Second, after truncating the domain, we need to ask what boundary values we
should pose at this now finite boundary. To take care of this, we use "put-call"
-parity @cite stoll1969relationship. A "pull option" is one in which I am
+parity @cite stoll1969relationship. A "pull option" is one in which we are
allowed, but not required, to *sell* a stock at price $K$ to someone at a future
time $T$. This says
@f{align*}{
The second complication of the Block-Scholes equation is that we are given a
final condition, and not an initial condition. This is because we know what the
-option is worth at time $t=T$: If the stock price at $T$ is $S<K$, then I have
-no incentive to use my option of buying a price $K$ because I can buy that stock
+option is worth at time $t=T$: If the stock price at $T$ is $S<K$, then we have
+no incentive to use our option of buying a price $K$ because we can buy that stock
for cheaper on the open market. So $V(S,T)=0$ for $S<K$. On the other hand, if
-at time $T$ we have $S>K$, then I can buy my stock at price $K$ via the option
+at time $T$ we have $S>K$, then we can buy the stock at price $K$ via the option
and immediately sell it again on the market for price $S$, giving me a profit of
$S-K$. In other words, $V(S,T)=S-K$ for $S>K$. So, we only know
values for $V$ at the *end time* but not the initial time -- in fact, finding
// @sect3{Solution Class}
// This section creates a class for the known solution when testing using the
- // MMS. Here I am using $v(\tau,S) = -\tau^2 -S^2 + 6$ for my solution. We
+ // MMS. Here we are using $v(\tau,S) = -\tau^2 -S^2 + 6$ for the solution. We
// need to include the solution equation and the gradient for the H1 seminorm
// calculation.
template <int dim>