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Finding extremals which arise from a non-linear third order ODE
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I'm trying to determine the extremals of the functional $F(x,y,y',y'') = (y')^2 + (y'')^2$.
I know that in this case the first integral satisfies
$$F - y' left( F_y' - fracmathrmdmathrmdxF_y''right)-y''F_y''=k$$
where $k$ is a constant.
For the aforementioned functional, I can easily see that $F_y' = 2y'$ and that $F_y'' = 2y''$, and furthermore $fracmathrmdmathrmdxF_y''=2y'''$.
If I put this into the equation satisfied by the extremal, we get
$$(y')^2 + (y'')^2 -y'(2y' - 2y''') - y''(2y'')=k$$
and it is then possible to re-write this as
$$(y')^2 + (y'')^2 = 2y'y''' - k$$
I am unsure what to do next. Is there a trick that I can use to solve this nonlinear third order ODE? I tried integrating both sides between 0 and 1 (as those are the two $x$-values that the boundary conditions specify) as I can see that the functional appears on the left hand side of the equation, but the exercise I am doing is asking for the extremals which implies that I need to find $y(x)$ rather than what the stationary value of the integral is.
differential-equations multivariable-calculus calculus-of-variations nonlinear-system
asked Jul 18 at 11:28
omegaSQU4RED
1936
add a comment |Â
up vote
2
down vote
favorite
I'm trying to determine the extremals of the functional $F(x,y,y',y'') = (y')^2 + (y'')^2$.
I know that in this case the first integral satisfies
$$F - y' left( F_y' - fracmathrmdmathrmdxF_y''right)-y''F_y''=k$$
where $k$ is a constant.
For the aforementioned functional, I can easily see that $F_y' = 2y'$ and that $F_y'' = 2y''$, and furthermore $fracmathrmdmathrmdxF_y''=2y'''$.
If I put this into the equation satisfied by the extremal, we get
$$(y')^2 + (y'')^2 -y'(2y' - 2y''') - y''(2y'')=k$$
and it is then possible to re-write this as
$$(y')^2 + (y'')^2 = 2y'y''' - k$$
I am unsure what to do next. Is there a trick that I can use to solve this nonlinear third order ODE? I tried integrating both sides between 0 and 1 (as those are the two $x$-values that the boundary conditions specify) as I can see that the functional appears on the left hand side of the equation, but the exercise I am doing is asking for the extremals which implies that I need to find $y(x)$ rather than what the stationary value of the integral is.
differential-equations multivariable-calculus calculus-of-variations nonlinear-system
asked Jul 18 at 11:28
omegaSQU4RED
1936
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to determine the extremals of the functional $F(x,y,y',y'') = (y')^2 + (y'')^2$.
I know that in this case the first integral satisfies
$$F - y' left( F_y' - fracmathrmdmathrmdxF_y''right)-y''F_y''=k$$
where $k$ is a constant.
For the aforementioned functional, I can easily see that $F_y' = 2y'$ and that $F_y'' = 2y''$, and furthermore $fracmathrmdmathrmdxF_y''=2y'''$.
If I put this into the equation satisfied by the extremal, we get
$$(y')^2 + (y'')^2 -y'(2y' - 2y''') - y''(2y'')=k$$
and it is then possible to re-write this as
$$(y')^2 + (y'')^2 = 2y'y''' - k$$
I am unsure what to do next. Is there a trick that I can use to solve this nonlinear third order ODE? I tried integrating both sides between 0 and 1 (as those are the two $x$-values that the boundary conditions specify) as I can see that the functional appears on the left hand side of the equation, but the exercise I am doing is asking for the extremals which implies that I need to find $y(x)$ rather than what the stationary value of the integral is.
differential-equations multivariable-calculus calculus-of-variations nonlinear-system
asked Jul 18 at 11:28
omegaSQU4RED
1936
I'm trying to determine the extremals of the functional $F(x,y,y',y'') = (y')^2 + (y'')^2$.
I know that in this case the first integral satisfies
$$F - y' left( F_y' - fracmathrmdmathrmdxF_y''right)-y''F_y''=k$$
where $k$ is a constant.
For the aforementioned functional, I can easily see that $F_y' = 2y'$ and that $F_y'' = 2y''$, and furthermore $fracmathrmdmathrmdxF_y''=2y'''$.
