# Exploring Laplace's Equation: An In-Depth Analysis in Physics

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Laplace’s equation is a fundamental partial differential equation encountered frequently in the fields of physics, particularly after the heat and wave equations:

The operator symbolized by ?² is referred to as the Laplacian operator. Previous discussions have focused on Laplace’s equation in the context of the relaxation algorithm, a numerical method for solving this equation. Now, we will shift our attention to analytical solutions.

The exploration of solutions to Laplace’s equation, along with the Poisson equation ?²?=<i>f</i>, is termed potential theory. This area of study is significant as these equations emerge in conservative vector fields, which can be expressed as the gradient of a scalar function known as potential.

## Potentials and Conservative Fields

Conservative vector fields derive their name from conservative force fields. In such fields, the total work done to move a particle from point A to B is independent of the path taken. Alternatively, a force field is considered conservative if no net work is done when a particle travels along any closed path within that field.

Let <b>F</b> represent a conservative vector field, <i>C</i> be any closed path, and d<b><i>l</i></b> the differential length element of <i>C</i> in the tangent direction at each point. Therefore:

Let <i>S</i> be any surface whose boundary is <i>C</i>, and d<b><i>a</i></b> denote the differential area element vector directed normal to the surface. According to Stokes’ theorem:

Consequently:

This must hold true for any choice of the path <i>C</i> and surface <i>S</i>, implying that ?×<b>F</b>=0. Thus, if <b>F</b> is a conservative force field, then ?×<b>F</b>=0. Conversely, if ?×<b>F</b>=0, then for any selection of <i>S</i> and <i>C</i>:

In summary, a force field <b>F</b> is conservative if and only if ?×<b>F</b>=0. Notably, the curl of a gradient is always zero. Focusing on the x-component of ?×?<i>f</i> for some scalar function <i>f</i> yields:

The last line follows from the equality of mixed partial derivatives. The proof for the other two components is analogous. This leads us to consider writing a conservative vector field as the gradient of a function, contingent upon proving that such a function exists.

Let <i>C<b> </b></i> be any path from A to B, <b>F</b> be a conservative force field, and U(A) and U(B) be the potential energies at points A and B respectively. According to the work-energy theorem, W=U(A)-U(B), thus:

Since the work only depends on points A and B, this integral can be expressed as:

By the fundamental theorem of calculus, we can express:

Let <b>t</b> represent the tangent vector to <i>C</i>, leading to d<b><i>l</i></b>=<b>t</b>d<i>l</i>. Thus:

Parameterizing the curve with <b>r</b>(<i>l</i>)=[x(<i>l</i>),y(<i>l</i>),z(<i>l</i>)], we express the tangent vector as:

Expanding dU/d<i>l</i> using the chain rule results in:

Finally, substituting into <b>F</b>?<b>t</b>=-dU/d<i>l</i> yields:

Thus, if <b>F</b> is indeed a conservative force field, it can be represented as a gradient.

While this has primarily focused on mechanical work, the same principle applies to any vector field <b>V</b>, regardless of its nature. The following statements are equivalent for any vector field <b>V</b>:

- <b>V</b> is a conservative vector field
- The line integral of <b>V</b> from A to B is path-independent
- The line integral of <b>V</b> around any closed path equals zero
- ?×<b>V</b>=0
- <b>V</b> can be expressed as the gradient of a scalar function, known as potential

In many cases, while it may be feasible to compute <b>V</b> directly for simpler physical problems, it is often more practical to determine the potential first, then derive <b>V</b> by calculating the gradient of the potential. This serves as the primary objective of Laplace’s equation.

## Laplace’s Equation

Even when <b>V</b> is unknown, the divergence of <b>V</b> is frequently known. A prime example is the electrostatic potential described by Laplace’s equation, which is directly derived from Gauss’s and Faraday’s laws:

Given that ?×<b>E</b>=0, we can express <b>E</b>=-??, where ? denotes the electrostatic potential. The negative gradient ensures the electric force is correctly oriented. This expression can then be substituted into Gauss’s law to yield Poisson’s equation:

If the potential results solely from charges located at the boundary or outside the area of interest, then ?=0, leading to Laplace’s equation ?²?=0.

Another significant example arises in the examination of incompressible and irrotational fluid flows. Let <b>V</b> signify the fluid's velocity field and ? its density. All fluid flows adhere to the continuity equation:

This equation implies that the amount of fluid at a given point can only change when fluid enters or exits that point.

A fluid is classified as incompressible when its density remains constant. For such fluids, ??/?t=0, allowing ? to be factored out of the divergence, resulting in ??<b>V</b>=0.

