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Are Green's functions generally symmetric?

  1. Nov 30, 2015 #1

    ShayanJ

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    In case of the Green's functions for the Laplace equation, we know that they're all symmetric under the exchange of primed and un-primed variables. But is it generally true for the Green's functions of all differential equations?
    Thanks
     
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  3. Dec 1, 2015 #2

    Geofleur

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    The Green's function will not always be symmetric.

    The Green's function formalism really involves considering two problems,

    ## \mathbf{L}_\mathbf{x}u = f(\mathbf{x}) ##

    and

    ## \mathbf{L}_\mathbf{x}^{\dagger} v = h(\mathbf{x}) ##,

    where ## \mathbf{L_x} ## and its adjoint, ## \mathbf{L}^\dagger_{\mathbf{x}} ##, are differential operators. These operators will satisfy the generalized Green's identity

    ## \int \left[ v^{*} \mathbf{L_x} u - u \left( \mathbf{L}^{\dagger}_{\mathbf{x}}v \right)^{*} \right] d^3 x = ## term that vanishes only if ## \mathbf{L_x} = \mathbf{L}_\mathbf{x}^{\dagger} ##.

    Let's assume that ## \mathbf{L_x} = \mathbf{L}_\mathbf{x}^{\dagger} ##, so that the right hand side really does vanish. Corresponding to the two problems we are trying to solve, there will be two different Green's functions, ## G(\mathbf{x},\mathbf{y}) ## and ## g(\mathbf{x},\mathbf{y}') ## that satisfy

    ## \mathbf{L}_\mathbf{x}G = \delta(\mathbf{x}-\mathbf{y}) ##

    and

    ## \mathbf{L}_\mathbf{x}^{\dagger} g = \delta(\mathbf{x}-\mathbf{y}') ##.

    If we set ## u = G(\mathbf{x},\mathbf{y}) ##, ## v = g(\mathbf{x},\mathbf{y}') ##, and put these into Green's identity (keeping the above two equations in mind), we get

    ## \int \left[ g^{*} \mathbf{L_x}G(\mathbf{x},\mathbf{y}) - G(\mathbf{x},\mathbf{y})\left( \mathbf{L}^{\dagger}_{\mathbf{x}}g(\mathbf{x},\mathbf{y}') \right)^{*} \right] d^3 x = \int \left[ g^{*}(\mathbf{x},\mathbf{y}') \delta(\mathbf{x}-\mathbf{y}) - G(\mathbf{x},\mathbf{y}) \delta(\mathbf{x} - \mathbf{y}') \right] d^3 x = 0##.

    Evaluating the integral in the last equality gives

    ## g^{*}(\mathbf{y},\mathbf{y}')-G(\mathbf{y}',\mathbf{y}) = 0 ##,

    so that

    ## g^{*}(\mathbf{y},\mathbf{y}') = G(\mathbf{y}',\mathbf{y}) ##.

    Because we have assumed that ## \mathbf{L_x} = \mathbf{L}_\mathbf{x}^{\dagger} ##, we really have ## g(\mathbf{y},\mathbf{y}') = G(\mathbf{y},\mathbf{y}') ##, so that

    ## G^{*}(\mathbf{y},\mathbf{y}') = G(\mathbf{y}',\mathbf{y}) ##.

    Finally, if the differential equations do not involve any complex coefficients, the complex conjugation will not change anything, and we will have

    ## G(\mathbf{y},\mathbf{y}') = G(\mathbf{y}',\mathbf{y}) ##.

    So only if the differential operator ## \mathbf{L}_{\mathbf{x}} ## is equal to its own adjoint and has no complex coefficients will the Green's function be symmetric. On the other hand, this is the case in a wide variety of circumstances.
     
  4. Dec 1, 2015 #3

    ShayanJ

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    Can you point to a book that treats Green's functions this way? I've seen Sadri-Hassani's but he considers equations with real coefficients and also doesn't consider the adjoint equation.
    Thanks
     
  5. Dec 2, 2015 #4

    Geofleur

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    I really like the way Dennery and Krzywicki talk about Green's functions in their book, Mathematics for Physicists. Also the first two chapters, on complex analysis and linear algebra, are wonderful.
     
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