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Find a function with a known gradient?

Posted 5 months ago
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Hello. How do I find a function with a known gradient? The problem I am trying to solve is in the attached file.

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This example give you how to solve :

ClearAll["`*"]
f[x_, y_] := Cos[x^2 - 3 y] + Sin[x^2 + y^2];
GRADf = Grad[f[x, y], {x, y}](*A Gradient*)
(*{2 x Cos[x^2 + y^2] - 2 x Sin[x^2 - 3 y], 2 y Cos[x^2 + y^2] + 3 Sin[x^2 - 3 y]} *)

f = Integrate[GRADf[[1]], x] // FullSimplify(* Yours function f *)
(* Cos[x^2 - 3 y] + Sin[x^2 + y^2] *)

Check:

Integrate[GRADf[[1]], x] - Integrate[GRADf[[2]], y] // FullSimplify(*OK*)
(* 0 *)

Integrating constant's in this case is zero !

If your gradient were not so complicated, I'd suggested converting the problem to a differential equation and solving it with DSolve. However, I don't think DSolve can handle such a complicated problem.

You mean F==Grad[f]

That topic is usually covered in in calculus (vector fields through stokes theorem sections).

As you said, if you have the known gradient only and want F, you have to undo the gradient, which is a partial differential equation, which - with what was posted, it would only be fair to offer a reward for solving.

] tfSin[\[Beta]\[CurlyTheta]]))/(a^2 (1 + f^2) + 2 gf t\[Beta]^2 + 
     f^2 g^2 \[Beta]^2 + t^2 \[Beta]^2 - 
     2 (1 + f^2) agSin[\[Beta]\[CurlyTheta]] + 
     2 (1 + f^2) atCos[\[Beta]\[CurlyTheta]] + (1 + 
        f^2) t^2 Cos[\[Beta]\[CurlyTheta]]^2 + 
     2 atSin[\[Beta]\[CurlyTheta]]^2 Cos[\[Beta]\[CurlyTheta]]^2 + (1 \
+ f^2) g^2 Sin[\[Beta]\[CurlyTheta]]^2 - 
     2 (1 + f^2) Cos[\[Beta]\[CurlyTheta]] \
tgSin[\[Beta]\[CurlyTheta]])) - (-at\[Beta]^2 f - at\[Beta]^2 f^3 + 
   a\[Beta]^2 g + a\[Beta]^2 gf^2 + 2 agSin[\[Beta]\[CurlyTheta]]^2 - 
   2 atfCos[\[Beta]\[CurlyTheta]]^2 - 
   a^2 f^3 Cos[\[Beta]\[CurlyTheta]] - 
   gf^2 t\[Beta]^2 Cos[\[Beta]\[CurlyTheta]] - 
   2 f^3 g^2 \[Beta]^2 Cos[\[Beta]\[CurlyTheta]] - 
   f^3 t^2 \[Beta]^2 Cos[\[Beta]\[CurlyTheta]] + 
   2 agfSin[\[Beta]\[CurlyTheta]] Cos[\[Beta]\[CurlyTheta]] - 
   2 atSin[\[Beta]\[CurlyTheta]] Cos[\[Beta]\[CurlyTheta]] - 
   2 atf^3 Cos[\[Beta]\[CurlyTheta]]^2 - 
   f^3 t^2 Cos[\[Beta]\[CurlyTheta]]^3 - 
   a^2 fCos[\[Beta]\[CurlyTheta]] - 
   2 g^2 \[Beta]^2 fCos[\[Beta]\[CurlyTheta]] - 
   t^2 \[Beta]^2 fCos[\[Beta]\[CurlyTheta]] - 
   t^2 fCos[\[Beta]\[CurlyTheta]]^3 - 
   g^2 Cos[\[Beta]\[CurlyTheta]] fSin[\[Beta]\[CurlyTheta]]^2 - 
   t\[Beta]^2 gCos[\[Beta]\[CurlyTheta]] - 
   t\[Beta]^2 gfSin[\[Beta]\[CurlyTheta]] - 
   a^2 Sin[\[Beta]\[CurlyTheta]] - a^2 f^2 Sin[\[Beta]\[CurlyTheta]] -
    gf^3 t\[Beta]^2 Sin[\[Beta]\[CurlyTheta]] - 
   g^2 \[Beta]^2 Sin[\[Beta]\[CurlyTheta]] - 
   f^2 g^2 \[Beta]^2 Sin[\[Beta]\[CurlyTheta]] - 
   2 t^2 \[Beta]^2 Sin[\[Beta]\[CurlyTheta]] - 
   2 f^2 t^2 \[Beta]^2 Sin[\[Beta]\[CurlyTheta]] + 
   2 agf^3 Cos[\[Beta]\[CurlyTheta]] Sin[\[Beta]\[CurlyTheta]] - 
   2 atf^2 Cos[\[Beta]\[CurlyTheta]] Sin[\[Beta]\[CurlyTheta]] - 
   t^2 Cos[\[Beta]\[CurlyTheta]]^2 Sin[\[Beta]\[CurlyTheta]] - 
   f^2 t^2 Cos[\[Beta]\[CurlyTheta]]^2 Sin[\[Beta]\[CurlyTheta]] + 
   2 tgf^3 C

The above is a fraction of the whole "Del f" this person has posted, and it involves many typos such as "fgSin" (instead of fgSin).

Posted 5 months ago
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