1
$\begingroup$

The whole procedure is much more complicated but the heart of the problem is the following. The variables are h and x. I define the function m containing h,x and the derivative of another function p:

m[h_, x_] := u[h, x, D[p[h, x], x]] + x

I need the derivatives of m:

mh[h_, x_] := D[m[h, x], h]; 
mx[h_, x_] := D[m[h, x], x];

From the derivatives of m I solve for the second derivatives of p:

solution = Solve[{mh[h, x] == 0, mx[h, x] == 0}, {D[p[h, x], h, x], D[p[h, x], x, x]}]; 
phx[h_, x_] := solution[[1]][[1, 2]] 
pxx[h_, x_] := solution[[1]][[2, 2]]

Finally I need the expression:

expression[h_, x_] := phx[h, x] pxx[h, x]

AT THIS POINT I write explicitly the dependence of u in terms of h, x and the derivative of p:

u[h, x, D[p[h, x], x]] = h D[p[h, x], x]^2;

mh[h,x] gives the expected result:

mh[h, x] p^(0,1)(h,x)^2+2 h p^(1,1)(h,x) p^(0,1)(h,x)

However, the expression[h, x] gives me an unexpected result:

 -(((-u^(0,1,0)(h,x,p^(0,1)(h,x))-1) u^(1,0,0)(h,x,p^(0,1)(h,x)))/u^(0,0,1)(h,x,p^(0,1)(h,x))^2)

I do not see the reason. Thanks for any help!

$\endgroup$

1 Answer 1

1
$\begingroup$
Clear["Global`*"]

m[h_, x_] := u[h, x, D[p[h, x], x]] + x
mh[h_, x_] := D[m[h, x], h];
mx[h_, x_] := D[m[h, x], x];

solution = 
  Solve[{mh[h, x] == 0, mx[h, x] == 0}, {D[p[h, x], h, x], 
    D[p[h, x], x, x]}];

phx[h_, x_] := solution[[1]][[1, 2]]
pxx[h_, x_] := solution[[1]][[2, 2]]

expression[h_, x_] := phx[h, x] pxx[h, x]

Replace u with a pure function

repl = u -> (#1*#3^2 &);

mh[h, x] /. repl

(* Derivative[0, 1][p][h, x]^2 + 
   2*h*Derivative[0, 1][p][h, x]*Derivative[1, 1][p][h, x] *)

expression[h, x] /. repl

(* 1/(4 h^2) *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.