11. If $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, then the minimum value of $|x|-|y|$ is $\qquad$
$\sqrt{3}$
11. 【Analysis and Solution】Notice
$$
\left\{\begin{array} { l } 
{ x + 2 y > 0 } \\
{ x - 2 y > 0 } \\
{ ( x + 2 y ) ( x - 2 y ) = 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x>2 \mid y \geqslant 0 \\
x^{2}-4 y^{2}=4
\end{array}\right.\right.
$$

By symmetry, we only consider $y \geqslant 0$, because $x>0$, so we only need to find the minimum value of $x-y$.
Let $x-y=u$ and substitute into $x^{2}-4 y^{2}=4$, we get $3 y^{2}-2 u y+\left(4-u^{2}\right)=0$. This quadratic equation in $y$ clearly has real roots, so $\Delta=16\left(u^{2}-3\right) \geqslant 0, \therefore u \geqslant \sqrt{3}$. When $x=\frac{4}{3} \sqrt{3}, y=\frac{\sqrt{3}}{3}$, $u=\sqrt{3}$. Therefore, the minimum value of $|x|-|y|$ is $\sqrt{3}$.