# Proving that zeros of a function are first-order

Let $c_{L},c_{T},\omega$ be positive constants with $c_{L}>c_{T}$. Define

$$p=\sqrt{\frac{\omega^{2}}{c_{L}^{2}}-\xi^{2}}\qquad q=\sqrt{\frac{\omega^{2}}{c_{T}^{2}}-\xi^{2}}$$

Consider the function $D_{S}\left(\xi\right)$ defined as follows:

$$D_{S}=4\xi^{2}pq\sin p\cos q+\left(\xi^{2}-q^{2}\right)^{2}\cos p\sin q$$

How can I prove that all the real zeros of $D_{S}\left(\xi\right)$ are first-order? Thanks.

-
I think this won't be too difficult, but I can't get my way around it :) BTW, I have not been able to prove that the zeros are first-order, but I'm quite sure that they are because of the way they are used in residue calculus in some papers I'm reading. – becko Sep 27 '11 at 5:00

Not sure your conjecture holds. For $\omega=\pi c_L=2\pi c_T$, $D_S(0)=0$ and $D_S(\xi)=8\pi^3\xi^2+O(\xi^4)$ when $\xi\to0$, hence $D_S$ has a second-order zero at $\xi=0$.
Thanks... But I think I have a new way to look at it now. This equation shows up in a physical problem, and the constants $c_T$, etc., are things we measure. It is highly unlikely (zero probability) that they turn out to have precisely the values that you gave them above, and with any other values, the zeros are first-order. What I'm trying to say is something like this: the set of values that we would have to assign to the constants to make the zeros higher order is of measure zero in the set of all possible values. Thus we may assume that the zeros are first-order. Am I right? – becko Sep 28 '11 at 15:47
Probably. Think of a zero of $D_S$ as an analytic function $p(x,y)$ of $x = \omega/c_L$ and $y = \omega/c_T$ (in a domain that stays away from the branch points). It's a first-order zero if $(D_S)'(p(x,y)) \ne 0$. But $(D_S)'(p(x,y))$ is analytic in $x$ and $y$. For each $x$, it will be 0 for only a discrete set of $y$ unless it is constant for that $x$. – Robert Israel Oct 3 '11 at 8:20