Mixing
Finding extremals of a functional with integral function as integrand
$begingroup$
Assume you want to find the extremals for the functional
$$
y rightarrow int_a^by(x)left[int_a^xy(xi), dxiright], dx
$$
where $[a,b]subset mathbb{R}$ and $yin mathcal{C}^1left([a,b],mathbb{R}right)$.
How can you write down the Euler Lagrange equations associated with this problem?
functional-analysis analysis optimization calculus-of-variations euler-lagrange-equation
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
$endgroup$
add a comment |
$begingroup$
Assume you want to find the extremals for the functional
$$
y rightarrow int_a^by(x)left[int_a^xy(xi), dxiright], dx
$$
where $[a,b]subset mathbb{R}$ and $yin mathcal{C}^1left([a,b],mathbb{R}right)$.
How can you write down the Euler Lagrange equations associated with this problem?
functional-analysis analysis optimization calculus-of-variations euler-lagrange-equation
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
$endgroup$
$begingroup$
As the function is linear in $y$, and there are no constraints, the problem is unsolvable.
$endgroup$
– daw
Jan 30 at 14:28
2
$begingroup$
You could try to substitute $z':=y$ or $z(x):=int_a^x y(s)ds$. Then derive Euler-Lagrange in $z$ and transform back.
$endgroup$
– daw
Jan 30 at 14:30
$begingroup$
@daw I am actually not focused on the problem itself, rather on the way you can write down the Euler-Lagrange equations for such a functional... if you like you can consider whatever additional constraint which guarantees the existence of a solution; Thanks for your second comment!
$endgroup$
– William Tomblin
Jan 30 at 14:32
add a comment |
$begingroup$
Assume you want to find the extremals for the functional
$$
y rightarrow int_a^by(x)left[int_a^xy(xi), dxiright], dx
$$
where $[a,b]subset mathbb{R}$ and $yin mathcal{C}^1left([a,b],mathbb{R}right)$.
How can you write down the Euler Lagrange equations associated with this problem?
functional-analysis analysis optimization calculus-of-variations euler-lagrange-equation
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
$endgroup$
Assume you want to find the extremals for the functional
$$
y rightarrow int_a^by(x)left[int_a^xy(xi), dxiright], dx
$$
where $[a,b]subset mathbb{R}$ and $yin mathcal{C}^1left([a,b],mathbb{R}right)$.
How can you write down the Euler Lagrange equations associated with this problem?
functional-analysis analysis optimization calculus-of-variations euler-lagrange-equation
functional-analysis analysis optimization calculus-of-variations euler-lagrange-equation
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
edited Jan 30 at 17:47
William Tomblin
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
asked Jan 30 at 14:14
William TomblinWilliam Tomblin
261112
261112
$begingroup$
As the function is linear in $y$, and there are no constraints, the problem is unsolvable.
$endgroup$
– daw
Jan 30 at 14:28
2
$begingroup$
You could try to substitute $z':=y$ or $z(x):=int_a^x y(s)ds$. Then derive Euler-Lagrange in $z$ and transform back.
$endgroup$
– daw
Jan 30 at 14:30
$begingroup$
@daw I am actually not focused on the problem itself, rather on the way you can write down the Euler-Lagrange equations for such a functional... if you like you can consider whatever additional constraint which guarantees the existence of a solution; Thanks for your second comment!
$endgroup$
– William Tomblin
Jan 30 at 14:32
add a comment |
$begingroup$
As the function is linear in $y$, and there are no constraints, the problem is unsolvable.
$endgroup$
– daw
Jan 30 at 14:28
2
$begingroup$
You could try to substitute $z':=y$ or $z(x):=int_a^x y(s)ds$. Then derive Euler-Lagrange in $z$ and transform back.
$endgroup$
– daw
Jan 30 at 14:30
$begingroup$
@daw I am actually not focused on the problem itself, rather on the way you can write down the Euler-Lagrange equations for such a functional... if you like you can consider whatever additional constraint which guarantees the existence of a solution; Thanks for your second comment!
