Why are these the chain rules for the functional derivatives?
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The chain rules are
$$frac{delta F}{delta f(x)}=frac{delta G[g(f(x))]}{delta g(f(x))}frac{dg(f(x))}{df(x)}$$
and
$$frac{delta F}{delta f(x)}=frac{dg(G[f(x)])}{dG[f(x)]}frac{delta G[f(x)]}{delta f(x)}$$
Is there some nice intuitive way to think about this?
The first one looks to make sense following the logic of the ODE chain rule but the second is less clear to me
ordinary-differential-equations functional-equations
asked Dec 19 '18 at 14:01
JamesJames
15310
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add a comment |
$begingroup$
The chain rules are
$$frac{delta F}{delta f(x)}=frac{delta G[g(f(x))]}{delta g(f(x))}frac{dg(f(x))}{df(x)}$$
and
$$frac{delta F}{delta f(x)}=frac{dg(G[f(x)])}{dG[f(x)]}frac{delta G[f(x)]}{delta f(x)}$$
Is there some nice intuitive way to think about this?
The first one looks to make sense following the logic of the ODE chain rule but the second is less clear to me
ordinary-differential-equations functional-equations
asked Dec 19 '18 at 14:01
JamesJames
15310
$endgroup$
$begingroup$
To me there is only two chain rules : $(f(x(t))' = f'(x(t)) x'(t)$ and the two variables generalization $(f(x(t),y(t)))' = x'(t)frac{partial f}{partial x}(x(t),y(t))+y'(t)frac{partial f}{partial y}(y(t),y(t))$. In functional calculus it happens to differentiate $F(u+ t v)$ with $u,v$ two functions and $F$ an operator
$endgroup$
– reuns
Dec 19 '18 at 15:08
add a comment |
$begingroup$
The chain rules are
$$frac{delta F}{delta f(x)}=frac{delta G[g(f(x))]}{delta g(f(x))}frac{dg(f(x))}{df(x)}$$
and
$$frac{delta F}{delta f(x)}=frac{dg(G[f(x)])}{dG[f(x)]}frac{delta G[f(x)]}{delta f(x)}$$
Is there some nice intuitive way to think about this?
The first one looks to make sense following the logic of the ODE chain rule but the second is less clear to me
ordinary-differential-equations functional-equations
asked Dec 19 '18 at 14:01
JamesJames
15310
$endgroup$
The chain rules are
$$frac{delta F}{delta f(x)}=frac{delta G[g(f(x))]}{delta g(f(x))}frac{dg(f(x))}{df(x)}$$
and
$$frac{delta F}{delta f(x)}=frac{dg(G[f(x)])}{dG[f(x)]}frac{delta G[f(x)]}{delta f(x)}$$
Is there some nice intuitive way to think about this?
The first one looks to make sense following the logic of the ODE chain rule but the second is less clear to me
ordinary-differential-equations functional-equations
ordinary-differential-equations functional-equations
asked Dec 19 '18 at 14:01
JamesJames
15310
asked Dec 19 '18 at 14:01
JamesJames
15310
asked Dec 19 '18 at 14:01
JamesJames
15310
asked Dec 19 '18 at 14:01
JamesJames
15310
asked Dec 19 '18 at 14:01
JamesJames
15310
15310
$begingroup$
To me there is only two chain rules : $(f(x(t))' = f'(x(t)) x'(t)$ and the two variables generalization $(f(x(t),y(t)))' = x'(t)frac{partial f}{partial x}(x(t),y(t))+y'(t)frac{partial f}{partial y}(y(t),y(t))$. In functional calculus it happens to differentiate $F(u+ t v)$ with $u,v$ two functions and $F$ an operator
$endgroup$
– reuns
Dec 19 '18 at 15:08
add a comment |
$begingroup$
To me there is only two chain rules : $(f(x(t))' = f'(x(t)) x'(t)$ and the two variables generalization $(f(x(t),y(t)))' = x'(t)frac{partial f}{partial x}(x(t),y(t))+y'(t)frac{partial f}{partial y}(y(t),y(t))$. In functional calculus it happens to differentiate $F(u+ t v)$ with $u,v$ two functions and $F$ an operator
$endgroup$
– reuns
Dec 19 '18 at 15:08
$begingroup$
To me there is only two chain rules : $(f(x(t))' = f'(x(t)) x'(t)$ and the two variables generalization $(f(x(t),y(t)))' = x'(t)frac{partial f}{partial x}(x(t),y(t))+y'(t)frac{partial f}{partial y}(y(t),y(t))$. In functional calculus it happens to differentiate $F(u+ t v)$ with $u,v$ two functions and $F$ an operator
$endgroup$
– reuns
Dec 19 '18 at 15:08
$begingroup$
To me there is only two chain rules : $(f(x(t))' = f'(x(t)) x'(t)$ and the two variables generalization $(f(x(t),y(t)))' = x'(t)frac{partial f}{partial x}(x(t),y(t))+y'(t)frac{partial f}{partial y}(y(t),y(t))$. In functional calculus it happens to differentiate $F(u+ t v)$ with $u,v$ two functions and $F$ an operator
$endgroup$
– reuns
Dec 19 '18 at 15:08
add a comment |
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$begingroup$
To me there is only two chain rules : $(f(x(t))' = f'(x(t)) x'(t)$ and the two variables generalization $(f(x(t),y(t)))' = x'(t)frac{partial f}{partial x}(x(t),y(t))+y'(t)frac{partial f}{partial y}(y(t),y(t))$. In functional calculus it happens to differentiate $F(u+ t v)$ with $u,v$ two functions and $F$ an operator
$endgroup$
– reuns
Dec 19 '18 at 15:08