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{  CONSTRAINT.PDE  

 

   This problem shows the use of CONSTRAINTS to resolve an ill-posed problem.

   There are no value boundary conditions in any of the three equations, so

   there are infinitely many solutions that satisfy the PDE's.  The constraints

   select from the family of solutions those which have a mean value of 1.

 

}

 

title 'Constraint Test'

 

variables

   u1 u2 u3

 

equations

   u1: div(grad(u1)) +x = 0

   u2: div(grad(u2)) +x+y = 0

   u3: div(grad(u3)) +y = 0

 

constraints

   integral(u1) = integral(1)

   integral(u2) = integral(1)

   integral(u3) = integral(1)

 

boundaries

  Region 1

    start(-1,-1) line to (1,-1) to (1,1) to (-1,1) to close

 

monitors

  contour(u1)

  contour(u2)

  contour(u3)

plots

  contour(u1) report(integral(u1)/integral(1)) as "Average"

  contour(u2) report(integral(u2)/integral(1)) as "Average"

  contour(u3) report(integral(u3)/integral(1)) as "Average"

  surface(u1) report(integral(u1)/integral(1)) as "Average"

  surface(u2) report(integral(u2)/integral(1)) as "Average"

  surface(u3) report(integral(u3)/integral(1)) as "Average"

 

end