邊界元法求解二維彈性問題

數據的輸入格式如下：

4,0.3,200.0,1  //分別爲：結點數、泊松比、楊氏模量和問題類型
0.0,1.0//依次爲四個結點的座標
1.0,1.0
1.0,0.0
0.0,0.0
1,2//單元編號
2,3
3,4
4,1
12,0.0,0.0//邊界條件
12,0.0,1.0
12,-1.0,0.0
12,0.0,-1.0

具體實現代碼如下，不敢保證絕對正確，目前只能算是Beta版本，最後輸出A和B矩陣爲AX=B方程的兩個矩陣，求解X可用Matlab用inv(A)*B得到。

#include <iostream>
#include <math.h>
#include <fstream>
using namespace std;
const double pi=3.14159265;
class DanYuan
{
public:
	int StartID,EndID;
};

class Condition
{
public:
	int Type;
	double GivenTan,GivenNormal;
};

int Number;
double *X,*Y;
DanYuan *Element;
Condition *BianJie;
int PlaneType;
double v,E,G;
double **A,*B;

void ReadData()
{
	char c;
	int i;
	ifstream read;
	read.open("Data.txt",ios::in);
	read>>Number>>c>>v>>c>>E>>c>>PlaneType;
	if (PlaneType==1)
	{
		v=v/(1+v);
	}
	G=E/(2+2*v);
	X=new double[Number+2];
	Y=new double[Number+2];
	Element=new DanYuan[Number+1];
	BianJie=new Condition[Number+1];
	A=new double *[2*Number+1];
	for (i=1;i<=2*Number;i++)
	{
		A[i]=new double[2*Number+1];
	}
	B=new double[2*Number+1];
	for (i=1;i<=Number;i++)
	{
		read>>X[i]>>c>>Y[i];
	}
	X[Number+1]=X[1];
	Y[Number+1]=Y[1];
	for (i=1;i<=Number;i++)
	{
		read>>Element[i].StartID>>c>>Element[i].EndID;
	}
	for (i=1;i<=Number;i++)
	{
		read>>BianJie[i].Type>>c>>BianJie[i].GivenTan>>c
			>>BianJie[i].GivenNormal;
	}
}

void ComputeGH(int i,int j,double Uknown[][3],double Known[][3])
{
	double ri,r,r1,r2,dist,n1,n2;
	double h11,h12,h21,h22,g11,g12,g21,g22;
	double xc[5],yc[5],ax,ay,bx,by;
	double guass[5],w[5];
	h11=h12=h21=h22=g11=g12=g21=g22=0.0;

	ax=X[Element[j].EndID]-X[Element[j].StartID];
	ay=Y[Element[j].EndID]-Y[Element[j].StartID];
	bx=X[Element[j].StartID]+X[Element[j].EndID];
	by=Y[Element[j].StartID]+Y[Element[j].EndID];

	w[1]=0.34785485;
	w[2]=w[1];
	w[3]=0.65214515;
	w[4]=w[3];

	guass[1]=0.86113631;
	guass[2]=-guass[1];
	guass[3]=0.33998104;
	guass[4]=-guass[3];

	if (i==j)
	{
		r=sqrt(ax*ax+ay*ay);
		r1=abs(ax/r);
		r2=abs(ay/r);
		double temp=r/(8*pi*G*(1-v));			
		
		h11=0.5;h12=0.0;
		h12=0.0;h22=0.5;

		g11=temp*((3-4*v)*(1-log(r/2))+r1*r1);
		g12=temp*r2*r1;
		g21=g12;
		g22=temp*((3-4*v)*(1-log(r/2))+r2*r2);

		switch(BianJie[j].Type)
		{
		case 12:
			Uknown[1][1]=g11;
			Uknown[1][2]=g12;
			Uknown[2][1]=g21;
			Uknown[2][2]=g22;

			Known[1][1]=h11;
			Known[1][2]=h12;
			Known[2][1]=h21;
			Known[2][2]=h22;
			break;
		}
	}
	else
	{
		double temp=1.0/(8.0*pi*G*(1-v));
		double a,b,c;
		a=ay;
		b=-ax;
		c=Y[Element[j].StartID]*ax-X[Element[j].StartID]*ay;
		r=sqrt(ax*ax+ay*ay);
		dist=abs(a*X[i]+b*Y[i]+c)/(sqrt(a*a+b*b));
		n1=abs(ax/r);
		n2=abs(ay/r);
		for (int k=1;k<=4;k++)
		{
			xc[k]=-ax*guass[k]+bx;
			yc[k]=-ay*guass[k]+by;
			ri=sqrt(pow(xc[k]-X[i],2)+pow(yc[k]-Y[i],2));
			r1=abs((xc[k]-X[i])/ri);
			r2=abs((yc[k]-Y[i])/ri);

			h11-=temp*2*dist*((1-2*v)+2*r1*r1)*(r/2)*w[k]/ri;
			h12-=temp*2*(2*dist*r1*r2-(1-2*v)*(r1*n2-r2*n1))*(r/2)*w[k]/ri;
			h21-=temp*2*(2*dist*r1*r2-(1-2*v)*(r2*n1-r1*n2))*(r/2)*w[k]/ri;
			h22-=temp*2*dist*((1-2*v)+2*r2*r2)*(r/2)*w[k]/ri;

			g11+=temp*((3-4*v)*log(1/ri)+r1*r1)*(r/2)*w[k];
			g12+=temp*r1*r2*(r/2)*w[k];
			g21+=temp*r1*r2*(r/2)*w[k];
			g22+=temp*((3-4*v)*log(1/ri)+r2*r2)*(r/2)*w[k];
		}
		switch(BianJie[j].Type)
		{
		case 12:
			Uknown[1][1]=g11;
			Uknown[1][2]=g12;
			Uknown[2][1]=g21;
			Uknown[2][2]=g22;

			Known[1][1]=h11;
			Known[1][2]=h12;
			Known[2][1]=h21;
			Known[2][2]=h22;
			break;
		}
	}
}

void AssignA(double a[][3],int r,int c)
{
	int i,j;
	for (i=1;i<=2;i++)
	{
		for (j=1;j<=2;j++)
		{
			A[r+i-1][c+j-1]=a[i][j];
		}
	}
}

void CreateAXF()
{
	int i,j;
	double Uknown[3][3],Known[3][3];
	for (i=1;i<=Number;i++)
	{
		B[2*i-1]=0.0;
		B[2*i]=0.0;
		for (j=1;j<=Number;j++)
		{
			ComputeGH(i,j,Uknown,Known);
			AssignA(Uknown,2*i-1,2*j-1);
			B[2*i-1]+=Known[1][1]*BianJie[j].GivenTan+
				Known[1][2]*BianJie[j].GivenNormal;
			B[2*i]+=Known[2][1]*BianJie[j].GivenTan+
				Known[2][2]*BianJie[j].GivenNormal;
		}
	}
}

int main()
{
	ReadData();
	CreateAXF();
	int i,j;
	ofstream outa,outb;
	outa.open("A.txt",ios::out);
	outb.open("B.txt",ios::out);
	for (i=1;i<=2*Number;i++)
	{
		for (j=1;j<=2*Number;j++)
		{
			outa<<A[i][j]<<" ";
		}
		outa<<endl;
	}
	for (j=1;j<=2*Number;j++)
	{
		outb<<B[j]<<endl;
	}
	return 0;
}