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I seen two variations of the Black-Scholes PDE with either $+{\frac {\partial V}{\partial t}}$ or $-{\frac {\partial V}{\partial t}}$, and wanted to ask why that is?

a) https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_equation#Solving_the_PDE $${\frac {\partial V}{\partial t}}+{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}+rS{\frac {\partial V}{\partial S}}-rV=0$$

b) https://www.quantstart.com/articles/C-Explicit-Euler-Finite-Difference-Method-for-Black-Scholes $$-{\frac {\partial V}{\partial t}}+{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}+rS{\frac {\partial V}{\partial S}}-rV=0$$

Please see below derivation of Black-Scholes PDE:

\begin{align*} \Pi &= C(S,t)+\Delta S\\ d\Pi &=dC(S,t)+\Delta dS\\ &= C_S dS+C_t dt + \frac{1}{2}C_{SS}d[S]+ \frac{1}{2}C_{tt}d[t]+ C_{St}d[S,t]+\Delta dS\\ &= C_S(S\mu dt+S\sigma dW)+C_tdt+\frac{1}{2}C_{SS}S^2\sigma^2dt+\Delta (S\mu dt+S\sigma dW)\\ &= (\Delta S\sigma+C_SS\sigma)dW+(C_SS\mu+C_t+\frac{1}{2}C_{SS}S^2\sigma^2+\Delta S\mu)dt \end{align*} Hedge: \begin{align*} d\Pi&\stackrel{!}{=}\Pi rdt\Rightarrow \Delta = -C_S:\\ &\Pi rdt= (C_t+\frac{1}{2}C_{SS}S^2\sigma^2)dt\\ &\Leftrightarrow(C-C_S S) rdt= (C_t+\frac{1}{2}C_{SS}S^2\sigma^2)dt \end{align*} $$\Rightarrow C_t+\frac{1}{2}C_{SS}S^2\sigma^2-r(C-C_S S)= 0$$

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    $\begingroup$ It depends whether you measure t forward (starts at 0 and increases to T) or backwards (time to maturity dwindles down to zero as maturity approaches). In the former case you have a minus sign in front of DV/dt. $\endgroup$ – Alex C Dec 8 '19 at 3:31
  • $\begingroup$ @AlexC I added a derivation of the PDE. Please show where the sign change would be observed? $\endgroup$ – emcor Dec 8 '19 at 3:41
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Your derivation is right and as Alex said, the only difference between the equations is how you measure time. There are two possibilities

  • Time going forward, $t\in[0,T]$,
  • Time going backwards, $\tau\in[T,0]$, i.e. $\tau=T-t$.

And clearly, after the change of variables, $\frac{\partial V}{\partial t}=-\frac{\partial V}{\partial\tau}$. Amongst others, this change of variables is employed to transform the Black-Scholes PDE into the heat (diffusion) equation.

Note that using $t\in[0,T]$ means that the corresponding PDE needs to be solved subject to a terminal payoff condition $V(t=T,S_t)=\varphi(S_T)$. On the other hand, using the time-to-maturity $\tau$, the PDE is solved subject to an initial condition $V(\tau=0,S_\tau)=\varphi(S_T)$.

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  • $\begingroup$ @AlexC said "In the former case you have a minus sign in front of DV/dt" but it seems to be the opposite? $\endgroup$ – emcor Dec 8 '19 at 11:58
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    $\begingroup$ Yes, sorry, it is the opposite. I could not edit the comment after 5 minutes. $\endgroup$ – Alex C Dec 8 '19 at 16:38

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