Understanding intuition behind time translation in classical mechanics In V.Arnold book "Mathematical Methods of Classical Mechanics" he says that invariance with respect to the time for isolated systems means that "the laws of nature remain constant", i.e., if $\phi(t)$ is a solution to differential equation that describes time evolution of isolated system then $\phi(t+s)$ is also a solution to the same differential equation where $s$ can be any real number. 
I am confused about how the physical intuition guides us to accept this mathematical requirement on the differential equation.
Specifically,


*

*Why does the mathematical statement mean that "the laws of nature remain constant"?

*Why should we assume that "laws of nature remain constant"? 

*Why after time shift we assume that the evolution of the system is governed by the same differential equation? 
I would appreciate your thoughts and comments. 
 A: 
1) Why does the mathematical statement mean that "the laws of nature remain constant"?

The "laws of nature" refer here to the differential equations derived from some presumably-fundamental Lagrangian (i.e. the equations of motion of a particular object). The solution to these equations is a particular trajectory $\phi(t)$. We impose the property that whenever $\phi(t)$ is a solution to a particular differential equation, then $\phi(t+s)$ is also a solution for any $s$. When we say that "the laws of nature remain constant," we mean that the equations of motion have no explicit time dependence. We will prove below that when you impose the property you described, the differential equation cannot have any explicit time dependence.
Suppose that $\phi(t)$ is the solution of a differential equation
$$F\left(\phi(t),\frac{\partial\phi(t)}{\partial t},\frac{\partial^2\phi(t)}{\partial t^2},...,t\right)=0$$
(you can always write a given differential equation in this way). Saying that the time-shifted $\phi$ is also a solution implies that
$$F\left(\phi(t+s),\frac{\partial\phi(t+s)}{\partial t},\frac{\partial^2\phi(t+s)}{\partial t^2},...,t\right)=0$$
Equivalently, we can relabel $t'=t+s$, so that
$$F\left(\phi(t'),\frac{\partial\phi(t')}{\partial t},\frac{\partial^2\phi(t')}{\partial t^2},...,t'-s\right)=0$$
But it doesn't really matter what we call the parameter $t'$; the equation will still be true if we simply label it as $t$ again. The differential equation doesn't really care what its parameter is called anyway:
$$F\left(\phi(t),\frac{\partial\phi(t)}{\partial t},\frac{\partial^2\phi(t)}{\partial t^2},...,t-s\right)=0$$
In case this is conceptually tricky, let's look carefully at what we're doing in each case (for clarity, we'll say $\phi$ refers to position):


*

*In the first version of the equation, we have replaced the position of the particle at one time with the position at a later time.

*In the last version, we have replaced the time which we associate with a particular position with an earlier time.


If you think carefully, these are exactly equivalent. If you think of position and time as two long parallel rulers, where the trajectory is determined by the relationship between the markings on the two rulers, then moving one ruler forward results in exactly the same relative positions of the markings as moving the other ruler backward.
With that said, we can now equate the last equation with the very first one:
$$F\left(\phi(t),\frac{\partial\phi(t)}{\partial t},\frac{\partial^2\phi(t)}{\partial t^2},...,t\right)=F\left(\phi(t),\frac{\partial\phi(t)}{\partial t},\frac{\partial^2\phi(t)}{\partial t^2},...,t-s\right)$$
The only difference between these two equations' inputs is in the explicit time parameter; the positions that they take in are exactly the same as a function of time. As such, this situation can be thought of in the same way as a function $f(x)$ with the property that $f(x)=f(x+k)$ for any $k$. For any $x$, you can always find a $k$ such that $f(x)=f(0)$ (namely, by setting $k=-x$).  Therefore, for any $x$, $f(x)=f(0)$, and $f$ must be constant. Another way to say this is that the output of $f$ cannot depend on $x$. 
Applying this to our situation, if the above is true for any $s$, then for any time $t$ there exists an $s$ (namely, $s=t$) such that
$$F\left(\phi(t),\frac{\partial\phi(t)}{\partial t},\frac{\partial^2\phi(t)}{\partial t^2},...,t\right)=F\left(\phi(t),\frac{\partial\phi(t)}{\partial t},\frac{\partial^2\phi(t)}{\partial t^2},...,0\right)$$
So, at any time $t$, the output of $F$ must be equal to its output if you had the same positions and set the explicit time parameter to $0$. In other words, the output of $F$ cannot depend on the value of the explicit time parameter $t$, and so $F$ cannot depend explicitly on time.

2) Why should we assume that "laws of nature remain constant"? 

Because we have observed, in the real world, that such an assumption leads us to correctly predict how nature behaves in an especially simple way for a variety of situations. That's the most fundamental explanation - physics is developed based on the outcome of observations. 
You can almost certainly construct a set of fundamental laws that don't rely on time-translation invariance. If you did, you would have to reconcile the predictions of that model with the reams and reams of experimental data that demonstrate no deviation from the current model. Once you do that, you'll have a model that, by its nature, is much more complicated than the current model (time-translation invariance is a rather strong constraint, after all, and loosening constraints leads to more complexity more often than not), and only deviates from the current model on scales that are small enough to be undetectable by current experiments. It's certainly possible that such a model could offer a better description of reality than the current one, if future experiments end up supporting it; but until then, Occam's Razor suggests that the simplest falsifiable model that explains the current results is the one to rely on.  

