 Solution

The smallest decay-rate eigenvalue is ${\lambda }_1=D{\pi }^2/L^2-\gamma$, so the threshold isBecause $0<x_0<L$, $a_1>0$. For every fixed interior point $0<x<L$, the first sine eigenfunction is positive and the remaining terms have strictly faster decay relative to it. HenceOne can justify the relative-error limit directly: after factoring out the first exponential, the absolute remainder is bounded by $(2{\rho }_0/L){\sum }_{n\ge 2}{e}^{-D{\pi }^2(n^2-1)t/L^2}$, which tends to zero. Thus $\boxed{\rho \to 0\text{ if }\gamma <{\gamma }_c}$, and $\boxed{\rho (x,t)\to +\infty \text{ for each interior }x\text{ if }\gamma >{\gamma }_c}$. The boundary values stay zero. At equality the first term is stationary and the density approaches its positive sine profile.