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!
(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 $y(x):=1-{x\over2}$.