CHAPTER 12

Partial Differential Equations (PDEs)

A PDE is an equation that contains one or more partial derivatives of an unknown function that depends on at least two variables. Usually one of these deals with time t and the remaining with space (spatial variable(s)). The most important PDEs are the wave equations that can model the vibrating string (Secs. 12.2, 12.3, 12.4, 12.12) and the vibrating membrane (Secs. 12.8, 12.9, 12.10), the heat equation for temperature in a bar or wire (Secs. 12.5, 12.6), and the Laplace equation for electrostatic potentials (Secs. 12.6, 12.10, 12.11). PDEs are very important in dynamics, elasticity, heat transfer, electromagnetic theory, and quantum mechanics. They have a much wider range of applications than ODEs, which can model only the simplest physical systems. Thus PDEs are subjects of many ongoing research and development projects.

Realizing that modeling with PDEs is more involved than modeling with ODEs, we take a gradual, well-planned approach to modeling with PDEs. To do this we carefully derive the PDE that models the phenomena, such as the one-dimensional wave equation for a vibrating elastic string (say a violin string) in Sec. 12.2, and then solve the PDE in a separate section, that is, Sec. 12.3. In a similar vein, we derive the heat equation in Sec. 12.5 and then solve and generalize it in Sec. 12.6.

We derive these PDEs from physics and consider methods for solving initial and boundary value problems, that is, methods of obtaining solutions which satisfy the conditions required by the physical situations. In Secs. 12.7 and 12.12 we show how PDEs can also be solved by Fourier and Laplace transform methods.

Prerequisites: Linear ODEs (Chap. 2), Fourier series (Chap. 11).

Sections that may be omitted in a shorter course: 12.7, 12.10–12.12.

References and Answers to Problems: App. 1 Part C, App. 2.

12.1 Basic Concepts of PDEs

A partial differential equation (PDE) is an equation involving one or more partial derivatives of an (unknown) function, call it u, that depends on two or more variables, often time t and one or several variables in space. The order of the highest derivative is called the order of the PDE. Just as was the case for ODEs, second-order PDEs will be the most important ones in applications.

Just as for ordinary differential equations (ODEs) we say that a PDE is linear if it is of the first degree in the unknown function u and its partial derivatives. Otherwise we call it nonlinear. Thus, all the equations in Example 1 are linear. We call a linear PDE homogeneous if each of its terms contains either u or one of its partial derivatives. Otherwise we call the equation nonhomogeneous. Thus, (4) in Example 1 (with f not identically zero) is nonhomogeneous, whereas the other equations are homogeneous.

A solution of a PDE in some region R of the space of the independent variables is a function that has all the partial derivatives appearing in the PDE in some domain D (definition in Sec. 9.6) containing R, and satisfies the PDE everywhere in R.

Often one merely requires that the function is continuous on the boundary of R, has those derivatives in the interior of R, and satisfies the PDE in the interior of R. Letting R lie in D simplifies the situation regarding derivatives on the boundary of R, which is then the same on the boundary as it is in the interior of R.

In general, the totality of solutions of a PDE is very large. For example, the functions

which are entirely different from each other, are solutions of (3), as you may verify. We shall see later that the unique solution of a PDE corresponding to a given physical problem will be obtained by the use of additional conditions arising from the problem. For instance, this may be the condition that the solution u assume given values on the boundary of the region R (“boundary conditions”). Or, when time t is one of the variables, u (u_t = ∂u/∂t or both) may be prescribed at t = 0 (“initial conditions”).

We know that if an ODE is linear and homogeneous, then from known solutions we can obtain further solutions by superposition. For PDEs the situation is quite similar:

The simple proof of this important theorem is quite similar to that of Theorem 1 in Sec. 2.1 and is left to the student.

Verification of solutions in Probs. 2–13 proceeds as for ODEs. Problems 16–23 concern PDEs solvable like ODEs. To help the student with them, we consider two typical examples.

