Good evening,

I'm currently working on the following problem and I would like an opinion on it,

Let's consider the Black-Scholes model with (time-varying) volatility, $\sigma = \sigma(t)$, and (time varying) risk free return rate,$r=r(t)$.

$$ V_t + \frac{\sigma^2(t)}{2}S^2 V_{SS} + r(t)V_S-r(t)V = 0 \space, \space S>0,\space 0<t<T $$

And the following final condition: $$V(S,T) = \phi(S)\space , \space S>0$$ where $\phi$ represents the option's payoff.

I started by considering the following variable change, $$ S = e^x$$ $$ t = T - \theta $$ This allowed me to consider the following functions:

$$ U(x,\theta) = V(e^x,T-\theta) \space,\space \hat\sigma(\theta) = \sigma(T-\theta) \space,\space \hat r(\theta) = r(T-\theta) $$ This also turned my final condition into an initial condition, $U(x,0) = \phi(e^x) $, and I derived the following transformation.

$$ U_{\theta} = \frac{\hat\sigma^2(\theta)}{2}U_{xx} + \Big(\hat r(\theta) - \frac{\hat\sigma^2(\theta)}{2}\Big)U_x - \hat r(\theta)U \space,\space x \in \mathbb{R} \space,\space 0 < \theta < T $$

Then, I introduced a new time variable, $$ \tau(\theta) = \frac{1}{2} \int_{0}^{\theta} \hat\sigma^2(\xi)d\xi$$ I managed to prove that this function is a bijection from an interval $[0,T]$ to an interval $[0,\Upsilon]$. Therefore, $\tau$ is invertible and we can have $\theta = \theta(\tau)$

With this new time variable, I defined the following functions, $$ R(\tau) = \hat r(\theta(\tau)) \space,\space \Sigma(\tau) = \hat\sigma(\theta(\tau))$$ Which then allowed me to to define $$ k(\tau) = 2 \frac{R(\tau)}{\Sigma^2(\tau)} $$

Given $u(x,\tau) = U(x,\theta(\tau))$, I derived the following new equation, $$u_{\tau} = u_{xx} + (k(t)-1)u_x -k(t)u \space,\space x \in \mathbb{R} \space,\space 0 < \tau < \Upsilon $$

I then defined the following "updating factor", $$d(\tau) = e^{{\int_{0}^{\tau}k(\xi)d\xi}} $$ and a new function $$ v(x,\tau) = d(\tau)u(x,\tau) $$ This new function allowed me to derive the following transformation, $$v_\tau = v_{xx} + (k(t)-1)v_x \space,\space x \in \mathbb{R} \space,\space 0 < \tau < \Upsilon $$

I then solved the following PDE problem,

$$ \psi_\tau = (k(t)-1)\psi_x \space,\space x \in \mathbb{R} \space,\space 0<\tau<\Upsilon $$ $$ \psi(x,0) = x $$

This problem has the following solution, $$\psi(x,\tau) = x + \int_{0}^{\tau} k(\xi)-1 d\xi $$

With this $\psi$ solution, with $\psi = y$, I made a new transform with the following function, $$v(x,\tau) = w(\psi(x,\tau),\tau) $$ This transformation allowed me to achieve the heat equation, $$w_\tau = w_{yy} $$ With the initial condition, $$w(y,0) = \phi(e^y)$$

Having all of these transforms and functions, my main goal is to solve the first problem, given all this information above.

$$ V_t + \frac{\sigma^2(t)}{2}S^2 V_{SS} + r(t)V_S-r(t)V = 0 \space, \space S>0,\space 0<t<T $$ $$V(S,T) = \phi(S)\space , \space S>0$$

My question here is the following: should I start by solving the heat equation and reversing each transform one by one? Or is there a simpler way to solve this Black-Scholes equation?

In order not to have this post being twice as long as it is, I won't explicit any reasoning behind these proofs.

I was looking forward into having some kind of clue in order to have a starting point, because I'm really lost in all of this "mess". I really appreciate if you have read this far, and I apologize for the long post.

