System of ODE - comparison with paper

  • #1
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Main Question or Discussion Point

I have the following system of differential equations, for the functions ##A(r)## and ##B(r)##:

##A'-\frac{m}{r}A=(\epsilon+1)B##

and

##-B' -\frac{m+1}{r}B=(\epsilon-1)A##

##m## and ##\epsilon## are constants, with ##\epsilon<1##. The functions ##A## and ##B## are the two components of a spinor.
By solving the first equation for ##B##, and substituting in the other one, we get a second-order differential for ##A##.Its solutions are modified Bessel functions, and due to the boundary conditions that I have, only those of the first kind are admissable solutions. Having solved for ##A##, going back to the first equation, we can find ##B##. Consequently, ##B=\frac{1}{\epsilon+1}...##. However, in a paper I am studying, the solution to this system is given by ##A=(\epsilon-1)(I_{m})## and ##B=I_{m+1}##. I understand the modified Bessel function part, but the prefactors seem odd to me. Even if we start by solving for ##B## instead of ##A## in the beginning, still ##A## ought to be equal to ##\frac{1}{\epsilon-1} *...## as is given by the second equation. Does anyone have any idea how to reach this result?
 

Answers and Replies

  • #2
pasmith
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The equation I get for [itex]B[/itex] is [tex]
r^2B'' + (1 + \epsilon m)rB' - ((m+1)^2 - \epsilon m(m+1) + (1 - \epsilon^2)r^2)B = 0
[/tex] which is not the modified Bessel equation [tex]
r^2I_\alpha'' + rI_\alpha' - (r^2 + \alpha^2)I_\alpha = 0
[/tex] except when [itex]\epsilon = 0[/itex], and if [itex]\epsilon = 0[/itex] then [itex]A = I_m[/itex].

I agree that if you substitute [itex]B = I_{m+1}[/itex] into [tex]
B' + \frac{m+1}r B = (1 - \epsilon) A[/tex] then you obtain [itex]I_m = (1 - \epsilon)A[/itex], so either the paper contains an error or you have misread it.
 
  • #3
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The equation I get for [itex]B[/itex] is [tex]
r^2B'' + (1 + \epsilon m)rB' - ((m+1)^2 - \epsilon m(m+1) + (1 - \epsilon^2)r^2)B = 0
[/tex] which is not the modified Bessel equation [tex]
r^2I_\alpha'' + rI_\alpha' - (r^2 + \alpha^2)I_\alpha = 0
[/tex] except when [itex]\epsilon = 0[/itex], and if [itex]\epsilon = 0[/itex] then [itex]A = I_m[/itex].

I agree that if you substitute [itex]B = I_{m+1}[/itex] into [tex]
B' + \frac{m+1}r B = (1 - \epsilon) A[/tex] then you obtain [itex]I_m = (1 - \epsilon)A[/itex], so either the paper contains an error or you have misread it.

How exactly did you derive this equation for ##B##?

From the first equation, you have that: ##B=\frac{1}{\epsilon+1}(A'-\frac{m}{r}A)##. When you substitute this expression into the second equation, you get:

## -(A''+\frac{m}{r^2}A-\frac{m}{r}A') -\frac{m+1}{r}(A'-\frac{m}{r}A)=(\epsilon+1)(\epsilon-1)A##.

Each term in parenthesis in the left-hand side, corresponds to each term of the left-hand side of the second equation, namely ##B'## and ##B##. Doing all the necessary calculations, you end up with this equation for the component #A#:

## A'' +\frac{1}{r}A' -\frac{m^2}{r^2}A=(1-\epsilon^2) A ## which is indeed the modified Bessel equation.

If you begin by solving the second equation for ##A## and then substitute back in the first one, you get a second-order differential equation for ##B## which is again modified Bessel equation, but of order ##m+1##. However you decide to solve it, you first have to solve either one of these two differential equations and then use the first or the second equation to solve for the other component, since ##A## and ##B## are coupled.
 
  • #4
pasmith
Homework Helper
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I appear to have made an error in obtaining the equation for [itex]B[/itex]; it should be [tex]
r^2B'' + rB' - \left( (m + 1)^2 + (1 - \epsilon^2)r^2\right)B = 0[/tex] and now changing variables to [itex]x = (1 - \epsilon^2)^{1/2}r[/itex] indeed recovers the modified Bessel equation so that [itex]B(r) = I_{m+1}((1 - \epsilon^2)^{1/2}r)[/itex].

It then follows that [itex]A(r) = \frac{1}{1 - \epsilon} I_m((1 - \epsilon^2)^{1/2}r)[/itex].
 

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