Solution of axisymmetric stress field using stress functions. Biharmonic equation in this case is:

Clear[r];
lapl = (∂_(r,r))ϕ[r]+(1/r)(∂_r)ϕ[r];
biharmonic = (∂_(r,r))lapl + (1/r)(∂_r)lapl//Simplify

This equation can be solved by doing the substitution r=(E^t), (Mathematica can solve it directly, but the exercise is valuable in itself)

Clear["Global`*"]
r = (E^t);
ϕ1r = ϕ'[t] (E^(-t))
ϕ2r = D[ϕ1r,t](E^(-t))//Simplify
ϕ3r = D[ϕ2r,t](E^(-t))//Simplify
ϕ4r = D[ϕ3r,t] (E^(-t)) //Simplify

lhs =(r^4)ϕ4r  + 2 (r^3) ϕ3r- (r^2) ϕ2r + r ϕ1r//Simplify
sol = DSolve[lhs==0, ϕ[t], t]//Simplify//Flatten

Clear[r]
ϕ[t] /. sol/. t-> Log[r]//FullSimplify

lhs

4 (∂_(t,t))ϕ-4 (∂_(t,t,t))ϕ+(∂_(t,t,t,t))ϕ/. ϕ-> c1+c2 t + c3 (E^(2t))+ c4 t (E^(2t))

ϕ =c1+c2 t + c3 (E^(2t))+ c4 t (E^(2t))/. t-> Log[r]

(Subscript[σ, rr]) = (1/r)(∂_r)ϕ //Simplify
(Subscript[σ, θθ])= (∂_(r,r))ϕ //Simplify

If there is no hole, c2 and c4 = 0, or the stresses ->∞. So the only possible solution is a constant stress field

The case of a cylinder subject to internal or external pressure follows from:c4=0
(Subscript[σ, rr])(Subscript[|, r=a])= (Subscript[p, i]), (Subscript[σ, rr])(Subscript[|, r=b])= (Subscript[p, e]);

bc1 = (Subscript[σ, rr])==(Subscript[p, i])/. {r-> a, c4-> 0}
bc2 = (Subscript[σ, rr])==(Subscript[p, e])/. {r-> b, c4-> 0}
sol = Solve[{bc1, bc2},{c2,c3}]//Simplify//Flatten

(Subscript[Pressureσ, rr])=(Subscript[σ, rr])/. sol /. c4-> 0
(Subscript[Pressureσ, θθ])=(Subscript[σ, θθ])/. sol /. c4-> 0
(Subscript[Pressureσ, rr])+(Subscript[Pressureσ, θθ]) (* Constant! *)