= Solution
{c}
The <trigonometric identity>
$$
\sin^3\theta=\frac34\sin\theta-\frac14\sin(3\theta)
$$
shows that only angular <Fourier modes> $1$ and $3$ are required. Differentiating the <harmonic expansion on a disk> at $r=1$ multiplies the coefficient of each nonconstant <Fourier mode> by $n$. Thus $B_1=3/4$ and $3B_3=-1/4$, giving
$$
\boxed{T(r,\theta)=C+\frac34r\sin\theta-\frac1{12}r^3\sin(3\theta),\qquad C\in\mathbb R}.
$$
Each term is a regular <harmonic function>, and the radial <derivative> is
$$
T_r(1,\theta)=\frac34\sin\theta-\frac14\sin(3\theta)=\sin^3\theta.
$$
The <Neumann compatibility condition for Laplace's equation> holds because
$$
\int_0^{2\pi}\sin^3\theta\,d\theta=0,
$$
as is necessary by the <divergence theorem> applied to the <gradient> of a <harmonic function>.
\b[The additive constant is undetermined.] This <Neumann boundary condition> is unchanged by $T\mapsto T+C$, so the paper's blanket uniqueness assertion requires a normalization in this case. The <uniqueness of the Neumann problem> gives uniqueness up to a constant; prescribing $T(0)=0$, or zero mean temperature, selects the displayed answer with $C=0$.
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