 Solution

Under the given risk-neutral measure, the option price is $V(t,S)=e^{-r(T-t)}\mathbb{E}[\varphi (S_T)\mid S_t=S]$. Thus $e^{-rt}V(t,S_t)$ is a martingale. This valuation also follows from delta hedging: cancel the Brownian exposure with $V_S$ units of stock, after which absence of arbitrage makes the locally riskless remainder earn $r$.

Set $f(t,b)=e^{-rt}V(t,S_0{e}^{\sigma b+(r-{\sigma }^2/2)t})$. The stated Brownian martingale identity and the chain rule giveThe drift vanishes because discounted option value is a martingale. Consequently the Black-Scholes equation isFor bounded measurable payoffs, the conditional-expectation solution is smooth for $t<T$; terminal values are interpreted in the usual limiting sense at continuity points, or in the bounded measurable pricing sense.