Mrthfx

NIntegrate on a solution of a matrix ODE

Searching extreme points of polyhedron

Why is there so little support for joining EFTA in the British parliament?

Dinosaur Word Search, Letter Solve, and Unscramble

Getting representations of the Lie group out of representations of its Lie algebra

What did Turing mean when saying that "machines cannot give rise to surprises" is due to a fallacy?

Inverse square law not accurate for non-point masses?

"Destructive power" carried by a B-52?

Does a random sequence of vectors span a Hilbert space?

French equivalents of おしゃれは足元から (Every good outfit starts with the shoes)

How to make triangles with rounded sides and corners? (squircle with 3 sides)

IC on Digikey is 5x more expensive than board containing same IC on Alibaba: How?

How do you cope with tons of web fonts when copying and pasting from web pages?

Should man-made satellites feature an intelligent inverted "cow catcher"?

malloc in main() or malloc in another function: allocating memory for a struct and its members

How could a hydrazine and N2O4 cloud (or it's reactants) show up in weather radar?

Does the universe have a fixed centre of mass?

calculator's angle answer for trig ratios that can work in more than 1 quadrant on the unit circle

Is there a verb for listening stealthily?

Is there night in Alpha Complex?

New Order #6: Easter Egg

Where did Ptolemy compare the Earth to the distance of fixed stars?

The test team as an enemy of development? And how can this be avoided?

Does the main washing effect of soap come from foam?

Proving that any solution to the differential equation of an oscillator can be written as a sum of sinusoids.

2

$begingroup$

Suppose you have a differential equation with n distinct functions of $t$ where

$frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$

.

.

.

$frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$

I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $

can be written as a linear combination of solutions of the form $(e^{iw_1t},...,e^{iw_1t}), (e^{iw_2t},...e^{iw_2t}), ...,(e^{iw_mt},...e^{iw_mt})$ where each $w_j$ is a real number.

i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.

linear-algebra ordinary-differential-equations physics

share|cite|improve this question

asked 4 hours ago

user446153user446153

1075

$endgroup$

add a comment | 

2

$begingroup$

Suppose you have a differential equation with n distinct functions of $t$ where

$frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$

.

.

.

$frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$

I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $

can be written as a linear combination of solutions of the form $(e^{iw_1t},...,e^{iw_1t}), (e^{iw_2t},...e^{iw_2t}), ...,(e^{iw_mt},...e^{iw_mt})$ where each $w_j$ is a real number.

i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.

linear-algebra ordinary-differential-equations physics

share|cite|improve this question

asked 4 hours ago

user446153user446153

1075

$endgroup$

add a comment | 

$begingroup$

Suppose you have a differential equation with n distinct functions of $t$ where

$frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$

.

.

.

$frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$

I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $

can be written as a linear combination of solutions of the form $(e^{iw_1t},...,e^{iw_1t}), (e^{iw_2t},...e^{iw_2t}), ...,(e^{iw_mt},...e^{iw_mt})$ where each $w_j$ is a real number.

i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.

linear-algebra ordinary-differential-equations physics

share|cite|improve this question

asked 4 hours ago

user446153user446153

1075

$endgroup$

share|cite|improve this question

asked 4 hours ago

user446153user446153

1075

share|cite|improve this question

asked 4 hours ago

user446153user446153

1075

asked 4 hours ago

user446153user446153

1075

asked 4 hours ago

user446153user446153

1075

1 Answer
1

active

oldest

votes

3

$begingroup$

The system of differential equations you wrote could be written as,

$$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$

$$ frac{d^2}{dt^2} vec{x} = K vec{x}$$

The matrix $K=[k_{ij}]$ acts on the indices of the functions $x_1,dots,x_n$.

We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.

Let $Lambda$ be the diagonal form of $K$.

Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^{-1} K V$.

We can now write the system of differential equations as,

$$
frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
$$

$$
V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
$$

$$
frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
$$

Let $vec{y} = V^{-1} vec{x}$, then we have $
frac{d^2}{dt^2} vec{y} = Lambda vec{y}
$. This corresponds to the following system of equations.

$$
frac{d^2 y_1}{dt^2} = lambda_1 y_1
$$
$$
frac{d^2 y_2}{dt^2} = lambda_2 y_2
$$
$$
vdots
$$
$$
frac{d^2 y_n}{dt^2} = lambda_n y_n
$$

Clearly the solutions are of the form,

$$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$

to obtain the $x_j$'s we just multiply by the $V$ matrix.

$$ x_j(t) = sum_i V_{ji} y_i(t)$$

Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.

share|cite|improve this answer

answered 3 hours ago

SpencerSpencer

8,76812156

$endgroup$

add a comment | 

Your Answer

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

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196600%2fproving-that-any-solution-to-the-differential-equation-of-an-oscillator-can-be-w%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

The system of differential equations you wrote could be written as,

$$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$

$$ frac{d^2}{dt^2} vec{x} = K vec{x}$$

The matrix $K=[k_{ij}]$ acts on the indices of the functions $x_1,dots,x_n$.

