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Find a function with a known gradient?

Meryem Aslı

Posted 4 years ago

Hello. How do I find a function with a known gradient? The problem I am trying to solve is in the attached file. Attachments: Gradient(f).nb

POSTED BY: Meryem Aslı

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Anonymous User

Posted 4 years ago

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 BY: Anonymous User

Meryem Aslı

Posted 4 years ago

POSTED BY: Meryem Aslı

Anonymous User

Posted 4 years ago

] 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

] 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

POSTED BY: Anonymous User

Anonymous User

Posted 4 years ago

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.

POSTED BY: Anonymous User

Frank Kampas

Frank Kampas, Physicist at Large Consulting

Posted 4 years ago

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.

POSTED BY: Frank Kampas

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 4 years ago

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 !*

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 !

POSTED BY: Mariusz Iwaniuk

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