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Calculate the extreme value of functional

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data
    We have functional ##I(y)=\int_{0}^{2}{y}'(2+e^x{y}')dx## where ##y\in C^1(\mathbb{R})## and ##y(0)=0##. Calculate the extreme value.


    2. Relevant equations



    3. The attempt at a solution

    I am having some troubles here... :/

    From Euler-Lagrange equation we get ##\frac{\partial L}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial L}{\partial {y}'}=0## where first part is obviously ##\frac{\partial L}{\partial y}=0## and the second part tells me that ##\frac{\partial L}{\partial {y}'}=konst.=C##.

    So ##(2+e^x{y}')+{y}'e^x=C##

    ##e^x{y}'=\frac{C}{2}-1=D##, where D is just another constant...

    ##{y}'=De^{-x}## so

    ##y=-De^{-x}+E##

    Now we know that ##y(0)=0## so ##y(0)=-D+E=0## therefore ##E=D##. The second condition says that ##\frac{\partial }{\partial {y}'}L(2)=0##.

    Than ##(2+e^x{y}')+{y}'e^x=0## so ##{y}'=0##.

    From ##y=-De^{-x}+E## we get that ##{y}'=De^{-x}##, therefore ##{y}'(2)=De^{-2}=0## and this gives me ##D=0## (remember that ##E=D##).

    So everything together tells that ##y## is actually ##y\equiv 0## and also ##I(y)=0##.


    I seriously doubt that this is right? :/
     
  2. jcsd
  3. Mar 6, 2014 #2

    pasmith

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    Homework Helper

    Where do you get this condition from? All that's stated in the problem is that [itex]y(0)= 0[/itex] and [itex]y[/itex] is once continuously differentiable. We know that [itex]\dfrac{\partial L}{\partial y'} = 2(1 + D)[/itex] is constant, but without more information we cannot determine that constant (although we can choose it such that [itex]dI/dD = 0[/itex]).

    Assuming you've omitted that condition from the problem, your error is here:
    If [itex]\left.\dfrac{\partial L}{\partial y'}\right|_{(x=2)} = 2(1 + e^2 y'(2)) = 0[/itex] then [itex]y'(2) = -e^{-2} \neq 0[/itex].
     
  4. Mar 6, 2014 #3
    That's something that should be true in general, at least my professor said so. So if one condition ##y(a)=c## is given the other one is ##\frac{\partial }{\partial {y}'}L=0## calculated for ##x=b##. We've probably also proven that.

    You are right. My mistake. Taking that into account, ##D=E=-1## so ##y(x)=e^{-x}-1##.
     
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