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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.
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
\[\#(u) = \lceil \theta(b)/\pi\rceil-\lfloor \theta(a)/\pi\rfloor-1.\]
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. \(\square\)