Iyfjky

Question

Let's assume we have a container, a jar, a can or whatever, which has a hole at its end. If there were water inside, via a differential equation we could calculate the time by which the container is empty.

But here is the thing: through the differential equation, with initial condition, I shall be able to know everything about the container: present past and future.

But let's assume I come and I find the container empty. Then

It could have always been empty

It could have been emptied in the past before my arrival

So this means I am not able to know, actually, all its story. Past present and future.

So it seems there is an absurdity in claiming that the solution of the differential equation is unique. Where am I wrong?

If you like this question you may also enjoy reading this Phys.SE post. — Sep 3 at 18:13

Schwern 3,05921022 · Answer 1 · 2018-09-03 19:15:20Z

You lack complete knowledge of the system you're asking about.

You only know about the jar. And if the complete system is the jar then yes, it has always been empty because there's nothing to interact with it.

But if you suppose there are things which may have been in the jar, now those are added to the system. That puddle of water on the floor, was it inside the jar? Now knowing only the state of the jar means your knowledge is incomplete. Now we need to include the state of the puddle of water. Also the floor its sitting on and the air in the room.

Here's a simpler example. Imagine an immovable box in space. A jar is floating in the center. There's also balls bouncing around in the box. The jar is stationary relative to the box. Has it always been so? We can't tell just from knowing the state of the jar. To answer that we'd need to know the state of the jar and anything it may have interacted with: the bouncing balls and the walls. Once we know the complete state of everything in the system, the box, the balls, the jar, and how they collide, we can know the history of the jar.

jim 2,417620 · Answer 2 · 2018-09-03 09:27:03Z

up vote
11
down vote

As you have noted, to solve the differential equation describing the system you need initial conditions. The condition "jar is empty" does not have a unique solution without specifying the conditions.

answered Sep 3 at 9:27

jim

2,417620

7

This is completely missing the point of the question. The initial condition in this question is "the jar is empty right now, at $t = 0$". It's just like saying "the temperature of the rod is $T_0$ at time $t = 0$" or "the particles position and velocity are $x$, $v$ at time $t = 0$". It is more subtle why this particular initial condition is not enough.
â€“Â knzhou
Sep 4 at 3:47

add a commentÂ |Â

akhmeteli 17.1k21739 · Answer 3 · 2018-09-03 09:39:40Z

The question is: how precise are your differential equations? Precise differential equations should also describe the state of the water after it left the container, and such equations can be reversible. So besides the empty container you would have falling water.

@Henry : Looks like you are talking about some specific equation, which describes the process only approximately. You have a similar situation when you approximate the exact dynamics by the Boltzmann equation and obtain irreversibility. — Sep 3 at 10:44

score 4 · Answer 4 · 2018-09-03 18:11:17Z

This is no different than saying "I found a rock on the ground. Maybe it fell, maybe it came rolling to a stop". Does that imply a lack of uniqueness in the equations of motion?

The answer is no. Mathematically, you get uniqueness in (some) differential equations when certain initial conditions are fixed. In the case of the equations of motion, that would be the initial position and the initial velocity.

Cort Ammon 21.9k34570 · Answer 5 · 2018-09-04 05:57:58Z

There's a slight adjustment needed. You don't need initial conditions, you need boundary conditions. For conservative systems, having a snapshot in time is enough, but you have a system that is intentionally losing mass. As you let this mass leave the system, it takes along with it the information you needed.

If you had the boundary conditions, including all of the information about the water as it leaves over time, then you could put together the history that you seek.

Of course, even in this case, it's worth considering funny corner cases such as Norton's Dome. The discussion of the validity of this model is nuanced, to this day.

