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

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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

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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 | 

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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