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Difficult integration by parts in deriving Euler-Lagrange equations
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I am doing some reading about the calculus of variations and I am finding it really difficult to see how the integrals are being manipulated. I sense it is due to an application of integration by parts (or some multivariable calculus) but I've been staring at this for some time and am not making any progress.
In this situation, I should say that $F = F(x,y,y',y'')in C^3(D)$ for some $D subseteq mathbbR^4$ and that $eta in C^4([a,b])$ is arbitrary, except that it satisfies $eta(a) = eta(b) = eta'(a) = eta'(b) = 0$.
The book I am reading (Differential and Integral Equations by P.J. Collins, pp 202) says
Because we are treating $x,y,y',y''$ as independent variables, I can see what happens to the last two terms inside the first integral - both the $eta$ and $eta'$ are integrated whilst the $F_y''$ is treated as a constant, explaining why those two terms come up in the first box on the second line. However, I am incredibly stumped what happens after that.
In particular, I am not sure how the integral on the second line arises.
Is it some application of a product/chain rule-type thing?
Any insight into this would help a lot. Thanks!
integration multivariable-calculus calculus-of-variations euler-lagrange-equation
asked Jul 17 at 16:41
omegaSQU4RED
1936
add a comment |Â
up vote
3
down vote
favorite
I am doing some reading about the calculus of variations and I am finding it really difficult to see how the integrals are being manipulated. I sense it is due to an application of integration by parts (or some multivariable calculus) but I've been staring at this for some time and am not making any progress.
In this situation, I should say that $F = F(x,y,y',y'')in C^3(D)$ for some $D subseteq mathbbR^4$ and that $eta in C^4([a,b])$ is arbitrary, except that it satisfies $eta(a) = eta(b) = eta'(a) = eta'(b) = 0$.
The book I am reading (Differential and Integral Equations by P.J. Collins, pp 202) says
Because we are treating $x,y,y',y''$ as independent variables, I can see what happens to the last two terms inside the first integral - both the $eta$ and $eta'$ are integrated whilst the $F_y''$ is treated as a constant, explaining why those two terms come up in the first box on the second line. However, I am incredibly stumped what happens after that.
In particular, I am not sure how the integral on the second line arises.
Is it some application of a product/chain rule-type thing?
Any insight into this would help a lot. Thanks!
integration multivariable-calculus calculus-of-variations euler-lagrange-equation
asked Jul 17 at 16:41
omegaSQU4RED
1936
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am doing some reading about the calculus of variations and I am finding it really difficult to see how the integrals are being manipulated. I sense it is due to an application of integration by parts (or some multivariable calculus) but I've been staring at this for some time and am not making any progress.
In this situation, I should say that $F = F(x,y,y',y'')in C^3(D)$ for some $D subseteq mathbbR^4$ and that $eta in C^4([a,b])$ is arbitrary, except that it satisfies $eta(a) = eta(b) = eta'(a) = eta'(b) = 0$.
The book I am reading (Differential and Integral Equations by P.J. Collins, pp 202) says
Because we are treating $x,y,y',y''$ as independent variables, I can see what happens to the last two terms inside the first integral - both the $eta$ and $eta'$ are integrated whilst the $F_y''$ is treated as a constant, explaining why those two terms come up in the first box on the second line. However, I am incredibly stumped what happens after that.
In particular, I am not sure how the integral on the second line arises.
Is it some application of a product/chain rule-type thing?
Any insight into this would help a lot. Thanks!
integration multivariable-calculus calculus-of-variations euler-lagrange-equation
asked Jul 17 at 16:41
omegaSQU4RED
1936
I am doing some reading about the calculus of variations and I am finding it really difficult to see how the integrals are being manipulated. I sense it is due to an application of integration by parts (or some multivariable calculus) but I've been staring at this for some time and am not making any progress.
In this situation, I should say that $F = F(x,y,y',y'')in C^3(D)$ for some $D subseteq mathbbR^4$ and that $eta in C^4([a,b])$ is arbitrary, except that it satisfies $eta(a) = eta(b) = eta'(a) = eta'(b) = 0$.
