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Homework Help: Extrems values on armonic function

  1. Dec 26, 2017 #1
    1. The problem statement, all variables and given/known data

    is an armonic function and C2(R2)
    a) Suposse a point P0 / fxx(P0) > 0. Prove that P0 is not a extreme value.
    b) consider D = {(x,y) / x2 + y2 < 1} and suposse fxx > 0 for all (x,y) in D. Prove that: if f(x,y) = 0 in x2 + y2 = 1, so f(x,y) = 0 for all (x,y) in D.

    2. Relevant equations

    In armonic funcions laplacian = 0

    3. The attempt at a solution

    I intuit that P0 is a saddle point, and I try to show this in the hessian matrix and I can´t
  2. jcsd
  3. Dec 26, 2017 #2


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  4. Dec 26, 2017 #3
    With the hessian i can classify the critical points, and i know that fxx > 0 and fyy < 0 (because f is an armonic function)
  5. Dec 26, 2017 #4


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    so you know about the diagonal components of your 2 x 2 Hessian, ##\mathbf H##. (and they cannot evaluate to zero, why?)

    1.) I can find some ##\mathbf x_1##, where ##\mathbf x_1^* \mathbf H \mathbf x_1 \gt 0##.
    2.) I can find some ##\mathbf x_2##, where ##\mathbf x_2^* \mathbf H \mathbf x_2 \lt 0##.

    Why? You should in fact be able to come up with these ##\mathbf x##'s.

    This quadratic form argument is enough to prove a saddle point. (Why?)
  6. Dec 26, 2017 #5
    and why evaluate on zero? if i can see that fxx = - fyy implies that is a saddle point I can do the second part... sory I am confused
  7. Dec 26, 2017 #6


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    I'm not sure what this means. Your posts in general are too light on details and a bit tough to figure out.

    Hmmm, if you're not familiar with quadratic forms: what tests are you aware of to prove something is a saddle point? Please list them, preferably with blurb on the intuition for why it makes sense. There are a lot of recipes you can follow, but having some intuition for why... makes things a lot easier. (Are you aware of what a taylor polynomial would look like here if you truncate it to have a quadratic remainder? This is probably the most direct way of making sense of these different saddle point tests.)
  8. Dec 26, 2017 #7
    I usually use the sylvester´s criterion (or see the eigenvalues from the hessian matrix) so I try to use this automatically, I thought it would be more simple. The problem is from a curse that not uses linear algebra knowledge.
    sorry my english is not good.
  9. Dec 26, 2017 #8


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    If you know about Sylvester's criterion, you're in good shape.

    You don't actually directly need the eigenvalues, but let's go that route.

    How does the ##trace\big(\mathbf H\big)## relate to the Laplacian operator? What does the trace tell you about the eigenvalues of ##\mathbf H##?
  10. Dec 26, 2017 #9
    The trace IS the laplacian operator, and is equal to zero, and if the trace is equal to zero so H have not eigenvalues...
    if H have not eigenvalues, what can i said about the saddle point?
  11. Dec 26, 2017 #10


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    Yes. This is a very important.

    Keep in mind we're dealing with scalars in ##\mathbb C##. In the case of finite dimensions there are always eigenvalues with complex numbers. Even better: with Hermitian (or in reals, symmetric) matrices the eigenvalues are always real. (Why?) This means if you have a real symmetric Hessian there are always eigenvalues. (Your Hessian should always have this kind of symmetry unless you have issues with Schwarz's / Clairaut's theorem.)

    so what you have is

    ##trace\big(\mathbf H\big) = \lambda_1 + \lambda_2 = 0##

    We know each eigenvalue is real. What does this tell you about the signs of the eigenvalues? If you know about the signs, what does that tell you about saddle points?

    - - - -
    I know you said this course doesn't use linear algebra here, but I think you would benefit from improving knowledge of spectral theory. In particular going through chapters 4-7 of Linear Algebra Done Wrong would be quite helpful. The book is outstanding and freely available from the author here: https://www.math.brown.edu/~treil/papers/LADW/LADW_2017-09-04.pdf
  12. Dec 26, 2017 #11
    Great, you really help me, the eigenvalues have diferent sign, and it means that is a undefinited cuadratic form, so, is a saddle point, its okey?
    and about the b) part, if f(x,y)=0, how fxx > 0 ?
    Last edited: Dec 26, 2017
  13. Dec 26, 2017 #12


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    I would say it's indefinite (which is what I think you said here albeit with a little language translation issues), i.e. we can find a case where ##\mathbf x_1^* \mathbf H \mathbf x_1 \gt 0## and ##\mathbf x_2^* \mathbf H \mathbf x_2 \lt 0##.

    So yes you have it right that it is a saddle point.

    The only remaining nit: how do you know for certain that you have one positive eigenvalue and one negative eigenvalue? Put differently, how do you know it isn't the case where ##0 = \lambda_1 = \lambda_2##?

    (Hint: real symmetric matrices, and complex Hermitian ones are always diagonalizable... and we know that the top left corner of ##\mathbf H ##, contains ##f_{xx} > 0##).
    Last edited: Dec 27, 2017
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