Functional Derivative for Specific Question
$begingroup$
Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$
$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$
calculus-of-variations quantum-field-theory functional-calculus
edited Mar 4 at 7:22
Andrews
1,2691421
asked Dec 25 '18 at 7:14
SAKSAK
83
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add a comment |
$begingroup$
Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$
$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$
calculus-of-variations quantum-field-theory functional-calculus
edited Mar 4 at 7:22
Andrews
1,2691421
asked Dec 25 '18 at 7:14
SAKSAK
83
$endgroup$
1
$begingroup$
It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
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– KReiser
Dec 25 '18 at 7:17
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Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32
add a comment |
$begingroup$
Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$
$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$
calculus-of-variations quantum-field-theory functional-calculus
edited Mar 4 at 7:22
Andrews
1,2691421
asked Dec 25 '18 at 7:14
SAKSAK
83
$endgroup$
Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$
$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$
calculus-of-variations quantum-field-theory functional-calculus
calculus-of-variations quantum-field-theory functional-calculus
edited Mar 4 at 7:22
Andrews
1,2691421
asked Dec 25 '18 at 7:14
SAKSAK
83
edited Mar 4 at 7:22
Andrews
1,2691421
asked Dec 25 '18 at 7:14
SAKSAK
83
edited Mar 4 at 7:22
Andrews
1,2691421
edited Mar 4 at 7:22
Andrews
1,2691421
edited Mar 4 at 7:22
Andrews
1,2691421
1,2691421
asked Dec 25 '18 at 7:14
SAKSAK
83
asked Dec 25 '18 at 7:14
SAKSAK
83
asked Dec 25 '18 at 7:14
SAKSAK
83
83
1
$begingroup$
It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
$endgroup$
– KReiser
Dec 25 '18 at 7:17
$begingroup$
Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32
add a comment |
1
$begingroup$
It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
$endgroup$
– KReiser
Dec 25 '18 at 7:17
$begingroup$
Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32
1
1
$begingroup$
It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
$endgroup$
– KReiser
Dec 25 '18 at 7:17
$begingroup$
It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
$endgroup$
– KReiser
Dec 25 '18 at 7:17
$begingroup$
Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32
$begingroup$
Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32
add a comment |
1 Answer
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$begingroup$
Use the generalized binomial theorem:begin{align}
frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
&= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
&= p[f(x)]^{p-1}phi(x)
end{align}
$phi$ is just an arbitrary function used to define the functional $J$.
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
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1 Answer
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$begingroup$
Use the generalized binomial theorem:begin{align}
frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
&= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
&= p[f(x)]^{p-1}phi(x)
end{align}
$phi$ is just an arbitrary function used to define the functional $J$.
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
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add a comment |
$begingroup$
Use the generalized binomial theorem:begin{align}
frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
&= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
&= p[f(x)]^{p-1}phi(x)
end{align}
$phi$ is just an arbitrary function used to define the functional $J$.
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
$endgroup$
add a comment |
$begingroup$
Use the generalized binomial theorem:begin{align}
frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
&= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
&= p[f(x)]^{p-1}phi(x)
end{align}
$phi$ is just an arbitrary function used to define the functional $J$.
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
$endgroup$
Use the generalized binomial theorem:begin{align}
frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
&= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
&= p[f(x)]^{p-1}phi(x)
end{align}
$phi$ is just an arbitrary function used to define the functional $J$.
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
answered Dec 25 '18 at 12:05
mechanodroidmechanodroid
28.6k62548
28.6k62548
add a comment |
add a comment |
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$begingroup$
It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
$endgroup$
– KReiser
Dec 25 '18 at 7:17
$begingroup$
Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32