Mixing
The “Energy transformation” of a second order differential equation
$begingroup$
After asking a question regarding the math myself, as well as finding some others, I am still puzzled by the opening sections of Jordan, D. and Smith, P. (2011). Nonlinear ordinary differential equations. We start off with the equation of motion for a simple pendulum, where $x$ denotes the angular displacement of the pendulum:
$$
ddot{x} + omega^2 sin{x} = 0.
$$
Here, $ddot{x} = mathop{mathrm{d}}^2x/mathop{mathrm{d}}t^2$, and similarily, $dot{x}= mathop{mathrm{d}}x / mathop{mathrm{d}}{t}$. Then, the first element that confuses me is the statement
$$
ddot{x} = frac{mathop{mathrm{d}}}{mathop{mathrm{d}}t} frac{mathop{mathrm{d}}x}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}x} frac{mathop{mathrm{d}x}}{mathop{mathrm{d}}t} = frac{mathrm{d}}{mathrm{d}x} left(frac{1}{2} dot{x}^2 right).
$$
According to the authors, this is known as the energy transformation. I haven't been able to find any other literature supporting this, so my first question is this: Is energy transformation a general term for ODEs, or is it a special name for this operation on the pendulum equation?
To continue, I understand the equation given that one may write $dot{x} = dot{x}(x)$, but that doesn't seem right to me, as the angular velocity may be both increasing and decreasing at any given $x$, depending upon if the pendulum is currently moving "up" or "down". I wouldn't know how to evaluate $mathop{mathrm{d}}dot{x} / mathop{mathrm{d}}x$ by itself. My second question is therefore: Is there a more rigorous way to arrive at the same result?
ordinary-differential-equations
asked Jan 10 at 14:26
Erik AndréErik André
857
$endgroup$
add a comment |
$begingroup$
After asking a question regarding the math myself, as well as finding some others, I am still puzzled by the opening sections of Jordan, D. and Smith, P. (2011). Nonlinear ordinary differential equations. We start off with the equation of motion for a simple pendulum, where $x$ denotes the angular displacement of the pendulum:
$$
ddot{x} + omega^2 sin{x} = 0.
$$
Here, $ddot{x} = mathop{mathrm{d}}^2x/mathop{mathrm{d}}t^2$, and similarily, $dot{x}= mathop{mathrm{d}}x / mathop{mathrm{d}}{t}$. Then, the first element that confuses me is the statement
$$
ddot{x} = frac{mathop{mathrm{d}}}{mathop{mathrm{d}}t} frac{mathop{mathrm{d}}x}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}x} frac{mathop{mathrm{d}x}}{mathop{mathrm{d}}t} = frac{mathrm{d}}{mathrm{d}x} left(frac{1}{2} dot{x}^2 right).
$$
According to the authors, this is known as the energy transformation. I haven't been able to find any other literature supporting this, so my first question is this: Is energy transformation a general term for ODEs, or is it a special name for this operation on the pendulum equation?
To continue, I understand the equation given that one may write $dot{x} = dot{x}(x)$, but that doesn't seem right to me, as the angular velocity may be both increasing and decreasing at any given $x$, depending upon if the pendulum is currently moving "up" or "down". I wouldn't know how to evaluate $mathop{mathrm{d}}dot{x} / mathop{mathrm{d}}x$ by itself. My second question is therefore: Is there a more rigorous way to arrive at the same result?
ordinary-differential-equations
asked Jan 10 at 14:26
Erik AndréErik André
857
$endgroup$
add a comment |
$begingroup$
After asking a question regarding the math myself, as well as finding some others, I am still puzzled by the opening sections of Jordan, D. and Smith, P. (2011). Nonlinear ordinary differential equations. We start off with the equation of motion for a simple pendulum, where $x$ denotes the angular displacement of the pendulum:
$$
ddot{x} + omega^2 sin{x} = 0.
$$
Here, $ddot{x} = mathop{mathrm{d}}^2x/mathop{mathrm{d}}t^2$, and similarily, $dot{x}= mathop{mathrm{d}}x / mathop{mathrm{d}}{t}$. Then, the first element that confuses me is the statement
$$
ddot{x} = frac{mathop{mathrm{d}}}{mathop{mathrm{d}}t} frac{mathop{mathrm{d}}x}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}x} frac{mathop{mathrm{d}x}}{mathop{mathrm{d}}t} = frac{mathrm{d}}{mathrm{d}x} left(frac{1}{2} dot{x}^2 right).
$$
According to the authors, this is known as the energy transformation. I haven't been able to find any other literature supporting this, so my first question is this: Is energy transformation a general term for ODEs, or is it a special name for this operation on the pendulum equation?
