Boundary integral method to solve poisson equation

**Suggest how to solve Poisson equation**

\begin{equation}

σ ∇^2 V = - I δ(x-x_s) δ(y-y_s) δ(z-z_s) \nonumber

\end{equation}

by using the boundary integration method to calculate the potential $V(r,z)$ with the help of changing the Poisson equation into cylindrical polar co ordinates?

Where V is electric Potential (scalar)[volts]. Solution depends only on $r$ and $σ$ is constant. $r$ is horizontal coordinate(direction) $0\leq r \leq \infty $ and $z$ is vertical co ordinate(direction) $-\infty\leq r \leq \infty $. What boundary conditions are appropriate?

=>

Firstly, is boundary integration method same as boundary element method?

if not what do you mean by boundary integration method?

To change the Poisson equation into cylindrical polar co ordinates:

$σ =(r,θ,z)$ where, $θ$ is not relevant

so, $σ=(r,z)$

cylindrical polar co ordinates is

\begin{equation}

σ ∇^2 V= \int ^{r= ∞}_ {r=0} \int^ {z=∞} _{ z= - ∞} {δ(r) δ(z-z_s)} dz dr.

\end{equation}

where $r_s$=0 at the top layer r=0.

Now, it suggest how to calculate the potential V(r,z) by using boundary integral method.

for the boundary conditions i thought to use green function in 3-d dimensional but pretending 2-d in our case.

\begin{equation}

( green function) G_{3D}= \frac{- 1}{4πσ r} = \frac{∂G}{∂r} + \frac{1}{r} ≈ 0

\end{equation}

my second question is can i use Neumann condition as a boundary condition under boundary integration method?

No current flux on a surface with normal n: $ σ \frac{∂V}{∂n} =0$ [ Neumann ]

my third question is how to solve the problem now?

can anyone please help me.

Re: Boundary integral method to solve poisson equation

Re: Boundary integral method to solve poisson equation

$I $ represents current and we take $I$ as $1.$

Re: Boundary integral method to solve poisson equation

With regard to changing your equation to cylindrical coordinates (and solving the problem) first consider re-writing your equation as follows: $\sigma\nabla^2V=I\delta(\mathbf{x}-\mathbf{x}_0) $ which of the variables in cylindrical coordinates could $ |\mathbf{x}-\mathbf{x}_0 |$ represent?

=> i don't know how did you get above equations and i think $ |\mathbf{x}-\mathbf{x}_0 |$ represents the $r$ variables in cylindrical coordinates.

What is the laplacian in cylindrical coordinates?

=> $r$ and $z$ are cylindrical coordinates in horizontal and vertical direction respectively.

Also what variable can you omit from the the laplacian i.e which variables are simply constants?

=>(conductiveity) $\sigma$ is constant. assume $\sigma = \sigma(z) $ or $\sigma = \sigma(r)$ for tractable solution.

Now consider the equation at different values of $ \delta(\mathbf{x}-\mathbf{x}_0)$ (Hint: There should be two equations that result from this analysis: one of them should be quite familiar to you).

=> i didn't get it.

Can you please clarify the first and last paragraphs a little bit further thank you.

I want to say one thing that : solution depends only on $r$.

Re: Boundary integral method to solve poisson equation

Quote:

Originally Posted by

**grandy** With regard to changing your equation to cylindrical coordinates (and solving the problem) first consider re-writing your equation as follows: $\sigma\nabla^2V=I\delta(\mathbf{x}-\mathbf{x}_0) $ which of the variables in cylindrical coordinates could $ |\mathbf{x}-\mathbf{x}_0 |$ represent?

=> i don't know how did you get above equations and i think $ |\mathbf{x}-\mathbf{x}_0 |$ represents the $r$ variables in cylindrical coordinates.

I got the above equation by simply recognizing that is the same thing as : The first form of the equation is saying that the PDE is inhomogeneous only at the point which is exactly the same thing that the form I put it into is saying: assuming and

Quote:

What is the laplacian in cylindrical coordinates?

=> $r$ and $z$ are cylindrical coordinates in horizontal and vertical direction respectively.

Yes that is true;however, it does not answer the question I asked. Since it won't give away too much of the problem: the Laplacian in cylindrical coordinates is:

Now which of the terms in the Laplacian become zero? i.e which derivatives of are

Quote:

Also what variable can you omit from the the laplacian i.e which variables are simply constants?

