Obtain first two terms in the asymptotic approximation $epsilon x^3+x^2+2x-3=0$
$begingroup$
Consider the singularly perturbed cubic equation
$$epsilon x^3+x^2+2x-3=0$$
where $epsilon >0$ is a small parameter. This equation has two regularly perturbed roots and one singularly perturbed root.
i. Obtain the first two terms in the asymptotic approximation of the regularly perturbed roots as $epsilon to 0$
ii. Use the substitution $x(epsilon) = frac{z(epsilon)}{epsilon}$ to obtain the first two terms of an asymptotic approximation of the singularyl perturbed solution of this equation as $epsilon to 0$
My solutions:
i. Look for the solution of the form
$$x(epsilon) = x_0 + epsilon x_1 + epsilon^2 x_2 +...$$
formally sub into equation
$$epsilon(x_0 + epsilon x_1 + epsilon^2 x_2 +...)^3 + (x_0 + epsilon x_1 + epsilon^2 x_2 +...)^2 + 2(x_0 + epsilon x_1 + epsilon^2 x_2 +...) - 3 = 0$$
then we have
$$epsilon(x^3_0+epsilon(3x^2_0x_1) + epsilon^2(3x_0x_1^2+3x_0^2x_2)+...)+(x^2_0+epsilon(2x_0x_1)+epsilon^2(x_1^2+2x_0x_2)+...)+2(x_0+epsilon x_1+epsilon^2x_2+...)-3=0$$
comparing coefficients of powers of $epsilon$
$O(epsilon^0) = x^2_0+2x_0-3=0$
$ implies x_0=3,1$
$O(epsilon^1) = x_0+3x^2_0x_1+2x_0x_1+2x_1=0$
$ implies $ if $x_0=3 space$then $x_1=frac{-3}{35}$ and if $x_0=-1$ then $x_1 = frac{1}{3}$
That is my current answer for i.
I have begun to answer part ii using the change of variable given, however im not sure if I am along the right lines by answering it in the same manner as part i. TIA
asymptotics perturbation-theory
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
$endgroup$
add a comment |
$begingroup$
Consider the singularly perturbed cubic equation
$$epsilon x^3+x^2+2x-3=0$$
where $epsilon >0$ is a small parameter. This equation has two regularly perturbed roots and one singularly perturbed root.
i. Obtain the first two terms in the asymptotic approximation of the regularly perturbed roots as $epsilon to 0$
ii. Use the substitution $x(epsilon) = frac{z(epsilon)}{epsilon}$ to obtain the first two terms of an asymptotic approximation of the singularyl perturbed solution of this equation as $epsilon to 0$
My solutions:
i. Look for the solution of the form
$$x(epsilon) = x_0 + epsilon x_1 + epsilon^2 x_2 +...$$
formally sub into equation
$$epsilon(x_0 + epsilon x_1 + epsilon^2 x_2 +...)^3 + (x_0 + epsilon x_1 + epsilon^2 x_2 +...)^2 + 2(x_0 + epsilon x_1 + epsilon^2 x_2 +...) - 3 = 0$$
then we have
$$epsilon(x^3_0+epsilon(3x^2_0x_1) + epsilon^2(3x_0x_1^2+3x_0^2x_2)+...)+(x^2_0+epsilon(2x_0x_1)+epsilon^2(x_1^2+2x_0x_2)+...)+2(x_0+epsilon x_1+epsilon^2x_2+...)-3=0$$
comparing coefficients of powers of $epsilon$
$O(epsilon^0) = x^2_0+2x_0-3=0$
$ implies x_0=3,1$
$O(epsilon^1) = x_0+3x^2_0x_1+2x_0x_1+2x_1=0$
$ implies $ if $x_0=3 space$then $x_1=frac{-3}{35}$ and if $x_0=-1$ then $x_1 = frac{1}{3}$
That is my current answer for i.
I have begun to answer part ii using the change of variable given, however im not sure if I am along the right lines by answering it in the same manner as part i. TIA
asymptotics perturbation-theory
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
$endgroup$
$begingroup$
Your $O(epsilon)$ equation is incorrect, you need the $O(1)$ terms of $x^3$, which is just $x_0^3$, and the $O(epsilon)$ terms from $x^2$ and $x$, so you should have $x_0^3+2x_0x_1+2x_1=0$. After you substitute for $z$ in ii., the process is the same as in i.
$endgroup$
– David
Dec 13 '18 at 19:10
add a comment |
$begingroup$
Consider the singularly perturbed cubic equation
$$epsilon x^3+x^2+2x-3=0$$
where $epsilon >0$ is a small parameter. This equation has two regularly perturbed roots and one singularly perturbed root.
