Wronskian of the confluent hypergeometric functions

[Domdamo]

According to [Erdely A,1953; Higher Transcendental Functions, Vol I, Ch. VI.] the confluent hypergeometric equation
$$\frac{d^2}{d x^2} y + \left(c - x \right) \frac{d}{d x} y - a y = 0$$
has got eight solutions, which are the followings:
$$y_1=M[a,c,x]$$
$$y_2=x^{1-c}M[a-c+1,2-c,x]$$
$$y_3=e^{x}M[c-a,c,-x]$$
$$y_4=x^{1-c}e^{x}M[1-a,2-c,-x]$$
$$y_5=U[a,c,x]$$
$$y_6=x^{1-c}U[a-c+1,2-c,x]$$
$$y_7=e^{x}U[c-a,c,-x]$$
$$y_8=x^{1-c}e^{x}U[1-a,2-c,-x]$$
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 $W(y_5,y_7)$.
$$W(y_1,y_5) = y_{1} y_{5}^{'} - y_{1}^{'} y_{5} = ? =-\frac{\Gamma(c)}{\Gamma(a)}$$
$$W(y_5,y_7) = y_{5} y_{7}^{'} - y_{5}^{'} y_{7} = ? = e^{j\cdot\pi\cdot \text{sign} \left[\Im(x)\right]\cdot (c-a)}$$
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?

 

[Domdamo]

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 $x\neq 0$):
$$W(y_m,y_n)(x)=\kappa_{m,n}e^{-\int\frac{c-x}{x}dx}=\kappa_{m,n}\cdot x^{-c}\cdot e^{x}$$
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] :
$$\kappa_{y_1,y_5}=-\frac{\Gamma(c)}{\Gamma(a)}$$
$$\kappa_{y_5,y_7}=e^{j\cdot\pi\cdot\text{sign}\left[\text{Im}(x)\right](c-a)}$$

Can anybody help me how to evaulate these constants in this given problem?

 

[rrogers]

W($y_{1},y_{5}$)

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/

 

[Domdamo]

Thanks Ray. :)

With this help , I think I am capable to solve my problem.

 

[Domdamo]

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?

Adam

 

[rrogers]

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.
$$-1=e^{-i\pi}=e^{i\pi}$$
The principal branch is
$$-\pi<arg(z)\leqq\pi$$
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 :)

 

[Domdamo]

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.

 

[rrogers]

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.