# Need solution of following eqn.

• Jan 19th 2011, 08:16 PM
bdcrown007
Need solution of following eqn.
I need a solution of this eqn.

∇.(ε∇φ)=0

if there is no ε, then φ=x^2-y^2 can easily be taken as solution but I am confused as ε is present. Here ε is electrical conductivity and φ is electrical potential.
• Jan 19th 2011, 08:36 PM
Prove It
Is $\displaystyle \displaystyle \varepsilon$ a constant or a function?
• Jan 19th 2011, 09:26 PM
bdcrown007
Actually I am trying to make simulation of flow induced by plasma actuator using PHOENICS CFD software. Here is the detailed image:

http://i47.tinypic.com/syxkr7.png

ε is a constant. In the solution domain upper part is air and bottom shaded part is dielectric material. Same equation for both domain except different dielectric constant/electrical conductivity.

for upper domain ε1 is 1 and bottom domain ε2 is 2.7. Possibly the solution should be made considering ε a function. because I have to define upper and bottom domain with 1 and 2.7 respectively.
• Jan 19th 2011, 09:31 PM
Prove It
A property of dot products is

$\displaystyle \displaystyle (c_1\mathbf{a}) \cdot (c_2\mathbf{b}) = (c_1c_2)(\mathbf{a}\cdot \mathbf{b})$.

So that means $\displaystyle \displaystyle \nabla \cdot (\varepsilon \nabla \varphi) = \varepsilon (\nabla \cdot \nabla \varphi)$.
• Jan 19th 2011, 09:48 PM
bdcrown007
that means: ε ∇^2φ=0?
• Jan 19th 2011, 09:53 PM
Prove It
Quote:

Originally Posted by bdcrown007
that means: ε ∇^2φ=0?

Yes
• Jan 19th 2011, 09:58 PM
bdcrown007
so, how to proceed for solution?
∇.(ε∇φ)=0
ε ∇^2φ=0
ε(d2φ/dx2+d2φ/dy2)=0

in this case ε=0 ?

>?
• Jan 19th 2011, 10:03 PM
Prove It
I think it's pretty clear that $\displaystyle \displaystyle \varepsilon \neq 0$...

How would you solve $\displaystyle \displaystyle \frac{\partial ^2 \varphi}{\partial x^2} + \frac{\partial ^2 \varphi}{\partial y^2} = 0$?
• Jan 19th 2011, 10:10 PM
bdcrown007
If I use http://www.mathhelpforum.com/math-he...be85ba44a6.png then there is no difference between two eqn for two domain (please check image of my 2nd reply).

confused :(
• Jan 19th 2011, 10:18 PM
bdcrown007
in this case of http://www.mathhelpforum.com/math-he...be85ba44a6.png
the solution becomes:$\displaystyle \varphi=x^2-y^2$
• Jan 19th 2011, 10:30 PM
bdcrown007
but $\displaystyle \varepsilon$ is necessary :( what should I do?