Transformation of a second order ODE
I have to transform a second order differential equation into a Bernoulli type differential equation, however, I am having some trouble.
The original equation is:
begin{equation}
ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0
end{equation}
By defining a new function:
begin{equation}
phi = frac{dt}{du}
end{equation}
The original equation has to be transformed into:
begin{equation}
frac{dphi}{du} + frac{1}{u}phi = (y-x_{0}beta u) phi^{2}
end{equation}
Would anyone be willing to suggest any ideas for how this might be achieved?
Thank you!
differential-equations
New contributor
add a comment |
I have to transform a second order differential equation into a Bernoulli type differential equation, however, I am having some trouble.
The original equation is:
begin{equation}
ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0
end{equation}
By defining a new function:
begin{equation}
phi = frac{dt}{du}
end{equation}
The original equation has to be transformed into:
begin{equation}
frac{dphi}{du} + frac{1}{u}phi = (y-x_{0}beta u) phi^{2}
end{equation}
Would anyone be willing to suggest any ideas for how this might be achieved?
Thank you!
differential-equations
New contributor
add a comment |
I have to transform a second order differential equation into a Bernoulli type differential equation, however, I am having some trouble.
The original equation is:
begin{equation}
ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0
end{equation}
By defining a new function:
begin{equation}
phi = frac{dt}{du}
end{equation}
The original equation has to be transformed into:
begin{equation}
frac{dphi}{du} + frac{1}{u}phi = (y-x_{0}beta u) phi^{2}
end{equation}
Would anyone be willing to suggest any ideas for how this might be achieved?
Thank you!
differential-equations
New contributor
I have to transform a second order differential equation into a Bernoulli type differential equation, however, I am having some trouble.
The original equation is:
begin{equation}
ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0
end{equation}
By defining a new function:
begin{equation}
phi = frac{dt}{du}
end{equation}
The original equation has to be transformed into:
begin{equation}
frac{dphi}{du} + frac{1}{u}phi = (y-x_{0}beta u) phi^{2}
end{equation}
Would anyone be willing to suggest any ideas for how this might be achieved?
Thank you!
differential-equations
differential-equations
New contributor
New contributor
edited Jan 4 at 22:52
nipohc88
New contributor
asked Jan 4 at 22:00
nipohc88nipohc88
82
82
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
We have
$$phi=frac{dt}{du}implies frac{dphi}{du}=frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}=-phifrac{frac{d^2u}{dt^2}}{left(frac{du}{dt}right)^2}=-phi^3frac{d^2u}{dt^2}$$
Then the original differential equation transforms from
$$ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0$$
to
$$-frac u{phi^3}frac{dphi}{du}-frac1{phi^2}+(gamma-x_0beta u)frac u{phi}=0$$
Multiplying through by $-frac{phi^3}u$, one gets to the required answer.
$$frac{dphi}{du}+frac1uphi-(gamma-x_0beta u)phi^2=0$$
(I think you have a couple of typos in the question)
add a comment |
Same solution, but with more details concerning the computations: Leibniz notation $frac{d}{dx}$ is very concise but you can lose track of what you really do, especially for high order derivatives.
