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For the Energy,

$$ Q(u)=\int_I(1+|u'(x)|^2)^{1/4} dx $$

where $u(0)=0$ and $u(1)=1$, $u$ is $C^1$ in $I$ and continuous up to the boundary, $I=(0,1)$. How to show the infimum of $Q$ is $1$?

Julián Aguirre
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Yourent
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1 Answers1

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It is clear that $Q(u)>1$. On the other hand, since $(a+b)^{1/4}\le a^{1/4}+b^{1/4}$ if $a,b\ge0$, we have $$ Q(u)\le 1+\int_I|u'(x)|^{1/2}dx. $$ Take $u_n(x)=x^n$. Then $$ 1<Q(u_n)\le1+n^{1/2}\int_I x^{(n-1)/2}dx=1+\frac{2n^{1/2}}{n+1}\implies \lim_{n\to\infty}Q(u_n)=1. $$

Julián Aguirre
  • 76,354
  • Thanks! What if we cannot assume any function to be u? – Yourent Oct 12 '13 at 19:05
  • Then you would have to solve the varíational equation and solve it to find the minimizing function. But there is no minimizing function. So the only way to find the infimum is to construct a minimizing sequence. – Julián Aguirre Oct 13 '13 at 05:51