# Lecture 11: Oscillation Theory
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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 $L=-d^2/dx^2+q(x)$ (and consequently by all self-adjoint regular Sturm-Liouville operators on $[a,b]$, by the reduction in lecture 10).

**Theorem.** Suppose $L=-d^2/dx^2+q(x)$ is defined on $C^2_B[a,b]$ with with separated boundary conditions. Then it has is a lowest eigenvalue $\lambda_1$, and if the eigenvalues are ordered as $\lambda_1<\lambda_2<\ldots$ 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 $-d^2/dx^2$ 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:
$$B_1u = \cos(\alpha) u(a) -\sin(\alpha) u'(a)=0, \quad B_2u = \cos(\beta) u(b) -\sin(\beta) u'(b)=0,$$
for $\alpha\in [0,\pi)$ and $\beta\in(0,\pi]$, without any loss of generality. The key idea is to look at the vector $(u(x),u'(x))\in \R^2$ in polar coordinates, as $(\rho(x),\theta(x))$. Note that $\theta:[a,b]\rightarrow\R$ 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 $\rho(x)=\sqrt{u(x)^2+u'(x)^2}=0$ 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:
$$\theta'=\cos(\theta)^2 + (\lambda -q)\sin(\theta)^2\quad(*), \qquad\rho' = \rho(1+q-\lambda)\frac{\sin(2\theta)}{2}.$$
Moreover, the boundary conditions are now simply $\theta(a)=\alpha (\mod \pi)$ and $\theta(b)=\beta(\mod \pi)$.

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 $u(x)$:
$$ \theta'(x) = \cos(\theta)^2=1$$
Thus, $\theta(x)$ must increase by exactly $\pi$ between two consecutive zeros. Letting $\#(u)$ denote the number of zeros of a solution $u$ in $(a,b)$, we have
$$\#(u) = \lceil \theta(b)/\pi\rceil-\lfloor \theta(a)/\pi\rfloor-1.$$

### 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 $$u_a(a,\lambda)=\sin(\alpha), u_a'(a,\lambda)=\cos(\alpha),$$ 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$\square$

**Lemma 2.** For every $x\in (a,b]$,
$$ \lim_{\lambda\rightarrow -\infty} \theta_a(x,\lambda)=0.$$
*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. $\square$