Lecture 11: Oscillation Theory

tags: 224a

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Recall that the solutions of

$u^{″}=\lambda u$ on

$[0,\pi ]$ were

$u_n=\sin (n\pi x)$, which has exactly

$n-1$ zeros in this interval. It turns out that this property is shared by the eigenfunctions of all operators of the more general type

$L=-d^2/d{x}^2+q(x)$ (and consequently by all self-adjoint regular Sturm-Liouville operators on

$[a,b]$, by the reduction in lecture 10).

Theorem. Suppose

$L=-d^2/d{x}^2+q(x)$ is defined on

$C_B^2[a,b]$ with with separated boundary conditions. Then it has is a lowest eigenvalue

${\lambda }_1$, and if the eigenvalues are ordered as

${\lambda }_1<{\lambda }_2<\dots$ then the eigenvector

$u_n$ has exactly

$n-1$ zeros in

$(a,b)$.

This theorem has a convenient added bonus: our proof in Lecture 9 of existence of an eigenbasis of solutions to

$Lu=\lambda u$ relied on the assumption that

$\ker \{L\}=\{0\}$. The above theorem implies that any such

$L$ has a least eigenvalue

${\lambda }_1\in \mathbb{R}$, so the assumption is automatically satisfied if we replace

$L$ by

$L-{\lambda }_1+1$; since this does not change the eigenvalues or eigenvectors, the conclusion of Lecture 9 still holds.

Remark. Note that the lower bound is not true in the case of periodic boundary conditions; this can be seen for example by considering

$-d^2/d{x}^2$ on

$[0,2\pi ]$ with periodic BC.

Change of Variables: Prufer Transformation

For the proof it will be convenient to write the boundary conditions slightly differently as:

$$B_{1}u=\cos (\alpha )u(a)-\sin (\alpha )u^{\prime }(a)=0,B_{2}u=\cos (\beta )u(b)-\sin (\beta )u^{\prime }(b)=0,$$
for

$\alpha \in [0,\pi )$ and

$\beta \in (0,\pi ]$, without any loss of generality. The key idea is to look at the vector

$(u(x),u^{\prime }(x))\in {\mathbb{R}}^2$ in polar coordinates, as

$(\rho (x),\theta (x))$. Note that

$\theta :[a,b]\to \mathbb{R}$ is only defined upto an additive multiple of

$2\pi$. Since

$u$ is twice differentiable we can choose

$\theta (x)$ to be differentiable, and uniqely determine this multiple by setting the value of

$\theta (c)$ for some fixed

$c\in [a,b]$. Moreover, if at any point

$\rho (x)=\sqrt{u(x{)}^2+u^{\prime }(x{)}^2}=0$ on

$[a,b]$, the uniqueness of the IVP implies that

$u\equiv 0$ on

$[a,b]$, which we will avoid, so we assume this is also not the case. Thus

$u(x)=0$ iff

$\theta (x)=0$ for such a parameterization.

Making this change of variables, the equation

$Lu=\lambda u$ is equivalent to the system of ODE:

$${\theta }^{\prime }=\cos (\theta {)}^2+(\lambda -q)\sin (\theta {)}^2(\ast ),{\rho }^{\prime }=\rho (1+q-\lambda )\frac{\sin (2\theta )}{2}.$$
Moreover, the boundary conditions are now simply

$\theta (a)=\alpha (\mathrm{mod}\pi )$ and

$\theta (b)=\beta (\mathrm{mod}\pi )$.

The second equation is a first order ODE in

$\rho (x)$ which always has a unique solution given a solution

$\theta (x)$ to the system. This means it in order to find a solution

$u(x)$ to the original system, it is both necessary and sufficient to find a solution

$\theta (x)$ to this one.

The crucial observation is that a solution

$\theta (x)$ of (*) is always increasing at its zeros, which correspond to the zeros of

$u(x)$:

$${\theta }^{\prime }(x)=\cos (\theta {)}^2=1$$
Thus,

$\theta (x)$ must increase by exactly

$\pi$ between two consecutive zeros. Letting

$\mathrm{\#}(u)$ denote the number of zeros of a solution

$u$ in

$(a,b)$, we have

$$\mathrm{\#}(u)=\lceil \theta (b)/\pi \rceil -\lfloor \theta (a)/\pi \rfloor -1.$$

Boundary Conditions, Monotonicity, and Eigenvalues

We now take the boundary conditions into account and study solutions of the eigenproblem. Let

$u_a(x,\lambda )$ be the solution of the IVP

$Lu=\lambda u$ with data

$$u_a(a,\lambda )=\sin (\alpha ),u_a^{\prime }(a,\lambda )=\cos (\alpha ),$$ clearly satisfying the left BC, and not identically zero by the uniqueness property. Let

${\theta }_a(x,\lambda )$ denote the corresponding angular function, defined so that

${\theta }_a(a,\lambda )=\alpha$. Observe that

$u_a(x,\lambda )$ satisfies the right BC if and only if

${\theta }_a(b,\lambda )=\beta$, which is therefore precisely the condition for

$\lambda$ to be an eigenvalue.

Lemma 1. For every

$x\in (a,b]$,

${\theta }_a(x,\lambda )$ is a strictly increasing function of

$\lambda$.
Proof. Teschl Theorem 1.3

$◻$

Lemma 2. For every

$x\in (a,b]$,

$$\underset{\lambda \to -\mathrm{\infty }}{lim}{\theta }_a(x,\lambda )=0.$$
Proof. See Coddington Chap 8 end of sec 2.

Proof of Theorem. By Lemma 2, there is some

${\lambda }_1$ below which

${\theta }_a(b)<\beta$ so all eigenvalues must be above this

${\lambda }_1$. On the other hand, each time

${\theta }_a(b)=\beta +n\pi$ we obtain an eigenvector of

$L$ with exactly one more zero. Since we proved in Lecture 9 that

$L$ has infinitely many eigenvalues, all simple, this must happen infinitely many times, as desired.

$◻$