In a previous article discussing the Navier-Stokes equations, I elaborated on how fluid motion can be characterized as an infinitesimal fluid parcel at each point, with each parcel's motion being decomposed into pure translation, rotation, and deformation:

The vector <b>?</b> denotes the rotational motion of the fluid parcel:

Here, u, v, and w represent the components of <b>V</b> in the x, y, and z directions, respectively. Thus, <b>?</b> is equivalent to half the curl of <b>V</b>. If this quantity equals zero, it signifies an absence of rotational motion, implying ?×<b>V</b>=0 for irrotational flows. Consequently, the velocity vector field is conservative, permitting us to express <b>V</b> as ?? for some function ?. It is important to note that the <i>flow</i> is irrotational while the <i>velocity field</i> is conservative. Given that ??<b>V</b>=0 for incompressible fluids, this indicates that the potential adheres to Laplace’s equation.

Since both irrotational and incompressible flows can be described through a potential, they are often referred to as potential flows.

## General Solution

The approach to solving Laplace’s equation mirrors the strategies utilized for the heat and wave equations: finding a complete set of orthogonal functions and determining the coefficients through inner products or alternative methods. For simplicity, we will focus on the two-dimensional case, although the ensuing discussion is equally applicable in three dimensions.

We will also bypass the formal proof of existence and uniqueness using a physical rationale. A harmonic potential represents a physical field, and the configuration of such a field is influenced by its sources. Since sources invariably create a field, a solution to Laplace’s equation must exist. The principles of physics dictate that a singular configuration of field sources cannot yield multiple distinct fields; the configuration of these sources determines the field by establishing boundary conditions (assuming the sources are at the boundary). Hence, solutions to Laplace’s equation are both existent and unique.

With this foundation established, we will now pursue the general solution. Utilizing separation of variables, we assume that ?=X(x)Y(y). Thus:

As X and Y are single-variable functions, the partial derivatives simplify to ordinary derivatives. Dividing through by XY leads to:

Because X?/X and Y?/Y depend on different variables, their sum can only equal zero if each is set to a constant:

As ? denotes a physical field, it must be real-valued, necessitating that <i>u</i>² and <i>v</i>² are real numbers. The condition <i>u</i>²+<i>v</i>²=0 implies that one of the constants must be real while the other is imaginary. The eigenfunctions are:

Thus, the general solution can be expressed as:

We do not necessarily assume that <i>u</i> and <i>v</i> are discrete, indicating that this summation may indeed be an integral.

We will also derive the general solution using polar coordinates. In polar coordinates, the Laplacian is expressed as:

Assuming that ?(r,?)=R(r)?(?), after expanding the derivatives, dividing by R?, and rearranging, we arrive at:

The left side is solely a function of r, while the right side is exclusively a function of ?. For both functions to be identically equal, they must both equate to the same constant <i>k</i>:

We have the freedom to select k as we wish, and the eigenfunctions take an aesthetically pleasing form when we assign k=n²:

Regarding the radial aspect, the differential equation resembles a second-order Cauchy-Euler equation:

Thus, the general solution is:

## Demonstration: A Charged Patch on a Conducting Plate

Two extensive (effectively infinite) parallel conducting plates are grounded and separated by a distance <i>d</i>. A square patch on one plate, with a side length of 2<i>a</i>, is maintained at a constant potential <i>V</i>. We aim to determine the potential between these plates.

The general solution is:

The potential should approach zero as <i>x</i> tends to infinity, which implies A?=0. The potential must also be zero outside the region between the plates, leading to C?=0. Since potentials are defined only up to an additive constant, we can assume the remaining constant term B?D? is also zero. This simplifies to:

Given that the boundary condition is an even function of <i>x</i>, ? must also be even, which can be achieved by setting <i>A</i>?=<i>B</i>?.

Consequently, <i>A</i>? can be factored out of the <i>x</i> portion, leading to a re-labeling of the constants:

It is crucial to remember that one of <i>u</i> and <i>v</i> is real while the other is imaginary. If <i>u</i> is real, the <i>x</i> component of the function becomes unbounded as <i>x</i> approaches infinity, which is not permissible. Thus, <i>u</i> must be a pure imaginary number, <i>u=ip</i>, and <i>v</i> is real. This results in:

For clarity, we have incorporated the factor of 2 into the <i>A</i>? coefficient. We can now define <i>v</i>=<i>p</i> since <i>u²+v²=0</i> and <i>u</i>=<i>ip</i>, simplifying the expression to:

Next, we need to apply the boundary conditions to determine <i>a</i>? and <i>b</i>?.

We can express the boundary condition at <i>y</i>=0 using the Heaviside function:

Thus, the boundary conditions are:

These conditions do not provide a straightforward solution. However, we can utilize a technique known as the method of images.

Let’s examine the potential field produced if the upper plate were absent:

Assuming we know the solution ?, and reflecting this system across the dashed line gives us:

It is intuitively clear that the solution for this mirrored system will simply be ? inverted. Denote this mirrored solution as ??. By symmetry, ??=? along the dashed line. If we now change <i>V</i> to -<i>V</i> in the reflected system, the only effect is to replace ?? with -??. While this can be mathematically proven, a physical understanding is more accessible. If the patch is at potential <i>+V</i>, the force on a positive test charge is identical to the force on a negative test charge of equal magnitude at potential -<i>V</i>. Thus, altering the patch’s potential sign solely changes the resulting electric field's direction, leaving the potential's sign unchanged.

