Maximally distance-unbalanced trees

For a graph $G$, and two distinct vertices $u$ and $v$ of $G$, let $n_G(u,v)$ be the number of vertices of $G$ that are closer in $G$ to $u$ than to $v$. Miklavič and Šparl (arXiv:2011.01635v1) define the distance-unbalancedness ${\rm uB}(G)$ of $G$ as the sum of $|n_G(u,v)-n_G(v,u)|$ over all unordered pairs of distinct vertices $u$ and $v$ of $G$. For positive integers $n$ up to $15$, they determine the trees $T$ of fixed order $n$ with the smallest and the largest values of ${\rm uB}(T)$, respectively. While the smallest value is achieved by the star $K_{1,n-1}$ for these $n$, which we then proved for general $n$ (Minimum distance-unbalancedness of trees, Journal of Mathematical Chemistry, DOI 10.1007/s10910-021-01228-4), the structure of the trees maximizing the distance-unbalancedness remained unclear. For $n$ up to $15$ at least, all these trees were subdivided stars. Contributing to problems posed by Miklavič and Šparl, we show $$\max\Big\{{\rm uB}(T):T\mbox{ is a tree of order }n\Big\} =\frac{n^3}{2}+o(n^3)$$ and $\max\Big\{{\rm uB}(S(n_1,\ldots,n_k)):1+n_1+\cdots+n_k=n\Big\} =\left(\frac{1}{2}-\frac{5}{6k}+\frac{1}{3k^2}\right)n^3+O(kn^2),$ where $S(n_1,\ldots,n_k)$ is the subdivided star such that removing its center vertex leaves paths of orders $n_1,\ldots,n_k$.

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