Eigenvalue

Last time, I was stuck by Sturm Liouville Problem. Finding the eigenvalue.

L u=-\displaystyle\frac{1}{\omega(x)}(p(x)u')'+q(x)u

And if:

\omega(x),p(x)\ge 0, and differentiable.

Given the boundary condition:

\alpha_1u(a)+\beta_1u'(a)=0;

\alpha_2u(b)+\beta_2u'(b)=0.

We know that \omega(x) defines some inner product in Hilbert Space L^2:

\displaystyle\langle f,g \rangle=\int \overline{f}(x) g(x) \omega(x)\mathrm{d}x.

and eigen-function \phi_k(x) satisfies that:

\displaystyle\int \overline{\phi_n}(x)\phi_m(x)\omega(x)\mathrm{d}x=0, if m\neq n.

The theory states that:

  1. The eigenvalues \lambda_1<\lambda_2<\dots<\lambda_n<\dots\rightarrow\infty
  2. Every eigenvalue is simple(algebraic multiplicity is one).
  3. There exists an orthogonal basis mentioned above.

Due to the inner product, we can easily solve one case, if we can do the integral by changing variables. If

L u=-\displaystyle\frac{1}{\omega(x)}(\frac{1}{\omega(x)}u')'

That is p(x)\omega(x)=1, or we can relax the situation by p(x)\omega(x)=const.

It is easy to see if we find another variable \mathrm{d}y=\omega\mathrm{d}x, then

Lu=-\displaystyle\frac{\mathrm{d}^2u}{\mathrm{d}y^2}

But for other cases, it won’t be easy to find out the eigenvalues easily. However, we ruled out the trivial cases that q(x)=const.

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4 thoughts on “Eigenvalue

  1. For regular cases that mentioned above, the eigenvalues are simple. which means the multiplicity is exactly one.
    For other cases, the second statement may not be valid. Check this article for the BVPs. http://www.math.niu.edu/SL2/papers/koze96b.pdf

    However, if you put Sturm Liouville in 2-d plane, with this form:
    Lu=\frac{1}{\omega}\nabla\cdot(p\nabla u)+qu, the problem turns out to be similar to Laplacian’s eigenvalue problem in 2-d domain. The eigen-space is not one-dimensional, certainly. But the first eigenvalue \lambda_1‘s multiplicity is one.

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