Solved Problems

9. Linear Systems

Problem 1

In each case, use Maxima to find the eigenvalues and eigenvectors of the system. Determine what kind of equilibrium point each of the systems have and plot their phase portraits.

  1.     
  2.     

(a) In Maxima

(%i1) vars: [x, y]$
(%i2) A: matrix ([1,1], [4,1])$
(%i3) eigenvectors (A);
(%o3)    
(%i4) plotdf (list_matrix_entries (A.vars), vars, [vectors,"blank"]);

And after tracing some evolution curves, the phase portrait is the following

Phase portrait of problem 9.1.a

the eigenvalues are 3, with eigenvector (1, 2), and −1, with eigenvector (1,-2). The equilibrium point is then a saddle point.

(b)

(%i5) A: matrix ([-3,sqrt(2)], [sqrt(2),-2])$
(%i6) eigenvectors (A);
(%o6)    
(%i7) plotdf (list_matrix_entries (A.vars), vars, [vectors,"blank"], [x,-2,2], [y,-2,2]);

The eigenvalues are −4, with eigenvector (1, -1/ ), and −1, with eigenvector (1, ). The equilibrium point is an attractive node.

Phase portrait of problem 9.1.b

(c)

(%i8) A: matrix ([1,-1], [1,3])$
(%i9) eigenvectors (A);
(%o9)    
(%i10) plotdf (list_matrix_entries (A.vars), vars, [vectors,"blank"], [x,-2,2], [y,-2,2]);

There is only one eigenvalue equal to 2, with eigenvector (1, −1). The equilibrium point is an improper repulsive node.

Phase portrait of problem 9.1.c

Problem 2

The figure shows the hypothetical evolution curve of a ball in free fall which after bouncing off the ground is thrown vertically up, in the case when there is no energy loss during the motion or the collision with the ground. The positive part of the plot corresponds to the vertical motion of a projectile with negligible air resistance. The negative part of the plot represents the elastic deformation of the ball during the collision with the ground; during the time of contact with the ground, it has been assumed that the vertical motion of the ball is that of a simple harmonic oscillator without energy dissipation.

Phase portrait of a bouncing ball

Knowing that the maximum height the ball reaches is  = 10 m and the maximum deformation of the ball during the collision with the ground is  = 1 cm, determine:

  1. The maximum speed of the ball during its motion.
  2. The angular frequency of the elastic deformation of the ball.
  3. The time that the ball remains in contact with the ground.

(a) At the point of maximum height, with coordinates (10, 0) in phase space, the mechanical energy is

and at the point (0, ), where the speed is maximum, the potential energy is zero and the mechanical energy is then equal to the kinetic energy

(b) At the point (−0.01,0), where the deformation of the ball is maximum, the kinetic energy is zero and the mechanical energy is equal to the potential energy of a harmonic oscillator with elastic constant

The angular frequency of the oscillation is then

(c) Since the evolution curve for the ball, during its time of contact with the ground, is half of an ellipse, the time it remains in contact with the ground is half of the period of the harmonic oscillator

Problem 4

A cylinder with mass is hanging on the vertical from a spring with elastic constant , as shown in figure 6.2; if is the height of the center of mass of the cylinder in the position where the spring is not deformed and neglecting the air resistance:

  1. Find the equation of motion, either from the Lagrange equation or from Newton's second law.
  2. Find the value of at the equilibrium point.
  3. Show that the system can be written as a linear system changing variable to a new variable and that the equation of motion in terms of is that of a simple harmonic oscillator with angular frequency .

(a) The kinetic energy and the potential energy, elastic plus gravitational, are

The Lagrange equation is

and the equation of motion is

(b) In the equilibrium point and are both zero; therefore,

(c) For the system to be linear, the constant term should not appear in the equation of motion. We can introduce a new variable such that

namely, and thus , leading to the equation of motion

which is the equation of a simple harmonic oscillator with angular frequency

Problem 5

A cylinder has circular base with area  cm2, height  cm and density  g/cm3. Since its density is less than that of water,  g/cm3, when the cylinder is immersed in water it floats on the surface, with a part of height outside the water, as shown in the figure ( ). If the cylinder is pushed downwards, it oscillates with changing in terms of time. Use the following procedure to analyze the oscillations of the cylinder:

  1. The buoyant force exerted by the water on the cylinder is equal to the weight of the water that the cylinder displaces; namely,
    Find the expression for the resultant force on the cylinder, in terms of (the force of resistance to the motion exerted by the water can be neglected compared to the weight of the cylinder and the buoyant force).
  2. Derive the equation of motion for the cylinder (expression of in terms of ).
  3. Determine the value of in the equilibrium position of the cylinder.
  4. Show that the dynamical system associated to the motion of the cylinder is linear and find its matrix.
  5. Show that the equilibrium point in phase space is a center, which means that the cylinder oscillates, and find the oscillation period of the cylinder.
Cylinder floating in water
  1. The resultant force is vertical and with value equal to (positive if it points upwards or negative if it points downwards):
    in grams times cm/s2 and in centimeters.
  2. The mass of the cylinder in grams is
    and the equation of motion is then
    (in cm/s2 and in cm).
  3. The value of for which the acceleration, , is zero is
  4. The two evolution equations are:
    Introducing a new variable , then and the evolution equations become
    which describe a linear dynamical system with matrix
  5. The characteristic equation for this matrix is . The two eigenvalues are then purely imaginary numbers
    That means that the equilibrium point at  cm is a center and the motion of the cylinder is oscillatory, with angular frequency equal to 8.250 and period (in seconds):

Problem 7

A transformer has two coils, the first coil has resistance and inductance and the second coil has resistance and inductance . When an oscillating voltage source is connected to the first coil, it produces a current , which induces a current in the second coil. When the source is disconnected from the first coil, the two currents gradually go down to zero, according to the following two equations:

where is the mutual inductance between the two coils and all the constants , , , and are positive.

