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Re: Integration of "Which"
*To*: mathgroup at smc.vnet.net
*Subject*: [mg32123] Re: [mg32113] Integration of "Which"
*From*: BobHanlon at aol.com
*Date*: Fri, 28 Dec 2001 02:41:39 -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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