11.  Given $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, the minimum value of $|x|-$ $|y|$ is $\qquad$ .
$11 \cdot \sqrt{3}$.
$$
\left\{\begin{array} { l } 
{ x + 2 y > 0 , } \\
{ x - 2 y > 0 , } \\
{ ( x + 2 y ) ( x - 2 y ) = 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x>2|y| \geqslant 0, \\
x^{2}-4 y^{2}=4 .
\end{array}\right.\right.
$$

By symmetry, we only need to consider $y \geqslant 0$. Since $x>0$, we only need to find the minimum value of $x-y$. Let $x-y=u$, 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. Therefore, $\Delta=16\left(u^{2}-3\right) \geqslant 0$, which gives $u \geqslant \sqrt{3}$. When $x=\frac{4 \sqrt{3}}{3}, y=\frac{\sqrt{3}}{3}$, we have $u=\sqrt{3}$.
Thus, the minimum value of $|x|-|y|$ is $\sqrt{3}$.