Classification of entire solutions to hypergeometric differential equation
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Question/Motivation
I am trying to classify all of the entire (i.e. holomorphic) solutions to a simple-looking ODE of the form
$$zfracpartial^2fpartial z^2 + (az+b)fracpartial fpartial z+(cz+d)f=0.,,,,,,,,,(1)$$
without directly solving the recursion (i.e. by applying transformations to simplify the ODE into a hypergeometric ODE, and then taking the standard holomorphic solutions). What indicates that this may be a difficult task is the limiting case $a=d=0$:
$$zfracpartial^2fpartial z^2 + bfracpartial fpartial z+czf=0.$$
$$$$
Mathematica/Wikipedia gets stuck:
In this limiting case, the standard derivation in Wikipedia, in addition to Mathematica's DSolve, both seem to miss a holomorphic solution. In particular, the Wikipedia derivation and Mathematica both spit out a set of solutions of the form
beginalign*
f_1&=,_0F_1(b;-cz^2/4)\
f_2&=(-cz^2/4)^1-b,_0F_1(2-b;-cz^2/4)\
endalign*
$f_2$ is not entire, so I usually just discard this solution, as I am looking to find only entire functions. What's disturbing about this, however, is that I have found a third linearly independent solution, which happens to be entire, and which is completely missed by Mathematica, and all of the references I find online:
beginalign*
f_3&:=sum_m=0^infty frac1(b)_mfrac(-cz^2/2)^m(2m)!!
endalign*
Here, $(b)_k:=Gamma(b+k)/Gamma(b)$ is the Pochammer symbol. I am deeply troubled by the sight of a solution slipping past my computer algebra program: what is happening here? Is the theory behind these equations incomplete, or is there a deep explanation behind the existence of this extra solution? Why is Mathematica unable to find this holomorphic solution?
complex-analysis differential-equations hypergeometric-function
asked Aug 15 at 8:53
David Roberts
63
add a comment |Â
up vote
1
down vote
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Question/Motivation
I am trying to classify all of the entire (i.e. holomorphic) solutions to a simple-looking ODE of the form
$$zfracpartial^2fpartial z^2 + (az+b)fracpartial fpartial z+(cz+d)f=0.,,,,,,,,,(1)$$
without directly solving the recursion (i.e. by applying transformations to simplify the ODE into a hypergeometric ODE, and then taking the standard holomorphic solutions). What indicates that this may be a difficult task is the limiting case $a=d=0$:
$$zfracpartial^2fpartial z^2 + bfracpartial fpartial z+czf=0.$$
$$$$
Mathematica/Wikipedia gets stuck:
In this limiting case, the standard derivation in Wikipedia, in addition to Mathematica's DSolve, both seem to miss a holomorphic solution. In particular, the Wikipedia derivation and Mathematica both spit out a set of solutions of the form
beginalign*
f_1&=,_0F_1(b;-cz^2/4)\
f_2&=(-cz^2/4)^1-b,_0F_1(2-b;-cz^2/4)\
endalign*
$f_2$ is not entire, so I usually just discard this solution, as I am looking to find only entire functions. What's disturbing about this, however, is that I have found a third linearly independent solution, which happens to be entire, and which is completely missed by Mathematica, and all of the references I find online:
beginalign*
f_3&:=sum_m=0^infty frac1(b)_mfrac(-cz^2/2)^m(2m)!!
endalign*
Here, $(b)_k:=Gamma(b+k)/Gamma(b)$ is the Pochammer symbol. I am deeply troubled by the sight of a solution slipping past my computer algebra program: what is happening here? Is the theory behind these equations incomplete, or is there a deep explanation behind the existence of this extra solution? Why is Mathematica unable to find this holomorphic solution?
complex-analysis differential-equations hypergeometric-function
asked Aug 15 at 8:53
David Roberts
63
It is a shame this question is not receiving much attention. I think it is interesting. BTW: The Wikipedia link is broken.
