I am trying to solve the eigenvalue problem $$\begin{cases} y''=\lambda y \\y(0)=y(1)=0\end{cases}$$ I use the finite difference to discretize the ODE with BVs. I get the following equation $$\frac{y_{i-1}+2y_i+y_{i+1}}{(\Delta x)^2}=\lambda_{\Delta x}y_i,$$ where $\Delta x =1/{(n+1)}.$

Then I get the following $n \times n$ tridiagonal matrix formulation

$\frac{1}{(\Delta x)^2}\begin{bmatrix}-2&1&~&~\\1&-2&1&~&~ \\~&~&\ddots\\ ~&~&1&-2&1\\~&~&~&1&-2\end{bmatrix}\begin{bmatrix}y_1 \\y_2\\\vdots\\ y_{n-1}\\y_n\end{bmatrix}=\lambda_{\Delta x}\begin{bmatrix}y_1 \\y_2\\\vdots\\ y_{n-1}\\y_n\end{bmatrix}$

I know that the ODE has infinitely many eigenvalue values and eigenfunction. We can compute the eigenvalue numerically by computing the eigenvalues of the matrix on the LHS.

I have two questions.

  1. Why can I only get the approximation of the first $n$th eigenvalue ($n\times n $ matrix) instead of the kth to $(k+n-1)$th eigenvalues?

  2. If I want to compute the $m$th eigenvalue, do I have to compute the eigenvalues of the $m\times m$ matrix ? When $m$ becomes very large, it needs a lot of computations. Do we have numerical method to compute the $m$th eigenvalue directively?

  • 1
    $\begingroup$ I think you would benefit from finding a closed formula for the eigenvalues and the eigenvectors of your $n$by $n$ matrix. Comparing the eigenvalues of the matrix to the eigenvalues of the differential operator would help settle some of your questions. $\endgroup$ – Carl Christian Nov 29 '17 at 17:48
  • $\begingroup$ I agree, and shouldn't $y = C e^{\sqrt{\lambda}x}$ work as a general solution? $\endgroup$ – I. Pittenger Jun 27 at 14:10

I'm not clear on what you're asking in your first question, but I can answer the second one for you (which I think will likely help with the first question, too).

First, let's move the $(\Delta x)^2$ over to the other side. The eigenvalues of the given matrix are well-known to be $$\lambda_j=2\left(\cos\left(\frac{\pi j}{N+1}\right)-1\right).$$ So, indeed, we have an explicit formula for each eigenvalue. Note that the largest eigenvalue (in magnitude) is $\lambda_N\approx -4,$ and the smallest is $\lambda_1,$ with $\lambda_j \approx -\left(\frac{j\pi }{N+1}\right)^2$ for small $j$. We also get unit (in $2$ norm) eigenvectors $z_j$ with entries $$z_j(k)=\sqrt{\frac{2}{N+1}}\sin\left(\frac{jk\pi}{N+1}\right),$$ which you will likely notice as being a discrete analogue of the eigenfuctions for the continuous problem.

I should also note that it is common to make the diagonal entries positive by multiplying both sides of your matrix equation by $-1$, providing you with an SPD matrix.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.