Feras> I almost proved the problem for k>2

Then you almost made a serious mistake :-), because for k>2 your differential equation has no periodic solution passing through y=0. The mechanical interpretation explained by Joachim is, in my opinion, the by far best approach to the problem.

For a periodic solution through y=0, the potential V(y) must have a maximum larger than V(0). It is easy to check that this only occurs for k0+, where the equation can be linearised to y''+ (2-k)y=0, with solution y(x) = ε sin(ωx), with (for consistency) ε->0+, and ω2=(2-k). For y(1)=0, we must have ω = π, i.e. k = 2 - π2. So far, this is just simple analysis, independent of intuition. The latter enters when considering finite amplitude corrections to the oscillation frequency. By inspecting the potential I conjecture that the oscillation period will increase monotonously and continuously with increasing amplitude (or initial kinetic energy), approaching infinity when K -> (-2k)-. This conjecture should be possible to prove from the integral expression for the quarter period of the oscillation, as indicated by Joachim.

So, for k < 2-π2, there is exactly one finite amplitude solution to your problem, otherwise none.

I would change the problem to the fixed point one for integral operator \int_0^1 G(t,s)f(y(s)) ds acting on the space of continuous functions and then use the Krasnosielski Cone compression/Expanding Cone Theorem.

G stands for the Green function G(t,s)=(1-t)s for st.

The given classical system with m=1, evolves driven by the following Hamiltonian:

H(y,p) = p2/2 + V(y),

where V(y)=k cos(y) - cos(2y)/2, p = v = \dot{y} = dy/dx.

Thus if the total (constant) energy equals E, then the time elapsed between the visits of points y1 and y2 equals

T(y1;y2) = \int_(y1;y2) dy/ {dy/dx} = \int_(y1;y2) dy/ \sqrt{ 2(E - V(y))},

whenever (E - V(y)) is nonnegative in the interval [y1,y2]. If y2>y1 and V(y2) = E, then T is half the time of duration of the motion from y1 to the maximal point y2 and back to y1, unless T is infinite.

In the case of the question, the integral to be computed is hyper-elliptic and probably ony in special cases can be expressed by values of elementary functions.

Consult, please the classical mechanics, e.g. Landau&Lifschitz's "Mechanics", paragraph 11 in Chapter III, formulas (11.3) and (11.4); for the English translation of the third Russian edition see:

http://renaissance.ucsd.edu/courses/mae207/mech.pdf

A "solvable" example with periodic potential (notation consistent with the one applied in the question and the completion introduced in my previous answer):

Let V(y) = |y| within |y| \le 1; AND V(y+2) = V(y), for all real y\in R.

1. Motion with the constant of total energy E < 0 does not exist.

2.The motion with total energy E > 1 is with a speed with constant, non-zero sign. Thus - no return to the initial position is possible.

3. The "typical" case with possible returns is 0 < E < 1. Then the moving particle (material point) remains all the time within [-E,E] or in a periodic shift by 2n , with any integer n \in Z. In particular, if it starts with y(x=0)=0, with positive speed, then \dot{y(0)} = \sqrt{2E}, AND at any other instant x the speed (velocity) equals \sqrt{2(E-|y(x)|)}. According to the integral formula for the time until the particle reaches the maximum E for the first time we obtain:

T(0;E) = \int_[0,E] dy / \sqrt{2(E-y)} = - \sqrt{2(E-y)}_0^E = \sqrt{2E}.

Hence the first return to zero appears after 2 T(0;E)= 2\sqrt{2E}, which will be 1 if E = 1/8. Motion with this energy is unique within solutions with positive value of y(x) for 0< x < 1 AND equal zero at the end-points.

A conjecture related to the given problem: If the maximum Vmax of the potential V(.) is greater than V(0), then for the energies E of the range (V(0), Vmax) the time of reaching the first positive point ymax with V(ymax)=E is a strictly increasing function varying from 0 to infinity.

