Extremum Function of a Functional

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SUMMARY

The discussion centers on solving the extremum function of a functional defined by the integral J(f)=∫(2xf−f′²+3f²f′)dx with boundary conditions f(0)=0 and f(1)=−1. The user attempts to derive the Euler-Lagrange equation, resulting in Ff=2x+6f f'' and Ff'=-2f' + 6f². The user identifies algebraic errors in their differentiation process, which led to an incorrect formulation of the equation. The resolution of these errors is critical for correctly applying the calculus of variations.

PREREQUISITES
  • Understanding of calculus of variations
  • Familiarity with Euler-Lagrange equations
  • Knowledge of functional derivatives
  • Basic algebra and differentiation skills
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equation in detail
  • Practice solving problems involving functionals and boundary conditions
  • Explore advanced topics in calculus of variations
  • Review common algebraic errors in differentiation
USEFUL FOR

Students and researchers in mathematics, particularly those focused on calculus of variations and functional analysis, will benefit from this discussion.

Shantih
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Homework Statement



J(f)=\int 2xf−f′2+3f2f′dx
f(0)=0,f(1)=−1.

Homework Equations



Ff-\frac{d}{dx}Ff'=0

The Attempt at a Solution



Ff=2x+6f f''
Ff'=-2f' + 6f2

Plugging in, I get:
2x+6f f''- \frac{d}{dx} (-2f&#039; + 6f<sup>2</sup>)<br /> <br /> 2x+6f f&#039;&#039;-12f f&#039;-2f&#039;&#039;=0<br /> <br /> Which doesn&#039;t look correct to me.<br /> I&#039;m guessing my mistake was in the differentiation, but I don&#039;t see it.
 
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I figured it out, algebra errors.
 

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