ENGI 9420 Engineering Analysis

Faculty of Engineering and Applied Science
2008 Fall

Final Examination Solutions

  1.  
    1. Find the inverse of the matrix

      [ 1 0 0 ] [-1 1 1 ] [-1 0 -1 ]

      [row reduction]
      [row reduction]
      [row reduction]

      inverse of A = [ 1 0 0 ] [ 2 1 1 ] [-1 0 -1 ]

    2. Hence (or otherwise) solve the following problem:
      The frictional force F (in units of Newtons = kg m s–2) on a spherical body of radius a (m) falling with speed v (m s–1) through a viscous liquid of viscosity h (kg m–1 s–1) is of the form F = k eta^x a^y v^z, where   k, x, y and z are dimensionless constants.   Find the values of   x, y and z.

      kg^x m^(-x+y+z) z^(-x-z) = kg m s^(-2)
      Ax = b
      x = A^(-1) b
      Therefore

      x = y = z = 1
      and
      F = k eta a v

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

    dx/dt = y ; dy/dt = x^2 - x - xy/2

    1. Find both critical points of this system of first order ODEs.
      y = 0 and x^2 - x - xy/2 = 0
      y = 0 and x(x-1) = 0
      Therefore the only critical points of this system are

        (0, 0) and (1, 0)  

    2. Determine the nature and stability of both critical points.

      The coefficient matrix for the linear approximation to the non linear system near any critical point is
      A = [ Px Py ; Qx Qy ]

      Near the critical point (0, 0):

      A = [ 0 1 ; -1 0 ] , D = -4
      lambda = ±j
      The eigenvalues are a pure imaginary pair.
      The linear approximation is a centre, (which is stable but not asymptotically stable).
      The non-linear system is equivalent to the single second order homogeneous ordinary differential equation
      x" + x x' / 2 + x - x^2 = 0
      A damping term is present and is positive whenever x is positive.
      There is some cause to suspect that the non-linear system may have a

        stable focus at (0, 0)  

      However, any focus will resemble a centre more and more as one approaches the singularity at (0, 0).

      Near the critical point (1, 0):

      A = [ 0 1 ; 1 -1/2 ] , D = 17/4
      lambda = (-1 ± sqrt(17)) / 4
      The eigenvalues are a real pair of opposite sign.
      The linear approximation is a[n unstable] saddle point.
      The non-linear system also has a

        saddle point at (1, 0)  

    3. Sketch the phase portrait for the linear approximation to the non-linear system in the neighbourhoods of both critical points.
      [Note:   the general solution is not required.]

      The sketches are aided by the fact that   dx/dt = y
      The trajectories must therefore move in a clockwise sense around the centre.
      This also determines which two of the four trajectories that pass through the saddle point do so in an inwards direction.

      [phase portraits of linear systems near (0, 0) and (1, 0)]

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

      [phase portrait of non-linear system]
      [The red trajectories are just outside the separatrix.]

  1. An initial value problem is defined by

    dy/dx = y - x , y(1) = 1

    1. Use the standard fourth order Runge-Kutta method, with a step size of h = 0.2, to find the value at x = 1.2 of the solution of this initial value problem.

      f(xn, yn) = yn - xn ; x0 = y0 = 1; h = 0.2
      k1 = 0
      k2 = 0.1
      k3 = 0.11
      k4 = 0.222
      y1 = 1 + 0.2 (0 - 0.2 - 0.22 - 0.222) / 6
      Therefore

      y(1.2) approx= 0.9786

    2. Verify your solution to part (a) by solving the initial value problem analytically.

      dy/dx + (-1)y = -x
      which is a linear first order ordinary differential equation.
      integrating factor e^h = e^(-x)
      Integral {e^h R} dx = (x + 1) e^(-x)
      The general solution is
      y = x + 1 + C e^x
      y(1) = 1 ==> C = -e^(-1)
      y = x + 1 - e^(x-1) ==> y(1.2) = 0.97859...
      The value for y(1.2) found by the RK4 algorithm is therefore correct to five significant figures.

