ENGI 9420 Engineering Analysis

Faculty of Engineering and Applied Science
2006 Fall

Final Examination Solutions

  1. Use the method of Gaussian elimination to find the inverse of the matrix

    [ 1 0 2 ] [ 1 1 1 ] [ 3 2 5 ]

    Reduce the augmented matrix   [A | I]   to   [I | A-1].
    [ 1 0 2 | 1 0 0 ] [ 1 1 1 | 0 1 0 ] [ 3 2 5 | 0 0 1 ]
    Row operations:     R 2 – R 1   and   R 3 – 3R 1:
    [reduced augmented matrix]
    R 3 – 2R 2:
    [reduced augmented matrix]
    R 1 – 2R 3   and   R 2 + R 3:
    [I | A^(-1)]
    Therefore

    A^(-1) = [ 3 4 -2 ] [ -2 -1 1 ] [ -1 -2 1 ]

  1. The second order ordinary differential equation

    x" + 3x' + 2x – x^2 = 0

    may be represented by the linked pair of first order ODEs

      dx/dt = y ; dy/dt = x^2 – 2x – 3y

      1. Find all of the critical points of this system of first order ODEs.

        P = dx/dt and Q = dy/dt
        y = 0 and x(x-2) = 0
        Therefore the critical points are

          (0, 0)   and   (2, 0)  

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

        Near any critical point (a, b), the coefficient matrix becomes
        [ 0 1 ] [ 2a-2 -3 ]

        Near (0, 0),   A = [ 0 1 ] [ -2 -3 ]
        Trace(A)   =   a + d   =   –3   <   0
        det(A)   =   ad – bc   =   2   >   0
        D   =   (a – d)2 + 4bc   =   9 – 8   =   1   >   0
        eigenvalues = – 2 or –1
        The eigenvalues are real, negative and distinct.   Therefore

          (0, 0)   is a[n asymptotically] stable node  

        Near (2, 0),   A = [ 0 1 ] [ +2 -3 ]
        Trace(A)   =   a + d   =   –3   <   0
        det(A)   =   ad – bc   =   –2   <   0
        D   =   (a – d)2 + 4bc   =   9 + 8   =   17   >   0
        eigenvalues = (–3 ± sqrt(17)) / 2
        The eigenvalues are real and of opposite sign.   Therefore

          (2, 0)   is a[n unstable] saddle point  

      3. Find the general solution to the linear approximation to the non-linear system in the neighbourhood of the origin (x, y) = (0, 0) only.

        Near (0, 0),   A = [ 0 1 ] [ -2 -3 ]
        Find the eigenvector for the eigenvalue   –2:
        eigenvector = c1 (1, –2)
        (where   c1   is any non-zero real number)

        Find the eigenvector for the eigenvalue   –1:
        eigenvector = c2 (1, –1)
        (where   c2   is any non-zero real number)

        Therefore the general solution near (0, 0) is

        (c1 exp(-2t) + c2 exp(-t), -2c1 exp(-2t) - c2 exp(-t))

      4. Sketch the phase portrait for the the linear approximation to the non-linear system in the neighbourhood of the origin (x, y) = (0, 0) only.

        The two straight-line trajectories are
        c1 = 0 leads to y = –x
        c2 = 0 leads to y = –2x
        The limiting behaviours of all other trajectories are
        t --> -oo leads to y/x = –2 at infinity
        t --> +oo leads to y/x = –1 at (0, 0)

        phase portrait of linear approximation near (0, 0)

      BONUS QUESTION

      1. Sketch the phase portrait for the non-linear system in a region that includes all critical points.

        Incorporating the influences of the stable node at (0, 0) and the saddle point at (2, 0), we obtain

        phase portrait of linear approximation near (0, 0)
        [Enhanced Maple® phase portrait]

      [A Maple file is also available for this question.]

    1. An initial value problem is defined by

      dy/dx = x + y , 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 = –0.8 of the solution of this initial value problem.

        x0 = -1 , y0 = 1 , f(x,y) = x+y , h = 0.2
        k1 = 0
        k2 = 0.1
        k3 = 0.11
        k4 = 0.222
        y1 = 1.0214
        Therefore

        y(-0.8) approx= 1.0214

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

        y' - y = x
        which is a linear ODE, with   P = –1   and   R = x.
        integrating factor = exp(-x)
        Integral (e^h R) dx = -(x+1) exp(-x)
        General solution
        Apply initial condition
        The complete solution is therefore

        y = exp(x+1) – (x+1)

        y(-0.8) = 1.021403
        The standard fourth order Runge-Kutta method has therefore provided an extremely accurate value, correct to better than four parts in a million.

