Lyapunov inequality for BVPs

Question

In some papers, it is said that Lyapunov [1, p. 406] proved the following result.

Let $p:[a,b]\to\mathbb{R}$ be a continuous nonnegative function. If the BVP \begin{align} y^{\prime\prime}+p(t)y=0,\quad{}t\in[a,b]\\ y(a)=0\quad\text{and}\quad{}y(b)=0 \end{align} has a nontrivial solution, then \begin{equation} (b-a)\int_{a}^{b}p(s)\mathrm{d}s>4. \end{equation} Further, $4$ is the best possible constant.

I have no problem with the inequality but the best possible constant $4$. Can you show me or redirect me to a proof showing that $4$ is the best possible?

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Reference

[1] A. Lyapunov, Probleme General de la Stabilite du Mouvement, Ann. Math. Studies 17, Princeton Univ. Press (1949) (reprinted from Ann. Fac. Sci. Toulouse, 9 (1907) 203--474, Translation of the original paper published in Comm. Soc. Math. Kharkow, 1892).

Answer 1

Consider the following function \begin{equation} y(t):= \begin{cases} \frac{1}{1-\delta}(1-|1-2t|),&0\leq{}t\leq\frac{1}{2}-\delta\ \text{or}\ \frac{1}{2}+\delta\leq{}t\leq1\\ 1-\frac{(1-2t)^{2}}{4\delta(1-\delta)},&\frac{1}{2}-\delta\leq{}t\leq\frac{1}{2}+\delta, \end{cases}\label{eqn1}\tag{1} \end{equation} where $\delta\in(0,\frac{1}{2})$ (see Figure 1).

From \eqref{eqn1}, we see that $y$ satisfies \begin{equation} \left\{ \begin{gathered} y^{\prime\prime}(t)+p(t)y(t)=0\quad\text{for}\ 0\leq{}t\leq1\\ y(0)=0\ \text{and}\ y(1)=0, \end{gathered} \right.\nonumber \end{equation} where \begin{equation} p(t):= \begin{cases} 0,&0\leq{}t\leq\frac{1}{2}-\delta\ \text{or}\ \frac{1}{2}+\delta\leq{}t\leq1\\ \frac{2}{t^{2}-t+(\frac{1}{2}-\delta)^{2}},&\frac{1}{2}-\delta\leq{}t\leq\frac{1}{2}+\delta. \end{cases}\label{eqn2}\tag{2} \end{equation} From \eqref{eqn2}, we compute that \begin{equation} \begin{aligned}[] 1\int_{0}^{1}p(s)\mathrm{d}s ={}&\frac{1}{\sqrt{\delta(1-\delta)}}\ln\Biggl(\frac{t-\frac{1}{2}\bigl(1-2\sqrt{\delta(1-\delta)}\bigr)}{t-\frac{1}{2}\bigl(1+2\sqrt{\delta(1-\delta)}\bigr)}\Biggr)\Biggr|_{\frac{1}{2}-\delta}^{\frac{1}{2}+\delta}\\ ={}&\frac{1}{\sqrt{\delta(1-\delta)}}\ln\Biggl(\frac{1+2\sqrt{\delta(1-\delta)}}{1-2\sqrt{\delta(1-\delta)}}\Biggr)=:\ell(\delta). \end{aligned}\nonumber \end{equation}

Readily, $\ell(\delta)>4$ for $\delta\in(0,\frac{1}{2})$, and $\ell(\delta)\searrow4$ as $\delta\searrow0$ (see Figure 2), which proves that $4$ is the best possible constant. $\blacksquare$