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202206081349 Exercise 4.1 (Q12)

The phase velocity v of a capillary wave with surface tension T and water density p is

\displaystyle{v=\sqrt{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}}}

where \lambda is the wavelength. Find the value of \lambda that minimizes v.

extracted from Michael Corral. (2020). Elementary Calculus.

Setup.

Let v=v(u(\lambda )) s.t.

\begin{aligned} v(u) &=\sqrt{u} \\ u(\lambda ) & =\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi } \\ \end{aligned}

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}u}\big( v(u)\big) & = \frac{1}{2}u^{-1/2}=\frac{1}{2}\Bigg(\sqrt{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}}\Bigg)^{-1} \\ \frac{\mathrm{d}}{\mathrm{d}\lambda}\big( u(\lambda )\big) & = -\frac{2\pi T}{\lambda^2p}+\frac{g}{2\pi} \\ \end{aligned}

Hence,

\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}\lambda} & = \bigg(\frac{\mathrm{d}v}{\mathrm{d}u}\bigg) \bigg(\frac{\mathrm{d}u}{\mathrm{d}\lambda}\bigg) \\ & = \frac{1}{2}\Bigg(\sqrt{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}}\Bigg)^{-1}\bigg(-\frac{2\pi T}{\lambda^2p}+\frac{g}{2\pi}\bigg) \\ \end{aligned}

Working.

For \displaystyle{\frac{\mathrm{d}v}{\mathrm{d}\lambda}}=0 requires:

\begin{cases} \displaystyle{\frac{2\pi T}{\lambda p}+\frac{\lambda g}{2\pi}} \neq 0 \\ \displaystyle{-\frac{2\pi T}{\lambda^2p}+\frac{g}{2\pi}}= 0 \\ \end{cases}

Answer.

We have the satisfying stationary point:

\lambda_0 =\displaystyle{2\pi\sqrt{\frac{T}{pg}}}.

It remains to be verified that this is a global minimum indeed.

(to be continued)