The book I am reading (Differential and Integral Equations by P.J. Collins, pp 202) says
Because we are treating $x,y,y',y''$ as independent variables, I can see what happens to the last two terms inside the first integral - both the $eta$ and $eta'$ are integrated whilst the $F_y''$ is treated as a constant, explaining why those two terms come up in the first box on the second line. However, I am incredibly stumped what happens after that.
In particular, I am not sure how the integral on the second line arises.
Is it some application of a product/chain rule-type thing?
Any insight into this would help a lot. Thanks!
integration multivariable-calculus calculus-of-variations euler-lagrange-equation
asked Jul 17 at 16:41
omegaSQU4RED
1936
edited Jul 17 at 16:55
asked Jul 17 at 16:41
omegaSQU4RED
1936
asked Jul 17 at 16:41
omegaSQU4RED
1936
asked Jul 17 at 16:41
omegaSQU4RED
1936
1936
add a comment |Â
add a comment |Â
3 Answers
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up vote
3
down vote
The OP understands that
$$eta'F_y'+eta''F_y''=fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxtag 1$$
Integrating both sides of $(1)$ reveals
$$beginalign
int_a^b eta'F_y'+eta''F_y'', dx&=int_a^b left(fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxright),dx\\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+eta'fracdF_y''dxright),dx \\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+fracddxleft(eta fracdF_y''dxright)-eta fracd^2F_y''dx^2right),dx\\
&=left.left(eta F_y'+eta' F_y''-eta fracdF_y''dxright)right|_a^b-int_a^b left(eta fracdF_y'dx-eta fracd^2F_y''dx^2right),dx
endalign$$
answered Jul 17 at 16:54
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
up vote
3
down vote
It's integration by-parts again. You have
$$int_a^bleft(-eta'fracddx,F_y''right)dx=-eta,fracddx,F_y''Bigg|_a^b+int_a^beta,fracd^2dx^2,F_y'',dx.$$
The goal of all this by-parts integration, incidentally, is to make $eta$ appear inside the integral without any derivatives. This allows us to invoke the Fundamental Lemma of the Calculus of Variations, and conclude that whatever's multiplying the $eta$ must be zero.
answered Jul 17 at 16:50
Adrian Keister
3,61721533
That makes sense - I see that now. However, how does one get from the first line to the second?
– omegaSQU4RED
Jul 17 at 16:53
As Davide Morgante mentioned in his answer, it's all by-parts. Recall how by-parts works: you get to yank the derivative off one term and slap it onto the other. You pick up a minus sign inside the integral and the boundary term (the term outside the integral, with neither term differentiated compared to what was inside the integral).
– Adrian Keister
Jul 17 at 17:11
add a comment |Â
up vote
2
down vote
It's all integration by parts: $$int_a^b (underbraceeta F_y_textfirst+underbraceeta' F_y'_textsecond+underbraceeta'' F_y''_textthird)dx$$
let's study the second and third integral with integration by parts
- (Second integral) Let $f'=eta'$ and $g=F_y'$ then $$int_a^beta' F_ydx = [eta F_y']_a^b-int_a^betafracddxF_y'dx$$
- (Third integral) Let $f' = eta''$ and $g=F_y''$ then $$int_a^beta'' F_y''dx = [eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx$$
Plugging all into the initial equation we get $$int_a^b(eta F_y+eta'F_y'+eta'' F_y'')dx = \ underbraceint_a^b eta F_ydx_textfirst + underbrace[eta F_y']_a^b-int_a^betafracddxF_y'_textsecond +underbrace[eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx_textthird$$
Then rearranging the terms we get $$[eta F_y'+eta' F_y'']_a^b + int_a^b left[eta(F_y-fracddxF_y')+eta'fracddxF_y''right]dx$$
Now I think you can get the last formula in the same manner
answered Jul 17 at 17:04
Davide Morgante
1,875220
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The OP understands that
$$eta'F_y'+eta''F_y''=fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxtag 1$$
Integrating both sides of $(1)$ reveals
$$beginalign
int_a^b eta'F_y'+eta''F_y'', dx&=int_a^b left(fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxright),dx\\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+eta'fracdF_y''dxright),dx \\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+fracddxleft(eta fracdF_y''dxright)-eta fracd^2F_y''dx^2right),dx\\
