This question already has an answer here:
How can one define in a functional way a 1st-order linear differential operator involving several independent variables that can then be applied to a function of that many variables?
Consider an example with two variables, where one wants to form
D[f[x, y], x] + 3 D[f[x, y], y] from a function
f. Of course one could define
diff[f_][x_, y_] := D[f[x, y], x] + 3 D[f[x, y], y]
so that for a function such as
g[x_, y_] := x^2 y + Cos[x + 2 y]
we just evaluate:
diff[g][x, y] 2 x y + 3 (x^2 - 2 Sin[x + 2 y]) - Sin[x + 2 y] (* desired final output *)
But how can such an operator be defined functionally, that is, without explicitly using variables initially?
We could try
diffOp[y_] := Derivative[1, 0][y] + 3 Derivative[0, 1][y]
diffOp[g] 3 (-2 Sin[#1 + 2 #2] + #1^2 &) + (-Sin[#1 + 2 #2] + 2 #1 #2 &)
But now how does one use such a combination of pure functions of several variables so as to produce the same result as from
The crux of the difficulty appears in the following simpler problem. Consider two functions of two variables:
g1 = (#1^2 + #2) & g2 = Cos[#1 #2] &
How can one produce the same result as, say,
g1[x, y] + 3 g2[x, y] (* x^2 + y + 3 Cos[x y] *)
directly from the functional linear combination
g1 + 3 g2 -- by forming an expression of the form
oper[g1 + 3 g2][x, y]?
By contrast with the single-variable situation, where a simple
Through would serve as the
oper',Through` will not work in the multi-variable situation here:
Through[(g1 + 3 g2)[x, y]] (* x^2 + y + (3 (Cos[#1 #2] &))[x, y] *)
A pure function embedded in that output.
Note that a simple sum
g1 + g2 instead of the linear combination
g1 + 3 g2,
Through will work (just as it does for a single variable):
Through[(g1 + g2)[x, y]] (* x^2 + y + Cos[x y] *)