Example 3 Find the minimum value of the function $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}$ with real numbers $x, y$ as variables.
(1991 "Hope Cup" Invitational Competition Question)
Solve $u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2$. From the above equation, we can consider points $P_{1}\left(x, \frac{q}{x}\right), P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_{1}$ is a hyperbola with the two coordinate axes as asymptotes; when $y \in \mathbf{R},|y| \leqslant \sqrt{2}$, the trajectory of $P_{2}$ is the lower half of a circle centered at the origin with a radius of $\sqrt{2}$. From the graph (omitted), it is evident that the minimum value of $\left|P_{1} P_{2}\right|$ is $3 \sqrt{2}-\sqrt{2}$. Therefore, $u(x, y)_{\text {min }}=(3 \sqrt{2}-\sqrt{2})^{2}-2=6$ is the desired result.

Note: A slight variation of this example is the 1983 Putnam Competition problem: Find the minimum value of $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}$.