I have a question about a first order non-linear differential equation. I have tried many method to solve this problem but not successful yet. Here is my question;

$$w^4 + (w')^2 = g(t)$$

$$w' = \left[g(t)- w^4\right]^{0.5}$$

I have tried runge kutta 4th order and 4-5 adaptive step size and cash-carp methods up to now, I think they are very good methods to solve this IVP but still not give expected values for some places.

Do you have another suggestions to solve this? Thanks in advance

  • $\begingroup$ Do you have any information on $g$? Does it fit in this direction? From the associated problem, do you know that there must be a real valued solution? $\endgroup$ – Nikolaj-K Feb 20 '13 at 14:43
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    $\begingroup$ Pleas use TeX-like formatting for math. For more information on this type of formatting, read the FAQ. $\endgroup$ – user41994 Feb 20 '13 at 14:46
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    $\begingroup$ From the two comments, you seem to have quite a bit more information that you provided in the initial question. I feel typing it out can't be a bad idea. $\endgroup$ – Nikolaj-K Feb 20 '13 at 14:57
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    $\begingroup$ Jacob, if $g$ falls abruptly, then the solution may cease to exist. Moreover, the numerical methods become unstable even before that. You may want to look at artofproblemsolving.com/Forum/viewtopic.php?f=73&t=216577 where my way out was not to solve it at all. Now, if you have full $g$ data (so you can look well ahead), there are better ways, but if you do it in real time, the difficulty is real, not just technical. $\endgroup$ – fedja Feb 20 '13 at 15:09
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    $\begingroup$ Yes, when you have a decent function that is far away from the breakdown regimes, the standard methods work. The problem there was that once the centrifuge was running, you had to do something even when the theory said that there was no solution, so you should either restrict the input and use high accuracy methods on the descents, or just "do something different" when the behavior gets close to critical. Actually, I'm happy to return to this question, so if you have some problem with the centrifuge equation that cannot be solved by mere increase in precision, let me know :). $\endgroup$ – fedja Feb 20 '13 at 17:33

OK, here are some (unpleasant, but apparently correct) observations. We are still trying to solve $w^4+\dot w^2=H$ in real time. I'll also assume that you want $w\ge 0$, i.e., you do not want to change the direction of your rotation in the process. Note that allowing the direction change won't help much in the sense that the final conclusion will stay the same but the argument will get more involved.

Recall (see my AoPS post) that invariant quantities are $\dot H/H^{5/4}$ and $\ddot H/H^{3/2}$. Assume that we want to create a control satisfying $w^4+\dot w^2\in [(1-\delta)H,(1+\delta)H]$ for all reasonable regimes $H$. Suppose that the pilot is not allowed to change $H$ or $\dot H$ abruptly but he can change $\ddot H$ any time and in any way within some reasonable limits, so $\ddot H$ does not need to be continuous. I believe, this is more or less what happens in reality but if I am wrong here, correct me.

Observation 1. Ascending regimes are asymptotically stable. What it means is that if $H$ is any ascending regime that starts at some distant past time moment $-T$ and equals some constant $H_0$ up to moment $0$ (we shall call such regimes canonical), then any non-negative solution of the equation $w^4+\dot w^2=H$ that is defined on the entire interval $[-T,T]$ has essentially the same values $w(t),\dot w(t)$ as the canonical solution whose initial condition is given by $w=H_0^{1/4}$ on $[-T,0]$.

Observation 2. Suppose that $H$ is a canonical ascending regime and $w$ is the corresponding canonical solution with $\dot w\ge 0$. Suppose also that $w_1$ satisfies $w_1^4+\dot w_1^2<H$ on $[-T,T]$. Then $w_1\le w$. Indeed, we clearly have $w_1\le w$ on $[-T,0]$. On the other hand, the graphs can never cross because at the first moment when $w_1=w$, we must have $\dot w_1<\dot w$, so a tiny bit before that we must have $w_1>w$, which contradicts the assumption that our crossing moment is the first one.

Observation 3. Suppose that $H$ is a canonical ascending regime and $w$ is the corresponding canonical solution with $\dot w\ge 0$. Suppose also that $c>0$ (it can be arbitrarily small) and $w_1$ satisfies $w_1^4+\dot w_1^2>(1+c)H$ with on $[-T,T]$. Then $w_1\ge w$ on $[0,T]$ if $T>T(c)$. Indeed, we have $w_1(0)\approx (1+c)^{1/4}H_0^{1/4}>w(0)$. Again, consider the first crossing moment if it exists. At that moment we must have $|\dot w_1|>\dot w$. If $\dot w_1>0$, we get the contradiction in the same way as before. However, if $\dot w_1<0$, $w_1$ must continue to head down and its derivative can only grow in absolute value in that process until the direction gets reversed so this situation corresponds to a prohibited regime.

Observation 4. The time in the equation is reversible, so instead of investigating the descending regimes, we can investigate the ascending ones. The question we'll ask will be the following. Take any canonical regime $H$. Suppose we have a clever control $v$ such that $v(t)$ depends only on $H(s)$ with $s\ge t$ (there are no assumptions about how you designed it: solving the ODE, writing some explicit formula, asking an oracle, whatever). All we know is that our control doesn't create relative error more than $\delta$ for every regime anywhere and it is not allowed to see the past (remember that the time is now running backwards). What can we say about the value $v(t)$ for some fixed $t>0$?

