Green's function for Bessel ODE I want to compute the Green's function for the Bessel ODE. Its from Arfken (7th ed, Problem # 10.1.5)
$ x^2y''(x) + xy'(x) + (k^2x^2 - 1)y(x) = 0 $, subject to the boundary condition, $y(0) = 0$, and $y(1) = 0$.
The solution are $J_1(kx)$ and $Y_1(kx)$. But how can I combine to form a linear combination to satisfy the boundary condition? And then compute the Wronskian, for computing A?
 A: Define $G(x,x')$ as the solution of the equation
$$x^2 \frac{d^2}{dx^2} G(x,x') + x \frac{d}{dx} G(x,x') + (k^2 x^2-1) G(x,x') = -\delta(x-x')\tag{1},$$
subject to $G(0,x')=G(1,x')=0$.  Then
$$G(x,x') = \begin{cases}A J_1(k x) + B Y_1(k x) & 0 \lt x \lt x'\\C J_1(k x) + D y_1(k x) & x' \lt x \lt 1 \end{cases} $$
There are four conditions from which we may fully deduce all of the coefficients. First, note that the boundary condition at $x=0$ requires that  $$B=0,\tag{2}$$ as $Y_1$ is not finite there.  Second, due to the boundary condition at $x=1$,$$C J_1(k) + D Y_1(k) = 0.\tag{3}.$$  Third, we impose continuity at $x=x'$. So,
$$A J_1(k x') = C J_1(k x') + D Y_1(k x')\tag{4}$$
Fourth, we impose a jump discontinuity on the derivative at $x=x'$.  This may be derived by integrating the defining differential equation, which is given in Equation~1, over $[x'-\epsilon,x'+\epsilon]$ for some small $\epsilon \gt 0$; and then be taking the limit as $\epsilon \rightarrow 0$.  The result is:
$$A \left [\frac{d}{dx} J_1(k x) \right ]_{x=x'} - C \left [\frac{d}{dx} J_1(k x) \right ]_{x=x'} - D \left [\frac{d}{dx} Y_1(k x) \right ]_{x=x'} = -\frac1{x'^2}\tag{5}.$$
The system of four equations (i.e., Equations 2, 3, 4, and 5) in four unknowns $A$, $B$, $C$, and $D$ must now be solved. Once done,  all the coefficients are determined and one obtains an expression for $G(x,x')$ that is a solution to the differential equation, and that satisfies the given boundary conditions.
A: The fact that the boundary conditions are Dirichlet (0 boundary conditions)
makes the problem simpler. Here is what we could do.


*

*Solve the homogeneous equation $Lu = 0$. Find two linearly independent
    solutions $u_1$, $u_2$.  These two functions should satisfy the boundary
    conditions, say $u_1$ the left boundary condition, and
    $u_2$ the right boundary condition. 

*Build the Green function as
\begin{align*}
      G(x,y) = \left \{
        \begin{array}{cc}
        C(y) u_1(x) u_2(y) &  0 < x  < y < 1 \\
        C(y) u_1(y) u_2(x) &  0  <  y  <  x < 1 \\
      \end{array}
      \right .   
 \end{align*}
Note that the zero boundary conditions at both ends of the interval are assured 
    from the product, and in addition the continuity of the Green function at $x=y$
    is guaranteed. If one of the solutions diverges to infinity at one end and the other
    does not converges to 0 faster than the first diverges, the Green function
    can not be constructed.

*Find $C(y)$ from the jump  discontinuity. That is solve the
    equation
\begin{eqnarray*}
  \lim_{x \to y^+} [G'(x,y)]- \lim_{x \to y^-} [G'(x,y)] =  \frac{1}{y^2}
\end{eqnarray*}
which provides
\begin{eqnarray}
  \label{thisC}
  C(y) = \frac{1}{ y^2 W(y)}
\end{eqnarray}
with
\begin{eqnarray}
    W(y) = u_1(y) u'_2(y) - u_2(y)  u'_1 (y) = 
  \left  |
  \begin{array}{cc} 
      u_1  & u_2 \\
      u_2  & u_1 
  \end{array}
  \right |.
  \label{wronskian}
\end{eqnarray}
The symbol $W$ stands for  Wronskian.
Let us see the details.
The solutions of the homogeneous equation are given by the Bessel functions of
first kind $J_{1}(kx)$  and the Bessel function of second kind
$Y_{1}(kx)$
The general solution is of the form
\begin{align}
  u(x) = c_1 J_{1}(k x) + c_2 Y_{1}(k x).  \quad \quad (1)
\end{align}
We consider the interval $[0,1]$ divided in two pieces with a source located at some
point $y$, $0 < y < 1$. Let us, for the moment, assume that there are two
solutions $u_1(x)$, and $u_2(x)$ such that $u_1(x)$ satisfies the left boundary
condition and $u_2(x)$ the right boundary condition. Then we claim that the Green function
has the form
\begin{align*}
  G(x,y, \mu) = \left \{
\begin{array}{cc}
C(y) u_1(x) u_2(y) & 0 < x < y < 1 \\ 
C(y) u_1(y)  u_2(x) & 0 < y < x < 1 
  \end{array}
  \right .
\end{align*}
Where $C(y)$  must found by assuring the jump discontinuity of
the first derivative of $G(x, y, k)$ at $x=y$.
To find $u_i(x)$ ($i=1,2$) we start with the general solution (1)
First, since at $x=0$, the Bessel function of the second kind diverges we see that 
on the first part of the interval $0 < x < y$, $c_2=0$.  We then
found a solution $u_1(x) = J_{1}(k x)$, which satisfies the left boundary condition,
since $J_{1}(0)=0$.
Now for right edge, $x=1$, we have that, from $u(1)=0$,
\begin{align}
  u(1) = c_1 J_{1}(1) + c_2 Y_{1}(1) = 0.
  \label{u1c1}
\end{align}
We are free to choose $c_1=Y_{1}(1)$ and $c_2=-J_{1}(1)$, so
\begin{eqnarray*}
  u_2(x) = Y_{1}(1) J_{1}(k x)  -  J_{1}(1) Y_{1}(k x).
\end{eqnarray*}
The jump discontinuity provides $C(y)$ as
\begin{eqnarray*}
  C(y) = \frac{1}{y^2 W(y)}
\end{eqnarray*}
where
\begin{eqnarray*}
W(y) &=& u_1(y) u_2'(y) - u_2(y) u_1'(y) \\
&=& J_{1}(y) [
   Y_{1}(1) J'_{1}(k y)  -  J_{1}(1) Y'_{1}(k y)
   - 
   J'_{1}(k y) [ Y_{1}(1) J_{1}(k x)  -  J_{1}(1) Y_{1}(k x)]
\end{eqnarray*}
Then the Green function is given by
\begin{align*}
  G(x,y, k) = \left \{
\begin{array}{cc}
  \displaystyle{\frac{J_{1}(k x)}{y^2 W(y)}}    \left [
  Y_{1}(1) J_{1}(y)  -  J_{1}(1) Y_{1}(y) \right ] & 0 < x < y < 1 \\
  \\
  \displaystyle{\frac{J_{1}(y)}{y^2 W(y)}}  \left [
Y_{1}(1) J_{1}(k x)  -  J_{1}(1) Y_{1}(k x) \right ] & 0 < y < x < 1 
  \end{array}
  \right .
\end{align*}
