ENGI 9420 Engineering Analysis

Faculty of Engineering and Applied Science
2007 Fall

Final Examination Solutions

  1. Use the method of Gaussian elimination to solve the linear system

    [ 1 1 1 | 0 ] [ 2 3 2 | 1 ] [ 3 2 -2 | 5 ] [ 6 6 1 | 6 ]

    [row reduction]
    [row reduction]
    [row reduction to echelon form]
    which obviously has the unique solution

    (1/5, 1, -6/5)

    [Note that one cannot simply drop the fourth row and reduce the resulting square 3×3 matrix.
    One must guard against the possibility of an inconsistent system (which has no solution at all).]

  1. A pair of simultaneous first order ODEs is defined by

    dx/dt = y ; dy/dt = x^2 + 4x

    1. Find both critical points of this system of first order ODEs.
      y = 0 and x(x+4) = 0

      (0, 0), (-4, 0) only

    2. Determine the nature and stability of both critical points.   In particular, provide a reason why it is likely that one of the critical points of the non-linear system is a centre (not a focus).

      Near any critical point (a, b), the non-linear system may be approximated by the linear system
      [x' | y'] = [ 0 1 | 2x+4 0 ] [ x-a | y-b ]

      Near (0, 0):
      A = [0 1 | 4 0 ]
      D = 16 > 0, lambda = ±2
      The eigenvalues are of opposite sign.   This critical point is therefore a

        saddle point  

      (which is automatically unstable).

      Near (-4, 0):
      A = [0 1 | -4 0 ]
      D = -16 < 0, lambda = ±2j
      The eigenvalues are a pure imaginary pair.   For the linear approximation, this critical point is a centre.
      The corresponding critical point in the original non-linear system could therefore be any of a stable focus, a centre or an unstable focus.
      However, the non-linear system of first order ODEs can be rewritten as the single second order ODE
      x" - (x^2 + 4x) = 0
      which has no damping term (first derivative).
      It is therefore likely that the critical point at (-4, 0), in the non-linear system, is also a

        centre  

      (which is automatically stable but not asymptotically stable).

    3. Sketch the phase portrait for the linear approximation to the non-linear system in the neighbourhoods of both critical points.

      For the centre, the orientation of the trajectories is easily determined to be clockwise, upon noting that
      y = dx/dt > 0 implies x is increasing
      For the saddle point, the exact equations of the asymptotes were not required.   Again, the directions of the trajectories are determined by the fact that all trajectories must move to the right above the x axis and to the left below it.

      [phase portraits of linear system near each critical point]

      The equations of the asymptotes can be found from the eigenvectors of the linear system:
      [ alpha | beta ] = [ ±2 | 4 ]
      The equations of the asymptotes are therefore   y = ±2x.

    4. Hence sketch the phase portrait for the original non-linear system.

      [phase portraits of non-linear system]

    BONUS QUESTION

    1. Find the equation of the separatrix and add it to your sketch in part (d).   Note on your sketch the exact values of the x axis intercepts for the separatrix.

      All trajectories are solutions to the differential equation
      dy/dx = (x^2 + 4x) / y
      Separating the variables,
      y^2 / 2 = x^3 / 3 + 2x^2 + C
      The separatrix divides the region of closed orbits about the centre from the open trajectories further away.   The separatrix passes through the saddle point.   Therefore the point (0, 0) must be on the separatrix.
      C = 0
      In simplified form, the equation of the separatrix is therefore

      y^2 = 2 x^3 / 3 + 4 x^2

      To find all x axis intercepts for the separatrix:
      2/3 x^2 (x + 6) = 0 ==> x = 0 or -6
      The intercepts are therefore at

      (0, 0), (-6, 0) only

  1. A function   f (x)   is defined by

    f(x) = x^3 + e^x

    1. Show that   f ' (x) > 0   for all x.
      It then follows that   f (x) = 0   has a unique solution.

      f'(x) = 3x^2 + e^x
      3x^2 > 0 and e^x > 0 for all x

    2. Use Newton’s method to find the solution to   f (x) = 0   correct to four decimal places.
      Justify your choice of an initial guess   x0.

      f (x) is monotonically increasing.
      The y axis intercept of   y = f (x)   is at f (0) = 1 > 0.
      f (–1) = –1 + e–1 < 0
      The unique solution must therefore be somewhere in (–1, 0).
      Choose as the initial guess   x0 = –0.5 .

