这是一份imperial帝国理工大学 H301作业代写的成功案例
If we know that
$$
\frac{d}{d t} \int_{\Omega_{t}^{\prime}} C d x=\int_{\Omega_{t}^{\prime}} f d x, \quad \forall \Omega_{t}^{\prime} \subset \Omega_{t}
$$
then
$$
\frac{\partial C}{\partial t}+\operatorname{div}(C U)=f \quad \text { in } \Omega_{t}^{i}, \quad i=1,2
$$
and
$$
\int_{\Sigma_{i}^{\prime}}[C V] \cdot N d \Gamma=0, \quad \forall \Sigma_{t}^{\prime} \subset \Sigma_{t}
$$
hence,
$$
[C V] \cdot N=0 \text { on } \Sigma_{t} .
$$
When $C$ is a scalar, setting $v=V \cdot N$, we rewrite the last relation as
$$
(C v){2}-(C v){1}=0
$$
G1GH COURSE NOTES ：
$$
[\rho v U]=[\sigma \cdot N] \text {, }
$$
and, setting $\rho_{1} v_{1}=\rho_{2} v_{2}=m$, we find
$$
m[U]=[\sigma \cdot N]=[T] .
$$
In the case of a perfect fluid, $\sigma_{i j}=-p \delta_{i j}$, and hence,
$$
\sigma_{i j} N_{j}=-p N_{i},
$$
which gives
$$
m[U]+[p N]=0,
$$
or else, by projecting onto $N$ and onto the tangent plane to $\Sigma_{t}$ :
$$
\left(U_{T}\right){1}=\left(U{T}\right){2}, $$ and $$ p{1}+\rho_{1} v_{1}^{2}=p_{2}+\rho_{2} v_{2}^{2}
$$