Hello everyone,
In designing the vector module, I have defined a class called 'Del', whose
instances will act as the Del operator <http://en.wikipedia.org/wiki/Del>
for the coordinate system it is assigned to.
An example of code using the operator -
>>> C = CoordSysCartesian('C')
>>> i, j, k = C.base_vectors()
>>> x, y, z = C.base_scalars()
>>> delop = C.delop
*For divergence*-
>>> v = x*y*z * (i + j + k)
>>> delop & v
x*y + y*z + z*x
*Curl *-
>>> delop ^ v #^ stands for 'cross'
(-x*y + x*z)*i + (x*y - y*z)*j + (-x*z + y*z)*k
*Gradient* -
>>> delop(x*y*z)
y*z*i + z*x*j + x*y*k
*Directional derivative*-
>>> (i & delop)(x)
1
Now the question is, is there a need/is it possible to code Del similar to
Derivative or Integral? Especially with respect to unevaluated expressions.
Jason suggested I look into it, though I am not sure how it will work, or
what a use case will be.
Any suggestions/thoughts?
On Friday, June 6, 2014 7:14:51 PM UTC+5:30, Sachin Joglekar wrote:
>
> Hello everyone,
> I have been working on a basic vector framework for sympy for some time
> now. I am still working on the classes for coordinate systems, so currently
> all operations occur in the _same_ frame. The PR for the code is here
> <https://github.com/sympy/sympy/pull/7566>.
>
> To get you started, here is how you would do some basic vector
> manipulations using the code (i, j, k are by default base vectors, and x,
> y, z are by default base scalars. Better naming ideas?)-
>
> >>> from sympy.vector import *
> >>> u = 4*i - 5*k
> >>> from sympy.vector.vector import VectorAdd
> >>> isinstance(u, VectorAdd)
> True
> >>> v = 8*k
> >>> u += v
> >>> u ^ v
> ...output...
>
> Now I am facing a few problems..
>
> 1. How do I ensure methods like 'trigsimp' work with these classes?
> Currently, passing a vector like *(sin(a)+cos(a))**2*i - j* to trigsimp
> returns the correct answer mathematically, _but_ the instance is of type
> Add (instead of the expected VectorAdd). How do I prevent this?
>
> 2. How do I get 'solve' to work with these classes? For example, consider
> a problem solved (in the future) using the module -
>
> *Do the following vectors form a basis for R^3: {(2,-3,1), (4,1,1),
> (0,-7,1)}?*
>
> To check if the vectors can form a basis, we check if the 3 conventional
> basis vectors can be expressed in terms of the given vectors. First, define
> b1, b2 and b3 as the vectors in question
>
> >>> b1 = 2*R.i - 3*R.j + R.k
> >>> b2 = 4*R.i + R.j + R.k
> >>> b3 = - 7*R.j + R.k
>
> Define three Symbols as the coefficients of the linear multiples
>
> >>> a, b, c = symbols('a b c')
>
> Now check if R.i = x*b1 + y*b2 + z*b3n returns a valid solution. If yes,
> do the same for R.j and R.z. If all three have some solution (not
> necessarily same), the vectors do form a basis.
>
> >>> solve(R.i - (a*b1 + b*b2 + c*b3), a, b, c)
> >>>
>
> solve returned None. Hence, solution does not exist. Therefore, the given
> vectors cannot define a basis for R^3.
>
> I think it would involve changing the code for 'solve' itself. Any ideas?
>
> I guess answering these questions may require a look at my code, so any
> help is appreciated.
>
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