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Solvability and adjoint equation?

  1. Aug 7, 2013 #1
    Hi all,

    I don't understand the following procedure discussing the solvability of the differential equation. Please have a look and comment.

    The operator [itex]\mathcal{L}[/itex] is defined as [itex]\mathcal{L}=\mathcal{L}(\phi_0;k,\omega,X)[/itex], where [itex]\phi_0[/itex] is the variable and [itex]k,\omega,X[/itex] are the parameters.

    Now take the derivative w.r.t. k of the homogeneous equation [itex]\mathcal{L}=0[/itex], we have [itex]\mathcal{L}(\partial_k\phi_0;k,\omega,X)=-\mathcal{L}_k(\phi_0;k,\omega,X)-\mathcal{L}_{\omega}(\phi_0;k,\omega,X)\partial_k \omega[/itex], where it assumes [itex]\omega [/itex] is dependent on [itex]k[/itex] while [itex]X[/itex] independent on [itex]k[/itex].

    Then the author writes that the solvability of the above differential equation requires that the right hand side be orthogonal to the homogeneous solution to the adjoint equation of [itex]\mathcal{L}=0[/itex].

    I've came across such solvability criterion now and then (I guess it is common knowledge), but never truly understand what it really means and where it comes from. Also, what's the role of the adjoint equation? Could you help me? Thanks a lot in advance.
     
    Last edited: Aug 7, 2013
  2. jcsd
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