In the derivation of Kelvin's circulation theorem, I take the material derivative of circulation, or
\begin{align}
\dfrac{D\Gamma}{Dt} = \dfrac{D}{Dt}\oint_C \vec{u} \cdot d\vec{\ell}.
\end{align}
Moving the material derivative inside the integral gives
\begin{align}
\dfrac{D\Gamma}{Dt} = \oint_C \dfrac{D\vec{u}}{Dt} \cdot d\vec{\ell} + \oint_C \vec{u} \cdot \dfrac{D(d\vec{\ell})}{Dt}.
\end{align}
Given that $d\vec{\ell} = ds \hspace{1mm} \vec{t}$, the second integral of the equation above becomes
\begin{align}
\oint_C \vec{u} \cdot d\vec{u}.
\end{align}
How does one arrive at this equation? My attempt would be to use the chain rule so that
\begin{align}
\dfrac{D(ds \vec{t})}{Dt} = ds\dfrac{D\vec{t}}{Dt} + \vec{t}\dfrac{D(ds)}{Dt}.
\end{align}
From here, I expand each material derivative as
\begin{align}
\dfrac{D\vec{t}}{Dt} = \dfrac{\partial \vec{t}}{\partial t} + u_t\dfrac{\partial \vec{t}}{\partial s}+ u_n\dfrac{\partial \vec{t}}{\partial n}\\
\dfrac{D(ds)}{Dt} = \dfrac{\partial ds}{\partial t} + u_t\dfrac{\partial ds}{\partial s}+ u_n\dfrac{\partial ds}{\partial n}.
\end{align}
I am unsure how these two relations will simplify so that I get $$\dfrac{D(d\vec{\ell})}{Dt} = d\vec{u}.$$
Best Answer
$\vec{l}$ is regarded as a position vector from an arbitrary origin to the contour. So, following a material particle along the contour, $\frac{D\vec{l}}{Dt}=\vec{u}$. The vector $d\vec{l}$ represents a differential position vector along the contour, and is given by $\vec{l}(s+ds)-\vec{l} (s)$. So the material derivative of $d\vec{l}$ is equal to $\vec{u}(s+ds)-\vec{u}(s)=d\vec{u}$.