Transforming the Black-Scholes PDE to the Heat Equation

From Pricing PDE to Diffusion Equation

The Black–Scholes PDE is already linear, but its coefficients depend on the stock price and it contains both a first spatial derivative and a discount term. Log-price makes the spatial coefficients constant. Time-to-maturity puts the payoff at the initial time, and an exponential rescaling removes the remaining first-derivative and zeroth-order terms. The result is the heat equation, with the call payoff carried along as its initial condition.

Use a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. In the Black–Scholes market, pricing is under the risk-neutral measure $\mathbb{Q}$, not necessarily the physical measure $\mathbb{P}$. Under $\mathbb{Q}$, the stock has drift $r$ and diffusion coefficient $\sigma$, where $r$ and $\sigma>0$ are constant. Its Brownian motion under $\mathbb{Q}$ is denoted $B_t$ when the working measure is understood. In integral notation, its stochastic term is $\int_0^t \sigma S_s\,dB_s$, with progressively measurable integrand $\sigma S_s$ in $\mathcal{L}^2$ on every finite horizon. The PDE derivation below is deterministic; it uses the Black–Scholes PDE and does not require a new stochastic-integral calculation.

Let $C(t,S)$ be the price of a European call with strike $K>0$ and maturity $T$. For $0\le t<T$ and $S>0$, its Black–Scholes PDE and terminal payoff are

C_t+\frac12\sigma^2S^2C_{SS}+rSC_S-rC=0, \qquad C(T,S)=(S-K)^+.

Here $r$ is the risk-free rate, $\sigma$ is the diffusion coefficient in the stock dynamics, and $(S-K)^+=\max(S-K,0)$. The transformation below assumes $K>0$ and $\sigma>0$. If $\sigma=0$, the spatial diffusion term vanishes and this change of variables does not produce a heat equation.

Log-Price and Time-to-Maturity

Set $\tau=T-t$, $x=\ln(S/K)$, and define the dimensionless function $u$ by $C(t,S)=K u(\tau,x)$. Thus the terminal payoff becomes an initial condition at $\tau=0$. The derivative calculations must account for both changes of variables: $d\tau/dt=-1$ and $dx/dS=1/S$.

Definition: The log-price and time-to-maturity representation of the call price is $u(\tau,x)=C(T-\tau,Ke^x)/K$, for $0\le\tau\le T$ and $x\in\mathbb{R}$.

For clarity, the derivatives of $C(t,S)=K u(T-t,\ln(S/K))$ are

C_t=-K u_\tau,\qquad C_S=\frac{K}{S}u_x,\qquad C_{SS}=\frac{K}{S^2}(u_{xx}-u_x).

In particular, the second derivative is not just $K u_{xx}/S^2$: differentiating $1/S$ in $C_S=Ku_x/S$ contributes the term $-Ku_x/S^2$. Substituting these derivatives in the PDE and dividing by $K>0$ gives

u_\tau =\frac12\sigma^2u_{xx} +\left(r-\frac12\sigma^2\right)u_x-r u.

Write $D=\sigma^2/2$ and $m=r-\sigma^2/2$. The equation is $u_\tau=D u_{xx}+m u_x-r u$. Its diffusion coefficient $D$ is positive. The next step chooses an exponential factor to cancel the drift term $m u_x$ and the discount term $-ru$.

Choosing the Exponential Rescaling

Let $a$ and $b$ be constants and set $u(\tau,x)=e^{ax+b\tau}v(\tau,x)$. The derivatives are

u_\tau=e^{ax+b\tau}(v_\tau+bv),\qquad u_x=e^{ax+b\tau}(v_x+av),\qquad u_{xx}=e^{ax+b\tau}(v_{xx}+2av_x+a^2v).

Substitute all three expressions into the equation for $u$, then divide by the strictly positive factor $e^{ax+b\tau}$. Rearranging for $v_\tau$ yields

v_\tau =Dv_{xx}+(2Da+m)v_x+(Da^2+ma-r-b)v.

The coefficient of $v_x$ vanishes when $a=-m/(2D)=-m/\sigma^2$. With this choice, $Da^2+ma=-m^2/(2\sigma^2)$. Therefore choosing $b=-r-m^2/(2\sigma^2)$ also cancels the coefficient of $v$.

Theorem: Suppose $r$ is constant, $\sigma>0$, and $C$ is a classical solution of the Black–Scholes PDE for $S>0$, $t<T$. Define $$ a=-\frac{r-\sigma^2/2}{\sigma^2},\qquad b=-r-\frac{(r-\sigma^2/2)^2}{2\sigma^2},\qquad D=\frac{\sigma^2}{2}. $$ Then $v$, defined by $u(\tau,x)=e^{ax+b\tau}v(\tau,x)$, satisfies $v_\tau=Dv_{xx}$. Conversely, any classical solution $v$ of this equation gives a solution of the Black–Scholes PDE through these changes of variables in reverse.

