Chain Rule for Black–Scholes Variable Transformations
Summary
The post explains why transforming the Black–Scholes partial differential equation requires the chain rule when a new spatial variable depends on time. The key is to track the dependencies explicitly: the transformed function can be viewed as U(ξ(τ), τ). Differentiating with respect to τ must then account for both the direct time dependence of U and the change in ξ as τ changes. Treating ξ as independent of τ omits that contribution and can produce an incorrect time derivative.
The answer also clarifies that the transformed time variable still has a derivative: changing variables does not make the function’s value constant in time. The discussion is conceptual and points to the chain rule rather than working through a full Black–Scholes-to-heat-equation derivation. It does not reproduce the original substitution or compare the resulting equations step by step, so readers need that context to apply the explanation to a specific transformation.
Key ideas
- When a transformed spatial variable depends on time, the time derivative must include its contribution through the chain rule.
- Writing a function with its variable dependencies visible helps distinguish direct time dependence from dependence through a transformed coordinate.
- A change of variables does not make the derivative with respect to the new time variable disappear.
- Ignoring a time-dependent coordinate can give an incorrect transformed partial differential equation.
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# When to use total derivative and when not to?
# When to use total derivative and when not to?
as I was trying to teach myself financial mathematics, I came across this topic on transforming black scholes pde to a heat equation. I had the exat same question as this post Black Scholes to Heat Equation - Substitution. The answer in that post answered the question perfectly well. However, I could not understand 2 parts.
Firsy is: Why was the total derivative used in the the transformation. When should we use total derivatives instead of just partial derivatives? Even the sign is written in partial derivatives?
For example, I do not understand why couldn't I just use $\frac{\partial U}{\partial \zeta} = \frac{\partial U}{\partial x} \frac{\partial x}{\partial \zeta}$ and then substitute in?
This approach seems to arrive for the same result for $\frac{\partial U}{\partial \zeta}$. But it arrives at different result for $\frac{\partial U}{\partial \tau}$, so I presume it is very wrong. I don't know what is wrong with it.
The second one is why would we need to take derivatives w.r.t $\tau$ at all. Since $\tau$ is just equal to $\tau$, i.e. $\tau = \tau$, what is the intuition behind $\frac{\partial U}{\partial \tau}$ changes? Why are we not just "copying down to the next line" $\frac{\partial U}{\partial \tau}$?
Thank you for any helps in advance. I hope I formulated my question clearly. I am sorry I tried to comment on the original post but I couldn't as I am a new joiner to the forum, so I had to duplicate the question a bit.
## Answer by THATS MY QUANT MY QUANTITATIVE (score 1)
https://quant.stackexchange.com/a/77123
If I understand your questions correctly, $\xi$ is a function of $\tau$ so you need to apply chain rule. Therefore, the rate of change of $U$ wrt $\xi$ also depends on the rate of change wrt $\tau$. Read more on the chain rule wikipedia page
To your 2nd question, we have done a change of variable of $\xi \to x$. But our $\xi$ is also dependent on $\tau$, that is $\xi(\tau)$.
A lot of your confusion can be solved if you write your variables with their dependencies. So here, $U(\xi(\tau),\tau)$.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.