$endgroup$
– William Tomblin
Jan 30 at 14:32
$begingroup$
As the function is linear in $y$, and there are no constraints, the problem is unsolvable.
$endgroup$
– daw
Jan 30 at 14:28
$begingroup$
As the function is linear in $y$, and there are no constraints, the problem is unsolvable.
$endgroup$
– daw
Jan 30 at 14:28
2
2
$begingroup$
You could try to substitute $z':=y$ or $z(x):=int_a^x y(s)ds$. Then derive Euler-Lagrange in $z$ and transform back.
$endgroup$
– daw
Jan 30 at 14:30
$begingroup$
You could try to substitute $z':=y$ or $z(x):=int_a^x y(s)ds$. Then derive Euler-Lagrange in $z$ and transform back.
$endgroup$
– daw
Jan 30 at 14:30
$begingroup$
@daw I am actually not focused on the problem itself, rather on the way you can write down the Euler-Lagrange equations for such a functional... if you like you can consider whatever additional constraint which guarantees the existence of a solution; Thanks for your second comment!
$endgroup$
– William Tomblin
Jan 30 at 14:32
$begingroup$
@daw I am actually not focused on the problem itself, rather on the way you can write down the Euler-Lagrange equations for such a functional... if you like you can consider whatever additional constraint which guarantees the existence of a solution; Thanks for your second comment!
$endgroup$
– William Tomblin
Jan 30 at 14:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
OP's functional is non-negative
$$ F[y]~:=~int_{[a,b]} !mathrm{d}x~y(x)int_{[a,x]}!mathrm{d}xi ~y(xi)
~=~iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~theta(x!-!xi) ~y(x)y(xi)$$
$$~=~frac{1}{2}iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~y(x)y(xi)
~stackrel{(2)}{=}~frac{G[y]^2}{2} ~geq~ 0,tag{1}$$
where
$$G[y]~:=~int_{[a,b]} !mathrm{d}x~y(x).tag{2}$$The functional/variational derivative is
$$frac{delta F[y]}{delta y(x)}~stackrel{(1)}{=}~G[y]frac{delta G[y]}{delta y(x)} ~stackrel{(2)}{=}~G[y]1_{[a,b]}(x),tag{3} $$
which is a more general notion than the Euler-Lagrange derivative.OP's sought-for equation is the vanishing of the functional derivative (3). Evidently, a stationary configuration $y:[a,b]to mathbb{R}$ has
$$G[y]~stackrel{(3)}{=}~0,tag{4}$$
which is clearly a minimum (and in particular an extremum) for OP's functional (1).
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
$endgroup$
$begingroup$
I would like to learn more about functional/variational derivative; do you have any textbook to recommend?
$endgroup$
– William Tomblin
Feb 1 at 12:02
add a comment |
$begingroup$
So the functional to be minimized is
$$int_a^by(x)left[int_a^xy(xi), dxiright]dx.$$
I think I would "shift the perspective" a bit from $y$ to $displaystyleint y(x),dx$. (This is daw's suggestion in the comments above.) That is, let
$$Y(x)=int_a^x y(xi),dxi.$$
Then the functional can be written as
$$int_a^bY(x) Y'(x),dx.$$
Now we let $L=YY',$ and the E-L equations are
$$frac{partial L}{partial Y}-frac{d}{dx}frac{partial L}{partial Y'}=Y'-frac{d}{dx}Y=Y'-Y'=0. $$
This procedure at least allows you to write down the E-L equations, even though, in this case, they tell you nothing. I suspect the minimization problem is still ill-defined.
Actually, we can say something more in this particular case:
$$int_a^b Y(x)Y'(x),dx=frac{Y^2(x)}{2}Bigg|_a^b=frac{Y^2(b)}{2}-frac{Y^2(a)}{2}=frac{Y^2(b)}{2},$$
by the definition of $Y$.
Minimizing this is tantamount to forcing $Y(b)=0,$ or as close to it as is feasible.
