Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $\vec{F}(x,y,z)=\vec{f}(x,y)\exp(ikz)$ and $\vec{F}$ satisfies the equations $\nabla \cdot \Re{\vec{F}}=0$ where $\Re{\vec{F}}$ is the real part of $\vec{F}$. It also satisfies $\nabla \times \Re\vec{F}=\Re\vec{G}$ for $\Re\vec{G}=\vec{g}(x,y)\exp(ikz)$

If I want to express these equations in terms of $\vec{f},\vec{g},k$, Is it possible to simplify do better than $\Re\left[\left({\partial \vec{f}\over \partial x}+{\partial \vec{f}\over \partial y}+ik\vec{f}\right)\exp(ikz)\right]=0$ for the first one? As for the second one I really have no idea how to express it in the desired form except trivially substituting the entire expression of $\vec{F}$ into the equation.

Could anyone help?

Added: $F,f$ are both 3-D vectors the $x,y,z$ are just their arguments.

Thanks.

share|improve this question
1  
The LHS suggests that F is a 3-dimensional vector field, however, on the RHS you have a 2-dimensional vector f in both real and imaginary parts, multiplied by a scalar. If you split real and imaginary parts and wrote the Laplacian clearly in components, perhaps it would become clearer –  Valentin May 27 '12 at 23:26
    
@Valentin: Thank you, actually $f$ is a 3-D vector but it only depends on $x,y$ –  hatch May 28 '12 at 7:03
add comment

1 Answer 1

up vote 0 down vote accepted

There are so many chances to get confused due to possibly ambiguous notation in this example (e.g. $i$ and $\vec{\boldsymbol{i}}$, $k$ and $\vec{\boldsymbol{k}}$) that I would suggest carefully writing out the components of each expression: $$\vec{F}=\vec{f}\cos kz+i\vec{f}\sin kz$$ $$\vec{f}=P\left(x,y\right)\vec{\boldsymbol{i}}+Q\left(x,y\right)\boldsymbol{\vec{j}+}R\left(x,y\right)\boldsymbol{\vec{k}}$$ $$\Re\vec{F}=\vec{f}\cos kz=P\left(x,y\right)\cos kz\vec{\boldsymbol{i}}+Q\left(x,y\right)\cos kz\boldsymbol{\vec{j}+}R\left(x,y\right)\cos kz\vec{\boldsymbol{k}}$$ $$\nabla\Re\vec{F}=\frac{\partial P}{\partial x}\cos kz+\frac{\partial Q}{\partial y}\cos kz-kR\left(x,y\right)\sin kz$$

Similar approach holds for the second one if you write $\nabla \times \vec{F}$ in components. If that is not what you are looking for, you have to specify the "desired form" more rigidly. Hope that helps.

share|improve this answer
    
Thank you,Valentin –  hatch May 28 '12 at 22:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.