$u$ is some function of $t$, $u=f(t)$. Let's denote $f^{-1}$ the reciprocal function (it must be assumed that $u$ is monotonic on a neighborhood of $t$ to make the change of function meaningful):
$$
fcirc f^{-1}(u)=u
$$
Using the chain rule we get:
$$
frac{d}{du}(fcirc f^{-1}(u))=f'(f^{-1}(u))(f^{-1})'(u)=1
$$
thus, by identification:
$$
phi(u)triangleq (f^{-1})'(u)=frac{1}{f'(f^{-1}(u))}triangleqfrac{dt}{du}
$$
which can be rewritten as (in a more concise form):
$$
frac{du}{dt}=frac{1}{phi}
$$
Now if you derivate once more:
begin{align}
frac{d}{du}phi(u)&=frac{d}{du}frac{1}{f'(f^{-1}(u))}\
&=-frac{(f^{-1})'(u)f''(f^{-1}(u))}{(f'(f^{-1}(u)))^2}\
&=-phi^3(u)f''(f^{-1}(u))
end{align}
which can be rewritten as (in a more concise form):
$$
frac{d^2u}{dt^2}=-frac{1}{phi^3}frac{dphi}{du}
$$
Now you can do direct substitution into your initial equation:
$$
ufrac{d^2u}{dt^2}−(frac{du}{dt})^2+(γ−x_0βu)ufrac{du}{dt}=0
$$
hence
$$
-frac{u}{phi^3}frac{dphi}{du}-(frac{1}{phi})^2+(γ−x_0βu)ufrac{1}{phi}=0
$$
If you multiply by $-frac{phi^3}{u}$ you get your expected result:
$$
frac{dphi}{du}+frac{phi}{u}=(γ−x_0βu)phi^2
$$
Extra comment concerning (critics of the) Leibniz notation:
I personally do not like Leibniz notation for high order derivatives. By example, it is clear that the chain rule can be written as:
$$
frac{dy}{dt}=frac{dy}{dx}frac{dx}{dt}
$$
however for higher order derivatives great care must be taken, as this notation can also be misleading:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}frac{dx^2}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2
$$
which is wrong, the right formula is:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2+frac{dy}{dx}frac{d^2x}{dt^2}
$$
You have not this problem with Lagrange notation:
$$
y(x(t))''=(y'(x(t))x'(t))'=y''(x(t))(x'(t))^2+y'(x(t))x''(t)
$$
So, personally I always have doubts when I write expressions like:
$$
frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}
$$
(again it is a matter of personal habits, I do not say it is a bad thing to do that (see post comments), it is just that personally I don't feel confident when I write these kind of expressions).
This problem (and others) with the Leibniz notation is well known when teaching differential calculus, see:
H. Poincaré, La Notation Différentielle et l'enseignement (pdf)
J. Hadamard, La notion de différentielle dans l'enseignement (pdf)
unfortunately both in French, however you can find an English translation of Hadamard's article here.
You can also see:
Differentials, higher-order differentials and the derivative in the Leibnizian calculus (pdf)
In your second equation, did you mean to differentiate with respect to $u$? Otherwise, I don't see how this can equal $1$. If that is not what you meant, then it would also mean you are missing a $frac{du}{dt}$ term. Also, I am not sure I agree that it is easier to keep track of what you are doing this way, it seems harder to follow for me (but of course, it can't hurt to have an alternative method posted). +1
– John Doe
Jan 5 at 0:44
@JohnDoe, thanks for the feedback, yes a typo, fixed!
– Picaud Vincent
Jan 5 at 0:47
@JohnDoe I have added two references concerning pros and cons of the Leibniz notation, sorry these two articles are in French. I know that one of the two also has an English translation I will try to find it (I do not remember where I have seen it)
– Picaud Vincent
Jan 5 at 0:55
@JohnDoe ok I just found the Hadamard's article English version. Links fixed.
– Picaud Vincent
Jan 5 at 1:05
1
@JohnDoe thanks for the link! Yes interesting debate. I also do agree that it is a personal preference, the most important thing is to know what there are behind the notations :)
– Picaud Vincent
Jan 5 at 1:22
|
show 3 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
nipohc88 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062165%2ftransformation-of-a-second-order-ode%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We have
$$phi=frac{dt}{du}implies frac{dphi}{du}=frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}=-phifrac{frac{d^2u}{dt^2}}{left(frac{du}{dt}right)^2}=-phi^3frac{d^2u}{dt^2}$$
Then the original differential equation transforms from
$$ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0$$
to
$$-frac u{phi^3}frac{dphi}{du}-frac1{phi^2}+(gamma-x_0beta u)frac u{phi}=0$$
Multiplying through by $-frac{phi^3}u$, one gets to the required answer.
$$frac{dphi}{du}+frac1uphi-(gamma-x_0beta u)phi^2=0$$
(I think you have a couple of typos in the question)
add a comment |
We have
$$phi=frac{dt}{du}implies frac{dphi}{du}=frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}=-phifrac{frac{d^2u}{dt^2}}{left(frac{du}{dt}right)^2}=-phi^3frac{d^2u}{dt^2}$$
Then the original differential equation transforms from
$$ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0$$
to
$$-frac u{phi^3}frac{dphi}{du}-frac1{phi^2}+(gamma-x_0beta u)frac u{phi}=0$$
Multiplying through by $-frac{phi^3}u$, one gets to the required answer.
