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.

Help me please.......

Printable View

- Jan 19th 2011, 08:16 PMbdcrown007Need 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.

Help me please....... - Jan 19th 2011, 08:36 PMProve It
Is $\displaystyle \displaystyle \varepsilon$ a constant or a function?

- Jan 19th 2011, 09:26 PMbdcrown007
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 PMProve 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 PMbdcrown007
that means: ε ∇^2φ=0?

- Jan 19th 2011, 09:53 PMProve It
- Jan 19th 2011, 09:58 PMbdcrown007
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 PMProve 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 PMbdcrown007
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 PMbdcrown007
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 PMbdcrown007
but $\displaystyle \varepsilon$ is necessary :( what should I do?