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The solution of the Newton equation $m\ddot{x} = F(t)$ (where $F(t)$ depends only on time $t$) can be expressed with an integral operator as $$x(t) = \int_{-\infty}^{\infty} G(t - t^{\prime})\; F(t^{\prime})\; dt^{\prime}.$$What is the kernel $G(t - t^{\prime})$ of the integral operator?

This problem came up in the Rudolf Ortvay 2020 Physics Competition and I can't help but think that it is not that complicated.

I simply proceeded by defining the correlated Green's function $G(t, t')$ of the homogenous equation $m\ddot x = 0$. As the Green's function follows $LG = \delta$ where $L$ is a linear differential operator, then the identity, $m\ddot G = \delta (t - t')$ holds. This tells us by Abel's identity that $R (t, t') = \frac{t - t'}{m}$ which means the associated Kernel follows $$G(t - t') = \begin{cases} 0 & t < t' \\ \frac{t - t'}{m} & t > t' \end{cases}.$$ Is my work correct? I can't help but think that this is too simple.

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  • $\begingroup$ No, it's not too simple. While solving the equation $\,m\,\ddot{x}=mv\,\delta(t)\,$ for the Green's function, think of hitting a billiard ball that flies away according to $\,x(t)=H(t)\,v\,t\,$ where $\,H(t)\,$ is the Heaviside step function. $\endgroup$ – Han de Bruijn Jun 11 at 12:17
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Two things, formulated slightly different as by the OP: $$ x(t) = \int_{-\infty}^{\infty} G(t - t')\; F(t')\; dt'. $$ $$ G(t-t') = \begin{cases} 0 & t' \gt t \\ (t - t')/m & t' \lt t \end{cases}. $$ Let's work the other way around to check this: $$ x(t) = \int_{-\infty}^t \frac{(t-t')}{m}\; F(t')\; dt' $$ Use the Leibniz integral rule for differentiating under the integral sign: $$ {\displaystyle {\frac {d}{dx}}\left(\int _{a}^{x}f(x,t)dt\right)= f{\big (}x,x{\big )}+\int _{a}^{x}{\frac {\partial }{\partial x}}f(x,t)dt,} \\ \mbox{with} \quad f(t,t')=\frac{(t-t')}{m}\; F(t') \quad \Longrightarrow \\ \dot{x} = \frac{dx}{dt} = 0 + \int_{-\infty}^t \frac{\partial}{\partial t}\frac{(t-t')}{m}\; F(t')\; dt' = \frac{1}{m}\int_{-\infty}^t F(t')\; dt' $$ $$ m\,\ddot{x} = F(t) $$ So it seems that the kernel is correct.
But not very useful, it seems. Take the simplest possible example, with $m=a=\mbox{constant}$: $$ F = \begin{cases} 0 & t \lt 0 \\ ma & t \gt 0 \end{cases}.\\ x(t) = \int_{-\infty}^t \frac{(t-t')}{m}\; F(t')\; dt'= a\left[t \int_0^t dt'- \int_0^t t'\; dt'\right]\\x(t) = a\,t^2 - \frac12 a\,t^2 = \frac12 a\,t^2 $$ So the kernel only solves for displacements without initial positions or initial velocities.

EDIT. The more obvious approach is to integrate the Newton force equation directly. Also take into account that physical events normally start at timestamp $t=0$ instead of $t=-\infty$. $$ \ddot{x} = \frac{F(t)}{m} \\ \dot{x}(t) = \int_0^t \frac{F(t')}{m}\; dt' \\ x(t) = \int_0^t \dot{x}(t')\; dt' $$ Now look what happens with partial integration: $$ \int_0^t \dot{x}(t')\; dt' = \left[t'\dot{x}(t')\right]_{t'=0}^{t'=t} - \int_0^t t'\ddot{x}(t')\; dt' = t\,\int_0^t \frac{F(t')}{m}\; dt' - \int_0^t t'\,\frac{F(t')}{m}\; dt' \\ \Longrightarrow \quad x(t) = \int_{-\infty}^{+\infty} G(t-t')F(t')\; dt' \quad \mbox{with} \quad G(t) = H(t)\frac{t}{m} $$ where it is assumed that $\,F(t\lt 0) = 0\,$; $\,H(t)\,$ is the Heaviside step function.
Hence, apart from some technicalities as mentioned above, the two approaches are the same.

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