Let \(x(t)\) solve \[ x''-2b x'+cx=0 \] for all real \(t\), where \(b,c\in\mathbb R\), and suppose \[ x(0)=x(1)=0. \] Prove that \[ x(n)=0 \] for every integer \(n\).
Solution
1. If \(x'(0)=0\), then \(x\) is identically zero.
The function \(x\) satisfies the second-order linear initial-value problem \[ x''-2bx'+cx=0, \qquad x(0)=0, \qquad x'(0)=0. \] The zero function satisfies the same initial-value problem. Uniqueness for linear ordinary differential equations therefore gives \(x\equiv0\).
2. Suppose \(x\) is not identically zero. Then there is a nonzero scalar\(\lambda\in\RR\) such that\[ x(t+1)=\lambda x(t) \qquad\text{for every }t\in\RR. \]
By step 1, \(x'(0)\neq0\). Define \[ y(t)\coloneqq x(t+1). \] Because the differential equation has constant coefficients, \(y\) satisfies the same equation as \(x\). Also \[ y(0)=x(1)=0. \] Set \[ \lambda\coloneqq\frac{y'(0)}{x'(0)} = \frac{x'(1)}{x'(0)}. \] Then \[ h\coloneqq y-\lambda x \] satisfies the same differential equation and has \[ h(0)=0, \qquad h'(0)=0. \] By uniqueness, \(h\equiv0\), so \[ x(t+1)=y(t)=\lambda x(t) \] for every real \(t\).
If \(\lambda=0\), then \(x(t+1)=0\) for every \(t\), hence \(x\equiv0\), contrary to the present assumption. Therefore \(\lambda\neq0\).
3. One has\[ x(n)=0 \]for every integer \(n\).
1. Otherwise, step 2gives\[ x(t+1)=\lambda x(t) \]with \(\lambda\neq0\). Starting from \(x(0)=0\) and iterating forward gives\(x(n)=0\) for every positive integer \(n\). Since\[ x(t)=\lambda^{-1}x(t+1), \]iterating backward from \(x(0)=0\) gives \(x(n)=0\) for every negative integer\(n\) as well.
4. Q.E.D.
Step 3 is the required conclusion.