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The harmonic oscillator coupled to a sinodial external source

$$\frac{\partial^2 x(t)}{\partial t^2}+\omega_0^2 x(t)=F_0\sin(\omega_\text{ext}\ t),$$

has the solution

$$x(t)=x(0)\cos(\omega_0 t)+C \sin(\omega_0 t) +\frac{F_0}{\omega_0^2-\omega_\text{ext}^2}\sin(\omega_\text{ext} t),$$

with a constant of integration $C$. So here we have a resonance for $\omega_\text{ext}=\omega_0$.

If we consider the Klein-Gordon equation with non-vanishing coupling and with a source function $F$

$$\frac{\partial^2 \phi(t,x)}{\partial t^2}-c^2\nabla_x^2\ \phi(t,x) +\omega_0^2\ \phi(t,x)=F(s,x,t),$$

(one dimensional, three dimension, whatever you can deal with)

is there also an family of functions $F(s,t,x)$, (family w.r.t. the parameter s), such that we have a singularity at $s=\omega_0$?

Maybe you go about similar as with the Helmholtz equation, except than you get a shifted coefficient $\omega_0+k^2$.

I think that could be the case as a damping term $\delta\cdot x'(t)$ also destroys the resonance, at least for a sinusoidal source.

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Your first SHO equation should have a full time derivative $d^2/dt^2$, not a partial one. Its solution, if the force starts acting at t = 0, is different - the amplitude grows with time. The Klein-Gordon equation should have either $\partial^2\varphi(x,t)/\partial t^2$ or $\omega_0^2 \varphi (x)$, not both of them. –  Vladimir Kalitvianski Apr 13 '13 at 9:21
    
@VladimirKalitvianski: You can edit the $\frac{\partial}{\partial t}$ to $\frac{d}{dt}$, but that woulnd't make much difference here. The $\omega_0$ in the Klein-Gordon equation is just the mass. Also, I don't see the growing amplitude thing for the oscillator, do you think of another fource? –  NikolajK Apr 13 '13 at 9:54
    
The solution in your post is not as general as one can imagine. The solution I speak about is $x(t) = x_0(t)+\int_0^tdt'F(t')\sin [(t-t')\omega_0]/\omega_0$. –  Vladimir Kalitvianski Apr 13 '13 at 10:21
    
@VladimirKalitvianski: In the harmonic oscillator intro, the source function (where you write $F(t')$) is given explicitly as $F_0 \sin(\omega_\text{ext}t)$, with $F_0$ being a constant. –  NikolajK Apr 13 '13 at 10:52
    
So make the integration with it and you will see the solution different from yours. If initially there is no free oscillations ($x_0(t)=0$), then the external force will give and take away energy to the oscillator. In a resonance case the energy (amplitude) will be growing. –  Vladimir Kalitvianski Apr 13 '13 at 11:12

2 Answers 2

Let us denote the field amplitude $\varphi$ (because $x(x)$ does not make sense).

If you drop the term $\omega_0^2\varphi(x,t)$, then your KG equation will look like a field sourced with an external source (force). The corresponding (space) Fourier harmonic $\varphi_{k(\omega_0)}(t)$ will have a growing solution if the force is periodic and starts acting at $t=0$.

EDIT: The mass term will shift the spectrum and it may make low-frequency force non resonant.

Edit2: The external source $F(s,x,t)$ can be represented as due to a point-like particle, i.e., $F(s,x,t)=\delta(x-x_0)F(s,t)$, then, after Fourier transformation, you will obtain driven oscillator equations for $\chi(k,t)$.

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Your intuition about the Helmholtz equation is correct in that it is the right way to handle the spatial dependence. If you are in free space, then you should be tackling this problem using a plane-wave basis by taking the Fourier transform of your equation: $$ \frac{\partial^2 \tilde\phi(t,\mathbf k)}{\partial t^2} +(c^2k^2+\omega_0^2)\tilde\phi(t,\mathbf k)=\tilde F(s,\mathbf k,t). $$ Thus the equation for the amplitude of each Fourier mode is exactly the forced, undamped harmonic oscillator you begin with (albeit with an increased natural frequency $\omega_0'=\sqrt{c^2k^2+\omega_0^2}$). Since the equation is the same, the conclusions are the same, including the existence of resonances.

The only real difference is that you have a bunch of independently oscillating amplitudes, and they need not all be excited at the same rate. It is the Fourier transform of $F$ that determines which modes get excited and which don't: in general, only modes whose wavenumber $k$ matches relevant spatial scales on which $F$ varies can couple significantly to the external driving.

On the other hand, if $\tilde F$ is more or less uniform and its frequency $s$ is bigger than $\omega_0$ then there will be some modes that will get resonantly excited.

So finally, to address what I think is your question: spatial coupling can impede resonances. If the driving frequency $s^2=c^2k^2+\omega_0^2$ matches the natural frequency of the field at a wavenumber $k$ such that none of the modes couple, i.e. $\tilde F (s,\mathbf k,t)\equiv 0$, then there will be no resonance. (In that case I would still expect interesting structure, as $\tilde F$ will likely be nonzero just off the resonant $k$. You will likely see large amplitudes but no singularities.)

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