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 $Q_{1}\left(x_{1}, y_{1}\right), Q_{2}\left(x_{2}, y_{2}\right)$, respectively. Find the maximum value of $y_{1}-y_{2}$.
Let $P(s, t)$, then $k_{P F_{1}}=\frac{t}{s+1} \Rightarrow P F_{1}: x=\frac{s+1}{t} y-1$,
Solving $\left\{\begin{array}{l}x=\frac{s+1}{t} y-1, \\ x^{2}+2 y^{2}=2\end{array} \Rightarrow \frac{2 s+3}{t^{2}} y^{2}-\frac{2(s+1)}{t} y-1=0 \Rightarrow y_{1}=-\frac{t}{3+2 s}\right.$;
Similarly, $P F_{2}: x=\frac{s-1}{t} y+1$, solving $\left\{\begin{array}{l}x=\frac{s-1}{t} y+1, \\ x^{2}+2 y^{2}=2\end{array}\right.$
$$
\Rightarrow \frac{-2 s+3}{t^{2}} y^{2}+\frac{2(s-1)}{t} y-1=0 \Rightarrow y_{2}=-\frac{t}{3-2 s} \text {. }
$$

Thus, $y_{1}-y_{2}=\frac{t}{3-2 s}-\frac{t}{3+2 s}=\frac{4 s t}{9-4 s^{2}}$.
Let $s=\sqrt{2} \cos \theta, t=\sin \theta, \theta \in\left(0, \frac{\pi}{2}\right)$,
then $y_{1}-y_{2}=\frac{4 \sqrt{2} \sin \theta \cos \theta}{9-8 \cos ^{2} \theta}=\frac{4 \sqrt{2} \sin \theta \cos \theta}{9 \sin ^{2} \theta+\cos ^{2} \theta} \leqslant \frac{2 \sqrt{2}}{3}$, 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 $y_{1}-y_{2}$ is $\frac{2 \sqrt{2}}{3}$.