eigenvalues and eigenvectors of the second derivative

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

:$\backslash lambda\_j\; =\; -\backslash frac\{j^2\; \backslash pi^2\}\{L^2\}$

:$v\_j(x)\; =\; \backslash sqrt\{\backslash frac\{2\}\{L\}\}\; \backslash sin\backslash left(\backslash frac\{j\; \backslash pi\; x\}\{L\}\backslash right)$

:$\backslash lambda\_j\; =\; -\backslash frac\{(j\; -\; 1)^2\; \backslash pi^2\}\{L^2\}$

:$$

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.

:$\backslash lambda\_j\; =$

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

:$v\_j(x)\; =\; \backslash begin\{cases\}$

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}

:$\backslash lambda\_j\; =\; -\backslash frac\{(2j\; -\; 1)^2\; \backslash pi^2\}\{4\; L^2\}$

:$v\_j(x)\; =\; \backslash sqrt\{\backslash frac\{2\}\{L\}\}\; \backslash sin\backslash left(\backslash frac\{(2j\; -\; 1)\; \backslash pi\; x\}\{2\; L\}\backslash right)$

:$\backslash lambda\_j\; =\; -\backslash frac\{(2j\; -\; 1)^2\; \backslash pi^2\}\{4\; L^2\}$

:$v\_j(x)\; =\; \backslash sqrt\{\backslash frac\{2\}\{L\}\}\; \backslash cos\backslash left(\backslash frac\{(2j\; -\; 1)\; \backslash pi\; x\}\{2\; L\}\backslash right)$

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.

:$\backslash lambda\_j\; =\; -\backslash frac\{4\}\{h^2\}\; \backslash sin^2\backslash left(\backslash frac\{\backslash pi\; j\}\{2(n\; +\; 1)\}\backslash right)$

:$v\_\{i,j\}\; =\; \backslash sqrt\{\backslash frac\{2\}\{n+1\}\}\; \backslash sin\backslash left(\backslash frac\{i\; j\; \backslash pi\}\{n+1\}\backslash right)$ F. Chung, S.-T. Yau, Discrete Green's Functions, Journal of Combinatorial Theory A 91, 191-214 (2000).

:$\backslash lambda\_j\; =\; -\backslash frac\{4\}\{h^2\}\; \backslash sin^2\backslash left(\backslash frac\{\backslash pi\; (j\; -\; 1)\}\{2n\}\backslash right)$

:$v\_\{i,j\}\; =\; \backslash begin\{cases\}$

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}

:$$

\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.)

:$v\_\{i,j\}\; =\; \backslash begin\{cases\}$

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}

:$\backslash lambda\_j\; =\; -\backslash frac\{4\}\{h^2\}\; \backslash sin^2\backslash left(\backslash frac\{\backslash pi\; (j-\backslash frac\{1\}\{2\})\}\{2n\; +\; 1\}\backslash right)$

:$v\_\{i,j\}\; =\; \backslash sqrt\{\backslash frac\{2\}\{n+0.5\}\}\; \backslash sin\backslash left(\backslash frac\{\backslash pi\; i\; (2j\; -\; 1)\}\{2n\; +\; 1\}\backslash right)$

:$\backslash lambda\_j\; =\; -\backslash frac\{4\}\{h^2\}\; \backslash sin^2\backslash left(\backslash frac\{\backslash pi\; (j-\backslash frac\{1\}\{2\})\}\{2n\; +\; 1\}\backslash right)$

:$v\_\{i,j\}\; =\; \backslash sqrt\{\backslash frac\{2\}\{n+0.5\}\}\; \backslash cos\backslash left(\backslash frac\{\backslash pi\; (i-0.5)\; (2j\; -\; 1)\}\{2n\; +\; 1\}\backslash right)$

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

:$\backslash frac\{v\_\{k+1\}\; -2v\_k\; +\; v\_\{k-1\}\}\{h^2\}\; =\; \backslash lambda\; v\_\{k\},\; \backslash \; k=1,...,n,\; \backslash \; v\_0\; =\; v\_\{n+1\}\; =\; 0.$

Rearranging terms, we get

:$v\_\{k+1\}\; =\; (2\; +\; h^2\; \backslash lambda)v\_k\; -\; v\_\{k-1\}.\; \backslash !$

Now let $2\; \backslash alpha\; =\; (2\; +\; h^2\; \backslash lambda)$. Also, assuming $v\_1\; \backslash neq\; 0$, we can scale eigenvectors by any nonzero scalar, so scale $v$ so that $v\_1\; =\; 1$.

Then we find the recurrence

:$$

v_0 = 0

\,\!

:$$

v_1 = 1.

\,\!

:$$

v_{k+1} = 2 \alpha v_{k} - v_{k-1}

\,\!

Considering $\backslash alpha$ as an indeterminate,

:$v\_\{k+1\}\; =\; U\_k\; (\backslash alpha)\; \backslash ,\backslash !$

where $U\_k$ is the kth Chebyshev polynomial of the 2nd kind.

