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Normal form of p.d.e?

  1. Jun 10, 2006 #1
    how do you change the following p.d.e to normal form and solve it?
    uxx -4uxy+3uyy=0?
     
  2. jcsd
  3. Jun 10, 2006 #2

    HallsofIvy

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    How would you eliminate the "xy" term in x2- 4xy+ 3y2= 0 in order to determine what type of conic it was?

    One method would be to write it as a matrix formula:
    [tex]\left( \begin{array} {cc} x & y \end{array} \right) \left( \begin{array} {cc} 1 & -2 \\-2 & 3 \end{array} \right) \left( \begin{array} {c} x \\ y \end{array}\right)= 0[/tex]
    and find the eigenvalues and eigenvectors. The eigenvector point in the "characteristic directions" and taking your axes in those directions reduces to "normal form".

    With a PDE, the same thing happens: using the characteristic directions as the new variables reduces the equation to normal form.

    This is the second problem in a row in which you have essentially said "how do I solve partial differential equations". What have you learned in class so far?
     
  4. Jun 10, 2006 #3
    why would you think of writing it as a matrix formula?
     
  5. Jun 10, 2006 #4

    HallsofIvy

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    I said that was one method.

    I'll ask again: "How would you eliminate the "xy" term in [itex]x^2- 4xy+ 3y^2= 0[/itex] in order to determine what type of conic it was?"
     
  6. Jun 11, 2006 #5
    shift the x and y values?
     
  7. Jun 11, 2006 #6

    HallsofIvy

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    I have no idea what you mean by "shift the x and y values". My point was that eliminating the uxy in a pde is essentially the same as eliminating the xy term in a conic (by rotating the coordinate system) and wanted to know if you knew how to do that.
     
  8. Jun 16, 2006 #7
    i've looked that part up~ thank you very much!
     
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