Poisson's inhomogeneous PDE and its solutions

[fog37]

TL;DR Summary

Poisson's equation solution and Green's function method

Hello,
Poisson equation and Laplace equation (which is the homogeneous version of Poisson PDE) are important equations in electrostatics where both the electric field $E$ and scalar potential $\phi$ don't depend on time. Poisson's equation is $$\nabla^2 \phi(x,y,z) = - \frac{\rho(x,y,z)} {\epsilon_0}$$ The inhomogeneous term, also called the source, is $- \frac{\rho(x,y,z)} {\epsilon_0}$. One of the possible methods (not the only one, correct?) to find a solution to Poisson's equation is the Green function method which involves finding the solution when the source is just a Diract delta function $\delta(x,y,z)$. By linearity, the solution when the source is $- \frac{\rho(x,y,z)} {\epsilon_0}$ is just a superposition integral adding all the responses from many delta sources.

• My dilemma: since Poisson's equation is a PDE and like every PDE it is a pointwise relation between partial derivatives at specific spatial points, i.e. the equation is valid all the spatial points where the source term $- \frac{\rho(x,y,z)} {\epsilon_0} \neq 0$ , correct? However, the Green's function method provides a solution $\phi(x,y,z)=\int \frac{\rho(x,y,z)} {\epsilon_0 r}dxdydx$ that is valid at spatial points where $- \frac{\rho(x,y,z)} {\epsilon_0} =0$, i.e. the solution $\phi(x,y,z)$ exists and has values at spatial points where the source is zero. How can the PDE solution $\phi(x,y,z)$ exist at spatial points where the PDE is not satisfied?
• Does the Green function method require the initial conditions (ICs) to be zero while he boundary conditions (BCs) can be arbitrary? If the PDE has nonzero ICs, then the Green function method provides only portion of the solution, correct?

Thank you!

 

[phyzguy]

The Green's function method provides a complete solution. You're missing a couple of points:
(1) Poisson's equation does not include time, so there are no "initial conditions". there are only "boundary conditions".
(2) The Green's function depends on the boundary conditions, so there will be a different Green's function if the boundary conditions change.
(3) The biggest thing you are missing is that the Green's function is a function of both (x,y,z) and (x',y',z'). In the case of Poisson's equation, you can think of it as the potential at (x,y,z) due to a point charge at (x',y',z'). Note that without this, the integral you wrote makes no sense. The LHS is a function of (x,y,z) and the RHS is a constant because you have integrated out x,y,and z. What you need to write is:
$$\phi(x,y,z) = \int \frac{\rho(x',y',z')}{\epsilon_0 (r-r')}dx'dy'dz'$$
where $(r-r') = \sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$
Then you can see that $\phi(x,y,z)$ can be non-zero even in regions where $\rho(x,y,z) = 0$, because the potential of regions where ρ is non-zero extend to infinity.

 

[Delta2]

About your dilemma: I don't know your exact reasoning behind the conclusion that the solution provided by the green function is valid only at points where the source is zero, but this conclusion is certainly wrong, the solution provided by the green function is valid at all points where the source is defined, regardless if the source is zero at those points or non-zero.

About your second question I think the boundary conditions might make the solution provided by the green function not being the only solution. The equation might have other solutions as well.

 

[fog37]

The Green's function method provides a complete solution. You're missing a couple of points:
(1) Poisson's equation does not include time, so there are no "initial conditions". there are only "boundary conditions".
(2) The Green's function depends on the boundary conditions, so there will be a different Green's function if the boundary conditions change.
(3) The biggest thing you are missing is that the Green's function is a function of both (x,y,z) and (x',y',z'). In the case of Poisson's equation, you can think of it as the potential at (x,y,z) due to a point charge at (x',y',z'). Note that without this, the integral you wrote makes no sense. The LHS is a function of (x,y,z) and the RHS is a constant because you have integrated out x,y,and z. What you need to write is:
$$\phi(x,y,z) = \int \frac{\rho(x',y',z')}{\epsilon_0 (r-r')}dx'dy'dz'$$
where $(r-r') = \sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$
Then you can see that $\phi(x,y,z)$ can be non-zero even in regions where $\rho(x,y,z) = 0$, because the potential of regions where ρ is non-zero extend to infinity.

