Local implicit function theorem and Global implicit function theorem
$begingroup$
Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?
implicit-function-theorem
asked Jan 3 '15 at 9:26
LCHLCH
381111
$endgroup$
add a comment |
$begingroup$
Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?
implicit-function-theorem
asked Jan 3 '15 at 9:26
LCHLCH
381111
$endgroup$
$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46
$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50
$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46
add a comment |
$begingroup$
Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?
implicit-function-theorem
asked Jan 3 '15 at 9:26
LCHLCH
381111
$endgroup$
Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,yinmathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $frac{partial f(x,y)}{partial y}>0$ for all $x,yinmathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $xinmathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $xinmathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?
implicit-function-theorem
implicit-function-theorem
asked Jan 3 '15 at 9:26
LCHLCH
381111
asked Jan 3 '15 at 9:26
LCHLCH
381111
edited Jan 3 '15 at 9:32
LCH
asked Jan 3 '15 at 9:26
LCHLCH
381111
asked Jan 3 '15 at 9:26
LCHLCH
381111
asked Jan 3 '15 at 9:26
LCHLCH
381111
381111
$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46
$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50
$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46
add a comment |
$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46
$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50
$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46
$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46
$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46
$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50
$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50
$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46
$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider
$$F(x,y) = - e^{-y} +x$$
Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
$endgroup$
add a comment |
$begingroup$
For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:
Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.
answered Apr 9 '18 at 6:10
RobertRobert
112
$endgroup$
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%2f1089431%2flocal-implicit-function-theorem-and-global-implicit-function-theorem%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
$begingroup$
I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider
$$F(x,y) = - e^{-y} +x$$
Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
$endgroup$
add a comment |
$begingroup$
I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider
$$F(x,y) = - e^{-y} +x$$
Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
$endgroup$
add a comment |
$begingroup$
I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider
$$F(x,y) = - e^{-y} +x$$
Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
$endgroup$
I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider
$$F(x,y) = - e^{-y} +x$$
Clearly, $frac{partial F}{partial y} = e^{-y}>0$ everywhere, but it is easy to see that $F$ is strictly negative for $xle0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
answered Jan 3 '15 at 15:12
ThomasThomas
16.9k21731
16.9k21731
add a comment |
add a comment |
$begingroup$
For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:
Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.
answered Apr 9 '18 at 6:10
RobertRobert
112
$endgroup$
add a comment |
$begingroup$
For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:
Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.
answered Apr 9 '18 at 6:10
RobertRobert
112
$endgroup$
add a comment |
$begingroup$
For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:
Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.
answered Apr 9 '18 at 6:10
RobertRobert
112
$endgroup$
For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely, $frac{partial f(x,y)}{partial y}geq d>0,$ for all $(x,y)inmathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:
Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.
answered Apr 9 '18 at 6:10
RobertRobert
112
answered Apr 9 '18 at 6:10
RobertRobert
112
answered Apr 9 '18 at 6:10
RobertRobert
112
answered Apr 9 '18 at 6:10
RobertRobert
112
112
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%2f1089431%2flocal-implicit-function-theorem-and-global-implicit-function-theorem%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
$begingroup$
maybe this is of interest for you: math.stackexchange.com/questions/678417/…
$endgroup$
– Thomas
Jan 3 '15 at 9:46
$begingroup$
Yes, that link helps much, thanks! I'm thinking what you said about using the mean value theorem to show that $varphi$ can be extended in $C^1$ to the boundary of any finite interval, hence to $mathbb{R}.$
$endgroup$
– LCH
Jan 3 '15 at 12:50
$begingroup$
Would you please give some hints?
$endgroup$
– LCH
Jan 3 '15 at 13:46