When solving an eigenvalue problem, I encounter the triconfluent Heun's function, $HeunT[-\lambda^2,0,-2\alpha,0,-b^2](x)$, which is the solution of $$u''-(b^2x^2+2\alpha)u'+\lambda^2u=0.$$ I need a $u(x)$ not divergent at $\infty$ and this condition hopefully can give us the eigenvalue $\lambda$. (This is just like Legendre function can be cut off to Legendre polynomial.) However, I can't find any clear info about this as far as I've searched. Maybe I missed something.

The numerical eigensolution to the original problem is very well-behaved. So I presume there must be some condition like this to help find the eigenvalue.



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