Eric Towers 86946 · Answer 6 · 2018-09-03 18:09:22Z

Picking a random (but entirely representative of those which assert uniqueness) example of a relevant theorem, what part of the Picard-LindelhÃƒÂ¶f theorem promises you can extend the solution from the initial condition to all of the "present past and future"? Certainly, you can extend to an open interval of times containing the time of the initial condition, but that interval may be surprisingly short and you may have no way to extend the interval further. (This is fairly likely to happen at (various order) discontinuities induced by additional constraints, for instance, the upper bound on the capacity of the can or the non-negativity constraint on the water quantity in the can.)

The_Sympathizer 2,510520 · Answer 7 · 2018-09-04 01:41:35Z

The differential equation, when used as a physical model, depend on a certain set of assumptions. For example, when we use the differential equation

$$h''(t) = -g$$

to describe the height of a thrown ball, we may set an initial height $h_0$, but the projection into the past would have the ball as coming from infinitely far below ($h = -infty$ at $t = -infty$), which makes no sense in real life. Clearly the ball did not surge up from below the ground, phasing through the matter, but rather was projected from your hand! The problem is that the equation has an assumption in it: namely that the only force acting is gravity, and when you project backwards you are projecting this assumption backwards. But in a likely reality, in the past, when you were getting ready to throw that ball there were other forces - namely from your hands, pockets, etc. - acting upon it, and they are not included in the equation.

The differential equation gives both a unique future and unique past history but that history, in both directions, only corresponds to reality so long as the equation's assumptions hold true, and the same also holds for the future as well, e.g. if a bird comes and knocks it while in flight then this equation won't hold either.

In your case, you cannot write down a differential equation that would be valid at all past times without knowing what assumptions are needed to correctly model the past behavior, that is, without some idea of what the past influences were. If the conditions have always been nothing has interacted with the jar then the differential equation will give that result. If that assumption is wrong, then of course, it won't work. The non-uniqueness comes out of effectively variation in the equation itself, rather than in any individual equation failing to prescribe a unique past history. You could say that in the past, the differential equation was different. If you're allowed to do that then of course multiple past histories can end up with the same starting point - the theorem they don't is dependent on using the same equation to go backward!

Abhimanyu Pallavi Sudhir 4,29142243 · Answer 8 · 2018-09-03 09:46:25Z

up vote
1
down vote

"The jar is empty at present" just tells you $f(0)$. You also need $f'(0)$, $f''(0)$, etc.

answered Sep 3 at 9:46

Abhimanyu Pallavi Sudhir

4,29142243

3

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â€“Â John Rennie
Sep 3 at 11:31

We also need to know if there was water at any time in the past, where is the water now. So even $f(0)$ is not completely known.
â€“Â Prakhar Gupta
Sep 3 at 15:38

2

@PrakharGupta You don't really need that. You can always take a cross-section of the function across the t-axis.
â€“Â Abhimanyu Pallavi Sudhir
Sep 3 at 19:01

3

@JohnRennie How does it not? The point is that even something as simple as a Taylor series does not let you predict the function based on just the value of the function at present.. If you have an objection to the answer, phrase it precisely, don't copy and paste standard templates please.
â€“Â Abhimanyu Pallavi Sudhir
Sep 3 at 19:03

2

@AbhimanyuPallaviSudhir that post is not copy pasted. It is auto generated when a delete vote is placed.
â€“Â The Great Duck
Sep 4 at 6:29

add a commentÂ |Â

gnasher729 23214 · Answer 9 · 2018-09-03 15:39:30Z

When you try to solve differential equations back in time, You usually run into the problem of unstable solutions. When dropping a stone, no matter where you start, the solution converges to Ã¢Â€Âœlies motionless on the groundÃ¢Â€Â. If you calculate backwards, the solution diverges. There are many different speeds that the stone could have had ten seconds ago if it lies still now.

Winther 58439 · Answer 10 · 2018-09-03 21:30:04Z

An ODE for some physical situation is just some mathematical model of reality. Such models always have limitations and almost always only holds in a probabilistic sense.