If this were true, then for every k such that k cos(y) - cos(2y)/2 >0 at least at one point y, there exists exactly one solution of the given boundary problem; the corresponding energy E(k) is the unique one for which the time T(0, ymax) = 1/2 (ymax is the first positive solution of V(ymax) = E(k).

Prof. Bogdan Przeradzki and Prof. Joachim Domsta many thanks for your assistance.

Actually, I almost proved the problem for k>2 by means of fixed point theorem on cone. However, Prof. Joachim Domsta's technique seems interesting.

Feras> I almost proved the problem for k>2

Then you almost made a serious mistake :-), because for k>2 your differential equation has no periodic solution passing through y=0. The mechanical interpretation explained by Joachim is, in my opinion, the by far best approach to the problem.

For a periodic solution through y=0, the potential V(y) must have a maximum larger than V(0). It is easy to check that this only occurs for k0+, where the equation can be linearised to y''+ (2-k)y=0, with solution y(x) = ε sin(ωx), with (for consistency) ε->0+, and ω2=(2-k). For y(1)=0, we must have ω = π, i.e. k = 2 - π2. So far, this is just simple analysis, independent of intuition. The latter enters when considering finite amplitude corrections to the oscillation frequency. By inspecting the potential I conjecture that the oscillation period will increase monotonously and continuously with increasing amplitude (or initial kinetic energy), approaching infinity when K -> (-2k)-. This conjecture should be possible to prove from the integral expression for the quarter period of the oscillation, as indicated by Joachim.

So, for k < 2-π2, there is exactly one finite amplitude solution to your problem, otherwise none.

Thank you Prof. Kåre Olaussen for your thoughtful comments.

Your clarification to the problem is convincing. I got similar results for the stability of these solutions. Thank you all for your time and consideration.

Kåre's intuition is ok for k=2, I just wonder, what is your proof of the problem for k>2 by means of fixed point theorem on cone. Did you also obtained absence of the solution? How does the method work?

Regards, Joachim

Joachim> the lower bound of the time is 0

How do you find the lower bound on the time to be 0? In the limit yk -> 0, the equation becomes linear with T_{k, E} = π/(2*sqrt(2-k)).

@ Kåre,

I beg your pardon! You are perfectly right! I have got into some blackout state, since you have stated this already in your previous answer. Thus, I am modifying my last answer and false claims put into double braces.

Thanks a lot!

Best regards, Joachim

Joachim> My proof was similar to the one in the article below with some adjustments:

http://www.ams.org/journals/proc/1994-120-03/S0002-9939-1994-1204373-9/S0002-9939-1994-1204373-9.pdf

But when I said k>2 my mistake I did not mentioned that f(y) was a bit different than the one given in the question. In fact, your proposed technique is quite new to me. I really appreciate your help and a special thanks goes to Kåre for his excellent explanation to your approach.

Joachim> I have got into some blackout state,

I assumed you were just temporarily misled by your "constant restoring force" (V = |y|) analysis :-)

Actually, the problem of how the return time varies with (kinetic) energy is quite interesting and complicated, and varied. Which is why I hid behind the intuition phrase. It is easy to construct potentials where almost any behaviour can occur. So, I just believe that the specific differential equation under discussion leads to the behaviour I conjectured; other cases will very likely be different.

The equations analysed in the AMS paper can be interpreted as one-dimensional motion of a point particle influenced by a time and position dependent force. I am rather unimpressed by the results in that paper; they look like mechanical trivialities dressed in abstract mathematical jargon, decorated with complicating concepts and notation. (I think I am allowed to say that, since I graduated from a department of mathematics :-D)

Note added: The problem at hand is formulated as a non-linear boundary value problem. Joachim and I have viewed it as an initial value problem, with an unknown initial velocity to be adjusted for return at the right time. In numerical analysis this is known as the shooting method. In many cases, but not here, the shooting method is numerically hopeless unstable. In the latter cases the differential equation must be replaced by a integral equation, more likely to have numerically stable solutions, but also less accessible to physical intuition.

Continuing the research, I am trying to prove the monotonicity of the period in the dependence on the total energy.

