European Options and the Heat Equation
September 21, 2026
1. The heat equation
Last time we derived the Black-Scholes PDE for a European call option. Unfortunately, PDEs are tough to solve! Luckily, we can fool around with the Black-Scholes equation and turn it into the heat equation after a few clever changes of variables.
The heat equation for a function \(u(x,t)\) is the famous PDE
\[\begin{equation} \label{eq:heat} \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}. \tag{1} \end{equation}\]It has the well-known solution
\[\begin{equation} \label{eq:heat-soln} u(x, \widetilde \tau) = \frac{1}{\sqrt{4\pi \widetilde\tau}}\int_{-\infty}^\infty \exp\left\{-\frac{(x -y)^2}{4\widetilde\tau}\right\} u_0(y)dy,\tag{2} \end{equation}\]where \(u_0(y) = u(y,0)\) is the initial condition at time 0. The heat equation was originally developed by Fourier in the 1800s to describe how heat diffuses in a given region. The quantity \(u(x,t)\) is the temperature at location \(x\) at time \(t\), and the equation says that how quickly the temperature is changing at point \(x\) is determined by the convexity or concavity of the temperature profile at that point. (This is an interesting phenomenon in and of itself, and not at all obvious.) In fact, there is nothing particularly special about heat: \eqref{eq:heat} models many diffusion processes after appropriate scaling.
Is it a surprise then that the Black-Scholes equation is related to the heat equation? Perhaps initially, but less and less as you think about it. The Black-Scholes PDE describes the price of a derivative whose underlying stock fluctuates as geometric Brownian motion—a specific type of diffusion process—and whose value is known at some time \(T\). While \(T\) happens to be the end of the time period instead of the beginning, this doesn’t matter much analytically. It just means that time is reversed.
2. Transforming Black-Scholes into the Heat Equation
Recall the Black-Scholes equation which we derived last time,
\[\begin{equation} \label{eq:black-scholes} \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS \frac{\partial V}{\partial S} - rV = 0, \tag{3} \end{equation}\]with terminal condition \(V(S,T) = \max\{0, S-K\}\) for some strike price \(K>0\). As we noted, this thing moves backwards in time. We know the value of \(V(S,T)\) and we’re trying to understand \(V(S,t)\) for \(t<T\). We can fix this by considering \(\tau = T-t\), the time until the payout. Reusing \(V\) for the function \(V(S,T-\tau)\), we have \(\partial V/\partial t= -\partial V/\partial \tau\), so \(\eqref{eq:black-scholes}\) becomes
\[\begin{equation} \label{eq:black-scholes-2} \frac{\partial V}{\partial \tau} = \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS \frac{\partial V}{\partial S} - rV. \tag{4} \end{equation}\]Next we would like to turn the coefficients into constants. The reason we have non-constant coefficients in \(\eqref{eq:black-scholes-2}\) is that the European option model considers geometric (i.e., multiplicative) Brownian motion, which suggests that we should instead work in log space.
Set \(x = \log S\). We want to write \(\eqref{eq:black-scholes-2}\) in terms of \(x\) instead of \(S\). The chain rule gives
\[\begin{equation} \label{eq:dvds} \frac{\partial V}{\partial S} = \frac{\partial V}{\partial x}\frac{\partial x}{\partial S} = \frac{\partial V}{\partial x}\frac{1}{S}, \tag{5} \end{equation}\]and
\[\begin{align} \frac{\partial^2 V}{\partial S^2} &= \frac{\partial}{\partial S}\left(\frac{\partial V}{\partial x}\frac{1}{S}\right) = \frac{\partial^2 V}{\partial S\partial x} \frac{1}{S} - \frac{1}{S^2}\frac{\partial V}{\partial x} \\ &= \frac{\partial^2 V}{\partial x^2}\frac{\partial x}{\partial S} \frac{1}{S} - \frac{1}{S^2}\frac{\partial V}{\partial x} = \frac{1}{S^2}\left(\frac{\partial^2 V}{\partial x^2} - \frac{\partial V}{\partial x}\right). \label{eq:dvds2} \tag{6} \end{align}\]Plugging \(\eqref{eq:dvds}\) and \(\eqref{eq:dvds2}\) back into \(\eqref{eq:black-scholes-2}\) we obtain