Proof. The log-price calculation gives $u_\tau=D u_{xx}+m u_x-r u$, where $m=r-\sigma^2/2$. For $u=e^{ax+b\tau}v$, substitution gives $v_\tau=Dv_{xx}+(2Da+m)v_x+(Da^2+ma-r-b)v$. The stated $a=-m/\sigma^2$ makes $2Da+m=\sigma^2a+m=0$. Also $Da^2+ma=(\sigma^2/2)(m^2/\sigma^4)-m^2/\sigma^2=-m^2/(2\sigma^2)$, so the stated $b$ makes $Da^2+ma-r-b=0$. Hence $v_\tau=Dv_{xx}$. Conversely, if $v_\tau=Dv_{xx}$, the same displayed derivative identities, substituted into the expression for $v_\tau$, give $u_\tau=D u_{xx}+m u_x-r u$; the log-price derivative identities then give the original Black–Scholes PDE. These implications hold wherever the functions have the derivatives used in the calculations. $\square$

To obtain the standard heat equation with coefficient one, define the scaled time $s=D\tau$ and write $V(s,x)=v(s/D,x)$. By the chain rule, $V_s=v_\tau/D=v_{xx}=V_{xx}$. Thus $V_s=V_{xx}$, with $0\le s\le DT$. The equivalent form $v_\tau=Dv_{xx}$ is also called the heat equation, with diffusivity $D$.

Transforming the Call Payoff

The payoff in log-price coordinates is $u(0,x)=(e^x-1)^+$, because $S=Ke^x$ and the price has been divided by $K$. At $\tau=0$, the factor $e^{b\tau}$ equals one. Since $v=e^{-ax-b\tau}u$, the transformed initial condition is $V(0,x)=v(0,x)=e^{-ax}(e^x-1)^+$.

Proposition: The heat-equation initial condition corresponding to the European call payoff is $$ V(0,x)= \begin{cases} 0,&x\le0,\\ e^{(1-a)x}-e^{-ax},&x>0, \end{cases} \qquad a=-\frac{r-\sigma^2/2}{\sigma^2}. $$ The value at $x=0$ is zero.

Proof. If $x\le0$, then $e^x\le1$, so $(e^x-1)^+=0$, and multiplying by $e^{-ax}$ still gives zero. If $x>0$, then $e^x-1>0$, so $e^{-ax}(e^x-1)=e^{(1-a)x}-e^{-ax}$. At $x=0$, the original payoff is $(1-1)^+=0$, so the piecewise definition assigns zero there. The exponential multiplier is finite and positive for each real $x$, so it does not change which side of the strike has zero payoff. $\square$

Worked Example: Transforming the PDE When the Log-Drift Vanishes

Take $r=0.08=2/25$ and $\sigma=0.4=2/5$. Compute the transformation parameters, transformed payoff, and heat-equation time horizon.

First, $\sigma^2=4/25$, so $D=\sigma^2/2=2/25$. The log-price drift is $m=r-D=2/25-2/25=0$. Consequently $a=-m/\sigma^2=0$, and $b=-r-m^2/(2\sigma^2)=-2/25$. The equation in $u$ is therefore $u_\tau=(2/25)u_{xx}-(2/25)u$, and $u=e^{-2\tau/25}v$ gives $v_\tau=(2/25)v_{xx}$.

The standard heat time is $s=D\tau=(2/25)\tau$, so the full horizon is $s_{\max}=(2/25)T$. Since $a=0$, the transformed initial payoff is $V(0,x)=(e^x-1)^+$. Check: at $x=0$ this equals $(1-1)^+=0$; at $x=\ln 2$ it equals $2-1=1$. The rescaling at $\tau=0$ is $e^{a x}=1$, as required for the payoff to be unchanged.

Worked Example: A Nonzero Exponential Price Factor

Take $r=0.05=1/20$ and $\sigma=0.3=3/10$. Find $a$, $b$, and the piecewise heat initial condition.

Here $\sigma^2=9/100$, $D=9/200$, and $m=r-D=1/20-9/200=1/200$. Thus $a=-m/\sigma^2=-(1/200)/(9/100)=-1/18$. Also $m^2/(2\sigma^2)=(1/40000)/(18/100)=1/7200$, so $b=-1/20-1/7200=-361/7200$. Substitution into the rescaling gives $u=e^{-x/18-(361/7200)\tau}v$.

For $x\le0$, $V(0,x)=0$. For $x>0$, $V(0,x)=e^{x/18}(e^x-1)=e^{19x/18}-e^{x/18}$. Check: at $x=\ln 2$, the value is $2^{19/18}-2^{1/18}=2^{1/18}(2-1)=2^{1/18}$, which also equals $e^{x/18}(e^x-1)$ evaluated directly. At $x=0$, the original positive-part payoff is zero, not the positive-$x$ expression extended without its case condition.

