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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.

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    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.2011-09-27

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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 $\xi=0$.

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    @RobertIsrael: That's what I had in mind.2011-10-03