The purpose of this post is to get an physical intuition of what a Green's function. Since we are examining couple oscillators and hence coupled differential equations we will not be able to get the solution but we shall understand what the Green function is doing. Here again are the couple ODEs were are considering:
\begin{eqnarray}
m_1 \ddot{x}= -k_1x - k_{12}(x-y)\\
m_2\ddot{y} = -k_{2}y + k_{12}(x-y)
\end{eqnarray}
m_1 \ddot{x}= -k_1x - k_{12}(x-y)\\
m_2\ddot{y} = -k_{2}y + k_{12}(x-y)
\end{eqnarray}
What we shall do is consider \(m_1\) as the system we are considering and then the driving force is given by \(m_2\). Now, at a specific moment in time ,t, the system fills an outside force called the impulse, this impulse could be described by a function over time or it could be one hit at a specific time.
In fact when we usually solve differential equations we can interpret the differential equation as a system, now if it is homogeneous equation then we can interpret that as hitting the system for a split second and observing it. System then moves according to what the differential equation says. If on the other hand the differential equation is in-homogeneous, we can imagine the resulting movement of the system as a linear superposition of these constant split second hits over a time period. In general these split second hits will not have the same force. To get the solution to the in-homogeneous equation all we need to do is to add the response of the system to these split second hits. The response to these split second hits is described by the Green's Function. These split second hits could be actual mechanical hits or charges which provide electric impulse or masses which provide gravitational impulse. So what we need to do first is solve the homogeneous equation in other words find the Green's Function.
It is interesting to note that all we have described can be understood in terms of convolutions in the following manner:
Let D be differential operator acting upon the system,y, and a driving force f(t). So Dy = f.
Now we know from the properties of delta function that \( \int_{-\infty}^{\infty}\delta(t'-t)f(t') dt' = f(t)\) but this is merely \(\delta \star f\). So we can re-write Dy = f as Dy =\( \delta \star f\). Let's suppose we can write 'y' as a convolution of two functions then D(\(A \star B\))= \(\delta \star f \). Using derivative properties of convolution we have \((DA) \star B \) = \(\delta \star f\). From this we see that B = f and DA = \(\delta\) . A here is the green's function which we will call G. Hence we have DG = \(\delta(t'-t)\), which is consistent with what I said about the green's function being the response of the system to a split second hit of the system. The split second hit here is given by the delta function and the magnitude of this hit is \( \int_{-\infty}^{\infty}\delta(t'-t)f(t') dt' = f(t)\) . From the above analysis we get that \( y = G \star f\).
At last going back to our coupled ODEs, we will look at the fir st one which we will write in this form
$$ m_1 \ddot{x} +k_1x + k_{12}x = ky \hspace{20mm }\text{eq.1} $$. Remembering that DG = \(\delta(t'-t)\) we shall write the following equation $$ m_1 \ddot{G_o} +k_1G + k_{12}G_o = \delta (t-t') \hspace{20mm} \text{eq.2} $$. This simplifies to $$\ddot{G_o} = - G_o\frac{k_1 + k_{12}}{m_1} \hspace{20mm}\text{eq.3} $$ we shall define \(\frac{k_1 + k_{12}}{m_1}\) as \(\omega_o ^2\)
The same procedure can be done with second ODE:
$$ m_2 \ddot{y} +k_2y + k_{12}y = kx \hspace{20mm }\text{eq.4} $$ . We arrive at $$\ddot{G_1} = - G_1\frac{k_2 + k_{12}}{m_2} \hspace{20mm} \text{eq.5} $$ and define \(\frac{k_2 + k_{12}}{m_2}\) as \(\omega_1 ^2\) .
For the next part of our analysis we shall continue with the first ODE and try to find the solution in terms of the force function.
