@acer Ah, that explains why implicitplot does not complain in nm's answer.

@nm Needn't we worry about the imaginary 7/5*Pi*I term?

@Earl You have asked an interesting question and I am glad to have been able to provide a solution that you have found satisfactory.  Cheers!

@C_R That's a pretty curve and works nicely in the context of this problem.  And you are correct in pointing out that the the curve's parametrization need not be arclength parametrization.

This is extracted from your worksheet:

Earl, in your worksheet you write:

    Both bobs slide down according to the downward tangential
    force of gravity and the opposing tangential force of friction.
    The latter is the product of the component of the gravitational
    force normal to the arc and the coefficient of friction.

That does not look correct.  The normal component of the reaction force should be the sum of two forces:

1. The component of the gravitational  force normal to the arc, as you have correctly stated;
2. The centripetal force due to the bob's curved path.

To see the significance of the #2 above, imagine turning off the force of gravity and letting the bob move solely due to imparted initial velocity.  According to your formulation, the force of friction will be zero.  But as the bob slides along the curve, the centripetal force will press it against the curve, making it feel the force of friction. That's where #2 comes in.

@Kitonum The help page on ListTools:-Collect says: 

    Note that the order of the output is session dependent, and is therefore not defined.

Isn't that a problem with implementing your suggestion?

Interesting subject and good work!

For whatever it's worth, here is a different take to what you have done.

Download EML.mw

@Aliocha Ah, I see the reason for the confusion.  The f(n,x) in my answer stands for the antiderivative of your f(n,x),  so I should have called it by some other name, like F:

F := (n,x) -> n*x + 2*sum((n-k)/k*sin(k*x), k=1..n-1);

You wish to integrate your f(n,x) over the interval (0,Pi).  That amounts to evaluating

F(n,Pi) - F(n,0);

But sin(k*x) is zero at 0 and Pi, therefore the summation evaluates to zero and we are left with n*Pi - n*0, that is n*Pi, as you had expected.
 

@Aliocha I don't have Maple 2026 but the formula works just fine in Maple 2024 and 2025.  Here it is:

Download mw.mw

@C_R My comment applies to your worksheet 2d_BVP_Haberman.mw where BC3 specifies zero temperature on the outer boundary.  That's where you need the compatibility condition that I have noted.  That wouldn't be an issue in the case of an insulated boundary.

@janhardo In the solutions that you have provided, you are taking q = q(t), but the original statement says q = q(r,t). 

To make the boundary conditions BC1 and BC3 mutually consistent, you need to have Q(a,t)=0 for all t.  Otherwise BC3 says that the temperature is zero on the outer wall while BC1 prescibes a nonzero rate of change of temperature at r=a and z=0.

Once you impose that condition, you may obtain a series solution in a straightforward way.  I can provide the details if that is of interest.

@dharr Thanks for the detailed explanation.  Voted up!

The additional option known=F fixes the issue:

dchange({tr}, DE, {xi, v(xi)}, known=F);

All is well now.

@sand15 Thanks for pointing out the presence of the "R" factor.  I didn't see it at the first glance.