simple but not obvious identity in calculus

Question

I'm having a little difficulty understanding this identity while studying for my calculus of variation exam.

\begin{equation} y'\delta y' = -y''\delta y \end{equation}

Can someone please explain this?

Answer 1

These sorts of statements have to be considered within the context of integrals. In the calculus of variations the variation itself is usually taken to vanish at the end points of the integration.

Consider the following expression,

$$ \int_{a}^b y' \delta y' \ dt ,$$

we will perform integration by parts. Recall that the formula is $\int_a^b u \ dv = (uv)_a^b - \int_a^b v du$. We will choose $u=y'$ and $dv=\delta y'\ dt$.

$$ \int_{a}^b y' \delta y' \ dt = (y' \delta y)_a^b - \int_a^b y'' \delta y \ dt$$

since the variation vanishes at $t=a$ and $t=b$ the boundary term vanishes leaving us with,

$$ \int_{a}^b y' \delta y' \ dt = \int_a^b \Big( -y'' \delta y \Big) \ dt,$$

from this we can see that in the appropriate context $y'\delta y'$ is equivalent to $-y'' \delta y$. As a short hand we say $y' \delta y' = -y'' \delta y$.