– Giuseppe Negro
Aug 17 at 18:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question/Motivation
I am trying to classify all of the entire (i.e. holomorphic) solutions to a simple-looking ODE of the form
$$zfracpartial^2fpartial z^2 + (az+b)fracpartial fpartial z+(cz+d)f=0.,,,,,,,,,(1)$$
without directly solving the recursion (i.e. by applying transformations to simplify the ODE into a hypergeometric ODE, and then taking the standard holomorphic solutions). What indicates that this may be a difficult task is the limiting case $a=d=0$:
$$zfracpartial^2fpartial z^2 + bfracpartial fpartial z+czf=0.$$
$$$$
Mathematica/Wikipedia gets stuck:
In this limiting case, the standard derivation in Wikipedia, in addition to Mathematica's DSolve, both seem to miss a holomorphic solution. In particular, the Wikipedia derivation and Mathematica both spit out a set of solutions of the form
beginalign*
f_1&=,_0F_1(b;-cz^2/4)\
f_2&=(-cz^2/4)^1-b,_0F_1(2-b;-cz^2/4)\
endalign*
$f_2$ is not entire, so I usually just discard this solution, as I am looking to find only entire functions. What's disturbing about this, however, is that I have found a third linearly independent solution, which happens to be entire, and which is completely missed by Mathematica, and all of the references I find online:
beginalign*
f_3&:=sum_m=0^infty frac1(b)_mfrac(-cz^2/2)^m(2m)!!
endalign*
Here, $(b)_k:=Gamma(b+k)/Gamma(b)$ is the Pochammer symbol. I am deeply troubled by the sight of a solution slipping past my computer algebra program: what is happening here? Is the theory behind these equations incomplete, or is there a deep explanation behind the existence of this extra solution? Why is Mathematica unable to find this holomorphic solution?
complex-analysis differential-equations hypergeometric-function
asked Aug 15 at 8:53
David Roberts
63
Question/Motivation
I am trying to classify all of the entire (i.e. holomorphic) solutions to a simple-looking ODE of the form
$$zfracpartial^2fpartial z^2 + (az+b)fracpartial fpartial z+(cz+d)f=0.,,,,,,,,,(1)$$
without directly solving the recursion (i.e. by applying transformations to simplify the ODE into a hypergeometric ODE, and then taking the standard holomorphic solutions). What indicates that this may be a difficult task is the limiting case $a=d=0$:
$$zfracpartial^2fpartial z^2 + bfracpartial fpartial z+czf=0.$$
$$$$
Mathematica/Wikipedia gets stuck:
In this limiting case, the standard derivation in Wikipedia, in addition to Mathematica's DSolve, both seem to miss a holomorphic solution. In particular, the Wikipedia derivation and Mathematica both spit out a set of solutions of the form
beginalign*
f_1&=,_0F_1(b;-cz^2/4)\
f_2&=(-cz^2/4)^1-b,_0F_1(2-b;-cz^2/4)\
endalign*
$f_2$ is not entire, so I usually just discard this solution, as I am looking to find only entire functions. What's disturbing about this, however, is that I have found a third linearly independent solution, which happens to be entire, and which is completely missed by Mathematica, and all of the references I find online:
beginalign*
f_3&:=sum_m=0^infty frac1(b)_mfrac(-cz^2/2)^m(2m)!!
endalign*
Here, $(b)_k:=Gamma(b+k)/Gamma(b)$ is the Pochammer symbol. I am deeply troubled by the sight of a solution slipping past my computer algebra program: what is happening here? Is the theory behind these equations incomplete, or is there a deep explanation behind the existence of this extra solution? Why is Mathematica unable to find this holomorphic solution?
complex-analysis differential-equations hypergeometric-function
asked Aug 15 at 8:53
David Roberts
63
edited Aug 15 at 23:12
asked Aug 15 at 8:53
David Roberts
63
asked Aug 15 at 8:53
David Roberts
63
asked Aug 15 at 8:53
David Roberts
63
63
It is a shame this question is not receiving much attention. I think it is interesting. BTW: The Wikipedia link is broken.
– Giuseppe Negro
Aug 17 at 18:24
add a comment |Â
It is a shame this question is not receiving much attention. I think it is interesting. BTW: The Wikipedia link is broken.
– Giuseppe Negro
Aug 17 at 18:24
It is a shame this question is not receiving much attention. I think it is interesting. BTW: The Wikipedia link is broken.
– Giuseppe Negro
Aug 17 at 18:24
It is a shame this question is not receiving much attention. I think it is interesting. BTW: The Wikipedia link is broken.
– Giuseppe Negro
Aug 17 at 18:24
add a comment |Â
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It is a shame this question is not receiving much attention. I think it is interesting. BTW: The Wikipedia link is broken.
– Giuseppe Negro
Aug 17 at 18:24