\(G_o\) we know to be \(A\sin \omega t + B\cos\omega t\) as a solution so then we apply the boundary condition that the system beings from rest at t = 0. Which means that B = 0. Thus our Green's function G = \(A\sin \omega_o t\). Now for t' < 0 the driving force is 0 and for t' > 0 the driving force is non-zero. This means that the full definition of our Green's function is $$G(t,t')= \begin{cases} 0 & t < t' \\ A\sin \omega_o t & t > t' \end{cases}$$ The remaining task is to find the value of our constant A. For that we go back to the original ODE in the following manner:
\begin{eqnarray}
\int_{t'-\epsilon}^ {t'+\epsilon}\left( m_1\ddot{G} + \omega_o G\right) dt = \int_{t'-\epsilon}^ {t'+\epsilon}\delta(t) dt\\
m_1\frac{dG}{dt}|_{t'-\epsilon}^{ t'+\epsilon} + \omega_o \int_{t'-\epsilon}^ {t+\epsilon} G dt = 1\\
m_1\frac{dG}{dt}|_{t'+\epsilon}- m_1\frac{dG}{dt}|_{t'- \epsilon} = 1 \hspace{20mm} \text{eq.6}
\end{eqnarray}
Here \( \epsilon\) is some positive infinitesimal real number and we are doing the integrals in the limit as \(\epsilon\) goes to 0. So technically I should have had the limit signs above. Plugging G into eq.6 and remembering that G = 0 for t' < 0, we get that \(A_o\)= \(\frac{1}{m_1\omega_o}\)
Thus our solution is $$ x(t) = k_{12}\int_0^t y(t)A_o\sin(\omega_o (t-t')) dt' \hspace{20mm} \text{eq.7}$$ or if we had started with the second ODE $$ y(t) = k_{12}\int_0^t x(t)A_1\sin(\omega_1 (t-t')) dt' \hspace{20mm} \text{eq.8}$$
If this was not a coupled equation we would have f(t) instead of the green's function and we would integrate in order to find the solution. What eq.7 and eq.8 tell us is that if we apply isotropic conditions i.e \(k_1= k_2 \)and \(m_1 = m_2\) the green function is the same for both ODE and the solution to the system would be the convolutions of each other. Eq. 7 and Eq.8 show us what Green's function is doing, we have a system feeling a force caused by the other mass coupled to our system , Green's function acts as a weighting function over time for the force we feel from the other system. Lastly, the fact that the argument in Green's function is t-t' shows causality. The solution up until time t is got by averaging the split second reactions of the system to the driving force up until that point. It matters where we are in time.
In fact when we usually solve differential equations we can interpret the differential equation as a system, now if it is homogeneous equation then we can interpret that as hitting the system for a split second and observing it. System then moves according to what the differential equation says. If on the other hand the differential equation is in-homogeneous, we can imagine the resulting movement of the system as a linear superposition of these constant split second hits over a time period. In general these split second hits will not have the same force. To get the solution to the in-homogeneous equation all we need to do is to add the response of the system to these split second hits. The response to these split second hits is described by the Green's Function. These split second hits could be actual mechanical hits or charges which provide electric impulse or masses which provide gravitational impulse. So what we need to do first is solve the homogeneous equation in other words find the Green's Function.
It is interesting to note that all we have described can be understood in terms of convolutions in the following manner:
Let D be differential operator acting upon the system,y, and a driving force f(t). So Dy = f.
Now we know from the properties of delta function that \( \int_{-\infty}^{\infty}\delta(t'-t)f(t') dt' = f(t)\) but this is merely \(\delta \star f\). So we can re-write Dy = f as Dy =\( \delta \star f\). Let's suppose we can write 'y' as a convolution of two functions then D(\(A \star B\))= \(\delta \star f \). Using derivative properties of convolution we have \((DA) \star B \) = \(\delta \star f\). From this we see that B = f and DA = \(\delta\) . A here is the green's function which we will call G. Hence we have DG = \(\delta(t'-t)\), which is consistent with what I said about the green's function being the response of the system to a split second hit of the system. The split second hit here is given by the delta function and the magnitude of this hit is \( \int_{-\infty}^{\infty}\delta(t'-t)f(t') dt' = f(t)\) . From the above analysis we get that \( y = G \star f\).
