5
$\begingroup$

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: http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative#Mixed_Dirichlet-Neumann_boundary_conditions

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

any help welcome

  • 2
    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?2012-04-30
  • 0
    @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.2012-04-30
  • 1
    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 $C\cos(\sqrt{-\lambda}x)$.2012-04-30
  • 0
    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.2012-04-30
  • 0
    @GiuseppeNegro Thank you! It confirms my first intuition then. So I guess we cannot solve $u=\Delta u$ with such B.C. using the usual separation of variable method... Because when doing so, we decompose the solution along the eigenvectors, and here without a discrete spectrum I do not know how to do it.2012-05-01
  • 0
    I guess it would be better to ask another question regarding the PDE you're approaching. Be sure to write down the boundary conditions and to link to this page. AFAIK, there are techniques to reduce non-homogeneous problems to homogeneous ones.2012-05-01
  • 1
    @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.2012-06-01
  • 0
    They aren't eigenfunctions: it is an inhomogeneous problem so the solutions are not actually eigenfunctions, they are just solutions to a PDE containing a parameter. The special thing about this situation comes from the Fredholm alternative: if $\lambda$ is an eigenvalue of the corresponding homogeneous problem, then there are zero or infinitely many solutions; if on the other hand $\lambda$ is not an eigenvalue, then there are zero or one solutions.2018-04-25

1 Answers 1

3

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 solution is: $$ X_{\lambda}(x) = \frac{a\cos(\sqrt{\lambda}x)}{\cos(\sqrt{\lambda}L)}. $$ Given the denominator is non-zero.

Thanks to the remark of Michael Seifert. This problem with a non-homogeneous boundary condition is like a Helmholtz equation, in order it has a unique solution $\lambda$ CANNOT be an eigenvalue of the eigenvalue problem below: i.e., for $j=1,2,3,\dots$ $$ \lambda \neq \frac{(2j - 1)^2 \pi^2}{4 L^2} $$ in order that (1) has a unique solution with boundary value $$ X'(0) = 0, \quad X(L)=a. $$

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.

  • 0
    Nitpick: $\lambda$ cannot be *any* positive value, since if $\sqrt{\lambda} = n \pi/2 L$ for odd $n$, then $\cos(\sqrt{\lambda} L) = 0$. In other words, the *complement* of the spectrum of $-\Delta$ for these boundary conditions is a countable set of points. (Weird.)2018-04-25
  • 0
    @MichaelSeifert Thanks for the remark. I revised my old answer.2018-04-25