Explicit formulas for eigenvalues and eigenvectors of the second derivative with different boundary conditions are provided both for the continuous and discrete cases.
In the discrete case, the standard central difference approximation of the second derivative is used on a uniform grid.
These formulas are used to derive the expressions for eigenfunctions of Laplacian in case of separation of variables, as well as to find eigenvalues and eigenvectors of multidimensional discrete Laplacian on a regular grid, which is presented as a Kronecker sum of discrete Laplacians in one-dimension.
The index j represents the jth eigenvalue or eigenvector and runs from 1 to
Assuming the equation is defined on the domain
, the following are the eigenvalues and normalized eigenvectors.
Notation: The index j represents the jth eigenvalue or eigenvector.
The index i represents the ith component of an eigenvector.
Both i and j go from 1 to n, where the matrix is size n x n. Eigenvectors are normalized.
(Note that eigenvalues are repeated except for 0 and the largest one if n is even.)
In the 1D discrete case with Dirichlet boundary conditions, we are solving Rearranging terms, we get Now let
, we can scale eigenvectors by any nonzero scalar, so scale
is the kth Chebyshev polynomial of the 2nd kind.
, we get that It is clear that the eigenvalues of our problem will be the zeros of the nth Chebyshev polynomial of the second kind, with the relation
These zeros are well known and are: Plugging these into the formula for
, And using a trig formula to simplify, we find In the Neumann case, we are solving In the standard discretization, we introduce
and define The boundary conditions are then equivalent to If we make a change of variables, we can derive the following: with
This is precisely the Dirichlet formula with
interior grid points and grid spacing
If we drop the assumption that
and this corresponds to eigenvalue
Relabeling the indices in the formula above and combining with the zero eigenvalue, we obtain, For the Dirichlet-Neumann case, we are solving where
We need to introduce auxiliary variables
We can also write Taking the correct combination of these three equations, we can obtain And thus our new recurrence will solve our eigenvalue problem when Solving for
again is the kth Chebyshev polynomial of the 2nd kind.
And combining with our Neumann boundary condition, we have A well-known formula relates the Chebyshev polynomials of the first kind,
, to those of the second kind by Thus our eigenvalues solve The zeros of this polynomial are also known to be And thus Note that there are 2n + 1 of these values, but only the first n + 1 are unique.
The (n + 1)th value gives us the zero vector as an eigenvector with eigenvalue 0, which is trivial.
This can be seen by returning to the original recurrence.
So we consider only the first n of these values to be the n eigenvalues of the Dirichlet - Neumann problem.