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
This trick works fine if you have something like I previously posted, i.e. $y rightarrow int_a^bleft[int_a^xy(xi), dxiright], dx$; but, honestly, i was looking for a method a bit more robust than this. This is the reason why I edited my original question: now the trick of changing the order of integration doesn't work anymore.
$endgroup$
– William Tomblin
Jan 30 at 17:49
$begingroup$
Hmm. Yeah, I see that. Hang on...
$endgroup$
– Adrian Keister
Jan 30 at 17:55
1
$begingroup$
Well, here's a way to handle the integral - surely this is more robust than what I had before, eh?
$endgroup$
– Adrian Keister
Jan 30 at 18:06
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
OP's functional is non-negative
$$ F[y]~:=~int_{[a,b]} !mathrm{d}x~y(x)int_{[a,x]}!mathrm{d}xi ~y(xi)
~=~iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~theta(x!-!xi) ~y(x)y(xi)$$
$$~=~frac{1}{2}iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~y(x)y(xi)
~stackrel{(2)}{=}~frac{G[y]^2}{2} ~geq~ 0,tag{1}$$
where
$$G[y]~:=~int_{[a,b]} !mathrm{d}x~y(x).tag{2}$$The functional/variational derivative is
$$frac{delta F[y]}{delta y(x)}~stackrel{(1)}{=}~G[y]frac{delta G[y]}{delta y(x)} ~stackrel{(2)}{=}~G[y]1_{[a,b]}(x),tag{3} $$
which is a more general notion than the Euler-Lagrange derivative.OP's sought-for equation is the vanishing of the functional derivative (3). Evidently, a stationary configuration $y:[a,b]to mathbb{R}$ has
$$G[y]~stackrel{(3)}{=}~0,tag{4}$$
which is clearly a minimum (and in particular an extremum) for OP's functional (1).
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
$endgroup$
$begingroup$
I would like to learn more about functional/variational derivative; do you have any textbook to recommend?
$endgroup$
– William Tomblin
Feb 1 at 12:02
add a comment |
$begingroup$
OP's functional is non-negative
$$ F[y]~:=~int_{[a,b]} !mathrm{d}x~y(x)int_{[a,x]}!mathrm{d}xi ~y(xi)
~=~iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~theta(x!-!xi) ~y(x)y(xi)$$
$$~=~frac{1}{2}iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~y(x)y(xi)
~stackrel{(2)}{=}~frac{G[y]^2}{2} ~geq~ 0,tag{1}$$
where
$$G[y]~:=~int_{[a,b]} !mathrm{d}x~y(x).tag{2}$$The functional/variational derivative is
$$frac{delta F[y]}{delta y(x)}~stackrel{(1)}{=}~G[y]frac{delta G[y]}{delta y(x)} ~stackrel{(2)}{=}~G[y]1_{[a,b]}(x),tag{3} $$
which is a more general notion than the Euler-Lagrange derivative.OP's sought-for equation is the vanishing of the functional derivative (3). Evidently, a stationary configuration $y:[a,b]to mathbb{R}$ has
$$G[y]~stackrel{(3)}{=}~0,tag{4}$$
which is clearly a minimum (and in particular an extremum) for OP's functional (1).
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
$endgroup$
$begingroup$
I would like to learn more about functional/variational derivative; do you have any textbook to recommend?
$endgroup$
– William Tomblin
Feb 1 at 12:02
add a comment |
$begingroup$
OP's functional is non-negative
$$ F[y]~:=~int_{[a,b]} !mathrm{d}x~y(x)int_{[a,x]}!mathrm{d}xi ~y(xi)
~=~iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~theta(x!-!xi) ~y(x)y(xi)$$
$$~=~frac{1}{2}iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~y(x)y(xi)
~stackrel{(2)}{=}~frac{G[y]^2}{2} ~geq~ 0,tag{1}$$
where
$$G[y]~:=~int_{[a,b]} !mathrm{d}x~y(x).tag{2}$$The functional/variational derivative is
$$frac{delta F[y]}{delta y(x)}~stackrel{(1)}{=}~G[y]frac{delta G[y]}{delta y(x)} ~stackrel{(2)}{=}~G[y]1_{[a,b]}(x),tag{3} $$
which is a more general notion than the Euler-Lagrange derivative.OP's sought-for equation is the vanishing of the functional derivative (3). Evidently, a stationary configuration $y:[a,b]to mathbb{R}$ has
$$G[y]~stackrel{(3)}{=}~0,tag{4}$$
which is clearly a minimum (and in particular an extremum) for OP's functional (1).
