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PDE with constant coefficient using orthogonal transformation

  1. Mar 27, 2006 #1
    Plz Help :(

    Hi

    I want 2 know how 2 solve 1st order partial differintial equation (PDE) with constant coefficient using orthogonal transformation

    example :
    solve: 2Ux + 2Uy + Uz = 0

    THnx :blushing:
     
  2. jcsd
  3. Mar 27, 2006 #2

    Tom Mattson

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    I would tackle this one geometrcially. You can express the left side of your PDE in terms of a directional derivative, as follows:

    [tex]D_{\vec{v}}u=0[/tex]
    [tex]\vec{v}\cdot\vec{\nabla}u=0[/tex]

    Now let me ask you something:

    Can you identify [itex]\vec{v}[/itex]? Can you figure out the lines along which [itex]u[/itex] is constant?
     
  4. Mar 30, 2006 #3
    Thnx sir

    but the dr. who teach me, didn't use this way with us

    when we have aUx+bUy+cU=f , where a,b,c,f are constants
    he take tan(r)=b/a
    then he find r
    after that take w=xcosr+ysinr , t=-xsinr+ycosr
    then find Ux , Uy
    that's it

    so if i have aUx+bUy+cUz+dU=f
    can i take same w, t and z=z ???
     
  5. Mar 30, 2006 #4

    Tom Mattson

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    My approach is not much different from your professor's.

    This equation:

    can be written in the following form.

    [tex]D_{\vec{v}}u+cu=f[/tex]

    where [itex]\vec{v}=<a,b>[/itex].

    What your instructor did was rotate the x-axis so that it coincides with the vector [itex]\vec{v}[/itex]. In that coordinate system there is only one independent variable, and so you effectively have an ODE. The rotation matrix that carries [itex]\{x,y\}[/itex] into [itex]\{w,t\}[/itex] is given by the following.

    [tex]R_{z}(r) = \left[\begin{array}{cc} \cos(r) & \sin(r)\\ -\sin(r) & \cos(r) \end{array}\right][/tex]

    No, that's not it. All that does is transform the equation. It still remains to solve it.

    You will want to do a rotation about some axis so that one coordinate axis is parallel to [itex]\vec{v}=<a,b,c>[/itex].
     
    Last edited: Mar 30, 2006
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