What is the intuition behind the Black-Scholes Equation?

Question

Consider the Black-Scholes equation for a European Call Option, \begin{equation} \begin{cases}\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + r\frac{\partial V}{\partial S} -rV = 0, \ &\text{for} \ (S,t)\in\mathbb{R}^+\times[0,T] \\ V(S,T) = \max(S-K,0), &\text{for} \ S\in\mathbb{R}^+ \\ V(0,t) = 0, &\text{for} \ t\in[0,T] \\ V(S,t) = S - Ke^{-r(T-t)}, &\text{as} \ S\rightarrow \infty, t\in[0,T] \end{cases} \end{equation} where $\sigma$ is the volatility of the underlying (the stock), $r$ is the interest rate, $K$ is the strike price, $T$ is the maturity time of the option, $S$ is the current stock price, and $V(S,t)$ is the value of the option.

I have few questions regarding the model. To start, why does the Black-Scholes model use a final condition at $t = T$, rather than using an initial condition, and why does it solve backward in time? For example, let us say we would like to find the value of a European Call option with a maturity time, $T=1$ year, and a strike price of $K = 10$. From my understanding, Black-Scholes should solve the value of $V(S,t)$, for all $t\in[0,T)$, for the current stock price $S$. My confusion is as to why can we let $S$ tend towards infinity if we do not know the future stock price from the get-go, and hence how would we know the value of $V(S,1 \ \text{year}) = \max(S-10,0), \text{for} \ S\in\mathbb{R}^+$? If we solve the equation numerically, we must set an upper bound to our stock price array, but how do we know this upper bound? Thus is there a formal way we may solve for this upper bound? If so how and why? If we solve the Black-Scholes equation, we will see that it solves backward in time. Hence we solve for the value of $V(S,t), \text{for} \ t\in[0,T)$. To me, this seems almost useless since European options may only be exercised at maturity time. Therefore why do we care to solve for the value of $V(S,t), \text{for} \ t\in[0,T)$?

Answer 1

Why would you know the value of the option at $T$? Because at that time there is no uncertainty anymore about its value. If at time $T$ the stock price has a value $S$, then the value of the Europen call option is by definition $\max(S-K,0)$. Because at that point you can hedge the option by selling the stock and receiving $K$ for it, effectively paying $S-K$ if the person having the option exercises it. If it doesn't exercise the option, you don't have to hedge it hence you pay $0$.

But at any time before $T$, you don't know what the value of the stock will be at $T$, hence an uncertainty exists that has to be factored into the value of the option, and that's what the Black-Scholes equation does, by weighing the price by the probabilities of all the possible paths toward a final stock price from the current stock price at $t$.