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
$endgroup$
OP's functional is non-negative
$$ F[y]~:=~int_{[a,b]} !mathrm{d}x~y(x)int_{[a,x]}!mathrm{d}xi ~y(xi)
~=~iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~theta(x!-!xi) ~y(x)y(xi)$$
$$~=~frac{1}{2}iint_{[a,b]^2} !mathrm{d}x~mathrm{d}xi ~y(x)y(xi)
~stackrel{(2)}{=}~frac{G[y]^2}{2} ~geq~ 0,tag{1}$$
where
$$G[y]~:=~int_{[a,b]} !mathrm{d}x~y(x).tag{2}$$The functional/variational derivative is
$$frac{delta F[y]}{delta y(x)}~stackrel{(1)}{=}~G[y]frac{delta G[y]}{delta y(x)} ~stackrel{(2)}{=}~G[y]1_{[a,b]}(x),tag{3} $$
which is a more general notion than the Euler-Lagrange derivative.OP's sought-for equation is the vanishing of the functional derivative (3). Evidently, a stationary configuration $y:[a,b]to mathbb{R}$ has
$$G[y]~stackrel{(3)}{=}~0,tag{4}$$
which is clearly a minimum (and in particular an extremum) for OP's functional (1).
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
edited Jan 31 at 22:33
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
answered Jan 30 at 21:42
QmechanicQmechanic
5,17711858
5,17711858
$begingroup$
I would like to learn more about functional/variational derivative; do you have any textbook to recommend?
$endgroup$
– William Tomblin
Feb 1 at 12:02
add a comment |
$begingroup$
I would like to learn more about functional/variational derivative; do you have any textbook to recommend?
$endgroup$
– William Tomblin
Feb 1 at 12:02
$begingroup$
I would like to learn more about functional/variational derivative; do you have any textbook to recommend?
$endgroup$
– William Tomblin
Feb 1 at 12:02
$begingroup$
I would like to learn more about functional/variational derivative; do you have any textbook to recommend?
$endgroup$
– William Tomblin
Feb 1 at 12:02
add a comment |
$begingroup$
So the functional to be minimized is
$$int_a^by(x)left[int_a^xy(xi), dxiright]dx.$$
I think I would "shift the perspective" a bit from $y$ to $displaystyleint y(x),dx$. (This is daw's suggestion in the comments above.) That is, let
$$Y(x)=int_a^x y(xi),dxi.$$
Then the functional can be written as
$$int_a^bY(x) Y'(x),dx.$$
Now we let $L=YY',$ and the E-L equations are
$$frac{partial L}{partial Y}-frac{d}{dx}frac{partial L}{partial Y'}=Y'-frac{d}{dx}Y=Y'-Y'=0. $$
This procedure at least allows you to write down the E-L equations, even though, in this case, they tell you nothing. I suspect the minimization problem is still ill-defined.
Actually, we can say something more in this particular case:
$$int_a^b Y(x)Y'(x),dx=frac{Y^2(x)}{2}Bigg|_a^b=frac{Y^2(b)}{2}-frac{Y^2(a)}{2}=frac{Y^2(b)}{2},$$
by the definition of $Y$.
Minimizing this is tantamount to forcing $Y(b)=0,$ or as close to it as is feasible.
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
This trick works fine if you have something like I previously posted, i.e. $y rightarrow int_a^bleft[int_a^xy(xi), dxiright], dx$; but, honestly, i was looking for a method a bit more robust than this. This is the reason why I edited my original question: now the trick of changing the order of integration doesn't work anymore.
