(我理解标题会让这听起来像是Sympy Can't differentiate wrt the variable的重复问题,但我相当肯定它是完全不同的。如果我弄错了,我会事先道歉)
我正在尝试使用哈密顿力学来解决双摆,但Sympy在使用其中一个衍生物时遇到了麻烦。代码如下,但更容易阅读iPython Notebook over at nbviewer.ipython.org
from __future__ import division
from sympy import *
init_session()
r_1,r_2,l_1,l_2, m_1, m_2, rdot_1, rdot_2,g = symbols("r_1 r_2 l_1 l_2 m_1 m_2 \dot{r_1} \dot{r_2} g")
phi_1, phi_2 = symbols("phi_1 phi_2", cls=Function)
phi_1 = phi_1(t)
phi_2 = phi_2(t)
r_1 = Matrix([l_1 * sin(phi_1), -l_1 * cos(phi_1)])
rdot_1 = r_1.diff(t)
r_2 = Matrix([r_1[0] + l_2 * sin(phi_2), r_1[1] -l_2 * cos(phi_2)])
rdot_2 = r_2.diff(t)
T = (1/2) * m_1 * rdot_1.T * rdot_1 + (1/2) * m_2 * rdot_2.T * rdot_2
T = T[0]
U = -g * ((m_1 * r_1[1]) + (m_2 * r_2[1]))
L = symbols("\mathcal{L}")
L = T - U
H = symbols("\mathcal{H}")
H = T + U
p_1, p_2 = symbols("p_1 p_2")
p_1 = L.diff(phi_1.diff(t))
p_2 = L.diff(phi_2.diff(t))
eq1_1 = p_1.diff(t) + H.diff(phi_1)
eq1_2 = p_2.diff(t) + H.diff(phi_2)
eq2_1 = phi_1.diff(t) - H.diff(p_1)
最后一行产生以下错误:
ValueError Traceback (most recent call last)
<ipython-input-96-83bfafadd08f> in <module>()
----> 1 eq2_1 = phi_1.diff(t) - H.diff(p_1)
2 eq2_2 = phi_2.diff(t) - H.diff(p_2)
/Library/Python/2.7/site-packages/sympy/core/expr.pyc in diff(self, *symbols, **assumptions)
2773 new_symbols = list(map(sympify, symbols)) # e.g. x, 2, y, z
2774 assumptions.setdefault("evaluate", True)
-> 2775 return Derivative(self, *new_symbols, **assumptions)
2776
2777 ###########################################################################
/Library/Python/2.7/site-packages/sympy/core/function.pyc in __new__(cls, expr, *variables, **assumptions)
1027 from sympy.utilities.misc import filldedent
1028 raise ValueError(filldedent('''
-> 1029 Can\'t differentiate wrt the variable: %s, %s''' % (v, count)))
1030
1031 if all_zero and not count == 0:
ValueError:
Can't differentiate wrt the variable:
1.0*l_1*(l_1*m_1*Derivative(phi_1(t), t) +
l_1*m_2*Derivative(phi_1(t), t) + l_2*m_2*cos(phi_1(t) -
phi_2(t))*Derivative(phi_2(t), t)), 1
表达式太复杂了吗?或者我误解了如何使用diff()?
答案 0 :(得分:1)
您是否无意中覆盖了p_1?
p_1, p_2 = symbols("p_1 p_2")
p_1 = L.diff(phi_1.diff(t))
你创建了一个符号,然后通过创建一个具有相同名称的Python变量来销毁它,所以当你试图区分wrt p_1时(正如Aaron指出的那样)区分表达式,而不是你创建的符号。 / p>
答案 1 :(得分:0)
这基本上是同样的问题。 diff
的第二个参数(或者作为方法调用的唯一参数)应该是单个符号,而不是表达式。在这种情况下,您的区分“变量”是一个完整的表达式,在数学上甚至没有意义。