Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving differential equation from variational principle

  1. Mar 26, 2013 #1
    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.
     
  2. jcsd
  3. Mar 27, 2013 #2
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook