1.5 Qualitative analysis of dynamics
Qualitative analysis of ODEs, deals with the determination of the general properties of the solutions of a given system, or of a given class of systems, for any initial condition, the explicit form of the solutions being not known (either because not interesting, or because impossible to compute, as usual). By general properties of the solutions one means, for example, stability of equilibria, asymptotic behavior for large times, existence of first integrals, and so on.
1.5.1 1D autonomous ODEs
The simplest ODE is of the form ˙x = f (x), where x ∈ R and f (x) is a smooth real function of a real variable. As is well known, such an equation can be solved by separating the variables, writing it as dx/f (x) = dt, for x such that f (x) 6= 0. Upon integration one gets F (x) = t + c, where F0(x) = 1/f (x), and by a local inversion one gets the general solution x(t) = F−1(t + c).
The arbitrary constant c is uniquely determined by the initial datum x(t0) = ξ: c = −t0 + F (ξ), and x(t) = F−1(t − t0 + F (ξ)). Particular solutions playing a fundamental role are the equilibrium, or stationary ones, defined as those constant functions x(t) = ¯x for any t, where ¯x is a zero of f (x), i.e. a solution of f (x) = 0. Check: ˙x(t) ≡ 0, and f (x(t)) = f (¯x) ≡ 0.
Such a method of solution works, in general, only in principle: both the explicit computation of a primitive function F (x) of 1/f (x), and its local inversion, are generally difficult when not impossible. More than this, were an explicit expression of x(t) available, it would be very difficult to extract interesting informations from it: one would have to study (with the standard tools of mathematical analysis) a complicated function.
On the other hand, one can understand everything about the behavior of the solutions of the equation ˙x = f (x) by interpreting x(t) as the position of a point moving on the real line.
The ODE is then interpreted as the law giving the velocity of the point when the position of the latter is known. One then concludes that ˙x(t) > 0, i.e. x(t) is a monotonically increasing function of time t, where f (x(t)) > 0. In other words, if the point occupies a position x(t) where f takes on a positive value, then the point is moving from left to right. The other way around, ˙x(t) < 0, i.e. x(t) is a monotonically decreasing function, where f (x(t) < 0, and the point occupying the position x(t) is moving from right to left. One can then draw the graph of f and point out the direction of motion of the point along the x axis. Such a picture is called the phase portrait, or diagram of the 1D ODE, or one dimensional dynamical system.
An example is reported in Figure 1.1 below.
Exercise 1.10. Plot the phase portrait of ˙x = f (x) for f (x) = −x, f (x) = −x3, f (x) = x(1 − x), f (x) = −x + x3, f (x) = sin(x).
Exercise 1.11. Plot the phase portrait of the equation for v of Exercise 1.4 on falling bodies in a gas.
The most important features of the phase diagram of ˙x = f (x) are the following.
• If ¯x is an isolated zero of f such that f0(x) < 0 for any x ∈ I \ {¯x}, then x(t) → ¯x as t → +∞ for any x(0) ∈ I \ {¯x}. The typical case of a simple zero, with f0(¯x) < 0, is
included here. One says that ¯x is an attracting equilibrium. Question: may x(t) reach ¯x in a finite time?
• If ¯x is an isolated zero of f such that f0(x) < 0 for any x ∈ I \ {¯x}, then x(t) exits I in a finite time for all x(0) ∈ I \ {¯x}. The typical case of a simple zero, withf0(¯x) > 0, is included. One says that ¯x is a repulsive equilibrium.
• If ¯x is an isolated zero of f such that f0(¯x) > 0 (< 0) for any x ∈ I \ {¯x}, then ¯x is attractive on the left (right) and repulsive on the right (left).
• If f (x) > 0 for all x ≥ ξ, then, the time of flight from ξ to +∞, or blow-up time with initial condition ξ, is given by the formula (deduce it)
T (ξ) = Z +∞
ξ
dx
f (x) := lim
u→+∞
Z u ξ
dx
f (x) . (1.31)
Thus, T (ξ) < +∞ if f (x) → +∞ as x → +∞ as fast as xα, with α > 1. Question: for f (x) = x ln x, is T (ξ) finite? Question: compute T (ξ) for f (x) = xα, α > 1.
