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Math Help - Derivative of Functional

  1. #1
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    Derivative of Functional

    hi there, i am actually studying about functional equation.
    I got stucked with some derivatives problem,
    and where i could find nowhere to refer or study from,
    because it seems it is out of university book level.

    my question is this :

    what does it means by taking derivative with respect to partial derivative?
    can anyone visualize this idea to me?
    because i couldn't figure out the term with respect to partial derivative,
    when it comes to a functional equation,
    of which is a differential equations.

    For e.g : F(x,y(x),y'(x),y''(x)) , find \frac{d}{dx} of  \partial F(x,y(x),y'(x),y''(x)) /  \partial y' .

    How can we write the full solution with partial derivative respect to y' ? and how bout y''?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by danong View Post
    hi there, i am actually studying about functional equation.
    I got stucked with some derivatives problem,
    and where i could find nowhere to refer or study from,
    because it seems it is out of university book level.

    my question is this :

    what does it means by taking derivative with respect to partial derivative?
    can anyone visualize this idea to me?
    because i couldn't figure out the term with respect to partial derivative,
    when it comes to a functional equation,
    of which is a differential equations.

    For e.g : F(x,y(x),y'(x),y''(x)) , find \frac{d}{dx} of  \partial F(x,y(x),y'(x),y''(x)) /  \partial y' .

    How can we write the full solution with partial derivative respect to y' ? and how bout y''?
    I assume you're talking about a standard Calculus of Variations problem with a functional of the form J=\int_a^b F(x,y,y',y'')dx. Now, for this specific case the variational derivative is of the form F_y-\frac{\partial }{\partial x}F_{y'}=0. Now, I agree this can be confusing at first but what that really means is call F(x,y,y',y'')=F(t,u,v,w). So, for example if F(x,y,y',y'')=x^2y+y'+\frac{y'}{y''} you would get F(t,u,v,w)=t^2u+v+\frac{v}{w}. Now, the Euler-Lagrange equation then says take F_u so in this case F_u=t^2 and F_v=1+\frac{1}{v}. But! You have to sub back in what that really means, i.e. F_y=x^2,F_{y'}=1+\frac{1}{y''}. Now, treating these again as actual functions of x you compute F_y-\frac{d}{dx}F_{y'}. The point is that F_y,F_{y'} intend you to treat F as a function of the independent variables x,y,y',y'' and differentiate with respect to the specified one. But! Once done with that the \frac{d}{dx} intends you to treat F_y,F_{y'} again as a function of functions of x.

    Write back if you need more clatification.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by danong View Post
    hi there, i am actually studying about functional equation.
    I got stucked with some derivatives problem,
    and where i could find nowhere to refer or study from,
    because it seems it is out of university book level.

    my question is this :

    what does it means by taking derivative with respect to partial derivative?
    can anyone visualize this idea to me?
    because i couldn't figure out the term with respect to partial derivative,
    when it comes to a functional equation,
    of which is a differential equations.

    For e.g : F(x,y(x),y'(x),y''(x)) , find \frac{d}{dx} of  \partial F(x,y(x),y'(x),y''(x)) /  \partial y' .

    How can we write the full solution with partial derivative respect to y' ? and how bout y''?
    It might be best to look at an example. Consider

    L(x,y,y') = \frac{\sqrt{1 + y'^2}}{\sqrt{y}}.

    Then

     <br />
\frac{\partial L}{\partial y'} = \frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2}}<br />
(treat x,y and y' as all independent variables). Then

     <br />
\frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) = \frac{d}{dx} \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2} } \right)

    = \frac{\partial}{\partial x} \underbrace{ \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2 }}\right)}_{\text{this will} = 0\, \text{- no}\,x } + \frac{\partial}{\partial y} \underbrace{ \left(\frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2 }}\right)}_{\text{hold}\, y'\, \text{fixed} } y'  + \frac{\partial}{\partial y'} \underbrace{ \left( \frac{1}{\sqrt{y}} \frac{y'}{\sqrt{1+y'^2}} \right)}_{\text{hold}\, y\, \text{fixed} } y''.

    For the last two derivatives, treat y and y' and independent.

    (A little slow)
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