Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider $$ F(u)=\int\limits_0^1 (1-u'(x)^2)^2+u(x)^2\, dx, u\in W^{1,4}(0,1), u(0)=u(1)=0. $$ Show, that $F$ is coercive. To do so use the Young inequality $$ 2ab\leq\varepsilon a^2+\frac{b^2}{\varepsilon}~\forall~a,b,\varepsilon > 0. $$


Solution we got:

The coercivity is fullfilled, because of the Young inqequality $$ u'^2\leq\frac{1}{4}u'^4 +c_{\varepsilon} $$

we get

$F(u)=\int\limits_0^1 u'^4-2u'^2+1+u^2\, dx$

$\geq \int\limits_0^1 u'^4 -(\frac{1}{2}u'^4+2c_{\varepsilon})+1+u^2\, dx$

$=\int\limits_0^1\frac{1}{2} u'^4\, dx + \int\limits_0^1 u^2+c\, dx$

$\geq \lVert u\rVert_{W^{1,4}(0,1)}$.

From the definition of coercivity we get

$$ \lim\limits_{\lVert u\rVert\to\infty}\frac{\langle F(u),u\rangle}{\lVert u\rVert}=\infty. $$

I do not really understand what is going on here... Can anybody please explain me that in rather easy words??

share|improve this question
    
Man that's the third time you post this question. This is impressive... –  Tomás Feb 1 '13 at 17:30
    
I delete the other two posts, sorry. –  math12 Feb 1 '13 at 17:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.