How do we get this equality about bilinear form

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Homework Help Overview

The discussion revolves around the properties of a bilinear form defined as \( B(u,u) = \int_{0}^{L} a \frac{du}{dx} \frac{du}{dx} dx \), where \( B(.,.) \) is bilinear and symmetric, and \( \delta \) is a variational operator. Participants are questioning the origin of the factor \( \frac{1}{2} \) in a specific expression involving the variational operator.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are examining the definition and properties of the bilinear form, particularly its symmetry and the application of the variational operator. There is a question regarding the commutativity of the variational operator and why it cannot simply be factored out. Additionally, one participant attempts to relate the factor \( \frac{1}{2} \) to a mathematical analogy involving differentials.

Discussion Status

The discussion is active, with participants exploring different interpretations of the bilinear form and seeking clarification on the mathematical reasoning behind the expressions presented. Some guidance has been offered regarding the structure of the bilinear form, but there is no explicit consensus on the questions raised.

Contextual Notes

There appears to be some confusion regarding the definition of the bilinear form and its application, as well as the assumptions related to the variational operator. The original poster's notation and reasoning are being scrutinized, indicating potential gaps in understanding.

omer21
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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 />
 
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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
?
 
omer21 said:

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
 
HallsofIvy said:
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?
 

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