微分方程数值解课程设计.docx
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微分方程数值解课程设计.docx
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微分方程数值解课程设计
利用五点差分格式求解拉普拉斯方程的混合边值问题
要求:
给出问题,计算格式,格式的矩阵形式,线性代数方程组求解的数值方法,稳定性,格式的截断误差;程序流程图;沿y=1画出计算结果与精确解的比较图形以及误差图形(在一个图上);分析两个迭代法的收敛速度。
E=
-1.10000.500000000000
0.5000-2.10000.50000000000
00.5000-2.10000.5000000000
000.5000-2.10000.500000000
0000.5000-2.10000.50000000
00000.5000-2.10000.5000000
000000.5000-2.10000.500000
0000000.5000-2.10000.50000
00000000.5000-2.10000.5000
000000000.5000-2.1000
h=0.1
Em=
-2.10001.000000000000
1.0000-4.00001.00000000000
01.0000-4.00001.0000000000
001.0000-4.00001.000000000
0001.0000-4.00001.00000000
00001.0000-4.00001.0000000
000001.0000-4.00001.000000
0000001.0000-4.00001.00000
00000001.0000-4.00001.0000
000000001.0000-4.0000
K=
0.5000000000000
01.000000000000
001.00000000000
0001.0000000000
00001.000000000
000001.00000000
0000001.0000000
00000001.000000
000000001.00000
0000000001.0000
A=[EKzeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)
KEmKzeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)
zeros(10)KEmKzeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)
zeros(10)zeros(10)KEmKzeros(10)zeros(10)zeros(10)zeros(10)zeros(10)
zeros(10)zeros(10)zeros(10)KEmKzeros(10)zeros(10)zeros(10)zeros(10)
zeros(10)zeros(10)zeros(10)zeros(10)KEmKzeros(10)zeros(10)zeros(10)
zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)KEmKzeros(10)zeros(10)
zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)KEmKzeros(10)
zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)KEmK
zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)zeros(10)KEm]
g=
0
-0.0100
-0.0400
-0.0900
-0.1600
-0.2500
-0.3600
-0.4900
-0.6400
-0.8100
0.0100
0
0
0
0
0
0
0
0
-9.9000
0.0400
0
0
0
0
0
0
0
0
-9.6000
0.0900
0
0
0
0
0
0
0
0
-9.1000
0.1600
0
0
0
0
0
0
0
0
-8.4000
0.2500
0
0
0
0
0
0
0
0
-7.5000
0.3600
0
0
0
0
0
0
0
0
-6.4000
0.4900
0
0
0
0
0
0
0
0
-5.1000
0.6400
0
0
0
0
0
0
0
0
-3.6000
5.8100
9.9000
9.6000
9.1000
8.4000
7.5000
6.4000
5.1000
3.6000
1.9000
U=zeros(100,1)
nm=100;
w=10^-2;
jacobi(A,h*g,U,nm,w)
迭代次数为
n=
44
方程组的解为
x=
0
0.0087
0.0350
0.0795
0.1437
0.2290
0.3368
0.4680
0.6228
0.8006
-0.0087
0
0.0260
0.0702
0.1338
0.2187
0.3261
0.4572
0.6121
0.7902
-0.0350
-0.0260
0
0.0439
0.1072
0.1915
0.2984
0.4289
0.5832
0.7607
-0.0795
-0.0702
-0.0439
-0.0000
0.0631
0.1469
0.2531
0.3826
0.5358
0.7120
-0.1437
-0.1338
-0.1072
-0.0631
-0.0000
0.0835
0.1891
0.3176
0.4693
0.6439
-0.2290
-0.2187
-0.1915
-0.1469
-0.0835
0
0.1051
0.2327
0.3831
0.5558
-0.3368
-0.3261
-0.2984
-0.2531
-0.1891
-0.1051
0
0.1271
0.2764
0.4476
-0.4680
-0.4572
-0.4289
-0.3826
-0.3176
-0.2327
-0.1271
0
0.1487
0.3189
-0.6228
-0.6121
-0.5832
-0.5358
-0.4693
-0.3831
-0.2764
-0.1487
0
0.1697
-0.8006
-0.7902
-0.7607
-0.7120
-0.6439
-0.5558
-0.4476
-0.3189
-0.1697
0
ans=
44
gaussseidel(A,h*g,U,nm,w)
迭代次数为
n=
29
方程组的解为
x=
-0.0045
0.0033
0.0282
0.0706
0.1309
0.2087
0.3020
0.4036
0.4908
0.4780
-0.0139
-0.0062
0.0185
0.0607
0.1209
0.1990
0.2941
0.4041
0.5271
0.6779
-0.0412
-0.0334
-0.0086
0.0339
0.0949
0.1747
0.2740
0.3935
0.5370
0.7170
-0.0875
-0.0795
-0.0542