  1. Write down the transformer equations as the evolution equations for a linear dynamical system and find the matrix of that system.
  2. In a real transformer, is less than . For the case , , , , (using some system of units in which those parameters are between 0 and 10), find out what kind of equilibrium point the system has.
  3. The values , , , and , describe a hypothetical case which cannot be a real transformer, since . Determine what kind of equilibrium point that system has and explain why it cannot describe a real transformer.

(a) The two equations of the transformer can be solved to obtain the expressions for and . Then Maxima's program coefmatrix can be used to extract the matrix of those two linear evolution equations; the input parameters for that program are the list of linear expressions and the list of variables.

(%i11) eq1: L1*dI1+M*dI2+R1*I1=0$
(%i12) eq2: L2*dI2+M*dI1+R2*I2=0$
(%i13) sys: solve ([eq1, eq2], [dI1, dI2]);
(%o13)
(%i14) vars: [I1, I2]$
(%i15) A: coefmatrix (map (rhs, sys[1]), vars);
(%o15)

(b) The values given for the parameters are substituted in the system matrix and the eigenvalues and eigenvectors of that matrix are then computed:

(%i16) A1: subst ([L1=2, L2=8, M=3, R1=1, R2=2], A);
(%o16)
(%i17) eigenvectors (A1)$
(%i18) float (%);
(%o18) [ [ [-1.527, -0.1871], [1.0, 1.0] ], [ [ [1.0, -0.4484] ], [ [1.0, 1.115] ] ] ]

The two real and negative eigenvalues mean that the equilibrium point is an attractive node.

(c) The components of the phase velocity are the right-hand sides of the equations stored in the list sys

(%i19) u1: subst ([L1=2, L2=8, M=3, R1=1, R2=2], map (rhs, sys[1]));
(%o19)
(%i20) plotdf (u1, vars)$

And after plotting some evolution curves, including those that approach the equilibrium points, the following phase portrait is obtained:

Phase portrait of problem 9.7.a

(d) Replacing the values given, the matrix of the system and its eigenvalues are,

(%i21) A2: subst ([L1=2, L2=8, M=5, R1=1, R2=2], A);
(%o21)
(%i22) eigenvectors (A2)$
(%i23) float (%);
(%o23) [ [ [-0.1498, 1.483], [1.0, 1.0] ], [ [ [1.0, 0.9348] ], [ [1.0, -0.5348] ] ] ]

Therefore, the equilibrium point would be a saddle point, which cannot describe a real transformer since it would be an unstable system in which the currents would grow to infinity.

Problem 8

A radioactive isotope A decays giving rise to another radioactive isotope B; this second isotope then decays into a stable isotope C.

If and stand for the number of A and B isotopes at any given instant , their derivatives with respect to time obey the following equations:

where is the decay constant for isotopes A (probability that an isotopes decays during a unit of time) and is the decay constant for isotopes B.

  1. Find the matrix of the system and its eigenvalues.
  2. Taking into account that the decay constants and are positive, explain what kind of equilibrium point the system can have for any possible values of those constants.
  3. If at an initial instant the number of isotopes of the three kinds A, B and C are respectively , and , where is Avogadro's number, what will be the values of , and after a long time?

(a) The system matrix and its eigenvalues are:

(%i24) A: matrix ([-k1, 0], [k1, -k2]);
(%o24)
(%i25) eigenvectors (A);
(%o25) [ [ [ , ], [1, 1] ], [ [ [0, 1] ], [ [1, ] ] ] ]

(b) There are two different cases. First, if and are different, there will be two real and negative eigenvalues; hence, the equilibrium point would be an attractive node. But if the two constants and are equal, there is only one eigenvalue, real and negative, and the equilibrium point would be an improper attractive node.

(c) Since the equilibrium point at the origin is attractive, after a long interval of time the system approaches the equilibrium point, namely, and . The extinction of the A and B isotopes mean that all the initial isotopes will become of kind C and, therefore, the final number of C isotopes, , equals the total number of isotopes at the initial instant, .

Problem 9

The evolution equations for a family of dynamical systems are:

where is a real parameter with any value between and . Determine for which values of the point ( , ) = (0,0) can be an attractive or repulsive node, attractive or repulsive focus, center or saddle point.

There are several ways to solve this problem. A simple method is as follows: the matrix of the system is,

which has trace and determinant equal to:

Thus, the relation between the trace and the determinant is . In a 2-d plot with in the domain axis and in the other axis, this is represented as a straight line with slope 1, passing through the point where the trace is and the determinant is zero.

The borderline between the regions where the system has a node or a focus is the parabola , which intersects the straight line at the two points where:

The following plot shows the straight line and the parabola.

Relation between determinant and trace

The equilibrium point is a saddle point when the trace of the matrix is less than , it is an attractive node if the trace is between and , an attractive focus if the trace is between and 0, a center if the trace is equal to zero, a repulsive focus if the trace is between 0 and or a repulsive node if the trace is greater than . Keeping in mind that is equal to the trace, we can then conclude:

Notice that for the equilibrium point is an improper node, which has already been include in the cases above. If , the determinant is zero, indicating that the system reduces to a single differential equation, , which corresponds to a system with only one state variable and an attractive equilibrium point. In that case, the second evolution equation shows that equals plus a constant.