  1.  

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

      I = Integral[0 to 1] { 12xy + (y')^2 } dx

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

      F(x, y, y') = 12xy + (y')^2
      The Euler equation for extremals becomes
      d/dx(partial dF/dy') - partial dF/dy = 0
      y = x^3 + Ax + B
      (0, 1) is on the curve     implies 1 = 0 + 0 + B     implies y = x3 + Ax + 1
      (1, 1) is on the curve     implies 1 = 1 + A + 1     implies A = –1
      Therefore the extremal path is

      y = x^3 - x + 1

      Evaluating the integral along the extremal path G0 :
      I(Gamma0) = 5.2
      I(Gamma0) = 5.2
      I(Gamma0) = 5.2
      I(Gamma0) = 5.2
      Evaluating the integral along the simplest path G1 between (0, 1) and (1, 1), namely the straight line y = 1:
      I(Gamma1) = 6
      I(Gamma0) < I(Gamma1).     Therefore the extremal is a

        minimum  

    BONUS QUESTION

    1. If the integrand   F (x, y, y' )   is a function of   y'   only, then it follows that the extremal path is the straight line joining the endpoints, unless partial d^2 F / d(y')^2 ident= 0.   Show that, if partial d^2 F / d(y')^2 ident= 0, then the integral I = Integral[x0 to x1] { F(y') } dx has the same value, no matter what path   y = f (x)   is taken between the fixed points (x0, y0) and (x1, y1)

      F   is a function of   y'   only and
      F = Ay' + B
      I = A Integral 1 dy + B Integral 1 dx
      I = A (y1-y0) + B (x1-x0)
      The value of the integral   I   for fixed endpoints (x0, y0) and (x1, y1) therefore does not depend on the path   y = f (x)   taken between those two points.

  1. Sketch the graph of the odd periodic extension of the function

    f(x) = x (pi – x) , (0 <= x &<= pi)

    and find its Fourier half-range sine series expansion.

    [Graph of the odd periodic extension of f(x)]

    Any odd function has only sine terms in its Fourier series         implies an = 0 for all n.
    bn = 2/pi Integral { x(pi - x) sin(nx) } dx
    [using integration by parts]
    [evaluating at upper and lower limits]
    bn = 4(1 - (-1)^n) / (pi n^3)     		[tabular integration by parts]

    Therefore the Fourier half-range sines series for   f (x)   is

    f(x) = 4/pi Sum { (1 - (-1)^n) sin(nx) / n^3 }

    A more compact form for the series is   f(x) = 8/pi Sum { sin((2k-1)x) / (2k-1)^3 }
    The first few terms are   f(x) = 8/pi (sin x + sin 3x / 27 + sin 5x / 125 + ...
    The convergence is quite rapid, as one would anticipate from the sketch of the function and its odd periodic extension.

  1. A cylinder containing a viscous fluid is rotating at a constant angular speed   wo . The fluid moves with a velocity (in the cylindrical polar coordinate system) of

    vector v = rho omega_o vector phiHat

    1. Find the divergence of the velocity vector.

      div v = (1/rho) (0+0+0)
      Therefore, everywhere (except possibly on r = 0, the z axis),

      divergence of the velocity vector is zero

    2. Find the curl of the velocity vector.

      curl v = 2 rho omega_o kHat
      curl v = 2 rho omega_o kHat
      Therefore, everywhere,

      curl v = 2 rho omega_o kHat

    3. Find d/dt curl v

      curl v is a constant vector everywhere.
      Therefore

      d/dt curl v ident= 0

      Note that div v is a scalar, but curl v is a vector.
      Therefore the answer to part (a) must be a scalar, while the answers to parts (b) and (c) must be vectors.

        Return to the index of assignments   [Index of Assignments]
        Return to your previous page   [Return to your previous page]
        Created 2008 12 17 and most recently modified 2008 12 17 by Dr. G.H. George.