3) Why after time shift we assume that the evolution of the system is governed by the same differential equation? 

In some cases this isn't true; there are certainly systems in nature that are most easily modeled using explicitly time-dependent equations of motion (for example, a harmonic oscillator with an external driving force, or a system that dissipates energy due to friction). Arnold's point is that the fundamental laws of nature are time-translation invariant, not that any emergent behavior will be. And as for the reason we assume that, the answer is the same as that for question 2 - we simply have no reason, at present, to adopt any more complicated assumptions.
A: Consider this series of physics problems:


*

*At the stroke of noon today, I drop a book from a height of 1m.  How long does the book fall before it hits the ground?

*At six in the evening, I drop a book from a height of 1m.  How long does the book fall before it hits the ground?

*At some time (which I will not tell you), I drop a book from a height of 1m.  How long does the book fall before it hits the ground?


If your intuition tells you that the answers to these questions are all the same, then you intuitively believe that the laws of physics are time symmetric (or at least they are when pertaining to falling books).
Let's say you solve the equations for the first problem.  You find that I drop the book at 12:00:00.00, and it hits the ground at 12:00:00.45.  You can write an equation for the path the book took as a function of time.  Call that equation $\phi(t)$.
Now consider the second problem.  I now drop the book at 18:00:00.00.  The equation for the path the book takes as a function of time isn't quite the same.  If it were exactly the same, we'd see that the book lands at 12:00:00.45, like it did in $\phi(t)$.  To fix this, we observe that I didn't just drop the book 6 hours later.  Everything in the system happened 6 hours later, including me dropping the book, the book hitting the ground, and the entire path inbetween.  We can represent that as $\phi(t+s)$ where $s$ is 6 hours.
The third question points to the general version of this.  We realize that the time I dropped the book was really unnecessary to answer the question.  Because the book dropping exhibits time symmetry, whatever time I drop the book, there is some $s$ such that $\phi(t+s)$ describes the path the book takes as it falls.

Why should we assume that "laws of nature remain constant"?

That is an good philosophical question to ask.  The best reason science can offer is "we are yet to observe any inkling of a suggestion that the laws of nature cannot be described as constant."  Every time we find something that looks non-constant, we find that we can identify a constant law of nature behind the changes.  So science assumes there exist laws of nature that are constant, and seeks them out.  This is why scientific experiments are always repeatable and reproducible.  If an experiment was not repeatable and reproducible, it would suggest the laws of nature change.
And we do have non-scientific human belief systems where the laws do change.  Typically these changes are called "miracles" in one way or another, and they represent something happening that "shouldn't" happen with the constant laws of nature.

Why after time shift we assume that the evolution of the system is governed by the same differential equation?

This is actually just a combination of the first two.  If you believe that there are only constant laws of nature, and adding $s$ to all time terms in an equation describing a system is the correct way to write time symmetry mathematically, then it is only natural to presume the evolution of the system will be governed by the same equation after the time shift.
A: Suppose you're performing an experiment. As an example, you launch a
ball with a gun. You record all data: inclination of the gun, quantity of powder, range observed. Or, if you have the necessary equipment, trajectory followed and times of ball passage in several points (I mean time passed from burst).
Then you repeat the experiment with the same initial data:
inclination, mass of powder. You observe that all other data (range,
trajectory) remain the same. Note that if things weren't so physics
would be impossible, since no experiment could be repeated to verify its outcome.
Let's translate all that into mathematics. If your experiment is
assumed to be governed by a differential equation, i.e. all variables $x$ of the experiment satisfy, as functions of time,
$$F(x, \dot x, \ldots, t) = 0.$$
Note that by writing this we're assuming a differential equation
exists, but are making no hypothesis on its dependence on $t$. If $F$ explicitly depended on $t$ the same experiment repeated at different times could give different results. 
What does your experiment say about $F\,$?
You gave initial data at time $t_0$: $x(t_0)=x_0$, etc. and observed
$x(t)$, say $x_1(t)$. Then you repeated the experiment at a different initial time $t_0+\tau$, with initial condition $x(t_0+\tau)=x_0$ and observed the at all times $x(t)$ is some $x_2(t)$ with 
$$x_2(t+\tau) = x_1(t).$$ 
This implies that $F$ doesn't explicitly depend on $t$.
A: ADD: I see there is more than one part to this question.

