Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 61 / 1 / c

The other roots of $w(\nu )=w(k)$ satisfyFor $k\in D_{+}$, select the unique other root $\nu (k)$ with negative imaginary part. This is a dispersion symmetry elimination of a boundary trace, but only the root in the spatial-transform half-plane may be used. The root is holomorphic throughout $D_{+}$: the upper branch point $2i/\sqrt{3}$ lies above its boundary. Near the real axis one other root is above and one below; a root can cross the real axis only when $\mathrm{Re}w=0$, so the lower-root choice persists throughout $D_{+}$.

Evaluate the global relation at $\nu$, using the same time transforms because $w(\nu )=w(k)$:Consequentlywhere the completely known spectral density isThe remaining term in the solution formula has integralIts transformed argument stays in the lower half-plane, it is holomorphic inside $D_{+}$, and $e^{w(T-t)}$ does not grow there. Spatial exponential decay closes the truncated domain. Thus the formula in part (b), with $F$ replaced by $H$, uses only the prescribed initial value and the two prescribed boundary traces. It contains no unknown $g_2$ or final-time solution transform. Keeping the inadmissible upper root instead would invalidate this elimination.