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

Quote:

Originally Posted by

**grandy** **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.

From what I have gathered: the answer to your first question is that the Boundary Integral method is *a* Boundary Element method. For the purposes of this problem they appear to be interchangeable terms.

Also a quick question: what does $\displaystyle I $ represent?

With regard to changing your equation to cylindrical coordinates (and solving the problem) first consider re-writing your equation as follows: $\displaystyle \sigma\nabla^2V=I\delta(\mathbf{x}-\mathbf{x}_0) $ which of the variables in cylindrical coordinates could $\displaystyle |\mathbf{x}-\mathbf{x}_0 $| represent? What is the laplacian in cylindrical coordinates? Also what variable can you omit from the the laplacian i.e which variables are simply constants?

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

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 $\displaystyle \delta(x-x_s)\delta(y-y_s)\delta(z-z_s)$is the same thing as $\displaystyle \delta(\mathbf{x}-\mathbf{x}_0)$: The first form of the equation is saying that the PDE is inhomogeneous only at the point $\displaystyle (x_s,y_s,z_s)$ which is exactly the same thing that the form I put it into is saying: assuming $\displaystyle \mathbf{x} = \begin{pmatrix} x\\y\\z\end{pmatrix} $ and $\displaystyle \mathbf{x}_0 = \begin{pmatrix} x_s\\y_s\\z_s\end{pmatrix} $

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:

$\displaystyle \nabla^2V = \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} $

Now which of the terms in the Laplacian become zero? i.e which derivatives of $\displaystyle V $ are $\displaystyle 0 $

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 $\displaystyle r$ or $\displaystyle z $. Unless what you meant to say was: $\displaystyle \sigma$ is constant with respect to one of the variables. Also I assume that $\displaystyle \sigma$ 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 $\displaystyle V $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 $\displaystyle V$ 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 $\displaystyle V $ only depends on $\displaystyle r $.

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 $\displaystyle \sigma$ depends on $\displaystyle r $ and then when $\displaystyle \sigma$ depends on $\displaystyle z$?

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.