Rewriting the Black–Scholes Equation in Divergence Form

Summary

The document explains how the standard Black–Scholes partial differential equation can be rearranged so that its second-order price term appears as the derivative of a product. Applying the product rule to the diffusion term creates an additional first-derivative term, which is offset by the corresponding derivative subtracted inside the adjusted drift coefficient. The rearrangement therefore preserves the original equation while expressing part of it in divergence form.

The evidence is an algebraic expansion of the product derivative, rather than a pricing example or empirical result. This identity can help readers compare equivalent formulations or recognize the structure used in some analytical and numerical treatments of differential equations. The note assumes the volatility parameter is constant with respect to the underlying price when differentiating. It does not discuss boundary conditions, solution methods, or how the reformulation affects numerical stability.

Key ideas

• Applying the product rule to the diffusion term produces an extra first-derivative contribution.
• The adjusted drift term cancels the contribution introduced by the product derivative.
• The two displayed forms are algebraically equivalent under the stated constant-volatility setup.
• The note gives an identity, not a new option-pricing result or numerical method.

Tags

# Black-Scholes differential equation rewritten


# Black-Scholes differential equation rewritten












I have seen that the Black-Scholes equation

$$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}+ rS\frac{\partial V}{\partial S}-rV=0$$

can also be written in the following form:

$$\frac{\partial V}{\partial t}+\frac{\partial}{\partial S} \left( \frac{1}{2}\sigma^2S^2\frac{\partial V}{\partial S}\right)+\left(rS-\frac{\partial}{\partial S}\left(\frac{1}{2}\sigma^2S^2\right)\right)\frac{\partial V}{\partial S}-rV=0. $$

How are the following two terms equal?

$$ \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}+ rS\frac{\partial V}{\partial S} = \frac{\partial}{\partial S} \left( \frac{1}{2}\sigma^2S^2\frac{\partial V}{\partial S}\right)+\left(rS-\frac{\partial}{\partial S}\left(\frac{1}{2}\sigma^2S^2\right)\right)\frac{\partial V}{\partial S} $$

## Answer by Daneel Olivaw (score 0, accepted)

https://quant.stackexchange.com/a/69953

Note that: $$ \frac{\partial}{\partial S} \left( \frac{1}{2}\sigma^2S^2\frac{\partial V}{\partial S}\right) =\sigma^2S\frac{\partial V}{\partial S}+ \frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial S^2} $$ The first term above cancels with the second term in the second bucket of brackets.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.