Eigenvalues and eigenvectors of the second derivative

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 continuous case

The index j represents the jth eigenvalue or eigenvector and runs from 1 to ${\displaystyle \infty }$. Assuming the equation is defined on the domain ${\displaystyle x\in [0,L]}$, the following are the eigenvalues and normalized eigenvectors. The eigenvalues are ordered in descending order.

Pure Dirichlet boundary conditions

${\displaystyle \lambda _{j}=-{\frac {j^{2}\pi ^{2}}{L^{2}}}}$

${\displaystyle v_{j}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {j\pi x}{L}}\right)}$

Pure Neumann boundary conditions

${\displaystyle \lambda _{j}=-{\frac {(j-1)^{2}\pi ^{2}}{L^{2}}}}$

${\displaystyle v_{j}(x)=\left\{{\begin{array}{lr}L^{-{\frac {1}{2}}}&j=1\\{\sqrt {\frac {2}{L}}}\cos \left({\frac {(j-1)\pi x}{L}}\right)&{\mbox{otherwise}}\end{array}}\right.}$

Periodic boundary conditions

${\displaystyle \lambda _{j}=\left\{{\begin{array}{lr}-{\frac {j^{2}\pi ^{2}}{L^{2}}}&{\mbox{j is even.}}\\-{\frac {(j-1)^{2}\pi ^{2}}{L^{2}}}&{\mbox{j is odd.}}\end{array}}\right.}$

(That is: ${\displaystyle 0}$ is a simple eigenvalue and all further eigenvalues are given by ${\displaystyle {\frac {j^{2}\pi ^{2}}{L^{2}}}}$, ${\displaystyle j=1,2,\ldots }$, each with multiplicity 2).

${\displaystyle v_{j}(x)={\begin{cases}L^{-{\frac {1}{2}}}&{\mbox{if }}j=1.\\{\sqrt {\frac {2}{L}}}\sin \left({\frac {j\pi x}{L}}\right)&{\mbox{ if j is even.}}\\{\sqrt {\frac {2}{L}}}\cos \left({\frac {(j-1)\pi x}{L}}\right)&{\mbox{ if j is odd.}}\end{cases}}}$

Mixed Dirichlet-Neumann boundary conditions

${\displaystyle \lambda _{j}=-{\frac {(2j-1)^{2}\pi ^{2}}{4L^{2}}}}$

${\displaystyle v_{j}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {(2j-1)\pi x}{2L}}\right)}$

Mixed Neumann-Dirichlet boundary conditions

${\displaystyle \lambda _{j}=-{\frac {(2j-1)^{2}\pi ^{2}}{4L^{2}}}}$

${\displaystyle v_{j}(x)={\sqrt {\frac {2}{L}}}\cos \left({\frac {(2j-1)\pi x}{2L}}\right)}$

The discrete case

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.

Pure Dirichlet boundary conditions

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

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

Pure Neumann boundary conditions

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

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

Periodic boundary conditions

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

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

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

Mixed Dirichlet-Neumann boundary conditions

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

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

Mixed Neumann-Dirichlet boundary conditions

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

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

Derivation of Eigenvalues and Eigenvectors in the Discrete Case

Dirichlet case

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

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

Neumann case

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,

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

Dirichlet-Neumann Case

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.

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

References