First, for the mathematical aspect:
The term "laws of nature" or "physical laws" refer to what a mathematician would call the evolution map of a dynamical system.
A "dynamical system", or DS, is basically a combination of a space $M$, typically some form of manifold, which is called the phase space and for which elements of represent possible present states or configurations of the dynamical system in question, a space $T$, called time space, which in this case is typically the real numbers $\mathbb{R}$ but in some cases is taken as a discrete space such as the natural numbers $\mathbb{N}$ or integers $\mathbb{Z}$, and represents the possible values that "time" can have, and finally a map $\Phi$, which is called the dynamics or "physical laws", if you want, of the system (also, time evolution map). The map $\Phi$ is just a mathematical function (though we might pose "reasonableness" stipulations like continuity, differentiability, etc.), and one which accepts both a time increment $\Delta t$ from $T$ and a configuration point $p$ from $M$, and returns what that point will evolve into after the lapse of $\Delta t$ worth of time. We typically write such as
$$\Phi^t(p)$$
or
$$\Phi(\Delta t, p)$$
depending. The key point is that also, this map can't just be arbitrary - in particular, it must satisfy two rules:
$$\Phi^0(p) = p$$
i.e. evolution by no time at all leaves the point unchanged, and
$$\Phi^{\Delta s}(\Phi^{\Delta t}(p)) = \Phi^{\Delta t + \Delta s}(p)$$
which basically says that if we evolve first by a time amount $\Delta t$, then evolve it some more by another amount $\Delta s$, that the total evolution produced should agree with that by evolving it straight off the bat by the sum-total of those amounts: $\Delta s + \Delta t$. The exponent notation is the one I personally prefer because the latter identity shows that this can be thought of as a sort of generalized compositional power or iteration of a function, so that the iteration count, or number of repeated applications of the function, may be a continuous parameter.
To formulate the concept of time symmetry in classical mechanics here requires a bit of care. For one, one should easily see from the above that an evolution map $\Phi$ as just defined will always respect a time translation, if you can also have negative times as well as positive. Thus, we need to be a bit choosey in what we call the manifold $M$. Generally, we should choose $M$ to also contain a time coordinate in itself - this can be understood by keeping in mind that the "time" in $\Phi$ actually is not a coordinate time, but rather a time increment given for evolution. If we want a time coordinate system, then we have to supply that in $M$. For example, for a single-particle classical system, we may take $M$ to be the set of all points
$$(t, \mathbf{r}, \mathbf{p})$$
where $\mathbf{r}$ and $\mathbf{p}$ are position and momentum and $t$ is the time. We may, say, define the dynamics of a free particle (i.e. no forces) thusly, where we take $p$ as having the form above, and $m$ is a pre-specified mass constant:
$$\Phi^{\Delta t}(p) := \left(t + \Delta t, \mathbf{r} + \frac{\mathbf{p}}{m} \Delta t, \mathbf{p}\right)$$
You should check that this satisfies the above laws to be a dynamical system. Basically, what it says is that a forward advance by $\Delta t$ time units should increase the time coordinate by $\Delta t$ (as should be expected), and it will move the particle by the amount given by its velocity maintained steadily over the given time.
Time symmetry now means this: If we take the time translation map
$$T_{\Delta t}(p) := (t + \Delta t, \mathbf{r}, \mathbf{p})$$
which shifts the spatial part of a given state $p$ up/down through time, then the evolution of that state, forward and backward, should be a time translation by the same map as the original state - that is, the evolution does not depend on the time coordinate assigned to the state, only the state itself:
$$\Phi^{\Delta s}(T_{\Delta t}(p)) = T_{\Delta t}(\Phi^{\Delta s}(p))$$
for every evolutionary increment, forward and backward, $\Delta s \in T$. Mathematically, the time translation and dynamical maps must commute. This is clearly a non-trivial statement since while that our evolution map ("law of physics") always must give the same evolution for any $M$-point and increment, a "point" now includes a time coordinate. For example, we may imagine that at a set time coordinate, instead we have that the particle in question receives a blow. The system is clearly not symmetric with regard to the time translation of the ball's evolution alone (i.e. ignoring the agent giving the external blow as being part of the system) and, of course, its energy is not conserved (which is accounted for in the real world by noting that an external blow makes it an "open system" and thus receives energy from that actor).

Second, for the philosophical aspect:
There is no reason we should "assume" it. We could, after all, live in a Universe where that, for example, on Mondays things decide to fall up for 1 second at the beginning of the day, then they drop back down again. That'd be very bad for us as we are now (though if this were our world, we'd have to be evolved/built in some fashion to take the "punishment", so then would not really be as "bad"), but it is an entirely logically consistent way that things could be. In this case, the dynamical map describing the physics would not translate as just mentioned - those "fall-ups" would act like the "blow" I talked about before, but they would not be the result of an extrinsic agent and rather intrinsic to the laws on which the Universe operated. If you shifted such an evolutionary path up/down on the time axis so that the "fall-ups" no longer coincided with the beginning of Monday, you'd have a situation that violated those laws because the laws say they must happen at that particular time.
The pure fact of the matter is we have no other reason, at least scientifically, to assume this other than that, from what we've observed, our Universe looks to be constructed to have no "preferred" points in time of this sort. We don't, at least as of yet, have any uncontested access to the mind of its Creator(s), or even can make an uncontestable case that any such do or do not exist, and/or what non-mental alternatives might be responsible for its existence in lieu of them, thus we have no way to answer this.

Finally, for the "differential equation" aspect:
This is really just a corollary of the mathematical definition. If the evolution map is given by a differential equation in the time coordinate, then for the map to yield the same trajectory after an arbitrary temporal translation, it will, logically have to be given by the same differential equation at all points in time.
