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.

  • $\begingroup$ Are you sure you typed your ODE in correctly? I can't verify that those are solutions. $\endgroup$ May 28, 2014 at 1:40
  • $\begingroup$ check $\lambda = -a^2$ for the first, $\lambda= \frac{1}{4} \pm a-a^2$ for the second and the third and $\lambda = 1-a^2$ for the last. (I leave out the fourth one, because this one has a rather complicated eigenvalue structure.) $\endgroup$
    – user37929
    May 28, 2014 at 6:57
  • $\begingroup$ I suspect there is an f(t) missing on the left hand side of the equation. $\endgroup$ May 28, 2014 at 14:23
  • $\begingroup$ @MichaelRenardy ah, how could I...thank you. $\endgroup$
    – user37929
    May 28, 2014 at 14:38

1 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;

The result is

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

  • $\begingroup$ interesting answer. this of course crushes the relationship between these functions and the Legendre polynomials. How did you actually get this solution with the $\sqrt(t^4...)$, what did you calculate? $\endgroup$
    – user37929
    May 30, 2014 at 10:18
  • 1
    $\begingroup$ In Maple, substitute the desired form of solution into the differential equation and then solve(identity(%),t). $\endgroup$ May 30, 2014 at 15:22
  • 1
    $\begingroup$ HeunC, by the way, is a solution to the Heun confluent equation: see maplesoft.com/support/help/Maple/view.aspx?path=Heun $\endgroup$ May 30, 2014 at 15:31
  • $\begingroup$ mhmm, I just got a Maple license in order to see this, but if I enter the ODE (x^2-1)*(diff(f(x), x, x))+x*(diff(f(x), x))*(8*ax^2-8*a+1)-(4*(a+a^2-2*ax^2+3*(2*a*x^2-a)+C))*f(x) = 0 and then with $f(x) = \sqrt{x^4+bx^2+d}$ and enter solve(identify(%),x) I don't get your answer. Rather, I get a messy expression with a lot of Root of...that does not make sense. $\endgroup$
    – user37929
    May 31, 2014 at 13:29
  • $\begingroup$ but maybe, i need to tkae a different function $f$ that I enter in Maple to get your result? $\endgroup$
    – user37929
    Jun 1, 2014 at 23:57

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