Stability of solutions to $y''+4=0$












0












$begingroup$


Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.



Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.



Here is my attempt at showing the $phi(t)$ is a stable solution.



Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.



This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.



A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
    $endgroup$
    – user539887
    Dec 8 '18 at 12:55
















0












$begingroup$


Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.



Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.



Here is my attempt at showing the $phi(t)$ is a stable solution.



Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.



This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.



A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
    $endgroup$
    – user539887
    Dec 8 '18 at 12:55














0












0








0





$begingroup$


Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.



Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.



Here is my attempt at showing the $phi(t)$ is a stable solution.



Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.



This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.



A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$










share|cite|improve this question











$endgroup$




Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.



Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.



Here is my attempt at showing the $phi(t)$ is a stable solution.



Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.



This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.



A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$







ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 0:59







Jungleshrimp

















asked Dec 8 '18 at 0:53









JungleshrimpJungleshrimp

314111




314111








  • 2




    $begingroup$
    In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
    $endgroup$
    – user539887
    Dec 8 '18 at 12:55














  • 2




    $begingroup$
    In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
    $endgroup$
    – user539887
    Dec 8 '18 at 12:55








2




2




$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55




$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55










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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030574%2fstability-of-solutions-to-y4-0%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030574%2fstability-of-solutions-to-y4-0%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Plaza Victoria

Puebla de Zaragoza

Musa