the heat equation is in spherical coordinates
$$ \frac{∂T}{∂t}=\frac{1}{r^2}\frac{∂}{∂r}(r^2\frac{∂T}{∂r}) $$
0<r<1, t>0
boundary condition
$$ \frac{∂T}{∂r}(1,T)=0 , 0<r<1 $$
initial condition $T(r,0)=r^2 , 0<r<1 $
I need to find numerically the smallest eigenvalue that corresponds to the first non constant term of the solution.
the solution is $ T = \frac{3}{4} + \sum_{n=1}^{\infty}C_n \frac{1}{r} sin((n-\frac{1}{2})r\pi)e^{-((n-\frac{1}{2})\pi)^2t} $
the eigenvalues are $λ_n = -((n-\frac{1}{2})\pi) $
wondering if I just need to set n = 1 to find the the smallest eigenvalue ?
I then need to approximate the steady state solution by truncating the solution to just one exponential term in t and then numerically evaluate a coefficient. I tried calculating $\frac{∂T}{∂t} $ and set n = 1 then set $\frac{∂T}{∂t}=0 $ , but then the coefficient C_n = 0. Anyone know what the question wants?