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Math Help - Linear Operators

  1. #1
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    Linear Operators

    To be linear, L(\alpha u+\alpha v)=\alpha (L(u)+ L(v)) \ \alpha\in\mathbb{R}.

    How is this equation (see below) linear?

    yu_{xxy}-e^xu_x+3=0

    yu_{xxy}-e^xu_x+3=0 \ \mbox{and} \ yv_{xxy}-e^xv_x+3=0

    L(u+v)=(yu_{xxy}-e^xu_x+yv_{xxy}-e^xv_x)+3\Rightarrow y(u_{xxy}+v_{xxy})-e^x(u_x+v_x)+3

    \neq L(u)+L(v)=y(u_{xxy}+v_{xxy})-e^x(u_x+v_x)+6
    Last edited by dwsmith; December 26th 2010 at 11:54 AM. Reason: Touching up
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  2. #2
    A Plied Mathematician
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    You have to distinguish between linear operators and linear equations. Typically, for the equation you've given, you'd throw the 3 over to the RHS and consider the operator to be everything else (that is, all the stuff with the dependent variable u in it). So the operator is

    y\partial_{x}\partial_{x}\partial_{y}-e^{x}\partial_{x}.

    This is linear. So a differential equation is considered linear if the associated operator is linear. The DE can still be linear even if it's non-homogeneous, as in your case. Does that make sense?
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  3. #3
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    I understand you can have linear non-homogeneous equation.

    Whenever there is a term that isn't associated with the partial derivatives, it is simply disregard?
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  4. #4
    A Plied Mathematician
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    I think you would ignore any term that doesn't have the dependent variable u in it.
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