Solving differential equation from variational principle

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SUMMARY

The discussion centers on solving the differential equation derived from the Euler-Lagrange variational principle: \(\frac{\partial}{\partial x}\left(\frac{1}{\sqrt{y}}\frac{dy}{dx}\right)=0\). The user seeks to express \(y\) as a function of \(x\) given the boundary conditions \(y(0)=y_{1}\) and \(y(D)=y_{2}\). It is concluded that integrating the equation leads to \(\frac{1}{\sqrt{y}}\frac{dy}{dx}=f(y)\), where \(f(y)\) remains unknown. Clarification is provided that if the equation is treated as an ordinary differential equation (ODE), it can be simplified and solved through integration by parts.

PREREQUISITES
  • Understanding of Euler-Lagrange equations
  • Knowledge of ordinary differential equations (ODEs)
  • Familiarity with boundary value problems
  • Basic calculus and integration techniques
NEXT STEPS
  • Explore methods for solving ordinary differential equations (ODEs)
  • Research boundary value problem techniques in differential equations
  • Learn about the Euler-Lagrange equation in classical mechanics
  • Investigate the role of function forms in differential equations
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Students and researchers in applied mathematics, physicists working with variational principles, and engineers dealing with differential equations in their projects.

JulieK
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I have the following differential equation which I obtained from Euler-Lagrange
variational principle
[itex]\frac{\partial}{\partial x}\left(\frac{1}{\sqrt{y}}\frac{dy}{dx}\right)=0[/itex]

I also have two boundary conditions: [itex]y\left(0\right)=y_{1}[/itex] and
[itex]y\left(D\right)=y_{2}[/itex] where [itex]D[/itex], [itex]y_{1}[/itex] and [itex]y_{2}[/itex] are known
numbers.
I assume that I should integrate once to get
[itex]\frac{1}{\sqrt{y}}\frac{dy}{dx}=f\left(y\right)[/itex]

where [itex]f\left(y\right)[/itex] is a function of [itex]y[/itex]. I would like to get
[itex]y[/itex] as a function of [itex]x[/itex] but the problem is that I don't know the
form of [itex]f\left(y\right)[/itex].
My question, how to solve this equation to get [itex]y[/itex] as a function
of [itex]x[/itex]. Is it possible to guess the form of [itex]f\left(y\right)[/itex], e.g.
from the bounday conditions.
 
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JulieK,
I am not sure I understand: why are you applying a partial differential operator to what looks like a function of one variable? If that is just a typo, then your equation gets much easier: you just equate the term inside the parenthesis to a constant and get an ODE that is integrable by part.
 

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