I just started using Mathematica so my apologies if this is a very basic question. I tried to find related questions on this forum but couldn't, so here goes. I have an equation,
$\quad \quad dp =\dfrac{\partial p}{\partial x}dx + \dfrac{\partial p}{\partial y}dy = P(x,y)dx + Q(x,y)dy$
where I know $P(x,y)$ and $Q(x,y)$. Then, is there a way that I could solve for $p$? I found some information on this when $P(x,y)dx + Q(x,y)dy = 0$, but this is not the case in which I am interested.
At a simple algebraic calculus level, one could do this:
exactQ[{P_, Q_}, {x_, y_}] := D[P, y] == D[Q, x] // Simplify;
exactQ[{P_, Q_, R_}, vars_] := Curl[{P, Q, R}, vars] == {0, 0, 0} // Simplify;
exactSolve[vf_, vars_] /; exactQ[vf, vars] :=
Fold[#1 + Integrate[#2[[1]] - D[#1, #2[[2]]], #2[[2]]] &,
0, Transpose[{vf, vars}]]
Example:
p = Exp[x] Cos[y] + Exp[y] + Sin[x] + y + 2 x
PQ = D[p, {{x, y}}]
(*
E^y + 2 x + y + E^x Cos[y] + Sin[x]
{2 + Cos[x] + E^x Cos[y], 1 + E^y - E^x Sin[y]}
*)
exactSolve[PQ, {x, y}]
(*
E^y + 2 x + y + E^x Cos[y] + Sin[x]
*)
Three-dimensional:
p2 = Exp[x y] Cos[z] + Sin[x z] - y^2 z
PQR = D[p2, {{x, y, z}}]
(*
-y^2 z + E^(x y) Cos[z] + Sin[x z]
{ E^(x y) y Cos[z] + z Cos[x z],
-2 y z + E^(x y) x Cos[z],
-y^2 + x Cos[x z] - E^(x y) Sin[z]}
*)
exactSolve[PQR, {x, y, z}]
(*
-y^2 z + E^(x y) Cos[z] + Sin[x z]
*)
Update - Extension to arbitrary dimensions.
Clear[exactQ];
exactQ[vf_List, vars_List] /; Length[vf] == Length[vars] :=
And @@ Map[Simplify[Curl @@ Transpose[#] == 0] &,
Subsets[Transpose[{vf, vars}], {2}]]
vf4 = D[w x y z + Exp[w^2 + x y] z, {{w, x, y, z}}];
exactSolve[vf4, {w, x, y, z}]
(* (E^(w^2 + x y) + w x y) z *)
The typical solution of such expressions proceeds as in the following example. Suppose
f[x, y] = Exp[x^2 + y^2] Cos[x + y] + x + y^3
(* x + y^3 + E^(x^2 + y^2)*Cos[x + y] *)
From these we construct p and q.
p[x, y] = D[f[x, y], x]
(* 1 + 2*E^(x^2 + y^2)*x*Cos[x + y] - E^(x^2 + y^2)*Sin[x + y] *)
q[x, y] = D[f[x, y], y]
(* 3*y^2 + 2*E^(x^2 + y^2)*y*Cos[x + y] - E^(x^2 + y^2)*Sin[x + y] *)
From p and q, we now wish to compute f. This is possible, only if
D[p[x, y], y] == D[q[x, y], x]
True
and it is. To proceed, we compute the integrals
fx[x, y] = FullSimplify[Integrate[p[x, y], x]] + cy[y]
(* x + E^(x^2 + y^2)*Cos[x + y] + cy[y] *)
where cy[y], an arbitrary function of y, has been added, because Integrate does not do so. Similarly,
fy[x, y] = FullSimplify[Integrate[q[x, y], y]] + cx[x]
(* y^3 + E^(x^2 + y^2)*Cos[x + y] + cx[x] *)
To determine cx[x] and cy[y], fx and fy are equated.
FullSimplify[fx[x, y] - fy[x, y]] == 0
(* x - y^3 - cx[x] + cy[y] == 0 *)
Hence, cx[x] = x + c and cy[y] = y^3 + c, with c an arbitrary constant. Substituting these into fx and fy give for both
f[x, y] = Exp[x^2 + y^2]*Cos[x + y] + x + y^3 + c
as desired. A more formal but equivalent approach is available here.
Addendum
DSolve has become more effective since this question first was posted and now can solve it directly. For instance, with the sample problem used earlier in this answer,
DSolveValue[
{D[f[x, y], x] == 1 + 2 E^(x^2 + y^2) x Cos[x + y] - E^(x^2 + y^2) Sin[x + y],
D[f[x, y], y] == 3 y^2 + 2 E^(x^2 + y^2) y Cos[x + y] - E^(x^2 + y^2) Sin[x + y]},
f[x, y], {x, y}] // FullSimplify
(* x + y^3 + C[1] + E^(x^2 + y^2) Cos[x + y] *)
as desired.
This should be able to be solved with the DifferentialForms.m package https://library.wolfram.com/infocenter/MathSource/482/. Here is a sample script using one of the above examples:
Clear["Global`*"]
<< DifferentialForms`
p1 = Exp[x] Cos[y] + Exp[y] + Sin[x] + y + 2 x
q = d[p1] (*Take exterior derivative*)
p2 = Simplify[HomotopyOperator[q]] (*Find antiderivative*)
r = Simplify[p1 - p2] (*Check p2 is antiderivative for q*)
s = d[r] (*Check r is closed and therefore exact*)
We get
q = (2 + Cos[x] + E^x Cos[y])dx + (1 + E^y - E^x Sin[y])dy
p2 = -2 + E^y + 2 x + y + E^x Cos[y] + Sin[x]
r = 2
s = 0
