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4 Solution using Green's Theorem

Green's theorem is based on integration by parts. We write

$\displaystyle{\frac{d^2x(t')}{dt'^2}}$ = - $\displaystyle\omega_{0}^{2}$x(t') + $\displaystyle{\textstyle\frac{1}{m}}$F(t')   
$\displaystyle{\frac{d^2G(t,t')}{dt'^2}}$ = - $\displaystyle\omega_{0}^{2}$G(t,t') + $\displaystyle{\frac{\lambda}{m}}$$\displaystyle\delta$(t - t') (7)
and multiply the first equation by G(t,t') the second by x(t') , subtract, and integrate over the range of t' from ti to tf corresponding to the range where we want the solution. Therefore ti < t < tf , and the result is

$\displaystyle\int_{t_i}^{t_f}$dt'G(t,t')$\displaystyle{\frac{d^2}{dt'^2}}$x(t') - $\displaystyle\int_{t_i}^{t_f}$dt'x(t')$\displaystyle{\frac{d^2}{dt'^2}}$G(t,t') = $\displaystyle\int_{t_i}^{t_f}$dt'G(t,t')$\displaystyle{\frac{F(t')}{m}}$ - $\displaystyle\int_{t_i}^{t_f}$dt'x(t')$\displaystyle{\frac{\lambda\delta(t-t') }{m}}$ (8)

and integrating the first two terms by parts or, equivalently, using Green's theorem, this becomes

$\displaystyle\left . G(t,t') \frac{d x(t')}{dt'} \right\vert _$t'=tit' = tf - $\displaystyle\left . x(t') \frac{d G(t,t')}{dt'} \right\vert _$t'=tit' = tf = $\displaystyle\int_{t_i}^{t_f}$dt'G(t,t')$\displaystyle{\frac{F(t')}{m}}$ - x(t)$\displaystyle{\frac{\lambda}{m}}$ . (9)

Notice that ti can be - $\infty$ and tf can be $\infty$ if the solution at all times is desired.

We want to solve for the case where the oscillator is at rest at the origin at time ti . In that case, the lower limits of the surface terms are zero since both x(ti) and dx(t)/dt|t = ti are zero. If we use the retarded Green's function, the surface terms will be zero since t < tf , and the retarded Green's functions are zero at the upper limits. We can also replace the upper limit of the integral containing G(t,t') with t , since the retarded Green's function is zero beyond that value. The solution in terms of the retarded Green's function is

x(t) = $\displaystyle\int_{t_i}^{t}$dt'G (R)(t,t')$\displaystyle{\frac{F(t')}{\lambda}}$   
  = $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t_i}^{t}$dt'sin($\displaystyle\omega_{0}^{}$[t - t'])F(t') . (10)

We can also use the advanced Green's function, however in this case, the surface terms at the upper limit do not cancel and the result is slightly more difficult to derive. The Green's function terms are

$\displaystyle\left . G^{(A)}(t,t') \right\vert _$t'=tf = - $\displaystyle{\frac{\lambda}{m \omega_0}}$sin($\displaystyle\omega_{0}^{}$[t - tf])   
$\displaystyle\left .\frac{d G^{(A)}(t,t')}{dt'} \right\vert _$t'=tf = $\displaystyle{\frac{\lambda}{m}}$cos($\displaystyle\omega_{0}^{}$[t - tf]) (11)
so that the result is
x(t) = $\displaystyle\int_{t}^{t_f}$dt'G (A)(t,t')$\displaystyle{\frac{F(t')}{\lambda}}$ + $\displaystyle{\textstyle\frac{1}{\omega_0}}$sin($\displaystyle\omega_{0}^{}$[t - tf])v(tf) - cos($\displaystyle\omega_{0}^{}$[t - tf])x(tf)   
  = $\displaystyle\int_{t}^{t_f}$dt'G (A)(t,t')$\displaystyle{\frac{F(t')}{\lambda}}$ + $\displaystyle{\textstyle\frac{1}{\omega_0}}$sin($\displaystyle\omega_{0}^{}$[t - tf])v(tf) - cos($\displaystyle\omega_{0}^{}$[t - tf])x(tf)   
  = - $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[t - t'])F(t') + $\displaystyle{\textstyle\frac{1}{\omega_0}}$sin($\displaystyle\omega_{0}^{}$[t - tf])v(tf) - cos($\displaystyle\omega_{0}^{}$[t - tf])x(tf) (12)
where we have written the velocity dx(t)/dt = v(t) . To use this, we need to find out the values of x(tf) and v(tf) that correspond to the desired boundary conditions x(ti) = v(ti) = 0 . Taking the limit t $\rightarrow$ ti , the result becomes

