Eigenvalues and eigenvectors of the second derivative

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.

${\displaystyle \lambda _{j}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi j}{2(n+1)}}\right)}$

${\displaystyle v_{i,j}={\sqrt {\frac {2}{n+1}}}\sin \left({\frac {ij\pi }{n+1}}\right)}$ ^[1]

${\displaystyle \lambda _{j}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi (j-1)}{2n}}\right)}$

${\displaystyle v_{i,j}={\begin{cases}n^{-{\frac {1}{2}}}&{\mbox{j = 1}}\\{\sqrt {\frac {2}{n}}}\cos \left({\frac {\pi (j-1)(i-0.5)}{n}}\right)&{\mbox{otherwise}}\end{cases}}}$

${\displaystyle \lambda _{j}={\begin{cases}-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi (j-1)}{2n}}\right)&{\mbox{ if j is odd.}}\\-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi j}{2n}}\right)&{\mbox{ if j is even.}}\end{cases}}}$

(Note that eigenvalues are repeated except for 0 and the largest one if n is even.)

${\displaystyle v_{i,j}={\begin{cases}n^{-{\frac {1}{2}}}&{\mbox{if }}j=1.\\n^{-{\frac {1}{2}}}(-1)^{i}&{\mbox{if }}j=n{\mbox{ and n is even.}}\\{\sqrt {\frac {2}{n}}}\sin \left({\frac {\pi (i-0.5)j}{n}}\right)&{\mbox{ otherwise if j is even.}}\\{\sqrt {\frac {2}{n}}}\cos \left({\frac {\pi (i-0.5)(j-1)}{n}}\right)&{\mbox{ otherwise if j is odd.}}\end{cases}}}$

${\displaystyle \lambda _{j}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi (j-{\frac {1}{2}})}{2n+1}}\right)}$

${\displaystyle v_{i,j}={\sqrt {\frac {2}{n+0.5}}}\sin \left({\frac {\pi i(2j-1)}{2n+1}}\right)}$

${\displaystyle \lambda _{j}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi (j-{\frac {1}{2}})}{2n+1}}\right)}$

${\displaystyle v_{i,j}={\sqrt {\frac {2}{n+0.5}}}\cos \left({\frac {\pi (i-0.5)(2j-1)}{2n+1}}\right)}$

In the 1D discrete case with Dirichlet boundary conditions, we are solving

${\displaystyle {\frac {v_{k+1}-2v_{k}+v_{k-1}}{h^{2}}}=\lambda v_{k},\ k=1,...,n,\ v_{0}=v_{n+1}=0.}$

Rearranging terms, we get

${\displaystyle v_{k+1}=(2+h^{2}\lambda )v_{k}-v_{k-1}.\!}$

Now let ${\displaystyle 2\alpha =(2+h^{2}\lambda )}$. Also, assuming ${\displaystyle v_{1}\neq 0}$, we can scale eigenvectors by any nonzero scalar, so scale ${\displaystyle v}$ so that ${\displaystyle v_{1}=1}$.

Then we find the recurrence

${\displaystyle v_{0}=0\,\!}$

${\displaystyle v_{1}=1.\,\!}$

${\displaystyle v_{k+1}=2\alpha v_{k}-v_{k-1}\,\!}$

Considering ${\displaystyle \alpha }$ as an indeterminate,

${\displaystyle v_{k+1}=U_{k}(\alpha )\,\!}$

where ${\displaystyle U_{k}}$ is the kth Chebyshev polynomial of the 2nd kind.

Since ${\displaystyle v_{n+1}=0}$, we get that

${\displaystyle U_{n}(\alpha )=0\,\!}$.

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 ${\displaystyle 2\alpha =(2+h^{2}\lambda )}$.

These zeros are well known and are:

${\displaystyle \alpha _{k}=\cos \left({\frac {k\pi }{n+1}}\right).\,\!}$

Plugging these into the formula for ${\displaystyle \lambda }$,

${\displaystyle 2\cos \left({\frac {k\pi }{n+1}}\right)=h^{2}\lambda _{k}+2\,\!}$

${\displaystyle \lambda _{k}=-{\frac {2}{h^{2}}}\left[1-\cos \left({\frac {k\pi }{n+1}}\right)\right].\,\!}$

And using a trig formula to simplify, we find

${\displaystyle \lambda _{k}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {k\pi }{2(n+1)}}\right).\,\!}$

In the Neumann case, we are solving

${\displaystyle {\frac {v_{k+1}-2v_{k}+v_{k-1}}{h^{2}}}=\lambda v_{k},\ k=1,...,n,\ v'_{0.5}=v'_{n+0.5}=0.\,\!}$

In the standard discretization, we introduce ${\displaystyle v_{0}\,\!}$ and ${\displaystyle v_{n+1}\,\!}$ and define

${\displaystyle v'_{0.5}:={\frac {v_{1}-v_{0}}{h}},\ v'_{n+0.5}:={\frac {v_{n+1}-v_{n}}{h}}\,\!}$

The boundary conditions are then equivalent to

${\displaystyle v_{1}-v_{0}=0,\ v_{n+1}-v_{n}=0.}$

If we make a change of variables,

${\displaystyle w_{k}=v_{k+1}-v_{k},\ k=0,...,n\,\!}$

we can derive the following:

