Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 5 / b / iii / Solution

For the stated inward heating, take $Q\ge 0$ and put $\tau ={\pi }^{2}kt/(16{\ell }^2)$. Since $|\cos ({\mu }_{n}x)|\le 1$, the transient from ii satisfiesEquality holds at $x=0$, where every cosine is one and all coefficients have the same sign. Thus the maximum is attained at the heated end, and its exact value and late-time form areThis is dominance of the slowest-decaying eigenfunction when $kt\gg {\ell }^2$. More explicitly, the omitted relative correction is nonnegative and bounded by $({\pi }^2/8-1)e^{-8\tau }$, using ${\sum }_{n\ge 0}(2n+1{)}^{-2}={\pi }^2/8$. For a signed imposed heat flux, the maximum-deviation formula uses $|Q|$.