Functional derivative of a functional that depends on antiderivative

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It is well known how to calculate functional derivative if a functional depends of the function and it's derivatives (Euler-Lagrange rule): $mathcalL=int F(x,dotx,t)dt$. There is also straight-forward generalization for a functional that depend on higher order function derivatives.



From now on I will use $x$ instead of $t$ and will denote functions by small english letters (e.g. $f$, $phi$. etc.).



In my problem, a functional depends on the function's antiderivative:



$$
mathcalL=int dx int_-1^x f(x')dx'.
$$



How can I calculate $delta mathcalL/delta f$ ? I tried to do it by definition:



$$
int fracdelta mathcalLdelta f phi(x) dx = left[fracdF[f+epsilonphi]depsilonright]_epsilon=0.
$$
However, after simplifying the right hand side I found out that the result cannot be factorized to $int (...)phi(x)dx$.



Denoting $F=int_-1^xf(x')dx'$, I was also trying to use the chain rule, but did not succeed.



In my actual problem I need to minimize $mathcalL=int dxint exp(-f(x'))dx'$.



Thanks,



Mikhail



Edit: Sorry, I should have been more specific from the beginning. The actual problem is find the functional gradient w.r.t. $f(x)$ of the following functional:



$$
mathcalL = int_-1^1 dx expleft[int_-1^x f(x')dx'right]beta(x).
$$



Note, that the second integral is entirely inside of the exponent. $beta(x)$ does not depend on f(x).




calculus-of-variations functional-calculus




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  • You could set $g(x)=int_-1^x f(x') dx'$ and then find $g$. Then $f(x)=g'(x)$. (But you say you tried that). For your concrete problem, there is probably no minimizer. (Hint: Consider $f(x)=n$ and let $ntoinfty$).
    – Kusma
    Aug 6 at 15:07










  • Hi Kusma, I just did not quite understand the chain rule for functional derivative. I am interested how can I solve problems like this one using the tools of functional calculus
    – Mikhail Genkin
    Aug 6 at 15:17










  • I don't see that you need a chain rule here. If we have $mathscrL=int L(f,int f(x')dx',x)dx$, we can set $F=int f(x')dx'$ then $mathscrL=int L(F',F,x)dx$. We can forget for the moment how $F$ was defined, solve for $F$ using functional derivative and then recover $f=F'$.
    – Kusma
    Aug 6 at 15:55










  • @Kusma, No, the way you describe will result in $delta mathcalL /delta F$ that described in terms of $f$. It is different from $delta mathcalL /delta f$. The relanshionship is manifested in the chain rule
    – Mikhail Genkin
    Aug 7 at 13:55















up vote
3
down vote

favorite
1












It is well known how to calculate functional derivative if a functional depends of the function and it's derivatives (Euler-Lagrange rule): $mathcalL=int F(x,dotx,t)dt$. There is also straight-forward generalization for a functional that depend on higher order function derivatives.



From now on I will use $x$ instead of $t$ and will denote functions by small english letters (e.g. $f$, $phi$. etc.).



In my problem, a functional depends on the function's antiderivative:



$$
mathcalL=int dx int_-1^x f(x')dx'.
$$



How can I calculate $delta mathcalL/delta f$ ? I tried to do it by definition:



$$
int fracdelta mathcalLdelta f phi(x) dx = left[fracdF[f+epsilonphi]depsilonright]_epsilon=0.
$$
However, after simplifying the right hand side I found out that the result cannot be factorized to $int (...)phi(x)dx$.



Denoting $F=int_-1^xf(x')dx'$, I was also trying to use the chain rule, but did not succeed.



In my actual problem I need to minimize $mathcalL=int dxint exp(-f(x'))dx'$.



