Are there multiple equilibria in the second price auction?
Sure. An example: if both valuations are drawn from the $[0,1]$ interval then the strategies
$$
b_1(v_1) = v_1
$$
and
$$
b_2(v_2) =
\left\{\begin{array}{cc}
v_2 &amp; \text{ if } v_2 &lt; 1 \\
5 &amp; \text{ if } v_2 = 1.
\end{array}\right.
$$
Another, slightly more annoying equilibrium for $v_1,v_2 \in [0,1]$:
$$
\begin{align*}
b_1(v_1) &amp; = 0 \\
b_2(v_2) &amp; = 2.
\end{align*}
$$



EDIT: The truth-telling bidding strategies are weakly dominant strategies. If we only allow trembling hand perfect equilibria, players assign positive probabilities (in this case densities) to all truncated strategy profiles. In such a setting, the weakly dominant strategies become strictly dominant. (Because of sequential rationality even zero measure deviations such as my first example are eliminated.) In a game where all players have strictly dominant strategies, there is only one equilibrium.