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You can see the solution here in the two-page "String with beads" (http://www.math.purdue.edu/~eremenko/dvi/beads.pdf). This is from a linear algebra course that I teach.

The assumption of smoothness of $t\mapsto p_t$ cannot imply monotonicity, as @Christian Remling noted in his comment. But if you assume that $t\mapsto p_t$ is monotone (pointwise) then $t\mapsto\lamb …

Your coefficients are piecewise constant, which allows you to write the general solution of your differential equation. First write the general solutions on each interval where they are constant. This …

The number of bound states is indeed finite. This can be proved as follows. First of all, for a bound state your eigenvalue $-k^2$ must be real. This is because your operator with real $u$ and zero bo …

I am not sure what's the exact meaning of "analytic" in "analytic methods". If you expand $|A-\lambda I|$ in the last row or column twice, you obtain a "three term recurrency" for the characteristic p …

A simpler formula is obtained using the Weierstrass notation for elliptic functions: Lame equation with $n=1$ in this form is $$w''=(2\wp(z)+\lambda)w,$$ and the general solution is $$w_{1,2}(z)=e^{\m …

Any $n$ orthogonal vectors are eigenvectors of some symmetric matrix. One example of a sufficient condition which implies that all coordinates of an eigenvector are non-zero is that the matrix has pos …

Solutions of this equation, normalized at $x\to\pm\infty$ are called Mathieu functions of the third kind. See, for example, D. Naylor, On a simplified asymptotic formula for the Mathieu function of th …

With this condition, all eigenvalues are real. … If the products of off diagonal elements are all positive, this quadratic form is positive definite, and we have all real eigenvalues. …

If $\mu$ is real, and $\sqrt{\mu+\lambda}$ is not pure imaginary, the zeros are real, because they are eigenvalues of one of the problems: $y(-\infty)=0,\; y(x^*)=0$ or $y(+\infty)=0,\; y(x^*)=0$, and …

Because all eigenvalues of an Hermitean matrix are real. Therefore, by the Schwarz symmetry principle, all branches of $\lambda$ have analytic continuations to a full neighborhood of the real line. …

Then there is a constant $c$ such that we have $P-c<V<P+c$, which implies that $\lambda_k^\prime-c<\lambda_k<\lambda^\prime_k+c$, where $\lambda_k^\prime$ are the eigenvalues of the potential $P$. … I suppose it is clear (from physical interpretation, or from the analogy with matrices) that bigger potential gives bigger eigenvalues, but for a formal proof of this one can refer on Sturm's comparison …

Simplicity of eigenvalues is proved in the first paragraph of Chapter 2. …

Woodward, Eigenvalues of products of unitary matrices and quantum Schubert calculus, math.AG/9712013, Math. Res. Lett. 5 (1998), 817–836. MR 2000a:14066 …