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Consider the problem \begin{equation} -u'' + u = 2 \sin(x) \mbox{ in } (0, \pi), \hspace{2mm} u(0) = 0, u'(\pi) = -1. \end{equation} I want to show that a solution of the problem satisifies the following: For every $b > 0$ there exists $K_b > 0$ such that \begin{equation} \inf_{v \in \mathcal{P}_n}(||u-v||_{L^{\infty}(0,\pi)} + ||(u-v)'||_{L^{\infty}(0,\pi)}) \leq K_b e^{-bn} \mbox{ for all } n \in \mathbb{N}_0. \end{equation} where $\mathcal{P}_n$ is the space of all polynomials with coefficients in $\mathbb{C}$ of degree $= n$. The solution is $u(x) = \sin(x) \in C^{\infty}(0,\pi)$. For the approximation I use the approximation at chebyshev nodes by chebyshev polynomials \begin{equation} |u(x) - p_{n-1}(x)| \leq \frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n \max_{\xi \in (0,\pi)}|u^{(n+1)}(\xi)| \leq \frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n . \end{equation} Therefore \begin{equation} ||u - p_{n-1}||_{L^{\infty}} \leq \frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n . \end{equation} I stuck at estimating $\frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n$ to get the result above for some given $b > 0$. I tried using the stirling formula for $n!$ but this didn't work for me.

Orb
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  • For your information, you can solve analytically the equation. – NN2 Nov 18 '22 at 14:39
  • u(x) = sin(x) would be the solution. – Orb Nov 18 '22 at 14:42
  • and with this solution, you can prove easily what you want to prove, right? – NN2 Nov 18 '22 at 14:44
  • Yes, by using the taylor series of $\sin(x)$. – Orb Nov 18 '22 at 14:46
  • But what if I consider a weak solution of the given problem. This doesn't have to be the analytical solution. – Orb Nov 18 '22 at 14:55
  • What do you mean by considering the ‘weak solution’? What is the weak solution here? – NN2 Nov 18 '22 at 15:01
  • The solution $u$ that satisfies the variational formulation $\int^{\pi}{0}u'\varphi' + \int^{\pi}{0}u\varphi = 2 \int^{\pi}_{0}{\sin(x) \varphi}$ for all $\varphi \in C^{\infty}(0,\pi)$ with $u(0) = 0$ and $u'(\pi) = 0$ where $u'$ is the weak derivative of $u$ and $u(0) = 0$ and $u'(\pi)$ means the evaluation of $u$ or $u'$ by using the trace operator and a continuous representative of $u$. – Orb Nov 18 '22 at 15:04
  • in this case you should rewrite the question and ask "I want to show that a weak solution of the problem written below satisifies the following" – NN2 Nov 18 '22 at 15:09
  • You can't replace $u$ by $u-v$ in the third equation from the top. – xpaul Nov 18 '22 at 20:19
  • Right. If I instead look at the analytical solution I would get $||u - I_h u||{L^{\infty}(0,\pi)} \leq \frac{\pi^2}{8}\sup{x \in (0,\pi)}|u''(x)|$ and using the cea lemma $\inf_{v \in \mathcal{P}n}||u-v||{H^1} \leq ||u - I_h u||{H^1}$ I can get estimates for $||u - v||{L^{\infty}}$ and $||(u - v)'||_{L^{\infty}}$. But I can't see how this leads to exponential convergence. Maybe there is some way to use the gronwall lemma. – Orb Nov 18 '22 at 23:30

1 Answers1

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If your problem is to know $n$ such that $$||u - p_{n-1}||_{L^{\infty}} \leq \frac{1}{2^{n-1}n!} \left( \frac{\pi}{2}\right)^n =10^{-k}$$ i means that you want to solve for $n$ the equation $$n!=4^{- n}\, \pi ^n\,(2\times10^k)$$

Have a look here and using @robjohn approximation $$n \sim \frac{e\pi }{4}\,e^{W(t)}-\frac 12 \qquad \text{where}\qquad t=\frac 1{e \pi}\,\log \left(\frac{64 }{\pi ^4}\, 10^{4k}\right)$$

Trying for $k=10$, the above expression gives $n=12.2985$ while the "exact" solution is $12.2997$.

Claude Leibovici
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