Thanks,



Mikhail



Edit: Sorry, I should have been more specific from the beginning. The actual problem is find the functional gradient w.r.t. $f(x)$ of the following functional:



$$
mathcalL = int_-1^1 dx expleft[int_-1^x f(x')dx'right]beta(x).
$$



Note, that the second integral is entirely inside of the exponent. $beta(x)$ does not depend on f(x).




calculus-of-variations functional-calculus




share|cite|improve this question





















  • You could set $g(x)=int_-1^x f(x') dx'$ and then find $g$. Then $f(x)=g'(x)$. (But you say you tried that). For your concrete problem, there is probably no minimizer. (Hint: Consider $f(x)=n$ and let $ntoinfty$).
    – Kusma
    Aug 6 at 15:07










  • Hi Kusma, I just did not quite understand the chain rule for functional derivative. I am interested how can I solve problems like this one using the tools of functional calculus
    – Mikhail Genkin
    Aug 6 at 15:17










  • I don't see that you need a chain rule here. If we have $mathscrL=int L(f,int f(x')dx',x)dx$, we can set $F=int f(x')dx'$ then $mathscrL=int L(F',F,x)dx$. We can forget for the moment how $F$ was defined, solve for $F$ using functional derivative and then recover $f=F'$.
    – Kusma
    Aug 6 at 15:55










  • @Kusma, No, the way you describe will result in $delta mathcalL /delta F$ that described in terms of $f$. It is different from $delta mathcalL /delta f$. The relanshionship is manifested in the chain rule
    – Mikhail Genkin
    Aug 7 at 13:55













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





It is well known how to calculate functional derivative if a functional depends of the function and it's derivatives (Euler-Lagrange rule): $mathcalL=int F(x,dotx,t)dt$. There is also straight-forward generalization for a functional that depend on higher order function derivatives.



From now on I will use $x$ instead of $t$ and will denote functions by small english letters (e.g. $f$, $phi$. etc.).



In my problem, a functional depends on the function's antiderivative:



$$
mathcalL=int dx int_-1^x f(x')dx'.
$$



How can I calculate $delta mathcalL/delta f$ ? I tried to do it by definition:



$$
int fracdelta mathcalLdelta f phi(x) dx = left[fracdF[f+epsilonphi]depsilonright]_epsilon=0.
$$
However, after simplifying the right hand side I found out that the result cannot be factorized to $int (...)phi(x)dx$.



Denoting $F=int_-1^xf(x')dx'$, I was also trying to use the chain rule, but did not succeed.



In my actual problem I need to minimize $mathcalL=int dxint exp(-f(x'))dx'$.



Thanks,



Mikhail



Edit: Sorry, I should have been more specific from the beginning. The actual problem is find the functional gradient w.r.t. $f(x)$ of the following functional:



$$
mathcalL = int_-1^1 dx expleft[int_-1^x f(x')dx'right]beta(x).
$$



Note, that the second integral is entirely inside of the exponent. $beta(x)$ does not depend on f(x).




calculus-of-variations functional-calculus




share|cite|improve this question













It is well known how to calculate functional derivative if a functional depends of the function and it's derivatives (Euler-Lagrange rule): $mathcalL=int F(x,dotx,t)dt$. There is also straight-forward generalization for a functional that depend on higher order function derivatives.



From now on I will use $x$ instead of $t$ and will denote functions by small english letters (e.g. $f$, $phi$. etc.).



In my problem, a functional depends on the function's antiderivative:



$$
mathcalL=int dx int_-1^x f(x')dx'.
$$



How can I calculate $delta mathcalL/delta f$ ? I tried to do it by definition:



$$
int fracdelta mathcalLdelta f phi(x) dx = left[fracdF[f+epsilonphi]depsilonright]_epsilon=0.
$$
However, after simplifying the right hand side I found out that the result cannot be factorized to $int (...)phi(x)dx$.



Denoting $F=int_-1^xf(x')dx'$, I was also trying to use the chain rule, but did not succeed.



In my actual problem I need to minimize $mathcalL=int dxint exp(-f(x'))dx'$.



