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1 文本格式
using System;
namespace Legalsoft.Truffer
{
/// <summary>
/// 求解一组m维线性Volterra方程组
/// Solves a set of m linear Volterra equations of the second kind using the
/// extended trapezoidal rule.On input, t0 is the starting point of the
/// integration and h is the step size.g(k, t) is a user-supplied function or
/// functor that returns gk(t), while ak(k, l, t, s) is another user- supplied
/// function or functor that returns the (k, l) element of the matrix K(t, s). The
/// solution is returned in f[0..m - 1][0..n - 1], with the corresponding abscissas
/// in t[0..n - 1], where n-1 is the number of steps to be taken.The value of m is
/// determined from the row-dimension of the solution matrix f.
/// </summary>
public abstract class Volterra
{
public abstract double g(int k, double t);
public abstract double ak(int k, int l, double t1, double t2);
public void voltra(double t0, double h, double[] t, double[,] f)
{
int m = f.GetLength(0);
int n = f.GetLength(1);
double[] b = new double[m];
double[,] a = new double[m, m];
t[0] = t0;
for (int k = 0; k < m; k++)
{
f[k, 0] = g(k, t[0]);
}
for (int i = 1; i < n; i++)
{
t[i] = t[i - 1] + h;
for (int k = 0; k < m; k++)
{
double sum = g(k, t[i]);
for (int l = 0; l < m; l++)
{
sum += 0.5 * h * ak(k, l, t[i], t[0]) * f[l, 0];
for (int j = 1; j < i; j++)
{
sum += h * ak(k, l, t[i], t[j]) * f[l, j];
}
if (k == l)
{
a[k, l] = 1.0 - 0.5 * h * ak(k, l, t[i], t[i]);
}
else
{
a[k, l] = -0.5 * h * ak(k, l, t[i], t[i]);
}
}
b[k] = sum;
}
LUdcmp alu = new LUdcmp(a);
alu.solve( b, b);
for (int k = 0; k < m; k++)
{
f[k, i] = b[k];
}
}
}
}
}
2 代码格式
using System;namespace Legalsoft.Truffer
{/// <summary>/// 求解一组m维线性Volterra方程组/// Solves a set of m linear Volterra equations of the second kind using the/// extended trapezoidal rule.On input, t0 is the starting point of the/// integration and h is the step size.g(k, t) is a user-supplied function or/// functor that returns gk(t), while ak(k, l, t, s) is another user- supplied/// function or functor that returns the (k, l) element of the matrix K(t, s). The/// solution is returned in f[0..m - 1][0..n - 1], with the corresponding abscissas/// in t[0..n - 1], where n-1 is the number of steps to be taken.The value of m is/// determined from the row-dimension of the solution matrix f./// </summary>public abstract class Volterra{public abstract double g(int k, double t);public abstract double ak(int k, int l, double t1, double t2);public void voltra(double t0, double h, double[] t, double[,] f){int m = f.GetLength(0);int n = f.GetLength(1);double[] b = new double[m];double[,] a = new double[m, m];t[0] = t0;for (int k = 0; k < m; k++){f[k, 0] = g(k, t[0]);}for (int i = 1; i < n; i++){t[i] = t[i - 1] + h;for (int k = 0; k < m; k++){double sum = g(k, t[i]);for (int l = 0; l < m; l++){sum += 0.5 * h * ak(k, l, t[i], t[0]) * f[l, 0];for (int j = 1; j < i; j++){sum += h * ak(k, l, t[i], t[j]) * f[l, j];}if (k == l){a[k, l] = 1.0 - 0.5 * h * ak(k, l, t[i], t[i]);}else{a[k, l] = -0.5 * h * ak(k, l, t[i], t[i]);}}b[k] = sum;}LUdcmp alu = new LUdcmp(a);alu.solve( b, b);for (int k = 0; k < m; k++){f[k, i] = b[k];}}}}
}
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