12.2 Modeling: Vibrating String, Wave Equation

In this section we model a vibrating string, which will lead to our first important PDE, that is, equation (3) which will then be solved in Sec. 12.3. The student should pay very close attention to this delicate modeling process and detailed derivation starting from scratch, as the skills learned can be applied to modeling other phenomena in general and in particular to modeling a vibrating membrane (Sec. 12.7).

We want to derive the PDE modeling small transverse vibrations of an elastic string, such as a violin string. We place the string along the x-axis, stretch it to length L, and fasten it at the ends x = 0 and x = L. We then distort the string, and at some instant, call it t = 0, we release it and allow it to vibrate. The problem is to determine the vibrations of the string, that is, to find its deflection u(x, t) at any point x and at any time t > 0; see Fig. 286.

u(x, t) will be the solution of a PDE that is the model of our physical system to be derived. This PDE should not be too complicated, so that we can solve it. Reasonable simplifying assumptions (just as for ODEs modeling vibrations in Chap. 2) are as follows.

Physical Assumptions

1. The mass of the string per unit length is constant (“homogeneous string”). The string is perfectly elastic and does not offer any resistance to bending.
2. The tension caused by stretching the string before fastening it at the ends is so large that the action of the gravitational force on the string (trying to pull the string down a little) can be neglected.
3. The string performs small transverse motions in a vertical plane; that is, every particle of the string moves strictly vertically and so that the deflection and the slope at every point of the string always remain small in absolute value.

Under these assumptions we may expect solutions that describe the physical reality sufficiently well.

Fig. 286. Deflected string at fixed time t. Explanation on p. 544

Derivation of the PDE of the Model (“Wave Equation”) from Forces

The model of the vibrating string will consist of a PDE (“wave equation”) and additional conditions. To obtain the PDE, we consider the forces acting on a small portion of the string (Fig. 286). This method is typical of modeling in mechanics and elsewhere.

Since the string offers no resistance to bending, the tension is tangential to the curve of the string at each point. Let T₁ and T₂ be the tension at the endpoints P and Q of that portion. Since the points of the string move vertically, there is no motion in the horizontal direction. Hence the horizontal components of the tension must be constant. Using the notation shown in Fig. 286, we thus obtain

In the vertical direction we have two forces, namely, the vertical components −T₁ sin α and T₂ sin β of T₁ and T₂; here the minus sign appears because the component at P is directed downward. By Newton's second law (Sec. 2.4) the resultant of these two forces is equal to the mass ρΔx of the portion times the acceleration ∂²u/∂t², evaluated at some point between x and x + Δx; here ρ is the mass of the undeflected string per unit length, and Δx is the length of the portion of the undeflected string. (Δ is generally used to denote small quantities; this has nothing to do with the Laplacian , which is sometimes also denoted by Δ) Hence

Using (1), we can divide this by T₂ cos β = T₁ cos α = T, obtaining

Now tan α and tan β are the slopes of the string at x and x + Δx:

Here we have to write partial derivatives because u also depends on time t. Dividing (2) by Δx, we thus have

If we let Δx approach zero, we obtain the linear PDE

This is called the one-dimensional wave equation. We see that it is homogeneous and of the second order. The physical constant T/ρ is denoted by c² (instead of c) to indicate that this constant is positive, a fact that will be essential to the form of the solutions. “One-dimensional” means that the equation involves only one space variable, x. In the next section we shall complete setting up the model and then show how to solve it by a general method that is probably the most important one for PDEs in engineering mathematics.

12.3 Solution by Separating Variables. Use of Fourier Series

We continue our work from Sec. 12.2, where we modeled a vibrating string and obtained the one-dimensional wave equation. We now have to complete the model by adding additional conditions and then solving the resulting model.

The model of a vibrating elastic string (a violin string, for instance) consists of the one-dimensional wave equation

for the unknown deflection u(x, t) of the string, a PDE that we have just obtained, and some additional conditions, which we shall now derive.

