How do we get this equality about bilinear form

[omer21]

Homework Statement

$B(u,u)=\int_{0}^{L}a\frac{du}{dx}\frac{du}{dx}dx$
B(.,.) is bilinear and symmetric, δ is variational operator.

In the following expression, where does $\frac{1}{2}$ come from? As i know variational operator is commutative why do not we just pull δ to the left?

$B(\delta u,u)=\int_{0}^{L}a\frac{d\delta u}{dx}\frac{du}{dx}dx=\delta\int_{0}^{L}\frac{a}{2}\left(\frac{du}{dx}\right)^{2}dx=\frac{1}{2}δ\int_{0}^{L}a\frac{du}{dx}\frac{du}{dx}dx=\frac{1}{2}δ\left[B(u,u)\right]$

 

[HallsofIvy]

First, the "binlinear form" you have written makes no sense since it does not operate on two things in order to be bilinear. Is it possible that the form is
$$B(u, v)= \int_0^L \frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dx$$
?

 

[Ray Vickson]

Homework Statement

$B(u,u)=\int_{0}^{L}a\frac{du}{dx}\frac{du}{dx}dx$
B(.,.) is bilinear and symmetric, δ is variational operator.

In the following expression, where does $\frac{1}{2}$ come from? As i know variational operator is commutative why do not we just pull δ to the left?

$B(\delta u,u)=\int_{0}^{L}a\frac{d\delta u}{dx}\frac{du}{dx}dx=\delta\int_{0}^{L}\frac{a}{2}\left(\frac{du}{dx}\right)^{2}dx=\frac{1}{2}δ\int_{0}^{L}a\frac{du}{dx}\frac{du}{dx}dx=\frac{1}{2}δ\left[B(u,u)\right]$

It comes from the same place as it does in $x\, dx = \frac{1}{2} d(x^2),$ and does so for exactly the same reason.

RGV

 

[omer21]

$$B(u, v)= \int_0^L \frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dx$$

Yes, it is.

@Ray vickson

Could you explain in more details?