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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