Since the string is fastened at the ends x = 0 and x = L (see Sec. 12.2), we have the two boundary conditions

Furthermore, the form of the motion of the string will depend on its initial deflection (deflection at time t = 0), call it f(x), and on its initial velocity (velocity at t = 0), call it g(x). We thus have the two initial conditions

where u_t = ∂u/∂t. We now have to find a solution of the PDE (1) satisfying the conditions (2) and (3). This will be the solution of our problem. We shall do this in three steps, as follows.

Step 1. By the “method of separating variables” or product method, setting u(x, t) = F(x)G(t), we obtain from (1) two ODEs, one for F(x) and the other one for G(t).

Step 2. We determine solutions of these ODEs that satisfy the boundary conditions (2).

Step 3. Finally, using Fourier series, we compose the solutions found in Step 2 to obtain a solution of (1) satisfying both (2) and (3), that is, the solution of our model of the vibrating string.

Step 1. Two ODEs from the Wave Equation (1)

In the method of separating variables, or product method, we determine solutions of the wave equation (1) of the form

which are a product of two functions, each depending on only one of the variables x and t. This is a powerful general method that has various applications in engineering mathematics, as we shall see in this chapter. Differentiating (4), we obtain

where dots denote derivatives with respect to t and primes derivatives with respect to x. By inserting this into the wave equation (1) we have

Dividing by c²FG and simplifying gives

The variables are now separated, the left side depending only on t and the right side only on x. Hence both sides must be constant because, if they were variable, then changing t or x would affect only one side, leaving the other unaltered. Thus, say,

Multiplying by the denominators gives immediately two ordinary DEs

and

Here, the separation constant k is still arbitrary.

Step 2. Satisfying the Boundary Conditions (2)

We now determine solutions F and G of (5) and (6) so that u = FG satisfies the boundary conditions (2), that is,

We first solve (5). If G ≡, then u = FG ≡ 0, which is of no interest. Hence G ≡ 0 and then by (7),

We show that k must be negative. For k = 0 the general solution of (5) is F = ax + b, and from (8) we obtain a = b = 0, so that F ≡ 0 and u = FG ≡ 0, which is of no interest. For positive k = μ² a general solution of (5) is

and from (8) we obtain F ≡ 0 as before (verify!). Hence we are left with the possibility of choosing k negative, say, k = −p². Then (5) becomes F″ + p²F = 0 and has as a general solution

From this and (8) we have

We must take B ≠ 0 since otherwise F ≡ 0. Hence sin pL = 0. Thus

Setting B = 1 we thus obtain infinitely many solutions F(x) = F_n(x) where

These solutions satisfy (8). [For negative integer n we obtain essentially the same solutions, except for a minus sign, because sin (−α) = −sin α.]

We now solve (6) with k = −p² = −(nπ/L²) resulting from (9), that is,

A general solution is

Hence solutions of (1) satisfying (2) are u_n(x, t) = F_n(x)G_n(t) = G_n(t)F_n(x), written out

These functions are called the eigenfunctions, or characteristic functions, and the values λ_n = cnπ/L are called the eigenvalues, or characteristic values, of the vibrating string. The set {λ₁, λ₂, …} is called the spectrum.

Discussion of Eigenfunctions. We see that each u_n represents a harmonic motion having the frequency λ_n/2π = cn/2L cycles per unit time. This motion is called the nth normal mode of the string. The first normal mode is known as the fundamental mode and the others are known as overtones; musically they give the octave, octave plus fifth, etc. Since in (11)

the nth normal mode has n − 1 nodes, that is, points of the string that do not move (in addition to the fixed endpoints); see Fig. 287.

Fig. 287. Normal modes of the vibrating string

Figure 288 shows the second normal mode for various values of t. At any instant the string has the form of a sine wave. When the left part of the string is moving down, the other half is moving up, and conversely. For the other modes the situation is similar.

Tuning is done by changing the tension T. Our formula for the frequency λ_n/2π = cn/2L of u_n with [see (3), Sec. 12.2] confirms that effect because it shows that the frequency is proportional to the tension. T cannot be increased indefinitely, but can you see what to do to get a string with a high fundamental mode? (Think of both L and ρ.) Why is a violin smaller than a double-bass?

