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MathGroup Archive 1992

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Re: Question on Integrate[ F . Dt[r], ?]

  • To: mathgroup at yoda.physics.unc.edu
  • Subject: Re: Question on Integrate[ F . Dt[r], ?]
  • From: John Lee <lee at math.washington.edu>
  • Date: Fri, 7 Feb 92 17:02:52 -0800

To compute a potential from a form such as u[x,y]Dt[x]+v[x,y]Dt[y], you
need to integrate u w.r.t. x AND v w.r.t. y.  More precisely, if V is the
potential, set intu = Integrate[u,x] and intv = Integrate[v,y].  Start with
the equations

  (1) D[V,x] = u
  (2) D[V,y] = v

Integrating the first with respect to x, you get

  V = intu + f[y]  (* the "constant of integration" can depend on y *)

Then, differentiating w.r.t. y and using (2),

  v = D[V,y] = D[intu,y] + f'[y]

Finally, integrating w.r.t. y and solving for f gives

  f[y] = intv - Int[ D[intu,y], y].

Plugging this in above, we get a formula for u.  Of course, this only works
if the vector field (u,v) is conservative, i.e. D[u,y] = D[v,x].  

Here is a simple Mathematica function that implements this algorithm.

  In[1]:= Literal[Potential[ u_ Dt[x_] + v_ Dt[y_] ]] := Module[{intu,intv},
    If[ Expand[D[u,y] - D[v,x] ] =!= 0,
       (*then*) Print[ "Error: ",u Dt[x] + v Dt[y]," is not conservative" ];
                Return[Null],
       (*else*) intu = Integrate[u,x];
                intv = Integrate[v,y];
                Return[ intu + intv - Integrate[ D[ intu, y ], y]]
      ]];

  In[2]:= Potential[x Dt[x] + y Dt[y]]

            2    2
           x    y
  Out[2]= -- + --
           2    2

  In[3]:= Potential[ x Dt[y] + y Dt[x]]

  Out[3]= x y

  In[4]:= Potential[ x y Dt[x] + y Dt[y]]

  Error: x y Dt[x] + y Dt[y] is not conservative

  In[5]:= dV = (x^2-y^2) Dt[x] - 2 x y Dt[y];

  In[6]:= Potential[dV]

            3
           x       2
  Out[6]= -- - x y
           3

You can probably find a fuller explanation of the mathematics in any
advanced calculus text.

Jack Lee
Dept. of Mathematics
Univ. of Washington
Seattle, WA




  





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