Solutions to Laplace's Equation in CartesianCoordinates

Having investigated some general properties of solutions to Poisson's equation, it is now appropriate to study specific methods of solution to Laplace's equation subject to boundary conditions. Exemplified by this and the next section are three standard steps often used in representing EQS fields. First, Laplace's equation is set up in the coordinate system in which the boundary surfaces are coordinate surfaces. Then, the partial differential equation is reduced to a set of ordinary differential equations by separation of variables. In this way, an infinite set of solutions is generated. Finally, the boundary conditions are satisfied by superimposing the solutions found by separation of variables.

In this section, solutions are derived that are natural if boundary conditions are stated along coordinate surfaces of a Cartesian coordinate system. It is assumed that the fields depend on only two coordinates,

xandy, so that Laplace's equation is (Table I)

This is a partial differential equation in two independent variables. One time-honored method of mathematics is to reduce a new problem to a problem previously solved. Here the process of finding solutions to the partial differential equation is reduced to one of finding solutions to ordinary differential equations. This is accomplished by the

method of separation of variables. It consists of assuming solutions with the special space dependence

In (2),

Xis assumed to be a function ofxalone andYis a function ofyalone. If need be, a general space dependence is then recovered by superposition of these special solutions. Substitution of (2) into (1) and division by then gives

Total derivative symbols are used because the respective functions

XandYare by definition only functions ofxandy.

In (3) we now have on the left-hand side a function of

xalone, on the right-hand side a function ofyalone. The equation can be satisfied independent ofxandyonly if each of these expressions is constant. We denote this "separation" constant byk, and it follows that^{2}

and

These equations have the solutions

If

k = 0, the solutions degenerate into

The product solutions, (2), are summarized in the first four rows of Table 5.4.1. Those in the right-hand column are simply those of the middle column with the roles of

xandyinterchanged. Generally, we will leave the prime off thekin writing these solutions. Exponentials are also solutions to (7). These, sometimes more convenient, solutions are summarized in the last four rows of the table.^{'}

The solutions summarized in this table can be used to gain insight into the nature of EQS fields. A good investment is therefore made if they are now visualized.

The fields represented by the potentials in the left-hand column of Table 5.4.1 are all familiar. Those that are linear in

xandyrepresent uniform fields, in thexandydirections, respectively. The potentialxyis familiar from Fig. 4.1.3. We will use similar conventions to represent the potentials of the second column, but it is helpful to have in mind the three-dimensional portrayal exemplified for the potentialxyin Fig. 4.1.4. In the more complicated field maps to follow, the sketch is visualized as a contour map of the potential with peaks of positive potential and valleys of negative potential.

On the top and left peripheries of Fig. 5.4.1 are sketched the functions

cos kxandcosh ky, respectively, the product of which is the first of the potentials in the middle column of Table 5.4.1. If we start out from the origin in either the+yor-ydirections (north or south), we climb a potential hill. If we instead proceed in the+xor-xdirections (east or west), we move downhill. An easterly path begun on the potential hill to the north of the origin corresponds to a decrease in thecos kxfactor. To follow a path of equal elevation, thecosh kyfactor must increase, and this implies that the path must turn northward.

Figure 5.4.1Equipotentials for =cos(kx)cosh (ky)and field lines. As an aid to visualizing the potential, the separate factorscos (kx)andcosh (ky)are, respectively, displayed at the top and to the left.A good starting point in making these field sketches is the identification of the contours of zero potential. In the plot of the second potential in the middle column of Table 5.4.1, shown in Fig. 5.4.2, these are the

yaxis and the lineskx = + /2, + 3 /2, etc. The dependence onyis now odd rather than even, as it was for the plot of Fig. 5.4.1. Thus, the origin is now on the side of a potential hill that slopes downward from north to south.

Figure 5.4.2Equipotentials for =cos (kx)sinh (ky)and field lines. As an aid to visualizing the potential, the separate factorscos (kx)andsinh (ky)are, respectively, displayed at the top and to the left.The solutions in the third and fourth rows of the second column possess the same field patterns as those just discussed provided those patterns are respectively shifted in the

xdirection. In the last four rows of Table 5.4.1 are four additional possible solutions which are linear combinations of the previous four in that column. Because these decay exponentially in either the+yor-ydirections, they are useful for representing solutions in problems where an infinite half-space is considered.

The solutions in Table 5.4.1 are nonsingular throughout the entire

x-yplane. This means that Laplace's equation is obeyed everywhere within the finitex-yplane, and hence the field lines are continuous; they do not appear or disappear. The sketches show that the fields become stronger and stronger as one proceeds in the positive and negativeydirections. The lines of electric field originate on positive charges and terminate on negative charges aty. Thus, for the plots shown in Figs. 5.4.1 and 5.4.2, the charge distributions at infinity must consist of alternating distributions of positive and negative charges of infinite amplitude.

Two final observations serve to further develop an appreciation for the nature of solutions to Laplace's equation. First, the third dimension can be used to represent the potential in the manner of Fig. 4.1.4, so that the potential surface has the shape of a membrane stretched from boundaries that are elevated in proportion to their potentials.

Laplace's equation, (1), requires that the sum of quantities that reflect the curvatures in the

xandydirections vanish. If the second derivative of a function is positive, it is curved upward; and if it is negative, it is curved downward. If the curvature is positive in thexdirection, it must be negative in theydirection. Thus, at the origin in Fig. 5.4.1, the potential is cupped downward for excursions in thexdirection, and so it must be cupped upward for variations in theydirection. A similar deduction must apply at every point in thex-yplane.

Second, because the

kthat appears in the periodic functions of the second column in Table 5.4.1 is the same as that in the exponential and hyperbolic functions, it is clear that the more rapid the periodic variation, the more rapid is the decay or apparent growth.

TABLE 5.4.1TWO-DIMENSIONAL CARTESIAN SOLUTIONS OF LAPLACE'S EQUATION

k = 0k^{2}0k^{2}0 (k jk^{'})Constant cos kxcoshkycosh kcos^{'}xk^{'}yycos kxsinhkycosh ksin^{'}xk^{'}yxsin kxcoshkysinh kcos^{'}xk^{'}yxysin kxsinhkysinh ksin^{'}xk^{'}ycos kxe^{ky}ecos^{k'x}k^{'}ycos kxe^{-ky}ecos^{-k'x}k^{'}ysin kxe^{ky}esin^{k'x}k^{'}ysin kxe^{-ky}esin^{-k'x}k^{'}y