Mixing
Eulerian to Lagrangian Coordinates Help: How to find constants? (Theoretical Mechanics)
$begingroup$
Consider material traveling in a two-dimensional space with its velocity at any given location in space (capital letters correspond to Eulerian, lower case to Lagrangian) given by $overrightarrow{V}(overrightarrow{X},t) =begin{pmatrix}X^1+t\(X^2)^2 end{pmatrix}$. (Vectors are written using Cartesian coordinates,$ overrightarrow{X}= overrightarrow{R}$,and $overrightarrow{x}=overrightarrow{r}$.)
Find the position of a parcel/particle that starts at the location$overrightarrow{x}=begin{pmatrix} x^1\x^2end{pmatrix}.$
I solved the system of ordinary differential equations $frac{d}{dt}(X^1) =X^1+t; frac{d}{dt}(X^2) = (X^2)^2$, but I'm having trouble using the initial conditions to plug stuff in? I feel like I need to find $overrightarrow{x}(overrightarrow{X},t)$ first, but I also have no clue how to do that. Do I need to convert from Eulerian to Lagrangian and then plug in the initial conditions? Given what I have, I'm not really sure how to do that? Thanks for any help!
Here is what I have:
$X^1=Ke^t-t-1$
$X^2=-frac{1}{t+C}$
classical-mechanics
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
$endgroup$
add a comment |
$begingroup$
Consider material traveling in a two-dimensional space with its velocity at any given location in space (capital letters correspond to Eulerian, lower case to Lagrangian) given by $overrightarrow{V}(overrightarrow{X},t) =begin{pmatrix}X^1+t\(X^2)^2 end{pmatrix}$. (Vectors are written using Cartesian coordinates,$ overrightarrow{X}= overrightarrow{R}$,and $overrightarrow{x}=overrightarrow{r}$.)
Find the position of a parcel/particle that starts at the location$overrightarrow{x}=begin{pmatrix} x^1\x^2end{pmatrix}.$
I solved the system of ordinary differential equations $frac{d}{dt}(X^1) =X^1+t; frac{d}{dt}(X^2) = (X^2)^2$, but I'm having trouble using the initial conditions to plug stuff in? I feel like I need to find $overrightarrow{x}(overrightarrow{X},t)$ first, but I also have no clue how to do that. Do I need to convert from Eulerian to Lagrangian and then plug in the initial conditions? Given what I have, I'm not really sure how to do that? Thanks for any help!
Here is what I have:
$X^1=Ke^t-t-1$
$X^2=-frac{1}{t+C}$
classical-mechanics
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
$endgroup$
$begingroup$
Mathematics is case-sensitive. Is your $X^1=x^1?$ If so, please edit your post so they're all lower-case (the more usual case for position variables).
$endgroup$
– Adrian Keister
Jan 31 at 20:50
$begingroup$
Nope! As sated above in parenthesis, the capital ones are Eulerian coordinates while the lowercase are Lagrangian.
$endgroup$
– sadsloth_96
Jan 31 at 20:51
add a comment |
$begingroup$
Consider material traveling in a two-dimensional space with its velocity at any given location in space (capital letters correspond to Eulerian, lower case to Lagrangian) given by $overrightarrow{V}(overrightarrow{X},t) =begin{pmatrix}X^1+t\(X^2)^2 end{pmatrix}$. (Vectors are written using Cartesian coordinates,$ overrightarrow{X}= overrightarrow{R}$,and $overrightarrow{x}=overrightarrow{r}$.)
Find the position of a parcel/particle that starts at the location$overrightarrow{x}=begin{pmatrix} x^1\x^2end{pmatrix}.$
I solved the system of ordinary differential equations $frac{d}{dt}(X^1) =X^1+t; frac{d}{dt}(X^2) = (X^2)^2$, but I'm having trouble using the initial conditions to plug stuff in? I feel like I need to find $overrightarrow{x}(overrightarrow{X},t)$ first, but I also have no clue how to do that. Do I need to convert from Eulerian to Lagrangian and then plug in the initial conditions? Given what I have, I'm not really sure how to do that? Thanks for any help!
Here is what I have:
$X^1=Ke^t-t-1$
$X^2=-frac{1}{t+C}$
classical-mechanics
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
$endgroup$
Consider material traveling in a two-dimensional space with its velocity at any given location in space (capital letters correspond to Eulerian, lower case to Lagrangian) given by $overrightarrow{V}(overrightarrow{X},t) =begin{pmatrix}X^1+t\(X^2)^2 end{pmatrix}$. (Vectors are written using Cartesian coordinates,$ overrightarrow{X}= overrightarrow{R}$,and $overrightarrow{x}=overrightarrow{r}$.)
Find the position of a parcel/particle that starts at the location$overrightarrow{x}=begin{pmatrix} x^1\x^2end{pmatrix}.$
I solved the system of ordinary differential equations $frac{d}{dt}(X^1) =X^1+t; frac{d}{dt}(X^2) = (X^2)^2$, but I'm having trouble using the initial conditions to plug stuff in? I feel like I need to find $overrightarrow{x}(overrightarrow{X},t)$ first, but I also have no clue how to do that. Do I need to convert from Eulerian to Lagrangian and then plug in the initial conditions? Given what I have, I'm not really sure how to do that? Thanks for any help!
Here is what I have:
$X^1=Ke^t-t-1$
$X^2=-frac{1}{t+C}$
classical-mechanics
classical-mechanics
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
edited Jan 31 at 21:04
sadsloth_96
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
asked Jan 31 at 19:28
sadsloth_96sadsloth_96
215
215
$begingroup$
Mathematics is case-sensitive. Is your $X^1=x^1?$ If so, please edit your post so they're all lower-case (the more usual case for position variables).
