律考虑固定容器中一定量的气体的压强、体积和温度之间，有如下定律：理想气体定律其中p为压强，V为体积，n为摩尔数，R为普适气体常数，T为绝对温度；非理想气体的van der Waals方程其中称为摩尔体积，a, b是与具体气体有关的经验常数。在某化工设计中，需要精确估计二氧化碳和氧气在不同温度和压强下的摩尔体积，以选择适当的容器，其中各参数分别为R=0.082054 L atm/(mol K)；二氧化碳：a=3.592，b=0.04267；氧 气：a=1.360，b=0.03183；设计压强分别为1，10，50，100 atm；温度分别为400，60