If I put this into the equation satisfied by the extremal, we get
$$(y')^2 + (y'')^2 -y'(2y' - 2y''') - y''(2y'')=k$$
and it is then possible to re-write this as
$$(y')^2 + (y'')^2 = 2y'y''' - k$$
I am unsure what to do next. Is there a trick that I can use to solve this nonlinear third order ODE? I tried integrating both sides between 0 and 1 (as those are the two $x$-values that the boundary conditions specify) as I can see that the functional appears on the left hand side of the equation, but the exercise I am doing is asking for the extremals which implies that I need to find $y(x)$ rather than what the stationary value of the integral is.
differential-equations multivariable-calculus calculus-of-variations nonlinear-system
asked Jul 18 at 11:28
omegaSQU4RED
1936
asked Jul 18 at 11:28
omegaSQU4RED
1936
asked Jul 18 at 11:28
omegaSQU4RED
1936
asked Jul 18 at 11:28
omegaSQU4RED
1936
1936
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Calling
$$
F(x,y,y',y'')= (y')^2+(y'')^2
$$
we have according to Euler-Lagrange conditions
$$
F_y-left(F_y'right)'+left(F_y''right)'' = -2(y^(4)-y'')=0
$$
hence
$$
y = C_1e^x+C_2e^-x+C_3 x+ C_4
$$
answered Jul 18 at 13:17
Cesareo
5,7722412
Thanks a lot for this! I ended up using the formula for the first integral to try and find the extremal rather than simply solving the Euler-Lagrange equation... now it seems so obvious. Cheers!
– omegaSQU4RED
Jul 18 at 16:49
Actually, you might want to double-check the Euler-Lagrange condition and plugging all the derivatives into it. I ended up getting $F_y = 0$, $(F_y')' = 2y''$ and $(F_y'')'' = 2y'''$ which, when put into the condition, ends up with $$2(y'''' - y'') = 0$$ which has general solution $$y = C_1 e^x + C_2 e^-x + C_3 x + C_4$$. This seems to make sense from the complete version of the problem in the book - the boundary conditions do not involve "nice" angular numbers in $x$. Still thanks though - your approache was nonetheless a lot more elegant than mine :-)
– omegaSQU4RED
Jul 18 at 17:06
@omegaSQU4RED Thanks for the correction. I had made a change of sign.
– Cesareo
Jul 18 at 17:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Calling
$$
F(x,y,y',y'')= (y')^2+(y'')^2
$$
we have according to Euler-Lagrange conditions
$$
F_y-left(F_y'right)'+left(F_y''right)'' = -2(y^(4)-y'')=0
$$
hence
$$
y = C_1e^x+C_2e^-x+C_3 x+ C_4
$$
answered Jul 18 at 13:17
Cesareo
5,7722412
Thanks a lot for this! I ended up using the formula for the first integral to try and find the extremal rather than simply solving the Euler-Lagrange equation... now it seems so obvious. Cheers!
– omegaSQU4RED
Jul 18 at 16:49
Actually, you might want to double-check the Euler-Lagrange condition and plugging all the derivatives into it. I ended up getting $F_y = 0$, $(F_y')' = 2y''$ and $(F_y'')'' = 2y'''$ which, when put into the condition, ends up with $$2(y'''' - y'') = 0$$ which has general solution $$y = C_1 e^x + C_2 e^-x + C_3 x + C_4$$. This seems to make sense from the complete version of the problem in the book - the boundary conditions do not involve "nice" angular numbers in $x$. Still thanks though - your approache was nonetheless a lot more elegant than mine :-)
– omegaSQU4RED
Jul 18 at 17:06
@omegaSQU4RED Thanks for the correction. I had made a change of sign.
– Cesareo
Jul 18 at 17:28
add a comment |Â
up vote
2
down vote
accepted
Calling
$$
F(x,y,y',y'')= (y')^2+(y'')^2
$$
we have according to Euler-Lagrange conditions
$$
F_y-left(F_y'right)'+left(F_y''right)'' = -2(y^(4)-y'')=0
$$
hence
$$
y = C_1e^x+C_2e^-x+C_3 x+ C_4
$$
answered Jul 18 at 13:17
Cesareo
5,7722412
Thanks a lot for this! I ended up using the formula for the first integral to try and find the extremal rather than simply solving the Euler-Lagrange equation... now it seems so obvious. Cheers!
– omegaSQU4RED
Jul 18 at 16:49
Actually, you might want to double-check the Euler-Lagrange condition and plugging all the derivatives into it. I ended up getting $F_y = 0$, $(F_y')' = 2y''$ and $(F_y'')'' = 2y'''$ which, when put into the condition, ends up with $$2(y'''' - y'') = 0$$ which has general solution $$y = C_1 e^x + C_2 e^-x + C_3 x + C_4$$. This seems to make sense from the complete version of the problem in the book - the boundary conditions do not involve "nice" angular numbers in $x$. Still thanks though - your approache was nonetheless a lot more elegant than mine :-)
– omegaSQU4RED
Jul 18 at 17:06
@omegaSQU4RED Thanks for the correction. I had made a change of sign.
