Mixing
Second variation and Legendre's condition for extremal path
$begingroup$
The form of the second variation of a functional $mathcal I [y] $ has been given as $$ begin{aligned} delta ^2 mathcal I &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} eta ^2 dfrac{partial ^2 F}{partial y ^2} + 2 eta eta' dfrac{partial ^2 F}{partial y partial y'} + eta'^2 dfrac{partial ^2 F}{partial y'^2} ; text d x\ \ &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} dfrac{partial^2 F}{partial y'^2} eta'^2 + left[ dfrac{partial ^2 F}{partial y^2} - dfrac{text d}{text d x} left( dfrac{partial ^2 F}{partial y partial y'} right)right] eta^2 ; text d x\ \ &stackrel{text{def}}{=} displaystyle int _{x_1}^{x_2} P eta'^2 + Q eta ^2 ; text d x .end{aligned}$$ where we have employed integration by parts on the "cross" term to obtain the second line. Legendre's condition states that if a curve $ y_0(x) $ is a minimizer, it must necessarily be the case that the second partial derivative of $ F $ w.r.t. $y'$ be negative at all points of $y_0(x)$. Seeking a contradiction, we suppose that this is not the case. Then by continuity, for some $ epsilon > 0$, $x_0 in (x_1,x_2) $, there exists an interval $ [x_0 - epsilon, x_0 + epsilon] $ over which $P$ is less than $ - alpha $, where $alpha > 0 $. In class, we then proceeded to define the function $$ eta(x) = begin{cases} sin^2 left[ dfrac{ pi ( x - x_0)}{epsilon} right] & text{for} ; x in [x_0 - epsilon, x_0 + epsilon] \ 0 & text{otherwise} end{cases}$$ and then substitute this into the initial expression for the second variation, resulting in $$ delta ^2 mathcal I = displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} underbrace{P dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right]}_{(spadesuit)} + underbrace{Q sin ^4 left[ dfrac{ pi (x - x_0)}{epsilon} right]}_{(clubsuit)} ; text d x. $$ From the initial conditions on $P$, we can bound the first integral by $$ ( spadesuit ) < displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} - alpha dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right] = - alpha dfrac{pi ^2}{epsilon}. $$ The next step is the one I am unclear on. We then proceeded to bound the second integral by using the fact that $ sin ^4 (x) < 1 $ and defining $ M = displaystyle max_{x in (x_1, x_2)} | Q | $ so we can say $ (clubsuit ) < 2M epsilon $. My professor then said that by taking the sum of these, we can show that for sufficiently small $epsilon$, it follows that $ delta ^2 mathcal I < 0 $ but I don't see how we can say that $ Q $ has a finite bound without some additional restrictions?
optimization proof-explanation calculus-of-variations
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
$endgroup$
add a comment |
$begingroup$
The form of the second variation of a functional $mathcal I [y] $ has been given as $$ begin{aligned} delta ^2 mathcal I &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} eta ^2 dfrac{partial ^2 F}{partial y ^2} + 2 eta eta' dfrac{partial ^2 F}{partial y partial y'} + eta'^2 dfrac{partial ^2 F}{partial y'^2} ; text d x\ \ &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} dfrac{partial^2 F}{partial y'^2} eta'^2 + left[ dfrac{partial ^2 F}{partial y^2} - dfrac{text d}{text d x} left( dfrac{partial ^2 F}{partial y partial y'} right)right] eta^2 ; text d x\ \ &stackrel{text{def}}{=} displaystyle int _{x_1}^{x_2} P eta'^2 + Q eta ^2 ; text d x .end{aligned}$$ where we have employed integration by parts on the "cross" term to obtain the second line. Legendre's condition states that if a curve $ y_0(x) $ is a minimizer, it must necessarily be the case that the second partial derivative of $ F $ w.r.t. $y'$ be negative at all points of $y_0(x)$. Seeking a contradiction, we suppose that this is not the case. Then by continuity, for some $ epsilon > 0$, $x_0 in (x_1,x_2) $, there exists an interval $ [x_0 - epsilon, x_0 + epsilon] $ over which $P$ is less than $ - alpha $, where $alpha > 0 $. In class, we then proceeded to define the function $$ eta(x) = begin{cases} sin^2 left[ dfrac{ pi ( x - x_0)}{epsilon} right] & text{for} ; x in [x_0 - epsilon, x_0 + epsilon] \ 0 & text{otherwise} end{cases}$$ and then substitute this into the initial expression for the second variation, resulting in $$ delta ^2 mathcal I = displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} underbrace{P dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right]}_{(spadesuit)} + underbrace{Q sin ^4 left[ dfrac{ pi (x - x_0)}{epsilon} right]}_{(clubsuit)} ; text d x. $$ From the initial conditions on $P$, we can bound the first integral by $$ ( spadesuit ) < displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} - alpha dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right] = - alpha dfrac{pi ^2}{epsilon}. $$ The next step is the one I am unclear on. We then proceeded to bound the second integral by using the fact that $ sin ^4 (x) < 1 $ and defining $ M = displaystyle max_{x in (x_1, x_2)} | Q | $ so we can say $ (clubsuit ) < 2M epsilon $. My professor then said that by taking the sum of these, we can show that for sufficiently small $epsilon$, it follows that $ delta ^2 mathcal I < 0 $ but I don't see how we can say that $ Q $ has a finite bound without some additional restrictions?