      [If an accurate sketch or plot is available, then
      an initial guess of   x0 = –0.8 becomes plausible.]

      [graph of y = f(x)]

      xn f(xn) = xn^3 + e^(xn) f'(xn) = 3xn^2 + e^(xn) -f(xn)/f'(xn)
      -0.500 000 000+0.481 530 660 +1.356 530 660-0.354 972 190
      -0.854 972 190-0.199 670 370 +2.618 227 358+0.076 261 662
      -0.778 710 528-0.013 204 851 +2.278 167 754+0.005 796 259
      -0.772 914 269-0.000 070 566 +2.253 855 090+0.000 031 309
      -0.772 882 960-0.000 000 002

      Therefore, correct to four decimal places,

      x = -0.7729

      [An Excel spreadsheet for this question is available here.   It allows you to experiment with various choices of   x0.]

  1. Find the path   y = f (x)   between the points (0, 1) and (1, 1) for which the integral

    I = Integral[0 to 1] { 2x + 3y + (y')^2 } dx

    has an extremum and determine whether the extremum is a maximum or a minimum.

    dF/dy = 3, dF/dy' = 2 y'
    The Euler-Lagrange equation   d/dx(dF/dy') - dF/dy = 0   becomes
    y" = 3/2
    y = 3/4 x^2 + Ax + B
    But this path must pass through both (0, 1) and (1, 1):
    [applying both conditions]     B = 1 and A = -3/4
    Therefore the extremal to the functional   I   is

    y = 3/4 x^2 - 3/4 x + 1

    Evaluating the integrand at the only extremum,
    [substitute y = 3/4 x^2 - 3/4 x + 1 into F]
    F = 9/2 x^2 - 5/2 x + 57/16
    [Evaluate integral]
    I = 61/16
    We need to compare this value of the integral to the value along any other path.
    The simplest path between two points is a straight line.
    The straight-line path through (0, 1) and (1, 1) is   y = 1.
    The value of   I   along this other path is
    I = 4 > 61/16
    Therefore the extremum of   I   is a

      minimum.  

  1. Find the Fourier series expansion of the function

    f(x) = 4 – x^2 , (abs(x) <= 2)

    f (x)   is an even function of   x   all b_n terms are zero
    a_n integral
    a0 = 16/3
    an = ...
    an = ...
    an = -(16/pi^2) (-1)^n / n^2
    		 table for integration by parts

    Therefore the Fourier series for   f (x)   is

    8/3 - 16/pi^2 Sum { (-1)^n / n^2 cos(n pi x / 2) }

  1. An arbitrary purely radial vector field F may be defined in spherical polar coordinates by F(r) = f(r) r^, where   f (r)   is some differentiable function of the distance   r   from the origin.

    1. Show that any purely radial vector field is irrotational (that is, curl F identically= vector 0).

      determinant form of curl F
      But   f (r)   is a function of   r   only.   Therefore   curl F identically= vector 0

    2. Show that, in order for the divergence of F to be zero everywhere,   f (r)   must be inversely proportional to the square of the distance from the origin.

      div F = 1/r^2 d/dr { r^2 f(r) }
      div F = 0 ==> f(r) = k/r^2
      (where   k   is a constant).   Therefore, in order for a purely radial vector field to be source free (zero divergence) everywhere (except possibly at the origin),   f (r)   must have the functional form

      f(r) = k/r^2

      [Note that the central force laws of nature, the gravitational and electrostatic forces, are both of this type.]

        Return to the index of solutions   [Index of Solutions]
        Return to your previous page   [Return to your previous page]

        Created 2007 12 13 and most recently modified 2007 12 15 by Dr. G.H. George.