      An Excel file is available to illustrate this question.

    1. Find the extremal   y(x)   for the integral

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

      that passes through the points (1, 1) and (2, 4).

      F = (1 + y')^2 / (2x + 1) dF/dy = 0
      The Euler equation for extremals becomes
      d/dx (2(1 + y') / (2x + 1)) = 0
      y' = c1(2x + 1)/2 - 1
      which leads to the general solution
      y' = c1(x^2 + x)/2 - x + c2
      But the points (1, 1) and (2, 4) both lie on this curve.
      [substitute into general solution]
      c1 = 2 , c2 = 0
      y' = 1(x^2 + x)/2 - x + 0
      Therefore the extremal for the integral   I   is

      y = x^2

    1. Find the Fourier series expansion of the function

      f(x) = 1 – abs(x) , (abs(x) <= 1)

      [graph of f(x), showing symmetry] bn = 0 for all n
      L = 1 .   Therefore (using symmetry)
      a0 = 1

      For   n > 0,

      an = 2 Integral[0 to 1] { (1-x) cos(n pi x) }
      [Integral]
      an = 2(1 - (-1)^n) / (n pi)^2

      Therefore the Fourier series of   f (x) is

      		 [tabular integration by parts of (1 - x) cos(n pi x) ]

      f(x) = 1/2 + (2/pi^2) Sum {(1-(-1)^n) cos(n pi x) / n^2}

      The first few terms of the series are
      f(x) = 1/2 + (4/pi^2) (cos(pi x) + cos(3 pi x)/9 + ...)
      The summation can be re-written as the equivalent form
      f(x) = 1/2 + (4/pi^2) Sum { cos((2k-1) pi x) / (2k-1)^2}

      [The graph of the partial sum of just the first three non-zero terms is displayed here.   The convergence is quite rapid, except near the non-differentiable points at all integer values of x:]

      [graph of f(x) and its Fourier series partial sum]

    1. For the vector field   F = –100 / r^2 rHat, where   r = < x, y, z >,

      1. find the curl of F.

        Cartesian method:

        First, note that   vector r = scalar r × unit vector rHat
        vector_F = -100 / r^3 vector_r
        Also note that
        partial dr/dx = x/r
        and, by symmetry,
        dr/dy = y/r and dr/dz = z/r
        By the chain rule,
        partial dF/dx = 300 x / r^5
        and, by symmetry,
        dF/dy = 300y / r^5 and dF/dz = 300z / r^5

        curl F = [determinant] = vector 0

        OR

        Spherical Polar Method:

        F   is a purely radial vector field:
        F = –100 / r^2 rHat
        The scale factors in spherical polar coordinates are:
        h_r = 1 , h_theta = r , h_phi = r sin theta
        curl F = vector 0
        Therefore

        curl F = vector 0

        Note that any spherically symmetric vector field, (that is, a vector field whose magnitude depends only on the distance from the origin and whose direction everywhere is directly towards or directly away from the origin only), is necessarily irrotational:
        vector F = F(r) rHat
        curl F = vector 0

      2. find the divergence of F.

        Cartesian method:

        By the results in part (a), together with the product rule of differentiation,
        partial d(xF)/dx = 300x^2 / r^5 - 100/r^3
        and, by symmetry,
        partial d(yF)/dy = 300y^2 / r^5 - 100/r^3 and partial d(zF)/dz = 300z^2 / r^5 - 100/r^3
        div F = 0

        OR

        Spherical Polar Method:

        For the general vector field
        F = Fr rHat + Ftheta thetaHat + Fphi phiHat
        div F = ...
        For the vector field in this question, we have
        div F = 0
        Therefore

        div F = scalar 0

        [Note that the inverse square law central force F = k / r^n rHat has zero divergence and is therefore free of sources and sinks everywhere (except possibly at the origin).]

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    Created 2006 12 14 and most recently modified 2006 12 15 by Dr. G.H. George.