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]<br />

 

[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]<br />

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?