1
$\begingroup$

Suppose I start with a differential operator in two variables x1 and x2 as follows:

dop[x1_, x2_] = a[x1,x2]*D[#, {x1, 2}] + b[x1,x2]*D[#, x1, x2] + c[x1,x2]*D[#, {x2, 2}] +
                d[x1,x2]*D[#, x1] + e[x1,x2]*D[#, x2] + f[x1,x2] &;  

Now, I want to transform this operator expressed as a set of two transformed variables: y1, y2 with the following definitions:

pdx1[y1_, y2_] = 
  1/D[x1[y1, y2], y1]*D[#, y1] + 1/D[x1[y1, y2], y2]*D[#, y2] &;
pdx2[y1_, y2_] = 
  1/D[x2[y1, y2], y1]*D[#, y1] + 1/D[x2[y1, y2], y2]*D[#, y2] &;

Now I just want to substitute these definitions for partial derivatives wrt x1 and x2 in terms of y1 and y2 into my linear differential operator. I try the following:

dop1[y1_, y2_] = 
  Replace[Replace[dop[x1, x2], 
     Derivative[order__][#][x1, x2] & -> 
      Nest[pdx2[y1, y2], 
       Nest[pdx1[y1, y2], f[y1, y2], {order}[[1]]], {order}[[2]]], 
     2] // ExpandAll, 
   Derivative[order__][f][y1, y2] -> Derivative[{order}][#][y1, y2] &,
    2];

The level specification up to 2 is to allow for Plus(level 0), Times(level 1) before hitting the derivative terms(level<=2) where I want to replace by those in terms of new coordinates.

However, this code is not working at all. What am I doing wrong?

Edit:
Following Kuba's suggestion, I am revising my code as follows:

ClearAll["Global`*"];
dop[x1_, x2_] = 
  a[x1[y1, y2], x2[y1, y2]]*D[#, {x1, 2}] + 
    b[x1[y1, y2], x2[y1, y2]]*D[#, x1, x2] + 
    c[x1[y1, y2], x2[y1, y2]]*D[#, {x2, 2}] + 
    d[x1[y1, y2], x2[y1, y2]]*D[#, x1] + e[x1[y1, y2], x2]*D[#, x2] + 
    f[x1[y1, y2], x2[y1, y2]] &;
dop1[y1_, y2_] = 
  Block[{f}, 
    Function[Evaluate[dop[x1, x2]@f[y1[x1, x2], y2[x1, x2]]]] /. 
     Derivative[n_, m_][f][_y1, _y2] :> D[#, {y1, n}, {y2, m}]] /. 
   Derivative[n_, m_][y : y1 | y2][x1, x2] ->  Nest[(1/\!\(
\*SubscriptBox[\(\[PartialD]\), \(y1\)]\(x2[y1, y2]\)\) \!\(
\*SubscriptBox[\(\[PartialD]\), \(y1\)]\((#)\)\) + 1/\!\(
\*SubscriptBox[\(\[PartialD]\), \(y2\)]\(x2[y1, y2]\)\) \!\(
\*SubscriptBox[\(\[PartialD]\), \(y2\)]\((#)\)\)) &, Nest[(1/\!\(
\*SubscriptBox[\(\[PartialD]\), \(y1\)]\(x1[y1, y2]\)\) \!\(
\*SubscriptBox[\(\[PartialD]\), \(y1\)]\((#)\)\) + 1/\!\(
\*SubscriptBox[\(\[PartialD]\), \(y2\)]\(x1[y1, y2]\)\) \!\(
\*SubscriptBox[\(\[PartialD]\), \(y2\)]\((#)\)\)) &, 1/\!\(
\*SubscriptBox[\(\[PartialD]\), \(y\)]\(x1[y1, y2]\)\), n], m];

Why is still the Nest function not working as desired? It is just supposed to determine the $\frac{\partial y}{\partial x}$'s in terms of $\frac{\partial x}{\partial y}$'s by repeated differentiation.

$\endgroup$
  • $\begingroup$ Is this what you need: Block[{f}, Function[Evaluate[ dop[x1, x2]@f[y1[x1, x2], y2[x1, x2]] ]] /. Derivative[n_, m_][f][_y1, _y2] :> D[#, {y1, n}, {y2, m}] ]? $\endgroup$ – Kuba Mar 3 '18 at 23:09
  • $\begingroup$ @Kuba This almost works, but not quite. I only know x's in terms of y's i.e. x1[y1,y2] and x2[y1,y2]. So I want to finally replace the partial derivative of y's wrt of x's in terms of partial derivatives of x's wrt y's(this can be done by taking reciprocal of the first order partial derivatives and then take higher derivatives). And also I have to put x1-> x1[y1,y2] , x2->x2[y1,y2] in the product terms. This will then yield the starting differential operator in terms of y1 and y2 only. $\endgroup$ – Subho Mar 4 '18 at 1:56
  • $\begingroup$ @Kuba So, how to implement this rule next? $\endgroup$ – Subho Mar 4 '18 at 2:53
  • $\begingroup$ I don't have time to focus on this now, will try later maybe, but your edit is not what I suggested, there is different dop definition than it was at the beginning. $\endgroup$ – Kuba Mar 5 '18 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.