如何将一个硬编码的微分方程通过龙格库塔4
本文关键字:微分方程 龙格库 编码 一个 | 更新日期: 2023-09-27 18:26:12
我正在尝试实现Runge-Kutta,例如问题dy/dt=y-t^2+1和dy/dt=t*y+t^3,我似乎无法得到我期望的输出。我把我的程序分成了几个班,试着单独看一下作品。我认为我的主要错误来自于试图使用委托将一个方法作为变量通过Runge-Kutta过程。
方程式类别:
namespace RK4
{
public class Eqn
{
double t;
double y;
double dt;
double b;
public Eqn(double t, double y, double dt, double b)
{
this.t = t;
this.y = y;
this.dt = dt;
this.b = b;
}
public void Run1()
{
double temp;
int step = 1;
RK4 n = new RK4();
while (t < b)
{
temp = n.Runge(t, y, dt, FN1);
y = temp;
Console.WriteLine("At step number {0}, t: {1}, y: {2}", step, t, y);
t = t + dt;
step++;
}
}
public void Run2()
{
int step = 1;
RK4 m = new RK4();
while (t < b)
{
y = m.Runge(t, y, dt, FN2);
Console.WriteLine("At step number {0}, t: {1}, y: {2}", step, t, y);
t = t + dt;
step++;
}
}
public static double FN1(double t, double y)
{
double x = y - Math.Pow(t, 2) + 1;
return x;
}
public static double FN2(double t, double y)
{
double x = t * y + Math.Pow(t, 3);
return x;
}
}
}
然后龙格库塔4级:
namespace RK4
{
class RK4
{
public delegate double Calc(double t, double y);
public double Runge(double t, double y, double dt, Calc yp)
{
double k1 = dt * yp(t, y);
double k2 = dt * yp(t + 0.5 * dt, y + k1 * 0.5 * dt);
double k3 = dt * yp(t + 0.5 * dt, y + k2 * 0.5 * dt);
double k4 = dt * yp(t + dt, y + k3 * dt);
return (y + (1 / 6) * (k1 + 2 * k2 + 2 * k3 + k4));
}
}
}
And my Program Class:
namespace RK4
{
class Program
{
static void Main(string[] args)
{
RunProgram();
}
public static void RunProgram()
{
Console.WriteLine("*******************************************************************************");
Console.WriteLine("************************** Fourth Order Runge-Kutta ***************************");
Console.WriteLine("*******************************************************************************");
Console.WriteLine("'nWould you like to implement the fourth-order Runge-Kutta on:");
string Fn1 = "y' = y - t^2 + 1";
string Fn2 = "y' = t * y + t^3";
Console.WriteLine("1) {0}", Fn1);
Console.WriteLine("2) {0}", Fn2);
Console.WriteLine("Please enter 1 or 2");
switch (Int32.Parse(Console.ReadLine()))
{
case 1:
Console.WriteLine("'nPlease enter beginning of the interval (a):");
double a = Double.Parse(Console.ReadLine());
Console.WriteLine("Please enter end of the interval (b):");
double b = Double.Parse(Console.ReadLine());
Console.WriteLine("Please enter the step size (h) to be used:");
double h = Double.Parse(Console.ReadLine());
Console.WriteLine("Please enter the inital conditions to satisfy y({0}) = d",a);
Console.WriteLine("d = ");
double d = Double.Parse(Console.ReadLine());
Console.Clear();
Console.WriteLine("Using the interval [{0},{1}] and step size of {2} and the inital condition of y({3}) = {4}:", a, b, h, a, d);
Console.WriteLine("With equation: {0}", Fn1);
Eqn One = new Eqn(a, d, h, b);
One.Run1();
Console.WriteLine("Press enter to exit.");
Console.ReadLine();
Environment.Exit(1);
break;
case 2:
Console.WriteLine("'nPlease enter beginning of the interval (a):");
a = Double.Parse(Console.ReadLine());
Console.WriteLine("Please enter end of the interval (b):");
b = Double.Parse(Console.ReadLine());
Console.WriteLine("Please enter the step size (h) to be used:");
h = Double.Parse(Console.ReadLine());
Console.WriteLine("Please enter the inital conditions to satisfy y({0}) = d",a);
Console.WriteLine("d = ");
d = Double.Parse(Console.ReadLine());
Console.Clear();
Console.WriteLine("Using the interval [{0},{1}] and step size of {2} and the inital condition of y({3}) = {4}:", a, b, h, a, d);
Console.WriteLine("With equation: {0}", Fn1);
Eqn Two = new Eqn(a, d, h, b);
Two.Run2();
Console.WriteLine("Press enter to exit.");
Console.ReadLine();
Environment.Exit(1);
break;
default:
Console.WriteLine("Improper input, please press enter to exit.");
Console.ReadLine();
Environment.Exit(1);
break;
}
}
}
}
无论如何,这都不是一个优雅的编程,但我没有工作知识来知道我在这一点上做错了什么。从我所读到的内容来看,我认为RK4类中的委托将能够通过我的硬编码diff eq.
您在RK4实现中犯了一个经典错误:有两个变量来定位与dt
的乘法,您同时使用这两个变量。
要么是
k2 = dt*f(t+0.5*dt, y+0.5*k1)
或
k2 = f(t+0.5*dt, y+0.5*dt*k1)
并且类似地在算法的其他行中。