如何使用龙格-库塔法求解牛顿冷却定律 (RK4)

Question

我需要帮助实现我的工作四阶龙格-库塔方法来解决牛顿冷却定律。由于时间 (t( 被引入此问题，我对给定条件的放置感到困惑。 以下是给出的内容：时间间隔开始 t = 0 到 t = 20(以秒为单位(，对象温度 = 300，环境温度 = 70，时间增量为 .1，比例常数 = 0.19

public class RungeKutta {
public static double NewtonsCoolingLaw(double objectTemp,double  ambientTemp)
     {
         double k = 0.19;       
         return -k * (objectTemp - ambientTemp);
     }
public static void main(String[] args) {
    double result = 0.0;  
    double  initialObjectTemp = 300.0, givenAmbientTemp = 70.0;
    double deltaX = (20.0 - 0)/10000;
    for(double t = 0.0; t <= 20.0; t += .1)
    {        
        double k1 = deltaX * NewtonsLaw(initialObjectTemp,givenAmbientTemp);
        double k2 = deltaX * NewtonsLaw(initialObjectTemp + (deltaX/2.0),givenAmbientTemp + (k1/2.0));
        double k3 = deltaX * NewtonsLaw(initialObjectTemp + (deltaX/2.0), givenAmbientTemp + (k2/2.0));
        double k4 = deltaX * NewtonsLaw(initialObjectTemp + deltaX, givenAmbientTemp + k3);
        givenAmbientTemp = givenAmbientTemp + (1.0/6.0) * (k1 + (2.0 * k2) + (2.0 * k3) + k4);
        result = givenAmbientTemp;
    }       
    System.out.println("The approx. object temp after 20 seconds is: " + result);
}
}

波纹管是我求解常微分方程的RK4方法。在下面的代码中，我求解 ODE y' = y - x 以近似 y(1.005(，假设 y(1( = 10 和增量 x = 0.001

public class RungeKutta {
public static double functionXnYn(double x,double  y)
     {
         return y-x;
     }
public static void main(String[] args) {
    double deltaX = (1.005 - 0)/10000;
    double y = 10.0;
    double result = 0.0;  
    for(double x = 1.0; x <= 1.005; x = x + deltaX)
    {
        double k1 = deltaX * functionXnYn(x,y);
        double k2 = deltaX * functionXnYn(x + (deltaX/2.0),y + (k1/2.0));
        double k3 = deltaX * functionXnYn(x + (deltaX/2.0), y + (k2/2.0));
        double k4 = deltaX * functionXnYn(x + deltaX, y + k3);
        y = y + (1.0/6.0) * (k1 + (2.0 * k2) + (2.0 * k3) + k4);
        result = y;
    }       
    System.out.println("The value of y(1.005) is: " + result);
}
}

根据公式 T(t( = Ts + (T0 - Ts( * e^(-k*t(，我应该有一个近似值 75.1 来求解牛顿的 DE。Ts = 环境温度，T0 = 物体初始温度，t = 20(经过的秒数(，k = .19 比例常数

Answer 1

我猜(但不是很难(您尝试解决的 ODE 是

dT(t)/dt = -k*(T(t)-T_amb)

如您所见，右侧并不直接取决于时间。由于您不尝试为系统编写代码，因此环境温度T_amb很可能是一个常数。因此，移动常量并使用一致的函数名称，并将 ODE 函数参数返回到格式time, state variable

public class RungeKutta {
    public static double CoolingLaw(double time, double objectTemp)
         {
             double k = 0.19, ambientTemp = 70.0;     
             return -k * (objectTemp - ambientTemp);
         }
    public static void main(String[] args) {
        double result = 0.0;  
        double  objectTemp = 300.0;
        double dt = 0.1
        for(double t = 0.0; t <= 20.0; t += dt)
        {        
            double k1 = dt * CoolingLaw(t, objectTemp);
            double k2 = dt * CoolingLaw(t + (dt/2.0), objectTemp + (k1/2.0));
            double k3 = dt * CoolingLaw(t + (dt/2.0), objectTemp + (k2/2.0));
            double k4 = dt * CoolingLaw(t + dt, objectTemp + k3);
            objectTemp = objectTemp + (1.0/6.0) * (k1 + (2.0 * k2) + (2.0 * k3) + k4);
            result = objectTemp;
        }       
        System.out.println("The approx. object temp after 20 seconds is: " + result);
    }

}