ProofWiki:Sandbox

Separation of variables means the assumption that a given function of two variables $$f(z_1,z_2)=\alpha(z_1) \beta(z_2) \ $$ for some functions $$\alpha, \beta \ $$.

We can use this method to arrive at a particular solution of the heat equation, $$u_t = k u_{xx} \ $$.

We assume the initial condition $$u(0,x) = f(x) \quad \forall x \in [0,L] \ $$ where the function f is given and the boundary conditions $$u(t,0) = 0 = u(t,L) \quad \forall  t > 0 \ $$.

Let us attempt to find a solution which is not identically zero satisfying the boundary conditions but with the assumptions of separation of variables, that is,

$$u(t,x) = X(x) T(t) \ $$

This gives

$$\frac{\partial}{\partial t} X(x)T(t) = k \frac{\partial^2}{\partial x^2} X(x)T(t) \ $$

or

$$\frac{T'(t)}{kT(t)} = \frac{X''(x)}{X(x)} \ $$

Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value − λ. Thus:


 * $$T'(t) = - \lambda kT(t) \quad $$

and


 * $$X''(x) = - \lambda X(x). \quad $$

We will now show that solutions for these for values of λ ≤ 0 cannot occur:

Suppose that λ < 0. Then there exist real numbers B, C such that

$$X(x) = B e^{\sqrt{-\lambda} \, x} + C e^{-\sqrt{-\lambda} \, x}.$$

We get

$$X(0) = 0 = X(L). \quad $$

and therefore B = 0 = C which implies u is identically 0.

Suppose that λ = 0. Then there exist real numbers B, C such that

$$X(x) = Bx + C. \quad $$

We conclude in the same manner as before that u is identically 0.

Therefore, it must be the case that λ > 0. Then there exist real numbers A, B, C such that

$$T(t) = A e^{-\lambda k t} \quad $$

and

$$X(x) = B \sin(\sqrt{\lambda} \, x) + C \cos(\sqrt{\lambda} \, x)$$

We get C = 0 and that for some positive integer n,

$$\sqrt{\lambda} = n \frac{\pi}{L}$$

This solves the heat equation in the special case that the dependence of u has the special form assumed in the separation of variables.