Since $v\_\{n+1\}\; =\; 0$, we get that

:$U\_n\; (\backslash alpha)\; =\; 0\; \backslash ,\backslash !$.

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 $2\; \backslash alpha\; =\; (2\; +\; h^2\; \backslash lambda)$.

These zeros are well known and are:

:$$

\alpha_k = \cos\left(\frac{k \pi}{n+1}\right).

\,\!

Plugging these into the formula for $\backslash lambda$,

:$$

2 \cos\left(\frac{k \pi}{n+1}\right) = h^2 \lambda_k + 2

\,\!

:$$

\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

:$$

\lambda_k = -\frac{4}{h^2}\sin^2\left(\frac{k \pi}{2(n+1)}\right).

\,\!

In the Neumann case, we are solving

:$\backslash frac\{v\_\{k+1\}\; -2v\_k\; +\; v\_\{k-1\}\}\{h^2\}\; =\; \backslash lambda\; v\_\{k\},\; \backslash \; k\; =\; 1,...,n,\; \backslash \; v\text{'}\_\{0.5\}\; =\; v\text{'}\_\{n+0.5\}\; =\; 0.$

\,\!

In the standard discretization, we introduce $v\_\{0\}\backslash ,\backslash !$ and $v\_\{n+1\}\backslash ,\backslash !$ and define

:$$

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

:$$

v_1 - v_0 = 0, \ v_{n+1} - v_n = 0.

If we make a change of variables,

:$$

w_k = v_{k+1} - v_k, \ k = 0,...,n

\,\!

we can derive the following:

:$$

\begin{alignat}{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{alignat}

with $w\_\{n\}\; =\; w\_\{0\}\; =\; 0$ being the boundary conditions.

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

:$$

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

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

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

:$$

\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

:$\backslash frac\{v\_\{k+1\}\; -2v\_k\; +\; v\_\{k-1\}\}\{h^2\}\; =\; \backslash lambda\; v\_\{k\},\; \backslash \; k=1,...,n,\; \backslash \; v\_0\; =\; v\text{'}\_\{n+0.5\}\; =\; 0.$,

where $v\text{'}\_\{n+0.5\}\; :=\; \backslash frac\{v\_\{n+1\}\; -\; v\_n\}\{h\}.$

We need to introduce auxiliary variables $v\_\{j\; +\; 0.5\},\; \backslash \; j\; =\; 0,...,n.$

Consider the recurrence

:$v\_\{k+0.5\}\; =\; 2\; \backslash beta\; v\_\{k\}\; -\; v\_\{k-0.5\},\; \backslash text\{\; for\; some\; \}\backslash beta\; \backslash ,\backslash !$.

Also, we know $v\_0\; =\; 0$ and assuming $v\_\{0.5\}\; \backslash neq\; 0$, we can scale $v\_\{0.5\}$ so that $v\_\{0.5\}\; =\; 1.$

We can also write

:$$

v_{k} = 2 \beta v_{k-0.5} - v_{k-1}

\,\!

:$$

v_{k+1} = 2 \beta v_{k+0.5} - v_{k}.

\,\!

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

:$v\_\{k+1\}\; =\; (4\; \backslash beta^2\; -\; 2)\; v\_\{k\}\; -\; v\_\{k-1\}.\; \backslash ,\backslash !$

And thus our new recurrence will solve our eigenvalue problem when

:$h^2\; \backslash lambda\; +\; 2\; =\; (4\; \backslash beta^2\; -\; 2).\; \backslash ,\backslash !$

Solving for $\backslash lambda$ we get

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

Our new recurrence gives

:$v\_\{n+1\}\; =\; U\_\{2n\; +\; 1\}(\backslash beta),\; \backslash \; v\_\{n\}\; =\; U\_\{2n\; -\; 1\}(\backslash beta),\; \backslash ,\backslash !$

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

And combining with our Neumann boundary condition, we have

:$U\_\{2n\; +\; 1\}(\backslash beta)\; -\; U\_\{2n\; -\; 1\}(\backslash beta)\; =\; 0.\; \backslash ,\backslash !$

A well-known formula relates the Chebyshev polynomials of the first kind, $T\_k(\backslash beta)$, to those of the second kind by

:$$

U_{k}(\beta) - U_{k - 2}(\beta) = T_k (\beta).

\,\!

Thus our eigenvalues solve

:$T\_\{2n\; +\; 1\}\; (\backslash beta)\; =\; 0,\; \backslash \; \backslash lambda\; =\; \backslash frac\{4\; (\backslash beta^2\; -\; 1)\}\{h^2\}.\; \backslash ,\backslash !$

The zeros of this polynomial are also known to be

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

And thus

:$$

\begin{alignat}{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{alignat}

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.

:$$

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

{{Reflist}}

Category:Operator theory

Category:Matrix theory

Category:Numerical differential equations

Category:Finite differences