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Math Help - Calculus of Variation - Euler Lagrange Question

  1. #1
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    Calculus of Variation - Euler Lagrange Question

    Find the minimum value taken by the following two integral expressions where y and y' are both functions of x

    (a) \int_{0}^{1}((y')^2-6y^2)e^{-5x}dx , y(0)=1, y'(0)=2

    (b) \int_{0}^{1}\frac{(y')^2}{(1+y)^2}dx, y(0)=0, y(1)=e^3-1

    I've had a few stabs at this but my answers are looking dodgy. Could someone please show me the way? ^^


    Here is what I've done so far:
    (a) Using Euler Lagrange Equation \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial F}{\partial y}, I get
    \frac{d}{dx}(2e^{-5x}y')=-12e^{-5x}y
    (2e^{-5x}y')=\frac{-12e^{-5x}y}{5}+A
    y'=\frac{6y}{5}+\frac{A}{2e^{-5x}}
    y(0)=1, y'(0)=2 so A=\frac{8}{5}
    y'=\frac{dy}{dx} so by integrating the last expression for y' I get y=\frac{6xy}{5}+\frac{4e^{5x}}{25}+B
    Applying boundary conditions again, I get B=\frac{21}{25}
    Substituting and rearranging, I end up with y(1-\frac{6x}{5})=\frac{4e^5x+21}{25}\rightarrow y(x)=\frac{4e^5x+21}{25-30x} as the extremal funcation.

    For part (b) I did something similar using F-y'\frac{\partial F}{\partial y'}=constant as the Euler Langrage equation and ended up with  y=\frac{x}{x-1}
    Here is my working out for part (b):
    I=\int_{0}^{1}\frac{(y')^2}{(1+y)^2}dx, y(0)=0, y(1)=e^3-1
    so using F-y'\frac{\partial F}{\partial y'}=constant, I get
    \frac{(y')^2-2(y')^2}{(1+y)^2}=k
    Taking square roots of both sides and multiplying the whole equation by -1, I get
    \frac{y'}{1+y}=A\rightarrow y'=A+Ay \rightarrow y=Ax+Axy+B
    Using boundary condition  y(0)=0, I get B=0
    Substituting and rearranging, I get
    y=\frac{Ax}{1+Ax}
    Using the second boundary condition, I get
    e^3-1=\frac{A}{1+A}
    (e^3-1)(1+A)=A
    A=\frac{e^3-2}{2-e^3}=-1
    Hence,  y=\frac{x}{x-1}
    Last edited by yaohui88; September 2nd 2011 at 09:28 AM.
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  2. #2
    Ted
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    Re: Calculus of Variation - Euler Lagrange Question

    Is y funcion of x?
    Otherwise, the two integral are just numbers.
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    Re: Calculus of Variation - Euler Lagrange Question

    Sorry, yes. y is a function of x
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    Re: Calculus of Variation - Euler Lagrange Question

    Quote Originally Posted by yaohui88 View Post
    Find the minimum value taken by the following two integral expressions where y and y' are both functions of x

    (a) \int_{0}^{1}((y')^2-6y^2)e^{-5x}dx, y(0)=1, y'(0)=2

    (b) \int_{0}^{1}\frac{(y')^2}{(1+y)^2}dx, y(0)=0, y(1)= e^3-1

    I've had a few stabs at this but my answers are looking dodgy. Could someone please show me the way? ^^


    Here is what I've done so far:
    (a) Using Euler Lagrange Equation \frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{\partial F}{\partial y}, I get
    \frac{d}{dx}(2e^{-5x}y')=-12e^{-5x}y
    Good so far, but the next step is, I think, incorrect. Try taking the LHS derivative again. You can't just integrate the way you did, because you don't know what y is.

    (2e^{-5x}y')=\frac{-12e^{-5x}y}{5}+A
    y'=\frac{6y}{5}+\frac{A}{2e^{-5x}}
    y(0)=1, y'(0)=2 so A= \frac{8}{5}
    y'=\frac{dy}{dx} so by integrating the last expression for y' I get y=\frac{6xy}{5}+\frac{4e^{5x}}{25}+B
    Applying boundary conditions again, I get B=\frac{21}{25}
    Substituting and rearranging, I end up with y(1-\frac{6x}{5})=\frac{4e^5x+21}{25}\rightarrow y(x)=\frac{4e^5x+21}{25-30x} as the extremal funcation.

    For part (b) I did something similar using F-y'\frac{\partial F}{\partial y'}=constant as the Euler Langrage equation and ended up with  y=\frac{x}{x-1}
    Hmm. That's not what I get. What is your differential equation?
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    Re: Calculus of Variation - Euler Lagrange Question

    Hmmm...okay. I've updated the original post to show you what I've done with part (b), and I'm pretty certain I have gone wrong somewhere...Is it the same issue with not being able to integrate like I did because I don't know what y is?

    As with part (a), I gave it another shot considering your advice. So, I differentiate again and end up with:
    2e^{-5x}(y''-5y'+6y)=0
    Which gives me a second order differential equation:
    y''-5y'+6y=0
    The solution I get from this is:
    y=e^{2x-2}+e^{3x} which I assume is the extremal function? Then I differentiate to get y'=2e^{2x-2}+3e^{3x}?
    Last edited by yaohui88; September 2nd 2011 at 09:03 AM.
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    Re: Calculus of Variation - Euler Lagrange Question

    Concerning part a):

    I don't know if I agree with your solution of the DE. You've got the right basic functions, but your constants are off. The solution looks like

    y=Ae^{2x}+Be^{3x}, and hence

    y'=2Ae^{2x}+3Be^{3x}. Plugging in the initial conditions yields the system of equations

    1=A+B
    2=2A+3B.

    Can you proceed?

    As for part b), I can see that you're using the equivalent of Troutman's Equation (4) on page 150 of Variational Calculus and Optimal Control, valid when x does not show up explicitly in the integrand of the functional.

    Again, your problem is in solving the DE. You cannot simply integrate the way you have done, when y = y(x) is unknown. Your DE is

    y'-Ay=A.

    Try using the integrating factor method.
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