 Black-Scholes reflection symmetry

For the dividend-free Black-Scholes equation, the displayed transform preserves solutions when $B>0$ and $\sigma >0$. Put $x=\log (S/B)$ and $v(x,t)=V(B{e}^x,t)$. The equation becomesSubstitution shows that $\hat{v}(x,t)=e^{qx}v(-x,t)$ solves the same equation when $q=-2b/{\sigma }^2=1-2r/{\sigma }^2$: its first-derivative coefficient becomes $-{\sigma }^{2}q-b=b$, and the additional zeroth-order term ${\sigma }^2{q}^2/2+bq$ vanishes. Subtracting the reflected solution enforces a zero boundary value at $S=B$. Its use in a killed barrier option is restricted to the alive side of the barrier; the algebraic extension to the other side is generally not a valid price.