# 2nd order derivative (Nabla^2)

1. Aug 4, 2014

### test1234

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

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

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,
$\nabla^2 E+\epsilon (w) k_0^{\phantom{0}2} E=0$

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

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

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

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

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

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

2. Aug 4, 2014

### SteamKing

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

∇$^{2}$f = ∂$^{2}$f/∂x$^{2}$ + ∂$^{2}$f/∂y$^{2}$ + ∂$^{2}$f/∂z$^{2}$

http://en.wikipedia.org/wiki/Del

3. Aug 5, 2014

### test1234

Thanks SteamKing!

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