• The qualitative form of the solutions can be understood by taking the time derivative of ˙x(t) = f (x(t)), which gives ¨x = f0(x) ˙x = f0(x)f (x). Thus, for example, x(t) is monotonically increasing where f (x) > 0, convex where f0(x) > 0, concave where f0(x) <
0. See Figure 1.1.
• Let ¯x be a simple zero of f : f0(¯x) 6= 0. Then, the change of variable x(t) = ¯x + y(t) transforms the equation ˙x = f (x) into ˙y = f (¯x + y) = f0(¯x)y + O(y2). Neglecting the remainder O(y2), under the hypothesis that y(t) be “small”, one gets the linearized equa-tion ˙y = f0(¯x)y. The solution of the latter equation is y(t) = ef0(¯x)ty(0). Thus, depending on the sign of f0(¯x), |y(t)| decreases or increases exponentially fast, in agreement with the qualitative analysis made above. However, in the case f0(¯x) > 0, the approximation made to linearize the original equation becomes wrong very quickly.
• In particular, if f0(¯x) < 0, |y(t)|/|y(0)| < ε (small) for
t > 1
|f0(¯x)|ln 1 ε
, (1.32)
whereas if f0(¯x) > 0, |y(t)|/|y(0)| > m (large) for t > 1
f0(¯x)ln(m) . (1.33)
In both cases, the time scale to amplify the initial condition by a factor ε or m, respec-tively, is 1/|f0(¯x)|. Notice that if ε passes from 10−1 to 10−10, or m passes from 10 to 1010, t passes from ln(10)/|f0(¯x)| to ten times the latter, in both cases.
1.5. QUALITATIVE ANALYSIS OF DYNAMICS 25
Figure 1.1: Typical phase portrait of a 1D autonomous ODE.
• One can rewrite the equation ˙x = f (x) in the gradient form ˙x = −V0(x), where V (x) =
−R f (x)dx. Isolated local minima of V (x) are attractive, isolated local maxima are repulsive (why?).
• Taking the derivative of V (x(t)) with respect to time yields ˙V = V0(x) ˙x = −(V0(x))2 ≤ 0.
For isolated minima, ˙V < 0 for any x ∈ I \ {¯x}, so that V (x(t)) is strictly monotonically decreasing and bounded from below, so that V (x(t)) → V (¯x) as t → +∞.
• The equation ˙x = f (x) does not admit any periodic solution. Indeed, suppose x(t) is a T -periodic solution, i.e. x(T ) = x(0). Then, integrating ˙V = −(V0(x))2 from 0 to T one gets 0 = V (x(T )) − V (x(0)) = −RT
0 (V0(x(t)))2dt < 0, absurd.
• Simple zeros of f are robust under small perturbations of f . If f (¯x) = 0, and f0(¯x) 6= 0, then f (x) − ε, with ε small enough, has a simple zero in ˆxε= ¯x + ε/f0(¯x) + O(ε2) (prove it). Thus, the phase portrait of ˙x = f (x), f displaying only simple zeroes, is robust under small perturbations of f . This is an example of structurally stable system.
Exercise 1.12. Plot the phase portrait of ˙x = x2 − ε, as ε varies from negative to positive values. Plot the graph of the equilibria of f as functions of ε.
Exercise 1.13. Study the phase portrait of the equation
˙n = a(1 − n)
n + 1
R0ln(1 − n)
, (1.34)
limiting the analysis to the interval 0 ≤ n ≤ 1. By a time rescaling you can set a = 1 (pass from t to t0 = at). In particular, prove that (1.34) admits a unique attracting equilibrium ¯n(R0), and compute its value approximately (e.g. by one step of the Newton-Raphson method of tangents).
Find the asymptotic expression of ¯n(R0) as R0 → +∞.
Equation (1.34) describes the time evolution of the fraction of total cases (infected plus removed) in the so-called SIR model for the propagation of infectious diseases. The parameter R0 > 1 is the initial reproduction number of the disease.
There is no specific theory for autonomous 1D ODEs, i.e. ˙x = f (x, t). The latter equation is equivalent to the 2D system ˙x = f (x, s), ˙s = 1, and as such it belongs to the class of 2D autonomous ODEs (or plane systems) for which there exists a complete classification of the possible motions. The key, qualitative idea here is that either motions are unbounded, and solutions may go to infinity, or are bounded in some compact region. In the latter case, (global) existence and uniqueness of the solutions prevents the phase curves to intersect. A single curve cannot then wander “chaotically” inside the region, which would cause a self intersection. One is left with two possibilities only: either the phase curve tends to an equilibrium point, or tends to (which includes is) a periodic solution (a limit cycle).
