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I am trying to solve the following equation; $$ U''+\left( \frac{1}{t}+\frac{3}{t-1}\right)U'+\left(\frac{1}{t}+C\right)\frac{U}{t(t-1)}=0 $$ where U is a function of t and C is constant. The above equation is similar to a form of Riemann equation.

Could anyone please provide any support on how the above equation can be solved. I am trying to obtain the solution in form of hyper-geometric series.

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  • $\begingroup$ Using Riemann P-function Transformation I am trying to transform this to a Solvable Hypergeoemtric Equation But I am facing problems in this, not sure if I am doing the right transformation. $\endgroup$
    – Ravinder Singh
    Commented Aug 7, 2015 at 14:13
  • $\begingroup$ The P-function of the above equation is $$ U=P\left\{ \begin{array}{ccc} 0 & \infty & 1 \\ +1 & \beta_1 & 0 & ;t \\ -1 & \beta_2 & -2 \end{array} \right\} $$ Where β1 and β2 depends on C and can be calcualted by solving $$\beta^2-3\beta+C=0$$ The above needs to be transformed to a P-function for a solvable hypergeometric form which is of form $$ w=P\left\{ \begin{array}{ccc} 0 & \infty & 1 \\ 0 & a & 0 & ;z \\ 1-c & b & c-a-b \end{array} \right\} $$ Please advise if this can be achieved. $\endgroup$
    – Ravinder Singh
    Commented Aug 7, 2015 at 15:38
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    $\begingroup$ As I said, this is NOT a hypergeometric equation. Forget about P-function. $\endgroup$
    – Alexandre Eremenko
    Commented Aug 7, 2015 at 18:13
  • $\begingroup$ Thanks Alexandre. I was just posting what I found. I am looking into what you have suggested and hope to find the answer soon. $\endgroup$
    – Ravinder Singh
    Commented Aug 7, 2015 at 18:41
  • $\begingroup$ Apologies, I missed a few terms in the equation. It does represent a confluent Heun equation. $\endgroup$
    – Ravinder Singh
    Commented Aug 7, 2015 at 19:45

4 Answers 4

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I'm not sure if this is more useful than the Maple solution given earlier, but Wolfram Mathematica finds a slightly different solution in terms of associated Legendre functions:

$$U(t) = \frac{1}{1-t} \left( k_1 P_\ell^2(2t-1) + k_2 Q_\ell^2(2t-1) \right)$$

Here, $k_1$ and $k_2$ are constants of integration, and $P_\ell^m$ and $Q_\ell^m$ are the associated Legendre functions of the first and second kinds, with $\ell = \frac{1}{2} \left(\sqrt{9-4 C}-1\right)$.

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  • $\begingroup$ Curious, it did not find this solution for me :( $\endgroup$
    – Igor Rivin
    Commented Aug 8, 2015 at 11:52
  • $\begingroup$ @IgorRivin Interesting, what solution does it find for you? I'm using version 10.1 on Windows. $\endgroup$
    – David Zhang
    Commented Aug 8, 2015 at 16:28
  • $\begingroup$ Ah, it works. That was pilot error (since the pilot was very under slept at the time...) $\endgroup$
    – Igor Rivin
    Commented Aug 8, 2015 at 16:37
  • $\begingroup$ The above solution can be obtained by the following two substitution $$ U=\frac{Y}{1-t} $$ and $$ w = 2t-1 $$ $\endgroup$
    – Ravinder Singh
    Commented Aug 10, 2015 at 17:56
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This is not a hypergeometric equation (singularity at infinity is irregular). It can be reduced to the prolate spheroid equation. (Also known as confluent Heun equation). See the references in Wikiedia,

https://en.wikipedia.org/wiki/Prolate_spheroidal_wave_function

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  • $\begingroup$ This may be another form of the equation, but I am unable to see how it can be solved. If possible could you point some literature in this direction. $\endgroup$
    – Ravinder Singh
    Commented Aug 7, 2015 at 14:09
  • $\begingroup$ Wikipedia article on "Prolate spheroidal wave" has literature. Solutions are NOT hypergeometric functions, they are much more complicated. $\endgroup$
    – Alexandre Eremenko
    Commented Aug 7, 2015 at 14:22
  • $\begingroup$ Thanks I shall look in to those and will update here once I get something. $\endgroup$
    – Ravinder Singh
    Commented Aug 7, 2015 at 14:24
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    $\begingroup$ Note that many references in wiki is not active/ Also there is not a reference to the best book on the subject in Russian: Komarov,Ponamarev,Slavyanov. Spheroidal and Coulomb Spheroidal Functions, 1976. $\endgroup$
    – Sergei
    Commented Aug 7, 2015 at 15:00
  • $\begingroup$ Yes, of course, wiki references are inactive. But this site is for professional mathematicians, and professional mathematicians usually can find the references when the key word is given:-) $\endgroup$
    – Alexandre Eremenko
    Commented Aug 7, 2015 at 18:11
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Maple's solution is $$ U \left( t \right) ={\frac {a{{\rm e}^{\sqrt {-C}t}}{\it HeunC} \left( 2\,\sqrt {-C},0,-2,1,-2,t \right) }{ \left( t-1 \right) ^{2}}} +{\frac {b{{\rm e}^{\sqrt {-C}t}}{\it HeunC} \left( 2\,\sqrt {-C},0,-2 ,1,-2,t \right) }{ \left( t-1 \right) ^{2}}\int \!{\frac { \left( t-1 \right) {{\rm e}^{-2\,\sqrt {-C}t}}}{t \left( {\it HeunC} \left( 2\, \sqrt {-C},0,-2,1,-2,t \right) \right) ^{2}}}\,{\rm d}t} $$

EDIT: Note that this was for an earlier version of the problem with the differential equation $$ U'' + \left(\dfrac{1}{t} + \dfrac{3}{t-1}\right) U' + \left(\dfrac{1}{t} + C\right) U = 0 $$ The equation has now been changed to $$ U'' + \left(\dfrac{1}{t} + \dfrac{3}{t-1}\right) U' + \left(\dfrac{1}{t} + C\right) \dfrac{U}{t(t-1)} = 0 $$ which does have hypergeometric solutions.

Please: In future, if you want to change a question, especially after answers have been posted, don't delete the original form of the question; rather, add a new paragraph with the change. Otherwise, the casual reader might think we've posted wrong answers.

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  • $\begingroup$ Thank you for your effort. I have to explore further on the above equation specially HeunC () function. $\endgroup$
    – Ravinder Singh
    Commented Aug 7, 2015 at 15:18
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    $\begingroup$ Information on the Heun DEs: B. D. Sleeman & V. B. Kuznetsov, Digital Library of Mathematical Functions. Chapter 31, Heun Functions. dlmf.nist.gov/31 "HeunC" is Maple-talk for a Heun confluent function. $\endgroup$
    – Gerald Edgar
    Commented Aug 7, 2015 at 15:40
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    $\begingroup$ See also maplesoft.com/support/help/maple/view.aspx?path=HeunC and in particular the References section. $\endgroup$
    – Robert Israel
    Commented Aug 7, 2015 at 22:40
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The given equation can be converted to a hyper-geometric equation by the following substitution, $$ U = tY $$ This gives, $$ t(1-t)Y''+(c-(a+b+1)t)Y'-abY=0 $$ with $$ c=3, a+b+1 = 6, ab=C+4 $$ The final solution Y(t) can then be obtained using the hyper-geometric series function and hence U(t) = t Y(t).

I appreciate all for their effort and help.

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