# How do we get this equality about bilinear form

1. May 6, 2012

### omer21

1. The problem statement, all variables and given/known data

$B(u,u)=\int_{0}^{L}a\frac{du}{dx}\frac{du}{dx}dx$
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?

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

2. May 6, 2012

### HallsofIvy

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

3. May 6, 2012

### Ray Vickson

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

4. May 7, 2012

### omer21

Yes, it is.

@Ray vickson

Could you explain in more details?