Thank you!

  • $\begingroup$ Please, explain between option and option payoff. $\endgroup$
    – Cloud Cho
    Apr 17, 2021 at 6:51
  • 1
    $\begingroup$ If the payoff is path-independent and has no early exercise feature you can use Carr-Madan decomposition where you don't even have to solve the PDE explicityly. $\endgroup$
    – user34971
    Apr 17, 2021 at 9:19
  • $\begingroup$ @CloudCho The option payoff is just the profatibility of the option under various price conditions. Meanwhile, options are just the contracts/derivatives based on underlying assets that give the buyer the option to either buy it, or sell it, but not the obligation. $\endgroup$
    – Daniel F.
    Apr 17, 2021 at 12:48
  • $\begingroup$ @FridoRolloos I've looked into the Carr-Madan before this post, and I believe I have to solve this without the Carr-Madan decomposition, since it isn't something that I learned during this semester. $\endgroup$
    – Daniel F.
    Apr 17, 2021 at 12:50
  • $\begingroup$ @DanielF According to your explanation, the difference between Option and Option Payoff won't make difference in equation, right? $\endgroup$
    – Cloud Cho
    Apr 19, 2021 at 18:31

2 Answers 2


Yours is the (backward Kolmogorov) PDE of a Black-Scholes model with time-varying short rate and volatility. Now, have you considered at all risk-neutral evaluation and Feynman-Kač representation? See e.g. Bjork chap 5.

Because, the infinitesimal generator of that PDE is the same of the following SDE:

$$ dS(t) = r(t) S(t)dt + \sigma(t) S(t) dW(t) $$

where $W(t)$ is standard Brownian motion. If your $\sigma(t)$ is deterministic (or at least adapted to the $W(t)$-filtration), you still get that $S(T)$ is conditionally (to $S(t)$) log-normally distributed with an effective volatility factor proportional to the integrated rate of variance

$$ \int^T_t\sigma^2(s) ds $$

See e.g. bjork eq. (26.33). Something similar for the time-varying interest rate contribution. Basically, I think you are trying to solve a PDE which in fact you don’t really need to solve, since once you know the terminal conditional distribution of the underlying $S(T)$, say $p(S)$, you can effectively price any (European exercised) payoff integrating it

$$ \int^{\infty}_0 \phi(S) p(S) dS $$

and the solution of that integral is guaranteed to be (a) solution of your PDE. Am I missing something here? Personally I still like your deduction of the heat equation.

  • $\begingroup$ I appreciate your answer, and I understand everything you said. However, this is a exercise for a class in introduction to PDE's in Finance, and I don't think that my professor doesn't want a solution that deep, especially since more than half of my class has almost zero math background. I did talk about the Feynman-Kac, but only in the context of heat equations. Given what you said, should I solve the heat equation with Feynman-Kac and reverse all transformations? $\endgroup$
    – Daniel F.
    Apr 17, 2021 at 15:27
  • 1
    $\begingroup$ Given what I said, it’s straightforward to use F-K to solve directly the original PDE. But it seems your professor want you to solve the heat eq.. and who I am to unplease him 😉 $\endgroup$ Apr 17, 2021 at 20:49

What do you mean by solving it? A heat equation can be solved by a simple sin(x) exp(-x t) function as it will satisfy the equation.

I think what you really mean is satisfy the boundary conditions. Without the boundary condition there are various solutions possible. Assuming you are looking to solve for call option where boundary values are determined using Max(S-K,0) then you have to transfer this also to the heat equation form.

Heat equation solution can of course be taken from fourier series and N number of terms will satisfy it but you will have to calibrate it at the boundary points.

  • $\begingroup$ I was thinking of simply solving the final heat equation with a Fourier transform and then reverse all of the transformations above. Would that work? $\endgroup$
    – Daniel F.
    Apr 17, 2021 at 12:54
  • $\begingroup$ As I said what is your boundary condition? $\endgroup$ Apr 17, 2021 at 14:31

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