这是一份southampton南安普敦大学FEEG2003W1-01作业代写的成功案例

Start with the following vector identity (see the table of vector identities at the back of the book)
$$
\frac{1}{2} \nabla(\mathbf{u} \cdot \mathbf{u})=\mathbf{u} \times \operatorname{curl} \mathbf{u}+(\mathbf{u} \cdot \nabla) \mathbf{u} .
$$
Substituting this into the equations of motion yields
$$
\frac{\partial \mathbf{u}}{\partial t}+\frac{1}{2} \nabla(\mathbf{u} \cdot \mathbf{u})-\mathbf{u} \times \operatorname{cur} \mathbf{u}=-\nabla w
$$
Taking the curl and using the identity $\nabla \times \nabla f=0$ gives
$$
\frac{\partial \xi}{\partial t}-\operatorname{curl}(\mathbf{u} \times \xi)=0
$$
Using the identity (also from the back of the book)
$$
\operatorname{curl}(\mathbf{F} \times \mathbf{G})=\mathbf{F} \operatorname{div} \mathbf{G}-\mathbf{G} \operatorname{div} \mathbf{F}+(\mathbf{G} \cdot \nabla) \mathbf{F}-(\mathbf{F} \cdot \nabla) \mathbf{G}
$$

FEEG2003W1-01 COURSE NOTES ：

vortex sheet. Let $V$ denote the region of the vortex tube between $C_{1}$ and $C_{2}$ and $\Sigma=S \cup S_{1} \cup S_{2}$ denote the boundary of $V$. By Gauss’ theorem,
$$
0=\int_{V} \nabla \cdot \xi d x=\int_{\Sigma} \xi \cdot d \mathbf{A}=\int_{S_{1} \cup S_{2}} \xi \cdot d \mathbf{A}+\int_{S} \xi \cdot d \mathbf{A} .
$$
By Stokes’ theorem
$$
\int_{C_{1}} \mathbf{u} \cdot d \mathbf{s}=\int_{S_{1}} \boldsymbol{\xi} \cdot d \mathbf{A} \quad \text { and } \quad \int_{C_{2}} \mathbf{u} \cdot d \mathbf{s}=-\int_{S_{2}} \boldsymbol{\xi} \cdot d \mathbf{A}
$$
so (a) holds. Part (b) now follows from Kelvin’s circulation theorem.