Transforming Black–Scholes into a Heat Equation
Summary
The document asks how logarithmic price and time-to-maturity substitutions change derivatives in the Black–Scholes partial differential equation, and how a further change of variables removes its first-order spatial term. The chain rule gives \(V_S=V_y/S\) and \(V_{SS}=(V_{yy}-V_y)/S^2\). Exponential rescaling removes the discount term, while shifting the log-price coordinate by a drift-dependent amount changes the time derivative through the chain rule. These substitutions are standard tools for converting option-pricing equations into the heat-equation form.
The answer correctly points to the chain rule and notes that the shifted coordinate has unit derivative with respect to log price, but its displayed second-derivative calculation is inconsistent: it arrives at a coefficient of minus two for \(V_y/S^2\), whereas the correct coefficient is minus one. The stated first-order transformed equation therefore follows from the correct formula, not from that answer's erroneous intermediate result. The response also does not show the final heat-equation form or discuss boundary conditions.
Key ideas
- The log-price substitution gives \(V_S=V_y/S\).\nApplying the chain rule twice yields \(V_{SS}=(V_{yy}-V_y)/S^2\).\nAn exponential rescaling removes the discount term from the transformed equation.\nA drift-adjusted coordinate shift changes the time derivative by the chain rule.\nThe answer contains an algebraic error in its displayed second-derivative calculation.
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# Question About Converting Black Scholes Differential Equation to Heat Equation
# Question About Converting Black Scholes Differential Equation to Heat Equation
I'm reading a book about converting Black Scholes equation to heat equation and I highlighted in bold for those I have doubts, and really appreciate your advice on it.
Let $S$,$T$,$V$ denote underlying asset price, maturity and option price separately. Here is the convert process:
Let $y=lnS$ since $(S=e^y)$ and $\tau_t=T-t$,then $\frac{\partial V}{\partial t}=-\frac{\partial V}{\partial \tau_t}$,$\frac{\partial V}{\partial S}=\frac{\partial V}{\partial y}\frac{\partial y}{\partial S}=\frac{1}{S}\frac{\partial V}{\partial y}$ and $\frac{\partial^2 V}{\partial S^2}=\frac{\partial }{\partial S}(\frac{\partial V}{\partial S})=\frac{\partial }{\partial S}(\frac{1}{S}\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S}\frac{\partial }{\partial S}(\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2}$,
here is my first doubt: why $\frac{1}{S}\frac{\partial V}{\partial S}(\frac{\partial V}{\partial y})=\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2}$ holds?
The Black Scholes equation $\frac{\partial V}{\partial t} + rS \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}-rV = 0$
can be converted to
$-\frac{\partial V}{\partial \tau_t} + (r-\frac{1}{2}\sigma^2) \frac{\partial V}{\partial y} + \frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial y^2}-rV = 0$
Let $u=e^{r\tau_t}V$,
the equation becomes
$-\frac{\partial u}{\partial \tau_t} + (r-\frac{1}{2}\sigma^2) \frac{\partial u}{\partial y} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial y^2} = 0$
Finally, let
$x=y+(r-\frac{1}{2}\sigma^2)\tau_t=lnS+(r-\frac{1}{2}\sigma^2)\tau_t$
and
$\tau=\tau_t$, then $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial x}$
and
$\frac{\partial u}{\partial \tau_t}=\frac{\partial u}{\partial \tau}+(r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x}$,
here is my second doubt: why $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial \tau_t}=\frac{\partial u}{\partial \tau}+(r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x}$ hold?
## Answer by Jónás Balázs (score 1, accepted)
https://quant.stackexchange.com/a/50526
The first part of your question:
- $\frac{\partial y}{\partial S} = \frac{\partial ln S}{\partial S} = \frac{1}{S}$
- $ \frac{\partial^2 V}{\partial S \partial y} = \frac{\partial}{\partial y} \frac{\partial V}{\partial S} = \frac{\partial}{\partial y} (\frac{\partial y}{\partial S}\frac{\partial V}{\partial y})= \frac{\partial}{\partial y} (\frac{1}{ S}\frac{\partial V}{\partial y}) = \frac{-1}{S^2} \frac{\partial S}{\partial y}\frac{\partial V}{\partial y} + \frac{1}{S}\frac{\partial^2 V}{\partial y^2} = \frac{-1}{S}\frac{\partial V}{\partial y} + \frac{1}{S}\frac{\partial^2 V}{\partial y^2} $
- $ \frac{\partial^2 V}{\partial S^2} = \frac{\partial}{\partial S} (\frac{\partial V}{\partial y}\frac{\partial y}{\partial S}) = \\ \frac{\partial^2 V}{\partial S \partial y} \frac{\partial y}{\partial S} + \frac{\partial V}{\partial y} \frac{\partial^2 y}{\partial S^2} = \\ \frac{\partial^2 V}{\partial S \partial y} \frac{1}{S} - \frac{1}{S^2}\frac{\partial V}{\partial y} = \\ \frac{-1}{S^2}\frac{\partial V}{\partial y} + \frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} -\frac{1}{S^2}\frac{\partial V}{\partial y} = \\ \frac{-2}{S^2}\frac{\partial V}{\partial y} + \frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} $
The key for the second part is that $\frac{\partial x}{\partial y}$ is 1.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.