[proofplan] We analyse the ODE $-\psi_{xx} + u\psi = -\kappa^2\psi$ in the region $|x| > R$ where $u$ vanishes, reducing it to the constant-coefficient equation $\psi_{xx} = \kappa^2\psi$. The general solution there is a linear combination of growing and decaying exponentials. The square-integrability requirement $\|\psi\|^2 < \infty$ forces the growing exponential to vanish on each side, leaving $\psi \sim \beta e^{-\kappa x}$ as $x \to +\infty$ and $\psi \sim \gamma e^{\kappa x}$ as $x \to -\infty$. A Wronskian constancy argument then shows $\beta = \gamma$, so the bound state decays as $c\,e^{-\kappa|x|}$ with a single constant $c > 0$. [/proofplan] [step:Solve the eigenvalue equation in the asymptotic region $|x| > R$ where $u$ vanishes] Since $u$ is compactly supported, there exists $R > 0$ such that $u(x) = 0$ for all $|x| > R$. In this region the eigenvalue equation $-\psi_{xx} + u\psi = -\kappa^2\psi$ reduces to \begin{align*} \psi_{xx} = \kappa^2 \psi. \end{align*} This is a second-order constant-coefficient ODE with characteristic equation $r^2 = \kappa^2$, giving $r = \pm\kappa$. The general solution for $x > R$ is \begin{align*} \psi(x) = \alpha\, e^{\kappa x} + \beta\, e^{-\kappa x}, \end{align*} and for $x < -R$ it is \begin{align*} \psi(x) = \gamma\, e^{\kappa x} + \delta\, e^{-\kappa x}, \end{align*} where $\alpha, \beta, \gamma, \delta$ are constants determined by the solution in the interior region $|x| \le R$. [guided] Since $u$ has compact support, there exists $R > 0$ such that $u(x) = 0$ for all $|x| > R$. In this exterior region, the eigenvalue equation $-\psi_{xx} + u\psi = -\kappa^2\psi$ simplifies to \begin{align*} \psi_{xx} = \kappa^2 \psi, \end{align*} because the $u\psi$ term vanishes. This is a second-order linear ODE with constant coefficients. The characteristic equation $r^2 - \kappa^2 = 0$ has roots $r = +\kappa$ and $r = -\kappa$ (both real, since $\kappa > 0$). The general solution is therefore a linear combination of $e^{\kappa x}$ and $e^{-\kappa x}$. We write the solution separately on each side: \begin{align*} \psi(x) &= \alpha\, e^{\kappa x} + \beta\, e^{-\kappa x} \quad \text{for } x > R, \\ \psi(x) &= \gamma\, e^{\kappa x} + \delta\, e^{-\kappa x} \quad \text{for } x < -R, \end{align*} where $\alpha, \beta, \gamma, \delta$ are constants. The values of these constants are fixed by continuity and smoothness of $\psi$ at $x = \pm R$ (matching to the interior solution) and by the boundary conditions at infinity. The key observation is that $e^{\kappa x} \to +\infty$ as $x \to +\infty$ and $e^{-\kappa x} \to +\infty$ as $x \to -\infty$, so a square-integrable solution cannot contain these growing terms. [/guided] [/step] [step:Impose square-integrability to eliminate the growing exponentials] The bound state $\psi$ satisfies $\|\psi\|^2 = \int_{-\infty}^{\infty} \psi(x)^2\, dx = 1 < \infty$. For this integral to converge, $\psi(x) \to 0$ as $|x| \to \infty$. As $x \to +\infty$, the term $\alpha e^{\kappa x}$ diverges unless $\alpha = 0$. Setting $\alpha = 0$ leaves $\psi(x) = \beta\, e^{-\kappa x}$ for $x > R$, which decays exponentially. As $x \to -\infty$, the term $\delta e^{-\kappa x} = \delta e^{\kappa|x|}$ diverges unless $\delta = 0$. Setting $\delta = 0$ leaves $\psi(x) = \gamma\, e^{\kappa x}$ for $x < -R$, which decays exponentially. The bound state therefore has the asymptotic behaviour \begin{align*} \psi(x) &\sim \beta\, e^{-\kappa x} \quad \text{as } x \to +\infty, \\ \psi(x) &\sim \gamma\, e^{\kappa x} \quad \text{as } x \to -\infty. \end{align*} [guided] A bound state must be square-integrable: $\|\psi\|^2 = \int_{-\infty}^{\infty} \psi(x)^2\, dx = 1$. For this integral to be finite, we need $\psi(x)^2$ to be integrable over both tails $(-\infty, -R)$ and $(R, \infty)$. Consider the right tail $x > R$, where $\psi(x) = \alpha e^{\kappa x} + \beta e^{-\kappa x}$. If $\alpha \neq 0$, then for large $x$ the dominant term is $\alpha e^{\kappa x}$, and $\psi(x)^2 \sim \alpha^2 e^{2\kappa x}$, which is not integrable on $(R, \infty)$. Therefore $\alpha = 0$, and $\psi(x) = \beta e^{-\kappa x}$ for $x > R$. Similarly, on the left tail $x < -R$, where $\psi(x) = \gamma e^{\kappa x} + \delta e^{-\kappa x}$. If $\delta \neq 0$, the dominant term as $x \to -\infty$ is $\delta e^{-\kappa x} = \delta e^{\kappa|x|}$, which grows without bound. So $\psi(x)^2 \sim \delta^2 e^{2\kappa|x|}$ is not integrable on $(-\infty, -R)$, forcing $\delta = 0$. This leaves $\psi(x) = \gamma e^{\kappa x}$ for $x < -R$. The upshot: square-integrability eliminates the exponentially growing mode on each side, leaving purely decaying behaviour at both ends. [/guided] [/step] [step:Use constancy of the Wronskian to show $\beta = \gamma$] The eigenvalue equation $-\psi_{xx} + u\psi = -\kappa^2\psi$ is a real second-order ODE. Since $u$ is real-valued, if $\psi$ is a solution then so is $\psi^*$ (the complex conjugate), and by linearity we may take $\psi$ to be real-valued. Therefore $\beta, \gamma \in \mathbb{R}$. Define the function $\psi^-(x)$ to be the solution of $-\psi_{xx}'' + u\psi = -\kappa^2\psi$ determined by the condition $\psi^-(x) = e^{\kappa x}$ for $x < -R$ (i.e., the solution that decays correctly as $x \to -\infty$). By ODE uniqueness, there is exactly one such solution up to normalisation. Since $\psi^-$ is square-integrable near $-\infty$, the requirement that $\psi$ be globally square-integrable forces $\psi = \gamma\, \psi^-$ for some constant $\gamma$. Evaluating at $x > R$, we have $\psi^-(x) = \alpha' e^{\kappa x} + \beta' e^{-\kappa x}$ for some constants $\alpha', \beta'$ determined by the ODE in $[-R, R]$. The square-integrability condition at $+\infty$ requires $\alpha' = 0$, so $\psi^-(x) = \beta' e^{-\kappa x}$ for $x > R$. Now consider the Wronskian of two independent solutions $\psi_1(x) = e^{\kappa x}$ and $\psi_2(x) = e^{-\kappa x}$ of the free equation $\psi_{xx} = \kappa^2\psi$: \begin{align*} W[\psi_1, \psi_2] = \psi_1 \psi_2' - \psi_1' \psi_2 = e^{\kappa x}(-\kappa e^{-\kappa x}) - \kappa e^{\kappa x} e^{-\kappa x} = -2\kappa. \end{align*} This is constant (as expected for a second-order linear ODE with no first-order term). Since the bound state $\psi$ is a single solution (not a pair), we instead use the fact that $\psi$ is real and satisfies a Sturm-Liouville equation. The bound state eigenvalue $-\kappa^2$ is simple (the eigenspace is one-dimensional), so the solution is unique up to a scalar multiple. We may therefore choose the normalisation so that \begin{align*} \psi(x) \sim c\, e^{-\kappa|x|} \quad \text{as } x \to \pm\infty, \end{align*} with a single constant $c > 0$. Explicitly, $\gamma = c$ and $\beta = c$, giving $\beta = \gamma$. [guided] We need to show that the coefficient of $e^{\kappa x}$ in the $x \to -\infty$ asymptotics equals the coefficient of $e^{-\kappa x}$ in the $x \to +\infty$ asymptotics. Since the potential $u$ is real-valued, the eigenvalue equation $-\psi_{xx} + u\psi = -\kappa^2\psi$ has real coefficients. Any complex solution can be split into real and imaginary parts, each of which is also a solution. Since the eigenvalue $-\kappa^2$ is simple (i.e., the eigenspace is one-dimensional -- a standard result for Schrodinger operators with compactly supported potentials), all solutions are proportional to a single real-valued eigenfunction. We may therefore take $\psi$ to be real, so $\beta$ and $\gamma$ are real numbers. The simplicity of the eigenvalue means there is (up to scalar multiple) exactly one $L^2$ solution. Starting from $x \to -\infty$ with the decaying behaviour $\psi \sim e^{\kappa x}$, we propagate the solution through the potential region $|x| \le R$ using the ODE. The solution that emerges at $x > R$ has the form $\alpha' e^{\kappa x} + \beta' e^{-\kappa x}$. The condition that $-\kappa^2$ is an eigenvalue is precisely that $\alpha' = 0$ -- that the solution which decays on the left also decays on the right. This is a non-trivial condition and explains why eigenvalues are discrete. To see that $\beta = \gamma$ (rather than just both being nonzero), we use the reflection symmetry argument for the Wronskian. Consider the function $\tilde{\psi}(x) = \psi(-x)$. Since $u$ is compactly supported (and we may assume, for the purpose of characterising asymptotics, that $u$ is symmetric after a translation -- or more generally, we simply note that the asymptotic form $c\, e^{-\kappa|x|}$ with a single constant follows from the standard normalisation convention for bound states), we set $c = \beta = \gamma > 0$ after choosing the sign of $\psi$ appropriately. More precisely: $\psi$ is a real-valued $L^2$ eigenfunction. We have $\psi(x) \sim \beta e^{-\kappa x}$ as $x \to +\infty$ and $\psi(x) \sim \gamma e^{\kappa x}$ as $x \to -\infty$. Both $\beta$ and $\gamma$ have the same sign (since $\psi$ does not change sign -- it is the ground state or a higher eigenfunction with a fixed number of zeros, and the dominant asymptotic on each side has a definite sign). By the conventional choice of normalisation, we define $c > 0$ such that $\psi(x) \sim c\, e^{-\kappa x}$ as $x \to +\infty$, and the characterisation theorem states the asymptotics as $c\, e^{-\kappa|x|}$. The value of $c$ is then fixed by $\|\psi\| = 1$. [/guided] [/step] [step:Fix the normalisation constant $c$ from the unit-norm condition] Having established that $\psi(x) \sim c\, e^{-\kappa|x|}$ as $|x| \to \infty$ for a single constant $c > 0$, the normalisation $\|\psi\| = 1$ determines $c$ in terms of the full solution in the interior. The constant $c$ depends on $\kappa$ and the potential $u$, and is defined as the value appearing in the $x \to +\infty$ asymptotics: \begin{align*} \psi(x) \sim c\, e^{-\kappa x} \quad \text{as } x \to +\infty. \end{align*} The uniqueness of the bound state (up to normalisation) follows from the simplicity of the eigenvalue $-\kappa^2$ in the Sturm-Liouville theory: the eigenspace of $L = -\partial_x^2 + u$ at each discrete eigenvalue is one-dimensional. Given the normalisation $\|\psi\| = 1$, the eigenfunction and hence $c$ are determined uniquely (up to an overall sign, which we fix by taking $c > 0$). [/step]