\[\begin{align} \frac{\partial V}{\partial \tau} &= \frac{1}{2}\sigma^2 S^2 \cdot \frac{1}{S^2}\left(\frac{\partial^2 V}{\partial x^2} - \frac{\partial V}{\partial x}\right) + rS \cdot \frac{1}{S}\frac{\partial V}{\partial x} - rV \\ &= \frac{1}{2}\sigma^2 \frac{\partial^2 V}{\partial x^2} + \left(r - \frac{1}{2}\sigma^2\right)\frac{\partial V}{\partial x} - rV. \label{eq:constant-coeff} \tag{7} \end{align}\]Ok, so now the coefficients on the derivatives are constant. But we still have too many terms floating around. The heat equation has no first-order term and no zeroth-order term, whereas \(\eqref{eq:constant-coeff}\) has both. To get rid of them, we’re going to define a new function \(u\) such that, for some \(\alpha,\beta\) to be determined,
\[\begin{equation} \label{eq:u-def} u(x,\tau) = e^{-\alpha x - \beta \tau} V(x,\tau). \tag{8} \end{equation}\]We can argue about whether this is an intuitive thing to do or not. The rough idea is that an exponential factor is precisely the sort of thing that spits out first-order and zeroth-order terms when you differentiate it, so we ought to be able to tune \(\alpha\) and \(\beta\) to cancel things we don’t want. But admittedly this is hard to see by just staring at \(\eqref{eq:u-def}\). Regardless, we have
\[\begin{align} \frac{\partial V}{\partial \tau} &= e^{\alpha x + \beta \tau}\left(\frac{\partial u}{\partial \tau} + \beta u\right), \\ \frac{\partial V}{\partial x} &= e^{\alpha x + \beta \tau}\left(\frac{\partial u}{\partial x} + \alpha u\right), \\ \frac{\partial^2 V}{\partial x^2} &= e^{\alpha x + \beta \tau}\left(\frac{\partial^2 u}{\partial x^2} + 2\alpha\frac{\partial u}{\partial x} + \alpha^2 u\right). \label{eq:u-derivs} \tag{9} \end{align}\]Substituting \(\eqref{eq:u-derivs}\) into \(\eqref{eq:constant-coeff}\) and dividing by \(e^{\alpha x + \beta\tau}\) (always nonzero), we obtain
\[\begin{align} \frac{\partial u}{\partial \tau} &= \frac{1}{2}\sigma^2 \frac{\partial^2 u}{\partial x^2} + \left(\sigma^2\alpha + r - \frac{1}{2}\sigma^2\right)\frac{\partial u}{\partial x} \\ &\quad + \left(\frac{1}{2}\sigma^2\alpha^2 + \left(r - \frac{1}{2}\sigma^2\right)\alpha - r - \beta\right)u. \label{eq:u-pde} \tag{10} \end{align}\]Now we just choose \(\alpha\) and \(\beta\) to kill the last two coefficients. Setting the coefficient on \(\partial u/\partial x\) to zero gives
\[\begin{equation} \label{eq:alpha} \alpha = \frac{1}{2} - \frac{r}{\sigma^2}. \tag{11} \end{equation}\]and then setting the coefficient on \(u\) to zero and simplifying gives
\[\begin{equation} \label{eq:beta} \beta = \frac{1}{2}\sigma^2\alpha^2 + \left(r - \frac{1}{2}\sigma^2\right)\alpha - r = -\frac{\sigma^2}{8}\left(1 + \frac{2r}{\sigma^2}\right)^2. \tag{12} \end{equation}\]With these choices, \(\eqref{eq:u-pde}\) becomes
\[\begin{equation} \label{eq:almost-heat} \frac{\partial u}{\partial \tau} = \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2}, \tag{13} \end{equation}\]and we can write it precisely as the heat equation by rescaling time as \(\widetilde\tau = \sigma^2\tau/2\). Since \(\partial/\partial\tau = (\sigma^2/2)\partial/\partial\widetilde\tau\), the factor of \(\sigma^2/2\) cancels off both sides and we’re left with \(\eqref{eq:heat}\) precisely.
To write the initial condition in terms of \(u_0\), note that \(\widetilde\tau = 0\) corresponds to \(\tau=0\), i.e., to \(t=T\), at which point \(V = \max\{0,S-K\} = \max\{0,e^x-K\}\). So by \(\eqref{eq:u-def}\),
\[\begin{equation} \label{eq:u0} u_0(y) = e^{-\alpha y}\max\{0, e^y - K\} = \max\left\{0, e^{(1-\alpha)y} - Ke^{-\alpha y}\right\}. \tag{14} \end{equation}\]The max is only positive for \(y > \log(K)\). Plugging this into the heat kernel solution \(\eqref{eq:heat-soln}\) and inverting \(\eqref{eq:u-def}\) to recover \(V\) gives
\[\begin{align} \label{eq:V-integral} V(S,t) &= \frac{e^{\alpha x + \beta \tau}}{\sqrt{4\pi\widetilde\tau}}\int_{\log K}^\infty \exp\left\{-\frac{(x-y)^2}{4\widetilde\tau}\right\} \left(e^{(1-\alpha)y} - Ke^{-\alpha y}\right)dy. \tag{15} \end{align}\]There are two ways to proceed from here. You can play around with this integral, complete the square in each term, and write it in terms of the standard normal CDF. The result is the famous Black-Scholes formula
\[\begin{equation} \label{eq:bs-formula} V(S,t) = S\Phi(d_1) - Ke^{-r(T-t)}\Phi(d_2), \tag{16} \end{equation}\]where \(\Phi\) is the standard normal CDF and
\[\begin{equation} \label{eq:d1d2} d_1 = \frac{\log(S/K) + \left(r + \frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}, \qquad d_2 = d_1 - \sigma\sqrt{T-t}. \tag{17} \end{equation}\]But there’s a second option that I much prefer, both because the solution looks a lot cleaner, and because it connects much more obviously with optimal stopping.
