# Why does it make sense that $S$ and $e^{rt}$ are solutions to the Black-Scholes PDE?

It's readily verified mathematically that $$V=S$$ and $$V=e^{rt}$$ are solutions to the Black-Scholes PDE $$\frac{\partial V}{\partial t} + \frac{\sigma^2 S^2}{2} \frac{\partial^2 V}{\partial S^2} + r S \frac{\partial V}{\partial S} - rV = 0$$. How can we motivate/explain this from a financial perspective?

## 2 Answers

Under the standard assumptions, generally speaking, any contract that depends on the current values of t and S, and which are paid for at the start satisfy this PDE. In the financial context, the boundary conditions would be different for the different contracts, so the solutions of the PDE would be different for the different contracts. Thus different contracts would be identified by different boundary conditions, and these conditions would determine the solution.

In summary, different payoffs, same PDE, different prices.

Re-comment, your commentary is correct; however for deterministic function of t only, the PDE reduces to $$\frac{\partial V}{\partial t}=rV$$ and this has the solution that you mentioned in your question. Anything deterministic must have this form, otherwise there is an arbitrage. In other words, risk free or deterministic must grow at this rate.

• Thanks for your reply, it makes more sense now, after all the only requirement for $V$ in the derivation of the BS PDE is that it depends on $t$ and $S(t)$ (as you pointed out), this requirement is met when constructing the Taylorseries to give an expression for $V(S(t),t)$. However, $V=t$ is not a solution, would this have to do with the fact that when we construct our delta hedged portfolio, it increases continuously with interest rate $r$, so it is dependent on $e^{rt}$ rather than $t$? – 6thsense Oct 18 '18 at 13:42
• Added text to the answer! Hope it is clearer now – Magic is in the chain Oct 18 '18 at 14:57

These are well known trivial solutions to the Black-Scholes PDE. The first one is just the price of the underlying stock and the second is interest bearing money in a bank. These are trivially true because there is no optionality involved (which is expressed in the boundary and terminal condition of the respective contract to price).