$$frac{dphi}{du}+frac1uphi-(gamma-x_0beta u)phi^2=0$$
(I think you have a couple of typos in the question)
add a comment |
We have
$$phi=frac{dt}{du}implies frac{dphi}{du}=frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}=-phifrac{frac{d^2u}{dt^2}}{left(frac{du}{dt}right)^2}=-phi^3frac{d^2u}{dt^2}$$
Then the original differential equation transforms from
$$ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0$$
to
$$-frac u{phi^3}frac{dphi}{du}-frac1{phi^2}+(gamma-x_0beta u)frac u{phi}=0$$
Multiplying through by $-frac{phi^3}u$, one gets to the required answer.
$$frac{dphi}{du}+frac1uphi-(gamma-x_0beta u)phi^2=0$$
(I think you have a couple of typos in the question)
We have
$$phi=frac{dt}{du}implies frac{dphi}{du}=frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}=-phifrac{frac{d^2u}{dt^2}}{left(frac{du}{dt}right)^2}=-phi^3frac{d^2u}{dt^2}$$
Then the original differential equation transforms from
$$ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0$$
to
$$-frac u{phi^3}frac{dphi}{du}-frac1{phi^2}+(gamma-x_0beta u)frac u{phi}=0$$
Multiplying through by $-frac{phi^3}u$, one gets to the required answer.
$$frac{dphi}{du}+frac1uphi-(gamma-x_0beta u)phi^2=0$$
(I think you have a couple of typos in the question)
answered Jan 4 at 22:25
John DoeJohn Doe
10.9k11238
10.9k11238
add a comment |
add a comment |
Same solution, but with more details concerning the computations: Leibniz notation $frac{d}{dx}$ is very concise but you can lose track of what you really do, especially for high order derivatives.
$u$ is some function of $t$, $u=f(t)$. Let's denote $f^{-1}$ the reciprocal function (it must be assumed that $u$ is monotonic on a neighborhood of $t$ to make the change of function meaningful):
$$
fcirc f^{-1}(u)=u
$$
Using the chain rule we get:
$$
frac{d}{du}(fcirc f^{-1}(u))=f'(f^{-1}(u))(f^{-1})'(u)=1
$$
thus, by identification:
$$
phi(u)triangleq (f^{-1})'(u)=frac{1}{f'(f^{-1}(u))}triangleqfrac{dt}{du}
$$
which can be rewritten as (in a more concise form):
$$
frac{du}{dt}=frac{1}{phi}
$$
Now if you derivate once more:
begin{align}
frac{d}{du}phi(u)&=frac{d}{du}frac{1}{f'(f^{-1}(u))}\
&=-frac{(f^{-1})'(u)f''(f^{-1}(u))}{(f'(f^{-1}(u)))^2}\
&=-phi^3(u)f''(f^{-1}(u))
end{align}
which can be rewritten as (in a more concise form):
$$
frac{d^2u}{dt^2}=-frac{1}{phi^3}frac{dphi}{du}
$$
Now you can do direct substitution into your initial equation:
$$
ufrac{d^2u}{dt^2}−(frac{du}{dt})^2+(γ−x_0βu)ufrac{du}{dt}=0
$$
hence
$$
-frac{u}{phi^3}frac{dphi}{du}-(frac{1}{phi})^2+(γ−x_0βu)ufrac{1}{phi}=0
$$
If you multiply by $-frac{phi^3}{u}$ you get your expected result:
$$
frac{dphi}{du}+frac{phi}{u}=(γ−x_0βu)phi^2
$$
Extra comment concerning (critics of the) Leibniz notation:
I personally do not like Leibniz notation for high order derivatives. By example, it is clear that the chain rule can be written as:
$$
frac{dy}{dt}=frac{dy}{dx}frac{dx}{dt}
$$
however for higher order derivatives great care must be taken, as this notation can also be misleading:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}frac{dx^2}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2
$$
which is wrong, the right formula is:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2+frac{dy}{dx}frac{d^2x}{dt^2}
$$
You have not this problem with Lagrange notation:
$$
y(x(t))''=(y'(x(t))x'(t))'=y''(x(t))(x'(t))^2+y'(x(t))x''(t)
$$
So, personally I always have doubts when I write expressions like:
$$
frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}
$$
(again it is a matter of personal habits, I do not say it is a bad thing to do that (see post comments), it is just that personally I don't feel confident when I write these kind of expressions).
