# Existence of $\mathrm{HE}_{11}$ mode in optical fibers for $v \to 0$

In a step-index optical fiber, the eigenvalue equation for the $$\mathrm{HE}_{11}$$ fundamental mode, under the weak guidance approximation, can be written as

$$\frac{J'_n (u)}{u J_n (u)} = - \frac{K'_n (w)}{w K_n (w)} - n \left( \displaystyle \frac{1}{u^2} + \frac{1}{w^2} \right)$$

It can be plotted with the same procedure shown in this answer. Refer to the link also for the notation used here.

This is the Left Hand Side with $$n = 1$$:

The $$\mathrm{HE}_{11}$$ arises when the first (leftmost) branch is intersected. This mode is well-known because it has no cutoff frequency, so even with a $$v = a \omega \sqrt{\mu_0 \epsilon_0} \sqrt{n_1^2 - n_2^2} \to 0$$, there should be an intersection. However, once $$v$$ has been fixed, the Right Hand Side plot is only available for the values of $$u$$ such that $$w = \sqrt{v^2 - u^2} \geq 0$$, that is $$u < v$$.

For example, with $$v = 0.1$$, this is the Right Hand Side:

It stops well before being able to intersect the first branch of the Left Hand Side. So, how can the $$\mathrm{HE}_{11}$$ mode exist?