达肯方程描述柯肯达尔效应的一个方程，是将物质的扩散和对流全部按照扩散的形式来表示，其思路类似于全电流定律。下面是其推导过程：

推导过程

假设存在一个AB两组分组成的一维扩散偶，考虑对流的因素
根据扩散第一定律，对于成分A

$$ J_A=J'_A + C_A \cdot v = -D_A \cdot \frac{\partial C_A}{\partial x} + C_A \cdot v \tag{1} $$

取其中一个微元体，由物质守恒可得

$$ \iint_S - \vec{J} \cdot d\vec{S} = \iint_{S} - \left ( - \vec{D}_A \cdot \frac{\partial C_A}{\partial x} + C_A \cdot \vec{v} \right ) \cdot d\vec{S} = \iiint_V \frac{\partial C_A}{\partial t} \cdot dV \tag{2} $$

由高斯定理可得

$$ \iint_S \left ( \vec{D}_A \cdot \frac{\partial C_A}{\partial x} - C_A \cdot \vec{v} \right ) \cdot d \vec{S} = \iiint_V \frac{\partial}{\partial x} \left ( D_A \cdot \frac{\partial C_A}{\partial x} - C_A \cdot v \right ) \cdot dV \tag{3} $$

联立$(2)、(3)$二式可得

$$ \frac{\partial C_A}{\partial t} = \frac{\partial}{\partial x} \left ( D_A \cdot \frac{\partial C_A}{\partial x} - C_A \cdot v \right ) \tag{4} $$

同理对于成分B可得

$$ \frac{\partial C_B}{\partial t} = \frac{\partial}{\partial x} \left ( D_B \cdot \frac{\partial C_B}{\partial x} - C_B \cdot v \right ) \tag{5} $$

由于晶体内部各处成分不变，即 $C_A+C_B = const$，两边求导可得

$$ \frac{\partial C}{\partial x} = \frac{\partial C_A}{\partial x} + \frac{\partial C_B}{\partial x} = 0 \tag{6} $$

将$(4)、(5)$式相加，并代入$(6)$式可得

$$ \left ( D_A - D_B \right ) \cdot \frac{\partial C_A}{\partial x} - C \cdot v = const = 0 $$

整理可得标记面的移动速度

$$ v = \frac{\left ( D_A - D_B \right )}{C} \cdot \frac{\partial C_A}{\partial x} = \left ( D_A - D_B \right ) \cdot \frac{\partial{x_A}}{\partial x} \tag{7} $$

将$(7)$式代入$(1)$式中可得

$$ J_A = -D_A \cdot \frac{\partial C_A}{\partial x} + C_A \cdot v = -\tilde {D} \cdot \frac{\partial C_A}{\partial x} \tag{8} $$

同理对于B组分有

$$ J_B = -D_B \cdot \frac{\partial C_B}{\partial x} + C_B \cdot v = -\tilde {D} \cdot \frac{\partial C_B}{\partial x} \tag{9} $$

其中$\tilde {D}$是互扩散系数，联立$(6)(7)(8)(9)$式可得

$$ \tilde {D}=\frac{C_B}{C_A+C_B} \cdot D_A + \frac{C_A}{C_A+C_B} \cdot D_B = x_B \cdot D_A + x_A \cdot D_B $$