=>(conductivity) $\sigma$ is constant. assume $\sigma = \sigma(z) $ or $\sigma = \sigma(r)$ for tractable solution.

You're last statement is contradictory:if sigma is just a constant it won't depend on or . Unless what you meant to say was: is constant with respect to one of the variables. Also I assume that is a surface charge density, correct?

Quote:

Now consider the equation at different values of $ \delta(\mathbf{x}-\mathbf{x}_0)$ (Hint: There should be two equations that result from this analysis: one of them should be quite familiar to you).

=> i didn't get it.

Think about the dirac delta function: what values does it take on at different points in the domain? How does the equation for look at those points?

Quote:

Can you please clarify the first and last paragraphs a little bit further thank you.

I want to say one thing that : solution depends only on $r$.

In my first paragraph I'm simply saying that for the purposes of this problem the *Boundary Integral Method* and the *Boundary Element Method* are the same thing

In my last paragraph I'm just trying to get you to pay attention to the inhomogeneous term on the RHS of the equation for because for a large part of the domain the equation simplifies to a special case of a poisson equation *hint* *hint*.

Re: Boundary integral method to solve poisson equation

\begin{equation}

- ∇^2 V= \frac{1}{r} \frac{\partial}{\partial r}[ r \frac{\partial v}{\partial r}] + \frac{1}{r^2}\frac{\partial^2 V}{\partial \theta^2}+\frac{\partial^2 V}{\partial z^2}

\end{equation}

$\theta$ is not relevant, so it goes to zero.

\begin{equation}

- ∇^2 V= \frac{1}{r} \frac{\partial}{\partial r}[ r \frac{\partial v}{\partial r}] +\frac{\partial^2 V}{\partial z^2}

\end{equation}

cylindrical polar co ordinates is

\begin{equation}

\int ^{r= ∞}_ {r=0} \int^ {z=∞} _{ z= - ∞} {δ(r) δ(z-z_s)} dz dr.

\end{equation}

where $r_s$ is $0$.

( green function) $G_{3D}= \frac{- 1}{4πσ r} = \frac{∂G}{∂r} + \frac{1}{r} ≈ 0$ on boundary $\Gamma_\infty$

where $r$ is $|X-X_0|$. $\sigma$ is constant with respect to one of the variables $r$ or $z$ but solution depend upon $r$ only.

No current flux on a surface with normal $n$: $ σ \frac{∂V}{∂n} =0$ [ Neumann boundary condition ]

now my question is how do I tackle the problem by using the boundary integral method with the help above equations.

Re: Boundary integral method to solve poisson equation

Quote:

Originally Posted by

**grandy** \begin{equation}

- ∇^2 V= \frac{1}{r} \frac{\partial}{\partial r}[ r \frac{\partial v}{\partial r}] + \frac{1}{r^2}\frac{\partial^2 V}{\partial \theta^2}+\frac{\partial^2 V}{\partial z^2}

\end{equation}

$\theta$ is not relevant, so it goes to zero.

\begin{equation}

- ∇^2 V= \frac{1}{r} \frac{\partial}{\partial r}[ r \frac{\partial v}{\partial r}] +\frac{\partial^2 V}{\partial z^2}

\end{equation}

Look a little closer. Are there any other derivatives that vanish since only depends on .

Quote:

cylindrical polar co ordinates is

\begin{equation}

\int ^{r= ∞}_ {r=0} \int^ {z=∞} _{ z= - ∞} {δ(r) δ(z-z_s)} dz dr.

\end{equation}

where $r_s$ is $0$.

( green function) $G_{3D}= \frac{- 1}{4πσ r} = \frac{∂G}{∂r} + \frac{1}{r} ≈ 0$ on boundary $\Gamma_\infty$

where $r$ is $|X-X_0|$. $\sigma$ is constant with respect to one of the variables $r$ or $z$ but solution depend upon $r$ only.

No current flux on a surface with normal $n$: $ σ \frac{∂V}{∂n} =0$ [ Neumann boundary condition ]

now my question is how do I tackle the problem by using the boundary integral method with the help above equations.

Just to clarify something: are you seeking a solution for both cases when depends on and then when depends on ?

Also you shouldn't need to apply a boundary condition once you have the Greene's function for the equation because the B.C is included in the Greene's function.