Thanks,



Mikhail



Edit: Sorry, I should have been more specific from the beginning. The actual problem is find the functional gradient w.r.t. $f(x)$ of the following functional:



$$
mathcalL = int_-1^1 dx expleft[int_-1^x f(x')dx'right]beta(x).
$$



Note, that the second integral is entirely inside of the exponent. $beta(x)$ does not depend on f(x).





calculus-of-variations functional-calculus





share|cite|improve this question












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edited Aug 7 at 14:56









Qmechanic

4,45811746




4,45811746









asked Aug 6 at 14:53









Mikhail Genkin

1457




1457











  • You could set $g(x)=int_-1^x f(x') dx'$ and then find $g$. Then $f(x)=g'(x)$. (But you say you tried that). For your concrete problem, there is probably no minimizer. (Hint: Consider $f(x)=n$ and let $ntoinfty$).
    – Kusma
    Aug 6 at 15:07










  • Hi Kusma, I just did not quite understand the chain rule for functional derivative. I am interested how can I solve problems like this one using the tools of functional calculus
    – Mikhail Genkin
    Aug 6 at 15:17










  • I don't see that you need a chain rule here. If we have $mathscrL=int L(f,int f(x')dx',x)dx$, we can set $F=int f(x')dx'$ then $mathscrL=int L(F',F,x)dx$. We can forget for the moment how $F$ was defined, solve for $F$ using functional derivative and then recover $f=F'$.
    – Kusma
    Aug 6 at 15:55










  • @Kusma, No, the way you describe will result in $delta mathcalL /delta F$ that described in terms of $f$. It is different from $delta mathcalL /delta f$. The relanshionship is manifested in the chain rule
    – Mikhail Genkin
    Aug 7 at 13:55

















  • You could set $g(x)=int_-1^x f(x') dx'$ and then find $g$. Then $f(x)=g'(x)$. (But you say you tried that). For your concrete problem, there is probably no minimizer. (Hint: Consider $f(x)=n$ and let $ntoinfty$).
    – Kusma
    Aug 6 at 15:07










  • Hi Kusma, I just did not quite understand the chain rule for functional derivative. I am interested how can I solve problems like this one using the tools of functional calculus
    – Mikhail Genkin
    Aug 6 at 15:17










  • I don't see that you need a chain rule here. If we have $mathscrL=int L(f,int f(x')dx',x)dx$, we can set $F=int f(x')dx'$ then $mathscrL=int L(F',F,x)dx$. We can forget for the moment how $F$ was defined, solve for $F$ using functional derivative and then recover $f=F'$.
    – Kusma
    Aug 6 at 15:55










  • @Kusma, No, the way you describe will result in $delta mathcalL /delta F$ that described in terms of $f$. It is different from $delta mathcalL /delta f$. The relanshionship is manifested in the chain rule
    – Mikhail Genkin
    Aug 7 at 13:55
















You could set $g(x)=int_-1^x f(x') dx'$ and then find $g$. Then $f(x)=g'(x)$. (But you say you tried that). For your concrete problem, there is probably no minimizer. (Hint: Consider $f(x)=n$ and let $ntoinfty$).
– Kusma
Aug 6 at 15:07




You could set $g(x)=int_-1^x f(x') dx'$ and then find $g$. Then $f(x)=g'(x)$. (But you say you tried that). For your concrete problem, there is probably no minimizer. (Hint: Consider $f(x)=n$ and let $ntoinfty$).
– Kusma
Aug 6 at 15:07












Hi Kusma, I just did not quite understand the chain rule for functional derivative. I am interested how can I solve problems like this one using the tools of functional calculus
– Mikhail Genkin
Aug 6 at 15:17




Hi Kusma, I just did not quite understand the chain rule for functional derivative. I am interested how can I solve problems like this one using the tools of functional calculus
– Mikhail Genkin
Aug 6 at 15:17












I don't see that you need a chain rule here. If we have $mathscrL=int L(f,int f(x')dx',x)dx$, we can set $F=int f(x')dx'$ then $mathscrL=int L(F',F,x)dx$. We can forget for the moment how $F$ was defined, solve for $F$ using functional derivative and then recover $f=F'$.
– Kusma
Aug 6 at 15:55