$endgroup$
– Adrian Keister
Jan 31 at 20:50
$begingroup$
Nope! As sated above in parenthesis, the capital ones are Eulerian coordinates while the lowercase are Lagrangian.
$endgroup$
– sadsloth_96
Jan 31 at 20:51
add a comment |
$begingroup$
Mathematics is case-sensitive. Is your $X^1=x^1?$ If so, please edit your post so they're all lower-case (the more usual case for position variables).
$endgroup$
– Adrian Keister
Jan 31 at 20:50
$begingroup$
Nope! As sated above in parenthesis, the capital ones are Eulerian coordinates while the lowercase are Lagrangian.
$endgroup$
– sadsloth_96
Jan 31 at 20:51
$begingroup$
Mathematics is case-sensitive. Is your $X^1=x^1?$ If so, please edit your post so they're all lower-case (the more usual case for position variables).
$endgroup$
– Adrian Keister
Jan 31 at 20:50
$begingroup$
Mathematics is case-sensitive. Is your $X^1=x^1?$ If so, please edit your post so they're all lower-case (the more usual case for position variables).
$endgroup$
– Adrian Keister
Jan 31 at 20:50
$begingroup$
Nope! As sated above in parenthesis, the capital ones are Eulerian coordinates while the lowercase are Lagrangian.
$endgroup$
– sadsloth_96
Jan 31 at 20:51
$begingroup$
Nope! As sated above in parenthesis, the capital ones are Eulerian coordinates while the lowercase are Lagrangian.
$endgroup$
– sadsloth_96
Jan 31 at 20:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think I figured it out!
Since we are given that the initial coordinates are $overrightarrow{x}$, we know that $X^1(0)=x^1$ and $X^2(0)=x^2$. So, we can solve for the constants in terms of $x^1$ and $x^2$. So, the end product looks like this:
$X^1=(x^1+1)e^t-t+1$
$X^2=frac{-1}{t-frac{1}{x^2}}$
(Note that $x^1, x^2, X^1, X^2$ are not powers; they are like x and y)
Then, finding $overrightarrow{x}(overrightarrow{X},t)$ is pretty simple!
Long story short, I made things way too hard.
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
$endgroup$
add a comment |
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
});
}
});
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%2f3095347%2feulerian-to-lagrangian-coordinates-help-how-to-find-constants-theoretical-mec%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think I figured it out!
Since we are given that the initial coordinates are $overrightarrow{x}$, we know that $X^1(0)=x^1$ and $X^2(0)=x^2$. So, we can solve for the constants in terms of $x^1$ and $x^2$. So, the end product looks like this:
$X^1=(x^1+1)e^t-t+1$
$X^2=frac{-1}{t-frac{1}{x^2}}$
(Note that $x^1, x^2, X^1, X^2$ are not powers; they are like x and y)
Then, finding $overrightarrow{x}(overrightarrow{X},t)$ is pretty simple!
Long story short, I made things way too hard.
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
$endgroup$
add a comment |
$begingroup$
I think I figured it out!
Since we are given that the initial coordinates are $overrightarrow{x}$, we know that $X^1(0)=x^1$ and $X^2(0)=x^2$. So, we can solve for the constants in terms of $x^1$ and $x^2$. So, the end product looks like this:
$X^1=(x^1+1)e^t-t+1$
$X^2=frac{-1}{t-frac{1}{x^2}}$
(Note that $x^1, x^2, X^1, X^2$ are not powers; they are like x and y)
Then, finding $overrightarrow{x}(overrightarrow{X},t)$ is pretty simple!
Long story short, I made things way too hard.
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
$endgroup$
add a comment |
$begingroup$
I think I figured it out!
Since we are given that the initial coordinates are $overrightarrow{x}$, we know that $X^1(0)=x^1$ and $X^2(0)=x^2$. So, we can solve for the constants in terms of $x^1$ and $x^2$. So, the end product looks like this:
$X^1=(x^1+1)e^t-t+1$
$X^2=frac{-1}{t-frac{1}{x^2}}$
(Note that $x^1, x^2, X^1, X^2$ are not powers; they are like x and y)
Then, finding $overrightarrow{x}(overrightarrow{X},t)$ is pretty simple!
Long story short, I made things way too hard.
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
$endgroup$
I think I figured it out!
Since we are given that the initial coordinates are $overrightarrow{x}$, we know that $X^1(0)=x^1$ and $X^2(0)=x^2$. So, we can solve for the constants in terms of $x^1$ and $x^2$. So, the end product looks like this:
$X^1=(x^1+1)e^t-t+1$
$X^2=frac{-1}{t-frac{1}{x^2}}$
(Note that $x^1, x^2, X^1, X^2$ are not powers; they are like x and y)
Then, finding $overrightarrow{x}(overrightarrow{X},t)$ is pretty simple!
Long story short, I made things way too hard.
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
answered Jan 31 at 21:00
sadsloth_96sadsloth_96
215
215
add a comment |
add a comment |
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%2f3095347%2feulerian-to-lagrangian-coordinates-help-how-to-find-constants-theoretical-mec%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
$begingroup$
Mathematics is case-sensitive. Is your $X^1=x^1?$ If so, please edit your post so they're all lower-case (the more usual case for position variables).
$endgroup$
– Adrian Keister
Jan 31 at 20:50
$begingroup$
Nope! As sated above in parenthesis, the capital ones are Eulerian coordinates while the lowercase are Lagrangian.
$endgroup$
– sadsloth_96
Jan 31 at 20:51