# Poincare's method in perturbation theory

1. Jul 10, 2008

### The-herod

First of all, I'm sorry about the last topic, accidentally I switched between the previous message and this one... Sorry about the troubles. I think it's the right forum (after reading a bit), sorry if I'm wrong...

My high school graduation project is about the perturbation theory, it's application in free fall and a comparison between the efficiency of several methods. The first method I'm learning is Poincare's method, and there are a few points I don't understand. I'll be much thankful if you could help me with it, or reference me to some textbook, article or book that explains it nice and clear (I couldn't find one...).

$${\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} = \mathbf{F}(\mathbf{x},t,\epsilon)$$
We say that
$$\mathbf{x}(t,\epsilon)=\sum_{k=0}^\infty \mathbf{x}_{(k)}(t)\cdot\epsilon^k$$
We substitute the last equation in the first one.
• Now comes the confusing part. Why do we compare between the coefficient of the same power of epsilon on both sides? It's just one solution of the equation, isn't it? I mean, theoretically, there are also other solutions for this equation, and we do it just for the comfort, to get a simple ODE of one variable for each x(k), am I wrong?
• The second question is - what is, exactly, epsilon? I know it's a small parameter, and that the accuracy of the solution after we find the nth first x(k) is $$O(\epsilon^n)$$, so the smaller epsilon is, the greater the accuracy of the solution is. So why not just choose epsilon=0 (it gives us the equation we had in the beginning)?
Which leads me to the last and most basic question...
• Why do we do these all from the beginning? We only get multiple ODEs instead of one, so how does it help us? If we know how to solve the equation for x(k), we'll probable know how to solve also the first equation, for x, don't we?
Am I missing anything...? (I guess I am...)

I know it's a lot of question, so as I said in the beginning, I'll much appreciate any reference to any good resource.

Thanks a lot! (and sorry for the long message)

2. Jul 10, 2008

### Crosson

Very good questions!

Consider this equation:

$$a_2 x^2 + a_1 x + a_0 = b_2 x^2 + b_1 x + b_0$$

The equality can only be satisfied when $a_i = b_i$ for $i = 1,2,3$. In other words, two polynomials over a variable $x$ are equal if and only if their corresponding coefficients are equal. This also hold true for power series, which are like infinite polynomials in an algebraic sense.

Epsilon is, literally, a perturbation. It represents a deviation away from a known problem. The situation is that we have a solution to the easy problem A, and we want a solution to a more complicated version of the same problem A*. If we can't solve A* exactly, the next best thing is to make the problem A look more like the problem A* by perturbing it in a sensible way.

Most nonlinear differential equations cannot be solved exactly, so the only options are to do numerical solutions on the computer or to perturb the simple linear problems that we can solve.

3. Jul 10, 2008

### The-herod

Thank you very much!
I understand it now theoretically (after two months with my bit lazy tutor :P). Though, even without get into the math (if it's possible), I still don't see how do I imply it to a practical problem, like free fall.
If we take a free fall as an example, I guess that the simple known problem is when the only fore is the gravity force, and the perturbation (at least in case we neglect the friction) is the Coriolis effect the the centrifugal force. So the main force is
$$\mathbf{F_m}=-G\frac{Mm}{\vert \mathbf{R} \vert}$$
and the perturbation is
$$\mathbf{F_p}= -m\mathbf{\omega} \times ( \mathbf{\omega} \times \mathbf{r}) -2m\mathbf{\omega} \times \mathbf{v}$$
So... what's next? we take Fp as epsilon? I'm sorry if it's a too basic question, but I'm a bit confused...:shy:
I used it just as an example, I don't ask for the calculations of this example or something. Just to understand how to imply it to a practical problem in general.

Thanks a lot again!

4. Jul 10, 2008

### Crosson

Instead of gravity, vectors, and rotation, let's look at a simple harmonic oscillator:

$$x'' + x = 0$$

We know the solution:

$$x(t) = C_1 Sin(t) + C_2 Cos(t)$$

Now consider a duffing oscillator:

$$x'' + x + x^3 = 0$$

The physical interpretation is that the spring in this oscillator does not obey Hooke's law exactly, we are including a nonlinear term to the spring force. This can be handled with perturbation theory:

$$x'' + x + \epsilon x^3 = 0$$

In this case we refer to it as a weakly-nonlinear duffing oscillator. Another example where this can be done is called the Van Der Pol oscillator:

$$x'' + x + \epsilon (x^2 - 1)x' = 0$$

5. Jul 10, 2008

### The-herod

Thank you very much! again:)
I think I understand, but just to make sure... only a few more questions, please.

This is the simple known problem.

So now we actually estimate the perturbation from the first problem, as a function of epsilon?

