问题描述
运行以下代码时,出现Derivative(Ksi(uix, uiy), uix))
和Derivative(Ksi(uix, uiy), uiy))
术语:
When the following code is run Derivative(Ksi(uix, uiy), uix))
and Derivative(Ksi(uix, uiy), uiy))
terms appear:
In [4]: dgN
Out[4]:
Matrix([
[-(x1x - x2x)*(-x1y + x2y)*((x1x - x2x)**2 + (-x1y + x2y)**2)**(-0.5)*Derivative(Ksi(uix, uiy), uix) + (-x1y + x2y)*(-(-x1x + x2x)*Derivative(Ksi(uix, uiy), uix) + 1)*((x1x - x2x)**2 + (-x1y + x2y)**2)**(-0.5)],
[-(-x1x + x2x)*(-x1y + x2y)*((x1x - x2x)**2 + (-x1y + x2y)**2)**(-0.5)*Derivative(Ksi(uix, uiy), uiy) + (x1x - x2x)*(-(-x1y + x2y)*Derivative(Ksi(uix, uiy), uiy) + 1)*((x1x - x2x)**2 + (-x1y + x2y)**2)**(-0.5)]])
我想用例如我知道的某个函数的导数的符号表达式来代替这个Derivative
项,我想设置Derivative(Ksi(uix,uiy), uix) = 2 * uix
.是否有一种巧妙的方法来执行此替换并在Derivative(Ksi(uix,uiy), uix)
设置为2 * uix
的情况下获取dgN
的符号表达式?这是我的代码:
I would like to replace this Derivative
terms by, let's say, the symbolic expression of the derivative of a function that I know for example, I would like to set Derivative(Ksi(uix,uiy), uix) = 2 * uix
.Is there a neat way to do this substitution and to get a symbolic expression for dgN
with Derivative(Ksi(uix,uiy), uix)
set to 2 * uix
? Here is my code:
import sympy as sp
sp.var("kPenN, Xix, Xiy, uix, uiy, Alpha, x1x, x1y, x2x, x2y, x3x, x3y ", real = True)
Ksi = sp.Function('Ksi')(uix,uiy)
Xi = sp.Matrix([Xix, Xiy])
ui = sp.Matrix([uix, uiy])
xix = Xix + uix
xiy = Xiy + uiy
xi = sp.Matrix([xix, xiy])
x1 = sp.Matrix([x1x, x1y])
x2 = sp.Matrix([x2x, x2y])
N = sp.Matrix([x2 - x1, sp.zeros(1)]).cross(sp.Matrix([sp.zeros(2,1) , sp.ones(1)]))
N = sp.Matrix(2,1, sp.flatten(N[0:2]))
N = N / (N.dot(N))**(0.5)
xp = x1 + (x2 - x1)*Ksi
# make it scalar (in agreement with 9.231)
gN = (xi - xp).dot(N)
dgN = sp.Matrix([gN.diff(uix), gN.diff(uiy)])
推荐答案
您想要的替代可以通过
dgN_subbed = dgN.subs(sp.Derivative(Ksi, uix), 2*uix)
此处Ksi没有参数(uix,uiy),因为这些参数是在创建Ksi时已经声明的.
Here Ksi is without arguments (uix,uiy) since those were already declared when Ksi was created.
如果将Ksi
定义为Ksi = sp.Function('Ksi')
,则语法会更直观一些,而无论参数如何,都可以在以后提供.那么sp.Derivative(Ksi(uix, uiy), uix)
将是引用导数的方式.
The syntax would be a little more intuitive if you defined Ksi
as Ksi = sp.Function('Ksi')
, leaving the arguments -- whatever they may be -- to be supplied later. Then sp.Derivative(Ksi(uix, uiy), uix)
would be the way to reference the derivative.
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