1.5. QUALITATIVE ANALYSIS OF DYNAMICS 27
1.5.2 1D conservative Newtonian motions
1D conservative Newtonian motions are ruled by the Newton equation m¨x = −U0(x), where U (x) is the potential energy. This is a special case of plane system, since it is equivalent to
˙x = v
˙v = −m1U0(x) (1.35)
The plane R2 3 (x, v), is called the phase space (or plane) (in this case) of the system. Since the x-space, i.e. the space of positions or configuration space of the system, has dimension one, this is a one-dimensional (1D in short) Newtonian system, or a Newtonian system with one degree of freedom. In practice, in mechanics, the number of degree of freedom of the system is half the dimension of the phase space, the latter dimension being the number of parameters identifying the initial condition.
Exercise 1.14. Consider the (undamped) harmonic oscillator equation m¨x = −kx, describing a particle of mass m attached to an ideal (spiral-shaped) spring, in absence of any resistance of the medium. Dividing by the mass one gets ¨x = −ω2x, or the corresponding system ˙x = v,
˙v = −ω2x, where ω :=pk/m is the pulsation of the oscillator.
1. Prove that the total energy H(x, v) = mv2/2+kx2/2 is a constant of motion of the system.
2. Write down the general solution x(t) of the Newton equation, and determine the unique solution corresponding to the initial condition x(0) = x0, ˙x(0) = v0. Hint: rescale time first, passing to the dimensionless time τ = ωt, which yields x00(τ ) = −x(τ ).
3. Look for a solution of the latter equation in the form of a power series x(τ ) =P
n≥0cnτn. Determine the two branches of solutions corresponding to c0 6= 0, c1 = 0, and to c0 = 0 and c1 6= 0. Finally, add them (why does it work?).
4. Introduce the complex variable
z(t) := mωx(t) + ımv(t)
√2mω ,
where ı is the imaginary unit. Deduce the equation satisfied by z(t) and solve it for any initial condition z(0). Which is the trace of the curve t 7→ z(t) ∈ C? Hint: compute
|z(t)|2 and relate it to the constant energy value E of H.
5. Deduce the expression of x(t) from that of z(t) and check that it coincides with that deduced above.
Fundamental particular solutions of system (1.35) are the equilibria, i.e. the zeros of the right hand side, or vector field of the system. These are the points (¯x, 0) in the phase plane, where ¯x is a critical point of U (x), i.e. U0(¯x) = 0: the equilibrium positions ¯x are those where the force f (x) = −U0(x)/m is zero. To the single equilibrium point (¯x, 0) there corresponds the
motion t 7→ (¯x, 0) for all t ∈ R, i.e. the unique solution of (1.35) corresponding to the initial condition (¯x, 0).
For system (1.35) the total energy H(x, v) is preserved, namely H(x, v) := mv2
2 + U (x) = E , (1.36)
where x(t) and v(t) is understood in the latter expression. The constant value E of the function H along the phase curve t 7→ (x(t), v(t)), i.e. the unique motion corresponding to a specific initial condition (x(0), v(0)), is determined by the the latter: E = H(x(0), v(0)). One thus concludes that the phase curves, i.e. the possible motions, belong to the E-level set of H, i.e.
H−1(E) := {(x, v) ∈ R2 : H(x, v) = E} . (1.37) The latter set consists, when not empty, of one or more connected curves, some of them possibly degenerating into single points. The picture of the level sets (1.37), where one distinguishes the traces and the directions of the possible phase curves, or motions, corresponding to all possible values of E, is named phase portrait, or diagram of system (1.35). This is the most complete description of the dynamics of the 1D Newtonian system.
Example 1.9. For the harmonic oscillator H = mv2/2 + kx2/2 = E ≥ 0. To E = 0 there corresponds the origin (0, 0) of the phase plane, the only equilibrium point. The corresponding motion is the phase curve t 7→ (0, 0) for all t ∈ R. The level set for any E > 0 is an ellipse of canonical equation x2/a2+ v2/b2 = 1, where a := p2E/k and b := p2E/m are the length of the two semi-axes. This corresponds to the periodic motions of the harmonic oscillator, namely x(t) = x(0) cos(ωt) + [v(0)/ω] sin(ωt), where T = 2π/ω is the period of the motion and ω :=pk/m the pulsation. The point mass oscillates on the real axis sweeping back and forth a segment of extremes x± = ±a = ±p2E/k, the two turning points, where the particle stops and inverts the motion. The ellipse in the phase plane is swept clockwise (why?) once per period T of motion.