Thanks phyzguy.

Ok on your point 1) and point 2). It makes sense that different BCs would lead to different Green's functions. The source remains a Dirac delta function but the BCs will affect the Green's function. Where, mathematically, would the BCs show up in shaping the resulting Green function? There is no presence of BCs in the (correct) integral for the solution $\phi(x,y,z)$ that you report.

On point 3: there are indeed general spatial points $(x,y,z)$ (I guess these are all the points where the $source =0$) and primed spatial points $(x',y',z')$ which is where $source \neq 0$. The LHS of the integral involves both with the primed variables being the integration variables.

What confused me was the fact that the PDE (Poisson's equation) is true only at all the primed spatial points $(x',y',z')$ but the solution $\phi(x,y,z)$ to the same equation is instead valid at all points $(x,y,z)$ other than $(x',y',z')$...

 

[phyzguy]

On point 3: there are indeed general spatial points $(x,y,z)$ (I guess these are all the points where the $source =0$) and primed spatial points $(x',y',z')$ which is where $source \neq 0$. The LHS of the integral involves both with the primed variables being the integration variables.

I don't think you're getting it. There are not "primed points" and "unprimed poins". Both (x,y,z) and (x',y',z') vary over all space. To find the potential at a point (x,y,z), you integrate the integral on the RHS with the (x,y,z) values fixed, and the (x',y',z') variables varying over all space.

What confused me was the fact that the PDE (Poisson's equation) is true only at all the primed spatial points $(x',y',z')$ but the solution $\phi(x,y,z)$ to the same equation is instead valid at all points $(x,y,z)$ other than $(x',y',z')$...

I don't understand what you are saying. Poisson's equation is true everywhere, both at points where ρ is zero and points where ρ is non-zero. The Green's function method is a way to find the solution $\phi(x,y,z)$, given $\rho(x,y,z)$. Once you have found the solution, Poisson's equation is satisfied everywhere. That's what it means to find the solution.

 

[phyzguy]

Where, mathematically, would the BCs show up in shaping the resulting Green function? There is no presence of BCs in the (correct) integral for the solution $\phi(x,y,z)$ that you report.

I missed this question. The Green's function I used $\frac{1}{\epsilon_0 (r-r')}$ is the Green's function for the boundary conditions where φ=0 at infinity, which is what we usually assume. If we have a different set of boundary conditions, like we are in a finite box with fixed potentials on the walls, then the Green's function will be different. You basically need to solve for the potential of a point charge with the given boundary conditions.

 

[phyzguy]

@fog37 , maybe one more thing might help. Apologies if this is obvious to you, but do you understand that the function $$\int_A^B f(x,y)dy$$ is a function of x only? You have integrated out the dependence on y. So when you integrate the Green's function over the primed variables (x',y',z'), only the dependence on (x,y,z) remains. Does this help? Green's functions are confusing when you first see them with the xs and x's.

Edit: I edited this to make clear that these are definite integrals. When doing the Green's function integral in Post #2, it is important to realize that the RHS involves definite integrals with (x',y',z') being integrated over all space.

 

[Delta2]

What confused me was the fact that the PDE (Poisson's equation) is true only at all the primed spatial points (x′,y′,z′)(x′,y′,z′)(x',y',z') but the solution ϕ(x,y,z)ϕ(x,y,z)\phi(x,y,z) to the same equation is instead valid at all points (x,y,z)(x,y,z)(x,y,z) other than (x′,y′,z′)(x′,y′,z′)(x',y',z')...

Poisson's equation is true in all $\mathbb{R}^3$ or to be more precise in the subset of $\mathbb{R}^3$ that is the domain of the source function. If we suppose that there is a bounded subset $V$ of $\mathbb{R}^3$ such that $$\rho(x,y,z)\neq 0 ,(x,y,z)\in V$$and $$\rho(x,y,z)=0, (x,y,z)\in \mathbb{R}^3-V$$ then poisson's equation holds in $\mathbb{R}^3-V$ too it is just that $$\nabla^2\phi=-\frac{\rho(x,y,z)}{\epsilon_0}=0, (x,y,z)\in\mathbb{R}^3-V$$.

