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2nd order derivative (Nabla^2)

  1. Aug 4, 2014 #1
    Hi there, I'm kind of rusty on some stuff, so hope someone can help enlighten me.

    I have an expression
    [itex]E(r,w-w0)=F(x,y) A(z,w-w0) \exp[i\beta_0 z][/itex]

    I need to substitute this into the Helmholtz equation and solve using separation of variables. However, I'm getting problems simplifying it to a form with can be separated... I reckon the problem lies with my understanding of the 2nd derivative, especially with more variables coming into play.

    From the Helmholtz equation,
    [itex]\nabla^2 E+\epsilon (w) k_0^{\phantom{0}2} E=0[/itex]

    Working out
    [itex]\nabla^2 E =\nabla(\nabla E)[/itex]
    [itex]=\nabla(A\exp[i\beta_0 z] \frac{\partial F}{\partial x}+A\exp[i\beta_0 z] \frac{\partial F}{\partial y}+ FA (i \beta_0 \exp[i\beta_0 z])+F \exp[i\beta_0 z] \frac{\partial A}{\partial z})
    [/itex]
    [itex]
    =A \exp[i\beta_0 z] \frac{\partial^2 F}{\partial x^2}+i \beta_0 A \exp[i\beta_0 z] \frac{\partial F}{\partial x}+ \exp[i\beta_0 z] \frac{\partial A}{\partial z} \frac{\partial F}{\partial x} [/itex]
    [itex]
    + A \exp[i\beta_0 z] \frac{\partial^2 F}{\partial y^2}+i \beta_0 A \exp[i\beta_0 z] \frac{\partial F}{\partial y}+\exp[i\beta_0 z] \frac{\partial A}{\partial z} \frac{\partial F}{\partial y}
    [/itex]
    [itex]
    +\frac{\partial F}{\partial x} \exp[i\beta_0 z] \frac{\partial A}{\partial z}+ \frac{\partial F}{\partial y} \exp[i\beta_0 z] \frac{\partial A}{\partial z}+ F A (i \beta_0)^2 \exp[i\beta_0 z]
    [/itex]
    [itex]
    +i \beta_0 F \exp[i\beta_0 z] \frac{\partial A}{\partial z}+F \frac{\partial A}{\partial z} (i \beta_0) \exp[i\beta_0 z]+F \exp[i\beta_0 z] \frac{\partial^2 A}{\partial z^2}
    [/itex]
    Which gives
    [itex]
    =\exp[i\beta_0 z][A \frac{\partial^2 F}{\partial x^2}+2i \beta_0 A \frac{\partial F}{\partial x}+ 2 \frac{\partial A}{\partial z} \frac{\partial F}{\partial x}+A \frac{\partial^2 F}{\partial y^2}+2i \beta_0 A \frac{\partial F}{\partial y}+ 2 \frac{\partial A}{\partial z} \frac{\partial F}{\partial y}]
    [/itex]


    It seems like the terms [itex]2 \frac{\partial A}{\partial z} \frac{\partial F}{\partial x}[/itex] and [itex]2 \frac{\partial A}{\partial z} \frac{\partial F}{\partial y}[/itex] need to vanish...

    ------------------------------------------------------------------------------------------

    Or have I done the derivative wrongly? Should it be the following instead?
    [itex]
    \nabla^2 E =\nabla_x ^{\phantom{0}2}E+\nabla_y ^{\phantom{0}2}E+\nabla_z ^{\phantom{0}2}E
    [/itex]

    where
    [itex]
    \nabla_x ^{\phantom{0}2}E = A \exp[i\beta_0 z] \frac{\partial^2 F}{\partial x^2}
    [/itex]

    [itex]
    \nabla_y ^{\phantom{0}2}E = A \exp[i\beta_0 z] \frac{\partial^2 F}{\partial x^y}
    [/itex]


    [itex]
    \nabla_z^{\phantom{0}2}E = \nabla_z [FA (i \beta_0) \exp[i\beta_0 z]+ F \exp[i\beta_0 z] \frac{\partial A}{\partial z}]
    [/itex]
    [itex]
    =[F (i \beta_0) \exp[i\beta_0 z] \frac{\partial A}{\partial z}+FA (i \beta_0)^2 \exp[i\beta_0 z]+
    F \exp[i\beta_0 z] \frac{\partial^2 A}{\partial z^2}]+ F \frac{\partial A}{\partial z} (i \beta_0) \exp[i\beta_0 z]
    [/itex]

    Thanks in advance!
     
  2. jcsd
  3. Aug 4, 2014 #2

    SteamKing

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    Staff Emeritus
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    The Laplacian operator ∇[itex]^{2}[/itex] is defined as ∇[itex]\cdot[/itex]∇, or in cartesian components

    ∇[itex]^{2}[/itex]f = ∂[itex]^{2}[/itex]f/∂x[itex]^{2}[/itex] + ∂[itex]^{2}[/itex]f/∂y[itex]^{2} [/itex] + ∂[itex]^{2}[/itex]f/∂z[itex]^{2}[/itex]

    http://en.wikipedia.org/wiki/Del
     
  4. Aug 5, 2014 #3
    Thanks SteamKing!

    I suppose my second intepretation should be the correct one instead.
     
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