Exercise 1.15. For the harmonic repulsor, or hyperbolic oscillator, defined by the Newton equation m¨x = +kx (k > 0), the energy function is H(x, v) = mv2/2 − kx2/2. Study the level set H = E as E varies on any real interval including E = 0.
Exercise 1.16. Consider the graph of the energy function H(x, v), i.e. the subset {(x, v, z) : z = H(x, v)} of R3. The level sets of H are then the intersections of the graph of H with the horizontal planes z = E. Sketch the graphs of H for the harmonic and hyperbolic oscillator, realizing that one has a paraboloid with elliptic section in the former case, and a mountain pass, or saddle-shaped graph in the latter. The origin (0, 0) of the phase plane (the only equilibrium point in both cases) is called a center for the harmonic oscillator, and a saddle for the hyperbolic one.
The particular structure of the energy function (1.36) imposes some restrictions on the structure of the E-level sets of H. These are the general rules to draw a phase portrait.
1.5. QUALITATIVE ANALYSIS OF DYNAMICS 29
Suggestion 1.2. Draw any graph of a function U (x) displaying local minima and maxima, and understand each item below on the picture. Draw a corresponding sketch of the phase portrait on the (x, v) phase plane reported just below the plane with the graph of U (x).
• Equilibria lay on the x-axis. The orientation of the phase curves is left-to-right in the upper half plane {(x, v) : v > 0}, and right to left in the lower half plane {(x, v) : v < 0}
(this follows from ˙x = v; why?). In particular, all closed curves are clockwise oriented.
• Since mv2/2 = E − U (x) ≥ 0, the possible values of x for motions of a given energy value E are those belonging to the E-sub-level set of U , i.e. the allowed values of x are such that U (x) ≤ E.
• If U (x) displays an absolute minimum value Um, then E cannot assume values lower than it: for E < Um there are no possible motions. If infxU (x) > −∞, then E must be greater than this. Finally, if U is unbounded from below, then E can take on any real value.
Remark: U is defined up to an arbitrary constant, that must be chosen once and for all.
• H is symmetric in v, i.e. H(x, −v) = H(x, v), which implies reflection symmetry of any level set with respect to the x-axis. More precisely, solving (1.36) for v, one gets
v±(x) = ± r2
m[E − U (x)] , (1.38)
which means that to any allowed x there correspond two values of v, v+ ≥ 0 and v− =
−v+ ≤ 0. The two branches v± may be connected to each other or disconnected.
• The even symmetry of H in v corresponds to the so-called time-reversal symmetry of system (1.35): the equations are invariant by changing (x, v, t) into (x, −v, −t)
• The simple zeros of the equation U (x) = E are called turning points. Indeed, if U (ξ) = E, and U0(ξ) > 0 (< 0), at the turning point ξ, the particle coming from the left (right) of ξ stops there (why?). Starting at the stop time with the initial condition (ξ, 0), the force on the particle is f (ξ) = −U0(ξ)/m < 0 (> 0), and the particle starts at rest and moves to the left (right).
• The previous point describes the elastic smooth reflection of a particle by a potential barrier: when the particle arrives at a turning point it rebounds inverting the direction of motion, assuming at the same positions equal and opposite velocities (this is implied by the v → −v symmetry of H).
• Taking the derivative of v+, from (1.38), one gets v+0 (x) = −U0(x)
p2m [E − U(x)] . (1.39)
From this formula one sees that v+ → −∞ as x → ξ−, where ξ is a turning point such that U0(ξ) > 0. Analogously, v+ → +∞ as x → ξ+, where ξ is a turning point such that U0(ξ) < 0. By the reflection symmetry v → −v it then follows that the tangent to a phase curve at the phase turning point (ξ, 0) is vertical, and the curve is locally smooth.
• Another implication of (1.39) is that the local maximum (minimum) points of v+(x) coincide with the local minimum (maximum) points of U (x), for values of E larger than the local minimum (maximum) value of U (x).
• A nondegenerate critical point ¯x of U (x) is a (nondegenerate) local minimum if U00(¯x) > 0, or maximum if U00(¯x) < 0. In the former case, for E = U (¯x) + ∆E, ∆E > 0, the phase curves opening around the equilibrium point (¯x, 0) are small ellipses of semi-axes p2∆E/m andp2∆E/U00(¯x) with center in the equilibrium point. This is the local phase portrait of a harmonic oscillator of elastic constant k = U00(¯x) (why?). In the latter case, in varying ∆E around zero, one gets branches of hyperbola and a local phase portrait of a hyperbolic oscillator of constant k = −U00(¯x) (why?).
