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Below is an operator

$\hat{D}=\sum_{i,j}^2 \frac{\partial}{\partial u^i}\sqrt{g}g^{ij}\frac{\partial}{\partial u^j}$

where $g_{ij}=\begin{pmatrix}p(x,y) & q(x,y)\\\ q(x,y) & r(x,y)\end{pmatrix}$ and $u^1,u^2$ are respectively $x$ and $y$.

In mathematica, how do I make it act on a function $F(x,y)$? Im having trouble understanding how to do the summation part. The functions $p,q,\& \, r$ will all be specified but $F$ is to remain arbitrary. What I actually then want to get is an equation

$\hat{D}F(x,y)=0$

This is easy to work out by hand if $i,j$ only run from 1 to 2, but I want to learn if there is a way to make it work in Mathematica if I replace 2 with 3 or more.

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Old

Do you mean something like this? (I did not know what you meant with the root of the matrix $g$, so I left it out)

g[x_, y_] := {{p[x, y], q[x, y]}, {q[x, y], r[x, y]}};
op[u1_, u2_] = (
   D[g[u1, u2][[1, 1]]*D[#, {u1, 1}], {u1, 1}]
    + D[g[u1, u2][[1, 2]]*D[#, {u2, 1}], {u1, 1}]
    + D[g[u1, u2][[2, 1]]*D[#, {u1, 1}], {u2, 1}]
    + D[g[u1, u2][[2, 2]]*D[#, {u2, 1}], {u2, 1}]
   ) &

enter image description here

Application on function

op[x, y][F[x, y]]

enter image description here

Update: Summation

op2[u_] = Sum[
   D[(g @@ u)[[i1, i2]]*D[#, {u[[i2]], 1}], {u[[i1]], 1}]
   , {i1, Length@u}
   , {i2, Length@u}
   ] &

op2[{x, y}][F[x, y]] == op[x, y][F[x, y]]

True

| improve this answer | |
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  • $\begingroup$ g was actually just the determinant of the matrix g_{ij}. Is there a way to write op[u1_,u2_] without explicitly writing code for all the terms? $\endgroup$ – jboy Sep 20 at 8:11
  • $\begingroup$ See update at the bottom $\endgroup$ – Mauricio Fernández Sep 20 at 8:19

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