Proving that any solution to the differential equation of an oscillator can be written as a sum of sinusoids. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How can the general Green's function of a linear homogeneous differential equation be derived?Proving a system of n linear equations has only one solutionThe pair $x_1$ , $x_2$ are Linearly IndependentName/Solution of this Differential EquationEvery solution of some linear differential equation of order 2 is boundedSolution of Matrix differential equation $textbfX'(t)=textbfAtextbfX(t)$How to relate the solutions to a Fuchsian type differential equation to the solutions to the hypergeometric differential equation?Difference between real and complex solution in differential equationFinding the general solution to a system of differential equations using eigenvaluesTwo systems of linear equations equivalent
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Proving that any solution to the differential equation of an oscillator can be written as a sum of sinusoids.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How can the general Green's function of a linear homogeneous differential equation be derived?Proving a system of n linear equations has only one solutionThe pair $x_1$ , $x_2$ are Linearly IndependentName/Solution of this Differential EquationEvery solution of some linear differential equation of order 2 is boundedSolution of Matrix differential equation $textbfX'(t)=textbfAtextbfX(t)$How to relate the solutions to a Fuchsian type differential equation to the solutions to the hypergeometric differential equation?Difference between real and complex solution in differential equationFinding the general solution to a system of differential equations using eigenvaluesTwo systems of linear equations equivalent
$begingroup$
Suppose you have a differential equation with n distinct functions of $t$ where
$fracd^2x_1dt^2=k_11x_1+...k_1nx_n$
.
.
.
$fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$
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
$endgroup$
add a comment |
$begingroup$
Suppose you have a differential equation with n distinct functions of $t$ where
$fracd^2x_1dt^2=k_11x_1+...k_1nx_n$
.
.
.
$fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$
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
$endgroup$
add a comment |
$begingroup$
Suppose you have a differential equation with n distinct functions of $t$ where
$fracd^2x_1dt^2=k_11x_1+...k_1nx_n$
.
.
.
$fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$
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
$endgroup$
Suppose you have a differential equation with n distinct functions of $t$ where
$fracd^2x_1dt^2=k_11x_1+...k_1nx_n$
.
.
.
$fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$
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
linear-algebra ordinary-differential-equations physics
asked 4 hours ago
user446153user446153
1075
1075
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1 Answer
1
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$begingroup$
The system of differential equations you wrote could be written as,
$$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$
$$ fracd^2dt^2 vecx = K vecx$$
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,
$$
fracd^2dt^2 vecx = V Lambda V^-1 vecx
$$
$$
V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
$$
$$
fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
$$
Let $vecy = V^-1 vecx$, then we have $
fracd^2dt^2 vecy = Lambda vecy
$. This corresponds to the following system of equations.
$$
fracd^2 y_1dt^2 = lambda_1 y_1
$$
$$
fracd^2 y_2dt^2 = lambda_2 y_2
$$
$$
vdots
$$
$$
fracd^2 y_ndt^2 = lambda_n y_n
$$
Clearly the solutions are of the form,
$$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$
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.
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
The system of differential equations you wrote could be written as,
$$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$
$$ fracd^2dt^2 vecx = K vecx$$
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,
$$
fracd^2dt^2 vecx = V Lambda V^-1 vecx
$$
$$
V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
$$
$$
fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
$$
Let $vecy = V^-1 vecx$, then we have $
fracd^2dt^2 vecy = Lambda vecy
$. This corresponds to the following system of equations.
$$
fracd^2 y_1dt^2 = lambda_1 y_1
$$
$$
fracd^2 y_2dt^2 = lambda_2 y_2
$$
$$
vdots
$$
$$
fracd^2 y_ndt^2 = lambda_n y_n
$$
Clearly the solutions are of the form,
$$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$
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.
$endgroup$
add a comment |
$begingroup$
The system of differential equations you wrote could be written as,
$$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$
$$ fracd^2dt^2 vecx = K vecx$$
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,
$$
fracd^2dt^2 vecx = V Lambda V^-1 vecx
$$
$$
V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
$$
$$
fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
$$
Let $vecy = V^-1 vecx$, then we have $
fracd^2dt^2 vecy = Lambda vecy
$. This corresponds to the following system of equations.
$$
fracd^2 y_1dt^2 = lambda_1 y_1
$$
$$
fracd^2 y_2dt^2 = lambda_2 y_2
$$
$$
vdots
$$
$$
fracd^2 y_ndt^2 = lambda_n y_n
$$
Clearly the solutions are of the form,
$$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$
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.
$endgroup$
add a comment |
$begingroup$
The system of differential equations you wrote could be written as,
$$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$
$$ fracd^2dt^2 vecx = K vecx$$
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,
$$
fracd^2dt^2 vecx = V Lambda V^-1 vecx
$$
$$
V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
$$
$$
fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
$$
Let $vecy = V^-1 vecx$, then we have $
fracd^2dt^2 vecy = Lambda vecy
$. This corresponds to the following system of equations.
$$
fracd^2 y_1dt^2 = lambda_1 y_1
$$
$$
fracd^2 y_2dt^2 = lambda_2 y_2
$$
$$
vdots
$$
$$
fracd^2 y_ndt^2 = lambda_n y_n
$$
Clearly the solutions are of the form,
$$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$
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.
$endgroup$
The system of differential equations you wrote could be written as,
$$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$
$$ fracd^2dt^2 vecx = K vecx$$
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,
$$
fracd^2dt^2 vecx = V Lambda V^-1 vecx
$$
$$
V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
$$
$$
fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
$$
Let $vecy = V^-1 vecx$, then we have $
fracd^2dt^2 vecy = Lambda vecy
$. This corresponds to the following system of equations.
$$
fracd^2 y_1dt^2 = lambda_1 y_1
$$
$$
fracd^2 y_2dt^2 = lambda_2 y_2
$$
$$
vdots
$$
$$
fracd^2 y_ndt^2 = lambda_n y_n
$$
Clearly the solutions are of the form,
$$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$
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.
answered 3 hours ago
SpencerSpencer
8,76812156
8,76812156
add a comment |
add a comment |
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