i. Obtain the first two terms in the asymptotic approximation of the regularly perturbed roots as $epsilon to 0$
ii. Use the substitution $x(epsilon) = frac{z(epsilon)}{epsilon}$ to obtain the first two terms of an asymptotic approximation of the singularyl perturbed solution of this equation as $epsilon to 0$
My solutions:
i. Look for the solution of the form
$$x(epsilon) = x_0 + epsilon x_1 + epsilon^2 x_2 +...$$
formally sub into equation
$$epsilon(x_0 + epsilon x_1 + epsilon^2 x_2 +...)^3 + (x_0 + epsilon x_1 + epsilon^2 x_2 +...)^2 + 2(x_0 + epsilon x_1 + epsilon^2 x_2 +...) - 3 = 0$$
then we have
$$epsilon(x^3_0+epsilon(3x^2_0x_1) + epsilon^2(3x_0x_1^2+3x_0^2x_2)+...)+(x^2_0+epsilon(2x_0x_1)+epsilon^2(x_1^2+2x_0x_2)+...)+2(x_0+epsilon x_1+epsilon^2x_2+...)-3=0$$
comparing coefficients of powers of $epsilon$
$O(epsilon^0) = x^2_0+2x_0-3=0$
$ implies x_0=3,1$
$O(epsilon^1) = x_0+3x^2_0x_1+2x_0x_1+2x_1=0$
$ implies $ if $x_0=3 space$then $x_1=frac{-3}{35}$ and if $x_0=-1$ then $x_1 = frac{1}{3}$
That is my current answer for i.
I have begun to answer part ii using the change of variable given, however im not sure if I am along the right lines by answering it in the same manner as part i. TIA
asymptotics perturbation-theory
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
$endgroup$
Consider the singularly perturbed cubic equation
$$epsilon x^3+x^2+2x-3=0$$
where $epsilon >0$ is a small parameter. This equation has two regularly perturbed roots and one singularly perturbed root.
i. Obtain the first two terms in the asymptotic approximation of the regularly perturbed roots as $epsilon to 0$
ii. Use the substitution $x(epsilon) = frac{z(epsilon)}{epsilon}$ to obtain the first two terms of an asymptotic approximation of the singularyl perturbed solution of this equation as $epsilon to 0$
My solutions:
i. Look for the solution of the form
$$x(epsilon) = x_0 + epsilon x_1 + epsilon^2 x_2 +...$$
formally sub into equation
$$epsilon(x_0 + epsilon x_1 + epsilon^2 x_2 +...)^3 + (x_0 + epsilon x_1 + epsilon^2 x_2 +...)^2 + 2(x_0 + epsilon x_1 + epsilon^2 x_2 +...) - 3 = 0$$
then we have
$$epsilon(x^3_0+epsilon(3x^2_0x_1) + epsilon^2(3x_0x_1^2+3x_0^2x_2)+...)+(x^2_0+epsilon(2x_0x_1)+epsilon^2(x_1^2+2x_0x_2)+...)+2(x_0+epsilon x_1+epsilon^2x_2+...)-3=0$$
comparing coefficients of powers of $epsilon$
$O(epsilon^0) = x^2_0+2x_0-3=0$
$ implies x_0=3,1$
$O(epsilon^1) = x_0+3x^2_0x_1+2x_0x_1+2x_1=0$
$ implies $ if $x_0=3 space$then $x_1=frac{-3}{35}$ and if $x_0=-1$ then $x_1 = frac{1}{3}$
That is my current answer for i.
I have begun to answer part ii using the change of variable given, however im not sure if I am along the right lines by answering it in the same manner as part i. TIA
asymptotics perturbation-theory
asymptotics perturbation-theory
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
asked Dec 13 '18 at 12:32
Ben JonesBen Jones
19111
19111
$begingroup$
Your $O(epsilon)$ equation is incorrect, you need the $O(1)$ terms of $x^3$, which is just $x_0^3$, and the $O(epsilon)$ terms from $x^2$ and $x$, so you should have $x_0^3+2x_0x_1+2x_1=0$. After you substitute for $z$ in ii., the process is the same as in i.
$endgroup$
– David
Dec 13 '18 at 19:10
add a comment |
$begingroup$
Your $O(epsilon)$ equation is incorrect, you need the $O(1)$ terms of $x^3$, which is just $x_0^3$, and the $O(epsilon)$ terms from $x^2$ and $x$, so you should have $x_0^3+2x_0x_1+2x_1=0$. After you substitute for $z$ in ii., the process is the same as in i.
$endgroup$
– David
Dec 13 '18 at 19:10
$begingroup$
Your $O(epsilon)$ equation is incorrect, you need the $O(1)$ terms of $x^3$, which is just $x_0^3$, and the $O(epsilon)$ terms from $x^2$ and $x$, so you should have $x_0^3+2x_0x_1+2x_1=0$. After you substitute for $z$ in ii., the process is the same as in i.
$endgroup$
– David
Dec 13 '18 at 19:10
$begingroup$
Your $O(epsilon)$ equation is incorrect, you need the $O(1)$ terms of $x^3$, which is just $x_0^3$, and the $O(epsilon)$ terms from $x^2$ and $x$, so you should have $x_0^3+2x_0x_1+2x_1=0$. After you substitute for $z$ in ii., the process is the same as in i.
$endgroup$
– David
Dec 13 '18 at 19:10
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%2f3037946%2fobtain-first-two-terms-in-the-asymptotic-approximation-epsilon-x3x22x-3-0%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
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%2f3037946%2fobtain-first-two-terms-in-the-asymptotic-approximation-epsilon-x3x22x-3-0%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
$begingroup$
Your $O(epsilon)$ equation is incorrect, you need the $O(1)$ terms of $x^3$, which is just $x_0^3$, and the $O(epsilon)$ terms from $x^2$ and $x$, so you should have $x_0^3+2x_0x_1+2x_1=0$. After you substitute for $z$ in ii., the process is the same as in i.
$endgroup$
– David
Dec 13 '18 at 19:10