[proofplan] We solve the eigenvalue equation $-\psi_{xx} + u\psi = -\kappa^2\psi$ in the asymptotic region $|x| > R$ where $u = 0$. The general solution is a combination of growing and decaying exponentials. Square-integrability eliminates the growing exponentials on each side, and a matching argument shows the normalisation constants agree. [/proofplan] [step:Solve the asymptotic equation for $|x| > R$] Since $u(x) = 0$ for $|x| > R$, the bound state equation reduces to \begin{align*} \psi_{xx} - \kappa^2\psi = 0 \end{align*} in both regions $x > R$ and $x < -R$. The general solution of this constant-coefficient ODE is $\psi = \alpha e^{\kappa x} + \beta e^{-\kappa x}$ for constants that may differ in the two regions. [/step] [step:Eliminate growing exponentials via square-integrability] As $x \to +\infty$, the term $e^{\kappa x}$ grows without bound. For $\psi$ to satisfy $\int_{-\infty}^\infty \psi(x)^2\, dx < \infty$, we need $\psi(x) \to 0$ as $x \to +\infty$, which forces the coefficient of $e^{\kappa x}$ to vanish. Similarly, as $x \to -\infty$, the term $e^{-\kappa x}$ grows, so its coefficient must vanish. We conclude: \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*} for constants $\beta, \gamma > 0$ (positivity can be arranged by replacing $\psi$ with $-\psi$ if needed, since the eigenvalue equation is linear). [/step] [step:Match the constants via Wronskian analysis] The bound state eigenvalue $-\kappa^2$ is simple (non-degenerate), meaning the eigenspace is one-dimensional. This is because the Schrödinger equation is a second-order ODE, and the boundary condition $\psi(x) \to 0$ as $x \to +\infty$ selects a one-dimensional subspace of solutions (the decaying exponential direction). Similarly, $\psi(x) \to 0$ as $x \to -\infty$ selects another one-dimensional subspace. For a bound state to exist, these two subspaces must coincide, yielding a unique (up to scalar) solution $\psi_\kappa$. Since $\psi_\kappa$ is unique up to normalisation and satisfies $\psi_\kappa(x) \sim \beta e^{-\kappa x}$ as $x \to +\infty$ and $\psi_\kappa(x) \sim \gamma e^{\kappa x}$ as $x \to -\infty$, both asymptotic coefficients are determined by the same function. Writing $\psi_\kappa(x) \sim c\, e^{-\kappa|x|}$ as $x \to \pm\infty$ with $c = \beta = \gamma$ captures both asymptotics in one expression. [/step]