2.5 * Green’s function approach

2.5.1 * Initial Conditions for the Homogeneous Case

The solution for the homogeneous equation f ≡ 0 was obtained above in Eq. (2.40),

\begin{eqnarray}{ y}_{h}(x)& =& {e}^{−{p\over 2} x}\left \{{c}_{1}\mathop{ cos}\nolimits (Ωx) + {c}_{2}\mathop{ sin}\nolimits (Ωx)\right \},\quad %& \\ {x}_{h}(t)& =& {e}^{−γt}\left \{{x}_{ 1}\mathop{ cos}\nolimits (Ωt) + {x}_{2}\mathop{ sin}\nolimits (Ωt)\right \}, %&(2.43) \\ \end{eqnarray}

with Ω = \sqrt{{ω}^{2 } − γ∕2}.

Specifying the initial conditions

\begin{eqnarray}{ x}_{h}(t = 0) = {x}_{0},\quad \dot{{x}}_{h}(t = 0) = {v}_{0},& & %&(2.44) \\ \end{eqnarray}

we find

\begin{eqnarray}{ x}_{h}(t)& =& {x}_{0}\left \{{e}^{−γt}\mathop{ cos}\nolimits (ωt) + {δ\over ω}{e}^{−γt}\mathop{ sin}\nolimits (ωt)\right \} + {v}_{ 0}{e}^{−γt}{\mathop{sin}\nolimits (ωt)\over ω} .%&(2.45) \\ \end{eqnarray}

Thus {x}_{h}(t) describes the motion of the harmonic oscillator for f ≡ 0 (homogeneous case). If we choose the initial time t = {t}_{0} instead of t = 0, we have

\begin{eqnarray}{ x}_{h}(t)& =& {x}_{0}\left \{{e}^{−γ[t−{t}_{0}]}\mathop{ cos}\nolimits (ω[t − {t}_{ 0}]) + {δ\over ω}{e}^{−γ[t−{t}_{0}]}\mathop{ sin}\nolimits (ω[t − {t}_{ 0}])\right \}%& \\ & +& {v}_{0}{e}^{−γ[t−{t}_{0}]}{\mathop{sin}\nolimits (ω[t − {t}_{0}])\over ω} , %&(2.46) \\ \end{eqnarray}

i.e. everything remains the same; only the ‘origin’ of time {t}_{0} is shifted, i.e the time scale is shifted by {t}_{0}.

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EXERCISE: check that Eq. (2.45) fulfills the correct initial conditions!

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2.5.2 * The Inhomogeneous Case: Effect of the External Force

Now let us discuss the additional effect of the external force, i.e. the inhomogeneous term f(t)∕m in Eq. (2.41). First of all, we recognize that f(t)∕m is an additional acceleration, a(t) = f(t)∕m, of the mass m due to the force f(t), using Newton’s law. What is the additional displacement, Δx(t), of the mass due to that acceleration? In a very short time interval from time t = t' to t = t' + δt', due to the acceleration a(t') the mass aquires the additional velocity

\begin{eqnarray} v(t') = a(t')δt' = {f(t')\over m} δt'.& & %&(2.47) \\ \end{eqnarray}

The subsequent additional displacement Δx(t > t') has to be proportional to that additional velocity and can be calculated using Eq.(2.46) with ‘initial’ additional shift {x}_{0} = 0 and ‘initial’ additional velocity {v}_{0} = v(t'),

\begin{eqnarray} Δx(t > t')& =& {e}^{−γ[t−t']}{\mathop{sin}\nolimits (ω[t − t'])\over ω} × v(t') %& \\ & =& {e}^{−γ[t−t']}{\mathop{sin}\nolimits (ω[t − t'])\over ω} ×{f(t')\over m} δt',%& \\ & =:& G(t − t') ×{f(t')\over m} δt', %&(2.48) \\ \end{eqnarray}

where in the last line we introduced an abbreviation for the term {e}^{−γ[t−t']}\mathop{sin}\nolimits (ω[t − t'])∕ω. The function G(t − t') is called response function (Green’s function) of the harmonic oscillator since it describes its response to an additional, infinitesimal acceleration f(t')δt'∕m. Note that we have made no additional assumptions on how this force f(t') actually behaves as a function of time.

The total additional shift {x}_{f}(t) at time t can be calculated from Eq.(2.48) by integrating the contributions from all times t' with {t}_{0} < t' < t,

\begin{eqnarray}{ x}_{f}(t)& =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{{t}_{0}}^{t}dt'Δx(t > t') ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{{t}_{0}}^{t}dt'G(t − t'){f(t')\over m} .%&(2.49) \\ \end{eqnarray}

The position x(t) at time x now is given by the contribution {x}_{h}(t) (force f = 0) plus the additional shift {x}_{f}(t) (force f\mathrel{≠}0),

\begin{eqnarray} x(t) = {x}_{h}(t) + {x}_{f}(t) = {x}_{h}(t) +{\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{{t}_{0}}^{t}dt'G(t − t'){f(t')\over m} .& & %&(2.50) \\ \end{eqnarray}

Putting everything together, we find a somewhat lengthy, but very convincing expression (we set the initial time {t}_{0} = 0 for simplicity),

\begin{eqnarray} x(t)& =& {x}_{0}\left \{{e}^{−γt}\mathop{ cos}\nolimits (ωt) + {δ\over ω}{e}^{−γt}\mathop{ sin}\nolimits (ωt)\right \} + {v}_{ 0}{e}^{−γt}{\mathop{sin}\nolimits (ωt)\over ω} %& \\ & +& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{t}dt'{e}^{−γ[t−t']}{\mathop{sin}\nolimits (ω[t − t'])\over ω} {f(t')\over m} . %&(2.51) \\ \end{eqnarray}