$endgroup$
– William Tomblin
Jan 30 at 17:49
$begingroup$
Hmm. Yeah, I see that. Hang on...
$endgroup$
– Adrian Keister
Jan 30 at 17:55
1
$begingroup$
Well, here's a way to handle the integral - surely this is more robust than what I had before, eh?
$endgroup$
– Adrian Keister
Jan 30 at 18:06
add a comment |
$begingroup$
So the functional to be minimized is
$$int_a^by(x)left[int_a^xy(xi), dxiright]dx.$$
I think I would "shift the perspective" a bit from $y$ to $displaystyleint y(x),dx$. (This is daw's suggestion in the comments above.) That is, let
$$Y(x)=int_a^x y(xi),dxi.$$
Then the functional can be written as
$$int_a^bY(x) Y'(x),dx.$$
Now we let $L=YY',$ and the E-L equations are
$$frac{partial L}{partial Y}-frac{d}{dx}frac{partial L}{partial Y'}=Y'-frac{d}{dx}Y=Y'-Y'=0. $$
This procedure at least allows you to write down the E-L equations, even though, in this case, they tell you nothing. I suspect the minimization problem is still ill-defined.
Actually, we can say something more in this particular case:
$$int_a^b Y(x)Y'(x),dx=frac{Y^2(x)}{2}Bigg|_a^b=frac{Y^2(b)}{2}-frac{Y^2(a)}{2}=frac{Y^2(b)}{2},$$
by the definition of $Y$.
Minimizing this is tantamount to forcing $Y(b)=0,$ or as close to it as is feasible.
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
This trick works fine if you have something like I previously posted, i.e. $y rightarrow int_a^bleft[int_a^xy(xi), dxiright], dx$; but, honestly, i was looking for a method a bit more robust than this. This is the reason why I edited my original question: now the trick of changing the order of integration doesn't work anymore.
$endgroup$
– William Tomblin
Jan 30 at 17:49
$begingroup$
Hmm. Yeah, I see that. Hang on...
$endgroup$
– Adrian Keister
Jan 30 at 17:55
1
$begingroup$
Well, here's a way to handle the integral - surely this is more robust than what I had before, eh?
$endgroup$
– Adrian Keister
Jan 30 at 18:06
add a comment |
$begingroup$
So the functional to be minimized is
$$int_a^by(x)left[int_a^xy(xi), dxiright]dx.$$
I think I would "shift the perspective" a bit from $y$ to $displaystyleint y(x),dx$. (This is daw's suggestion in the comments above.) That is, let
$$Y(x)=int_a^x y(xi),dxi.$$
Then the functional can be written as
$$int_a^bY(x) Y'(x),dx.$$
Now we let $L=YY',$ and the E-L equations are
$$frac{partial L}{partial Y}-frac{d}{dx}frac{partial L}{partial Y'}=Y'-frac{d}{dx}Y=Y'-Y'=0. $$
This procedure at least allows you to write down the E-L equations, even though, in this case, they tell you nothing. I suspect the minimization problem is still ill-defined.
Actually, we can say something more in this particular case:
$$int_a^b Y(x)Y'(x),dx=frac{Y^2(x)}{2}Bigg|_a^b=frac{Y^2(b)}{2}-frac{Y^2(a)}{2}=frac{Y^2(b)}{2},$$
by the definition of $Y$.
Minimizing this is tantamount to forcing $Y(b)=0,$ or as close to it as is feasible.
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
So the functional to be minimized is
$$int_a^by(x)left[int_a^xy(xi), dxiright]dx.$$
I think I would "shift the perspective" a bit from $y$ to $displaystyleint y(x),dx$. (This is daw's suggestion in the comments above.) That is, let
$$Y(x)=int_a^x y(xi),dxi.$$
Then the functional can be written as
$$int_a^bY(x) Y'(x),dx.$$
Now we let $L=YY',$ and the E-L equations are
$$frac{partial L}{partial Y}-frac{d}{dx}frac{partial L}{partial Y'}=Y'-frac{d}{dx}Y=Y'-Y'=0. $$
This procedure at least allows you to write down the E-L equations, even though, in this case, they tell you nothing. I suspect the minimization problem is still ill-defined.
