11. As shown in the figure, in the Cartesian coordinate system, the ellipse $\Gamma$ : $\frac{x^{2}}{2}+y^{2}=1$ has left and right foci $F_{1}, F_{2}$, respectively. Let $P$ be a point on $\Gamma$ in the first quadrant, and the extensions of $P F_{1}, P F_{2}$ intersect $\Gamma$ at points $Q_{1}, Q_{2}$, respectively. Let $r_{1}, r_{2}$ be the inradii of $\triangle P F_{1} Q_{2}, \triangle P F_{2} Q_{1}$, respectively. Find the maximum value of $r_{1}-r_{2}$.
Let $P\left(x_{1}, y_{1}\right)$, then $k_{P F_{1}}=\frac{y_{1}}{x_{1}+1} \Rightarrow P F_{1}: x=\frac{x_{1}+1}{y_{1}} y-1$,
Solving $\left\{\begin{array}{l}x=\frac{x_{1}+1}{y_{1}} y-1, \\ x^{2}+2 y^{2}=2\end{array} \Rightarrow \frac{2 x_{1}+3}{y_{1}^{2}} y^{2}-\frac{2\left(x_{1}+1\right)}{y_{1}} y-1=0 \Rightarrow y_{Q_{1}}=-\frac{y_{1}}{3+2 x_{1}}\right.$;
Similarly, $P F_{2}: x=\frac{x_{1}-1}{y_{1}} y+1$, solving $\left\{\begin{array}{l}x=\frac{x_{1}-1}{y_{1}} y+1, \\ x^{2}+2 y^{2}=2\end{array}\right.$
$\Rightarrow \frac{-2 x_{1}+3}{y_{1}^{2}} y^{2}+\frac{2\left(x_{1}-1\right)}{y_{1}} y-1=0 \Rightarrow y_{Q_{2}}=-\frac{y_{1}}{3-2 x_{1}}$.
Since $S_{\triangle P F_{1} Q_{2}}=\left|y_{1}-y_{Q_{2}}\right|=2 \sqrt{2} r_{1} \Rightarrow r_{1}=\frac{\sqrt{2} y_{1}}{4} \cdot\left(1+\frac{1}{3-2 x_{1}}\right)$,
Similarly, $r_{2}=\frac{\sqrt{2} y_{1}}{4} \cdot\left(1+\frac{1}{3+2 x_{1}}\right) \Rightarrow r_{1}-r_{2}=\frac{\sqrt{2} y_{1}}{4} \cdot\left(\frac{1}{3-2 x_{1}}-\frac{1}{3+2 x_{1}}\right)$
$=\frac{\sqrt{2} y_{1}}{4} \cdot \frac{4 x_{1}}{9-4 x_{1}^{2}}=\frac{\sqrt{2} x_{1} y_{1}}{9-4 x_{1}^{2}}$. Let $x_{1}=\sqrt{2} \cos \theta, y_{1}=\sin \theta, \theta \in\left(0, \frac{\pi}{2}\right)$,
then $r_{1}-r_{2}=\frac{2 \sin \theta \cos \theta}{9-8 \cos ^{2} \theta}=\frac{2 \sin \theta \cos \theta}{9 \sin ^{2} \theta+\cos ^{2} \theta} \leqslant \frac{1}{3}$, equality holds when
$9 \sin ^{2} \theta=\cos ^{2} \theta \Rightarrow \sin \theta=\frac{\sqrt{10}}{10}, \cos \theta=\frac{3 \sqrt{10}}{10} \Rightarrow P\left(\frac{3 \sqrt{5}}{5}, \frac{\sqrt{10}}{10}\right)$.
In summary, the maximum value of $r_{1}-r_{2}$ is $\frac{1}{3}$.