And... what's next? if we define x as a series, as before, we get some... interesting results.

$$x=x_0+\epsilon x_1+\epsilon^2 x_2+\epsilon^3 x_3+\ldots$$
$$x''_0+\epsilon x''_1+\epsilon^2 x''_2+\epsilon^2 x''_2+\ldots + x_0+\epsilon x_1+\epsilon^2 x_2+\epsilon^2 x_2+\ldots + \epsilon x^3_0 + \epsilon^3x_0 x^2_1 \ldots$$
so when we compare between the coefficient of epsilon3, for example, we have nonlinear terms, like
$$x_0 x^2_1$$

Where did I go wrong...?:-/

Thanks a lot!

6. Jul 10, 2008

### Crosson

You didn't go wrong, perturbation theory is an iterative process. By the time you are calculating $x_3$ you have already solved for $x_0,x_1,x_2$ as functions of time, they are not unknown functions at that point.

When you substitute the series into the equation, don't include terms higher then first order in epsilon (just truncate them, no ellipsis '...', that's why it is an approximation). Then see what equation you get for x_1.

7. Jul 11, 2008

### The-herod

Sorry I didn't answer earlier, I was asleep...

I think I got it now.
For every problem it's necessary to find it's own "develop" in perturbation theory, so it will fit to the method (means, it will be a linear equation each step). In other words, you have to choose what you consider as a perturbation right (Obviously...), otherwise this method does not ensure you only linear equations.

Two more, hm... technical question, please:)
In my previous example (this is my task now, after I finally got it!:) ), of free fall, is this what I should work with?
$$\Sigma\mathbf{F}=\mathbf{F_g}+\epsilon(\mathbf{F_{Coriolis}}+\mathbf{F_{Centrifugal}})$$
The second question is, after I multiple in the last two forces (Coriolis and the centrifugal), I get a function of the components of r, instead a function of r, because
$$\mathbf{a}\times\mathbf{b}=(a_y b_z-a_z b_y, a_z b_x - a_x b_z, a_x b_y - a_y b_x)$$
so after I compare between the same powers of epsilon, I should also compare between the same component (left side x component against right side x component etc), and get a system of three linear equations?

It's a bit sad I owe you much more then I owe to my tutor...:uhh:

Once again...
Thanks a lot!

8. Jul 12, 2008

### The-herod

I'm sorry to annoy.
But all I need is a V/X, please, so I'll know if what I'm doing is right...

Sorry for being a bit bothersome... And thanks a lot again.

Last edited: Jul 12, 2008
9. Jul 12, 2008

### Crosson

Don't worry, it's not a bother, I just hadn't seen your most recent question.

That is definitely a reasonable start, although you may want to work on the coriolis force and the centrifugal force separately. For example on the earth we have:

$$\frac{F_{Cor}}{F_{cf}} \approx \frac{v}{V}$$

Where $v$ is the velocity of the object, and $V$ is the velocity of a point on the earth's equator, due to the rotation of the earth.

Yes, exactly. In general an n-dimensional vector equation is equivalent to n scalar equations. In 3D you will always get 3 equations.

I want to make a comment, since you are talking about the coriolis and centrifugal forces, that usually calculations done with these forces are in spherical coordinates. That means that instead of x,y,z we have a radius coordinate and two angular coordinates to describe positions in space.

I also want to comment that even with the coriolis and centrifugal forces, Newton's law equation is still linear, and so it is possible to solve exactly for these types of problems. That doesn't mean that perturbation theory won't work, however, and it gives you a chance to compare the perturbation solution with the exact solution. As a fine example of this, you could attempt perturbation theory on the following equation:

$$x'' + 2 \epsilon x' + x = 0$$

The factor of 2 is only there for convenience, and this is of course a simple harmonic oscillator with weak damping. The solution from simple perturbation theory compares well with the exact solution for short times, but eventually the approximation becomes so inaccurate as to be useless.

10. Jul 12, 2008

### The-herod

Thanks a lot!

Actually, this is the mission that my tutor gave me for now. I already calculated by hand, the Jacobian of this problem, to simplify the solution with a linearization. Found out to be a wrong one, after I actually understood the material, thanks to you.

That means that for start, I can use it so simplify the problem only to one perturbation, right? (by expressing one force with the other)

I'm a bit familiar with such coordinates from Astrophysics. But how can it be used in cross product?
Simply build the vector for the product, from terms like r*cos(a)*sin(b) (where a,b are the angular coordinates), or is there a more sophisticated method?

Never heard it from my tutor before. Actually, he told it can't be solved analytically... (Have I mentioned he's not the best tutor already?;):P)
I'll ask my physics teacher about that, maybe he could show me how to solve it, so I could compare it to my solutions of the perturbations' method, as you suggest. (I guess it's too big, and requires too much background for my level, to be discussed here...).

