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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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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 $textbf{X}'(t)=textbf{A}textbf{X}(t)$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
$frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$
.
.
.
$frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
linear-algebra ordinary-differential-equations physics
asked 4 hours ago
user446153user446153
1075
1075
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
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.
$endgroup$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
answered 3 hours ago
SpencerSpencer
8,76812156
8,76812156
add a comment |
add a comment |
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