Fig. 288. Second normal mode for various values of t

Step 3. Solution of the Entire Problem. Fourier Series

The eigenfunctions (11) satisfy the wave equation (1) and the boundary conditions (2) (string fixed at the ends). A single u_n will generally not satisfy the initial conditions (3). But since the wave equation (1) is linear and homogeneous, it follows from Fundamental Theorem 1 in Sec. 12.1 that the sum of finitely many solutions u_n is a solution of (1). To obtain a solution that also satisfies the initial conditions (3), we consider the infinite series (with λ_n = cnπ/L as before)

Satisfying Initial Condition (3a) (Given Initial Displacement). From (12) and (3a) we obtain

Hence we must choose the B_n*'s so that u(x, 0) becomes the Fourier sine series of f(x). Thus, by (4) in Sec. 11.3,

Satisfying Initial Condition (3b) (Given Initial Velocity). Similarly, by differentiating (12) with respect to t and using (3b), we obtain

Hence we must choose the so that for t = 0 the derivative ∂u/∂t becomes the Fourier sine series of g(x). Thus, again by (4) in Sec. 11.3,

Since λ_n = cnπ/L, we obtain by division

Result. Our discussion shows that u(x, t) given by (12) with coefficients (14) and (15) is a solution of (1) that satisfies all the conditions in (2) and (3), provided the series (12) converges and so do the series obtained by differentiating (12) twice termwise with respect to x and t and have the sums ∂²u/∂x² and ∂²u/∂t², respectively, which are continuous.

Solution (12) Established. According to our derivation, the solution (12) is at first a purely formal expression, but we shall now establish it. For the sake of simplicity we consider only the case when the initial velocity g(x) is identically zero. Then the B_n* are zero, and (12) reduces to

It is possible to sum this series, that is, to write the result in a closed or finite form. For this purpose we use the formula [see (11), App. A3.1]

Consequently, we may write (16) in the form

These two series are those obtained by substituting x − ct and x + ct, respectively, for the variable x in the Fourier sine series (13) for f(x). Thus

where f* is the odd periodic extension of f with the period 2L (Fig. 289). Since the initial deflection f(x) is continuous on the interval 0 x L and zero at the endpoints, it follows from (17) that u(x, t) is a continuous function of both variables x and t for all values of the variables. By differentiating (17) we see that u(x, t) is a solution of (1), provided f(x) is twice differentiable on the interval 0 < x < L, and has one-sided second derivatives at x = 0 and x = L, which are zero. Under these conditions u(x, t) is established as a solution of (1), satisfying (2) and (3) with g(x) ≡ 0.

Fig. 289. Odd periodic extension of f(x)

Generalized Solution. If f′(x) and f″(x) are merely piecewise continuous (see Sec. 6.1), or if those one-sided derivatives are not zero, then for each t there will be finitely many values of x at which the second derivatives of u appearing in (1) do not exist. Except at these points the wave equation will still be satisfied. We may then regard u(x, t) as a “generalized solution,” as it is called, that is, as a solution in a broader sense. For instance, a triangular initial deflection as in Example 1 (below) leads to a generalized solution.

Physical Interpretation of the Solution (17). The graph of f* (x − ct) is obtained from the graph of f*(x) by shifting the latter ct units to the right (Fig. 290). This means that f*(x − ct)(c > 0) represents a wave that is traveling to the right as t increases. Similarly, f*(x + ct) represents a wave that is traveling to the left, and u(x, t) is the superposition of these two waves.

