# Black-Scholes方程的等價形式（轉換為熱方程）

$\ frac {\ partial C} {\ partial t}（S，t）+ rS \ frac {\ partial C} {\ partial S}（S，t）+ \ frac {1} {2} \ sigma ^ 2 S ^ 2 \ frac {\ partial ^ 2 C} {\ partial S ^ 2}（S，t）-rC = 0$

$\ frac {\ partial C} {\ partial t}（S，t）+（r- \ frac {1} {2} \ sigma ^ 2）S \ frac {\ partial C} {\ partial S}（S ，t）+ \ frac {1} {2} \ sigma ^ 2（S \ frac {\ partial} {\ partial S}）^ 2 C -rC = 0$，

\begin{align*} & \frac{1}{2}\sigma^2(S \frac{\partial}{\partial S})^2 C(S,t) \\ &= \frac{1}{2} \sigma^2 \frac{\partial}{\partial S} [\frac{\partial}{\partial S} (S^2 C(S,t))] \\ &= \frac{1}{2}\sigma^2 \frac{\partial}{\partial S} [2S\cdot C(S,t) + S^2 \frac{\partial C(S,t)}{\partial S}] \\ &= \frac{1}{2}\sigma^2 [2C(S,t) + 2S \frac{\partial C(S,t)}{\partial S} + 2S \frac{\partial C(S,t)}{\partial S} + S^2 \frac{\partial^2 C(S,t)}{\partial S^2} \\ &= \frac{1}{2} \sigma^2 [2C(S,t) + 4S \frac{\partial C(S,t)}{\partial S}] + \frac{1}{2} \sigma^2 S^2 \frac{\partial ^2 C(S,t)}{\partial S^2} \end{align*}

## 最佳答案

$$（S \ frac {\ partial} {\ partial S}）^ 2 C =（S \ frac {\ partial} {\ partial S}）（S \ frac {\ partial} {\ partial S}）C$$

$$（S \ frac {\ partial} {\ partial S}）^ 2 C = S \ frac {\ partial} {\ partial S}（S \ frac {\ partial C} {\ partial S}）$$

$$S（\ frac {\ partial S} {\ partial S} \ frac {\ partial C} {\ partial S} + S \ frac {\ partial ^ 2 C} {\ partial S ^ 2}）$$

（順便說一下，我們可以看到，正是產品規則使得衍生品無法通過乘以$S$進行通勤）。所以

$$（S \ frac {\ partial} {\ partial S}）^ 2 C = S \ frac {\ partial C} {\ partial S} + S ^ 2 \ frac {\ partial ^ 2 C} {\ partial S ^ 2 }$$