I don't see that you need a chain rule here. If we have $mathscrL=int L(f,int f(x')dx',x)dx$, we can set $F=int f(x')dx'$ then $mathscrL=int L(F',F,x)dx$. We can forget for the moment how $F$ was defined, solve for $F$ using functional derivative and then recover $f=F'$.
– Kusma
Aug 6 at 15:55












@Kusma, No, the way you describe will result in $delta mathcalL /delta F$ that described in terms of $f$. It is different from $delta mathcalL /delta f$. The relanshionship is manifested in the chain rule
– Mikhail Genkin
Aug 7 at 13:55





@Kusma, No, the way you describe will result in $delta mathcalL /delta F$ that described in terms of $f$. It is different from $delta mathcalL /delta f$. The relanshionship is manifested in the chain rule
– Mikhail Genkin
Aug 7 at 13:55











2 Answers
2






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up vote
3
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  1. OP first considers a functional $Sequiv cal L$ of the form
    $$beginalignS[f]~=&~int_a^b!mathrmdx int_a^x!mathrmdx^prime ~gcirc f(x^prime) cr
    ~=&~iint_[a,b]^2!mathrmdx ~mathrmdx^prime
    ~theta(x-x^prime)~gcirc f(x^prime) cr
    ~=&~ (b-x^prime) int_a^b!mathrmdx^prime~gcirc f(x^prime),
    endalign
    tag1a$$
    where $theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^-y$ and $a=-1$.)



    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ (b-x^prime) ~g^primecirc f(x^prime) .tag1b$$



  2. In an edit OP considers a functional $Sequiv cal L$ of the form
    $$S[f]~=~int_a^b!mathrmdx ~g(x,F(x)), qquad
    F(x)~:=~int_a^x!mathrmdx^prime~f(x^prime)
    ~=~ int_a^b!mathrmdx^prime~theta(x-x^prime)~f(x^prime).
    tag2a$$
    Let $g_F=fracpartial gpartial F$ denote the partial derivative wrt. the second argument of the $g$-function.
    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ int_a^b!mathrmdx~theta(x-x^prime)~g_F(x,F(x))
    ~=~ int_x^prime^b!mathrmdx~g_F(x,F(x)).tag2b$$






share|cite|improve this answer























  • Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit.
    – Mikhail Genkin
    Aug 7 at 14:26











  • I updated the answer.
    – Qmechanic
    Aug 7 at 14:52










  • Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ?
    – Mikhail Genkin
    Aug 7 at 15:04

















up vote
1
down vote













I am not convinced that calculus of variations will help you find a minimiser $f$ for your problem, as no minimiser exists. Consider
$$ mathcalL[f]=int_-1^1 int_-1^x exp(-f(x'))dx'dx
$$
Clearly $mathcalL[f]ge 0$ for all functions $f$. However, choosing $f_nequiv ln n$ (constant function), we have
$$mathcalL[f_n]=int_-1^1 int_-1^x frac1n dx' dx = frac1nint_-1^1 (x+1)dx = frac2n,
$$
which tends to $0$ for $ntoinfty$. So if there is a minimiser, the value of the minimum must be zero. Hence we must have
$int_-1^x exp(-f(x'))dx'=0
$ for almost every $x$, which implies $exp(-f(y))=0$ for almost every $y$. The exponential is never zero, so no such function exists.






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  • _Well, I omitted another function. My actual problem $int_-1^1 int_-1^x exp(-f(x')dx' ) beta(x) dx $. Function $beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific
    – Mikhail Genkin
    Aug 7 at 14:20










  • Please see the edit, you are actually absolutely right
    – Mikhail Genkin
    Aug 7 at 14:24










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2 Answers
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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes








up vote
3
down vote



accepted











  1. OP first considers a functional $Sequiv cal L$ of the form
    $$beginalignS[f]~=&~int_a^b!mathrmdx int_a^x!mathrmdx^prime ~gcirc f(x^prime) cr
    ~=&~iint_[a,b]^2!mathrmdx ~mathrmdx^prime
    ~theta(x-x^prime)~gcirc f(x^prime) cr
    ~=&~ (b-x^prime) int_a^b!mathrmdx^prime~gcirc f(x^prime),
    endalign
    tag1a$$
    where $theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^-y$ and $a=-1$.)



