Suppose we have two participants. They play the game with utilities: $u_1 = v_1 - b_2$ if $b_2 \le b_1$ and zero otherwise. For the second participant $u_2 = v_2 - b_2$ if $b_2 \ge b_1$ and zero otherwise. Where $v_1, v_2$ - known non-negative values. We want to determine does there a weakly dominant strategy for both players?
For player $1$ there is a weakly dominant strategy: $b_1 = v_1$ (it's easy to show by varying $b_1$).
For the second player, I proved that there is no weakly dominant strategy. Suppose w.d.s. $b_2^* > v_2$ then $0 > u_2(b_2^*, v_2) < u_2(v_2, v_2) = 0$ - contradiction. Now suppose that $0 < b_2^* < v_2$, then since there is a gap between $b_1^*$ and $v_2$ we can find $\epsilon_1 < \epsilon_2$ such that: $b_2^* + \epsilon_1 \in (b_2^*, v_2)$ and $b_2^* + \epsilon_2 \in (b_2^*, v_2)$. For these two points we have: $0 = u_2(b_2^*, b_2^* + \epsilon_1) < u_2(b_2^* + \epsilon_2, b_2^* + \epsilon_1)$, the last term is greater than zero, so we have contradiction. $b_2^* = 0$ is also not weakly dominant strategic, since for all points $u_2(0, b_1) = 0$ and $u_2 \ge 0$ for all $v_2 \ge 0$. So we need to check only one case: $b_2^* = v_2$. But in this case: $0 = u_2(b_2^*, b_1) < u_2(b_1 + \epsilon, b_1) = (v_2 - b_1 - \epsilon)$ for $b_1^* < v_2$ and $\epsilon: b_1^* + \epsilon < v_2$. So there is no weakly dominant strategic for second player.
Am I right? There is a problem. There is also a question: find all N.E. of form $(v_1, v_2)$, but I'm not sure whether there is at least one N.E., since there is no weakly dominant one. Any ideas?
OK, suppose the equilibrium strategy of 1 is $b_1=b_1$ - you already showed there is no strictly better strategy as it is weakly dominant, so this is a good candidate. To find a Nash equilibrium, we need the best response of 2 to that strategy.
That is, we need $$b_2=\arg \max_b 1_{\{b\ge b_1 \}}(v_2-b),$$ where $1_{\{b\ge b_1 \}}$ is the indicator function with the case where 2 has the larger bid. First, clearly $b_2>v_2$ is strictly dominated (e.g., by $b_2=-\infty$), so this cannot be an equilibrium strategy. Second, $b_2=v_2$ yields a profit of 0, again not optimal, as we should be able to get a positive payoff in some cases.
From this we can already conclude that, in equilibrium, 1 will get the item if $v_1>v_2$, since the above implies $v_1=b_1>v_2>b_2$. So we focus on the interesting case $v_2>v_1=b_1$.
And we need to clarify what happens in case of $b_1=b_2$, i.e., we need a tie breaking rule. Your example above is inconsistent, as it gives both bidders the item in case of equality (which is not feasible). So I will adopt the standard tie breaking rule that the item is given to each with 1/2 probability in case of a tie.
Moreover, we need to clarify whether the bidding grid is discrete or continuous. If discrete, let $\epsilon$ be the smallest increment.
Now, if $v_2>v_1$ and a discrete grid, then I claim the best response is $b_2=v_1+\epsilon$. Obviously, a larger bid is never better, as this is already enough to get the item, and a large bid only reduces the payoff. A smaller bid of $b_2=v_1$ expects a profit of $(v_2-v_1)/2$, since the item is obtained only with probability 1/2. This is smaller than $(v_2-v_1-\epsilon)$ (payoff from my strategy) for $\epsilon>0$ small enough (i.e., the discrete grid fine enough). Finally, an even smaller bid $b_2<v_1$ is also inferior, as the item is not obtained and the profit is 0. So, for $\epsilon>0$ small enough, this is a Nash equilibrium.
Moreover, if $v_2>v_1$ and a continuous grid, then there exists no Nash equilibrium. Why? Intuitively, the best response is just above $b_1=v_1$. However, there exists no smallest $b_2$ such that $b_2>b_1$ - you can always find a smaller one. Hence, the definition of a best response is not fulfilled and this is not a Nash equilibrium.
Final note: A Nash equilibrium can exist without weakly dominant strategies. A dominant strategy equilibrium is always a Nash equilibrium, but a Nash equilibrium need not be a dominant strategy equilibrium. Hence, finding that no weakly dominant strategy exists is not sufficient to conclude that no equilibrium exists.