求偏微分方程代码！！

grapepaul 2005-01-14 11:29:03

如体，多谢！！

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redex 2005-01-14

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看看：
http://mathforum.org/library/toc.html
http://www.math.niu.edu/~rusin/known-math/index/35-XX.html#COMP

laocai8 2005-01-14

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运行结果：
sor Solution:

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 -0.02 -0.04 -0.06 -0.08 -0.09 -0.08 -0.06 -0.04 -0.02 0.00
0.00 -0.04 -0.09 -0.13 -0.17 -0.19 -0.17 -0.13 -0.09 -0.04 0.00
0.00 -0.06 -0.13 -0.20 -0.28 -0.32 -0.28 -0.20 -0.13 -0.06 0.00
0.00 -0.08 -0.17 -0.28 -0.41 -0.55 -0.41 -0.28 -0.17 -0.08 0.00
0.00 -0.09 -0.19 -0.32 -0.55 -1.05 -0.55 -0.32 -0.19 -0.09 0.00
0.00 -0.08 -0.17 -0.28 -0.41 -0.55 -0.41 -0.28 -0.17 -0.08 0.00
0.00 -0.06 -0.13 -0.20 -0.28 -0.32 -0.28 -0.20 -0.13 -0.06 0.00
0.00 -0.04 -0.09 -0.13 -0.17 -0.19 -0.17 -0.13 -0.09 -0.04 0.00
0.00 -0.02 -0.04 -0.06 -0.08 -0.09 -0.08 -0.06 -0.04 -0.02 0.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Test that sulotion satisfies Difference Eqns:

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00
0.00 0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00 0.00
0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00
0.00 0.00 -0.00 0.00 2.00 0.00 -0.00 0.00 0.00
0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00
0.00 0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00 0.00
0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00 -0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

grapepaul 2005-01-14

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多谢！

laocai8 2005-01-14

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你是搞什么的，我是作BI的，你呢？

laocai8 2005-01-14

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例子：
import java.text.*;
public class d16r1
{
public static void main (String[] args)
{
//program d16r1
//driver for routine sor
int jmax, i, j, midl;
double pi, rjac, aaa;
double a[][] = new double[12][12];
double b[][] = new double[12][12];
double c[][] = new double[12][12];
double d[][] = new double[12][12];
double e[][] = new double[12][12];
double f[][] = new double[12][12];
double u[][] = new double[12][12];
d16r1F g = new d16r1F();
DecimalFormat form = new DecimalFormat("0.00");
jmax = 11;
pi = 3.1415926;
for (i = 1; i <= jmax; i++)
{
for (j = 1; j <= jmax; j++)
{
a[i][j] = 1.0;
b[i][j] = 1.0;
c[i][j] = 1.0;
d[i][j] = 1.0;
e[i][j] = -4.0;
f[i][j] = 0.0;
u[i][j] = 0.0;
}
}
midl = jmax / 2 + 1;
f[midl][midl] = 2.0;
rjac = Math.cos(pi / jmax);
g.sor(a, b, c, d, e, f, u, jmax, rjac);
System.out.println();
System.out.println("sor Solution:");
System.out.println();
for (i = 1; i <= jmax; i++)
{
for (j = 1; j <= jmax; j++)
System.out.print(form.format(u[i][j]) + " ");
System.out.println();
}
System.out.println();
System.out.println("Test that sulotion satisfies Difference Eqns:");
System.out.println();
for (i = 2; i <= jmax - 1; i++)
{
for (j = 2; j <= jmax - 1; j++)
{
aaa = u[i + 1][j] + u[i - 1][j] + u[i][j + 1] + u[i][j - 1];
f[i][j] = aaa - 4.0 * u[i][j];
}
for (j = 2; j <= jmax - 1; j++)
System.out.print(form.format(f[i][j]) + " ");
System.out.println();
}
}
}

laocai8 2005-01-14

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方法是：解边值问题的松弛法
方法：
public class d16r1F
{
void sor(double a[][], double b[][], double c[][],
double d[][], double e[][], double f[][],
double u[][], int jmax, double rjac)
{
int n, j, l;
double maxits, eps, zero, half, qtr, one, anormf, omega, aaa, bbb;
double resid, anorm;
maxits = 1000;
eps = 0.00001;
zero = 0.0;
half = 0.5;
qtr = 0.25;
one = 1.0;
anormf = zero;
for (j = 2; j <= jmax - 1; j++)
for (l = 2; l <= jmax - 1; l++)
anormf = anormf + Math.abs(f[j][l]);
omega = one;
for (n = 1; n <= maxits; n++)
{
anorm = zero;
for (j = 2; j <= jmax - 1; j++)
{
for (l = 2; l <= jmax - 1; l++)
{
if (((j + l) % 2) ==( n % 2))
{
aaa = a[j][l] * u[j + 1][l] + b[j][l] * u[j - 1][l];
bbb = c[j][l] * u[j][l + 1] + d[j][l] * u[j][l - 1];
resid = aaa + bbb + e[j][l] * u[j][l] - f[j][l];
anorm = anorm + Math.abs(resid);
u[j][l] = u[j][l] - omega * resid / e[j][l];
}
}
}
if (n == 1 )
omega = one / (one - half * rjac * rjac);
else
omega = one / (one - qtr * rjac * rjac* omega);
if ((n > 1) && (anorm < eps * anormf) ) return;
}
System.out.println(" maxits exceeded");
System.exit(1);;
}
}

grapepaul 2005-01-14