My staring point is the idea to change the variables of integration in the formula for T(E), with the notation, that E = V(ymax), where V is an increrasing continuous bounded below function on [0, \infty), for E>0:

T(E) = \int_0^{ymax} dy/ \sqrt{ E - V(y) } =

\int_0^1 du / \sqrt{ W(1,ymax) - W(u,ymax) }

where W(u,ymax) := [V(u ymax) ] / ymax^2

A promissing example is provided when V(y) = y^2 + a^2 y^4. Then

W(u,y) = u^2 + a^2 u^4 y^2, which implies that the subintegral function equals 1/ \sqrt{ (1- u^2) + a^2 y^2 (1- u^4) }, thus being strictly decreasing with increasing y>0. Consequently, the period is strictly decreasing with increasing the total energy. Obviously, with the trigonometric functions the staff is much more difficult.

Kåre > I assumed you were just temporarily misled by your "constant restoring force" (V = |y|) analysis :-)

You are again right, at least partially. Thanks for further comments. I like the mentioned shooting method!

Regards, Joachim

Continuing the research, I have found a simple sufficient condition for the monotonicity of T(E): namely, it sufficies that V'(y)/y is montonic in the interval of strict increase of the potential V. Probably, this condition can be verified positively for the following example by Feras

V'(y) = - k sin(y) + sin(2y).

Can someone check this conjecture, please.

The details of the proof of the sufficiency are in preparation and I plan to release it soon.

Joachim

Let me share a sufficient condition for strict monotonicity of the period of the autonomous Hamiltonian systems. The method is based on the following

PROPOSITION. Let $V: [0,Y) \to [0,M)$, with $0< Y,M \le \infty$, be a $C^1$ homeomorphism, and let us consider the following positive function (possibly assuming the value infinity):

$$(1) T(y) := \int_0^y \, \frac{dx}{\sqrt{ V(y) - V(x)} }.$$

If the ratio $ \left( \frac{V'(y)}{y}; \, y \in (0,Y) \right) $ is strictly decreasing (increasing), then $T$ is strictly increasing (decreasing, respectively).

Proof (the main step, only): By substitution $ x = u y$ one easily obtains an equivalent formula for $T$:

$$ (2) T(y) = \int_0^1\, \frac{ du }{ \sqrt{W(1|y) - W(u|y) }} , $$

where $W(u|y) := V(u y ) / y^2$. Now it sufficies to note, that the difference in the denominator equals the following integrals

$$ \int_u^1\, frac{\partial W(w|y)}{\partial u} \, dw =

\int_u^1 \, w \, \frac{ V'(wy)}{ wy} \, dw $$

where the last factor is strictly decreasing (increasing, respectively) with respect to $y>0$, for every $w>0$. Thus the denominator in (2) is also strictly decreasing (increasing, respectively). $\Box$

Also the continuity of $T$ can be proved by standard method of the analysis, which is omitted.

Regards, Joachim

This is very nice! I think the crucial idea is the introduction of the scaled integration variable, x = u y (since a direct differentiation of the original integral leads to two terms, both of which diverges). After that the criterium may be written in seemingly different forms, perhaps because a product of two increasing functions is also increasing.

I did a direct differentiation of the scaled integral, to find

dT/dy = ∫01 du [N(y)- N(uy)]/[V(y) - V(uy)]3/2

with N(y) = V(y) - ½ y V'(y). So I find that dT/dy is positive if N(y) is increasing, and dT/dy is negative if N(y) is decreasing. This, at first sight, looks different than the condition that^* M(y) ≡ V'(y)/y being strictly decreasing (in the first case) or strictly increasing (in the second case). However, assuming differentiability on finds that the condition that N'(y) being positive (negative) is equivalent to M'(y) being negative (positive).

I will proceed to check the explicit potential of this thread.

Thank you, Kåre, for cleaning up my mistates; the one discoverd by you has been corrected. It helps in spreading only true facts! With respect to your sufficient condition - I shall check it soon.

Best regards, Joachim

I did some numerical checks for the case V(y) = -k cos(y) + (1/2) cos(2y), with k < 2. Let ymax = π when k