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Maximizing a functional when the Euler-Lagrange equation's solution violates ICs

  1. Oct 4, 2009 #1


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    I am trying to minimize:


    by choice of [tex]f[/tex], subject to [tex]f(0)=1[/tex] and [tex]f'(x)>0[/tex] for all [tex]x[/tex].

    The (real) solution to the Euler-Lagrange differential equation is:


    rather unsurprisingly. However, this violates [tex]f(0)=1[/tex].

    If we constrain to solutions of the form:


    then numerical optimisation puts [tex]p=1.2848\cdots[/tex] at which point the target is [tex]0.6327\cdots[/tex].

    Is it possible to solve this problem without restricting to a particular solution form?

    Thanks in advance,

  2. jcsd
  3. Oct 4, 2009 #2
    I found the solution to be c1 + c2t.
  4. Oct 4, 2009 #3
    Oh wait, i was wrong. But there has to be another solution, since the euler lagrange equation will be second order
  5. Oct 4, 2009 #4
    Do you have a complex solution as well? That solution will be accompanied by a complex constant so you cannot disregard it
  6. Oct 4, 2009 #5


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    Yeah there are two constants and the second one multiplies something complex (and messy). There's no reason you can't set the second constant to zero to the best of my knowledge.

    In any case, whatever the values of those constants, it won't be the case that f(0)=1.
  7. Oct 4, 2009 #6
    You cannot set it equal to zero. You must solve it from the initial values. Try it, youll see that it works

  8. Oct 4, 2009 #7


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    I get that the limit of the general solution as t -> 0 is signum(C2)*infinity where C2 is the constant on the messy complex part. Clearly you can't satisfy the initial conditions...
  9. Oct 4, 2009 #8
    Please state the general solution you have. If you cannot satisfy the initial conditions with that, you will have to settle with some constrained function.
  10. Oct 5, 2009 #9


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    The full solution is:


    where [tex]Ei(a,z)=z^{(a-1)} \,\Gamma{(1-a,z)}[/tex].
  11. Oct 5, 2009 #10


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    No solution can satisfy the initial condition f(0) = 1. The differential equation you're trying to solve is

    [tex]t^2 f''(t) - (t-2)(t f'(t) - f(t)) = 0,[/tex]

    right? At t=0, this reduces to -2 f(0) = 0, and hence it can only be the case that a solution to the DE which includes the point t = 0 will necessarily have f(0) = 0.

    Compare to the case of Bessel's Equation:

    [tex]x^2 y'' + xy' + (x^2-n^2)y = 0[/tex]
    has a regular singular point at x = 0. Note that for [itex]n \neq 0[/itex] all the solutions [itex]J_n(x)[/itex] have [itex]J_n(0) = 0[/itex], as at x = 0 the DE reduces to [itex]-n^2y(0) = 0[/itex]. At n = 0 we get 0 = 0 so [itex]J_0(x)[/itex] may take on any finite value at x = 0. The other independent solution of course diverges at the origin and may not be included in a solution which contains x = 0 as a valid point.
    Last edited: Oct 5, 2009
  12. Oct 5, 2009 #11


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    Yeah I know the differential equation doesn't have a solution satisfying the initial condition.

    The question I was asking in the first message is how you go about minimizing that functional given this.
  13. Oct 5, 2009 #12


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    You could enforce your initial condition by adding a Lagrange multiplier, perhaps. You would then need to minimize the functional

    [tex]S[f(t)] = \int_{0-}^\infty dt e^{-t} (tf'(t) - f(t))^2 - \lambda \left[f(0) - 1].[/tex]

    Writing [itex]f(0) - 1 = \int_{0-}^\infty dt (f(t) - 1)\delta(t)[/itex] you can combine terms into one integral. Minimizing then gives you

    [tex]e^{-t}\left[t^2 f''(t) + (2-t)(t f'(t) - f(t))\right] + \frac{\lambda}{2}\delta(t) = 0,[/tex]

    which you can now solve. Note that I specified that your integration limit is actually slightly less than zero in order for the delta function to give you a non-zero value when integrated over that range. However, if t is restricted to being zero or greater this may be a shifty thing to do; you could get around it by having the delta function contribute to an integral at some other point, but the cost of doing so is that you would then have a differential equation for f at different points in time, which I believe would be much harder to solve.

    Plus, I don't know if you could get a unique solution, as differentiating the functional with respect to lambda will only give you your desired initial condition.
  14. Oct 6, 2009 #13
    Nice idea to enforce the initial condition. I suspect however that would result in a discontinious solution at the origin, with -infinity slope, while a positive slope is required. The curve f=Ct will make the always positive integrand zero, so you might want to consider a D1 curve such in connecting the initial value to f = Ct in some small interval.

  15. Oct 6, 2009 #14


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    Yes, I'm also not sure it would give a satisfactory solution due to the discontinuity being a boundary point. If one sets

    [tex]f(0)-1 = \int_{0}^\infty dt (f(t-\xi) - 1)\delta(t-\xi),[/tex]
    to get around that problem then I don't even think one could combine the multiplier term with the original integral to get one differential equation - the two terms would have to vanish separately, which of course just reduces to the original attempt at the problem.
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