• To an interval of allowed values of x with extrema two turning points x±(E), two consec-utive simple zeros of U (x) = E, with at least a local minimum point of U (x) in between them (why?), there corresponds a closed phase curve CE, i.e. a periodic motion. From the law of conservation of energy it follows that the period of such a motion T (E) is given by
T (E) = I
CE
dx v = 2
Z x+(E) x−(E)
dx q2
m[E − U (x)]
(1.40)
• The area enclosed by the closed phase curve CE (periodic motion) in the (x, p) phase plane, where p := mv is the momentum, is
A(E) = I
CE
p dx = 2
Z x+(E) x−(E)
p2m [E − U (x)] dx . (1.41)
The latter quantity is related to the period of motion (1.40) by the following fundamental relation:
T (E) = d
dEA(E) (1.42)
Prove it: pay attention to the energy dependence of the turning points in (1.41).
• To an interval of allowed values of x with extrema a turning point and a nondegenerate local maximum ¯x of U (x) there corresponds a closed curve connecting the equilibrium point (¯x, 0) to itself. Such a closed curve is called homoclinic connection, or loop. The time to reach the equilibrium along such a phase curve is infinite (why?). The possible motions of the system in this case are two: the equilibrium one and the homoclinic motion.
• To an interval of allowed values of x with extrema two nondegenerate local maxima ¯x1 and ¯x2 of U (x), there correspond two symmetric phase curves, one above and one below the x-axis, connecting the equilibria (¯x1, 0) and (¯x2, 0). Such curves are called heteroclinic connections. The possible motions of the system in this case are four: the two equilibria and the two connections.
1.5. QUALITATIVE ANALYSIS OF DYNAMICS 31
• Suppose that U (x) → −∞ as x → +∞. Let ξ be such that E − U (x) is positive on [ξ, +∞[ (such a ξ exists; why?). The time the particle takes to reach +∞, starting at ξ, is
τ∞(E) = Z +∞
ξ
dx q2
m[E − U (x)]
. (1.43)
Prove it, and find a sufficient condition on the divergence of U for the convergence of the integral (which means off to infinity in a finite time).
Exercise 1.17. A “physical pendulum” of length ` is defined by the Newton equation
m¨x = −mgˆe2+ R , (1.44)
where x = x1eˆ1 + x2eˆ2 ∈ R2, (ˆe1, ˆe2) being the canonical basis, whereas R is the unknown reaction force exerted by the ideal massless rod of length ` to which the particle of mass m is suspended. The problem is conveniently studied passing to the co-moving polar basis (ˆer, ˆeθ), defined by
eˆr= sin θ ˆex− cos θ ˆey
ˆ
eθ = cos θ ˆex+ sin θ ˆey ,
where θ is the angle between the rod and the y-axis, counter clockwise oriented. Starting with x = `ˆer and R = Rreˆr+ Rθeˆθ, prove that the vector equation (1.44) is equivalent to the system
m`¨θ = −mg sin θ + Rθ
−m` ˙θ2 = mg cos θ + Rr
. (1.45)
The hypothesis of ideal constraint Rθ ≡ 0, i.e. of reaction force orthogonal to the manifold (the circle) on which the particle must move, allows to completely solve the problem. Set ω :=pg/`.
1. Draw the phase portrait of the “mathematical” pendulum equation ¨θ = −ω2sin θ.
2. Linearize the equation, solve for the small oscillations and compute the reaction force Rr. 3. Determine the rotation frequency of the pendulum for E → +∞, and determine the
corresponding reaction Rr. Hint: use the conservation of energy.
Exercise 1.18. A common potential mimicking certain chemical bonds in molecules (e.g. the hydrogen bonds linking complementary bases in DNA molecules) is the Morse one:
U (x) = D
2(e−αx− 1)2 . (1.46)
Draw the phase portrait of m¨x = −U0(x). Pay attention to the phase curve at E = D/2, separating bounded from unbounded motions. Which is the frequency of the small oscillations of the system as E → 0+?
Exercise 1.19. A particle in a box is described by the sequence of smooth potentials of the form
Un(x) =
x
`
2n
, n = 1, 2, . . . (1.47)
1. Compute the limit U∞:= limn→∞Un(x) for any x ∈ R.
2. Draw the phase portrait of m¨x = −Un0(x) for increasing values of n. Which is the limit phase portrait as n → ∞? Which are the corresponding physical motions of the particle?
