Existence of trees with prescribed maximum degrees and spectral radii

It is well known that the spectral radius $\rho (T)$ of a tree $T$ with at least $3$ vertices has the property that $\frac{1}{4}\rho (T{)}^2+1<\mathrm{\Delta }(T)\le \rho (T{)}^2$, where $\mathrm{\Delta }(T)$ is the maximum degree of $T$. Let $\mathbb{P}$ denote the set of spectral radii of all non-trivial trees. In this article, we study the inverse problem that for any $\alpha \in \mathbb{P}$ and integer $r$ satisfying the condition $\frac{1}{4}{\alpha }^2+1<r\le {\alpha }^2$, is there a tree $T$ such that $\mathrm{\Delta }(T)=r$ and $\rho (T)=\alpha$? For any positive integer $r$ and positive number $\alpha$, let ${\mathcal{W}}_r(\alpha )$ denote a set of non-negative real numbers defined as follows: $\alpha \in {\mathcal{W}}_r(\alpha )$, and for any multi-set $\{q_i\in {\mathcal{W}}_r(\alpha ):q_i>0,1\le i\le s\}$, if $\beta :=\alpha -{\sum }_{i=1}^s{q}_i^{-1}\ge 0$ and $s\le r-\lceil \frac{\beta }{\beta +1}\rceil$, then $\beta \in {\mathcal{W}}_r(\alpha )$. We first show that $0\in {\mathcal{W}}_r(\alpha )$ if and only if there exists a tree $T$ with $\mathrm{\Delta }(T)\le r$ and $\rho (T)=\alpha$. It follows directly that $\mathbb{P}$ is exactly the set of positive numbers $\alpha$ such that $0\in {\mathcal{W}}_{\lfloor {\alpha }^2\rfloor }(\alpha )$. Applying this conclusion, we prove that for any two positive integers $r\ge 2$ and $k$, there exists a tree $T$ with $\mathrm{\Delta }(T)=r$ and $\rho (T)=\sqrt{k}$ if and only if $\frac{1}{4}k+1<r\le k$.