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.

I wish to find the critical point of the functional $F[X]=\int_0^1(f(t))^2dt$ subject to $\int_0^1f(t)dt=k$ for a constant $k$. I read something about using a Lagrange multiplier to convert it to a free problem, but really have no idea how this would work. I haven't got a brilliant understanding of the calculus of variations so a slow explanation would be very useful.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

$(xf(t)-1)^2\ge 0,x\in R$

$\Rightarrow x^2f(t)^2-2xf(t)+1^2\ge0$

$\displaystyle\Rightarrow \int_0^1(x^2f(t)^2-2xf(t)+1)dt\ge0$

$\displaystyle\Rightarrow x^2\int_0^1f(t)^2dt-2x\int_0^1f(t)dt+1\int_0^1dt\ge0$

This is a quadratic in x which is always greater than or equal to zero. So we must have its discriminant to be less than or equal to zero (otherwise it will have two distinct real roots implying that there are points x in which this expression assumes negative values)

S we have, $\displaystyle (2\int_0^1f(t)dt)^2-4\int_0^1f(t)^2dt\le 0$

$\displaystyle \Rightarrow \int_0^1f(t)^2dt\ge k^2$

We have equality if botth the roots are equal ,

$(xf(t)-1)^2=0$

$\Rightarrow f(t)=\frac{1}{x}$

Using the given condition we have, $k=\frac{1}{x}$

$f(t)=k,\forall t\in[0,1]$

Note that this is a special case of Holder's Inequality:http://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.