Integral and differential forms of field equations

[sergiokapone]

As is known, electrostatics studies the interaction of stationary charges and the fields created by them. The basis of electrostatics is a system of equations describing the properties of the electrostatic field. These equations are formulated in integral and differential forms. The integral form of the equations is more general, while to determine the fields using differential equations it is necessary to know the boundary conditions at the interfaces.

I have a question, is there some other fundamental difference in the integral and differential form of the field equations that is missed here and should have been mentioned?

 

[Orodruin]

I disagree that the integral form is more general. The integral and differential forms are equivalent, as is easily determined by the divergence and curl theorems.

 

[sergiokapone]

Yes, but boundary conditions are derived from integral theorems, and are needed to solve differential equations. If we have symmetry and integral equations we will solve the problem, but if we have symmetry and differential equations we will not solve the problem, we need to know what will be on the boundaries.

Let's say, for example, we can take the problem of finding the field of a uniformly charged ball. Here it is obvious that integral equations and symmetry are enough. But if you solve differential equations, you would need to know how to glue the solutions together and define constants.

 

[Orodruin]

Yes, but boundary conditions are derived from integral theorems, and are needed to solve differential equations. If we have symmetry and integral equations we will solve the problem, but if we have symmetry and differential equations we will not solve the problem, we need to know what will be on the boundaries.

Let's say, for example, we can take the problem of finding the field of a uniformly charged ball. Here it is obvious that integral equations and symmetry are enough. But if you solve differential equations, you would need to know how to glue the solutions together and define constants.

That’s simply incorrect.

 

[hutchphd]

That’s simply incorrect.

Could you be a bit more specific?

 

[Orodruin]

Could you be a bit more specific?

It is perfectly possible to find the solutions directly from the differential form of the field equations, including the field discontinuites across a surface charge.

 

[sergiokapone]

How?

 

[Orodruin]

How?

Uhmm ... solve the differential equation.

Note that the charge density of surface charges and similar are described by delta distributions, but these can be easily handled (you can go into formal distribution theory if you need to, but the basics are fine). These surfaces are not really boundaries per se, they are part of the volume in which you are solving the differential equation.

You can use any method available to you to solve the differential equation, although the most straight forward one is likely the use of Green's functions.

 

[Orodruin]

So to make that more explicit. Take the case of the uniformly charged ball:

Let's say, for example, we can take the problem of finding the field of a uniformly charged ball. Here it is obvious that integral equations and symmetry are enough. But if you solve differential equations, you would need to know how to glue the solutions together and define constants.

... or actually ... let's spice it up ... uniformly charged spherical shell so we actually have some delta distributions in the PDE. The field is then described by the differential equation$$
\nabla \cdot \vec E = \frac{Q}{4\varepsilon_0\pi R^2} \delta(r- R).
$$ Here $R$ is the radius of the sphere. The spherical symmetry of the problem results in $\vec E = E(r) \vec e_r$ and so $$
\nabla \cdot \vec E = \frac{1}{r^2}\frac{d(r^2 E(r))}{dr} = E'(r) + \frac{2E(r)}{r} = \frac{Q}{4\varepsilon_0\pi R^2} \delta(r- R).
$$ This is now an ODE that we need to solve. For brevity of notation, let's introduce $C = Q/4\varepsilon_0\pi R^2$. Since the inhomogeneity on the RHS contains a delta function at $r = R$ and $\delta(r-R) = \theta'(r-R)$, we make the ansatz $E(r) = f(r) \theta(r-R) + g(r) \theta(R-r)$. Clearly $f(r)$ describes the behaviour of $E(r)$ for $r > R$ and $g(r)$ for $r < R$. Both $f$ and $g$ are assumed to be smooth functions although it really does not matter what $g(r)$ is for $r > R$ etc. The derivative $E'(r)$ is now given by
\begin{align*}
E'(r) &= f'(r)\theta(r-R) + f(r) \delta(r-R) + g'(r) \theta(R-r) - g(r)\delta(r-R) \\
&= f'(r) \theta(r-R) + \delta(r-R)[f(R) - g(R)] + g'(r) \theta(R-r).
\end{align*} Insertion into the differential equation leads to $$
[f'(r) + 2f(r)/r] \theta(r-R) + \delta(r-R)[f(R) - g(R)] + [g'(r) + 2g(r)/r] \theta(R-r) = C \delta(r-R)
$$Three things are necessary to satisfy this:
\begin{align*}
f(R) - g(R) &= C \\
f'(r) + 2f(r)/r &= 0 \quad (r > R) \\
g'(r) + 2g(r)/r &= 0 \quad (r < R).
\end{align*}
For $f$ and $g$ we have the same form of differential equation, resulting in the same functional form of the general solution:
$$
f(r) = \frac{B_f}{r^2}, \quad g(r) = \frac{B_g}{r^2}.
$$
For $g$ we must require that $B_g = 0$ in order to avoid the resulting field $\vec E(r)$ having a delta divergence at $r = 0$*. We now have $f(r) = B_f/r^2$ and $g(r) = 0$. This implies that $$
f(R) - g(R) = f(R) = \frac{B_f}{R^2} = C = \frac{Q}{4\varepsilon_0\pi R^2}$$ and therefore $$
B_f = \frac{Q}{4\varepsilon_0 \pi} \quad \Longrightarrow \quad \vec E = \frac{Q}{4\varepsilon_0 \pi r^2} \theta(r-R) \vec e_r$$

