我有一个类似下面的matlab格式的微分方程。
syms x y m g r l J
% x,y are variables, the others are constant
1: 0.5*m*(r^2*x^2+l^2*(Dx-Dy)^2+2*r*l*Dx*(Dx-Dy)*cos(y))+0.5*J*(Dx-Dy)^2=m*g*
(l*sin(x-y)-r*(1-cos(x)));
2: J*(D2x-D2y)+l^2*(D2x-D2y)-r*l*(Dx)^2*sin(y)+r*l*D2x*cos(y)-m*g*l*cos(x-y)=0,
3: x(0)=pi/2,y(0)=pi/2,
我试着解决最大化函数的问题。14个约束,15个变量。有些是线性的,有些则不是。它们都是方程(没有不等式)。
我试过使用'fsolve‘和’求解‘,使用拉格朗日(最终得到29个方程和30个变量)
我转到了fmincon。我在一个名为objectfun.m的文件中设置了一个带有目标函数的脚本:
function f = objectfun(x,I,rho)
% SWAPPING VARIABLE NAMES FOR READABILITY:
w = x(1);
t = x(2);
beta = x(3);
r = x(4);
% VALUE FUNCTION TO BE MINIMIZ
由于R对函数参数的求值,可以指定一组一致的输入参数,并自动计算其他参数。
考虑下面的函数,在化学中将浓度、质量、体积和摩尔重量联系起来,
concentration <- function(c = m / (M*V), m = c*M*V, V = m / (M*c), M = 417.84){
cat(c("c=", c*1e6, "micro.mol/L\n",
"m=", m*1e3, "mg\n",
"M=", M, "g/mol\n",
我有以下方程,并希望R为L求解。 有什么想法吗? Average = 370.4
m = 2
p = 0.2
n = 5
#L = ?
log10(Average) = 0.379933834 -0.107509315* m + 0.104445717 * p + 0.016517169 * n -0.025566689* L + 0.014393465 * m * p + 0.001601271 * m * n - 0.014250365 * n * L + 0.002523518 * m^2 + 0.237090759 * L^2
我需要解决一个非线性方程系统,如下所示:
def trySolveEquation(V, L):
#The equation to solve is:
#{ (V . Ct) ^ 2 = 1
#{ (L + u) . Ct = 0
#C and u are the unknowns, C is a vector and u is a scalar, Ct is a vector transposed from C.
#V is a vector with dimension equal to Ct, L is a square matri
我在下面的代码中定义了两个方程f1(b,bb) =0和f2(b,bb) =0。我试图找到满足这两个条件的(b,bb)集。 import sympy as sp
from sympy import symbols, simplify, factor
a, b, aa, bb, q, l, h = symbols('a b aa bb q l h')
pb = l*q+h*(1-q)
pb0 = (l*(1-l)*q+h*(1-h)*(1-q))/((1-l)*q+(1-h)*(1-q))
pb1 = (l**2*q+h**2*(1-q))/(l*q+h*(1-q))
a =