    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ (b-x^prime) ~g^primecirc f(x^prime) .tag1b$$



  2. In an edit OP considers a functional $Sequiv cal L$ of the form
    $$S[f]~=~int_a^b!mathrmdx ~g(x,F(x)), qquad
    F(x)~:=~int_a^x!mathrmdx^prime~f(x^prime)
    ~=~ int_a^b!mathrmdx^prime~theta(x-x^prime)~f(x^prime).
    tag2a$$
    Let $g_F=fracpartial gpartial F$ denote the partial derivative wrt. the second argument of the $g$-function.
    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ int_a^b!mathrmdx~theta(x-x^prime)~g_F(x,F(x))
    ~=~ int_x^prime^b!mathrmdx~g_F(x,F(x)).tag2b$$






share|cite|improve this answer























  • Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit.
    – Mikhail Genkin
    Aug 7 at 14:26











  • I updated the answer.
    – Qmechanic
    Aug 7 at 14:52










  • Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ?
    – Mikhail Genkin
    Aug 7 at 15:04














up vote
3
down vote



accepted











  1. OP first considers a functional $Sequiv cal L$ of the form
    $$beginalignS[f]~=&~int_a^b!mathrmdx int_a^x!mathrmdx^prime ~gcirc f(x^prime) cr
    ~=&~iint_[a,b]^2!mathrmdx ~mathrmdx^prime
    ~theta(x-x^prime)~gcirc f(x^prime) cr
    ~=&~ (b-x^prime) int_a^b!mathrmdx^prime~gcirc f(x^prime),
    endalign
    tag1a$$
    where $theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^-y$ and $a=-1$.)



    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ (b-x^prime) ~g^primecirc f(x^prime) .tag1b$$



  2. In an edit OP considers a functional $Sequiv cal L$ of the form
    $$S[f]~=~int_a^b!mathrmdx ~g(x,F(x)), qquad
    F(x)~:=~int_a^x!mathrmdx^prime~f(x^prime)
    ~=~ int_a^b!mathrmdx^prime~theta(x-x^prime)~f(x^prime).
    tag2a$$
    Let $g_F=fracpartial gpartial F$ denote the partial derivative wrt. the second argument of the $g$-function.
    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ int_a^b!mathrmdx~theta(x-x^prime)~g_F(x,F(x))
    ~=~ int_x^prime^b!mathrmdx~g_F(x,F(x)).tag2b$$






share|cite|improve this answer























  • Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit.
    – Mikhail Genkin
    Aug 7 at 14:26











  • I updated the answer.
    – Qmechanic
    Aug 7 at 14:52










  • Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ?
    – Mikhail Genkin
    Aug 7 at 15:04












up vote
3
down vote



accepted







up vote
3
down vote



accepted







  1. OP first considers a functional $Sequiv cal L$ of the form
    $$beginalignS[f]~=&~int_a^b!mathrmdx int_a^x!mathrmdx^prime ~gcirc f(x^prime) cr
    ~=&~iint_[a,b]^2!mathrmdx ~mathrmdx^prime
    ~theta(x-x^prime)~gcirc f(x^prime) cr
    ~=&~ (b-x^prime) int_a^b!mathrmdx^prime~gcirc f(x^prime),
    endalign
    tag1a$$
    where $theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^-y$ and $a=-1$.)



    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ (b-x^prime) ~g^primecirc f(x^prime) .tag1b$$



