Converting Second Order ODE to Hypergeometric Function

In summary: I get##\begin{align}&z(1-z)\frac{d^2a}{dz^2}+(z-2)\frac{da}{dz} + \frac{1}{4m^2v^2z}(4E^2m^2+m^2v^2-4E^2m^2z+4Ek_{y}mv^2z+m^2v^2z\nonumber \\&-4E^2m^2z^2-4E^2k_{y}^2v^2z^2+4E^2m^2z^3+4E^
  • #1
thatboi
133
18
I believe it is the case that any linear second order ode with at most 3 regular singular points can be transformed into a hypergeometric function. I am trying to solve the following equation for a(x):
1656684197353.png

where E, m, v, k_{y} are all constants and I believe turning it into hypergeometric form will help me solve it. Any help would be appreciated!
 
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  • #2
Have you located all the singular points? Just a quick look, ##x=0## and ##x= -\frac{E}{m}## look like candidates. What about at ##x=\infty##? It would really help to reduce and simplify the constants.
 
  • #3
Paul Colby said:
Have you located all the singular points? Just a quick look, ##x=0## and ##x= -\frac{E}{m}## look like candidates. What about at ##x=\infty##? It would really help to reduce and simplify the constants.
It appears that ##x=\infty## is a singular point but it is not regular so I am not sure if we can use it here. Nevertheless it should still be possible to turn it into hypergeometric form right?
Also I am not sure what you mean by reducing and simplifying the constants here, I do not see anything that can be further cleaned up.
 
  • #4
Okay, so the hypergeometric equation is,

##
z(1-z)\frac{d^2w}{dz^2} +[c-(a+b+1)z]\frac{dw}{dz} - abw = 0
##

which has three regular singular points in the extended ##z##-plane. These are at ##z=0,1,\infty##. For starters you should transform your equations singular points to the same places. Two out out of three are already where they need to be. The last may be done by the change of variable,

##
z = -\frac{m}{E}x
##

if I've done things correctly. After this is done and you've multiplied through by the ##z(1-z)## factor, if you get the hypergeometric equation, you're done. If not, then your ##a(x)## isn't a hypergeometric function. That said, it may still be related to one with the right transformations.
 
  • #5
Right, but for my equation, x = ##\infty## isn't a regular singular point so are we still able to match it to that equation?
Paul Colby said:
Okay, so the hypergeometric equation is,

##
z(1-z)\frac{d^2w}{dz^2} +[c-(a+b+1)z]\frac{dw}{dz} - abw = 0
##

which has three regular singular points in the extended ##z##-plane. These are at ##z=0,1,\infty##. For starters you should transform your equations singular points to the same places. Two out out of three are already where they need to be. The last may be done by the change of variable,

##
z = -\frac{m}{E}x
##

if I've done things correctly. After this is done and you've multiplied through by the ##z(1-z)## factor, if you get the hypergeometric equation, you're done. If not, then your ##a(x)## isn't a hypergeometric function. That said, it may still be related to one with the right transformations.
 
  • #6
I don't think it's likely. I'd have to do the work to find out. This is your question. If you transform the equation as suggested, what's the result?
 
  • #7
Paul Colby said:
I don't think it's likely. I'd have to do the work to find out. This is your question. If you transform the equation as suggested, what's the result?
I get
##\begin{align}
&z(1-z)\frac{d^2a}{dz^2}+(z-2)\frac{da}{dz} + \frac{1}{4m^2v^2z}(4E^2m^2+m^2v^2-4E^2m^2z+4Ek_{y}mv^2z+m^2v^2z\nonumber \\
&-4E^2m^2z^2-4E^2k_{y}^2v^2z^2+4E^2m^2z^3+4E^2k_{y}^2v^2z^3)a(z) = 0 \nonumber
\end{align}
##
which I guess is not the form we wanted.
 
  • #8
No, this is clearly not the hypergeometric equation. If I assume ##E## is energy, ##m## as mass, ##v## as velocity, the units between terms seems off. Oh, I'm also assuming this is an equation arising from a physics problem. I'd recheck your work.
 

FAQ: Converting Second Order ODE to Hypergeometric Function

1. How do I convert a second order ODE to a hypergeometric function?

To convert a second order ODE to a hypergeometric function, you will need to use the method of Frobenius. This involves assuming a solution in the form of a power series and then solving for the coefficients using recurrence relations.

2. What is the significance of converting a second order ODE to a hypergeometric function?

Converting a second order ODE to a hypergeometric function can make it easier to solve and analyze the equation. Hypergeometric functions have many useful properties and can be expressed in terms of simpler functions, making them a valuable tool in mathematical and scientific research.

3. Can any second order ODE be converted to a hypergeometric function?

No, not all second order ODEs can be converted to a hypergeometric function. The equation must have certain properties, such as being linear and having regular singular points, in order for the conversion to be possible.

4. Are there any limitations to using hypergeometric functions to solve ODEs?

While hypergeometric functions are useful for solving certain types of ODEs, they may not be the most efficient or accurate method for all equations. In some cases, other methods such as numerical approximation or special functions may be more suitable.

5. Can converting a second order ODE to a hypergeometric function help in real-world applications?

Yes, hypergeometric functions have many practical applications in fields such as physics, engineering, and economics. They can be used to model and solve various real-world problems, making them a valuable tool for scientists and researchers.

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