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Wronskian of the confluent hypergeometric functions

  1. Jul 16, 2014 #1
    According to [Erdely A,1953; Higher Transcendental Functions, Vol I, Ch. VI.] the confluent hypergeometric equation
    [tex]\frac{d^2}{d x^2} y + \left(c - x \right) \frac{d}{d x} y - a y = 0[/tex]
    has got eight solutions, which are the followings:
    where ##M[a,c,x]## is the confluent hypergeometric function of the first kind
    and ##U[a,c,x]## is the confluent hypergeometric function of the second kind.

    I would like to know how to evaluate exactly (step by step) the wronskian of the solutions. For example the ##W(y_1,y_5)## or ##W(y_5,y_7)##.
    [tex] W(y_1,y_5) = y_{1} y_{5}^{'} - y_{1}^{'} y_{5} = ? =-\frac{\Gamma(c)}{\Gamma(a)} [/tex]
    [tex] W(y_5,y_7) = y_{5} y_{7}^{'} - y_{5}^{'} y_{7} = ? = e^{j\cdot\pi\cdot \text{sign} \left[\Im(x)\right]\cdot (c-a)}[/tex]
    The results are given in the referred book but the calculation is missing.
    Can anybody help me or suggest a hint? Can anybody offer a article for this problem?
    Last edited: Jul 16, 2014
  2. jcsd
  3. Jul 19, 2014 #2
    I did not define my problem clearly:
    We know from the Abel's theorem that the wronskian of two solutions for the confluent hypergeometic equation equal with this (if ##x\neq 0##):
    [tex] W(y_m,y_n)(x)=\kappa_{m,n}e^{-\int\frac{c-x}{x}dx}=\kappa_{m,n}\cdot x^{-c}\cdot e^{x} [/tex]
    where ##\kappa_{m,n} ## only depend on the choice of the ##y_m##, ##y_n##, but not on ##x##.
    I would like to know how to calculate/determine for instance step by step the ##\kappa_{1,5}## or ##\kappa_{5,7}##, which are equal the followings according to [Erdélyi A, 1953, Higher transcendental functions, Ch. VI] :
    [tex]\kappa_{y_1,y_5}=-\frac{\Gamma(c)}{\Gamma(a)} [/tex]
    [tex]\kappa_{y_5,y_7}=e^{j\cdot\pi\cdot\text{sign}\left[\text{Im}(x)\right](c-a)} [/tex]

    Can anybody help me how to evaulate these constants in this given problem?
    Last edited: Jul 19, 2014
  4. Jul 19, 2014 #3

    This case is fairly easy. Here is a simplified form.
    Move the x terms to the left and then evaluate as $$z \rightarrow +\inf $$
    First some preliminary asymptotic calculations/definitions from "Handbook of Mathematical Functions" 13,5,1
    Replacing c by b to conform with the Handbook.
    $$ M_{z\rightarrow
    +\inf} : \Gamma\left( b\right) \,\left( \frac{{z}^{a−b}\,{e}^{z}}{\Gamma\left( a\right) }+\frac{{e}^{\mp \pi \,a\,i}}{\Gamma\left( b−a\right) \,{z}^{a}}\right) $$
    $$ DM : \frac{dM}{dz} = \Gamma\left( b\right) \,\left( \frac{{z}^{a−b}\,{e}^{z}}{\Gamma\left( a\right) }+\frac{\left( a−b\right) \,{z}^{−b+a−1}\,{e}^{z}}{\Gamma\left( a\right) }−\frac{a\,{e}^{\mp \pi \,a\,i}\,{z}^{−a−1}}{\Gamma\left( b−a\right) }\right) $$
    $$ U_{z\rightarrow
    +\inf} : \frac{1}{z^{a}}$$
    $$ DU : \frac{dU}{dz} = -a*\frac{1}{z^{1+a}}$$
    Evaluating for the constant.
    $$ e^{-z} \cdot z^{b} \cdot (−\frac{\Gamma\left( b\right) \,{e}^{z}}{\Gamma\left( a\right) \,{z}^{b}}+\frac{b\,\Gamma\left( b\right) \,{z}^{−b−1}\,{e}^{z}}{\Gamma\left( a\right) }−\frac{2\,a\,\Gamma\left( b\right) \,{z}^{−b−1}\,{e}^{z}}{\Gamma\left( a\right) })$$
    Taking the dominant term in the third term
    $$ k_{1,5}=−\frac{\Gamma\left( b\right) }{\Gamma\left( a\right) }$$
    Recovering the z dependent factors
    $$ W(y_{1},y_{5})=-z^{-b}e^{z}\frac{\Gamma\left( b\right) }{\Gamma\left( a\right) }$$

    There are a lot of simplifications that could be done; but I used wxmaxima to make sure I didn't slip up.
    Doing $$W(y_{1},y_{2})$$ or $$W(y_{5},y_{7})$$is a little more subtle but basically the same.

