Editing Separation of variables (section)

=== Example: homogeneous case ===

Consider the one-dimensional [[heat equation]]. The equation is

{{NumBlk|:|<math>\frac{\partial u}{\partial t} - \alpha\frac{\partial^{2}u}{\partial x^{2}} = 0</math>|{{EqRef|1}}}}

The variable ''u'' denotes temperature. The boundary condition is homogeneous, that is
{{NumBlk|:|<math>u\big|_{x=0}=u\big|_{x=L}=0 </math>|{{EqRef|2}}}}

Let us attempt to find a [[Triviality_(mathematics)#Trivial_and_nontrivial_solutions|nontrivial solution]] satisfying the boundary conditions but with the following property: ''u'' is a product in which the dependence of ''u'' on ''x'', ''t'' is separated, that is:

{{NumBlk|:|<math> u(x,t) = X(x) T(t).</math>|{{EqRef|3}}}}

Substituting ''u'' back into equation {{EqNote|1}} and using the [[product rule]],

{{NumBlk|:|<math>\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)}= -\lambda,</math>|{{EqRef|4}}}}

where ''λ'' must be constant since the right hand side depends only on ''x'' and the left hand side only on ''t''. Thus:

{{NumBlk|:|<math>T'(t) = - \lambda \alpha T(t),</math>|{{EqRef|5}}}}

and

{{NumBlk|:|<math>X''(x) = - \lambda X(x).</math>|{{EqRef|6}}}}

−''λ'' here is the [[eigenvalue]] for both differential operators, and ''T''(''t'') and ''X''(''x'') are corresponding [[eigenfunction]]s.

We will now show that solutions for ''X''(''x'') for values of ''λ'' ≤ 0 cannot occur:

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

:<math>X(x) = B e^{\sqrt{-\lambda} \, x} + C e^{-\sqrt{-\lambda} \, x}.</math>

From {{EqNote|2}} we get

{{NumBlk|:|<math>X(0) = 0 = X(L),</math>|{{EqRef|7}}}}

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

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

:<math>X(x) = Bx + C.</math>

From {{EqNote|7}} we conclude in the same manner as in 1 that ''u'' is identically 0.

Therefore, it must be the case that ''λ'' > 0. Then there exist real numbers ''A'', ''B'', ''C'' such that
:<math>T(t) = A e^{-\lambda \alpha t},</math>
and
:<math>X(x) = B \sin(\sqrt{\lambda} \, x) + C \cos(\sqrt{\lambda} \, x).</math>

From {{EqNote|7}} we get ''C'' = 0 and that for some positive integer ''n'',

:<math>\sqrt{\lambda} = n \frac{\pi}{L}.</math>

This solves the heat equation in the special case that the dependence of ''u'' has the special form of {{EqNote|3}}.

In general, the sum of solutions to {{EqNote|1}} which satisfy the boundary conditions {{EqNote|2}} also satisfies {{EqNote|1}} and {{EqNote|3}}. Hence a complete solution can be given as

:<math>u(x,t) = \sum_{n = 1}^{\infty} D_n \sin \frac{n\pi x}{L} \exp\left(-\frac{n^2 \pi^2 \alpha t}{L^2}\right),</math>

where ''D''<sub>''n''</sub> are coefficients determined by initial condition.

Given the initial condition

:<math>u\big|_{t=0}=f(x),</math>

we can get

:<math>f(x) = \sum_{n = 1}^{\infty} D_n \sin \frac{n\pi x}{L}.</math>

This is the [[Fourier sine series]] expansion of ''f''(''x'') which is amenable to [[Fourier analysis]]. Multiplying both sides with <math display="inline">\sin \frac{n\pi x}{L}</math> and integrating over {{closed-closed|0, ''L''}} results in

:<math>D_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n\pi x}{L} \, dx.</math>

This method requires that the eigenfunctions ''X'', here <math display="inline">\left\{\sin \frac{n\pi x}{L}\right\}_{n=1}^{\infty}</math>, are [[orthogonal]] and [[Schauder basis|complete]]. In general this is guaranteed by [[Sturm–Liouville theory]].