11 If $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, then the minimum value of $|x|-|y|$ is $\qquad$ .
Solution: First, from
$$
\left\{\begin{array}{l}
x + 2 y > 0, \\
x - 2 y > 0, \\
(x + 2 y)(x - 2 y) = 4
\end{array} \text { we get } \quad \left\{\begin{array}{l}
x > 2|y| \geqslant 0, \\
x^{2} - 4 y^{2} = 4 .
\end{array}\right.\right.
$$

By symmetry, we only need to consider the case when $y \geqslant 0$. Since $x > 0$, we only need to find the minimum value of $x - y$.
Let $u = x - y$, substitute into $x^{2} - 4 y^{2} = 4$, we get
$$
3 y^{2} - 2 u y + (4 - u^{2}) = 0,
$$

For the equation $(*)$ to have real solutions for $y$, we have
$$
\Delta = 4 u^{2} - 12(4 - u^{2}) \geqslant 0,
$$

Thus,
$$
u \geqslant \sqrt{3}.
$$

Furthermore, when $x = \frac{4}{3} \sqrt{3}, y = \frac{\sqrt{3}}{3}$, we have $u = \sqrt{3}$.
Therefore, the minimum value of $|x| - |y|$ is $\sqrt{3}$.