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Nature of stationary points three variable functions

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Find and characterise the stationary points for F(x,y,z) = x2 + xy + y2 - 2z2 +3x -2y +z






    3. The attempt at a solution
    I found fx, fy, fz and let them equal to 0. Solving gives me the critical point (-8/3,7/3,14). From here I'm not sure how to determine the nature of this critical point. I know how to check given a two variable function, but for 3 I am a bit stuck. I think I have to find the determinant of the Hessian with all second derivative entries but is there an easier way than this?
     
  2. jcsd
  3. Sep 8, 2012 #2

    HallsofIvy

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    (Was that z value a typo? It is certainly not "14".)

    The best way to find the "nature" of the critical points of a general function of three or more variables is to first write the "true" second derivative. Not the partials. Just as the "true" first derivative of a function of three variables is a vector, the gradient,
    [tex]\begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{bmatrix}[/tex]
    so the "true" second derivative is the matrix
    [tex]\begin{bmatrix}\frac{\partial^2f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial x\partial z}\\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y\partial z} \\ \frac{\partial^2 f}{\partial z\partial x} & \frac{\partial^2 f}{\partial z\partial y} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}[/tex]

    Set the x, y, and z equal to the coordinates of the critical point and find the eigenvalues. If they are all negative, that is a maximum, if all postive, a minimum, and, in any other case a "saddle point"- though you can characterize more fully by the number of positive and negative eigenvalues.
    For this particular function, that is relatively straight forward.
     
    Last edited: Sep 8, 2012
  4. Sep 8, 2012 #3

    Ray Vickson

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    Finding the determinant is not enough: you need to test if the Hessian H is positive definite or positive semi-definite, negative definite or negative semi-definite, or indefinite. The second-order optimality conditions at a *stationary point* are:
    (1) Necessary for a local min: H is positive semi-definite (psd).
    Sufficient for a strict local min: H is positive definite (pd).
    (2) Necessary for a local max: H is negative semi-definite.
    Sufficient for a strict local max: H is negative-definite.
    Note: if H is indefinite, the point is neither a max nor a min---it is a "saddle point".
    The easiest way to test for negative (semi-) definiteness of H is to test the positive (semi-) definiteness of K = -H.

    There are some simple tests that allow you to bypass the details in some cases: (i) in order that H be psd it is necessary that all diagonal elements H(i,i) >= 0; (ii) if H is psd and H(i,i) = 0 for some i, then all elements in row and column i of H must also be zero (that is, we need H(i,j) = H(j,i) = 0); if this fails, H is indefinite. (iii) In order that H be pd it is necessary that all diagonal elements H(i,i) > 0. (Thus, if H has non-zero diagonal elements of opposite signs it is automatically indefinite.) All these things can be determined without additional work. If all these tests are "passed", then to test for psd or pd the standard way nowadays is to apply Cholesky factorization (which DOES apply to both psd and pd matrices, unlike what it says in some articles). We usually do not bother trying to find the eigenvalues unless we need them for some other purposes. (Cholesky factorization is so easy that I used to have it programmed into my HP hand-held calculator; you can easily do by hand matrices up to about 8x8.)

    RGV
     
    Last edited: Sep 8, 2012
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