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A simple harmonic oscillator whose Action is given by:

$S=\displaystyle\int dt\left(\frac{1}{2}\cdot m\left(\frac{dx}{dt}\right)^2-\frac{1}{2}m\omega x^2\right)$

here x is a function of time i.e $x(t)$ . Use the above equation and by doing integration by parts show that : $S=\displaystyle\int dt\left(-\frac{1}{2}\cdot m\cdot x\left(\frac{d^2x}{dt^2}\right)-\frac{1}{2}m\omega x^2\right)$ (here the upper limit is $t_1 = x_1$ and lower limit is $t_2 = x_2$)

I know what integration by parts is which is by defining a $u$ and $v$ functions according to simplicity then integrate according to it. Here I have to calculate everything with respect to Time how will I handle variable here which doesn't depends on time will I take them as constant ? and how did the second part$\left(-\frac{1}{2}m\omega x^2\right)$ remained the same? Please help me with the procedure.

My solution from @Y Tong hint ,

$=\int \frac{1}{2}\cdot m \frac{dx}{dt} dx-\int \frac {1}{2}m\omega x^2 dt$

taking , $\frac{1}{2}m \frac{dx}{dt}=u$ and $dx=v$

$=\int\frac{1}{2}m \frac{dx}{dt} x \biggr\rvert_{\mathbf{t_1}}-\int \frac{1}{2}m \frac{d}{dt}(\frac{dx}{dt})x $

Eventually I got ,

$=-\int \frac{1}{2}m \frac{d}{dt}(\frac{dx}{dt}) x -\int \frac{1}{2} m \omega x^2 dt$

And I am stuck here to generalize it into proven part . Can anyone help with the next steps ?

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  • $\begingroup$ Yes, everything except $x$ is to be treated as a constant. $\endgroup$
    – Tavish
    Dec 25 '20 at 17:51
  • $\begingroup$ @Tavish can you tell me what does that t1=x1 mean in the limit part ? $\endgroup$
    – Hoppo
    Dec 25 '20 at 18:59
  • $\begingroup$ I’m not sure, but it could mean the upper limit of the integral, i.e. the integral is from $x_2$ to $x_1$. $\endgroup$
    – Tavish
    Dec 25 '20 at 19:29
  • $\begingroup$ The stated upper limits don't make sense given the context. I would instead expect $x_1=x(t_1)$ and $x_2=x(t_2)$, i.e., $x_1,x_2$ are the initial and final positions. $\endgroup$ Dec 25 '20 at 19:33
  • $\begingroup$ @Tavish i really have no idea how to use Integration by parts here! If i use integration by parts here, from (1/2 m x'') i should get two expression ... $\endgroup$
    – Hoppo
    Dec 25 '20 at 19:46
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You want to use $\int udv=uv|_{t_1}^{t_2}-\int v du$ which is due to $(uv)'=u'v+uv'$.

Take $u=\frac12 m\frac{dx}{dt}$ and $dv=dx$, you have $\int_{t_1}^{t_2}\frac12 m\frac{dx}{dt} dx=\frac12 m\frac{dx}{dt} x|_{t_1}^{t_2}-\int_{t_1}^{t_2} \frac12 mx \frac{d^2x}{dt^2}dt.$ The variation should be there but for typical applications of the least action principle, you will fix the ends, so it doesn't matter. However, the potential term should have been $\frac12m\omega^2 x^2$ instead of $\frac12 m\omega x^2,$ or you won't even have matching units.

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  • $\begingroup$ Do we have to touch the second part ? coz we are getting exactly what we want from the potential part ! $\endgroup$
    – Hoppo
    Dec 26 '20 at 8:19
  • $\begingroup$ Whoever designed the question just got some typos. The action has the unit of energylength (the unit of Planck's constant). You need masslength^2/time^2. So the first term is missing an m. That's also why I said the second term also has a typo. $\endgroup$
    – C Tong
    Dec 26 '20 at 12:36
  • $\begingroup$ I am sorry i made a correction where there is m ...:) $\endgroup$
    – Hoppo
    Dec 26 '20 at 16:48
  • $\begingroup$ Oh, I see what you were wondering about. I made a typo in my $vdu$ part. $v$ is just x$, and $du=\frac12 m \frac{d^2 x}{dt^2} dt$. So the integration by part was already complete. $\endgroup$
    – C Tong
    Dec 26 '20 at 18:41

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