values $l=2,3,\cdots$. Hence, if we require that $H_T$ be bounded at infinity, which is a weaker condition than , we have $B=0$. However, from the general formula we find that Therefore, there exists no stationary tensor perturbation of the Schwarzschild-Tangherlini solution that is bounded and regular outside the horizon.

Contrastingly, for the case in which $\K^n$ is a generic Einstein space, the above general solution always falls off at infinity if the Lichnerowicz operator has an eigentensor for which $k_T^2<-2$. In this case, because the above equation for $H_T$ always has a solution that is regular at $x=1$, there exists a regular and bounded stationary tensor perturbation, and uniqueness does not hold.

Vector perturbations {#vector-perturbations-1}
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Because there exists no harmonic vector for which $k_V^2<n-1$, $l$ defined by can be assumed to satisfy $l\ge1$. Because the solutions for $l=1$, i.e. the exceptional modes, have been discussed in §§\[subsec:BasicEqs:Vector\] and \[subsec:ExceptionalMode2\], we need only consider modes with $l>1$.

For a generic mode, the Einstein equations are equivalent to where $x=(r_0/r)^{n-1}$, as previously. The general solution to this equation can be expressed in terms of hypergeometric functions as where Note that when $a+b$ is a non-negative integer, $F(-a,1-b,-a-b;x)$ should be replaced by 