3. Which is the period of the motion at a given energy E in the limit n → ∞? Hint: use the previous results.
4. Write down in detail the formula (1.40) for the period of the motions Tn(E) for any finite n. Determine the exact dependence of Tn(E) on E. Finally, compute the limit T∞(E) = limn→∞Tn(E).
5. The homogeneity of the potential (in the Euler sense) implies scale invariance of the problem. Rescaling space and time variables as x = λx0, t = λαt0, determine the value of α (depending on n) such that the rescaled Newton equation does not depend on λ. Find the consequent rescaling of the energy, i.e. the exponent β such that E = λβE0. Deduce from this the scaling of the time with energy. Hint: from the scaling laws of t and E, find the exponent γ such that t/Eγ = t0/E0γ.
Exercise 1.20. A potential barrier is modeled by the potential U (x) = V0Θ(x) := 0 , x < 0
V0 , x > 0 , (1.48)
where Θ(x) is the Heaviside step function, jumping from the constant value zero to the constant value one at x = 0 (where it is either not defined, or it can be completed assigning any value).
Draw the corresponding phase portrait. Hint: make use of physical intuition and conservation of energy; approximate graphically (no matter the formulas) the barrier with steeper and steeper smooth sigmoids.
Exercise 1.21. A model of periodic potential (e.g. that experienced by a conduction electron in a metal) is that of Kronig-Penney, which consists in the periodic repetition (or translation) of the finite width barrier
B(x) = 0 , |x| > a
V0 , |x| < a (1.49)
with space period b > 2a. The Kronig-Penney potential U (x) is given by the formula U (x) = P
n∈ZB(x − nb). Draw the corresponding phase portrait.
Exercise 1.22. Study the phase portraits corresponding to the potentials U (x) = kx2/2 + λx4/4 as k and λ vary in R (in an interval around zero).
1.5. QUALITATIVE ANALYSIS OF DYNAMICS 33 Exercise 1.23. An important partial differential equation (PDE) in physics is the nonlinear Klein-Gordon equation
utt = c2uxx− ω2u − λu3 , (1.50) where u(x, t) is a certain real function of the two variables (x, t) (i.e. a scalar field defined on a 2-dimensional space-time). The equation is ruled by the three parameters c, ω and λ. Solving equation (1.50) in general is impossible, unless λ = 0 (i.e. the equation is linear). However, in many applications one can be interested in finding special solutions called solitary, or traveling waves, namely special solutions of the form u(x, t) = ϕ(x − vt), where ϕ(ξ) is a function of one real variable and v is a parameter. Such solutions describe profiles that translate at the constant velocity v.
1. Determine the equation satisfied by ϕ(ξ), where ξ := x − vt.
2. Distinguish the three cases v2 > c2, v2 = c2, and v2 < c2; call m = v2− c2 in the first case, m = 0 in the second, and m = c2− v2 in the third case. Interpret ϕ(ξ) as the abscissa at
“time” ξ of a particle of mass m moving on the real line and subject to a certain force.
3. Draw the phase portraits in the various cases (let λ vary around zero) and determine the possible forms of the traveling waves of (1.50).
1.5.3 Quantization of the action and energy levels
In the first 20 years of the past century, in trying to establish a new theory in agreement with many experimental results that classical electrodynamics was unable to explain, Planck, Ein-stein, Bohr, Sommerfeld and Born [8] developed what is known today as the “old quantum theory”. Quantum mechanics in its modern form was developed starting from 1925, by Heisen-berg, Schr¨odinger, Pauli and Dirac, with the fundamental and definitive contribution of Von Neumann on the mathematical formulation of the theory.
The basis of the first quantum theory is the following. Consider a 1D conservative Newto-nian system. For bounded periodic motions, consider the formulas (1.40), (1.41) and (1.42), expressing the period of the motion T (E), the area A(E) inside the closed phase curve CE, and their relation, respectively. Then one observes that A(E) = H
CEpdx has the dimension of an action, i.e. energy times time, or momentum times space. The principle of quantization set up by the founders of quantum physics requires that for bounded periodic motions A(E) takes on
CEpdx has the dimension of an action, i.e. energy times time, or momentum times space. The principle of quantization set up by the founders of quantum physics requires that for bounded periodic motions A(E) takes on