This problem (and others) with the Leibniz notation is well known when teaching differential calculus, see:
H. Poincaré, La Notation Différentielle et l'enseignement (pdf)
J. Hadamard, La notion de différentielle dans l'enseignement (pdf)
unfortunately both in French, however you can find an English translation of Hadamard's article here.
You can also see:
Differentials, higher-order differentials and the derivative in the Leibnizian calculus (pdf)
In your second equation, did you mean to differentiate with respect to $u$? Otherwise, I don't see how this can equal $1$. If that is not what you meant, then it would also mean you are missing a $frac{du}{dt}$ term. Also, I am not sure I agree that it is easier to keep track of what you are doing this way, it seems harder to follow for me (but of course, it can't hurt to have an alternative method posted). +1
– John Doe
Jan 5 at 0:44
@JohnDoe, thanks for the feedback, yes a typo, fixed!
– Picaud Vincent
Jan 5 at 0:47
@JohnDoe I have added two references concerning pros and cons of the Leibniz notation, sorry these two articles are in French. I know that one of the two also has an English translation I will try to find it (I do not remember where I have seen it)
– Picaud Vincent
Jan 5 at 0:55
@JohnDoe ok I just found the Hadamard's article English version. Links fixed.
– Picaud Vincent
Jan 5 at 1:05
1
@JohnDoe thanks for the link! Yes interesting debate. I also do agree that it is a personal preference, the most important thing is to know what there are behind the notations :)
– Picaud Vincent
Jan 5 at 1:22
|
show 3 more comments
Same solution, but with more details concerning the computations: Leibniz notation $frac{d}{dx}$ is very concise but you can lose track of what you really do, especially for high order derivatives.
$u$ is some function of $t$, $u=f(t)$. Let's denote $f^{-1}$ the reciprocal function (it must be assumed that $u$ is monotonic on a neighborhood of $t$ to make the change of function meaningful):
$$
fcirc f^{-1}(u)=u
$$
Using the chain rule we get:
$$
frac{d}{du}(fcirc f^{-1}(u))=f'(f^{-1}(u))(f^{-1})'(u)=1
$$
thus, by identification:
$$
phi(u)triangleq (f^{-1})'(u)=frac{1}{f'(f^{-1}(u))}triangleqfrac{dt}{du}
$$
which can be rewritten as (in a more concise form):
$$
frac{du}{dt}=frac{1}{phi}
$$
Now if you derivate once more:
begin{align}
frac{d}{du}phi(u)&=frac{d}{du}frac{1}{f'(f^{-1}(u))}\
&=-frac{(f^{-1})'(u)f''(f^{-1}(u))}{(f'(f^{-1}(u)))^2}\
&=-phi^3(u)f''(f^{-1}(u))
end{align}
which can be rewritten as (in a more concise form):
$$
frac{d^2u}{dt^2}=-frac{1}{phi^3}frac{dphi}{du}
$$
Now you can do direct substitution into your initial equation:
$$
ufrac{d^2u}{dt^2}−(frac{du}{dt})^2+(γ−x_0βu)ufrac{du}{dt}=0
$$
hence
$$
-frac{u}{phi^3}frac{dphi}{du}-(frac{1}{phi})^2+(γ−x_0βu)ufrac{1}{phi}=0
$$
If you multiply by $-frac{phi^3}{u}$ you get your expected result:
$$
frac{dphi}{du}+frac{phi}{u}=(γ−x_0βu)phi^2
$$
Extra comment concerning (critics of the) Leibniz notation:
I personally do not like Leibniz notation for high order derivatives. By example, it is clear that the chain rule can be written as:
$$
frac{dy}{dt}=frac{dy}{dx}frac{dx}{dt}
$$
however for higher order derivatives great care must be taken, as this notation can also be misleading:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}frac{dx^2}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2
$$
which is wrong, the right formula is:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2+frac{dy}{dx}frac{d^2x}{dt^2}
$$
You have not this problem with Lagrange notation:
$$
y(x(t))''=(y'(x(t))x'(t))'=y''(x(t))(x'(t))^2+y'(x(t))x''(t)
$$
So, personally I always have doubts when I write expressions like:
$$
frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}
$$
(again it is a matter of personal habits, I do not say it is a bad thing to do that (see post comments), it is just that personally I don't feel confident when I write these kind of expressions).