If furthermore you have the question that in this case how can it be that the value of the solution $$\phi(x,y,z)=\iiint_V\frac{\rho(x',y',z')}{\epsilon_0\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}dx'dy'dz'$$ at a point $(x,y,z)\in \mathbb{R}^3-V$ depends only on the values that $\rho(x,y,z)$ takes on V (because the integral is simply zero in the region outside of V, because the source is simply zero there), then my answer would be:

1. Look at divergence theorem
2. Welcome to the magic of the Green's functions. Somehow when we do the math to prove that the above integral is the solution to Poisson's equation, dirac delta function plays an important role and we end up that the above integral is indeed the solution for all (x,y,z) and not only for (x,y,z) in V.

 

[wrobel]

like every PDE it is a pointwise relation between partial derivatives at specific spatial points,

be care: in general this is not so

 

[fog37]

hello phyzguy! no worries about anything.

I see how that integral works and why the result $\phi(x,y,z,)$ does not depend on the integration variables $(x',y',z')$.

Maybe I am thinking too hard and wrongly about this :) Conceptually, I guess I am wondering if, given an equation (ODE, PDE, algebraic, etc.) which is valid (if that is the right word) at certain specific spatial points, the solution(s) of such equation can exist and have values at spatial points where the equation is not valid...
In our situation, the equation is Poisson equation which is valid/applicable/satisfied at those spatial points where the source term is nonzero. But the solution to the equation, i.e. the function $\phi(x,y,z)$, exist at spatial points that are different from the points where the source term is nonzero and where the equation itself is satisfied. If the equation is not satisfied there, can a solution to the equation still exist there?

Thanks for the patience : )

 

[fog37]

be care: in general this is not so

Hello wrobel, interesting point. Could you briefly elaborate? I tend to think that a PDE or ODE is an equation relating time and space derivatives at specific points in space.

Are you saying that there could be derivative related to certain spatial points and derivatives and terms related to different points?

 

[phyzguy]

Maybe I am thinking too hard and wrongly about this :) Conceptually, I guess I am wondering if, given an equation (ODE, PDE, algebraic, etc.) which is valid (if that is the right word) at certain specific spatial points, the solution(s) of such equation can exist and have values at spatial points where the equation is not valid...
In our situation, the equation is Poisson equation which is valid/applicable/satisfied at those spatial points where the source term is nonzero. But the solution to the equation, i.e. the function $\phi(x,y,z)$, exist at spatial points that are different from the points where the source term is nonzero and where the equation itself is satisfied. If the equation is not satisfied there, can a solution to the equation still exist there?

Thanks for the patience : )

Again, Poisson's equation is satisfied everywhere, at all spatial points. At points where the source term is non-zero, then $\nabla^2 \phi = \rho$, and at points where the source term is zero, then $\nabla^2 \phi = 0$.

 

[wrobel]

Hello wrobel, interesting point. Could you briefly elaborate?

Standard point. Generalized solution is a substantially globally defined object , in general it is not reduced to the pointwise dependencies .

 

[fog37]

Thank you wrobel.
I guess I never fully appreciated this fundamental point.

If I may, here a general review of some very basic concepts to see if I understand:

• Functional equations are equations that have functions as solutions.
• Functional equations can algebraic, ODE or PDE. They are defined over a specific region of space $D$ (finite or infinite in extent).
• Functional equations must have both initial conditions (ICs) and boundary conditions (BCs). BCs specify what the solution must be at the edge of the region of interest $D$ only. The BCs can be either fixed or even time-dependent. The ICs, on the other hand, specify the solution at time $t=0$ over the entire region $D$, including the edge, correct? So ICs and BCs must be consistent at the edge.
• ODEs and PDE sare pointwise equations specifying pointwise dependencies between the monomials but their solutions are globally defined meaning that the value of the solution at a particular point $P$ is not only determined by the pointwise dependencies specified by the equations at that point $P$ only but also by the equations at other spatial points.