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MathGroup Archive 2001

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Re: Integration of "Which"

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32124] Re: [mg32113] Integration of "Which"
  • From: BobHanlon at aol.com
  • Date: Fri, 28 Dec 2001 02:41:40 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 12/27/01 5:03:29 AM, klepachd at yahoo.com writes:

>This is as a follow up to my last question about integration of Which .
>My problem is alittle more complicated:
>
>W[i_, x_]:= Which[[i == 1,Which[0 <= x <=  1/2, 0, 1/2 <  x <=  1,2(x -
>1)],
>         i == 2,Which[0 <=  x <=  1/2, 2x, 1/2 <  x <=  1, 2(1 - x)]
>         i == 3,Which[0 <=  x <=\ 1/2, 1 - 2x, 1/2 <  x <= 1,0 ]];
>k[x_] := 1 + x;
>
>K[i_, j_, x_] := 
>  N[-(Integrate[D[W[j, x], x]*
>       D[W[i, x], x]*k[x], 
>      {x, 0, 1/2}] + 
>     Integrate[D[W[j, x], x]*
>       D[W[i, x], x]*k[x], 
>      {x, 1/2, 1}])]
>so I can`t use UnitStep , maybe ramp will work?
>Anyway I can`t integrate this , is there a way to solve this ?
>

Needs["Graphics`Colors`"];

The three components of W are

Plot[{2(x-1)UnitStep[x-1/2], 
      2(x+(1-2x)UnitStep[x-1/2]), 
      (1-2x)(1-UnitStep[x-1/2])}, 
    {x,0,1}, PlotStyle -> {Red, Green, Blue}, 
    Frame -> True, Axes -> False];

The index i merely functions as a selector switch

W[i_, x_]  := 
    (i-2)(i-3)(x-1)UnitStep[x-1/2] - 
      (i-1)(i-3)2(x + (1-2x)UnitStep[x-1/2]) +
      (i-1)(i-2)/2(1-2x)(1-UnitStep[x-1/2]);

Plot[{W[1,x], W[2,x], W[3,x]}, 
    {x,0,1}, PlotStyle -> {Red, Green, Blue}, 
    Frame -> True, Axes -> False];

k[x_] := 1+x;

Since you integrate out the dependence on x, K is only a function of i and j

K[i_,j_]:= Evaluate[
      Module[{dj = Simplify[D[W[j,x],x]], 
          di = Simplify[D[W[i,x],x]]}, -Integrate[
            Simplify[dj*di*k[x]],{x,0,1}]]];

The case i=j=1 is handled separately

K[1,1] = -Integrate[D[2(x-1), x]^2*k[x], {x,1/2,1}];

Table[K[i,j],{i,3},{j,3}]

{{-(7/2), 1/2, 0}, {1/2, -6, 5/2}, 
  {0, 5/2, -(5/2)}}


Bob Hanlon
Chantilly, VA  USA


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