This problem (and others) with the Leibniz notation is well known when teaching differential calculus, see:
H. Poincaré, La Notation Différentielle et l'enseignement (pdf)
J. Hadamard, La notion de différentielle dans l'enseignement (pdf)
unfortunately both in French, however you can find an English translation of Hadamard's article here.
You can also see:
Differentials, higher-order differentials and the derivative in the Leibnizian calculus (pdf)
In your second equation, did you mean to differentiate with respect to $u$? Otherwise, I don't see how this can equal $1$. If that is not what you meant, then it would also mean you are missing a $frac{du}{dt}$ term. Also, I am not sure I agree that it is easier to keep track of what you are doing this way, it seems harder to follow for me (but of course, it can't hurt to have an alternative method posted). +1
– John Doe
Jan 5 at 0:44
@JohnDoe, thanks for the feedback, yes a typo, fixed!
– Picaud Vincent
Jan 5 at 0:47
@JohnDoe I have added two references concerning pros and cons of the Leibniz notation, sorry these two articles are in French. I know that one of the two also has an English translation I will try to find it (I do not remember where I have seen it)
– Picaud Vincent
Jan 5 at 0:55
@JohnDoe ok I just found the Hadamard's article English version. Links fixed.
– Picaud Vincent
Jan 5 at 1:05
1
@JohnDoe thanks for the link! Yes interesting debate. I also do agree that it is a personal preference, the most important thing is to know what there are behind the notations :)
– Picaud Vincent
Jan 5 at 1:22
|
show 3 more comments
Same solution, but with more details concerning the computations: Leibniz notation $frac{d}{dx}$ is very concise but you can lose track of what you really do, especially for high order derivatives.
$u$ is some function of $t$, $u=f(t)$. Let's denote $f^{-1}$ the reciprocal function (it must be assumed that $u$ is monotonic on a neighborhood of $t$ to make the change of function meaningful):
$$
fcirc f^{-1}(u)=u
$$
Using the chain rule we get:
$$
frac{d}{du}(fcirc f^{-1}(u))=f'(f^{-1}(u))(f^{-1})'(u)=1
$$
thus, by identification:
$$
phi(u)triangleq (f^{-1})'(u)=frac{1}{f'(f^{-1}(u))}triangleqfrac{dt}{du}
$$
which can be rewritten as (in a more concise form):
$$
frac{du}{dt}=frac{1}{phi}
$$
Now if you derivate once more:
begin{align}
frac{d}{du}phi(u)&=frac{d}{du}frac{1}{f'(f^{-1}(u))}\
&=-frac{(f^{-1})'(u)f''(f^{-1}(u))}{(f'(f^{-1}(u)))^2}\
&=-phi^3(u)f''(f^{-1}(u))
end{align}
which can be rewritten as (in a more concise form):
$$
frac{d^2u}{dt^2}=-frac{1}{phi^3}frac{dphi}{du}
$$
Now you can do direct substitution into your initial equation:
$$
ufrac{d^2u}{dt^2}−(frac{du}{dt})^2+(γ−x_0βu)ufrac{du}{dt}=0
$$
hence
$$
-frac{u}{phi^3}frac{dphi}{du}-(frac{1}{phi})^2+(γ−x_0βu)ufrac{1}{phi}=0
$$
If you multiply by $-frac{phi^3}{u}$ you get your expected result:
$$
frac{dphi}{du}+frac{phi}{u}=(γ−x_0βu)phi^2
$$
Extra comment concerning (critics of the) Leibniz notation:
I personally do not like Leibniz notation for high order derivatives. By example, it is clear that the chain rule can be written as:
$$
frac{dy}{dt}=frac{dy}{dx}frac{dx}{dt}
$$
however for higher order derivatives great care must be taken, as this notation can also be misleading:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}frac{dx^2}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2
$$
which is wrong, the right formula is:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2+frac{dy}{dx}frac{d^2x}{dt^2}
$$
You have not this problem with Lagrange notation:
$$
y(x(t))''=(y'(x(t))x'(t))'=y''(x(t))(x'(t))^2+y'(x(t))x''(t)
$$
So, personally I always have doubts when I write expressions like:
$$
frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}
$$
(again it is a matter of personal habits, I do not say it is a bad thing to do that (see post comments), it is just that personally I don't feel confident when I write these kind of expressions).