  2. In an edit OP considers a functional $Sequiv cal L$ of the form
    $$S[f]~=~int_a^b!mathrmdx ~g(x,F(x)), qquad
    F(x)~:=~int_a^x!mathrmdx^prime~f(x^prime)
    ~=~ int_a^b!mathrmdx^prime~theta(x-x^prime)~f(x^prime).
    tag2a$$
    Let $g_F=fracpartial gpartial F$ denote the partial derivative wrt. the second argument of the $g$-function.
    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ int_a^b!mathrmdx~theta(x-x^prime)~g_F(x,F(x))
    ~=~ int_x^prime^b!mathrmdx~g_F(x,F(x)).tag2b$$






share|cite|improve this answer
















  1. OP first considers a functional $Sequiv cal L$ of the form
    $$beginalignS[f]~=&~int_a^b!mathrmdx int_a^x!mathrmdx^prime ~gcirc f(x^prime) cr
    ~=&~iint_[a,b]^2!mathrmdx ~mathrmdx^prime
    ~theta(x-x^prime)~gcirc f(x^prime) cr
    ~=&~ (b-x^prime) int_a^b!mathrmdx^prime~gcirc f(x^prime),
    endalign
    tag1a$$
    where $theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^-y$ and $a=-1$.)



    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ (b-x^prime) ~g^primecirc f(x^prime) .tag1b$$



  2. In an edit OP considers a functional $Sequiv cal L$ of the form
    $$S[f]~=~int_a^b!mathrmdx ~g(x,F(x)), qquad
    F(x)~:=~int_a^x!mathrmdx^prime~f(x^prime)
    ~=~ int_a^b!mathrmdx^prime~theta(x-x^prime)~f(x^prime).
    tag2a$$
    Let $g_F=fracpartial gpartial F$ denote the partial derivative wrt. the second argument of the $g$-function.
    The functional/variational derivative is then
    $$forall x^prime~in~[a,b]:~~fracdelta S[f]delta f(x^prime)
    ~=~ int_a^b!mathrmdx~theta(x-x^prime)~g_F(x,F(x))
    ~=~ int_x^prime^b!mathrmdx~g_F(x,F(x)).tag2b$$







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share|cite|improve this answer



share|cite|improve this answer








edited Aug 7 at 14:50


























answered Aug 6 at 20:45









Qmechanic

4,45811746




4,45811746











  • Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit.
    – Mikhail Genkin
    Aug 7 at 14:26











  • I updated the answer.
    – Qmechanic
    Aug 7 at 14:52










  • Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ?
    – Mikhail Genkin
    Aug 7 at 15:04
















  • Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit.
    – Mikhail Genkin
    Aug 7 at 14:26











  • I updated the answer.
    – Qmechanic
    Aug 7 at 14:52










  • Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ?
    – Mikhail Genkin
    Aug 7 at 15:04















Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit.
– Mikhail Genkin
Aug 7 at 14:26





Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit.
– Mikhail Genkin
Aug 7 at 14:26













I updated the answer.
– Qmechanic
Aug 7 at 14:52




I updated the answer.
– Qmechanic
Aug 7 at 14:52












Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ?
– Mikhail Genkin
Aug 7 at 15:04




Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ?
– Mikhail Genkin
Aug 7 at 15:04










up vote
1
down vote













I am not convinced that calculus of variations will help you find a minimiser $f$ for your problem, as no minimiser exists. Consider
$$ mathcalL[f]=int_-1^1 int_-1^x exp(-f(x'))dx'dx
$$
Clearly $mathcalL[f]ge 0$ for all functions $f$. However, choosing $f_nequiv ln n$ (constant function), we have
$$mathcalL[f_n]=int_-1^1 int_-1^x frac1n dx' dx = frac1nint_-1^1 (x+1)dx = frac2n,
$$
which tends to $0$ for $ntoinfty$. So if there is a minimiser, the value of the minimum must be zero. Hence we must have
$int_-1^x exp(-f(x'))dx'=0
$ for almost every $x$, which implies $exp(-f(y))=0$ for almost every $y$. The exponential is never zero, so no such function exists.






share|cite|improve this answer





















  • _Well, I omitted another function. My actual problem $int_-1^1 int_-1^x exp(-f(x')dx' ) beta(x) dx $. Function $beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific
    – Mikhail Genkin
    Aug 7 at 14:20