Let's take a very simple example and assume the container is filled with atoms of some radioactive material that decays with a half-life $T$. The ODE describing the expected number of atoms in the jar is $N'(t) = -fracN(t)T$ with solution $N(t) = N(0)e^-t/T$ which is uniquely determined by specifying $N(0)$. If we wait for a long time then eventually $N(t) ll 1$ for which we don't expect to find any radioactive atoms in the jar. So even if we start of with $N(0) = 1000$ or $N(0) = 2000$ atoms we will in both cases end up with an empty jar for large $t$.

This does not contradict uniqueness of the solution to the respective ODE since the ODE describes the expected number (a probabilistic quantity) not the actual number of atoms. The solution to the ODEs will have a non-zero value for any time $t$ even if the jar is empty (but we can't access this value with our one observation of the system).

Aaron Stevens 2,689319 · Answer 11 · 2018-09-03 19:34:14Z

To helpfully summarize some answers here and give my own thoughts (even though there are a lot of answers right now): Basically your system is not just the jar. Let's look at some steps of what happens.

You have your jar of water.

You put a hole in it.

The water leaks out of the jar.

Air is pushed out of the way as water falls to the floor. Air also
fills the jar.

The floor absorbs energy from the water as it hits the ground.

The water is now a puddle on the floor.

Other things probably happen too

Now you leave and someone else comes in an sees a puddle of water on the floor. If this person was able to take into account everything that happened, then they could discern that the water started in the jar and did not come from another place (a leak in the roof for example). If this person could see the trajectories of all of the water molecules, the air molecules, the floor molecules, the jar molecules, etc. and knew how each of these evolved, then they could "play back" everything in time and see that the water did in fact start in the jar.

Of course this is impossible. We do not have the capabilities to do this. We must work with limited knowledge and limited equations. So if we are dealing with just a simple rate equation that describes how fast water leaves the jar and nothing else, then there are in fact multiple scenarios that lead to an empty jar (for example, we could have started with different volumes of water).

This is not a physical issue. This is an issue in our knowledge of the system and the equations that govern this system. As stated above, with perfect knowledge of the system at some time we would know exactly how the water left the jar and there would not be multiple "solutions". In choosing a model for a system we must take this into account. Are the simplifications we have made justified for the questions we are asking of the model? If we just want to know about water leaving a jar, then a simple model is great. However, if we want to know where water came from whenever we see a puddle below a jar with a hole in it, then we better think of a different model.

The Great Duck 1095 · Answer 12 · 2018-09-04 06:34:10Z

Everyone keeps saying all these things but really the issue here is so much more succinct and I think it has a lot more to do with science itself. The purpose of science is to construct models that attempt to predict future behavior based on previously observed behavior. So while yes the differential equation might be able to predict past motion the problem here is that the universe nor its model is necessarily reversible. Nobody has proven nor claimed afaik that any given state of the universe has a unique previous state. In fact, I would claim there isn't such a state. Therefore while your bucket is an analogy I would say that it shows there is unique future behavior and NOT unique past behavior. Of course, there is also the issue that you aren't modelling everything perfectly. There would be evidence to suggest the puddle came from the bucket or whatever such as ripples or the bucket being wet or whatever else would indicate such things.

Pete Kirkham 29717 · Answer 13 · 2018-09-04 12:26:48Z

So it seems there is an absurdity in claiming that the solution of the differential equation is unique. Where am I wrong?

You seem to be taking that the fact that an equation has a unique solution to imply that that equation is the only one with that solution.

An even simpler example:

The solution of $x=1+2$ is unique - there is only one value of $x$ which satisfies the equation, and that value is $3$.

The solution of $x=4-1$ is also unique, and its also has a unique solution where the value of $x$ is $3$.

Given only the statement that the value of $x$ is $3$, you do not know which equation this was a solution of.

The fact that an equation has a unique solution does not imply that that particular equation is the only one which yields that solution; there will be infinite such equations for which the same state is their unique solution.

