[proofplan] We prove isospectrality of the Lax pair $(L, A)$: differentiating the eigenvalue equation $L\psi = \lambda\psi$ in time and substituting the Lax equation $L_t = [L,A]$ yields $(L - \lambda)\tilde{\psi} = \dot{\lambda}\psi$, where $\tilde{\psi} = \psi_t + A\psi$. Taking the $L^2$ inner product with $\psi$ and using self-adjointness of $L$ forces $\dot{\lambda} = 0$. With $\dot{\lambda} = 0$, the relation $(L - \lambda)\tilde{\psi} = 0$ shows $\tilde{\psi}$ is an eigenfunction of $L$ at the same eigenvalue. Discreteness and finiteness of the bound states is then a consequence of the compact-support assumption on $u$ and the analytic structure of the Jost function. [/proofplan] [step:Differentiate the eigenvalue equation $L\psi = \lambda\psi$ with respect to time] Let $\psi(\cdot, t)$ be an eigenfunction of $L(t)$ with eigenvalue $\lambda(t)$, so that \begin{align*} L(t)\psi(\cdot, t) = \lambda(t)\psi(\cdot, t). \end{align*} Differentiating both sides with respect to $t$: \begin{align*} L_t \psi + L\psi_t = \dot{\lambda}\psi + \lambda\psi_t. \end{align*} Since $(L, A)$ is a Lax pair, we have $L_t = LA - AL$. Substituting: \begin{align*} (LA - AL)\psi + L\psi_t = \dot{\lambda}\psi + \lambda\psi_t. \end{align*} Using $L\psi = \lambda\psi$ to replace $LA\psi = L(A\psi)$ and $AL\psi = A(\lambda\psi) = \lambda A\psi$: \begin{align*} L(A\psi) - \lambda(A\psi) + L\psi_t - \lambda\psi_t = \dot{\lambda}\psi. \end{align*} Factoring: \begin{align*} (L - \lambda)(\psi_t + A\psi) = \dot{\lambda}\psi. \end{align*} Define $\tilde{\psi} := \psi_t + A\psi$. Then \begin{align*} (L - \lambda)\tilde{\psi} = \dot{\lambda}\psi. \end{align*} [guided] We start from the eigenvalue equation $L(t)\psi(\cdot, t) = \lambda(t)\psi(\cdot, t)$, where both $L$ and $\lambda$ may depend on $t$, and differentiate in $t$. The product rule on the left gives two terms: $L_t\psi$ from differentiating $L$, and $L\psi_t$ from differentiating $\psi$. The Lax equation $L_t = [L, A] = LA - AL$ lets us replace $L_t$. The key algebraic manipulation is to collect all terms involving $\psi_t$ and $A\psi$ together. Observe that $AL\psi = A(\lambda\psi) = \lambda A\psi$ (since $\lambda$ is a scalar and $A$ is linear). So the $LA - AL$ term contributes $L(A\psi) - \lambda(A\psi)$, which combines naturally with $L\psi_t - \lambda\psi_t$ to give $(L - \lambda)(\psi_t + A\psi)$. This factoring is the essential step -- it shows that the combination $\tilde{\psi} = \psi_t + A\psi$ is the natural object to study, since it satisfies $(L - \lambda)\tilde{\psi} = \dot{\lambda}\psi$. [/guided] [/step] [step:Use self-adjointness of $L$ to show $\dot{\lambda} = 0$] Take the $L^2$ inner product of both sides of $(L - \lambda)\tilde{\psi} = \dot{\lambda}\psi$ with $\psi$: \begin{align*} \dot{\lambda}\,(\psi, \psi)_{L^2} = (\psi,\, (L - \lambda)\tilde{\psi})_{L^2}. \end{align*} Since $L$ is self-adjoint, we transfer $L - \lambda$ to the first argument: \begin{align*} (\psi,\, (L - \lambda)\tilde{\psi})_{L^2} = ((L - \lambda)\psi,\, \tilde{\psi})_{L^2} = 0, \end{align*} because $(L - \lambda)\psi = 0$ by the eigenvalue equation. Since $\|\psi\|^2 = (\psi, \psi)_{L^2} > 0$ (eigenfunctions are nonzero), we conclude \begin{align*} \dot{\lambda} = 0. \end{align*} The eigenvalue $\lambda$ is constant in time. [guided] The inner product trick exploits the self-adjointness hypothesis on $L$. Self-adjointness means $(L\psi, \varphi)_{L^2} = (\psi, L\varphi)_{L^2}$ for all $\psi, \varphi$ in the domain of $L$. Since $\lambda$ is real (all eigenvalues of a self-adjoint operator are real), the operator $L - \lambda$ is also self-adjoint. By taking the inner product with $\psi$, we can move $L - \lambda$ from the right slot to the left slot. But in the left slot, it acts on $\psi$, which is an eigenfunction with eigenvalue $\lambda$ -- so $(L - \lambda)\psi = 0$ identically. This annihilates the entire right-hand side, forcing $\dot{\lambda}\|\psi\|^2 = 0$. Why is self-adjointness essential? If $L$ were not self-adjoint, we could not move $(L - \lambda)$ across the inner product. The adjoint $L^*$ would differ from $L$, and $L^*\psi \neq \lambda\psi$ in general, so the right-hand side would not vanish. Non-self-adjoint operators can have eigenvalues that move under perturbation, and indeed isospectrality fails generically without this hypothesis. [/guided] [/step] [step:Conclude that $\tilde{\psi} = \psi_t + A\psi$ is an eigenfunction at the same eigenvalue] Since $\dot{\lambda} = 0$, the relation $(L - \lambda)\tilde{\psi} = \dot{\lambda}\psi$ reduces to \begin{align*} (L - \lambda)\tilde{\psi} = 0, \end{align*} which says that $\tilde{\psi} = \psi_t + A\psi$ is an eigenfunction of $L$ with the same eigenvalue $\lambda$ (provided $\tilde{\psi} \neq 0$; if $\tilde{\psi} = 0$, then $\psi_t = -A\psi$, which gives the evolution equation for the eigenfunction directly). This establishes both claims of the [Isospectral Flow Theorem](/theorems/???): the eigenvalues of $L$ are time-independent, and the operator $\psi \mapsto \psi_t + A\psi$ maps eigenfunctions to eigenfunctions at the same eigenvalue. [/step] [step:Establish discreteness and finiteness of the bound states] The bound state eigenvalues $\{-\kappa_n^2\}$ of the Schrodinger operator $L = -\partial_x^2 + u$ with compactly supported potential $u$ are discrete and finite in number. This is a standard result in spectral theory. The essential spectrum of $L$ is $[0, \infty)$, since $u$ has compact support and $L$ is a compact perturbation (in the resolvent sense) of $-\partial_x^2$. By Weyl's theorem, the essential spectrum is unchanged by such perturbations. Any eigenvalues below the essential spectrum must be discrete (isolated with finite multiplicity). The finiteness of the number of negative eigenvalues follows from the fact that $u$ has compact support. The Jost function $a(k)$, defined via the scattering solution asymptotics, extends to an analytic function in the upper half $k$-plane. The bound state eigenvalues correspond to zeros of $a(k)$ on the positive imaginary axis $k = i\kappa$. Since $a(k)$ is analytic on the upper half-plane, continuous up to the boundary, and satisfies $a(k) \to 1$ as $|k| \to \infty$ in the upper half-plane, it can have only finitely many zeros. Therefore the set of bound state eigenvalues $\{-\kappa_1^2, \ldots, -\kappa_N^2\}$ is finite, with $\kappa_1 > \kappa_2 > \cdots > \kappa_N > 0$. [guided] The discreteness and finiteness claim has two components, each relying on different tools from spectral theory. **Discreteness.** The Schrodinger operator $L = -\partial_x^2 + u$ on $L^2(\mathbb{R})$ with compactly supported $u$ is a self-adjoint operator. Its essential spectrum -- the part of the spectrum that is stable under compact perturbations -- equals $[0, \infty)$. This follows from Weyl's essential spectrum theorem: since $u$ has compact support, the operator $u \cdot$ (multiplication by $u$) is a relatively compact perturbation of $-\partial_x^2$, so $\sigma_{\text{ess}}(L) = \sigma_{\text{ess}}(-\partial_x^2) = [0, \infty)$. Any eigenvalues below $0$ are therefore isolated and of finite multiplicity -- this is precisely the definition of the discrete spectrum. **Finiteness.** That there are only finitely many such eigenvalues (rather than a sequence accumulating at $0$) requires more: it uses the compact support of $u$. The scattering coefficient $a(k)$ is the Wronskian of the Jost solutions, and the compact support of $u$ ensures that $a(k)$ extends to an entire function of $k$ (not just analytic in the upper half-plane). The bound state eigenvalues $-\kappa_n^2$ correspond to the zeros of $a(i\kappa)$ for $\kappa > 0$. Since $a(k) \to 1$ as $|k| \to \infty$ in the closed upper half-plane, and $a$ is analytic there, the zeros form a bounded discrete set -- and a bounded discrete subset of $\mathbb{R}$ is finite. Alternatively, one can use the Bargmann bound: for a compactly supported potential, the number of bound states satisfies $N \le 1 + \frac{1}{2}\int_{-\infty}^{\infty} |x|\,|u(x)|\, dx < \infty$. [/guided] [/step]