Actually, we can say something more in this particular case:
$$int_a^b Y(x)Y'(x),dx=frac{Y^2(x)}{2}Bigg|_a^b=frac{Y^2(b)}{2}-frac{Y^2(a)}{2}=frac{Y^2(b)}{2},$$
by the definition of $Y$.
Minimizing this is tantamount to forcing $Y(b)=0,$ or as close to it as is feasible.
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
edited Jan 30 at 18:09
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 14:43
Adrian KeisterAdrian Keister
5,26971933
5,26971933
$begingroup$
This trick works fine if you have something like I previously posted, i.e. $y rightarrow int_a^bleft[int_a^xy(xi), dxiright], dx$; but, honestly, i was looking for a method a bit more robust than this. This is the reason why I edited my original question: now the trick of changing the order of integration doesn't work anymore.
$endgroup$
– William Tomblin
Jan 30 at 17:49
$begingroup$
Hmm. Yeah, I see that. Hang on...
$endgroup$
– Adrian Keister
Jan 30 at 17:55
1
$begingroup$
Well, here's a way to handle the integral - surely this is more robust than what I had before, eh?
$endgroup$
– Adrian Keister
Jan 30 at 18:06
add a comment |
$begingroup$
This trick works fine if you have something like I previously posted, i.e. $y rightarrow int_a^bleft[int_a^xy(xi), dxiright], dx$; but, honestly, i was looking for a method a bit more robust than this. This is the reason why I edited my original question: now the trick of changing the order of integration doesn't work anymore.
$endgroup$
– William Tomblin
Jan 30 at 17:49
$begingroup$
Hmm. Yeah, I see that. Hang on...
$endgroup$
– Adrian Keister
Jan 30 at 17:55
1
$begingroup$
Well, here's a way to handle the integral - surely this is more robust than what I had before, eh?
$endgroup$
– Adrian Keister
Jan 30 at 18:06
$begingroup$
This trick works fine if you have something like I previously posted, i.e. $y rightarrow int_a^bleft[int_a^xy(xi), dxiright], dx$; but, honestly, i was looking for a method a bit more robust than this. This is the reason why I edited my original question: now the trick of changing the order of integration doesn't work anymore.
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– William Tomblin
Jan 30 at 17:49
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This trick works fine if you have something like I previously posted, i.e. $y rightarrow int_a^bleft[int_a^xy(xi), dxiright], dx$; but, honestly, i was looking for a method a bit more robust than this. This is the reason why I edited my original question: now the trick of changing the order of integration doesn't work anymore.
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– William Tomblin
Jan 30 at 17:49
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Hmm. Yeah, I see that. Hang on...
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– Adrian Keister
Jan 30 at 17:55
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Hmm. Yeah, I see that. Hang on...
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– Adrian Keister
Jan 30 at 17:55
1
1
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Well, here's a way to handle the integral - surely this is more robust than what I had before, eh?
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– Adrian Keister
Jan 30 at 18:06
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Well, here's a way to handle the integral - surely this is more robust than what I had before, eh?
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– Adrian Keister
Jan 30 at 18:06
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As the function is linear in $y$, and there are no constraints, the problem is unsolvable.
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– daw
Jan 30 at 14:28
2
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You could try to substitute $z':=y$ or $z(x):=int_a^x y(s)ds$. Then derive Euler-Lagrange in $z$ and transform back.
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– daw
Jan 30 at 14:30
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@daw I am actually not focused on the problem itself, rather on the way you can write down the Euler-Lagrange equations for such a functional... if you like you can consider whatever additional constraint which guarantees the existence of a solution; Thanks for your second comment!
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– William Tomblin
Jan 30 at 14:32