java - 在 Java 中返回方程的问题

Question

我的函数调用遇到问题。我试图找到边界条件为 x(0) + x(1) 的两点边界值问题 x' = f(t,x) = x + 0.09 x 2 + cos(10 t) 的解 - 3.0 = 0 使用割线和三阶龙格-库塔方法，无论出于何种原因，我的方程方法在一行上给出了四个错误，这些错误是它不是一个陈述，缺少两个分号，并且缺少一个结束括号。

public class BoundaryValueProblem 
{
    public static double f(double t, double x) 
    {
        return x + 0.09x^2 + Math.cos(10t);
    }

    public static void findZero
    {
        double x4; 
        double x5 = .7; 
        double x6 = 1.0;
        int n = 1; 
        double fx; 
        double f1;
        double f2; 
        double Error;
        BoundaryValueProblem FZ = new BoundaryValueProblem();
        f1 = FZ.f(1.0, 1.0);
        f2 = FZ.f(1.0, 1.0);
        System.out.println("Secant Method");
        System.out.println("n: \t\t x1: \t\t x2: \t\t Error:");
        while(abs(x5 + x6 - 3.0) < 1e-5)
        {
            x4 = x6 - f2 * ((x6 - x5)/(f2 - f1));         
            fx = FZ.f(1.0, 1.0);
            x5 = x6;
            x6 = x4;
            f1 = f2;
            f2 = fx;         
            Error = x5 + x6 - 3.0;              
            System.out.println(n + "\t\t" + x5 + "\t\t" + x6 + "\t\t" + Error);
            n++;
        }
        System.out.println();
    }

    public void rkm(double x0, double t0, double h)
    {
        double x1, x2, x3;
        int i=0;
        double a1 = 0.5;
        double a2 = 0.25;
        double c0 = 2.0/3.0;
        double c1 = 5.0/3.0;
        double c2 = -4.0/3.0;
        double b21 = -.25;
        double b10 = .5;
        double b20 = .5;   
        double stepsize = .025;
        System.out.println("Runge-Kutta Method:");
        System.out.println("i: \t\t h: \t\t t0: \t\t x0:");
        System.out.println(i + "\t\t " + h + "\t\t " + t0 + "\t\t " + x0 );  
        for(i = 0; i < 40; i++)
        {            
            x1 = x0 + h * b10 * f(t0, x0);
            x2 = x0 + h * (b20 * f(t0, x0) + b21 * f(t0 + a1 * h, x1));
            x3 = x0 + h * (c0 * f(t0, x0) + c1 * f(t0 + a1 * h, x1) + c2 * f(t0 + a2 * h, x2));           
            t0 = t0 + stepsize;  
            x0 = x3;
            System.out.println(i + "\t\t " + h + "\t\t " + t0 + "\t\t " + x0 );   
        }
        System.out.println();
    }

    public static void main(String[] args) 
    {
        BoundaryValueProblem FZ1 = new BoundaryValueProblem();
        FZ1.findZero();
        BoundaryValueProblem RKM1 = new BoundaryValueProblem();
        RKM1.rkm(1.0, 0.0, 0.0);
    }     
}

我已经让它可以处理更简单的方程，但这个对我不起作用。我的其余代码大部分都是正确的(我相信)，但我不认为这是错误的原因，因为我将其注释掉并且错误仍然存​​在，因此我不关注这一点。任何建议或帮助将不胜感激。

Answer 1

下面是固定代码。请记住，与 Octave 或 Matlab 不同，java 需要乘法运算符 (*) 并且不支持 pow 运算符 (^) - 您需要使用 Math.pow () 代替。

public class BoundaryValueProblem {
    public static double f(double t, double x) {
        return x + 0.09 * x * x + Math.cos(10 * t);
    }

    public static void findZero() {
        double x4;
        double x5 = .7;
        double x6 = 1.0;
        int n = 1;
        double fx;
        double f1;
        double f2;
        double Error;
        f1 = f(1.0, 1.0);
        f2 = f(1.0, 1.0);
        System.out.println("Secant Method");
        System.out.println("n: \t\t x1: \t\t x2: \t\t Error:");
        while (Math.abs(x5 + x6 - 3.0) < 1e-5) {
            x4 = x6 - f2 * ((x6 - x5) / (f2 - f1));
            fx = f(1.0, 1.0);
            x5 = x6;
            x6 = x4;
            f1 = f2;
            f2 = fx;
            Error = x5 + x6 - 3.0;
            System.out.println(n + "\t\t" + x5 + "\t\t" + x6 + "\t\t" + Error);
            n++;
        }
        System.out.println();
    }

    public void rkm(double x0, double t0, double h) {
        double x1, x2, x3;
        int i = 0;
        double a1 = 0.5;
        double a2 = 0.25;
        double c0 = 2.0 / 3.0;
        double c1 = 5.0 / 3.0;
        double c2 = -4.0 / 3.0;
        double b21 = -.25;
        double b10 = .5;
        double b20 = .5;
        double stepsize = .025;
        System.out.println("Runge-Kutta Method:");
        System.out.println("i: \t\t h: \t\t t0: \t\t x0:");
        System.out.println(i + "\t\t " + h + "\t\t " + t0 + "\t\t " + x0);
        for (i = 0; i < 40; i++) {
            x1 = x0 + h * b10 * f(t0, x0);
            x2 = x0 + h * (b20 * f(t0, x0) + b21 * f(t0 + a1 * h, x1));
            x3 = x0 + h * (c0 * f(t0, x0) + c1 * f(t0 + a1 * h, x1) + c2 * f(t0 + a2 * h, x2));
            t0 = t0 + stepsize;
            x0 = x3;
            System.out.println(i + "\t\t " + h + "\t\t " + t0 + "\t\t " + x0);
        }
        System.out.println();
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        BoundaryValueProblem FZ1 = new BoundaryValueProblem();
        FZ1.findZero();
        BoundaryValueProblem RKM1 = new BoundaryValueProblem();
        RKM1.rkm(1.0, 0.0, 0.0);
    }
}