To continue, I understand the equation given that one may write $dot{x} = dot{x}(x)$, but that doesn't seem right to me, as the angular velocity may be both increasing and decreasing at any given $x$, depending upon if the pendulum is currently moving "up" or "down". I wouldn't know how to evaluate $mathop{mathrm{d}}dot{x} / mathop{mathrm{d}}x$ by itself. My second question is therefore: Is there a more rigorous way to arrive at the same result?
ordinary-differential-equations
asked Jan 10 at 14:26
Erik AndréErik André
857
$endgroup$
After asking a question regarding the math myself, as well as finding some others, I am still puzzled by the opening sections of Jordan, D. and Smith, P. (2011). Nonlinear ordinary differential equations. We start off with the equation of motion for a simple pendulum, where $x$ denotes the angular displacement of the pendulum:
$$
ddot{x} + omega^2 sin{x} = 0.
$$
Here, $ddot{x} = mathop{mathrm{d}}^2x/mathop{mathrm{d}}t^2$, and similarily, $dot{x}= mathop{mathrm{d}}x / mathop{mathrm{d}}{t}$. Then, the first element that confuses me is the statement
$$
ddot{x} = frac{mathop{mathrm{d}}}{mathop{mathrm{d}}t} frac{mathop{mathrm{d}}x}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}t} = frac{mathop{mathrm{d}}dot{x}}{mathop{mathrm{d}}x} frac{mathop{mathrm{d}x}}{mathop{mathrm{d}}t} = frac{mathrm{d}}{mathrm{d}x} left(frac{1}{2} dot{x}^2 right).
$$
According to the authors, this is known as the energy transformation. I haven't been able to find any other literature supporting this, so my first question is this: Is energy transformation a general term for ODEs, or is it a special name for this operation on the pendulum equation?
To continue, I understand the equation given that one may write $dot{x} = dot{x}(x)$, but that doesn't seem right to me, as the angular velocity may be both increasing and decreasing at any given $x$, depending upon if the pendulum is currently moving "up" or "down". I wouldn't know how to evaluate $mathop{mathrm{d}}dot{x} / mathop{mathrm{d}}x$ by itself. My second question is therefore: Is there a more rigorous way to arrive at the same result?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 10 at 14:26
Erik AndréErik André
857
asked Jan 10 at 14:26
Erik AndréErik André
857
asked Jan 10 at 14:26
Erik AndréErik André
857
asked Jan 10 at 14:26
Erik AndréErik André
857
asked Jan 10 at 14:26
Erik AndréErik André
857
857
add a comment |
add a comment |
1 Answer
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$begingroup$
Simply multiply the equation with $dot x$ and integrate for the time to get
$$
int (dot xddot x+ω^2sin(x)dot x)dt=frac12dot x^2 + ω^2(1-cos x) = E(x,dot x)=const.
$$
The second part can be substituted as $int sin x,dx$. Forcing the same substitution on the first term instead of integrating directly as product $int vdot v,dt$ gives rise to the cited contortion of the formalism.
Note that in the Lagrange formalism you get in this context
$$
ddot x =frac{d}{dt}frac{partial(frac12dot x^2)}{partial dot x}.
$$
as part of the Euler-Lagrange equations
$$
0=frac{d}{dt}frac{∂L}{∂dot x}-frac{∂L}{∂x}.
$$
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
$endgroup$
$begingroup$
Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $int left( dot{x} ddot{x} + omega^2 sin{(x)} dot{x} right) dt = frac{1}{2} dot{x}^2 - omega^2 cos{x}$?
$endgroup$
– Erik André
Jan 11 at 19:59
1
$begingroup$
There is a free integration constant. I choose to have the energy at rest to be zero, $E=frac12dot x^2+frac{ω^2}2(2sinfrac x2)^2.$ Your choice gives a minimality of terms.
$endgroup$
– LutzL
Jan 11 at 20:50
add a comment |
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1 Answer
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$begingroup$
Simply multiply the equation with $dot x$ and integrate for the time to get
$$
int (dot xddot x+ω^2sin(x)dot x)dt=frac12dot x^2 + ω^2(1-cos x) = E(x,dot x)=const.
$$
The second part can be substituted as $int sin x,dx$. Forcing the same substitution on the first term instead of integrating directly as product $int vdot v,dt$ gives rise to the cited contortion of the formalism.
Note that in the Lagrange formalism you get in this context
$$
ddot x =frac{d}{dt}frac{partial(frac12dot x^2)}{partial dot x}.
$$
as part of the Euler-Lagrange equations
$$
0=frac{d}{dt}frac{∂L}{∂dot x}-frac{∂L}{∂x}.
$$
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
$endgroup$
$begingroup$
Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $int left( dot{x} ddot{x} + omega^2 sin{(x)} dot{x} right) dt = frac{1}{2} dot{x}^2 - omega^2 cos{x}$?
$endgroup$
– Erik André
Jan 11 at 19:59
1
$begingroup$
There is a free integration constant. I choose to have the energy at rest to be zero, $E=frac12dot x^2+frac{ω^2}2(2sinfrac x2)^2.$ Your choice gives a minimality of terms.
$endgroup$
– LutzL
Jan 11 at 20:50
add a comment |
$begingroup$
Simply multiply the equation with $dot x$ and integrate for the time to get
$$
int (dot xddot x+ω^2sin(x)dot x)dt=frac12dot x^2 + ω^2(1-cos x) = E(x,dot x)=const.
$$
The second part can be substituted as $int sin x,dx$. Forcing the same substitution on the first term instead of integrating directly as product $int vdot v,dt$ gives rise to the cited contortion of the formalism.
Note that in the Lagrange formalism you get in this context
$$
ddot x =frac{d}{dt}frac{partial(frac12dot x^2)}{partial dot x}.
$$
as part of the Euler-Lagrange equations
$$
0=frac{d}{dt}frac{∂L}{∂dot x}-frac{∂L}{∂x}.
$$
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
$endgroup$
$begingroup$
Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $int left( dot{x} ddot{x} + omega^2 sin{(x)} dot{x} right) dt = frac{1}{2} dot{x}^2 - omega^2 cos{x}$?
$endgroup$
– Erik André
Jan 11 at 19:59
1
$begingroup$
There is a free integration constant. I choose to have the energy at rest to be zero, $E=frac12dot x^2+frac{ω^2}2(2sinfrac x2)^2.$ Your choice gives a minimality of terms.
$endgroup$
– LutzL
Jan 11 at 20:50
add a comment |
$begingroup$
Simply multiply the equation with $dot x$ and integrate for the time to get
$$
int (dot xddot x+ω^2sin(x)dot x)dt=frac12dot x^2 + ω^2(1-cos x) = E(x,dot x)=const.
$$
The second part can be substituted as $int sin x,dx$. Forcing the same substitution on the first term instead of integrating directly as product $int vdot v,dt$ gives rise to the cited contortion of the formalism.
Note that in the Lagrange formalism you get in this context
$$
ddot x =frac{d}{dt}frac{partial(frac12dot x^2)}{partial dot x}.
$$
as part of the Euler-Lagrange equations
$$
0=frac{d}{dt}frac{∂L}{∂dot x}-frac{∂L}{∂x}.
$$
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
$endgroup$
Simply multiply the equation with $dot x$ and integrate for the time to get
$$
int (dot xddot x+ω^2sin(x)dot x)dt=frac12dot x^2 + ω^2(1-cos x) = E(x,dot x)=const.
$$
The second part can be substituted as $int sin x,dx$. Forcing the same substitution on the first term instead of integrating directly as product $int vdot v,dt$ gives rise to the cited contortion of the formalism.
Note that in the Lagrange formalism you get in this context
$$
ddot x =frac{d}{dt}frac{partial(frac12dot x^2)}{partial dot x}.
$$
as part of the Euler-Lagrange equations
$$
0=frac{d}{dt}frac{∂L}{∂dot x}-frac{∂L}{∂x}.
$$
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
edited Jan 10 at 15:33
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
answered Jan 10 at 15:27
LutzLLutzL
57.9k42054
57.9k42054
$begingroup$
Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $int left( dot{x} ddot{x} + omega^2 sin{(x)} dot{x} right) dt = frac{1}{2} dot{x}^2 - omega^2 cos{x}$?
$endgroup$
– Erik André
Jan 11 at 19:59
1
$begingroup$
There is a free integration constant. I choose to have the energy at rest to be zero, $E=frac12dot x^2+frac{ω^2}2(2sinfrac x2)^2.$ Your choice gives a minimality of terms.
$endgroup$
– LutzL
Jan 11 at 20:50
add a comment |
$begingroup$
Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $int left( dot{x} ddot{x} + omega^2 sin{(x)} dot{x} right) dt = frac{1}{2} dot{x}^2 - omega^2 cos{x}$?
$endgroup$
– Erik André
Jan 11 at 19:59
1
$begingroup$
There is a free integration constant. I choose to have the energy at rest to be zero, $E=frac12dot x^2+frac{ω^2}2(2sinfrac x2)^2.$ Your choice gives a minimality of terms.
$endgroup$
– LutzL
Jan 11 at 20:50
$begingroup$
Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $int left( dot{x} ddot{x} + omega^2 sin{(x)} dot{x} right) dt = frac{1}{2} dot{x}^2 - omega^2 cos{x}$?
$endgroup$
– Erik André
Jan 11 at 19:59
$begingroup$
Thank you, this made me understand it. Is your integral correct, by the way? Shouldn't it be $int left( dot{x} ddot{x} + omega^2 sin{(x)} dot{x} right) dt = frac{1}{2} dot{x}^2 - omega^2 cos{x}$?
$endgroup$
– Erik André
Jan 11 at 19:59
1
1
$begingroup$
There is a free integration constant. I choose to have the energy at rest to be zero, $E=frac12dot x^2+frac{ω^2}2(2sinfrac x2)^2.$ Your choice gives a minimality of terms.
$endgroup$
– LutzL
Jan 11 at 20:50
$begingroup$
There is a free integration constant. I choose to have the energy at rest to be zero, $E=frac12dot x^2+frac{ω^2}2(2sinfrac x2)^2.$ Your choice gives a minimality of terms.
$endgroup$
– LutzL
Jan 11 at 20:50
add a comment |
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