But if we take, lets say... 5th order approximation, instead of first order, it should fit well to the exact solution, for a longer period, doesn't it? So we can fit the approximation to our needs, by choosing the order of approximation, right?

By the way, if you'll agree, I'll ask you for some details (you name, mainly) in the future (I'll probably lose them now), to add you to the "with Thanks to..." list when I'll finish my work (it supposes to be a big work... about 60 pages). Unless formalism etc, you are suppose to be the first one in the list, for being the only one who actually helped me with this work until now...

Thanks a lot again

Last edited: Jul 12, 2008
11. Jul 12, 2008

### Crosson

Remember that the relationship I gave between the forces is only approximate, so its probably better to continue working with the exact expressions. The reason I showed you that relationship is to see that, in normal conditions on the earth, the centrifugal force is orders of magnitude smaller then the Coriolis force. This means that you could either (1) ignore the centrifugal force as a first approximation or (2) include the centrifugal force as a 'second-order perturbation', which means:

$$\Sigma\mathbf{F}=\mathbf{F_g}+\epsilon\mathbf{F_{Coriolis }}+ \epsilon^2\mathbf{F_{Centrifugal}}$$

If the velocity of the rotating frame is $\Omega$, then the centrifugal force is proportional to $\Omega^2$ and the Coriolis force to $\Omega$ so the setup above is equivalent to doing a perturbation expansion in powers of $\Omega$, which is probably what your tutor/teacher is thinking.

Your right, spherical coordinates may be more useful for a problem involving planetary motion (in this case you procedure for finding the cross product is correct). For motion in the Earth's rotating frame, it might be best to continue using rectangular coordinates (it looks like you have a specific problem in mind).

Motion in a constantly rotating frame with Coriolis and Centrifugal forces lead to a system of 3 linear differential equations with constant coefficients. We can solve any system of equations of this form in terms of sines, cosines, and exponentials. Unfortunately, just to write down the solution in the general case takes about one page of dense expressions.

Unless your teacher is savvy with the computer, he won't be able to solve it either, since it takes too long and involves too much writing. The method is, however, straightforward, and you can learn it later on involving matrices and eigenvalues. In my case I avoid writing with my hand due to injuries, so I just use Mathematica on the computer to do these kinds of things.

Not necessarily, perturbation theory is more limited then that. You cannot always get arbitrary accuracy by increasing the number of terms, the way that you can with a Taylor or Fourier series.

The truth is that perturbation theory is an art form, by which I mean there is no prescribed method and lots of creativity is useful. For example, in the weakly damped harmonic oscillator, notice that there are two time scales. The fast time scale covers the oscillations. The slow time scale covers the eventual decrease in amplitude (remember the damping is weak). We can improve the approximation by accounting for these two time scales (the method is called two-timing) moreso then by increasing the number of terms with the basic method.

Thank you for your appreciation, it has been a pleasure to help you so far since you ask good questions and take the time to understand the answers. Let me know when you want the info.

12. Jul 13, 2008

### The-herod

You're right, I didn't noticed that in that context. I guess I'll develop sensitivity for that with time... However, wouldn't it be a bad approximation in such a case? because $$\Omega$$ is not a very small parameter, definitely bigger than 1, so the series supposes to diverge, no? (because in opposite to cases when $$\epsilon<1$$, when it's bigger than one, $$f(n)=\epsilon^n$$ is a monotonic increasing function)

I have a specific problem just for now. I want to finish it so my tutor will be satisfied and get off my back, and I could find what most interest me. It supposes to be my main aim, to explore some free fall problems, and show that the approximations usually done, of a homogeneous gravitational field etc are OK, but I rather to try it in addition to other things, and see what most interest me.

Actually, both of them are savvy with computers. One with Matlab (His doctorate is about neuron science. A lot of simulations...) , and the other one with Mathematica, so I guess they could handle it. It worths at least a try ;)

Is this because what I asked about above, that sometime the series diverges?

Great. Now I know what to learn next (after I'll finish with Poincare's method). Thanks.

Thank you for you help. I was still trying to substitute the series of epsilon in x, without a perturbation in the equation without you help... :P
So thanks again:)

13. Jul 13, 2008

### Crosson

My bad, I meant that $\Omega$ was the angular velocity, expressed in rad/sec.

Sometimes this is the reason, but not always. For example, the two-timed solution to the weakly damped harmonic oscillator does not diverge, but it is eventually so inaccurate as to be useless for the following reason: the exact solution has a slightly modified frequency from that of the undamped oscillator. Over time the error due to difference in frequency becomes significant, which is a third extremely long time scale.

