What does this ODE have to do with the associated Legendre polynomials?

Question

I am currently struggeling with the following differential equation:

$$(t^2-1)f''(t)+tf'(t)(1-8a+8at^2)-4(a+a^2-2at^2+\phi (-a+2at^2))f(t)= 4\lambda f(t),$$

where $a \in \mathbb{R}$ constant, $\phi \in \mathbb{N}$ is a parameter and $\lambda$ is the eigenvalue.

Now, I noticed the following:

for $\phi = 1$ we have that $f(t)=1$ is a solution (remember $P_0^0=1$)

for $\phi = 2$ we have that $f(t)= t$ is a solution (remember $P_1^0=t$)

for $\phi = 2$ we have that $f(t)=\sqrt{1-t^2}$ is a solution(remember $P_1^1 = -\sqrt{1-x^2}$)

for $\phi = 3$ we have that $f(t) = t^2+\alpha$ for some $\alpha \in \mathbb{R}$ is a solution (remember $P_2^0= \frac{1}{2}(3x^2-1)$)

for $\phi = 3$ we have that $f(t)=t\sqrt{1-t^2}$ is a solution (remember $P_2^1 = -3x\sqrt{1-x^2}$)

well, by is a solution I mean that we find an eigenvalue $\lambda$ such that this equation is fulfilled.

But the thing is the following: I noticed that there is a great resemble between a few solutions I found and the associated Legendre polynomials. I would like to understand this deeper. I mean, they are not the same, but the structure is the same. Is there any way to understand this or to see whether this will go on like that? I just don't see the relationship that explains this structure. I am sorry, if you find my question vague, but the thing is that I want to understand this symmetry which I currently don't.

EDIT: Since nobody answered so far. If you need additional information, please let me know. Also, even if you cannot exactly answer the question, you may be able to say something about this differential equation. My idea was, that maybe this similarity is not particularly extraordinary, but I am not very familar with this type of differential equation. Maybe it is also possible to derive some kind of recurrence relations similar to the ones for the Legendre polynomials. In particular, one would expect to find a third polynomial with $\phi=3$ that looks like $(1-\beta x^2)^{\frac{3}{2}}$. The thing is that I cannot find it ( in particular, because you need to manage it that the eigenvalue is chosen appropriately), but I guess that this would be a huge step forward, because then it may be possible to derive recurrence relations for the solution.

Answer 1

According to Maple, the general solution is in terms of HeunC functions: $$ f \left( t \right) ={\it \_C1}\,{\it HeunC} \left( 4\,a,-1/2,-1/2,-2\, a\phi,3/8-{a}^{2}+a\phi-\lambda,{t}^{2} \right) +{\it \_C2}\,{\it HeunC} \left( 4\,a,1/2,-1/2,-2\,a\phi,3/8-{a}^{2}+a\phi-\lambda,{t}^{2 } \right) t $$ For $\phi = 3$ (with $a \ne 0$) there is no solution of the form $f(t) = P(t)^{3/2}$ where $P$ is a polynomial of degree $\le 5$. There are, however, solutions of the form $$\sqrt {{t}^{4}+{\frac { \left( -4\,a+1+\sqrt {16\,{a}^{2}+1} \right) {t}^{2}}{4a}}-{\frac {4\,\sqrt {16\,{a}^{2}+1}a-16\,{a}^ {2}-\sqrt {16\,{a}^{2}+1}+4\,a-1}{32{a}^{2}}}} $$

EDIT: The Maple commands are:

(x^2-1)*(diff(f(x), x, x))+x*(diff(f(x), x))*(8*a*x^2-8*a+1)  
   -(4*(a+a^2-2*a*x^2+3*(2*a*x^2-a)+C))*f(x) = 0;
eval(%,f(x)=sqrt(x^4+b*x^2+d));
solve(identity(%,x));

The result is

$$ \left\{ C=-{a}^{2}+1,a=a,b=-1,d=0 \right\} $$ $$ \left\{ C=-{a}^{2}+1/2-1/2\,\sqrt {16\,{a}^{2}+1},a=a,b=1/4\,{\frac { -4\,a+1+\sqrt {16\,{a}^{2}+1}}{a}},d=-1/8\,{\frac {1}{a} \left( -4\,a+ \sqrt {16\,{a}^{2}+1}-1/4\,{\frac {-4\,a+1+\sqrt {16\,{a}^{2}+1}}{a}} \right) } \right\} $$ $$ \left\{ C=-{a}^{2}+1/2+1/2\,\sqrt {16\,{a}^{2}+1},a=a,b=-1/4\,{\frac {4\,a-1+\sqrt {16\,{a}^{2}+1}}{a}},d=-1/8\,{\frac {1}{a} \left( -4\,a- \sqrt {16\,{a}^{2}+1}+1/4\,{\frac {4\,a-1+\sqrt {16\,{a}^{2}+1}}{a}} \right) } \right\} $$