This problem (and others) with the Leibniz notation is well known when teaching differential calculus, see:
H. Poincaré, La Notation Différentielle et l'enseignement (pdf)
J. Hadamard, La notion de différentielle dans l'enseignement (pdf)
unfortunately both in French, however you can find an English translation of Hadamard's article here.
You can also see:
Differentials, higher-order differentials and the derivative in the Leibnizian calculus (pdf)
Same solution, but with more details concerning the computations: Leibniz notation $frac{d}{dx}$ is very concise but you can lose track of what you really do, especially for high order derivatives.
$u$ is some function of $t$, $u=f(t)$. Let's denote $f^{-1}$ the reciprocal function (it must be assumed that $u$ is monotonic on a neighborhood of $t$ to make the change of function meaningful):
$$
fcirc f^{-1}(u)=u
$$
Using the chain rule we get:
$$
frac{d}{du}(fcirc f^{-1}(u))=f'(f^{-1}(u))(f^{-1})'(u)=1
$$
thus, by identification:
$$
phi(u)triangleq (f^{-1})'(u)=frac{1}{f'(f^{-1}(u))}triangleqfrac{dt}{du}
$$
which can be rewritten as (in a more concise form):
$$
frac{du}{dt}=frac{1}{phi}
$$
Now if you derivate once more:
begin{align}
frac{d}{du}phi(u)&=frac{d}{du}frac{1}{f'(f^{-1}(u))}\
&=-frac{(f^{-1})'(u)f''(f^{-1}(u))}{(f'(f^{-1}(u)))^2}\
&=-phi^3(u)f''(f^{-1}(u))
end{align}
which can be rewritten as (in a more concise form):
$$
frac{d^2u}{dt^2}=-frac{1}{phi^3}frac{dphi}{du}
$$
Now you can do direct substitution into your initial equation:
$$
ufrac{d^2u}{dt^2}−(frac{du}{dt})^2+(γ−x_0βu)ufrac{du}{dt}=0
$$
hence
$$
-frac{u}{phi^3}frac{dphi}{du}-(frac{1}{phi})^2+(γ−x_0βu)ufrac{1}{phi}=0
$$
If you multiply by $-frac{phi^3}{u}$ you get your expected result:
$$
frac{dphi}{du}+frac{phi}{u}=(γ−x_0βu)phi^2
$$
Extra comment concerning (critics of the) Leibniz notation:
I personally do not like Leibniz notation for high order derivatives. By example, it is clear that the chain rule can be written as:
$$
frac{dy}{dt}=frac{dy}{dx}frac{dx}{dt}
$$
however for higher order derivatives great care must be taken, as this notation can also be misleading:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}frac{dx^2}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2
$$
which is wrong, the right formula is:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2+frac{dy}{dx}frac{d^2x}{dt^2}
$$
You have not this problem with Lagrange notation:
$$
y(x(t))''=(y'(x(t))x'(t))'=y''(x(t))(x'(t))^2+y'(x(t))x''(t)
$$
So, personally I always have doubts when I write expressions like:
$$
frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}
$$
(again it is a matter of personal habits, I do not say it is a bad thing to do that (see post comments), it is just that personally I don't feel confident when I write these kind of expressions).
This problem (and others) with the Leibniz notation is well known when teaching differential calculus, see:
H. Poincaré, La Notation Différentielle et l'enseignement (pdf)
J. Hadamard, La notion de différentielle dans l'enseignement (pdf)
unfortunately both in French, however you can find an English translation of Hadamard's article here.
You can also see:
Differentials, higher-order differentials and the derivative in the Leibnizian calculus (pdf)
edited Jan 5 at 10:39
answered Jan 4 at 23:28
Picaud VincentPicaud Vincent
1,33439
1,33439
In your second equation, did you mean to differentiate with respect to $u$? Otherwise, I don't see how this can equal $1$. If that is not what you meant, then it would also mean you are missing a $frac{du}{dt}$ term. Also, I am not sure I agree that it is easier to keep track of what you are doing this way, it seems harder to follow for me (but of course, it can't hurt to have an alternative method posted). +1
– John Doe
Jan 5 at 0:44
@JohnDoe, thanks for the feedback, yes a typo, fixed!