14. Jul 13, 2008

### The-herod

Mine too. I thought you meant to it, but I thought about it as velocity (instead of angular velocity...). OK, so it is small, I'll use it, thanks:)

I see now. Which is why the two-timing works much better.

I have no more questions at the moment, but I warn, I'll probably be back in the near future;)

15. Jul 19, 2008

### The-herod

And... I'm back again.
I think, technically, this message should be in the h.w. forum, but consider the fact that it's a continuance of the previous messages... I thought I should put it here. If I should still move it to the h.w. forums, please give me a note.

I tried to solve the question I asked about last time, a gravitational field with Coriolis and centrifugal forces. I calculated only the first order, using Poincaré's method, but I got non linear differential equations, which I think isn't very good...
For example, I got the following equation. r0 is a parameter, a vector from the center of earth to some point on its surface. Phi is the latitude. Omega is earth's angular velocity.
$$-\ddot r_x = \frac{GM}{\sqrt{r_x^2+r_z^2+r_y^2+2r_yr_0+r_0^2}}r_x+2\Omega \sin(\varphi) \dot r_z + \Omega^2(\frac{\sin(2\varphi)}{2}(r_y+r_0)-\sin^2(\varphi)r_x)$$
From this I got
$$-GMr_x=\sqrt{r_x^2+r_z^2+r_y^2+2r_yr_0+r_0^2}\cdot[\ddot r_x + 2\Omega \sin(\varphi) \dot r_z + \Omega^2(\frac{\sin(2\varphi)}{2}(r_y+r_0)-\sin^2(\varphi)r_x)]$$
Now canceling the square root -
$$(GM)^2r^2_x=(r_x^2+r_z^2+r_y^2+2r_yr_0+r_0^2)\cdot[\ddot r_x + 2\Omega \sin(\varphi) \dot r_z + \Omega^2(\frac{\sin(2\varphi)}{2}(r_y+r_0)-\sin^2(\varphi)r_x)]^2$$
Now if we substitute for rx,y,z its series ($$r_0+\epsilon r_1 + ...$$), and take the first order (no epsilon), we'll get
$$G^2M^2r^2_{x,0}=(r^2_{x,0}+r^2_{z,0}+r^2_{y,0}+2r_{y,0}r_0+r^2_0)\cdot \ddot r^2_{x,0}$$
As far as I understood, we should get linear equations with the perturbation theory... But I can't find where did I go wrong... :-/

Thanks a lot! (and sorry again if I should have write it in the h.w. forum).

Last edited: Jul 19, 2008
16. Jul 20, 2008

### The-herod

After thinking about it a little bit more, I understand that even without the perturbation, the equation is non-linear, though I as much as I know, it is possible to solve it. So I should obviously get nonlinear equations. Does that mean that equation should be able to be solved, though it is nonlinear?

I also found a mistake, the root should be the power of 3/2, not 1/2... so the fixed equation is:
$$G^2M^2r^2_{x,0}=(r^2_{x,0}+r^2_{z,0}+r^2_{y,0}+2r_{y,0}r_0+r^2_0)^3\cdot \ddot r^2_{x,0}$$

Thanks!

Last edited: Jul 20, 2008
17. Jul 22, 2008

### Crosson

Hi Herod,

I was absent from the forum for a week, I'm sorry that I missed your question. Anyway, the answer to your problem is to use the following equation:

$F_g = -m g$

for projectile motion, instead of using Newton's equation of gravitation. Remember that the perturbation you are interested in comes from the coriolis + centrifugal forces. It is a different problem entirely to calculate a solution to projectile motion using Newton's equation.

18. Jul 22, 2008

### The-herod

So I guess you deserve a welcome back!

No problem... I should be the one thanking you for answer me in such consistency.

Thanks a lot:) That's what I did for in the meanwhile. Now I know that's all I had to do.
I don't really see the point of using perturbation here (as you said, it's easy to solve it analytically). Oh well, I think maybe my next task will be to add friction, so it won't be pointless...

Oh, and please, you can call me Gal (the nickname is kept for sentimental reasons anyhow...)

19. Jul 25, 2008

### The-herod

Hi again!

I tried to read a book about perturbation theory. I understood, more or less, at the beginning. But from some point, I hardly understand. I think the main reason is that I don't have a sufficient background in solving differential equations. I'm not sure though.

So what I'm doing now looking for a good resource about DE, and after that I'll have to learn, also from other resources (seems like this book is a bit poor) about different methods etc in perturbation theory (I think I can say that basic usage of Poincaré's method I know now).

I have no problem to find resources for DE, though if you have a recommended one, I'll be glad to know. However, my main problem is resources for perturbation, perhaps someone know about good resources, to learn about it (more or less) from the beginning?

Thanks a lot!