The catch is that now we can glue any canonical regime on $[-T,t]$ which has correct values $H(t)$ and $\dot H(t)$ to the piece we have on $[t,T]$ and $v(t)$ is determined by that piece alone. Thus, $v(t)$ should make sense for any piece we glue.

Observation 5. Let us glue a canonical piece $H=w^4+\dot w^2$ with $\dot w\ge 0$ on $[-T,t]$ (note that we can choose $w$ instead of choosing $H$). Then the solution $(1+\delta)^{1/2}w$ will give us a canonical regime that is at least $(1+\delta)H$. Similarly, $(1-\delta)^{1/2}w$ will give us a canonical regime that is at most $(1-\delta)H$. Thus, our clever control must satisfy $(1-\delta)^{1/2}w(t)\le v(t)\le (1+\delta)^{1/2} w(t)$. Note that it should hold for every admissible $w$. Now you, probably, smell trouble: we have many different $w$ to try and, if they don't give the same value, the exact solution is impossible. Moreover, if those values may be noticeably different, $\delta$ cannot be too small.

Observation 6. We still have two conditions to satisfy at $t$, which, in terms of $w$, can be written as $w^4+\dot w^2=H$, $2w^3\dot w(2+\psi)=\dot H$ where $\psi=\ddot w/w^3$. Note that we have 3 free parameters $w,\dot w,\ddot w$ and only 2 equations. Now it remains to find the quantitative estimates for the freedom we have. To this end, we will compare the scale invariant quantities $I=w^4/H$ for two regimes with the same $\dot H/H^{5/4}$. Since the controller I gave creates dismal results near the bottom of the cosine wave $2+\cos kt$ for $k>2$, it will be a good idea to look at the "quadratic departures" from the ground state and compute the maximal uncertainty along the way. For technical reasons, it'll be convenient to look at $w(t)=1+at^2$, i.e., $H(t)=(1+at^2)^4+4a^2t^2$. The value $k=2$ corresponds to $a\approx 0.35$ here. This regime gives the invariant derivative quantity $\frac{8at(1+at^2)^3+a}{[(1+at^2)^4+4a^2t^2]^{5/4}}$. This is to be compared with the time/scale invariant regime $w(t)=b/t$, i.e., $H(t)=\frac{b^4+b^2}{t^4}$ with the invariant derivative quantity $\frac{4}{(b^4+b^2)^{1/4}}$. The ratio of the corresponding $I$-values is $$ \frac{(1+at^2)^4(b^4+b^2)}{[(1+at^2)^4+4a^2t^2]b^4} $$ where $b$ is determined from the equation $$ b^4+b^2=\frac{[(1+at^2)^4+4a^2t^2]^{5}}{[8at(1+at^2)^3+a]^4} $$ I will leave it to you to run a simple script maximizing over $t$ ($t=0.2$ already gives you something to be unhappy with for $a\in(0,1)$) and to recast the corresponding "inevitable uncertainty ratio" into the $\delta$-values. The table is like that:

$a=0.1, \delta=0.006$

$a=0.2, \delta=0.018$

$a=0.3, \delta=0.036$

$a=0.4, \delta=0.058$

$a=0.5, \delta=0.085$

$a=0.6, \delta=0.116$

$a=0.7, \delta=0.150$

$a=0.8, \delta=0.188$

$a=0.9, \delta=0.228$

As you can see, at the "breakdown point" $a=0.35$, the error of $4\%$ in the $H$-profile approximation is just inevitable and my controller is about that accurate. If $a$ gets to $0.6$, which corresponds to $k=\sqrt{8(a+a^2)}=2.8$ in the cosine model, you are going to be $10\%$ off somewhere no matter what you do. So, the version of the problem you are trying to solve is not just hard but unsolvable. You have to restrict the input to something reasonable and be happy with approximate control. However, you can try to investigate what features of the current control you dislike and design a better control in some respects. Just don't waste your time trying to achieve the impossible :).

  • $\begingroup$ P.S. I suggest you show this all to ETCKerry as well. As usual, you both are welcome to ask as many questions as you want :). $\endgroup$ – fedja Feb 24 '13 at 17:29
  • $\begingroup$ Thank you fedja for your valuable time and effort again. I will compare your observations with the metdod I have tried to solve the equation. I would send the g regime data but after your observations I have decided to combine my data with your observations. With regard to my next step, I will try how I can reduce the error for omega and g as much as possible. Thanks again. You are right imposible is imposible but once upon a time it was impossible to find electricity :) $\endgroup$ – yakupc Feb 25 '13 at 9:28
  • $\begingroup$ As it became clear now, the first step in reducing the error as much as possible is to understand clearly what exactly are the restrictions on the $H$ (or, if it is easier for you to understand, on the $G$) profiles. The point is that there cannot be any good or even decent controller for harsh regimes in principle but, once you restrict harshness and tell the objectives in a precise way, we can try to find the optimal controller for those limits. Send the data anyway: I'd like to look at what you have there no matter what :). $\endgroup$ – fedja Feb 25 '13 at 12:12
  • $\begingroup$ Ok, I'll provide what I have soon.Thanks $\endgroup$ – yakupc Feb 25 '13 at 12:28

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