– Picaud Vincent
Jan 5 at 0:47
@JohnDoe I have added two references concerning pros and cons of the Leibniz notation, sorry these two articles are in French. I know that one of the two also has an English translation I will try to find it (I do not remember where I have seen it)
– Picaud Vincent
Jan 5 at 0:55
@JohnDoe ok I just found the Hadamard's article English version. Links fixed.
– Picaud Vincent
Jan 5 at 1:05
1
@JohnDoe thanks for the link! Yes interesting debate. I also do agree that it is a personal preference, the most important thing is to know what there are behind the notations :)
– Picaud Vincent
Jan 5 at 1:22
|
show 3 more comments
In your second equation, did you mean to differentiate with respect to $u$? Otherwise, I don't see how this can equal $1$. If that is not what you meant, then it would also mean you are missing a $frac{du}{dt}$ term. Also, I am not sure I agree that it is easier to keep track of what you are doing this way, it seems harder to follow for me (but of course, it can't hurt to have an alternative method posted). +1
– John Doe
Jan 5 at 0:44
@JohnDoe, thanks for the feedback, yes a typo, fixed!
– Picaud Vincent
Jan 5 at 0:47
@JohnDoe I have added two references concerning pros and cons of the Leibniz notation, sorry these two articles are in French. I know that one of the two also has an English translation I will try to find it (I do not remember where I have seen it)
– Picaud Vincent
Jan 5 at 0:55
@JohnDoe ok I just found the Hadamard's article English version. Links fixed.
– Picaud Vincent
Jan 5 at 1:05
1
@JohnDoe thanks for the link! Yes interesting debate. I also do agree that it is a personal preference, the most important thing is to know what there are behind the notations :)
– Picaud Vincent
Jan 5 at 1:22
In your second equation, did you mean to differentiate with respect to $u$? Otherwise, I don't see how this can equal $1$. If that is not what you meant, then it would also mean you are missing a $frac{du}{dt}$ term. Also, I am not sure I agree that it is easier to keep track of what you are doing this way, it seems harder to follow for me (but of course, it can't hurt to have an alternative method posted). +1
– John Doe
Jan 5 at 0:44
In your second equation, did you mean to differentiate with respect to $u$? Otherwise, I don't see how this can equal $1$. If that is not what you meant, then it would also mean you are missing a $frac{du}{dt}$ term. Also, I am not sure I agree that it is easier to keep track of what you are doing this way, it seems harder to follow for me (but of course, it can't hurt to have an alternative method posted). +1
– John Doe
Jan 5 at 0:44
@JohnDoe, thanks for the feedback, yes a typo, fixed!
– Picaud Vincent
Jan 5 at 0:47
@JohnDoe, thanks for the feedback, yes a typo, fixed!
– Picaud Vincent
Jan 5 at 0:47
@JohnDoe I have added two references concerning pros and cons of the Leibniz notation, sorry these two articles are in French. I know that one of the two also has an English translation I will try to find it (I do not remember where I have seen it)
– Picaud Vincent
Jan 5 at 0:55
@JohnDoe I have added two references concerning pros and cons of the Leibniz notation, sorry these two articles are in French. I know that one of the two also has an English translation I will try to find it (I do not remember where I have seen it)
– Picaud Vincent
Jan 5 at 0:55
@JohnDoe ok I just found the Hadamard's article English version. Links fixed.
– Picaud Vincent
Jan 5 at 1:05
@JohnDoe ok I just found the Hadamard's article English version. Links fixed.
– Picaud Vincent
Jan 5 at 1:05
1
1
@JohnDoe thanks for the link! Yes interesting debate. I also do agree that it is a personal preference, the most important thing is to know what there are behind the notations :)
– Picaud Vincent
Jan 5 at 1:22
@JohnDoe thanks for the link! Yes interesting debate. I also do agree that it is a personal preference, the most important thing is to know what there are behind the notations :)
– Picaud Vincent
Jan 5 at 1:22
|
show 3 more comments
nipohc88 is a new contributor. Be nice, and check out our Code of Conduct.
nipohc88 is a new contributor. Be nice, and check out our Code of Conduct.
nipohc88 is a new contributor. Be nice, and check out our Code of Conduct.
nipohc88 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062165%2ftransformation-of-a-second-order-ode%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown