Optimizing y(x) for \int_a^b y^2(1+(y')^2) \, dx with given boundary conditions

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Homework Help Overview

The discussion revolves around finding the extremal of the functional \(\int_a^b y^2(1+(y')^2) \, dx\) under specific boundary conditions \(y(a)=y_{0}\) and \(y(b)=y_{1}\). The subject area pertains to calculus of variations and differential equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the Euler-Lagrange equation and its implications for the problem. There are attempts to derive a first-order ordinary differential equation (ODE) from the functional, with some questioning the correctness of the approach taken. Others suggest that additional parameters may arise from the integration process.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the Euler-Lagrange equation and its application to the problem. Some guidance has been offered regarding the nature of the ODE derived from the functional, but there is no explicit consensus on the correct approach or solution at this stage.

Contextual Notes

There is a mention of boundary conditions and the need to determine limits of integration, which may influence the parameters involved in the solution. The original poster expresses uncertainty about handling these limits.

jimmycricket
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Homework Statement


Find the extremal for the case
\int_a^b y^2(1+(y')^2) \, dx
where y(a)=y_{0}, y(b)=y_{1}

Homework Equations

The Attempt at a Solution


Using the Euler-Lagrange equation for a functional that doesn't depend on x I get
F-y'\frac{\partial F}{\partial y'}=c
\Leftrightarrow y^2(1-(y')^2)=c
\Leftrightarrow \int \frac{1}{\sqrt{1-\frac{c}{y^2}}}dy=\int dx
\Leftrightarrow y=\frac{-c}{x^2-1}
Now I have to sub this y(x) into the original integral and I am comfortable doing the integral apart from what to do for the upper and lower limits of integration.
 
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There should be another free parameter from the integration.
You can use the limits of the integration to fix those two parameters.
 
That doesn't look like any version of Euler Lagrange I know. I'd try looking it up again. You should get a 2nd order ODE.
 
In this case F does not depend on x so the E-L equation is reduced to
F-y'\frac{\partial F}{\partial y'}=c and you are left with a 1st order ode.
 

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