The above essentially derives Newton's shell theorem. The field for a uniformly charged ball can be obtained by integrating over all infinitesimally thick spherical shells out to the ball radius $R_1$, noting that the charge on each is $Q = 4\pi \rho R^2 dR$, where $\rho$ is the constant charge density: $$
E_{\rm ball}(r) = \frac{\rho}{\varepsilon_0 r^2} \int_0^{R_1} \theta(r-R)R^2 dR =
\begin{cases}
\frac{\rho R_1^3}{3\varepsilon_0 r^2} & \qquad (r > R_1) \\
\frac{\rho r}{3\varepsilon_0} & \qquad (r < R_1)
\end{cases}
$$
Note that $\rho = 3 Q_{\rm tot}/4\pi R_1^3$ and so this may also be written$$
E_{\rm ball}(r) =
\begin{cases}
\frac{Q_{\rm tot}}{4\pi \varepsilon_0 r^2} & \qquad (r > R_1) \\
\frac{Q_{\rm tot} r}{4\pi \varepsilon_0 R_1^3} & \qquad (r < R_1)
\end{cases}
$$ which may be more familiar.

Further notice that the integral form of the field equations are nowhere in sight. In fact, you can easily derive the integral form from the differential form (and vice versa). In the differential form to integral form direction:

Assume that $\nabla \cdot \vec E = \rho/\varepsilon_0$. Then for any volume $V$ with boundary surface $S$, the flux of the electric field through $S$ is given by $$
\Phi = \oint_S \vec E \cdot d\vec S$$ By the divergence theorem, we therefore find that$$
\Phi = \int_V \nabla\cdot \vec E \, dV = \frac{1}{\varepsilon_0} \int_V \rho\, dV \equiv \frac{Q_{\rm enc}}{\varepsilon_0}$$ where $Q_{\rm enc}$ by definition is the total charge in the volume $V$.

In true textbook fashion, I will leave the other direction for the interested reader.


* Note that $r = 0$ is not really a boundary of the physical region in which we are solving the PDE! It is common to argue that $r = 0$ is a boundary and that hand-wavingly state that the solution must be finite in that regime. That is not a very good argument though. The better argument comes from the $1/r$ field having a delta-divergence at $r = 0$ meaning that it does not actually solve the $\nabla \cdot \vec E$ at $r = 0$.

Edit: Fixed an error in the solution. (I am too used to solving for the potential rather than field strength apparently…)

 

[sergiokapone]

Well, what if we solve the equation for the ball right away:

$$\nabla \cdot \vec E = \frac{Q}{4\varepsilon_0\pi R^2} \theta(R- r).$$

I can't get solution without condition of continuity of E_r on the boundary of r=R.

 

[Orodruin]

Well, what if we solve the equation for the ball right away:

$$\nabla \cdot \vec E = \frac{Q}{4\varepsilon_0\pi R^2} \theta(R- r).$$

I can't get solution without condition of continuity of E_r on the boundary of r=R.

Of course you can. You make the same assumption of $E(r) = f(r)\theta(r-R) + g(r) \theta(R-r)$. However, instead of $f(R) - g(R) = C$ you will obtain $f(R) - g(R) = 0$.

 

[sergiokapone]

So then it goes like this:
the differential form makes sense only when charge is distributed in space with finite density $\rho$. If $\rho$ goes to infinity at individual points, on lines or surfaces, the differential form becomes inapplicable, whereas the integral form is applicable in such cases.

By using the generalized functions, the differential form of Gauss's theorem can be extended to these cases.

 

[Orodruin]

So then it goes like this:
the differential form makes sense only when charge is distributed in space with finite density $\rho$. If $\rho$ goes to infinity at individual points, on lines or surfaces, the differential form becomes inapplicable, whereas the integral form is applicable in such cases.

By using the generalized functions, the differential form of Gauss's theorem can be extended to these cases.

The above is true also for the integral form. To handle divergences in the field (think point or line charge) you must still consider the field a generalized function.

The results of both forms are the same and the forms are equivalent.

 

[sergiokapone]

Well, look, if I don't use generalized functions, how can I find a solution to a differential equation? I have to involve boundary conditions, which are derived from the integral theorem. Here's what I don't understand.

 

[Orodruin]

Well, look, if I don't use generalized functions, how can I find a solution to a differential equation? I have to involve boundary conditions, which are derived from the integral theorem. Here's what I don't understand.

You need to use generalized functions because generally the solutions to these kinds of problems are generalized functions.

The integral form argument kind of hides this buy only considering the field where it has a well defined smooth behaviour, but it doesn’t change the facts. There are however a couple of tricks you can use also in the differential formulation, such as integrating the differential equation over a small interval across the boundary.

 

[Orodruin]

As an example, take the boundary between the ball and outside of the ball at $r = R$. The differential equation is effectively on the form:$$
(r^2 E(r))’= cr^2 \theta(R-r)
$$ Integrating this from $r = R-\epsilon$ to $R+\epsilon$ yields: $$
(R+\epsilon)^2 E(R+\epsilon) - (R -\epsilon)^2 E(R-\epsilon) = c [R^3 - (R-\epsilon)^3]/3
$$ This may be written $$
R^2[ E(R^+) - E(R^-)] + \mathcal O(\epsilon) = 0
$$ where $E(R^\pm)$ are the right/left limits as $r\to R$. Taking the $\epsilon \to 0$ limit therefore results in $$
E(R^+) = E(R^-)
$$
In other words, $E$ is continuous across the interface.

 

[Gavran]

The integral form is not more general than the differential form and the differential form is not more general than the integral form.
The main difference between them is the case where a particular form will be applied. The integral form is more suitable for dealing with symmetric cases with fields on lines or surfaces whereas the differential form is more suitable for dealing with asymmetric cases with fields at points.