Poincaré inequality in dimension one with mixed boundary condition

Question

Suppose that $u$ is $C^1$ in $[0,\pi/2]$ and $u(0)=u’(\pi/2)=0$. I want to derive the following Poincaré inequality $$ \int_0^{\pi/2} u(x)^2\,dx \leq \int_0^{\pi/2} u’(x)^2\,dx. $$

Since $u(0)=0$, we have that $u(x) = \int_0^x u’(s)\,ds$. Hence \begin{align} u^2(x) & \leq \biggl( \int_0^x u’(s)\,dx\biggr)^2 \\ & \leq x\int_0^x u’(s)^2\,ds \end{align} by Cauchy-Schwarz inequality. It follows that \begin{align} \int_0^{\pi/2} u^2(x)\,dx & \leq \int_0^{\pi/2} \int_0^x xu’(s)^2\,ds\,dx \\ & = \int_0^{\pi/2} \int_s^{\pi/2} xu’(s)^2\,dx\,ds \\ & = \int_0^{\pi/2} \biggl(\frac{\pi^2}{8} - \frac{s^2}{2}\biggr) u’(s)^2\,ds \\ & \leq \frac{\pi^2}{8} \int_0^{\pi/2} u’(s)^2\,ds. \end{align}

As you can see, the constant $1$ (better than $\pi^2/8$) can not be obtained by using only the condition $u(0)=0$.

My question is how to utilize the other boundary condition $u’(\pi/2)=0$ to derive this better constant. I have seen this conclusion in one paper and the author said that the constant can be obtained by considering the eigenvalue problem

\begin{cases} -u’’ = \lambda u \\ u(0)=u’(\pi/2)=0 \end{cases}

and calculating that $1$ is the minimal eigenvalue of the above equation. What’s the principle behind this?

Answer 1

We want to show that $$\int_0^{\pi/2} u(x)^2\,dx \le \int_0^{\pi/2} u'(x)^2\,dx \tag{1}\label{1}$$ for all $u\in C^1[0,\pi/2]$ such that $u(0)=0$ and $u'(\pi/2)=0$.

Let us show that \eqref{1} holds even without any condition on $u'(\pi/2)$.

We have the Poincaré inequality $$\int_0^\pi w(x)^2\,dx \le \int_0^\pi w'(x)^2\,dx \tag{2}\label{2}$$ for all $w\in W_{0}^{{1,2}}(0,\pi)$.

Take any $u\in C^1[0,\pi/2]$ such that $u(0)=0$. For $x\in[0,\pi]$, let $$w(x):=u(x)\,1(x\le\pi/2)+u(\pi-x)\,1(x>\pi/2). \tag{3}\label{3}$$ Then $w\in W_{0}^{{1,2}}(0,\pi)$. So, by \eqref{3} and \eqref{2}, $$\int_0^{\pi/2} u(x)^2\,dx =\frac12\int_0^\pi w(x)^2\,dx \le \frac12\int_0^\pi w'(x)^2\,dx=\int_0^{\pi/2} u'(x)^2\,dx,$$ which proves \eqref{1}. $\quad\Box$

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Remark 1: That the condition $u'(\pi/2)=0$ can be dropped should be clear from the very beginning, because any $u\in C^1[0,\pi/2]$ such that $u(0)=0$ can be however closely approximated in $ W^{{1,2}}(0,\pi/2)$ by a function $v\in C^1[0,\pi/2]$ such that $v(0)=0$ and $v'(\pi/2)=0$.

Remark 2: In view of what has been said here, your statement "As you can see, the constant $1$ (better than $\pi^2/8$) can not be obtained by using only the condition $u(0)=0$" cannot be true. The "reason" why this statement of yours does not follow from your reasoning is that the Cauchy--Schwarz inequality in your reasoning and the inequality resulting from dropping $-\frac{s^2}2$ in the integrand cannot both turn into the corresponding equalities, for any nonzero $u\in C^1[0,\pi/2]$. Indeed, the equality in the Cauchy--Schwarz inequality in your reasoning holds only if $u'$ is almost everywhere equal to a constant $c$, and $c=0$ together with the condition $u(0)=0$ would imply $u=0$.