Statement

Fix A>1/2A>1/2, c,ν>0c,\nu>0, and a time TT. Put s=r2/2s=r^2/2, τ=T−t\tau=T-t, and

K(r,t)=c s−AFA+1/2,A−1/2(2ντ/s),r>0, t≤T.K(r,t)=c\,s^{-A}F_{A+1/2,A-1/2}(2\nu\tau/s), \qquad r>0,\ t\le T.

Then KK is positive, smooth up to t=Tt=T from below for every r>0r>0, and solves the

∂tK=ν(Krr+r−1Kr−r−2K).\partial_tK=\nu(K_{rr}+r^{-1}K_r-r^{-2}K).
Verification

Set z=2ντ/sz=2\nu\tau/s and F=FA+1/2,A−1/2F=F_{A+1/2,A-1/2}. The two sides are respectively −2νcs−A−1F′-2\nu c s^{-A-1}F' and

2νcs−A−1[z2F′′+(2A+1)zF′+(A2−1/4)F].2\nu c s^{-A-1}\bigl[z^2F''+(2A+1)zF'+(A^2-1/4)F\bigr].

The Gamma-average equation makes them equal. Moreover −zF′/F-zF'/F is the average of (A−1/2)zv/(1+zv)(A-1/2)zv/(1+zv) against a positive normalized density. It lies in [0,A−1/2)[0,A-1/2), hence rKr/(2K)=−A−zF′/F<−1/2rK_r/(2K)=-A-zF'/F<-1/2, proving Kr<0K_r<0.

The statement is on r>0r>0. It does not give a regular field on the axis or solve a general backward heat initial-value problem. At the terminal time its value is the power law c(r2/2)−Ac(r^2/2)^{-A}.