&=left.left(eta F_y'+eta' F_y''-eta fracdF_y''dxright)right|_a^b-int_a^b left(eta fracdF_y'dx-eta fracd^2F_y''dx^2right),dx
endalign$$
answered Jul 17 at 16:54
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
up vote
3
down vote
The OP understands that
$$eta'F_y'+eta''F_y''=fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxtag 1$$
Integrating both sides of $(1)$ reveals
$$beginalign
int_a^b eta'F_y'+eta''F_y'', dx&=int_a^b left(fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxright),dx\\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+eta'fracdF_y''dxright),dx \\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+fracddxleft(eta fracdF_y''dxright)-eta fracd^2F_y''dx^2right),dx\\
&=left.left(eta F_y'+eta' F_y''-eta fracdF_y''dxright)right|_a^b-int_a^b left(eta fracdF_y'dx-eta fracd^2F_y''dx^2right),dx
endalign$$
answered Jul 17 at 16:54
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The OP understands that
$$eta'F_y'+eta''F_y''=fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxtag 1$$
Integrating both sides of $(1)$ reveals
$$beginalign
int_a^b eta'F_y'+eta''F_y'', dx&=int_a^b left(fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxright),dx\\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+eta'fracdF_y''dxright),dx \\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+fracddxleft(eta fracdF_y''dxright)-eta fracd^2F_y''dx^2right),dx\\
&=left.left(eta F_y'+eta' F_y''-eta fracdF_y''dxright)right|_a^b-int_a^b left(eta fracdF_y'dx-eta fracd^2F_y''dx^2right),dx
endalign$$
answered Jul 17 at 16:54
Mark Viola
126k1172167
The OP understands that
$$eta'F_y'+eta''F_y''=fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxtag 1$$
Integrating both sides of $(1)$ reveals
$$beginalign
int_a^b eta'F_y'+eta''F_y'', dx&=int_a^b left(fracddxleft(eta F_y'+eta' F_y''right)-eta fracdF_y'dx-eta' fracdF_y''dxright),dx\\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+eta'fracdF_y''dxright),dx \\
&=left.left(eta F_y'+eta' F_y''right)right|_a^b-int_a^b left(eta fracdF_y'dx+fracddxleft(eta fracdF_y''dxright)-eta fracd^2F_y''dx^2right),dx\\
&=left.left(eta F_y'+eta' F_y''-eta fracdF_y''dxright)right|_a^b-int_a^b left(eta fracdF_y'dx-eta fracd^2F_y''dx^2right),dx
endalign$$
answered Jul 17 at 16:54
Mark Viola
126k1172167
answered Jul 17 at 16:54
Mark Viola
126k1172167
answered Jul 17 at 16:54
Mark Viola
126k1172167
answered Jul 17 at 16:54
Mark Viola
126k1172167
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:15
add a comment |Â
up vote
3
down vote
It's integration by-parts again. You have
$$int_a^bleft(-eta'fracddx,F_y''right)dx=-eta,fracddx,F_y''Bigg|_a^b+int_a^beta,fracd^2dx^2,F_y'',dx.$$
The goal of all this by-parts integration, incidentally, is to make $eta$ appear inside the integral without any derivatives. This allows us to invoke the Fundamental Lemma of the Calculus of Variations, and conclude that whatever's multiplying the $eta$ must be zero.
answered Jul 17 at 16:50
Adrian Keister
3,61721533
That makes sense - I see that now. However, how does one get from the first line to the second?
– omegaSQU4RED
Jul 17 at 16:53
As Davide Morgante mentioned in his answer, it's all by-parts. Recall how by-parts works: you get to yank the derivative off one term and slap it onto the other. You pick up a minus sign inside the integral and the boundary term (the term outside the integral, with neither term differentiated compared to what was inside the integral).
– Adrian Keister
Jul 17 at 17:11
add a comment |Â
up vote
3
down vote
It's integration by-parts again. You have
$$int_a^bleft(-eta'fracddx,F_y''right)dx=-eta,fracddx,F_y''Bigg|_a^b+int_a^beta,fracd^2dx^2,F_y'',dx.$$
The goal of all this by-parts integration, incidentally, is to make $eta$ appear inside the integral without any derivatives. This allows us to invoke the Fundamental Lemma of the Calculus of Variations, and conclude that whatever's multiplying the $eta$ must be zero.
answered Jul 17 at 16:50
Adrian Keister
3,61721533
That makes sense - I see that now. However, how does one get from the first line to the second?
