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In the notation of the original Black-Scholes paper, let $w(x, t)$ be the price of an option with underlying priced at $x$, and let $w_1$ denote the derivative of $w$ w.r.t. to $x$ and $w_2$ denote the derivative of $w$ w.r.t. to $t$. In Black-Scholes, they derive the differential equation for the value of an option as:

$$ w_2 = rw - rxw_1 - \frac{1}{2} v^2 x^2 w_{11} $$

But in the Black paper, the differential equation is

$$ w_2 = rw - \frac{1}{2} v^2 x^2 w_{11} $$

And Black's justification is:

Note that this is like the differential equation for an option on a security, but with one term missing. The term is missing because the value of a futures contract is zero, while the value of a security position is positive.

Notice that the term that is missing is, slightly more verbose notation:

$$ \frac{\partial w}{\partial x} x r \stackrel{???}{=} 0. $$

where now $x$ refers to the forward price.

My question is: why is this quantity zero? I understand that one does not pay for a futures contract up front, and that since it settles into the underlying, it is worth the underlying at expiration. Formally, this can't be that $x = 0$, though, because we see $x^2$ elsewhere! I also don't think delta of the Black model (here $w_1$) is zero. Is it that $xr = 0$?

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$w_1$ is also known as $\Delta$. And $x \Delta$ is the value of the hedging position, so $r x \Delta$ is the instantaneous cost of financing the hedging position. Since as you say futures have no financing cost, this particular $r$ can be set to zero (but the other $r$ which has a different interpretation cannot). Thus the whole term $-r x w_1$ can be left out of the equation.

In other words the term disappears because we set the appropriate interest rate to 0.

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