I am currently reading Reddy's An Introduction to the Finite Element Method 3rd. ed. Section 2.4.3, and I have stumbled upon the expression: $B(\delta u, u) = \frac{1}{2} \delta[B(u,u)]$, where B is bilinear and symmetric. $\delta u$ in this context is called the variation of u.
Why is there a $\frac{1}{2}$ term in the equation? Where did it come from?
Later in the section, it was shown that:
[EDIT: Included the whole solution as shown in the book]
$$ \begin{align*} B(\delta u, u) &= \int_{0}^{L} a \frac{d\delta u}{dx} \frac{du}{dx} \,dx = \delta \int_{0}^{L} \frac{a}{2} \left( \frac{du}{dx} \right) ^ 2 \,dx \\&=\frac{1}{2}\delta \int_{0}^{L} a \frac{du}{dx} \frac{du}{dx} \,dx = \frac{1}{2} \delta[B(u,u)] \end{align*} $$
However, I still do not see where the $\frac{1}{2}$ term came from.
Thanks.
[EDIT: To add additional context regarding $\delta u$]
Excerpt from Section 2.3.2 of Reddy's An Introduction to the Finite Element Method:
Consider the function F(x, u, u'). For an arbitrarily fixed value of the independent variable x, F depends on u and u'. The change $\epsilon v$ in u, where $\epsilon$ is a constant and v is a function, is called the variation of u and is denoted by $\delta u$: $$\delta u \equiv \epsilon v$$ The operator $\delta$ is called the variational operator. The variation $\delta u$ of a function u represents an admissible change in the function u(x) at a fixed value of the independent variable x.
Example, considering the integral expression of the form
$$I(u) = \int_{a}^{b} F(x, u, u') \,dx, \hspace{0.5cm}u=u(x), \hspace{0.5cm}u'=\frac{du}{dx}$$
The first variation of I(u) is
$$\delta I(u) = \delta \int_{a}^{b} F(x, u, u') \,dx = \int_{a}^{b} \delta F(x, u, u') \,dx$$ $$ = \int_{a}^{b} \left( \frac{\partial F}{\partial u} \delta u + \frac{\partial F}{\partial u'} \delta u' \right) \,dx$$
The factor of $\frac{1}{2}$ comes from the chain rule "run backwards", so roughly speaking it comes from:
$$\delta (u^2)=2u \delta (u)$$
from which we get:
$$u\delta (u)=\frac{1}{2}\delta (u^2)$$
In more detail, when you write this as a bilinear form with some linear operator (integral or matrix) $A$ you may write it as (using a bit sloppy notation to make it more clear what is going on):
$$\delta(B(u,u))=\delta(u^*Au)=\delta(u^*)Au+u^*A\delta(u)$$
But since it is a symmetric form you have:
$$B(\delta(u),u)=\delta(u^*)Au=u^*A\delta(u)=B(u,\delta(u))$$ so that:
$$\delta(B(u,u))=\delta(u^*Au)=\delta(u^*)Au+u^*A\delta(u)=2\delta(u^*)Au=2B(\delta(u),u)$$
which gives us:
$$\frac{1}{2}\delta(B(u,u))=B(\delta(u),u)$$
Note that here we used the product rule. Say we have two different functions $u$ and $v$: $$\delta(u^*Av)=\delta(u^*)Av+u^*A\delta(v)$$ Now, in this relationship just set $v=u$.