# Bessel function divergence

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 $K_0$.

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: $y_-(x) = 0$, $y_+(x) = 0$, $y_0 = 0$.

• Can I dodge the issue by using a different fundamental set of solutions to the ODE? (I expect the answer is no, but it's worth a shot.)
– Ian
Commented Oct 23, 2015 at 23:09
• Hmm...I've been going in circles, but I think your answer has shown me the problem with my formulation. I can't require continuity and also force there to be no singularity at zero. The lack of a singularity at zero is already a "boundary condition", in the sense that it specifies one of the four independent coefficients that I can work with. So I can only impose three other conditions, and the other three are more important. Thanks!
– Ian
Commented Oct 23, 2015 at 23:20

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 $(4)$.

• Indeed, you've formally demonstrated what I described in my comment to Robert Israel's answer. Good work! Except it turns out that the condition that I can drop for physical reasons is actually the first one in (4); the other two are crucial.
– Ian
Commented Oct 23, 2015 at 23:31
• I didn't read that comment when typing my answer. But it seems you are right! :) Commented Oct 23, 2015 at 23:33
• Can I ask that what is the physical background of your problem? :) I mean what is the physical problem? Commented Oct 23, 2015 at 23:38
• You have a 2D crystal surface with a circular island on it. There are molecules diffusing on the surface. Molecules can attach or detach from the boundary, with the overall tendency being towards an equilibrium density $y_0$ of molecules near the boundary. In the formulation above, the diffusion has no time dependence because it is considered to be so fast that it is at equilibrium at all times. The motion of the boundary then has a time dependence (which I haven't described here).
– Ian
Commented Oct 23, 2015 at 23:50