2
$\begingroup$

Could anyone help me find the extremals of I[y]=\int_0^1(y')^2 \mathrm dx+\{y(1)\}^2 subject to $y(0)=1$

Most crucially I can't work out how to find the boundary $x=1$. I'm trying to go back to first principles and letting $y \rightarrow y+\alpha \eta$. Here the normal step would be to differentiate and set $\alpha=0$ and hence derive the Euler-Lagrange equation.

If anyone can explain to me how to deal with this case i'd be very grateful!

  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/2235/discussion-between-davide-giraudo-and-lhs)2012-01-17

1 Answers 1

2

(Disregard this answer if the question is a homework problem in variational calculus.)

It is obvious that $I(y)$ can be made arbitrarily large, so we only have to look for the minimum. The situation here is so simple that we can do with Schwarz' inequality. Put $y(1)=:y_1$. Then (y_1-1)^2=\Bigl(\int_0^1 1\cdot y'(x)\ dx\Bigr)^2\leq \int_0^1 1^2\ dx\cdot \int_0^1 y'^2(x)\ dx with equality iff y'(x) is constant. It follows that I(y):=\int_0^1 y'^2(x)\ dx + y_1^2\geq 2y_1^2-2y_1+1=2\Bigl(y_1-{1\over2}\Bigr)^2 +{1\over2}\ . Therefore $I(y)\geq{1\over2}$, and the minimum is attained for the function $y(x):=1-{x\over2}$.

  • 0
    @DavideGiraudo Ah ok, I see what's happened now. Thanks so much anyway!!2012-01-18