x(ti) = 0 = - $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t_i}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[ti - t'])F(t') + $\displaystyle{\textstyle\frac{1}{\omega_0}}$sin($\displaystyle\omega_{0}^{}$[ti - tf])v(tf) - cos($\displaystyle\omega_{0}^{}$[ti - tf])x(tf) (13)

and taking the derivative of the equation

v(ti) = 0 = - $\displaystyle{\textstyle\frac{1}{m}}$$\displaystyle\int_{t_i}^{t_f}$dt'cos($\displaystyle\omega_{0}^{}$[ti - t'])F(t') + cos($\displaystyle\omega_{0}^{}$[ti - tf])v(tf) + $\displaystyle\omega_{0}^{}$sin($\displaystyle\omega_{0}^{}$[ti - tf])x(tf) . (14)

We have two linear equations which we can solve for x(tf) and v(tf) ,
x(tf) = $\displaystyle{\textstyle\frac{1}{m\omega_0}}$sin($\displaystyle\omega_{0}^{}$[ti - tf])$\displaystyle\int_{t_i}^{t_f}$dt'cos($\displaystyle\omega_{0}^{}$[ti - t'])F(t') - $\displaystyle{\textstyle\frac{1}{m\omega_0}}$cos($\displaystyle\omega_{0}^{}$[ti - tf])$\displaystyle\int_{t_i}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[ti - t'])F(t')   
  = $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t_i}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[tf - t'])F(t')   
v(tf) = $\displaystyle{\textstyle\frac{1}{m}}$sin($\displaystyle\omega_{0}^{}$[ti - tf])$\displaystyle\int_{t_i}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[ti - t'])F(t') + $\displaystyle{\textstyle\frac{1}{m}}$cos($\displaystyle\omega_{0}^{}$[ti - tf])$\displaystyle\int_{t_i}^{t_f}$dt'cos($\displaystyle\omega_{0}^{}$[ti - t'])F(t')   
  = - $\displaystyle{\textstyle\frac{1}{m}}$$\displaystyle\int_{t_i}^{t_f}$dt'cos($\displaystyle\omega_{0}^{}$[tf - t'])F(t') (15)
and substituting back gives the result
x(t) = - $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[t - t'])F(t') - $\displaystyle{\textstyle\frac{1}{m\omega_0}}$sin($\displaystyle\omega_{0}^{}$[t - tf])$\displaystyle\int_{t_i}^{t_f}$dt'cos($\displaystyle\omega_{0}^{}$[tf - t'])F(t')   
     - $\displaystyle{\textstyle\frac{1}{m\omega_0}}$cos($\displaystyle\omega_{0}^{}$[t - tf])$\displaystyle\int_{t_i}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[tf - t'])F(t')   
  = - $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[t - t'])F(t') + $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t_i}^{t_f}$dt'sin($\displaystyle\omega_{0}^{}$[t - t'])F(t')   
  = $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{t_i}^{t}$dt'sin($\displaystyle\omega_{0}^{}$[t - t'])F(t') (16)
which is identical to the result we obtained using the retarded Green's function.

So we see that while using the Green's function optimized for the boundary conditions at hand makes the calculation simpler, any Green's function can be used to get the correct answer.


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