– omegaSQU4RED
Jul 17 at 16:53
As Davide Morgante mentioned in his answer, it's all by-parts. Recall how by-parts works: you get to yank the derivative off one term and slap it onto the other. You pick up a minus sign inside the integral and the boundary term (the term outside the integral, with neither term differentiated compared to what was inside the integral).
– Adrian Keister
Jul 17 at 17:11
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It's integration by-parts again. You have
$$int_a^bleft(-eta'fracddx,F_y''right)dx=-eta,fracddx,F_y''Bigg|_a^b+int_a^beta,fracd^2dx^2,F_y'',dx.$$
The goal of all this by-parts integration, incidentally, is to make $eta$ appear inside the integral without any derivatives. This allows us to invoke the Fundamental Lemma of the Calculus of Variations, and conclude that whatever's multiplying the $eta$ must be zero.
answered Jul 17 at 16:50
Adrian Keister
3,61721533
It's integration by-parts again. You have
$$int_a^bleft(-eta'fracddx,F_y''right)dx=-eta,fracddx,F_y''Bigg|_a^b+int_a^beta,fracd^2dx^2,F_y'',dx.$$
The goal of all this by-parts integration, incidentally, is to make $eta$ appear inside the integral without any derivatives. This allows us to invoke the Fundamental Lemma of the Calculus of Variations, and conclude that whatever's multiplying the $eta$ must be zero.
answered Jul 17 at 16:50
Adrian Keister
3,61721533
edited Jul 17 at 19:13
answered Jul 17 at 16:50
Adrian Keister
3,61721533
answered Jul 17 at 16:50
Adrian Keister
3,61721533
answered Jul 17 at 16:50
Adrian Keister
3,61721533
3,61721533
That makes sense - I see that now. However, how does one get from the first line to the second?
– omegaSQU4RED
Jul 17 at 16:53
As Davide Morgante mentioned in his answer, it's all by-parts. Recall how by-parts works: you get to yank the derivative off one term and slap it onto the other. You pick up a minus sign inside the integral and the boundary term (the term outside the integral, with neither term differentiated compared to what was inside the integral).
– Adrian Keister
Jul 17 at 17:11
add a comment |Â
That makes sense - I see that now. However, how does one get from the first line to the second?
– omegaSQU4RED
Jul 17 at 16:53
As Davide Morgante mentioned in his answer, it's all by-parts. Recall how by-parts works: you get to yank the derivative off one term and slap it onto the other. You pick up a minus sign inside the integral and the boundary term (the term outside the integral, with neither term differentiated compared to what was inside the integral).
– Adrian Keister
Jul 17 at 17:11
That makes sense - I see that now. However, how does one get from the first line to the second?
– omegaSQU4RED
Jul 17 at 16:53
That makes sense - I see that now. However, how does one get from the first line to the second?
– omegaSQU4RED
Jul 17 at 16:53
As Davide Morgante mentioned in his answer, it's all by-parts. Recall how by-parts works: you get to yank the derivative off one term and slap it onto the other. You pick up a minus sign inside the integral and the boundary term (the term outside the integral, with neither term differentiated compared to what was inside the integral).
– Adrian Keister
Jul 17 at 17:11
As Davide Morgante mentioned in his answer, it's all by-parts. Recall how by-parts works: you get to yank the derivative off one term and slap it onto the other. You pick up a minus sign inside the integral and the boundary term (the term outside the integral, with neither term differentiated compared to what was inside the integral).