3. The solution as an expectation
Go back to \(\eqref{eq:V-integral}\), but keep \(u_0\) in the unexpanded form \(e^{-\alpha y}\max\{0,e^y-K\}\) from \(\eqref{eq:u0}\), and substitute \(z = y - x\). Since \(e^x = S\), the two factors of \(e^{\alpha x}\) cancel, leaving
\[\begin{align} V(S,t) &= \frac{e^{\beta\tau}}{\sqrt{4\pi\widetilde\tau}}\int_{-\infty}^\infty \exp\left\{-\frac{z^2}{4\widetilde\tau} - \alpha z\right\}\max\{0, Se^z - K\}\,dz \\ &= \frac{e^{\beta\tau + \widetilde\tau\alpha^2}}{\sqrt{4\pi\widetilde\tau}}\int_{-\infty}^\infty \exp\left\{-\frac{(z + 2\widetilde\tau\alpha)^2}{4\widetilde\tau}\right\}\max\{0, Se^z - K\}\,dz, \label{eq:tilted} \tag{18} \end{align}\]where the second line just completes the square in the exponent. The factor
\[\frac{1}{\sqrt{4\pi\widetilde\tau}}\exp\left\{-\frac{(z + 2\widetilde\tau\alpha)^2}{4\widetilde\tau}\right\},\]is the density of a Gaussian (in \(z\)) with mean \(m = -2\widetilde\tau\alpha\) and variance \(\upsilon^2 = 2\widetilde\tau\). Recalling that \(\widetilde\tau = \sigma^2\tau/2\) and \(\alpha = \frac{1}{2} - r/\sigma^2\), we can rewrite those as
\[\begin{equation} \label{eq:moments} \upsilon^2 = 2\widetilde\tau = \sigma^2\tau, \qquad m = -\sigma^2\tau\alpha = \left(r - \frac{1}{2}\sigma^2\right)\tau. \tag{19} \end{equation}\]Now, keeping in mind the definition of \(\alpha\) in \(\eqref{eq:alpha}\), note that \(\beta\tau + \widetilde\tau \alpha^2 = -r\tau\), so \(\exp(\beta\tau + \widetilde\tau\alpha^2) = \exp(-r\tau ) = \exp(-r(T-t))\). Therefore, \(\eqref{eq:tilted}\) simplifies to
\[\begin{align} V(S,t) &= e^{-r(T-t)}\,\mathbb{E}_{Y\sim N(m,\upsilon^2)}\left[\max\{0, Se^Y - K\}\right]\\ &= e^{-r(T-t)}\,\mathbb{E}_{Z\sim N(0,1)}\left[\max\{0, Se^{\upsilon Z + m} - K\}\right]. \tag{20} \end{align}\]This is commonly written as
\[\begin{equation} \label{eq:expectation} V(S,t) = e^{-r(T-t)}\,\mathbb{E}_{Z\sim N(0,1)}\left[\max\{0, S_T - K\}\mid S_t=S\right], \tag{21} \end{equation}\]where, conditional on \(S_t=S\),
\[\begin{equation} \label{eq:ST} S_T = S\exp\left\{\left(r - \frac{1}{2}\sigma^2\right)(T-t) + \sigma\sqrt{T-t}\,Z\right\}, \qquad Z \sim N(0,1). \tag{22} \end{equation}\]Why is this nicer? For one, it expresses the price as a discounted expected payoff, which is easier to think about intuitively. Second, it relates directly to the stopping-time formulation of the problem, which is what will come next. In particular, an American (as opposed to European) option can be exercised at any time up to \(T\). If \(\mathcal{T}_{t,T}\) denotes the set of stopping times taking values in \([t,T]\), its price is
\[\begin{equation} \label{eq:american} V(S,t) = \sup_{\theta\in\mathcal{T}_{t,T}} \mathbb{E}_{\mathbb{Q}}\left[e^{-r(\theta - t)}\max\{0, S_\theta - K\}\mid S_t=S\right]. \tag{23} \end{equation}\]Finally, \(\eqref{eq:expectation}\) has a nice relationship with the stock price process \(S_t\) that we started with. As we saw when solving geometric Brownian motion, solving \(dS_t = \mu S_t\,dt + \sigma S_t\,dB_t^{\mathbb{P}}\) conditional on \(S_t=S\) gives
\[S_T = S\exp\left\{\left(\mu - \frac{1}{2}\sigma^2\right)(T-t) + \sigma\sqrt{T-t}\,Z\right\}.\]This is the same expression as \eqref{eq:ST} but with \(\mu\) in place of \(r\). So \eqref{eq:expectation} can be interpreted as the expected payoff of a fictitious stock whose drift has been replaced by the “risk-free rate” \(r\), and then discounted by time (see the previous post for more on risk-free stocks).