Edgar Vardanyan 1 · Answer 14 · 2018-09-05 13:29:37Z

Now imagine you have a jar, and there is a drop of water moving vertically behind the hole. Can you solve this one provided you have the coordinates and the velocity of the drop. Yes, you can. The only difference is that the initial state of the jar is not enough for solving the (jar, water) system, you need the information about water.

Schwern 3,05921022 · Answer 15 · 2018-09-03 19:15:20Z

You lack complete knowledge of the system you're asking about.

You only know about the jar. And if the complete system is the jar then yes, it has always been empty because there's nothing to interact with it.

But if you suppose there are things which may have been in the jar, now those are added to the system. That puddle of water on the floor, was it inside the jar? Now knowing only the state of the jar means your knowledge is incomplete. Now we need to include the state of the puddle of water. Also the floor its sitting on and the air in the room.

Here's a simpler example. Imagine an immovable box in space. A jar is floating in the center. There's also balls bouncing around in the box. The jar is stationary relative to the box. Has it always been so? We can't tell just from knowing the state of the jar. To answer that we'd need to know the state of the jar and anything it may have interacted with: the bouncing balls and the walls. Once we know the complete state of everything in the system, the box, the balls, the jar, and how they collide, we can know the history of the jar.

jim 2,417620 · Answer 16 · 2018-09-03 09:27:03Z

up vote
11
down vote

As you have noted, to solve the differential equation describing the system you need initial conditions. The condition "jar is empty" does not have a unique solution without specifying the conditions.

answered Sep 3 at 9:27

jim

2,417620

7

This is completely missing the point of the question. The initial condition in this question is "the jar is empty right now, at $t = 0$". It's just like saying "the temperature of the rod is $T_0$ at time $t = 0$" or "the particles position and velocity are $x$, $v$ at time $t = 0$". It is more subtle why this particular initial condition is not enough.
â€“Â knzhou
Sep 4 at 3:47

add a commentÂ |Â

akhmeteli 17.1k21739 · Answer 17 · 2018-09-03 09:39:40Z

The question is: how precise are your differential equations? Precise differential equations should also describe the state of the water after it left the container, and such equations can be reversible. So besides the empty container you would have falling water.

@Henry : Looks like you are talking about some specific equation, which describes the process only approximately. You have a similar situation when you approximate the exact dynamics by the Boltzmann equation and obtain irreversibility. — Sep 3 at 10:44

score 4 · Answer 18 · 2018-09-03 18:11:17Z

This is no different than saying "I found a rock on the ground. Maybe it fell, maybe it came rolling to a stop". Does that imply a lack of uniqueness in the equations of motion?

The answer is no. Mathematically, you get uniqueness in (some) differential equations when certain initial conditions are fixed. In the case of the equations of motion, that would be the initial position and the initial velocity.

Cort Ammon 21.9k34570 · Answer 19 · 2018-09-04 05:57:58Z

There's a slight adjustment needed. You don't need initial conditions, you need boundary conditions. For conservative systems, having a snapshot in time is enough, but you have a system that is intentionally losing mass. As you let this mass leave the system, it takes along with it the information you needed.

If you had the boundary conditions, including all of the information about the water as it leaves over time, then you could put together the history that you seek.

Of course, even in this case, it's worth considering funny corner cases such as Norton's Dome. The discussion of the validity of this model is nuanced, to this day.

Eric Towers 86946 · Answer 20 · 2018-09-03 18:09:22Z

Picking a random (but entirely representative of those which assert uniqueness) example of a relevant theorem, what part of the Picard-LindelhÃƒÂ¶f theorem promises you can extend the solution from the initial condition to all of the "present past and future"? Certainly, you can extend to an open interval of times containing the time of the initial condition, but that interval may be surprisingly short and you may have no way to extend the interval further. (This is fairly likely to happen at (various order) discontinuities induced by additional constraints, for instance, the upper bound on the capacity of the can or the non-negativity constraint on the water quantity in the can.)