${\displaystyle {\begin{alignedat}{2}{\frac {v_{k+1}-2v_{k}+v_{k-1}}{h^{2}}}&=\lambda v_{k}\\v_{k+1}-2v_{k}+v_{k-1}&=h^{2}\lambda v_{k}\\(v_{k+1}-v_{k})-(v_{k}-v_{k-1})&=h^{2}\lambda v_{k}\\w_{k}-w_{k-1}&=h^{2}\lambda v_{k}\\&=h^{2}\lambda w_{k-1}+h^{2}\lambda v_{k-1}\\&=h^{2}\lambda w_{k-1}+w_{k-1}-w_{k-2}\\w_{k}&=(2+h^{2}\lambda )w_{k-1}-w_{k-2}\\w_{k+1}&=(2+h^{2}\lambda )w_{k}-w_{k-1}\\&=2\alpha w_{k}-w_{k-1}.\end{alignedat}}}$

with ${\displaystyle w_{n}=w_{0}=0}$ being the boundary conditions.

This is precisely the Dirichlet formula with ${\displaystyle n-1}$ interior grid points and grid spacing ${\displaystyle h}$. Similar to what we saw in the above, assuming ${\displaystyle w_{1}\neq 0}$, we get

${\displaystyle \lambda _{k}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {k\pi }{2n}}\right),\ k=1,...,n-1.}$

This gives us ${\displaystyle n-1}$ eigenvalues and there are ${\displaystyle n}$. If we drop the assumption that ${\displaystyle w_{1}\neq 0}$, we find there is also a solution with ${\displaystyle v_{k}=\mathrm {constant} \ \forall \ k=0,...,n+1,}$ and this corresponds to eigenvalue ${\displaystyle 0}$.

Relabeling the indices in the formula above and combining with the zero eigenvalue, we obtain,

${\displaystyle \lambda _{k}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {(k-1)\pi }{2n}}\right),\ k=1,...,n.}$

For the Dirichlet-Neumann case, we are solving

${\displaystyle {\frac {v_{k+1}-2v_{k}+v_{k-1}}{h^{2}}}=\lambda v_{k},\ k=1,...,n,\ v_{0}=v'_{n+0.5}=0.}$,

where ${\displaystyle v'_{n+0.5}:={\frac {v_{n+1}-v_{n}}{h}}.}$

We need to introduce auxiliary variables ${\displaystyle v_{j+0.5},\ j=0,...,n.}$

Consider the recurrence

${\displaystyle v_{k+0.5}=2\beta v_{k}-v_{k-0.5},{\text{ for some }}\beta \,\!}$.

Also, we know ${\displaystyle v_{0}=0}$ and assuming ${\displaystyle v_{0.5}\neq 0}$, we can scale ${\displaystyle v_{0.5}}$ so that ${\displaystyle v_{0.5}=1.}$

We can also write

${\displaystyle v_{k}=2\beta v_{k-0.5}-v_{k-1}\,\!}$

${\displaystyle v_{k+1}=2\beta v_{k+0.5}-v_{k}.\,\!}$

Taking the correct combination of these three equations, we can obtain

${\displaystyle v_{k+1}=(4\beta ^{2}-2)v_{k}-v_{k-1}.\,\!}$

And thus our new recurrence will solve our eigenvalue problem when

${\displaystyle h^{2}\lambda +2=(4\beta ^{2}-2).\,\!}$

Solving for ${\displaystyle \lambda }$ we get

${\displaystyle \lambda ={\frac {4(\beta ^{2}-1)}{h^{2}}}.}$

Our new recurrence gives

${\displaystyle v_{n+1}=U_{2n+1}(\beta ),\ v_{n}=U_{2n-1}(\beta ),\,\!}$

where ${\displaystyle U_{k}(\beta )}$ again is the kth Chebyshev polynomial of the 2nd kind.

And combining with our Neumann boundary condition, we have

${\displaystyle U_{2n+1}(\beta )-U_{2n-1}(\beta )=0.\,\!}$

A well-known formula relates the Chebyshev polynomials of the first kind, ${\displaystyle T_{k}(\beta )}$, to those of the second kind by

${\displaystyle U_{k}(\beta )-U_{k-2}(\beta )=T_{k}(\beta ).\,\!}$

Thus our eigenvalues solve

${\displaystyle T_{2n+1}(\beta )=0,\ \lambda ={\frac {4(\beta ^{2}-1)}{h^{2}}}.\,\!}$

The zeros of this polynomial are also known to be

${\displaystyle \beta _{k}=\cos \left({\frac {\pi (k-0.5)}{2n+1}}\right),\ k=1,...,2n+1\,\!}$

And thus

${\displaystyle {\begin{alignedat}{2}\lambda _{k}&={\frac {4}{h^{2}}}\left[\cos ^{2}\left({\frac {\pi (k-0.5)}{2n+1}}\right)-1\right]\\&=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi (k-0.5)}{2n+1}}\right).\end{alignedat}}}$

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.

${\displaystyle \lambda _{k}=-{\frac {4}{h^{2}}}\sin ^{2}\left({\frac {\pi (k-0.5)}{2n+1}}\right),\ k=1,...,n.}$