Derivation of the Black-Scholes-Merton PDE › Worked Example: Verifying the PDE

Let's verify that a proposed solution satisfies the Black-Scholes PDE. This exercise builds intuition for what it means to "satisfy" a partial differential equation and demonstrates how to check whether a candidate pricing formula is valid. Consider a simple derivative whose price is $V(S, t) = S$, representing a forward contract with zero delivery price and zero interest rates.

We need to check whether $V = S$ satisfies:

$$\frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} = rV$$

where:

• $V$: price of the derivative
• $S$: stock price
• $r$: risk-free interest rate
• $\sigma$: volatility
• $t$: time

Computing the partial derivatives is straightforward since $V = S$ is a simple function:

$$\frac{\partial V}{\partial t} = 0, \quad \frac{\partial V}{\partial S} = 1, \quad \frac{\partial^2 V}{\partial S^2} = 0$$

where:

• $V$: derivative price ($S$)

The time derivative is zero because $V = S$ has no explicit dependence on time. The first derivative with respect to $S$ is one because $V$ increases linearly with $S$. The second derivative is zero because the relationship is perfectly linear with no curvature.

Substituting into the PDE:

$$0 + rS(1) + \frac{1}{2}\sigma^2 S^2(0) = rS$$

where:

• $r, S, \sigma$: model parameters

The left side equals $rS$, and the right side is $rV = rS$. The equation is satisfied, confirming that the stock price itself is a trivial solution to the BSM PDE. This makes economic sense: holding the stock is a valid "derivative strategy" that trivially satisfies no-arbitrage pricing.

Now consider a more interesting example: a derivative that pays $S^2$ at maturity $T$. Such a derivative might seem exotic, but it illustrates how the PDE handles nonlinear payoffs. We'll verify that a solution of the following form satisfies the PDE:

$$V(S, t) = S^2 e^{(r + \sigma^2)(T-t)}$$

where:

• $V(S, t)$: price of the derivative
• $S$: stock price
• $r$: risk-free interest rate
• $\sigma$: volatility
• $T$: maturity time
• $t$: current time

Notice that this formula has an interesting structure: it equals $S^2$ at maturity (when $T - t = 0$), satisfying the terminal condition, and it includes an exponential factor that accounts for the time value and volatility effects.

Computing the partial derivatives requires careful application of the chain rule:

$$\begin{aligned} \frac{\partial V}{\partial t} &= -S^2(r + \sigma^2)e^{(r + \sigma^2)(T-t)} \\ \frac{\partial V}{\partial S} &= 2Se^{(r + \sigma^2)(T-t)} \\ \frac{\partial^2 V}{\partial S^2} &= 2e^{(r + \sigma^2)(T-t)} \end{aligned}$$

where:

• $V$: derivative price ($S^2 e^{(r + \sigma^2)(T-t)}$)
• $S, t, r, \sigma, T$: model parameters

The time derivative is negative because the exponential factor decreases as we approach maturity (when $T - t$ shrinks). The first spatial derivative reflects that the price increases with $S$, at a rate proportional to $S$ itself. The second spatial derivative captures the constant curvature of the quadratic relationship.

Substituting these derivatives into the left-hand side (LHS) of the PDE and simplifying:

$$\begin{aligned} \text{LHS} &= -S^2(r + \sigma^2)e^{(r+\sigma^2)(T-t)} + rS \cdot 2Se^{(r+\sigma^2)(T-t)} + \frac{1}{2}\sigma^2 S^2 \cdot 2e^{(r+\sigma^2)(T-t)} \\ &= S^2 e^{(r+\sigma^2)(T-t)} \left[ -(r + \sigma^2) + 2r + \sigma^2 \right] && \text{(factor out common term)} \\ &= S^2 e^{(r+\sigma^2)(T-t)} \left[ -r - \sigma^2 + 2r + \sigma^2 \right] && \text{(expand terms)} \\ &= S^2 e^{(r+\sigma^2)(T-t)} \cdot r && \text{(simplify)} \\ &= rV && \text{(definition of V)} \end{aligned}$$

where:

• LHS: value of left-hand side of PDE
• $V$: derivative price

The algebraic simplification is satisfying: the terms involving $\sigma^2$ cancel (one from the time derivative and one from the gamma term), and the terms involving $r$ combine to give exactly what we need.

The right side of the PDE is:

$$rV = r \cdot S^2 e^{(r+\sigma^2)(T-t)}$$

where:

• $r$: risk-free rate
• $V$: derivative price

Both sides are equal, confirming our solution. This example illustrates that derivatives with nonlinear payoffs require appropriate adjustments to their prices, captured by factors that depend on volatility as well as interest rates.