The_Sympathizer 2,510520 · Answer 21 · 2018-09-04 01:41:35Z

The differential equation, when used as a physical model, depend on a certain set of assumptions. For example, when we use the differential equation

$$h''(t) = -g$$

to describe the height of a thrown ball, we may set an initial height $h_0$, but the projection into the past would have the ball as coming from infinitely far below ($h = -infty$ at $t = -infty$), which makes no sense in real life. Clearly the ball did not surge up from below the ground, phasing through the matter, but rather was projected from your hand! The problem is that the equation has an assumption in it: namely that the only force acting is gravity, and when you project backwards you are projecting this assumption backwards. But in a likely reality, in the past, when you were getting ready to throw that ball there were other forces - namely from your hands, pockets, etc. - acting upon it, and they are not included in the equation.

The differential equation gives both a unique future and unique past history but that history, in both directions, only corresponds to reality so long as the equation's assumptions hold true, and the same also holds for the future as well, e.g. if a bird comes and knocks it while in flight then this equation won't hold either.

In your case, you cannot write down a differential equation that would be valid at all past times without knowing what assumptions are needed to correctly model the past behavior, that is, without some idea of what the past influences were. If the conditions have always been nothing has interacted with the jar then the differential equation will give that result. If that assumption is wrong, then of course, it won't work. The non-uniqueness comes out of effectively variation in the equation itself, rather than in any individual equation failing to prescribe a unique past history. You could say that in the past, the differential equation was different. If you're allowed to do that then of course multiple past histories can end up with the same starting point - the theorem they don't is dependent on using the same equation to go backward!

Abhimanyu Pallavi Sudhir 4,29142243 · Answer 22 · 2018-09-03 09:46:25Z

up vote
1
down vote

"The jar is empty at present" just tells you $f(0)$. You also need $f'(0)$, $f''(0)$, etc.

answered Sep 3 at 9:46

Abhimanyu Pallavi Sudhir

4,29142243

3

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â€“Â John Rennie
Sep 3 at 11:31

We also need to know if there was water at any time in the past, where is the water now. So even $f(0)$ is not completely known.
â€“Â Prakhar Gupta
Sep 3 at 15:38

2

@PrakharGupta You don't really need that. You can always take a cross-section of the function across the t-axis.
â€“Â Abhimanyu Pallavi Sudhir
Sep 3 at 19:01

3

@JohnRennie How does it not? The point is that even something as simple as a Taylor series does not let you predict the function based on just the value of the function at present.. If you have an objection to the answer, phrase it precisely, don't copy and paste standard templates please.
â€“Â Abhimanyu Pallavi Sudhir
Sep 3 at 19:03

2

@AbhimanyuPallaviSudhir that post is not copy pasted. It is auto generated when a delete vote is placed.
â€“Â The Great Duck
Sep 4 at 6:29

add a commentÂ |Â

gnasher729 23214 · Answer 23 · 2018-09-03 15:39:30Z

When you try to solve differential equations back in time, You usually run into the problem of unstable solutions. When dropping a stone, no matter where you start, the solution converges to Ã¢Â€Âœlies motionless on the groundÃ¢Â€Â. If you calculate backwards, the solution diverges. There are many different speeds that the stone could have had ten seconds ago if it lies still now.

Winther 58439 · Answer 24 · 2018-09-03 21:30:04Z

An ODE for some physical situation is just some mathematical model of reality. Such models always have limitations and almost always only holds in a probabilistic sense.

Let's take a very simple example and assume the container is filled with atoms of some radioactive material that decays with a half-life $T$. The ODE describing the expected number of atoms in the jar is $N'(t) = -fracN(t)T$ with solution $N(t) = N(0)e^-t/T$ which is uniquely determined by specifying $N(0)$. If we wait for a long time then eventually $N(t) ll 1$ for which we don't expect to find any radioactive atoms in the jar. So even if we start of with $N(0) = 1000$ or $N(0) = 2000$ atoms we will in both cases end up with an empty jar for large $t$.

