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Let $I$ be the open interval $(0, 1)$. I am trying to solve a problem in Brezis' Functional Analysis

Exercise 8.31 Consider the Sturm-Liouville operator $A u=-u^{\prime \prime}+u$ on $I$ with Neumann boundary condition $u' (0) = u' (1)=0$.

  1. Compute the eigenvalues of $A$ and the corresponding eigenfunctions.
  2. Given $f \in L^2 (I)$ with $\int_I f=0$, let $u$ be the weak solution of $$ (1) \quad \begin{cases} -u'' + u = f \quad \text {on} \quad I, \\ u'(0)=u'(1)=0. \end{cases} $$ Prove that $$ \|u\|_{L^2(I)} \leq \frac{1}{(1+\pi^2)} \|f\|_{L^2(I)} . $$

I am trying to solve question (2.). In below attempt, I get the inequality with a bigger constant and have not utilized $\int_I f=0$. Could you elaborate on how to obtain the desired constant $\frac{1}{\left(1+\pi^2\right)}$?


It follows from (1) that $$ \int_I [ -u''v + uv ] = \int_I f v, \quad \forall v \in H^2(I), $$ which (by integration by parts) implies $$ \int_I [ u'v' + uv ] = \int_I f v, \quad \forall v \in H^2(I), $$

Substituting $v=u$, we get $$ \int_I |u|^2 \le \int_I fu, $$ which (by Cauchy-Schwarz inequality) implies the desired inequality.

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  • $\begingroup$ The problem gave a hint no? [Apply Question 6 in Problem 49]. $\endgroup$
    – Chee Han
    Jan 5 at 21:31
  • $\begingroup$ @CheeHan I would like to see if there is a more direct approach without appealing to Question 6 of Problem 49. $\endgroup$
    – Akira
    Jan 5 at 21:33

1 Answer 1

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Clearly the corresponding eigen-problem $$ \begin{cases} -u'' + u = \lambda u\quad \text {on} \quad I, \\ u'(0)=u'(1)=0, \end{cases} \tag{*} $$ has eigenvalues $\lambda_n=(n\pi)^2+1$ with the corresponding eigen-functions $u_n=\cos(n\pi t)$, $n=0,1,2,\cdots$. Observe $$\int_Iu_n=\int_Iu_mu_n=0 \text{ for }m\neq n,\int_Iu_n^2=\frac12, m,n=1,2,3,\cdots.$$ Integrating (1) and using $\int_If=0$, one has $$ \int_Iu=0 $$ and hence one has to rule out the eigenvalue $1$. Let $$ u=\sum_{n=1}^\infty u_n\cos(n\pi t), f=\sum_{n=1}^\infty f_n\cos(n\pi t)$$ in (1). Then $$ \sum_{n=1}^\infty((n\pi)^2+1) u_n\cos(n\pi t)=\sum_{n=1}^\infty f_n\cos(n\pi t)$$ and hence $$ ((n\pi)^2+1) u_n=f_n\text{ or }u_n=\frac{f_n}{(n\pi)^2+1}. $$ So $$ \|u\|^2_{L^2}= \frac12\sum_{n=1}^\infty|u_n|^2=\frac12\sum_{n=1}^\infty\frac1{((n\pi)^2+1)^2}|f_n|^2\le \frac12\frac1{(\pi^2+1)^2}\sum_{n=1}^\infty|f_n|^2=\frac1{(\pi^2+1)^2}\|f\|_{L^2}^2 $$ which gives the desired the estimate.

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