How to prove $fequiv 0$ $forall xin [a,b]$?$quad$($f''＋pf'＋qf＝0$ with $f(a)＝f(b)＝0$)
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Define $f in C^{2}[a,b]$ satisfying $f''＋pf'＋qf＝0$ with $f(a)＝f(b)＝0$ , where $pin C^{0}[a,b]$ and $qin C^{0}[a,b]$ are two functions. If $qleq0$ , can we prove $fequiv 0$ $forall xin [a,b]$ ? My try: If $fnotequiv 0$ , without loss of generality, we assume that the maximum of $f$ on $[a,b]$ is greater than zero, while notating $f(x_0)＝displaystylemax_{[a,b]} f$ . Then we have $f(x_0) > 0$ , $f'(x_0) ＝ 0$ and $f''(x_0) leq 0$ . I figured out that if we alter the condition $qleq0$ into $q(x)<0$ there evidently exists contradiction. But how to analyze further with the condition $qleq0$ ? Can we still find contradiction if $q(x_0)＝0$ and $f''(x_0)＝0$ ? Any ideas would be highy appreciated!
real-analysis diffe