热传导方程的数值模拟

 1 public class HeatConduction {
 2     public double[][] conduction(double x, double t,int i,int j) {//i,j表示x,t分别分割的离散点
 3         double [][] XT = new double[i][j];
 4         double dx=x/(i-1),dt=t/(j-1);
 5         double c=dt/(dx*dx);
 6         for (int k = 0; k < i; k++) {//u(x,0)的初值
 7             XT[k][0]=k*dx*(x-k*dx)/x/x;
 8         }
 9         for (int k = 0; k < j; k++) {//u(0,t)的初值
10             XT[0][k]=0;
11         }
12         for (int k = 0; k < j; k++) {//u(x,t)的初值
13             XT[i-1][k]=0;
14         }
15         for (int k = 1; k < j; k++) {
16             for (int l = 1; l < i-1; l++) {
17                 XT[l][k]=c*(XT[l-1][k-1]+XT[l+1][k-1])+(1-2*c)*XT[l][k-1];
18             }
19         }
20         return XT;
21     }
22     
23     public static void main(String[] args) {
24         HeatConduction heatConduction=new HeatConduction();
25         double[][] XT = heatConduction.conduction(1.0, 1.0, 10, 500);
26         for (int j = 0; j < XT[0].length; j++) {
27             for (int i = 0; i < XT.length; i++) {
28                 if (i==XT.length-1) {
29                     System.out.println(XT[i][j]);
30                 }else {
31                     System.out.print(XT[i][j]+" ");
32                 }    
33             }
34         }
35     }
36 }