I know the general Black-Scholes formula for Option pricing theory (for calls and puts), however I want to know the other solutions to the Black-Scholes PDE and its various boundary conditions. Can someone start from the B-S PDE and derive its various solutions based on different boundary conditions? Even if you could provide some links/sources where it is done, I'll appreciate that. The point is that I want to know various other solutions and their boundary conditions which are derived from Black-Scholes PDE. Thank you.
[Math] Black Scholes PDE and its many solutions
financepartial differential equationsreference-request
Related Solutions
I think what you say seems correct. You just need to compute the conditional expectation (under the risk neutral measure) of $(S(T)-K)_+ 1_{\{S(T_0)>K_0\}}$. Now depending on whether $t<T_0$ or $t>T_0$ you can take out the indicator function out of the expectation or not by measurability. In the case that you can not things get trickier but not a big deal since you can separate $(S(T)-K)_+$ as $$(S(T)-K)_+ =(S(T)-K) 1_{\{S(T)>K\}}$$ And now the product of indicator functions is the indicator that both events occur:
$$1_{\{S(T)>K\}}1_{\{S(T_0)>K_0\}}=1_{\{S(T)>K,S(T_0)>K_0\}}$$ From here you may use the same trick that $S(T)/S(t)$ and $S(T_0)/S(t)$ are independent of $S(t)$. But, careful, they are not independent of each other. So you may need to condition your expectation on both $S(T_0)$ and $S(t)$ (the latter only if you want to find the price at any time $t$). Maybe also by writing: $S(T)=S(T)/S(T_0)* S(T_0)/S(t)*S(t)$ and use independence to compute the expectation by conditioning. Hope it helped.
I add more details here but I assume $r=0$ for notational simplicity, you have to add $e^{-t(T-t)}$ for discounting: If $0<T_0<t<T$ then $1_{\{S(T_0)>K_0\}}$ has been observed and we have under the risk neutral measure $Q$: $$p(t) = E_Q [(S(T)-K)_+ 1_{\{S(T_0)>K_0\}}|\mathcal{F}_t]=1_{\{S(T_0)>K_0\}}E_Q [(S(T)-K)_+|\mathcal{F}_t]$$ which is either 0 or equal to the usual Black-Scholes price if $S(T_0)>K_0$ obviously. Hence, the interesting part is when $0<t<T_0<T$ and $S(T_0)$ has yet not been observed. Observe that $$\frac{S(T)}{S(t)} = e^{\left(\mu-\frac{1}{2}\sigma^2\right)(T-t)+\sigma (W_T-W_t)},$$ where $W$ is the driving Brownian noise in $S$. In the above you can see that $W_T-W_t$ is independent of $W_t$ and hence $S(T)/S(t)$is independent of $S(t)$ for any $0\leq t<T$. We can thus condition on $S(T_0)$ and use this fact:
\begin{align*} p(t) &= E_Q [(S(T)-K)_+ 1_{\{S(T_0)>K_0\}}|\mathcal{F}_t]=E_Q [(S(T)-K) 1_{\{S(T)>K,S(T_0)>K\}}|\mathcal{F}_t]\\ &=E_Q [S(T)1_{\{S(T)>K,S(T_0)>K_0\}}|\mathcal{F}_t]-KE_Q [ 1_{\{S(T)>K,S(T_0)>K_0\}}|\mathcal{F}_t] \\ \end{align*} I explain the first one a little bit then you can do the second term: now writing $S(T)=\frac{S(T)}{S(t)}S(t)$ and $S(T_0)=\frac{S(T_0)}{S(t)}S(t)$ we can write: \begin{align*} E_Q [S(T)1_{\{S(T)>K,S(T_0)>K\}}|\mathcal{F}_t] = S(t) E_Q [\frac{S(T)}{S(t)}1_{\{\frac{S(T)}{S(t)}>\frac{K}{S(t)},\frac{S(T_0)}{S(t)}>\frac{K_0}{S(t)}\}}|\mathcal{F}_t] \\ \end{align*} where I took $S(t)$ out of the expectation since $S(t)$ is $\mathcal{F}_t$-measurable. Now everything in the expectation is independent of $\mathcal{F}_t$ so we can just compute: \begin{align*} S(t) E_Q [\frac{S(T)}{S(t)}1_{\{\frac{S(T)}{S(t)}>\frac{K}{S(t)},\frac{S(T_0)}{S(t)}>\frac{K_0}{S(t)}\}}] \end{align*} which is a usual expectation. The rest is analogous to the derivation of the Black-Scholes formula via expectation: change from $Q$to $P$ (Girsanov) and use that $\log(\frac{S(T)}{S(0)})$ are normally distributed \begin{align*} S(t) E_Q [\frac{S(T)}{S(t)}1_{\{\frac{S(T)}{S(t)}>\frac{K}{S(t)},\frac{S(T_0)}{S(t)}>\frac{K_0}{S(t)}\}}]=S(t) E_Q [\frac{S(T)}{S(t)}1_{\{\log(S(T)/S(0))>\log(K/S(0)),\log(S(T_0)/S(0))>\log(K_0/S(0))\}}] \end{align*} and now you can proceed, but! careful here, because the normally distributed random variables $\log(S(T)/S(0))$ and $\log(S(T_0)/S(0))$ are not independent, so you may need to derive the bivariate density of $(\log(S(T)/S(0)),\log(S(T_0)/S(0)) )$ or use the trick $\log(S(T)/S(T_0)*S(T_0)/S(0))$ and $\log(S(T_0)/S(0))$ and here $S(T)/S(T_0)$ is independent of $S(T_0)/S(0)$ so you can condition on $S(T_0)/S(0)$ first, compute the expectation w.r.t. $S(T)/S(T_0)$ first letting $S(T_0)/S(0)$ be untouched, then average further $S(T_0)/S(0)$.
Olver writes on p. 129 that the time reversed heat equation having a negative coefficient in front of the $\partial^2_xu$ term is ill-posed. In fact, he explains very nicely in this earlier chapter that such an equation is numerically unstable: very small changes in the intitial data can produce arbitrarily large changes in the solution at an arbitrarily small time. The BS PDE (having a negative coefficient in front of the $\partial^2_xu$ term) is analogous to the ill-posed time reveresed heat equation.
For more than 20 years I was aware of the fact that BS is solved backwards in time, i.e., you start with final data and solve for the initial option price. And yes, final data are smoothed out quickly. Olver's chapter 4.1 is indeed marvelous. Will buy that book.
Best Answer
The Black-Scholes formula involving the standard normal distribution is specific to call or put options. The Black-Scholes formalism, relating the prices to random walks and PDE, works for pricing a European option with arbitrary payoff. For any boundary condition (except some artificial ones with incredibly rapid growth that makes the random walk expectations diverge) the price is the expected value of the option value at the time of maturity, the expectation taken over price paths of the underlying security whose "risk-adjusted" drift and volatility are calculated from the parameters assumed to govern the "physical" price movement. So one does not need the whole theory of random walks, just the probability density of where the risk-adjusted path will land at the time of maturity, and integrating the European option payoff against this density will give the answer. The Black-Scholes theory shows that the result satisfies the PDE.
Alternatively, Black Scholes is, after a change of variables, equivalent to a "backwards" heat diffusion equation -- one whose time parameter is $(-t)$, or better, $(T_{m} - t)$ where $T_m$ is the maturity time of the option. So your question is the same as asking for all solutions of the heat equation. I assume that means heat diffusion with arbitrary boundary conditions. As long as the boundary conditions do not grow extremely fast the expectation of a random walk gives the solution.