Suppose \(f:\mathbb R\to\mathbb R\) satisfies \[ f''(x)=(x^2-1)f(x) \] for every \(x\in\mathbb R\), with \[ f(0)=1, \qquad f'(0)=0. \] Show that \[ f(x)\longrightarrow0 \qquad(x\to\infty). \]
Solution
1. The function\[ g(x)\coloneqq e^{-x^2/2} \]satisfies the same differential equation and initial conditions as \(f\).
Differentiating gives \[ g'(x)=-x e^{-x^2/2} \] and \[ g''(x) =(x^2-1)e^{-x^2/2} =(x^2-1)g(x). \] Also \[ g(0)=1, \qquad g'(0)=0. \] Thus \(g\) satisfies precisely the initial-value problem stated for \(f\).
2. One has\[ f(x)=e^{-x^2/2} \]for every \(x\in\RR\).
The equation \[ y''=(x^2-1)y \] is a linear second-order differential equation with continuous coefficient \(x^2-1\). The uniqueness theorem for its initial-value problem says that two solutions with the same values of \(y(0)\) and \(y'(0)\) coincide. Step 1 shows that \(f\) and \(g\) have the same equation and initial data, so \(f=g\).
3. Therefore\[ \boxed{\lim_{x\to\infty}f(x)=0}. \]
By step 2, \[ f(x)=e^{-x^2/2}. \] Since \(x^2/2\to\infty\) as \(x\to\infty\), the right-hand side tends to \(0\).
4. Q.E.D.
Step 3 is the required limit.