Bessel function divergence

Question

I have two functions $y_+,y_-$. The former is defined on $[L,\infty)$, the latter is defined on $[0,L]$, where $L>0$. I want them to both satisfy the following differential equation on their separate domains:

$$y''+\frac{1}{x}y'-y=0$$

I then want them to satisfy the system of boundary conditions:

$$\lim_{x \to \infty} y_+(x)=0 \\ y_+(L)=y_-(L) \\ y_+'(L)=y_+(L)-y_0 \\ y_-'(L)=y_0-y_-(L).$$

where $y_0$ is a parameter of indefinite sign. When I do this, $y_+$ is a multiple of the modified Bessel function of the second kind with imaginary argument $K_0$, while $y_-$ is a linear combination of $K_0$ and the modified Bessel function of the first kind with imaginary argument $I_0$. The first of these is fine; this happens in a nicer boundary condition that I've already handled as well.

My problem is the second one: the $K_0$ term in $y_-$ will have a logarithmic divergence at zero, which should not occur in my problem for physical reasons. The $I_0$ term, being bounded at zero, does not help us to remove it. Any idea how to fix this? If need be I can show my working, but it is mostly just awkward symbolic linear algebra for the system of boundary conditions, initially assuming $y_-$ and $y_+$ are combinations of $I_0$ and $K_0$.

Answer 1

If you want a linear combination of $I_0(x)$ and $K_0(x)$ to not have a singularity at $x=0$, the coefficient of $K_0(x)$ must be $0$. So $y_-(x) = c I_0(x)$, and the boundary condition at $x=L$ must have $y_-(L)/I_0(L) = {y'}_-(L)/I_0'(L)$.

EDIT:

Since $y_-(L) = y_+(L)$, your conditions on the derivatives become

$${y_-}'(L) = -{y_+}'(L) = y_-(L) - y_0$$

In particular, if $c \ne 0$ you need $$ \dfrac{K_0'(L)}{K_0(L)} = - \dfrac{I_0'(L)}{I_0(L)} $$ i.e. $$ \dfrac{K_1(L)}{K_0(L)} = \dfrac{I_1(L)}{I_0(L)} $$

Unfortunately, this seems to be impossible: $K_1(L) I_0(L) - K_0(L) I_1(L) > 0 $ for real $L$. The conclusion is that the only solution is the trivial one: $y_-(x) = 0$, $y_+(x) = 0$, $y_0 = 0$.

Answer 2

You have enforced your solution to have the following properties:

$$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} {y_\_}\left( x \right) \ne \pm \infty \cr & \mathop {\lim }\limits_{x \to + \infty } {y_ + }\left( x \right) = 0 \cr} \tag{1}$$

So, the most general form that they can take due to these conditions are

$$\eqalign{ & {y_\_}\left( x \right) = A{I_0}(x) \cr & {y_ + }(x) = B{K_0}(x) \cr}\tag{2}$$

The remaining boundary conditions that are to be satisfied are

$$\eqalign{ & {y_ + }(x) = {y_\_}\left( x \right) \cr & {{y'}_ + }(L) = {y_ + }(x) - {y_0} \cr & {{y'}_ - }(L) = {y_0} - {y_ + }(x) \cr}\tag{3}$$

And consequently you have

$$\eqalign{ & B{K_0}(L) = A{I_0}(L) \cr & - B{K_1}(L) = B{K_0}(L) - {y_0} \cr & A{I_1}(L) = {y_0} - A{I_0}(L) \cr}\tag{4}$$

As you can see $(4)$ is an over determined linear algebraic system of equations which has no solutions. So if you want to obtain a solution you should use a least square method which minimizes the error or you can simply ignore one of the equations in $(4)$. What I suggest is to solve for

$$\eqalign{ & B{K_0}(L) = A{I_0}(L) \cr & A{I_1}(L) - B{K_1}(L) = - A{I_0}(L) + B{K_0}(L) \cr}\tag{5}$$

where I just summed the second and third equations in $(4)$.