By combining the original and mirrored systems with the potential sign reversed, we ensure the potential along the dashed line equals zero:

This presents an equivalent set of boundary conditions in a more manageable form:

We will use these boundary conditions to generate two equations involving the unknowns <i>a</i>? and <i>b</i>?:

First, we eliminate <i>b</i>?:

Next, we simplify the expression for ?(<i>x</i>,0):

The term in the brackets is solely a function of <i>p</i>, allowing us to incorporate it into the <i>a</i>? coefficient. Therefore:

This function is non-periodic, and since the domain is the entire real line, the Fourier series will manifest as an integral, leading to <i>a</i>? being a continuous function of <i>p</i>. Thus:

Fortunately, a continuous adaptation of the orthogonality rule from the discrete scenario exists:

The ?(p-q) function is identified as the Dirac delta function, characterized by the sampling property:

We can now determine <i>a</i>(<i>p</i>):

Now that we have derived the coefficients:

The resulting solution is:

Integrals involving terms like sin(ap)/p are categorized as sine integrals, and these integrals cannot be computed in terms of elementary functions. Thus, we will leave it in its current form. Fortunately, this problem can be easily solved numerically using the relaxation algorithm, making it straightforward to visualize the resulting solution. The outcome is illustrated below, as well as at the top of this article:

The two plates are depicted at the top and bottom of the image, with the patch located at the center of the lower plate. The plot illustrates the contour lines of the potential, with warmer colors indicating higher potential levels. The electric field lines, shown in red, overlay the contour plot, emanating from the potential patch and terminating at one of the plates. The sharply curved field lines at the edges of the graph are not artifacts; they illustrate the transition of field lines from returning to the lower plate to heading toward the upper plate.

## Demonstration: Potential Flow Around a Cylinder

The potential flow surrounding a circular cylinder represents a fundamental problem in potential theory.

Consider a cylinder of radius <i>a</i> centered at the origin, with its axis perpendicular to the xy-plane. An incompressible, inviscid fluid flows toward the cylinder from the left with uniform velocity <b>V</b>?=<i>u</i>?<b>x</b>. We seek to analyze the flow near the cylinder by determining the fluid's velocity through its potential and gradient. As usual, we will begin by establishing our boundary conditions.

At points far from the cylinder, the velocity should equal <b>V</b>?=<i>u</i>?<b>x</b>, indicating that the potential approaches ??=<i>u</i>?<i>x</i> far from the origin. In polar coordinates, this is expressed as ?=<i>u</i>?<i>r</i>cos?. Though this may initially appear to contradict the requirement for the potential to be finite everywhere, we are positing that the flow's sources are located at a finite distance from the cylinder, far enough that the cylinder's presence does not significantly alter the fluid's motion in the vicinity of these sources. Therefore, our first boundary condition states that when <i>r</i> is much larger than <i>a</i>, the potential limits to <i>u</i>?<i>r</i>cos?.

The cylinder’s boundary prevents fluid from passing through, so the flow's velocity component normal to the boundary surface must equal zero. The normal vector is the radial unit vector, leading to <b>r?V</b>=??/?<i>r</i>=0 when <i>r</i>=<i>a</i>.

When <i>r</i> is significantly large, the <i>r</i>?? terms in the general solution vanish, yielding:

To equate this to <i>u</i>?<i>r</i>cos?, we must set <i>a</i>?=<i>b</i>?=<i>c</i>?=<i>d</i>?=0, while also ensuring <i>a</i>?=0 for all <i>n</i>>1, leading to:

Thus, we need to set <i>a</i>?<i>c</i>?=<i>u</i>? and <i>d</i>?=0. Plugging this back into the general solution while setting <i>A</i>?=<i>b</i>?<i>c</i>? and <i>B</i>?=<i>b</i>?<i>d</i>? provides:

This results in a velocity field whose radial component is:

If we set <i>A</i>?=<i>u</i>?a² and <i>A</i>?=<i>B</i>?=0 for all <i>n</i>>1, then ??/?<i>r </i> will equal zero for all ? when <i>r</i>=<i>a</i>. Therefore, the solution is:

The velocity is simply the gradient of this function. Below are the velocity vectors illustrated:

The background contour plot of the vector plot depicts the pressure distribution.

The following animation illustrates the flow itself, with the streamlines and pressure also displayed:

## Copyright Information

I have credited all content that is not my own original work, and I am utilizing such content under fair use guidelines. The problem in the first demonstration is a modification of one found in the textbook *Modern Electrodynamics* by Zangwill (2013).