We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.

Let $Lambda$ be the diagonal form of $K$.

Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^{-1} K V$.

We can now write the system of differential equations as,

$$
frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
$$

$$
V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
$$

$$
frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
$$

Let $vec{y} = V^{-1} vec{x}$, then we have $
frac{d^2}{dt^2} vec{y} = Lambda vec{y}
$. This corresponds to the following system of equations.

$$
frac{d^2 y_1}{dt^2} = lambda_1 y_1
$$
$$
frac{d^2 y_2}{dt^2} = lambda_2 y_2
$$
$$
vdots
$$
$$
frac{d^2 y_n}{dt^2} = lambda_n y_n
$$

Clearly the solutions are of the form,

$$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$

to obtain the $x_j$'s we just multiply by the $V$ matrix.

$$ x_j(t) = sum_i V_{ji} y_i(t)$$

Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.

share|cite|improve this answer

answered 3 hours ago

SpencerSpencer

8,76812156

$endgroup$

add a comment | 

3

$begingroup$

The system of differential equations you wrote could be written as,

$$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$

$$ frac{d^2}{dt^2} vec{x} = K vec{x}$$

The matrix $K=[k_{ij}]$ acts on the indices of the functions $x_1,dots,x_n$.

We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.

Let $Lambda$ be the diagonal form of $K$.

Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^{-1} K V$.

We can now write the system of differential equations as,

$$
frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
$$

$$
V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
$$

$$
frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
$$

Let $vec{y} = V^{-1} vec{x}$, then we have $
frac{d^2}{dt^2} vec{y} = Lambda vec{y}
$. This corresponds to the following system of equations.

$$
frac{d^2 y_1}{dt^2} = lambda_1 y_1
$$
$$
frac{d^2 y_2}{dt^2} = lambda_2 y_2
$$
$$
vdots
$$
$$
frac{d^2 y_n}{dt^2} = lambda_n y_n
$$

Clearly the solutions are of the form,

$$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$

to obtain the $x_j$'s we just multiply by the $V$ matrix.

$$ x_j(t) = sum_i V_{ji} y_i(t)$$

Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.

share|cite|improve this answer

answered 3 hours ago

SpencerSpencer

8,76812156

$endgroup$

add a comment | 

$begingroup$

The system of differential equations you wrote could be written as,

$$ frac{d^2}{dt^2} left[begin{array}{c} x_1 \ vdots \ x_n end{array}right] = left[begin{array}{ccc} k_{11} & cdots & k_{1n} \ vdots & ddots & vdots \ k_{1n} & cdots & k_{nn} end{array}right] left[ begin{array}{c} x_1 \ vdots \ x_nend{array}right]$$

$$ frac{d^2}{dt^2} vec{x} = K vec{x}$$

The matrix $K=[k_{ij}]$ acts on the indices of the functions $x_1,dots,x_n$.

We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.

Let $Lambda$ be the diagonal form of $K$.

Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^{-1} K V$.

We can now write the system of differential equations as,

$$
frac{d^2}{dt^2} vec{x} = V Lambda V^{-1} vec{x}
$$

$$
V^{-1}frac{d^2}{dt^2} vec{x} = Lambda V^{-1} vec{x}
$$

$$
frac{d^2}{dt^2} V^{-1}vec{x} = Lambda V^{-1} vec{x}
$$

Let $vec{y} = V^{-1} vec{x}$, then we have $
frac{d^2}{dt^2} vec{y} = Lambda vec{y}
$. This corresponds to the following system of equations.

$$
frac{d^2 y_1}{dt^2} = lambda_1 y_1
$$
$$
frac{d^2 y_2}{dt^2} = lambda_2 y_2
$$
$$
vdots
$$
$$
frac{d^2 y_n}{dt^2} = lambda_n y_n
$$

Clearly the solutions are of the form,

$$y_j(t) = C_1 e^{sqrt{lambda_j} t} + C_2 e^{-sqrt{lambda_j} t},$$

to obtain the $x_j$'s we just multiply by the $V$ matrix.

$$ x_j(t) = sum_i V_{ji} y_i(t)$$

Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.

share|cite|improve this answer

answered 3 hours ago

SpencerSpencer

8,76812156

$endgroup$

share|cite|improve this answer

answered 3 hours ago

SpencerSpencer

8,76812156

answered 3 hours ago

SpencerSpencer

8,76812156

answered 3 hours ago

SpencerSpencer

8,76812156