Hello, 

I might be wrong about this, but assuming that the first eigenvalue comes along with the most symmetric eigenfunction (often the case), one can calculate the eigenvalue as follows. 

We search for eigenfunctions f (r) that only depends on r, a radial coordinate defined as 

r= sqrt (sum_{i=1}^d   x_i x_i) 

Here d is the dimension of space. This function is an eigenfunction of the Laplacian, when it satisfies the ODE

f'' + (d-1) r^{-1} f' + lambda  f = 0  

Here lambda is the minimal eigenvalue we search for, derivatives are noted with primes. You can check that this formula is correct with d=1,2,3 (1D cartesian, 2D cylindrical and axisymmetric, 3D spherical and spherically symmetric).  To solve this ODE, we transform it into a Bessel equation using the following trick. We search for 

f(r)  = r^a g(r)

Injected in the ode we then get 

g'' + (2 a + d- 1)  r^{-1} g' + a ( a + d -2 ) r^{-2} g + lambda g = 0    

This is a Bessel ODE if 

2a + d - 1 = 1   =>    a  = (2-d)/2 

Since we then have

g'' + r^{-1} g' - a^2 r^{-2} g + lambda g = 0 

The solution can be expressed using Bessel functions

g(r) = A J_{|a|} (sqrt(lambda) r) + B   Y_{|a|} (sqrt(lambda) r) 

Here A, B are arbitrary constants and so f(r) = r^a  g(r) is known.  If the domain is the inside of a ball with radius 1, then regularity at the origin demands  B=0. Dirichlet BC at r=1 requires

J_{|a|} (sqrt(lambda) ) = 0

If k_1 \neq 0 is the smallest zero of the Bessel function, we have

J_{|a|} (k_1) = 0

and so lambda = k_1^2 as the smallest eigenvalue. For d=4, a =1 and so we need to find the first zero of J_1 (k_1)= 0. You can easily find this zeros numerically, I measure k_1 = 3.8317, and so lambda = 14.68.. The eigenfunction is 

f(r) = r^{-1} J_1 (k_1 r) 

Hope this helps. 

Wietze Herreman, Associate Professor, Université Paris-Sud, France

I had hoped someone could come out with a nice and simple formula... But this is really helpful! Thank you very much!!