II. (25 points) If $2 x+y \geqslant 1$. Try to find the minimum value of the function $u=y^{2}-2 y$ $+x^{2}+4 x$.


$\therefore$ From $u=y^{2}-2 y+x^{2}+4 x$ completing the square, we get
$$
(x+2)^{2}+(y-1)^{2}=u+5 \text {. }
$$

Since $(x+2)^{2}+(y-1)^{2}$ can be regarded as the square of the distance from point $P(x, y)$ to the fixed point $(-2,1)$. The constraint $2 x+y \geqslant 1$ indicates that point $P$ is in the region $G$ which is the right upper area of the line $l: 2 x+y=1$. Therefore, the problem is transformed into finding the shortest distance from point $A(-2,1)$ to the region $G$.
From the right figure, it can be seen that the distance from point $A$ to

line $I$ is the minimum value of the distance from point $A$ to any point in region $G$.
$$
\begin{aligned}
d & =\frac{|2 \cdot(-2)+1-1|}{\sqrt{2^{2}+1^{2}}} \\
& =\frac{4}{\sqrt{5}}, \\
& \therefore d^{2}=\frac{16}{5} \text {, i.e., } u+5=\frac{16}{5}
\end{aligned}
$$
$\therefore u=-\frac{9}{5}$. That is, the minimum value of $u$ is $-\frac{9}{5}$.