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Integrating a Piecewise function four times but different results

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121770] Integrating a Piecewise function four times but different results
  • From: "Rob Y. H. Chai" <yhchai at ucdavis.edu>
  • Date: Sat, 1 Oct 2011 03:08:00 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

Hello everyone,

 

I tried to integrate a piecewise function 4 times but got different results
for two supposedly identical cases. The integration was done in the Mathematica
version 8.0.1.0. I am interested in what I did wrong, so any comments will
be appreciated. My code is below.

 

Sincerely

 

Rob Y. H. Chai

UC Davis

 

 

(* Introduction p1[x] as piecewise funtion *)

p1[x_] = Piecewise[{{2*wmax*x/L, x <= L/2}, {2*wmax*(1 - x/L), x > L/2}}];

 

 

(* p2[x] is a specialized case of p1[x] with wmax = 1 and L = 1 *)

p2[x_] = p1[x] /. wmax -> 1 /. L -> 1; 

 

 

(* now integrate p1[x] 4 times and introduce integration constants *)

soln1[x_] = Integrate[Integrate[Integrate[Integrate[p1[x], x], x], x], x] +
a*x^3 + b*x^2 + c*x + d ;

 

(* solve for integration constants in the solution of p1[x] *)

eq1a = soln1[0] == 0;

eq1b = (D[soln1[x], {x, 2}] /. x -> 0) == 0;

eq1c = soln1[L] == 0;

eq1d = ( D[soln1[x], {x, 2}] /. x -> L) == 0;

soln1A[x_] = soln1[x] /. Simplify[Solve[{eq1a, eq1b, eq1c, eq1d}, {a, b, c,
d}], L > 0][[1]];

(* check case 1 solution at x = L/2 *)

soln1A[L/2] /. wmax -> 1 /. L -> 1

17/1920

 

 

(* similarly integrate p2[x] 4 times and introduce integration constants *)

soln2[x_] = Integrate[Integrate[Integrate[Integrate[p2[x], x], x], x], x] +
r*x^3 + s*x^2 + u*x + v;

(* solve for integration constants in the solution of p2[x] *)

eq2a = soln2[0] == 0;

eq2b = (D[soln2[x], {x, 2}] /. x -> 0) == 0;

eq2c = soln2[1] == 0;

eq2d = ( D[soln2[x], {x, 2}] /. x -> 1) == 0;

soln2A[x_] = soln2[x] /. Simplify[Solve[{eq2a, eq2b, eq2c, eq2d}, {r, s, u,
v}]][[1]];

soln2A[1/2] (* case 2 solution at x = 1/2 *)

1/120

 

Plot[{soln1A[x] /. wmax -> 1 /. L -> 1, soln2A[x]}, {x, 0, 1}]

 

Question - why doesn't case 1 solution converge to case 2 solution when case
2 is simply a special case of case 1?


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