# Separation of heat equation

1. Mar 31, 2012

### volog

Hi,

So in my book they separate out a solution the heat equation into a function of x and a function of t and then separate the t and x dependencies, giving

$$-\frac{T''}{c^2T} = - \frac {X''}{X} = \lambda$$

The book says that the proof is because

$$\frac {\partial \lambda}{\partial x} = 0$$

and

$$\frac {\partial \lambda}{\partial t} = 0$$

but if X = X(x) and T = T(t), how do you get zero out when you take d/dx and d/dt? Why don't you use the chain rule on -X''/X to get something like

$$\frac{\partial}{\partial x} (\frac{-X''}{X}) = (-\frac{X'''}{X} + \frac{X''(X')}{X^2})$$

thanks