Worked Example: Reversing the Transformation for a Heat Solution

Verify the reverse transformation on an exact solution. Take $r=0$, $\sigma=\sqrt{2}$, and $v(\tau,x)=e^{x+\tau}$, and show that the resulting $u$ solves the log-price Black–Scholes equation.

Here $D=\sigma^2/2=1$, $m=r-D=-1$, $a=-m/\sigma^2=1/2$, and $b=-r-m^2/(2\sigma^2)=-1/4$. The proposed $v$ satisfies $v_\tau=e^{x+\tau}$ and $v_{xx}=e^{x+\tau}$, so $v_\tau=Dv_{xx}$. Reversing the exponential rescaling gives $u=e^{x/2-\tau/4}e^{x+\tau}=e^{3x/2+3\tau/4}$.

Check directly against $u_\tau=D u_{xx}+m u_x-r u=u_{xx}-u_x$. The derivatives are $u_\tau=(3/4)u$, $u_x=(3/2)u$, and $u_{xx}=(9/4)u$. Thus $u_{xx}-u_x=(9/4-3/2)u=(3/4)u=u_\tau$. This verifies the transformed equation. This particular heat solution has its own initial data; it is an algebraic check of the transformation, not the call-payoff solution.

Why the Transformation Helps—and a Common Error

The transformation converts a variable-coefficient pricing equation into a constant-coefficient diffusion equation, while converting the terminal payoff into initial data. This makes heat-equation methods available for analysing the pricing problem. The transformed payoff still has a kink at $x=0$, and its growth as $x$ increases is modified by the exponential factor. Boundary or growth conditions must therefore be carried through the same changes of variables when solving a particular pricing problem. The transformation itself does not assert uniqueness or supply a heat-equation solution.

Common mistake: losing the chain-rule term in $C_{SS}$. A tempting but incorrect calculation is $C_{SS}=Ku_{xx}/S^2$. Differentiating $C_S=Ku_x/S$ gives, instead, $C_{SS}=K(u_{xx}-u_x)/S^2$. For the parameters $r=1/20$, $\sigma=3/10$ in the second example, the correct coefficient of $u_x$ in the log-price equation is $r-\sigma^2/2=1/20-9/200=1/200$, not $r=1/20$. The omitted term changes that coefficient and therefore gives the wrong $a$ in the exponential rescaling.

A second sign check is to substitute the selected constants into both coefficients after rescaling. The $v_x$ coefficient must be $2Da+m=0$, and the $v$ coefficient must be $Da^2+ma-r-b=0$. Checking both prevents an incorrect sign in $b$ from leaving a residual discount term.

Stage	Variables or function	Equation or data
Pricing problem	$t,S,C$	Black–Scholes PDE with terminal call payoff
Log-price form	$\tau=T-t,\ x=\ln(S/K),\ u=C/K$	$u_\tau=D u_{xx}+m u_x-r u$
Exponential rescaling	$u=e^{ax+b\tau}v$	$v_\tau=Dv_{xx}$
Standard heat form	$s=D\tau,\ V(s,x)=v(s/D,x)$	$V_s=V_{xx}$, with $V(0,x)=e^{-ax}(e^x-1)^+$

Check Your Understanding

Use the definitions and derivative calculations above. When specifying payoff cases, include the value at the strike.

For $r=0.06$ and $\sigma=0.2$, calculate $D$, $m$, $a$, and $b$ exactly.
Starting from $C(t,S)=Ku(T-t,\ln(S/K))$, derive $C_t$, $C_S$, and $C_{SS}$.
For general $a,b$, calculate the coefficients of $v_x$ and $v$ after substituting $u=e^{ax+b\tau}v$.
Explain why the transformed call payoff is zero at $x=0$, and give its expression for $x>0$.
If $v_\tau=Dv_{xx}$ and $s=D\tau$, use the chain rule to verify $V_s=V_{xx}$ for $V(s,x)=v(s/D,x)$.

Stage	Variables or function	Equation or data
Pricing problem	\(t,S,C\)	Black–Scholes PDE with terminal call payoff
Log-price form	\(\tau=T-t,\ x=\ln(S/K),\ u=C/K\)	\(u_\tau=D u_{xx}+m u_x-r u\)
Exponential rescaling	\(u=e^{ax+b\tau}v\)	\(v_\tau=Dv_{xx}\)
Standard heat form	\(s=D\tau,\ V(s,x)=v(s/D,x)\)	\(V_s=V_{xx}\), with \(V(0,x)=e^{-ax}(e^x-1)^+\)