Fig. 290. Interpretation of (17)

Fig. 291. Solution u(x, t) in Example 1 for various values of t (right part of the figure) obtained as the superposition of a wave traveling to the right (dashed) and a wave traveling to the left (left part of the figure)

12.4 D'Alembert's Solution of the Wave Equation. Characteristics

It is interesting that the solution (17), Sec. 12.3, of the wave equation

can be immediately obtained by transforming (1) in a suitable way, namely, by introducing the new independent variables

Then u becomes a function of v and w. The derivatives in (1) can now be expressed in terms of derivatives with respect to v and w by the use of the chain rule in Sec. 9.6. Denoting partial derivatives by subscripts, we see from (2) that v_x = 1 and w_x = 1. For simplicity let us denote u(x, t) as a function of v and w, by the same letter u. Then

We now apply the chain rule to the right side of this equation. We assume that all the partial derivatives involved are continuous, so that u_wv = u_vw. Since v_x = 1 and w_x = 1 we obtain

Transforming the other derivative in (1) by the same procedure, we find

By inserting these two results in (1) we get (see footnote 2 in App. A3.2)

The point of the present method is that (3) can be readily solved by two successive integrations, first with respect to w and then with respect to v. This gives

Here h(v) and ψ(w) are arbitrary functions of v and w, respectively. Since the integral is a function of v, say, φ(v), the solution is of the form u = φ(v) + ψ(w). In terms of x and t, by (2), we thus have

This is known as d'Alembert's solution¹ of the wave equation (1).

Its derivation was much more elegant than the method in Sec. 12.3, but d'Alembert's method is special, whereas the use of Fourier series applies to various equations, as we shall see.

D'Alembert's Solution Satisfying the Initial Conditions

These are the same as (3) in Sec. 12.3. By differentiating (4) we have

where primes denote derivatives with respect to the entire arguments x + ct and x − ct, respectively, and the minus sign comes from the chain rule. From (4)–(6) we have

Dividing (8) by c and integrating with respect to x, we obtain

If we add this to (7), then ψ drops out and division by 2 gives

Similarly, subtraction of (9) from (7) and division by 2 gives

In (10) we replace x by x + ct; we then get an integral from to x₀ to x + ct. In (11) we replace x by x − ct and get minus an integral from x₀ to x − ct or plus an integral from x − ct to x₀. Hence addition of φ(x + ct) and ψ(x − ct) gives u(x, t) [see (4)] in the form

If the initial velocity is zero, we see that this reduces to

in agreement with (17) in Sec. 12.3. You may show that because of the boundary conditions (2) in that section the function f must be odd and must have the period 2L.

Our result shows that the two initial conditions [the functions f(x) and g(x) in (5)] determine the solution uniquely.

The solution of the wave equation by the Laplace transform method will be shown in Sec. 12.11.

Characteristics. Types and Normal Forms of PDEs

The idea of d'Alembert's solution is just a special instance of the method of characteristics. This concerns PDEs of the form

(as well as PDEs in more than two variables). Equation (14) is called quasilinear because it is linear in the highest derivatives (but may be arbitrary otherwise). There are three types of PDEs (14), depending on the discriminant AC − B², as follows.

Note that (1) and (2) in Sec. 12.1 involve t, but to have y as in (14), we set y = ct in (1), obtaining u_tt − c²u_xx = c²(u_yy − u_xx) = 0. And in (2) we set y = c²t, so that u_t − c²u_xx = c²(u_y − u_xx).

A, B, C may be functions of x, y, so that a PDE may be of mixed type, that is, of different type in different regions of the xy-plane. An important mixed-type PDE is the Tricomi equation (see Prob. 10).

Transformation of (14) to Normal Form. The normal forms of (14) and the corresponding transformations depend on the type of the PDE. They are obtained by solving the characteristic equation of (14), which is the ODE

where y′ = dy/dx (note −2B, not + 2B). The solutions of (15) are called the characteristics of (14), and we write them in the form Φ(x, y) = const and Ψ(x, y) = const. Then the transformations giving new variables v, w instead of x, y and the normal forms of (14) are as follows.

