For a given solution x (t), the total energy E can be expressed as,

E = 12mx˙2 + V(x)

Here, V(x) is the potential energy and 12mx˙2  is the kinetic energy.

This kinetic energy is also called a conserved quantity because it is constant as a function of time.

The relativistic equation for the orbit of a planet around the sun is given as,

d2udθ2 + u = α + εu2

Here, u is 1r, α is any parameter having a positive value, εu2 is Einstein’s correction, ε is any parameter having a very small positive value and r , θ are the polar coordinates of the planet.

The objective is to find the equation of the given system is to be rewritten in the (u, v) phase plane. It is given that,

v = dudθ

Differentiate it with respect to θ,

dvdθ=d2udθ2

Substitute, α + εu2- u for d2udθ2 in the above equation and solve,

dvdθ=α + εu2- u.

The objective is to find the system can be rewritten as,

u˙ = v

And,

v˙ = (α + εu2- u)

Fixed points occur for the following conditions,

u˙ = 0

And,

v˙ = 0

Substitute, 0 for v˙ in (α + εu2- u)=0 and solve,

(α + εu2- u) = 0

The roots of the equation can be calculated as,

u = - b ± b2- 4ac2a

Substitute, -1 for ε for a and α for c and solve,

u = 1 ± b2- 4αε2ε

Thus, the equilibrium points come out to be (1 ± b2- 4αε2ε,0)

The objective is to be shown that one of the equilibrium is a center in the (u, v) phase plane, according to the linearization.

The Jacobian matrix at a general point (u, v) is given by,

A = (∂u˙∂u∂u˙∂v∂v˙∂u∂v˙∂v) Substitute, v for u˙, (α + εu2- u) for v˙ and solve,

A = (012uε0)

Evaluate the matrix at the fixed point (1 + 1- 4αε2ε,0),

Substitute, 1 + 1- 4αε2ε for x and 0 for y in the Jacobian matrix,

A(1 + 1 - 4αε2ε,0) = (011 - 4αε0)

For the above Jacobian matrix determinant, Δ comes out to be - 1 - 4αε and trace, τ comes out to be 0. This implies that this fixed point is a saddle point.

Evaluate the matrix at the fixed point (1 ± 1- 4αε2ε,0),

Substitute, 1 - 1- 4αε2ε for x and 0 for y in the Jacobian matrix,

A(1 - 1 - 4αε2ε,0) = (01−1 - 4αε0)

For the above Jacobian matrix determinant, Δ comes out to be 1 - 4αε and trace, τ comes out to be 0. This implies that this fixed point is a linear center.

Thus, it is proved that one of the equilibrium is a center.

The theorem for non-linear centers states that,

A system is considered as x˙ = f (x). Here, x = (x, y)∈R2 and f is continuously differentiable. Suppose a conserved quantity E (x) exists and x* is an isolates fixed point.

If a local minimum of E is x*, then all trajectories close to x* are closed.

According to this above theorem, the center is a non-linear center provided that the fixed point is a local minimum of a conserved quantity.

The energy equation for the system can be expressed as,

E = 12 x˙2 - ∫y˙dx

Substitute, α + εu2- x for y˙ and solve,

E = 12 x˙2 - ∫(α + εu2 - x) dx

The divergence of energy, ∇E for the above equation, will come out to be,

∇E = (α + εu2- x , x˙)

For the calculated fixed point, the divergence is,

∇E(1 - 1 -  4αε2ε,0)=(0,0)

Determine the matrix for the above energy equation

|E|=|ExxEyxExyEyy|

Here, the subscripts determine the partial derivatives.

Solve the above matrix by substituting the above value of E,

|E|=|     -2εx˙-α-εx2+xx˙-α-εx2+x       1|

For a fixed point (1 - 1- 4αε2ε,0)

|E|(1 - 1- 4αε2ε,0)= - 2ε

Since, (−2ε<0), then (1 - 1- 4αε2ε,0) is a local minimum.

Thus, the center is a non-linear center.

The objective is to be shown that the equilibrium point (1 - 1- 4αε2ε,0) corresponds to a planetary orbit. It can be done by showing that the radius of this orbit is constant.

It is given that,

u = 1r

r = 1u

Substitute, 1 - 1- 4αε2ε for u and solve,

r = 2ε1 - 1 - 4αε

This generated value is a constant as both the parameters ε and α are constants.

Thus, the equilibrium point (1 - 1- 4αε2ε,0) corresponds to a planetary orbit.