This does not contradict uniqueness of the solution to the respective ODE since the ODE describes the expected number (a probabilistic quantity) not the actual number of atoms. The solution to the ODEs will have a non-zero value for any time $t$ even if the jar is empty (but we can't access this value with our one observation of the system).

Aaron Stevens 2,689319 · Answer 25 · 2018-09-03 19:34:14Z

To helpfully summarize some answers here and give my own thoughts (even though there are a lot of answers right now): Basically your system is not just the jar. Let's look at some steps of what happens.

You have your jar of water.

You put a hole in it.

The water leaks out of the jar.

Air is pushed out of the way as water falls to the floor. Air also
fills the jar.

The floor absorbs energy from the water as it hits the ground.

The water is now a puddle on the floor.

Other things probably happen too

Now you leave and someone else comes in an sees a puddle of water on the floor. If this person was able to take into account everything that happened, then they could discern that the water started in the jar and did not come from another place (a leak in the roof for example). If this person could see the trajectories of all of the water molecules, the air molecules, the floor molecules, the jar molecules, etc. and knew how each of these evolved, then they could "play back" everything in time and see that the water did in fact start in the jar.

Of course this is impossible. We do not have the capabilities to do this. We must work with limited knowledge and limited equations. So if we are dealing with just a simple rate equation that describes how fast water leaves the jar and nothing else, then there are in fact multiple scenarios that lead to an empty jar (for example, we could have started with different volumes of water).

This is not a physical issue. This is an issue in our knowledge of the system and the equations that govern this system. As stated above, with perfect knowledge of the system at some time we would know exactly how the water left the jar and there would not be multiple "solutions". In choosing a model for a system we must take this into account. Are the simplifications we have made justified for the questions we are asking of the model? If we just want to know about water leaving a jar, then a simple model is great. However, if we want to know where water came from whenever we see a puddle below a jar with a hole in it, then we better think of a different model.

The Great Duck 1095 · Answer 26 · 2018-09-04 06:34:10Z

Everyone keeps saying all these things but really the issue here is so much more succinct and I think it has a lot more to do with science itself. The purpose of science is to construct models that attempt to predict future behavior based on previously observed behavior. So while yes the differential equation might be able to predict past motion the problem here is that the universe nor its model is necessarily reversible. Nobody has proven nor claimed afaik that any given state of the universe has a unique previous state. In fact, I would claim there isn't such a state. Therefore while your bucket is an analogy I would say that it shows there is unique future behavior and NOT unique past behavior. Of course, there is also the issue that you aren't modelling everything perfectly. There would be evidence to suggest the puddle came from the bucket or whatever such as ripples or the bucket being wet or whatever else would indicate such things.

Pete Kirkham 29717 · Answer 27 · 2018-09-04 12:26:48Z

So it seems there is an absurdity in claiming that the solution of the differential equation is unique. Where am I wrong?

You seem to be taking that the fact that an equation has a unique solution to imply that that equation is the only one with that solution.

An even simpler example:

The solution of $x=1+2$ is unique - there is only one value of $x$ which satisfies the equation, and that value is $3$.

The solution of $x=4-1$ is also unique, and its also has a unique solution where the value of $x$ is $3$.

Given only the statement that the value of $x$ is $3$, you do not know which equation this was a solution of.

The fact that an equation has a unique solution does not imply that that particular equation is the only one which yields that solution; there will be infinite such equations for which the same state is their unique solution.

Edgar Vardanyan 1 · Answer 28 · 2018-09-05 13:29:37Z

Now imagine you have a jar, and there is a drop of water moving vertically behind the hole. Can you solve this one provided you have the coordinates and the velocity of the drop. Yes, you can. The only difference is that the initial state of the jar is not enough for solving the (jar, water) system, you need the information about water.

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