-0.0111
0.0510
0.1328
0.2355
0.3607
0.5114
0.6932
-0.1537
-0.1455
-0.1197
-0.0758
-0.0126
0.0708
0.1756
0.3032
0.4556
0.6347
-0.2411
-0.2325
-0.2062
-0.1613
-0.0971
-0.0126
0.0933
0.2216
0.3736
0.5500
-0.3501
-0.3412
-0.3142
-0.2685
-0.2033
-0.1179
-0.0116
0.1165
0.2673
0.4417
-0.4809
-0.4718
-0.4441
-0.3974
-0.3312
-0.2451
-0.1386
-0.0114
0.1372
0.3094
-0.6333
-0.6239
-0.5954
-0.5477
-0.4804
-0.3934
-0.2867
-0.1608
-0.0164
0.1488
-0.8066
-0.7969
-0.7676
-0.7188
-0.6502
-0.5620
-0.4544
-0.3290
-0.1909
-0.0580
ans=
29
x=linspace(0,1,100)
u=x.^2-1
X=linspace(0,1,10)
Y=(-0.8066-0.7969-0.7676-0.7188-0.6502-0.5620-0.4544-0.3290-0.1909-0.0580)
plot(x,u)
>>holdon
>>plot(X,Y)
截断误差分析:
h=0.2
e=
-1.20000.5000000
0.5000-2.20000.500000
00.5000-2.20000.50000
000.5000-2.20000.5000
0000.5000-2.2000
k=
0.50000000
01.0000000
001.000000
0001.00000
00001.0000
em=
-2.20001.0000000
1.0000-4.00001.000000
01.0000-4.00001.00000
001.0000-4.00001.0000
0001.0000-4.0000
k=
0.50000000
01.0000000
001.000000
0001.00000
00001.0000
a=[ekzeros(5)zeros(5)zeros(5)
kemkzeros(5)zeros(5)
zeros(5)kemkzeros(5)
zeros(5)zeros(5)kemk
zeros(5)zeros(5)zeros(5)kem]
v=
0
-0.0080
-0.0320
-0.0720
-0.6280
0.0080
0
-0.9600
0.0320
0
-0.8400
0.0720
0
-0.6400
1.1280
0.9600
0.8400
0.6400
0
u=zeros(25,1)
nm=100;
w=10^-2;
jacobi(a,v,u,nm,w)
迭代次数为
n=15
方程组的解为
x=
-0.0141
0.0258
0.1454
0.3488
0.6330
-0.0597
-0.0214
0.1030
0.3061
0.5927
-0.2023
-0.1546
-0.0246
0.1854
0.4722
-0.4618
-0.3847
-0.2323
-0.0166
0.2732
-0.9327
-0.6955
-0.5145
-0.2929
-0.0049
x=linspace(0,1,100)
u=x.^2-1
X=linspace(0,1,5)
Y=[-0.9327-0.6955-0.5145-0.2929-0.0049]
plot(x,u)
holdon
plot(X,Y)
画图如下:
h=0.1的时候的解的一部分:
u1=
0
0.0350
0.1437
0.3368
0.6228
-0.0350
0
0.1072
0.2984
0.5832
-0.1437
-0.1072
-0.0000
0.1891
0.4693
-0.3368
-0.2984
-0.1891
0
0.2764
-0.6228
-0.5832
-0.4693
-0.2764
0
u2=
-0.0141
0.0258
0.1454
0.3488
0.6330
-0.0597
-0.0214
0.1030
0.3061
0.5927
-0.2023
-0.1546
-0.0246
0.1854
0.4722
-0.4618
-0.3847
-0.2323
-0.0166
0.2732
-0.9327
-0.6955
-0.5145
-0.2929
-0.0049
ur=[0
0.04
0.16
0.36
0.64
-0.04
0
0.12
0.32
0.6
-0.16
-0.12
0
0.2
0.48
-0.36
-0.32
-0.2
0
0.28
-0.64
-0.6
-0.48
-0.28
0]
ur-u1
ans=
0
0.0050
0.0163
0.0232
0.0172
-0.0050
0
0.0128
0.0216
0.0168
-0.0163
-0.0128
0
0.0109
0.0107
-0.0232
-0.0216
-0.0109
0
0.0036
-0.0172
-0.0168
-0.0107
-0.0036
0
ur-u2
ans=
0.0141
0.0142
0.0146
0.0112
0.0070
0.0197
0.0214
0.0170
0.0139
0.0073
0.0423
0.0346
0.0246
0.0146
0.0078
0.1018
0.0647
0.0323
0.0166
0.0068
0.2927
0.0955
0.0345
0.0129
0.0049
a./b
ans=
0
0.3521
1.1164
2.0714
2.4571
-0.2538
0
0.7529
1.5540
2.3014
-0.3853
-0.3699
0
0.7466
1.3718
-0.2279
-0.3338
-0.3375
0
0.5294
-0.0588
-0.1759
-0.3101
-0.2791
0
为了计算log2()将0去掉变为20个点将负数取绝对值
使ans=
0.3521
1.1164
2.0714
2.4571
0.2538
0.7529
1.5540
2.3014
0.3853
0.3699
0.7466
1.3718
0.2279
0.3338
0.3375
0.5294
0.0588
0.1759
0.3101
0.2791
>>log2(ans)
ans=
-1.5059
0.1589
1.0506
1.2970
-1.9782
-0.4095
0.6360
1.2025
-1.3759
-1.4348
-0.4216
0.4561
-2.1335
-1.5829
-1.5670
-0.9176
-4.0880
-2.5072
-1.6892
-1.8411
n=norm(ans,2)
n=7.4125
n/(20^0.5)
ans=1.6575
所以截断误差阶为2阶
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