  • Please see the edit, you are actually absolutely right
    – Mikhail Genkin
    Aug 7 at 14:24














up vote
1
down vote













I am not convinced that calculus of variations will help you find a minimiser $f$ for your problem, as no minimiser exists. Consider
$$ mathcalL[f]=int_-1^1 int_-1^x exp(-f(x'))dx'dx
$$
Clearly $mathcalL[f]ge 0$ for all functions $f$. However, choosing $f_nequiv ln n$ (constant function), we have
$$mathcalL[f_n]=int_-1^1 int_-1^x frac1n dx' dx = frac1nint_-1^1 (x+1)dx = frac2n,
$$
which tends to $0$ for $ntoinfty$. So if there is a minimiser, the value of the minimum must be zero. Hence we must have
$int_-1^x exp(-f(x'))dx'=0
$ for almost every $x$, which implies $exp(-f(y))=0$ for almost every $y$. The exponential is never zero, so no such function exists.






share|cite|improve this answer





















  • _Well, I omitted another function. My actual problem $int_-1^1 int_-1^x exp(-f(x')dx' ) beta(x) dx $. Function $beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific
    – Mikhail Genkin
    Aug 7 at 14:20










  • Please see the edit, you are actually absolutely right
    – Mikhail Genkin
    Aug 7 at 14:24












up vote
1
down vote










up vote
1
down vote









I am not convinced that calculus of variations will help you find a minimiser $f$ for your problem, as no minimiser exists. Consider
$$ mathcalL[f]=int_-1^1 int_-1^x exp(-f(x'))dx'dx
$$
Clearly $mathcalL[f]ge 0$ for all functions $f$. However, choosing $f_nequiv ln n$ (constant function), we have
$$mathcalL[f_n]=int_-1^1 int_-1^x frac1n dx' dx = frac1nint_-1^1 (x+1)dx = frac2n,
$$
which tends to $0$ for $ntoinfty$. So if there is a minimiser, the value of the minimum must be zero. Hence we must have
$int_-1^x exp(-f(x'))dx'=0
$ for almost every $x$, which implies $exp(-f(y))=0$ for almost every $y$. The exponential is never zero, so no such function exists.






share|cite|improve this answer













I am not convinced that calculus of variations will help you find a minimiser $f$ for your problem, as no minimiser exists. Consider
$$ mathcalL[f]=int_-1^1 int_-1^x exp(-f(x'))dx'dx
$$
Clearly $mathcalL[f]ge 0$ for all functions $f$. However, choosing $f_nequiv ln n$ (constant function), we have
$$mathcalL[f_n]=int_-1^1 int_-1^x frac1n dx' dx = frac1nint_-1^1 (x+1)dx = frac2n,
$$
which tends to $0$ for $ntoinfty$. So if there is a minimiser, the value of the minimum must be zero. Hence we must have
$int_-1^x exp(-f(x'))dx'=0
$ for almost every $x$, which implies $exp(-f(y))=0$ for almost every $y$. The exponential is never zero, so no such function exists.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 7 at 14:09









Kusma

1,127112




1,127112











  • _Well, I omitted another function. My actual problem $int_-1^1 int_-1^x exp(-f(x')dx' ) beta(x) dx $. Function $beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific
    – Mikhail Genkin
    Aug 7 at 14:20










  • Please see the edit, you are actually absolutely right
    – Mikhail Genkin
    Aug 7 at 14:24
















  • _Well, I omitted another function. My actual problem $int_-1^1 int_-1^x exp(-f(x')dx' ) beta(x) dx $. Function $beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific
    – Mikhail Genkin
    Aug 7 at 14:20










  • Please see the edit, you are actually absolutely right
    – Mikhail Genkin
    Aug 7 at 14:24















_Well, I omitted another function. My actual problem $int_-1^1 int_-1^x exp(-f(x')dx' ) beta(x) dx $. Function $beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific
– Mikhail Genkin
Aug 7 at 14:20




_Well, I omitted another function. My actual problem $int_-1^1 int_-1^x exp(-f(x')dx' ) beta(x) dx $. Function $beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific
– Mikhail Genkin
Aug 7 at 14:20












Please see the edit, you are actually absolutely right
– Mikhail Genkin
Aug 7 at 14:24




Please see the edit, you are actually absolutely right
– Mikhail Genkin
Aug 7 at 14:24












 

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