Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

1D forced harmonic oscillator. Quick solution of non-homogeneous problem.

Posted by peeterjoot on February 19, 2010

[Click here for a PDF of this post with nicer formatting]Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

Motivation.

In [1] equation (25) we have a forced harmonic oscillator equation

\begin{aligned}m \ddot{x} + m \omega^2 x = \gamma(t).\end{aligned} \hspace{\stretch{1}}(1.1)

The solution of this equation is provided, but for fun lets derive it.

Guts

Writing

\begin{aligned}\omega u = \dot{x},\end{aligned} \hspace{\stretch{1}}(2.2)

we can rewrite the second order equation as a first order linear system

\begin{aligned}\dot{u} + \omega x &= \gamma(t)/m \omega \\ \dot{x} - \omega u &= 0,\end{aligned} \hspace{\stretch{1}}(2.3)

Or, with X = (u, x), in matrix form

\begin{aligned}\dot{X} + \omega \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}X&=\begin{bmatrix}\gamma(t)/m \omega \\ 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.5)

The two by two matrix has the same properties as the complex imaginary, squaring to the identity matrix, so the equation to solve is now of the form

\begin{aligned}\dot{X} + \omega i X &= \Gamma.\end{aligned} \hspace{\stretch{1}}(2.6)

The homogeneous part of the solution is just the matrix

\begin{aligned}X &= e^{-i \omega t} A \\ &= \left( \cos(\omega t) \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}- \sin(\omega t)\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}\right) A,\end{aligned}

where A is a two by one column matrix of constants. Assuming for the specific solution X = e^{-i \omega t} A(t), and substuiting we have

\begin{aligned}e^{-i \omega t} \dot{A} = \Gamma(t).\end{aligned} \hspace{\stretch{1}}(2.7)

This integrates directly, fixing the unknown column vector function A(t)

\begin{aligned}A(t) = A(0) + \int_0^t e^{i \omega \tau} \Gamma(\tau).\end{aligned} \hspace{\stretch{1}}(2.8)

Thus the non-homogeneous solution takes the form

\begin{aligned}X = e^{-i \omega t} A(0) + \int_0^t e^{i \omega (\tau - t)} \Gamma(\tau).\end{aligned} \hspace{\stretch{1}}(2.9)

Note that A(0) = (\dot{x}_0/\omega, x_0). Multiplying this out, and discarding all but the second row of the matrix product gives x(t), and Feynman’s equation (26) follows directly.

References

[1] L.M. Brown, G.D. Carson, L.F. Locke, W.W. Spirduso, S.J. Silverman, D. Holtom, E. Fisher, J.E. Mauch, J.W. Birch, K.L. Turabian, et al. Feynman’s thesis: A New approach to quantum theory. Houghton Mifflin, 1954.

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