At last going back to our coupled ODEs, we will look at the fir st one which we will write in this form
$$ m_1 \ddot{x} +k_1x + k_{12}x = ky \hspace{20mm }\text{eq.1} $$. Remembering that DG = \(\delta(t'-t)\) we shall write the following equation $$ m_1 \ddot{G_o} +k_1G + k_{12}G_o = \delta (t-t') \hspace{20mm} \text{eq.2} $$. This simplifies to $$\ddot{G_o} = - G_o\frac{k_1 + k_{12}}{m_1} \hspace{20mm}\text{eq.3} $$ we shall define \(\frac{k_1 + k_{12}}{m_1}\) as \(\omega_o ^2\)
The same procedure can be done with second ODE:
$$ m_2 \ddot{y} +k_2y + k_{12}y = kx \hspace{20mm }\text{eq.4} $$ . We arrive at $$\ddot{G_1} = - G_1\frac{k_2 + k_{12}}{m_2} \hspace{20mm} \text{eq.5} $$ and define \(\frac{k_2 + k_{12}}{m_2}\) as \(\omega_1 ^2\) .
For the next part of our analysis we shall continue with the first ODE and try to find the solution in terms of the force function.
\(G_o\) we know to be \(A\sin \omega t + B\cos\omega t\) as a solution so then we apply the boundary condition that the system beings from rest at t = 0. Which means that B = 0. Thus our Green's function G = \(A\sin \omega_o t\). Now for t' < 0 the driving force is 0 and for t' > 0 the driving force is non-zero. This means that the full definition of our Green's function is $$G(t,t')= \begin{cases} 0 & t < t' \\ A\sin \omega_o t & t > t' \end{cases}$$ The remaining task is to find the value of our constant A. For that we go back to the original ODE in the following manner:
\begin{eqnarray}
\int_{t'-\epsilon}^ {t'+\epsilon}\left( m_1\ddot{G} + \omega_o G\right) dt = \int_{t'-\epsilon}^ {t'+\epsilon}\delta(t) dt\\
m_1\frac{dG}{dt}|_{t'-\epsilon}^{ t'+\epsilon} + \omega_o \int_{t'-\epsilon}^ {t+\epsilon} G dt = 1\\
m_1\frac{dG}{dt}|_{t'+\epsilon}- m_1\frac{dG}{dt}|_{t'- \epsilon} = 1 \hspace{20mm} \text{eq.6}
\end{eqnarray}
Here \( \epsilon\) is some positive infinitesimal real number and we are doing the integrals in the limit as \(\epsilon\) goes to 0. So technically I should have had the limit signs above. Plugging G into eq.6 and remembering that G = 0 for t' < 0, we get that \(A_o\)= \(\frac{1}{m_1\omega_o}\)
Thus our solution is $$ x(t) = k_{12}\int_0^t y(t)A_o\sin(\omega_o (t-t')) dt' \hspace{20mm} \text{eq.7}$$ or if we had started with the second ODE $$ y(t) = k_{12}\int_0^t x(t)A_1\sin(\omega_1 (t-t')) dt' \hspace{20mm} \text{eq.8}$$
If this was not a coupled equation we would have f(t) instead of the green's function and we would integrate in order to find the solution. What eq.7 and eq.8 tell us is that if we apply isotropic conditions i.e \(k_1= k_2 \)and \(m_1 = m_2\) the green function is the same for both ODE and the solution to the system would be the convolutions of each other. Eq. 7 and Eq.8 show us what Green's function is doing, we have a system feeling a force caused by the other mass coupled to our system , Green's function acts as a weighting function over time for the force we feel from the other system. Lastly, the fact that the argument in Green's function is t-t' shows causality. The solution up until time t is got by averaging the split second reactions of the system to the driving force up until that point. It matters where we are in time.
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