Here, Φ = Φ(x, y), Ψ = Ψ(x, y), F₁ = F₁(u, w, u, u_v, u_w), etc., and we denote u as function of v, w again by u, for simplicity. We see that the normal form of a hyperbolic PDE is as in d'Alembert's solution. In the parabolic case we get just one family of solutions Φ = Ψ. In the elliptic case, , and the characteristics are complex and are of minor interest. For derivation, see Ref. [GenRef3] in App. 1.

12.5 Modeling: Heat Flow from a Body in Space. Heat Equation

After the wave equation (Sec. 12.2) we now derive and discuss the next “big” PDE, the heat equation, which governs the temperature u in a body in space. We obtain this model of temperature distribution under the following.

Physical Assumptions

1. The specific heat σ and the density ρ of the material of the body are constant. No heat is produced or disappears in the body.
2. Experiments show that, in a body, heat flows in the direction of decreasing temperature, and the rate of flow is proportional to the gradient (cf. Sec. 9.7) of the temperature; that is, the velocity v of the heat flow in the body is of the form

   

   where u(x, y, z, t) is the temperature at a point (x, y, z) and time t.

3. The thermal conductivity K is constant, as is the case for homogeneous material and nonextreme temperatures.

Under these assumptions we can model heat flow as follows.

Let T be a region in the body bounded by a surface S with outer unit normal vector n such that the divergence theorem (Sec. 10.7) applies. Then

is the component of v in the direction of n. Hence |v • n ΔA| is the amount of heat leaving T (if v • n > 0 at some point P) or entering T (if v • n < 0 at P) per unit time at some point P of S through a small portion ΔS of s of area ΔA. Hence the total amount of heat that flows across S from T is given by the surface integral

Note that, so far, this parallels the derivation on fluid flow in Example 1 of Sec. 10.8.

Using Gauss's theorem (Sec. 10.7), we now convert our surface integral into a volume integral over the region T. Because of (1) this gives [use (3) in Sec. 9.8]

Here,

is the Laplacian of u.

On the other hand, the total amount of heat in T is

with σ and ρ as before. Hence the time rate of decrease of H is

This must be equal to the amount of heat leaving T because no heat is produced or disappears in the body. From (2) we thus obtain

or (divide by −σρ)

Since this holds for any region T in the body, the integrand (if continuous) must be zero everywhere. That is,

This is the heat equation, the fundamental PDE modeling heat flow. It gives the temperature u(x, y, z, t) in a body of homogeneous material in space. The constant c² is the thermal diffusivity. K is the thermal conductivity, σ the specific heat, and ρ the density of the material of the body. is the Laplacian of u and, with respect to the Cartesian coordinates x, y, z, is

The heat equation is also called the diffusion equation because it also models chemical diffusion processes of one substance or gas into another.

12.6 Heat Equation: Solution by Fourier Series. Steady Two-Dimensional Heat Problems. Dirichlet Problem

We want to solve the (one-dimensional) heat equation just developed in Sec. 12.5 and give several applications. This is followed much later in this section by an extension of the heat equation to two dimensions.

Fig. 294. Bar under consideration

As an important application of the heat equation, let us first consider the temperature in a long thin metal bar or wire of constant cross section and homogeneous material, which is oriented along the x-axis (Fig. 294) and is perfectly insulated laterally, so that heat flows in the x-direction only. Then besides time, u depends only on x, so that the Laplacian reduces to u_xx = ∂²u = ∂²u/∂x², and the heat equation becomes the one-dimensional heat equation

This PDE seems to differ only very little from the wave equation, which has a term u_tt instead of u_t, but we shall see that this will make the solutions of (1) behave quite differently from those of the wave equation.

We shall solve (1) for some important types of boundary and initial conditions. We begin with the case in which the ends x = 0 and x = L of the bar are kept at temperature zero, so that we have the boundary conditions

Furthermore, the initial temperature in the bar at time t = 0 is given, say, f(x), so that we have the initial condition

Here we must have f(0) = 0 and f(L) = 0 because of (2).