optimization proof-explanation calculus-of-variations
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
$endgroup$
add a comment |
$begingroup$
The form of the second variation of a functional $mathcal I [y] $ has been given as $$ begin{aligned} delta ^2 mathcal I &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} eta ^2 dfrac{partial ^2 F}{partial y ^2} + 2 eta eta' dfrac{partial ^2 F}{partial y partial y'} + eta'^2 dfrac{partial ^2 F}{partial y'^2} ; text d x\ \ &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} dfrac{partial^2 F}{partial y'^2} eta'^2 + left[ dfrac{partial ^2 F}{partial y^2} - dfrac{text d}{text d x} left( dfrac{partial ^2 F}{partial y partial y'} right)right] eta^2 ; text d x\ \ &stackrel{text{def}}{=} displaystyle int _{x_1}^{x_2} P eta'^2 + Q eta ^2 ; text d x .end{aligned}$$ where we have employed integration by parts on the "cross" term to obtain the second line. Legendre's condition states that if a curve $ y_0(x) $ is a minimizer, it must necessarily be the case that the second partial derivative of $ F $ w.r.t. $y'$ be negative at all points of $y_0(x)$. Seeking a contradiction, we suppose that this is not the case. Then by continuity, for some $ epsilon > 0$, $x_0 in (x_1,x_2) $, there exists an interval $ [x_0 - epsilon, x_0 + epsilon] $ over which $P$ is less than $ - alpha $, where $alpha > 0 $. In class, we then proceeded to define the function $$ eta(x) = begin{cases} sin^2 left[ dfrac{ pi ( x - x_0)}{epsilon} right] & text{for} ; x in [x_0 - epsilon, x_0 + epsilon] \ 0 & text{otherwise} end{cases}$$ and then substitute this into the initial expression for the second variation, resulting in $$ delta ^2 mathcal I = displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} underbrace{P dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right]}_{(spadesuit)} + underbrace{Q sin ^4 left[ dfrac{ pi (x - x_0)}{epsilon} right]}_{(clubsuit)} ; text d x. $$ From the initial conditions on $P$, we can bound the first integral by $$ ( spadesuit ) < displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} - alpha dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right] = - alpha dfrac{pi ^2}{epsilon}. $$ The next step is the one I am unclear on. We then proceeded to bound the second integral by using the fact that $ sin ^4 (x) < 1 $ and defining $ M = displaystyle max_{x in (x_1, x_2)} | Q | $ so we can say $ (clubsuit ) < 2M epsilon $. My professor then said that by taking the sum of these, we can show that for sufficiently small $epsilon$, it follows that $ delta ^2 mathcal I < 0 $ but I don't see how we can say that $ Q $ has a finite bound without some additional restrictions?
optimization proof-explanation calculus-of-variations
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
$endgroup$
The form of the second variation of a functional $mathcal I [y] $ has been given as $$ begin{aligned} delta ^2 mathcal I &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} eta ^2 dfrac{partial ^2 F}{partial y ^2} + 2 eta eta' dfrac{partial ^2 F}{partial y partial y'} + eta'^2 dfrac{partial ^2 F}{partial y'^2} ; text d x\ \ &= dfrac{1}{2} displaystyle int _{x_1}^{x_2} dfrac{partial^2 F}{partial y'^2} eta'^2 + left[ dfrac{partial ^2 F}{partial y^2} - dfrac{text d}{text d x} left( dfrac{partial ^2 F}{partial y partial y'} right)right] eta^2 ; text d x\ \ &stackrel{text{def}}{=} displaystyle int _{x_1}^{x_2} P eta'^2 + Q eta ^2 ; text d x .end{aligned}$$ where we have employed integration by parts on the "cross" term to obtain the second line. Legendre's condition states that if a curve $ y_0(x) $ is a minimizer, it must necessarily be the case that the second partial derivative of $ F $ w.r.t. $y'$ be negative at all points of $y_0(x)$. Seeking a contradiction, we suppose that this is not the case. Then by continuity, for some $ epsilon > 0$, $x_0 in (x_1,x_2) $, there exists an interval $ [x_0 - epsilon, x_0 + epsilon] $ over which $P$ is less than $ - alpha $, where $alpha > 0 $. In class, we then proceeded to define the function $$ eta(x) = begin{cases} sin^2 left[ dfrac{ pi ( x - x_0)}{epsilon} right] & text{for} ; x in [x_0 - epsilon, x_0 + epsilon] \ 0 & text{otherwise} end{cases}$$ and then substitute this into the initial expression for the second variation, resulting in $$ delta ^2 mathcal I = displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} underbrace{P dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right]}_{(spadesuit)} + underbrace{Q sin ^4 left[ dfrac{ pi (x - x_0)}{epsilon} right]}_{(clubsuit)} ; text d x. $$ From the initial conditions on $P$, we can bound the first integral by $$ ( spadesuit ) < displaystyleint_{x_0 - epsilon}^{x_0 + epsilon} - alpha dfrac{pi ^2}{epsilon^2} sin^2 left[ dfrac{2 pi (x - x_0)}{epsilon} right] = - alpha dfrac{pi ^2}{epsilon}. $$ The next step is the one I am unclear on. We then proceeded to bound the second integral by using the fact that $ sin ^4 (x) < 1 $ and defining $ M = displaystyle max_{x in (x_1, x_2)} | Q | $ so we can say $ (clubsuit ) < 2M epsilon $. My professor then said that by taking the sum of these, we can show that for sufficiently small $epsilon$, it follows that $ delta ^2 mathcal I < 0 $ but I don't see how we can say that $ Q $ has a finite bound without some additional restrictions?
optimization proof-explanation calculus-of-variations
optimization proof-explanation calculus-of-variations
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
edited Jan 21 at 21:01
backstrapp
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
asked Mar 14 '18 at 22:53
backstrappbackstrapp
12819
12819
add a comment |
add a comment |
0
active
oldest
votes
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%2f2691545%2fsecond-variation-and-legendres-condition-for-extremal-path%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2691545%2fsecond-variation-and-legendres-condition-for-extremal-path%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
