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. The eigenvalues are ordered in descending order.
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. The eigenvalues are ordered in descending order.
In the 1D discrete case with Dirichlet boundary conditions, we are solving Rearranging terms, we get Now let. Also, assuming, we can scale eigenvectors by any nonzero scalar, so scale so that. Then we find the recurrence Considering as an indeterminate, where is the kth Chebyshev polynomial of the 2nd kind. Since, 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
Neumann case
In the Neumann case, we are solving In the standard discretization, we introduce and and define The boundary conditions are then equivalent to If we make a change of variables, we can derive the following: with being the boundary conditions. This is precisely the Dirichlet formula with interior grid points and grid spacing. Similar to what we saw in the above, assuming, we get This gives us eigenvalues and there are. If we drop the assumption that, we find there is also a solution with and this corresponds to eigenvalue. Relabeling the indices in the formula above and combining with the zero eigenvalue, we obtain,
Dirichlet-Neumann Case
For the Dirichlet-Neumann case, we are solving where We need to introduce auxiliary variables Consider the recurrence Also, we know and assuming, we can scale so that 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 we get Our new recurrence gives where 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 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.
