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Conic formula Eigenvalues

  1. Mar 27, 2013 #1
    1. The problem statement, all variables and given/known data
    We have the following conic formula ##ax^2 + 2bxy + cy^2 + dx + ey = ## constant which corresponds to a ellipse, hyperbola or parabola. The second order terms of the corresponding PDE $$ a\frac{\partial^2 u}{\partial x_1^2} + 2b\frac{\partial^2 u}{\partial x_1\partial x_2} + c\frac{\partial^2 u}{\partial x_2^2} + d\frac{\partial u}{\partial x_1} + e\frac{\partial u}{\partial x_2} + gu = f(x_1,x_2) $$ can be written as $$ \sum_{i,j=1}^2 a_{ij} \frac{\partial^2 u}{\partial x_i\partial x_j} $$ where ##a_{ij}## are the entries of the symmetric matrix $$ A = \begin{pmatrix}
    a & b \\
    b & c
    \end{pmatrix}. $$ Show that the eigenvalues have the same sign if ##b^2-ac > 0##, opposite signs if ##b^2-ac < 0## and one is zero if ##b^2-ac = 0##.


    2. Relevant equations



    3. The attempt at a solution
    I know that the eigenvalues of a symmetric matrix are always real. Also I know that the PDE is said to be elliptic, hyperbolic or parabolic depending on whether ##b^2-ac## is positive, negative or zero. Not too sure what to do next. Please help. The determinant of ##A## is ##ac-b^2##.
     
    Last edited: Mar 27, 2013
  2. jcsd
  3. Mar 27, 2013 #2
    Given matrix A, how are its eigenvalues computed?
     
  4. Mar 27, 2013 #3

    HallsofIvy

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    A 2 by 2 matrix can always be put in "Jordan Normal form" or "diagonalized" with its eigenvalues on the diagonal. That is, the derterminant of a matrix is always the product of its eigenvalues.

    You are aware that [itex]b^2- ac= -(ac- b^2)[/itex] aren't you?
     
  5. Mar 28, 2013 #4
    So we have $$ (\lambda - a)(\lambda - c) -b^2 = \lambda^2 -(a+c)\lambda + ac - b^2$$ which is the characteristic polynomial with the coefficient of ##\lambda## being the trace of ##A## and the last term being the determinant of ##A##.
     
    Last edited: Mar 28, 2013
  6. Mar 28, 2013 #5

    vela

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    Do you understand what HallsofIvy is getting at?
     
  7. Mar 28, 2013 #6
    I do. But not in showing the signs of eigenvalues in each case for ##b^2-ac##.
     
  8. Mar 28, 2013 #7

    vela

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    If ##\lambda_1 \lambda_2 > 0##, what can you say about the relative signs of ##\lambda_1## and ##\lambda_2##?
     
  9. Mar 28, 2013 #8
    Then ##\lambda_1## and ##\lambda_2## must either be both positive or negative, i.e. the same sign. Same argument for < 0 and = 0. Cheers.
     
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