– Cesareo
Jul 18 at 17:28
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Calling
$$
F(x,y,y',y'')= (y')^2+(y'')^2
$$
we have according to Euler-Lagrange conditions
$$
F_y-left(F_y'right)'+left(F_y''right)'' = -2(y^(4)-y'')=0
$$
hence
$$
y = C_1e^x+C_2e^-x+C_3 x+ C_4
$$
answered Jul 18 at 13:17
Cesareo
5,7722412
Calling
$$
F(x,y,y',y'')= (y')^2+(y'')^2
$$
we have according to Euler-Lagrange conditions
$$
F_y-left(F_y'right)'+left(F_y''right)'' = -2(y^(4)-y'')=0
$$
hence
$$
y = C_1e^x+C_2e^-x+C_3 x+ C_4
$$
answered Jul 18 at 13:17
Cesareo
5,7722412
edited Jul 18 at 17:25
answered Jul 18 at 13:17
Cesareo
5,7722412
answered Jul 18 at 13:17
Cesareo
5,7722412
answered Jul 18 at 13:17
Cesareo
5,7722412
5,7722412
Thanks a lot for this! I ended up using the formula for the first integral to try and find the extremal rather than simply solving the Euler-Lagrange equation... now it seems so obvious. Cheers!
– omegaSQU4RED
Jul 18 at 16:49
Actually, you might want to double-check the Euler-Lagrange condition and plugging all the derivatives into it. I ended up getting $F_y = 0$, $(F_y')' = 2y''$ and $(F_y'')'' = 2y'''$ which, when put into the condition, ends up with $$2(y'''' - y'') = 0$$ which has general solution $$y = C_1 e^x + C_2 e^-x + C_3 x + C_4$$. This seems to make sense from the complete version of the problem in the book - the boundary conditions do not involve "nice" angular numbers in $x$. Still thanks though - your approache was nonetheless a lot more elegant than mine :-)
– omegaSQU4RED
Jul 18 at 17:06
@omegaSQU4RED Thanks for the correction. I had made a change of sign.
– Cesareo
Jul 18 at 17:28
add a comment |Â
Thanks a lot for this! I ended up using the formula for the first integral to try and find the extremal rather than simply solving the Euler-Lagrange equation... now it seems so obvious. Cheers!
– omegaSQU4RED
Jul 18 at 16:49
Actually, you might want to double-check the Euler-Lagrange condition and plugging all the derivatives into it. I ended up getting $F_y = 0$, $(F_y')' = 2y''$ and $(F_y'')'' = 2y'''$ which, when put into the condition, ends up with $$2(y'''' - y'') = 0$$ which has general solution $$y = C_1 e^x + C_2 e^-x + C_3 x + C_4$$. This seems to make sense from the complete version of the problem in the book - the boundary conditions do not involve "nice" angular numbers in $x$. Still thanks though - your approache was nonetheless a lot more elegant than mine :-)
– omegaSQU4RED
Jul 18 at 17:06
@omegaSQU4RED Thanks for the correction. I had made a change of sign.
– Cesareo
Jul 18 at 17:28
Thanks a lot for this! I ended up using the formula for the first integral to try and find the extremal rather than simply solving the Euler-Lagrange equation... now it seems so obvious. Cheers!
– omegaSQU4RED
Jul 18 at 16:49
Thanks a lot for this! I ended up using the formula for the first integral to try and find the extremal rather than simply solving the Euler-Lagrange equation... now it seems so obvious. Cheers!
– omegaSQU4RED
Jul 18 at 16:49
Actually, you might want to double-check the Euler-Lagrange condition and plugging all the derivatives into it. I ended up getting $F_y = 0$, $(F_y')' = 2y''$ and $(F_y'')'' = 2y'''$ which, when put into the condition, ends up with $$2(y'''' - y'') = 0$$ which has general solution $$y = C_1 e^x + C_2 e^-x + C_3 x + C_4$$. This seems to make sense from the complete version of the problem in the book - the boundary conditions do not involve "nice" angular numbers in $x$. Still thanks though - your approache was nonetheless a lot more elegant than mine :-)
– omegaSQU4RED
Jul 18 at 17:06
Actually, you might want to double-check the Euler-Lagrange condition and plugging all the derivatives into it. I ended up getting $F_y = 0$, $(F_y')' = 2y''$ and $(F_y'')'' = 2y'''$ which, when put into the condition, ends up with $$2(y'''' - y'') = 0$$ which has general solution $$y = C_1 e^x + C_2 e^-x + C_3 x + C_4$$. This seems to make sense from the complete version of the problem in the book - the boundary conditions do not involve "nice" angular numbers in $x$. Still thanks though - your approache was nonetheless a lot more elegant than mine :-)
– omegaSQU4RED
Jul 18 at 17:06
@omegaSQU4RED Thanks for the correction. I had made a change of sign.
– Cesareo
Jul 18 at 17:28
@omegaSQU4RED Thanks for the correction. I had made a change of sign.
– Cesareo
Jul 18 at 17:28
add a comment |Â
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