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I want to minimize the functional $$I=\int_{-1}^1u^2(x)|2x-u'(x)|^2dx$$

Here i applied and found the euler langrange equation and found the differential equation

$$u'^2+2uu'-4u=4x^2$$ given is $u\in C^1 $ and $u(-1)=0 , u(1)=1$

but the minimizer is given as $$u(x)=0:x\in[-1,0], x^2:x\in[0,1]$$

Can anyone help me how to go about with this problem .

I need some idea on this functional as well to see that if we choose approperiate $h\in (-1,1)$ i get a circle arc as a minimizer with radium $\frac{1}{h}$

Functional is $$I=\int_0^1\sqrt{1+y'^2} +h y dx$$

Thank you for your kind guidance .

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over which set are you minimizing $I$? –  Mercy Oct 29 '12 at 9:23
    
@Mercy : Over the $C^2$ class for the 2nd one and $C^1$ for the first one . –  Theorem Oct 29 '12 at 9:27
    
Concerning the first functional: either the Euler-Lagrange is false, or the function you claim to be the minimizer isn't the minimizer. In fact, for $0 < x \le 1$ you have $4x^2=0$, which is false. –  Mercy Oct 29 '12 at 13:32
    
@Mercy : Sorry i had missed a square . now may be you can guide me . –  Theorem Oct 29 '12 at 14:22
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1 Answer

up vote 1 down vote accepted

For the first functional. Since the integrand is non-negative, we have that $I\ge0$. If $u'=2\,x$ then $I=0$. Thus $$ u=x^2+C $$ is a minimizer for any $C\in\mathbb{R}$. Are these the only ones? No. A function $u$ will be a minimizer if $u^2\,|2\,x-u'|\equiv0$. This, and the regularity condition gives two new minimizers: $$ u(x)=\begin{cases} 0 & \text{if }-1\le x<0 \\ x^2 &\text{if }0\le x\le1,\end{cases}\text{ and } u(x)=\begin{cases} x^2 & \text{if }-1\le x<0,\\ 0 & \text{if }0\le x\le1.\end{cases} $$ Only the first one satisfies the boundary conditions $u(-1)=0$, $u(1)=1$.

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Sir , But that implies that $u=x^2 $ is the minimizer in $[-1,1]$ , don't quite understand why the minimizer should be $0$ in [-1,0]. Uniqueness follows from the fact that the integrat is non negative and any other function will have $I\ge 0$ so its clear. Am i right ? –  Theorem Oct 29 '12 at 14:34
    
@Theorem If the minimizer is $0$ in $[-1,0]$ and $x^2$ in $[0,1]$, I think there is an information you did not provide. I would say that the minimization is over $C^1$ functions with certain conditions, for instance $u(-1)=0$ and $u(1)=1$. –  Mercy Oct 29 '12 at 14:50
    
I have edited my answer. –  Julián Aguirre Oct 29 '12 at 14:56
    
@Mercy : Yes exactly , there is a boundary condition which says $u(-1)=0$, and $u(1)=1$ . But mercy how would that change ? I was thinking that i can make the minimizer $0$ in arbitrary interval $[-1,1]$ ? –  Theorem Oct 29 '12 at 21:19
    
@JuliánAguirre : Sir i think you meant to say that $u^2|2x-u'|$ identically zero ? –  Theorem Oct 29 '12 at 21:29
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