Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to find the eigenvalues and eigenvectors of the Laplacian with mixed boundary conditions on $[0,L]$:

More precisely:

$$X''(x) = \lambda X(x)$$ with $X'(0)=0$ and $X(L)=a$.

I know how to do it with pure Dirichlet or pure Neumann, but not for this mixture.

Could you help me or point me to the right reference ?

Thanks folk

Just found part of the answer here:

but I am not sure how to relate it to parameter $a$ in the question.

any help welcome

share|cite|improve this question
What's the problem in proceeding "by hand"? Take the general integral of $X''-\lambda X =0$ and find conditions on the arbitrary constants so that it fulfils boundary conditions. Doesn't it work? – Giuseppe Negro Apr 30 '12 at 17:39
@GiuseppeNegro Hi, thanks for the hint; I tried this already. I find a zero eigenvalue (ok) but then I cannot exclude positive eigenvalues, and I do not get a discrete spectrum. So I believe I may be wrong. – mellow Apr 30 '12 at 17:59
This is not surprising: if $a \ne 0$ this problem is not homogeneous. For $\lambda < 0$ you have exactly one eigenfunction if $\cos(\sqrt{-\lambda}L)\ne 0$, that is $[a/ \cos(\sqrt{-\lambda}L)]\cos (\sqrt{-\lambda}x)$, and a one-parameter family of eigenfunctions if $\cos(\sqrt{-\lambda}L)=0$, that is $C\cos(\sqrt{-\lambda}x)$. – Giuseppe Negro Apr 30 '12 at 18:52
If the problem is homogeneous ($a=0$), then $[a/ \cos(\sqrt{-\lambda}L)]\cos (\sqrt{-\lambda}x)=0$. That's why you get a discrete spectrum. Indeed, I am not sure it is correct to speak of "spectrum" in the non-homogeneous case, but I'm far from being an expert. – Giuseppe Negro Apr 30 '12 at 18:55
@mellow, if you're interested in solving $u = \Delta u$ with mixed BC you might be interested in Fokas' book "A Unified Approach To Boundary Value Problems" (or the research papers that underpin the book). The introduction speaks about the failure of classical methods (Fourier transforms, separation of variables) for PDEs with mixed BC. I believe in the later sections he applies a new method to solve a number of elliptic equations with mixed BC on convex polygonal domains. – in_wolframAlpha_we_trust Jun 1 '12 at 7:08

For the $-\Delta$ operator is positive definite if the Dirichlet boundary is non-empty, here your problem is actually to find $\lambda >0$ such that: $$ -X'' = \lambda X.\tag{1} $$ General solution to (1) is $$ X = A\cos(\sqrt{\lambda}x)+ B\sin(\sqrt{\lambda}x), $$ and $$ X' = -A\sqrt{\lambda}\sin(\sqrt{\lambda}x)+ B\sqrt{\lambda}\cos(\sqrt{\lambda}x). $$ The boundary condition $$ X'(0)=0\implies B = 0, $$ and $$ X(L) = a\implies A\cos(\sqrt{\lambda}L) = a\implies A = \frac{a}{\cos(\sqrt{\lambda}L)}.\tag{2} $$ Therefore the eigenfunction for $\lambda$ is: $$ X_{\lambda}(x) = \frac{a\cos(\sqrt{\lambda}x)}{\cos(\sqrt{\lambda}L)}. $$

Some remarks: This tells us that this mixed boundary eigenvalue problem on an interval can have any positive $\lambda $ as $-\Delta$'s eigenvalue when the Neumann boundary value is $0$ and Dirichlet boundary is non-zero. Notice that for Dirichlet eigenvalue problem on an interval, you could only have complete squares $k^2$ (un-normalized) as $\lambda$. The wikipedia pages mellow posted has normalized eigenfunctions for mixed Neumann-Dirichlet boundary value problems as: $$ X_j(x) = \sqrt{\frac{2}{L}} \cos\left(\frac{(2j - 1) \pi x}{2 L}\right), $$ with eigenvalue $$ \lambda_j = \frac{(2j - 1)^2 \pi^2}{4 L^2}. $$ This is because the wikipedia page uses homogeneous boundary condtions on both Neumann and Dirichlet boundaries: $$ X'(0) = 0\quad \text{and}\quad X(L) = 0. $$ Notice the second Dirichlet boundary condition will change (2) to $$ A\cos(\sqrt{\lambda}L) = 0\implies \sqrt{\lambda}L = \frac{(2j-1)\pi}{2}. $$ That's why wikipedia's page has that solution.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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