Extremum of functional via functional derivative

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I have the following functional:
$$
H[lambda(T)]
= intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$



with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:



$$
int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
$$

In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.










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    $begingroup$


    I have the following functional:
    $$
    H[lambda(T)]
    = intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$



    with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:



    $$
    int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
    $$

    In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have the following functional:
      $$
      H[lambda(T)]
      = intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$



      with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:



      $$
      int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
      $$

      In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.










      share|cite|improve this question









      $endgroup$




      I have the following functional:
      $$
      H[lambda(T)]
      = intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$



      with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:



      $$
      int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
      $$

      In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.







      functional-analysis derivatives optimization






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      asked Nov 26 '18 at 12:47









      zodiaczodiac

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