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Linearity of Partial Differential Equations

  1. Feb 15, 2010 #1
    Is this linear homogeneous, linear inhomogeneous etc...
    [tex]u_{t}-u_{xx}+xu=0[/tex]

    From that first one I get this
    [tex]\frac{u_{t}-u_{xx}}{u}=-x[/tex]
    which I'm not sure is linear.

    Edit:
    Similar questions involve the following equations:
    [tex]iu_{t}-u_{xx}+\frac{u}{x}=0[/tex]

    and

    [tex]u_{x}+e^{y}u_{y}=0[/tex]

    Another Edit:

    I think I see the answer. I can rewrite the first equation like this:
    [tex](u_{t}-u_{xx}+xu)(\frac{1}{x})=0(\frac{1}{x})[/tex]

    and get a linear equation:

    [tex]\frac{u_{t}}{x}-\frac{u_{xx}}{x}+u=0[/tex]



    UGH... Another Edit..

    I'm not sure which approach is correct, the first or the second...
     
    Last edited: Feb 15, 2010
  2. jcsd
  3. Feb 15, 2010 #2

    LCKurtz

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    Science Advisor
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    Gold Member

    What makes such an equation linear is the fact that it is linear in the dependent variable u. That means u and its derivatives appear to the first degree. No u2 or uux and the like.

    More specifically, if your differential operator is denoted L(u), it has the property that:

    L(u + v) = L(u) + L(v) and L(cu) = cL(u) for constant c.

    For example, given the equation

    c(x,y)uxx+ f(x,y)u + g(x,y)uyy = h(x,y)

    if you call L(u) = c(x,y)uxx+ f(x,y)u + g(x,y)uyy the equation becomes:

    L(u) = h(x,y)

    If you try the above two conditions on L, you will see that it satisfies them because of the linearity of taking derivatives. It wouldn't if it had terms like u2, sin(u) etc.
     
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