    Abramowitz, Milton, and Irene A. Stegun. Handbook of mathematical functions. Vol. 1046. New York: Dover, 1965.
    Digital copy at ; http://people.math.sfu.ca/~cbm/aands/
  5. Jul 20, 2014 #4
    Thanks Ray. :)

    With this help , I think I am capable to solve my problem.
  6. Sep 16, 2014 #5
    Dear Ray,

    I do not to find out where the ##\epsilon=\text{sign}\left[\mathfrak{Im}(x)\right]## come from, when I evaluated the ##k_{5,7}##.

    My calculation for ##|x|## large:
    $$y_5=U[a,c,x]\propto x^{-a}$$
    $$y_5^{'}=-a\cdot U[a+1,c+1,x]\propto -a\cdot x^{-a-1}$$
    $$y_7=e^x\cdot U[c-a,c,-x]\propto e^x\cdot (-x)^{a-c}$$
    $$y_7^{'}=e^x\cdot U[c-a,c,-x]+(c-a) e^x \cdot U[c-a+1,c+1,-x] \propto e^x\cdot (-x)^{a-c} + (c-a)\cdot e^x \cdot (-x)^{a-c-1}$$
    Putting these asymptotic identities into the following equation:

    $$k_{5,7}=x^c \cdot e^x\cdot\left( y_{5}\cdot y_{7}^{'} - y_{5}^{'}\cdot y_{7}\right)$$
    the result will be this:
    $$k_{5,7}=x^{c-a}\cdot (-x)^{a-c}\cdot \left( 1+\frac{2a-c}{x}\right)$$
    Since ##x\rightarrow\infty##

    $$k_{5,7}=x^{c-a}\cdot (-x)^{a-c}=\left(\frac{x}{-x}\right)^{c-a}$$

    $$k_{5,7}=\left(-1\right)^{c-a} = e^{\text{ln}\left[(-1)^{c-a}\right]}=e^{j\pi(c-a)}$$

    My problem is that how the ##\epsilon=\text{sign}\left[\mathfrak{Im}(x)\right]## enter to this calculation?
    Could you give me some hint?

  7. Oct 3, 2014 #6
    Sorry I didn't answer earlier. Apparently I got dropped off the email list?
    I think the situation is a lot clearer in DLMF 13.2.38 . But basically it means you have arbitrarily selected the sign of ln(-1) i.e.
    The principal branch is
    In the division you threw away the direction; like the sign of tan(x) loses some quadrant information that has to be kept around separately if you want to do atan() correctly.
    The reason I like the DLMF version is that the coordination of the direction of z and the Wronskian is explicitly pointed out. Let me point out the two choice lead to two separate solutions which have the advantage of the Wronskian having no poles (except z=0) .
    In addition examing DLMF 13.2.41 shows how the imaginary parts cancel out when z is real. As a matter of fact (well as I recall) you can use 13.2.41 to prove that
    U(complex conjugate z)= complex conjugate U(z) . But I have taken up to much room already.
    In the problem I had, the poles of the other Wronskian's kept getting in the way of numerically finding the eigenvalues; until I realized that 13.2.38 was "clean" like sine/cos in trig. As a matter of fact one could sum and difference the two solutions and find the real and complex parts; or use Eulers expansion.
    Sorry to ramble on but these realizations cost me some time :)
  8. Dec 23, 2014 #7
    Dear Ray!

    Thanks Ray for the information. With your explanation I have understood how to come in the ##\epsilon##.

    I also got dropped off the email list. (Maybe because of the new face of the physics forums.) That's why I just saw your answer.
  9. Dec 23, 2014 #8
    No Problem; I am getting better at Confluent Hypergeometric functions; a somewhat narrow specialization :) but one that seems to occur and a lot of people ( like me) were put off by the intricacies. BTW: I did calculate the asymptotic via Sage and not just copied it (double checking is good); If you like I could try to retrieve the code.
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