Question

我一直在使用势流方程

pSI：0 = 10 * y + 23.8732414638 *（pi + atan（abs（（x-12）/ x）））+ 23.8732414638 *（pi - atan（abs（（y + 12）/ x））） -234.882642242

V：0 =（（10 + 23.8732414638 * x /（x * 2 +（y - 12） * 2）+ 23.8732414638 * x /（x * 2 +（y + 12） * 2））** 2 +（23.8732414638 *（y - 12）/（x * 2 +（y - 12） * 2）+ 23.8732414638 *（y + 12）/（x * 2 +（y + 12） * 2））* 2） * 0.5-8

使用下一个代码来解决它们：

# Prototype of N-R for a system of two non-linear equations
# evaluating  functions of two variables
from math import *
eq1 = raw_input('Enter the equation 1: ')
eq2 = raw_input('Enter the equation 2: ')
x0 = float(input('Enter x: '))
y0 = float(input('Enter y: '))

def f(x,y):
    return eval(eq1)

def g(x,y):
    return eval(eq2)

x = x0
y = y0

for n in range(1, 8):

    a = (f(x + 1e-06, y) - f(x,y)) / 1e-06
    b = (f(x, y + 1e-06) - f(x,y)) / 1e-06
    c =  - f(x,y)
    d = (g(x + 1e-06, y) - g(x,y)) / 1e-06
    eE = (g(x, y + 1e-06) - g(x,y)) / 1e-06
    fF =  - g(x,y)

    print "f(x, y)= ", eq1
    print "g(x, y)= ", eq2
    print """x   y """
    print x, y
    print """a   b   c   d   e   f """
    print a, b, c, d, e, fF

    print """
    a * x + b * y = c
    d * x + e * y = f
    """

    print a," * x  +  ",b," * y  =  ",c
    print d," * x  +  ",eE," * y  =  ",fF

    _Sy = (c - a * fF / d) / (b - a * eE / d)
    _Sx = (fF / d) - (eE / d) * _Sy

    Ea_X = (_Sx ** 2 + _Sy ** 2)**0.5

    x = x + _Sx
    y = y + _Sy

    print "Sx = ", _Sx
    print "Sy = ", _Sy

    print "x = ", x
    print "y = ", y

    print "|X_1 - X_0| = ", Ea_X

并给我零分错误

Traceback (most recent call last):
  File "C:\Documents and Settings\Principal\Mis documentos\Finished_Non_lin_N_R.py", line 38, in <module>
    d = (g(x + 1e-06, y) - g(x,y)) / 1e-06
  File "C:\Documents and Settings\Principal\Mis documentos\Finished_Non_lin_N_R.py", line 27, in g
    return eval(eq2)
  File "<string>", line 1, in <module>
ZeroDivisionError: float division by zero

然后我尝试了一个我在一个已解决的问题中找到的那个：

from scipy.optimize import fsolve
from math import *

def equations(p):
    x, y = p
    return ( ((10 + 23.8732414638 * x / (x**2 + (y - 12)**2) + 23.8732414638 * x / (x**2 + (y + 12)**2))**2 + (23.8732414638 * (y -12) / (x**2 + (y - 12)**2) + 23.8732414638 * (y + 12) / (x**2 + (y + 12)**2))**2)**0.5-8, 10 * y + 23.8732414638 * (pi + atan(abs((x - 12) / x))) + 23.8732414638 * (pi - atan(abs((y + 12) / x)))-234.882642242)

x, y =  fsolve(equations, (0, 18))

print equations((x, y))

还有下一个问题：

<module2>:19: RuntimeWarning: divide by zero encountered in long_scalars
<module2>:19: RuntimeWarning: divide by zero encountered in double_scalars
(-1.3374190643844486e-11, -2.8308022592682391e-11)

这个故事的最后一部分是我有一些实际可行的代码但是太笨拙而且有点不准确，我想提出一些建议，以简化：

from math import *
U = 10
b = 15
h = 12

cF = 0.5 * U * b / pi

pSI0 = 234.882642242
VP = 12.0

X0 = 0
Y0 = 40

sX = 0.01
sY = 0.01

print cF

def FirstFor(A,B,C,D):
    for n in range(1,808):
        for m in range(1,4002):
            X = A - B*n
            Y = C - D*m

            pSI = U * Y + cF * (pi + atan(abs((Y - h) / X))) + cF * (pi - atan(abs((Y + h) / X)))
            Vaprox = ((10 + 23.8732414638 * X / (X**2 + (Y - 12)**2) + 23.8732414638 * X / (X**2 + (Y + 12)**2))**2 +(23.8732414638 * (Y - 12) / (X**2 + (Y - 12)**2) + 23.8732414638 * (Y + 12) / (X**2 + (Y + 12)**2))**2)**0.5

            EA1 = abs(pSI0 - pSI)
            EA2 = abs(VP - Vaprox)

            if EA1 < 0.05 and EA2 < 0.05:
                X1f = X
                Y1f = Y
                H3 = [X1f,Y1f]
                print n,m,X,Y,"GG son",EA1,EA2
                print H3

    for n in range(1,808):
        for m in range(1,4002):
            X = H3[0] - B*n*0.001
            Y = H3[1] - D*m*0.001

            pSI = U * Y + cF * (pi + atan(abs((Y - h) / X))) + cF * (pi - atan(abs((Y + h) / X)))
            Vaprox = ((10 + 23.8732414638 * X / (X**2 + (Y - 12)**2) + 23.8732414638 * X / (X**2 + (Y + 12)**2))**2 +(23.8732414638 * (Y - 12) / (X**2 + (Y - 12)**2) + 23.8732414638 * (Y + 12) / (X**2 + (Y + 12)**2))**2)**0.5

            EA1 = abs(pSI0 - pSI)
            EA2 = abs(VP - Vaprox)

            if EA1 < 0.0001 and EA2 < 0.0001:
                X2f = X
                Y2f = Y
                I3 = [X2f,Y2f]
                print n,m,X,Y,"GG son",EA1,EA2
                print I3


    for n in range(1,808):
        for m in range(1,4002):
            X = H3[0] + B*n*0.001
            Y = H3[1] + D*m*0.001

            pSI = U * Y + cF * (pi + atan(abs((Y - h) / X))) + cF * (pi - atan(abs((Y + h) / X)))
            Vaprox = ((10 + 23.8732414638 * X / (X**2 + (Y - 12)**2) + 23.8732414638 * X / (X**2 + (Y + 12)**2))**2 +(23.8732414638 * (Y - 12) / (X**2 + (Y - 12)**2) + 23.8732414638 * (Y + 12) / (X**2 + (Y + 12)**2))**2)**0.5

            EA1 = abs(pSI0 - pSI)
            EA2 = abs(VP - Vaprox)

            if EA1 < 0.0001 and EA2 < 0.0001:
                X2f = X
                Y2f = Y
                I3 = [X2f,Y2f]
                print n,m,X,Y,"GG son",EA1,EA2
                print I3


# starting with the FirstFor
FirstFor(X0,sX,Y0,sY)

print "\n This should be good"
raw_input()

求解2个非线性方程组

0 个答案: