What are the eigenfunctions of the Laplace operator in 1D?

Answer 1

In general, the Laplace operator in n dimensions is

∇2 = (∂/∂x1)2 + (∂/∂x2)2 + ... + (∂/∂xn)2,

and the eigenfunctions are the solutions f(x1, x2, ..., xn) of the partial differential equation:

∇2f = -λf,

where the eigenvalues -λ are to be determined. Often, the set of solutions will be constrained by given boundary conditions (which limits the possible values of λ), but for the purposes of this question that does not matter.

In one dimension this gives a simple linear differential equation with constant coefficients:

d2f/dx2 = -λf

which may be solved using standard, elementary techniques. For λ > 0 the solutions may be written:

f(x) = A cos((√λ) x) + B sin((√λ) x)

and for λ < 0:

f(x) = A exp((√-λ) x) + B exp(-(√-λ) x)

where in each case A and B are arbitrary constants. Using Euler's formula

exp(ia) = cos(a) + i sin(a)

the solutions in both cases can be written as linear combinations of the exponential functions exp((±i√λ) x).

In the case that λ = 0, the solutions are straight lines:

f(x) = Ax + B.

Answer 2

The eigenfunctions of the Laplace operator in 1D are sine and cosine functions. Specifically, the eigenfunctions are sin(nx) and cos(nx), where n is an integer. These functions satisfy the equation ∂^2u/∂x^2 = -λu, where λ is the eigenvalue.