# Can one use the Greeks (delta,gamma,theta) to show that the Black-Scholes call formula satisfies the Black-Scholes PDE?

If so, is there a derivation anywhere that shows this? I was told that this could be done in a class but I don't see how it's possible.

It's quite straightforward when you plug in the formulae for the greeks into the PDE.

Preliminaries:

$\Delta =\frac{\partial c_t}{\partial S_t}=\Phi(d_1)$

$\Gamma=\frac{\partial^2 c_t}{\partial S_t^2}=\frac{\phi(d1)}{S_t\sigma\sqrt{u}}$

$\Theta=\frac{\partial c_t}{\partial t}=-rKe^{ru}\Phi(d_2)-S_t\phi(d_1)\frac{\sigma}{2\sqrt{u}}$

The Black Scholes PDE:

\begin{eqnarray} rc_t&=&\Theta+rS_t\Delta + \frac{1}{2}S_t^2\sigma^2\Gamma\\ RHS&=&-rKe^{ru}\Phi(d_2)-S_t\phi(d_1)\frac{\sigma}{2\sqrt{u}}+rS_t\Phi(d_1)+\frac{1}{2}S_t^2\sigma^2 \frac{\phi(d1)}{S_t\sigma\sqrt{u}}\\ &=&-rKe^{ru}\Phi(d_2)-\frac{S_t\phi(d_1)\sigma}{2\sqrt{u}}+rS_t\Phi(d_1)+ \frac{S_t\sigma\phi(d1)}{2\sqrt{u}}\\ &=&rS_t\Phi(d_1)-rKe^{ru}\Phi(d_2)\\ &=&rc_t\\ &=&LHS \end{eqnarray}