5. If $P$ is a moving point on the side $A F$ of a regular hexagon $A B C D E F$ with side length 1, then the minimum value of $\overrightarrow{P D} \cdot \overrightarrow{P E}$ is $\qquad$ .
5. $\frac{3}{2}$.

Let the midpoint of $D E$ be $M$.
Notice that,
$$
\overrightarrow{P D} \cdot \overrightarrow{P E}=|\overrightarrow{P M}|^{2}-|\overrightarrow{M E}|^{2}=|\overrightarrow{P M}|^{2}-\frac{1}{4} \text {. }
$$

By geometric relations, the minimum value of $|\overrightarrow{P M}|^{2}$ is
$$
\begin{array}{l}
F M^{2}=E F^{2}+E M^{2}-2 E F \cdot E M \cos 120^{\circ} \\
=1+\frac{1}{4}-2 \times 1 \times \frac{1}{2}\left(-\frac{1}{2}\right)=\frac{7}{4} .
\end{array}
$$

Therefore, the desired result is $\frac{7}{4}-\frac{1}{4}=\frac{3}{2}$.