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Can mathematica perform symbolic vector calculus? Say I want to find

$$\mathbf{v} = \mathbf{a} \exp(\mathbf{b}\cdot\mathbf{x})$$ $$\nabla \cdot \mathbf{v} = \mathbf{b}\cdot \mathbf{v}$$

If I try

v = a Exp[b . x]
Div[v, x]

It doesn't evaluate the divergence. I can find the result I want by doing

v = {a1, a2, a3} Exp[{b1, b2, b3} . {x1, x2, x3}]
Div[v, {x1, x2, x3}]
% == {b1, b2, b3}.v

But this is highly tedious. Is it possible to symbolically find the divergence?

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    $\begingroup$ The add-on package xAct could do what you want, but it would probably be overkill for this purpose alone. (It's designed for calculations in differential geometry.) $\endgroup$ Aug 26, 2021 at 14:02

1 Answer 1

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The divergence cannot be calculated unless one did not explicitly define v as a 3D vector. This, however, can be easily done.

Below I will use the capital letters for vectors, while the small ones I reserve for their projections.

Try the following. Let us first define vectors A, B and X:

Clear[a, b, v, V, A, B];

{A, B, X} = Transpose@Table[j[i], {i, 3}, {j, {a, b, x}}]

(* {{a[1], a[2], a[3]}, {b[1], b[2], b[3]}, {x[1], x[2], x[3]}}  *)

Now let us define the vector V according to your formula:

V = A*Exp[B . X]

(*  {E^(b[1] x[1] + b[2] x[2] + b[3] x[3]) a[1], 
 E^(b[1] x[1] + b[2] x[2] + b[3] x[3]) a[2], 
 E^(b[1] x[1] + b[2] x[2] + b[3] x[3]) a[3]}  *)

After that the divergence calculates automatically:

Div[V, X]

(*  E^(b[1] x[1] + b[2] x[2] + b[3] x[3]) a[1] b[1] + 
 E^(b[1] x[1] + b[2] x[2] + b[3] x[3]) a[2] b[2] + 
 E^(b[1] x[1] + b[2] x[2] + b[3] x[3]) a[3] b[3]  *)

You did not ask about it, however, sometimes people claim that the resulting form is much too clumsy. If you want to transform it to a more usual form it can be done in several ways. For example, this:

Div[V, X] /. z_[k_] -> Subscript[z, k]

yielding the following:

enter image description here

Have fun!

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