# Upper bounds for the second largest eigenvalue in terms of degree?

I am looking for upper bounds on the second largest eigenvalue, $$\lambda_2(G)$$ of a given graph $$G$$, with respect to minimum/maximum degrees of the graph. I looked around for some existing bounds most I found were with respect to the order of $$G$$.

Can anyone point me to something?

There exists arbitrarily large $$d$$-regular graphs with a cut edge. A direct inspection shows that the expansion tends to $$0$$. By the Cheeger inequality, $$λ_2$$ is arbitrarily close to $$d=λ_1$$.
There could be no bounds except $$λ_2(G)<λ_1(G)$$; in other words, $$λ_2(G)$$ can be arbitrarily close to $$λ_1(G)$$.
If $$G$$ is a d-regular graph, then the second-largest eigenvalue of $$A(G)$$ is at least $$2\sqrt{d-1} + o(1)$$, where the $$o(1)$$ term goes to 0 as the number of vertices gets large.
If the number of vertices isn't much larger than $$d$$ then the second-largest eigenvalue can be much smaller. If $$G$$ is a complete graph on $$d$$ vertices where each vertex is adjacent to itself, then the second-largest eigenvalue of $$A(G)$$ is 0.
An upper bound for the second-largest eigenvalue is $$d$$ itself (if $$G$$ is $$d$$-regular but not connected).
ETA: I can add a bit more to my answer. The second-largest eigenvalue f a $$d$$-regular graph gives some sort of measure as far as the expansion of the graph i.e., how many vertices need to be cut to disconnect a large set of vertices from the rest of the graph. A $$d$$-regular graph that is disconnected will have $$d$$ as an eigenvalue with multiplicity at least 2. A d-regular connected graph with a cut vertex that bisects the graph nearly in half, will have a second-largest eigenvalue quite close to $$d$$ something like $$d-O(1/n)$$ where $$n$$ is the number of vertices. A $$d$$-regular expander where $$d$$ is $$\theta(1)$$ will have the second-largest eigenvalue no larger than $$1-\epsilon)d$$ for some $$\epsilon \in \theta(1)$$.