– Adrian Keister
Jul 17 at 17:11
add a comment |Â
up vote
2
down vote
It's all integration by parts: $$int_a^b (underbraceeta F_y_textfirst+underbraceeta' F_y'_textsecond+underbraceeta'' F_y''_textthird)dx$$
let's study the second and third integral with integration by parts
- (Second integral) Let $f'=eta'$ and $g=F_y'$ then $$int_a^beta' F_ydx = [eta F_y']_a^b-int_a^betafracddxF_y'dx$$
- (Third integral) Let $f' = eta''$ and $g=F_y''$ then $$int_a^beta'' F_y''dx = [eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx$$
Plugging all into the initial equation we get $$int_a^b(eta F_y+eta'F_y'+eta'' F_y'')dx = \ underbraceint_a^b eta F_ydx_textfirst + underbrace[eta F_y']_a^b-int_a^betafracddxF_y'_textsecond +underbrace[eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx_textthird$$
Then rearranging the terms we get $$[eta F_y'+eta' F_y'']_a^b + int_a^b left[eta(F_y-fracddxF_y')+eta'fracddxF_y''right]dx$$
Now I think you can get the last formula in the same manner
answered Jul 17 at 17:04
Davide Morgante
1,875220
add a comment |Â
up vote
2
down vote
It's all integration by parts: $$int_a^b (underbraceeta F_y_textfirst+underbraceeta' F_y'_textsecond+underbraceeta'' F_y''_textthird)dx$$
let's study the second and third integral with integration by parts
- (Second integral) Let $f'=eta'$ and $g=F_y'$ then $$int_a^beta' F_ydx = [eta F_y']_a^b-int_a^betafracddxF_y'dx$$
- (Third integral) Let $f' = eta''$ and $g=F_y''$ then $$int_a^beta'' F_y''dx = [eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx$$
Plugging all into the initial equation we get $$int_a^b(eta F_y+eta'F_y'+eta'' F_y'')dx = \ underbraceint_a^b eta F_ydx_textfirst + underbrace[eta F_y']_a^b-int_a^betafracddxF_y'_textsecond +underbrace[eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx_textthird$$
Then rearranging the terms we get $$[eta F_y'+eta' F_y'']_a^b + int_a^b left[eta(F_y-fracddxF_y')+eta'fracddxF_y''right]dx$$
Now I think you can get the last formula in the same manner
answered Jul 17 at 17:04
Davide Morgante
1,875220
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's all integration by parts: $$int_a^b (underbraceeta F_y_textfirst+underbraceeta' F_y'_textsecond+underbraceeta'' F_y''_textthird)dx$$
let's study the second and third integral with integration by parts
- (Second integral) Let $f'=eta'$ and $g=F_y'$ then $$int_a^beta' F_ydx = [eta F_y']_a^b-int_a^betafracddxF_y'dx$$
- (Third integral) Let $f' = eta''$ and $g=F_y''$ then $$int_a^beta'' F_y''dx = [eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx$$
Plugging all into the initial equation we get $$int_a^b(eta F_y+eta'F_y'+eta'' F_y'')dx = \ underbraceint_a^b eta F_ydx_textfirst + underbrace[eta F_y']_a^b-int_a^betafracddxF_y'_textsecond +underbrace[eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx_textthird$$
Then rearranging the terms we get $$[eta F_y'+eta' F_y'']_a^b + int_a^b left[eta(F_y-fracddxF_y')+eta'fracddxF_y''right]dx$$
Now I think you can get the last formula in the same manner
answered Jul 17 at 17:04
Davide Morgante
1,875220
It's all integration by parts: $$int_a^b (underbraceeta F_y_textfirst+underbraceeta' F_y'_textsecond+underbraceeta'' F_y''_textthird)dx$$
let's study the second and third integral with integration by parts
- (Second integral) Let $f'=eta'$ and $g=F_y'$ then $$int_a^beta' F_ydx = [eta F_y']_a^b-int_a^betafracddxF_y'dx$$
- (Third integral) Let $f' = eta''$ and $g=F_y''$ then $$int_a^beta'' F_y''dx = [eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx$$
Plugging all into the initial equation we get $$int_a^b(eta F_y+eta'F_y'+eta'' F_y'')dx = \ underbraceint_a^b eta F_ydx_textfirst + underbrace[eta F_y']_a^b-int_a^betafracddxF_y'_textsecond +underbrace[eta'F_y'']_a^b-int_a^beta'fracddxF_y''dx_textthird$$
Then rearranging the terms we get $$[eta F_y'+eta' F_y'']_a^b + int_a^b left[eta(F_y-fracddxF_y')+eta'fracddxF_y''right]dx$$
Now I think you can get the last formula in the same manner
answered Jul 17 at 17:04
Davide Morgante
1,875220
answered Jul 17 at 17:04
Davide Morgante
1,875220
answered Jul 17 at 17:04
Davide Morgante
1,875220
answered Jul 17 at 17:04
Davide Morgante
1,875220
1,875220
add a comment |Â
add a comment |Â
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