How to Solve Problems in Nonlinear Conservative Systems

How to Solve Problems in Nonlinear Conservative Systems

Let us consider a dynamical system in which the dissipation og energy of so slow that it may be neglected. This neglecting this slow dissipation, we assume the law of conservation of energy, namely, that the some of the kinetic energy of the’ kinetic energy and the potential energy is a constant .In others world, the system is regarded as conservative. Specifically, we consider a particle of mass m in rectilinear motion under the action of restoring force F which is a function of the displacements x only, the differential equation of motion is thus

m (d^2 x)/(dt^2 )=F(x)

Where we assume that F is the different equation pf the paths defined by the solution of (13.62) in the xy phase plane:

dx/dt=y

dy/dt=(F(x))/m

Eliminating we obtain the differential equation of the paths defined by the solution of (13.62) in the xy phase plan:

dy/dx=(F(x))/my

Separating variables in this equation, we obtain

my dy=F(x)dx.

Suppose x=x_0 and y(-dx⁄dt)=y_0 at t=t_0. Then integrating we find

1/2 〖my〗^2-1/2 〖my〗^2=∫_0^x▒F(x)dx

or

1/2 〖my〗^2-∫_0^x▒F(x)dx=1/2 〖my〗_0^2-∫_0^(x_0)▒F(x)dx

But 1/2 m(dx⁄dt)=1/2 〖my〗^2 is the kynetic energy of the system and

V(x)=-∫_0^x▒F(x)d

is the potential energy, Thus Equation(13.64)takes the from

1/2 〖my〗^2+V(x)=h

where the constant

h=1/2 〖my〗_0^2-∫_0^(x_0)▒F(x)dx=1/2 〖my〗_0^2+V(x_0)≥0

is the total energy of the system

Since Equation (13.65) was obtained by integrating (13.63) we see that (13,65) gives the family of paths in the xy phase plane.For a given value of it,the path given by (13.65)is a curve of constant energy in this plane .In others world along a particular path the total energy of the system is a constant the expresses the law of conservation of energy.

The critical points of the system (13.62) are the point with coordinates where are the roots of the equation . From the differential equation (13.63)we see that the path cross the x axis at right angles and have horizontal tangents along the lines .From Equation (13.65)we observe that the paths are symmetrical with respect to the x axis. From the equation (13.65)of the paths we find at once

y=±√(█(2/m [h-V(x) ]@))

This may be used to construct the paths in the following convenient way.

On a rectangular x Y plane construct the graph of Y= V (x) and the lines Y=h, for various of h (see figure 13.17a, where one such line is shown). For each fixed value h this shows the difference h- V (x) for each (x)

Directly below the x Y place draw the xy phase plane, with the y axis on the same vertical line as the axis [see figure 13.17b]. For each x multiply the difference h-V (x) obtained from the graph in step (1) by the constant factor 2/m and use (13.66) to plot the corresponding y value on the phase planc.

We observed that the critical point of the system (13.62) are the points (x,0) such that F(x)=0. Now note that F(x)=-V(x), and recall that V(x_t )=0 implies that V(x) has either a relative extremum or a horizontal inflection point at x=x_t. We thus conclude that all the critical points of (13.62) lie on the x axis and correspond to points where the potential energy V(x) has a relative, maximum, or a horizontal infection point. We now discus each of these three possibilities, using the graphical technique just introduced.

Suppose V(x) has a relative minimum at x=x_t (see Figure 13.18)

Let V(x_(t))=h_0≥0, and consider the path corresponding to a typical value h_1 of h such that h_1> h_0 For h=h_1, the values of y given by symmetry of the paths, we see that the path corresponding to h= h_1 is a closed ellipse –like curve surrounding the critical point (x_c,0). Thus, we conclude that if the potential energy V(x) has a relative minimum at x=x_c, then the critical point 〖(x〗_c,0) is a center and so is stable.

Suppose V (x) has a relative maximum at x=x_c(See Figure 13.19).

For h=h_2 the value of y given by(13.66) are real for all x sufficiently close to x_c .Moreover,the positive value of y has a realative minimum(say m) corresponding to x=x_c, and the negative value has a relative maximum (which would then be –m) corresponding to x=x_c. Thus for all x sufficiently close to x_c. the paths corresponding to h=h_2 are a pair of hyperbolae-like curves ,one opening up world above (x_c,m) and the other opening downward below (x_c-m) ,

Finally consider the paths corresponding to h=h_0 itself, For h=h_0,the value of y given by (13.66)for x=x_c is zero. The value of y given by (13.66) for x≠x_c (and sufficiently close to x=x_c) are and opposite sign. Thus there are two geometric curves with cross the x axis at x=x_c ,and these two curves provide four paths which approach and enter the critical point (x_(c,)0), two as t→ + ∞. From the discussion in this and the two preceding paragraphs, we conclude the following. 

1.     Suppose  has a horizontal inflection point at (see Figure 13.20)

To be define, suppose  is strictly increasing for all  (and sufficiently close to ); and let . Then, proceeding as in the discussions of cases 1 and 2, we find that the paths (for  sufficiently close to  ) have the appearance shown in the phase plane part of Figure 13.20. For typical ( ), the paths merely have - axis symmetry and intersect that axis at right angles. For , the symmetry remains, but the path has a so-called cups as it passes through the critical point ( ,0) (see Figure 13.20 again).

the critical point ( ,0) is a “degenerate” type of critical point and is unstable .The case in which  is strictly decreasing (instead of increasing ) is analogous.

We have presented a graphical treatment of the situation .We point out that an analytic treatment ,using principles of elementary  calculus, could also be employed to obtain the same results .We summarize these result as the following theorem.