Why an Exponentially Growing Value Solves the Black–Scholes PDE
Summary
The note checks whether a value of the form A times the exponential of the risk-free rate and time satisfies the Black–Scholes partial differential equation. Since the value has no dependence on the underlying price, its first and second derivatives with respect to that price are zero. Its time derivative is the risk-free rate multiplied by the value, matching the equation’s right-hand side. This direct substitution establishes the PDE solution claim.
The note also considers an intuition based on a zero-strike call and a stock whose volatility is set to zero, so its price grows at the risk-free rate. That picture is illustrative, but it is not needed to verify the PDE and relies on a degenerate zero-volatility case. The PDE check alone does not specify a complete option-pricing problem: a terminal payoff or boundary conditions are needed to identify the relevant solution. The discussion therefore explains a mathematical solution form rather than deriving a general European option price.
Key ideas
- A value independent of the underlying has zero first and second price derivatives.
- Its time derivative equals the risk-free rate times the value.
- Substitution verifies that the exponential form satisfies the Black–Scholes PDE.
- A zero-strike, zero-volatility stock analogy offers intuition but is a special case.
- A PDE solution alone does not determine an option price without terminal or boundary conditions.
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Full text
# Show that $Ae^{rt}$ is a solution of the Black-Scholes equation. Why should this be so?
# Show that $Ae^{rt}$ is a solution of the Black-Scholes equation. Why should this be so?
The following is taken from Mark Joshi's Concepts and Practice of Mathematical Finance, second edition, exercise $5.6$.
> Question: Show that $Ae^{rt}$ is a solution of the Black-Scholes equation. Why should this be so?
Recall that the Black-Scholes equation is $$\frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2} S^2\sigma^2 \frac{\partial^2 V}{\partial S^2} = rV$$ where $V=V(t,S_t)$ is either European call or put option value, $r$ is risk-free interest rate and $\sigma$ is volatility.
It can be verified easily that $Ae^{rt}$ satisfies the equation above. In terms of explanation on why this is so, I guess we need to concoct a European option whose payoff is $Ae^{rt}$ to justify it.
Since $S_t$ follows geometric Brownian motion with respect to risk-neutral probability measure, so $$S_t = S_0 e^{(r-\frac{1}{2}\sigma^2)t + \sigma W_t}.$$ I think in this case, we take $\sigma = 0$ to obtain that $$S_t = S_0 e^{r t}.$$ So $A=S_0.$ Therefore, $V(t,S_t)=S_0 e^{rt} = S_t$ is a European call option with zero strike price. This justifies why $Ae^{rt}$ satisfies the Black-Scholes equation.
Is it correct?
## Answer by Crushh (score 4, accepted)
https://quant.stackexchange.com/a/50316
While your approach is correct, generally what people would do is that find derivative of the equation for example $V(t,S_t)=Ae^{rt}$. \begin{eqnarray} &\frac{dV}{dt}=rA e^{rt}\\ &\frac{dV}{dS}=0 \\ &\frac{d^2V}{dS^2}=0 \end{eqnarray} Then you plug in the derivatives above to the left hand side of your Black-Scholes equation. Then you will get $rAe^{rt}=rV$ which is equals to the right hand side of your Black-Scholes equation. Therefore, $V(t,S_t)=Ae^{rt}$ is a solution to the Black Scholes equation!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.