We shall determine a solution u(x, t) of (1) satisfying (2) and (3)—one initial condition will be enough, as opposed to two initial conditions for the wave equation. Technically, our method will parallel that for the wave equation in Sec. 12.3: a separation of variables, followed by the use of Fourier series. You may find a step-by-step comparison worthwhile.

Step 1. Two ODEs from the heat equation (1). Substitution of a product u(x, t) = F(x)G(t) into (1) gives with and F″ = d²F/dx². To separate the variables, we divide c²FG by obtaining

The left side depends only on t and the right side only on x, so that both sides must equal a constant k (as in Sec. 12.3). You may show that for k = 0 or k > 0 the only solution u = FG satisfying (2) is u ≡ 0. For negative k = −p² we have from (4)

Multiplication by the denominators immediately gives the two ODEs

and

Step 2. Satisfying the boundary conditions (2). We first solve (5). A general solution is

From the boundary conditions (2) it follows that

Since G ≡ 0 would give u ≡ 0, we require F(0) = 0, F(L) = 0 and get F(0) = A = 0 by (7) and then F(L) = B sin pL = 0, with B ≠ 0 (to avoid F ≡ 0); thus,

Setting B = 1, we thus obtain the following solutions of (5) satisfying (2):

(As in Sec. 12.3, we need not consider negative integer values of n.)

All this was literally the same as in Sec. 12.3. From now on it differs since (6) differs from (6) in Sec. 12.3. We now solve (6). For p = nπ/L, as just obtained, (6) becomes

It has the general solution

where B_n is a constant. Hence the functions

are solutions of the heat equation (1), satisfying (2). These are the eigenfunctions of the problem, corresponding to the eigenvalues λ_n = cnπ/L.

Step 3. Solution of the entire problem. Fourier series. So far we have solutions (8) satisfying the boundary conditions (2). To obtain a solution that also satisfies the initial condition (3), we consider a series of these eigenfunctions,

From this and (3) we have

Hence for (9) to satisfy (3), the B_n's must be the coefficients of the Fourier sine series, as given by (4) in Sec. 11.3; thus

The solution of our problem can be established, assuming that f(x) is piecewise continuous (see Sec. 6.1) on the interval 0 x L and has one-sided derivatives (see Sec. 11.1) at all interior points of that interval; that is, under these assumptions the series (9) with coefficients (10) is the solution of our physical problem. A proof requires knowledge of uniform convergence and will be given at a later occasion (Probs. 19, 20 in Problem Set 15.5).

Because of the exponential factor, all the terms in (9) approach zero as t approaches infinity. The rate of decay increases with n.

The solution (9) is

Also, 100e^−0.001785t = 50 when t = (ln 0.5)/(−0.001785) = 388[sec] ≈ 6.5 [min]. Does your guess, or at least its order of magnitude, agree with this result?

Fig. 295. Example 3. Decrease of temperature with time t for and L = π and c = 1

Steady Two-Dimensional Heat Problems. Laplace's Equation

We shall now extend our discussion from one to two space dimensions and consider the two-dimensional heat equation

for steady (that is, time-independent) problems. Then ∂u/∂t = 0 and the heat equation reduces to Laplace's equation

(which has already occurred in Sec. 10.8 and will be considered further in Secs. 12.8–12.11). A heat problem then consists of this PDE to be considered in some region R of the xy-plane and a given boundary condition on the boundary curve C of R. This is a boundary value problem (BVP). One calls it:

Fig. 296. Rectangle R and given boundary values

Dirichlet Problem in a Rectangle R (Fig. 296). We consider a Dirichlet problem for Laplace's equation (14) in a rectangle R, assuming that the temperature u(x, y) equals a given function f(x) on the upper side and 0 on the other three sides of the rectangle.

We solve this problem by separating variables. Substituting into u(x, y) = F(x)G(y) into (14) written as u_xx = −u_yy, dividing by FG, and equating both sides to a negative constant, we obtain

From this we get

and the left and right boundary conditions imply

This gives k = (nπ/a)² and corresponding nonzero solutions

The ODE for G with k = (nπ/a)² then becomes

Solutions are

Now the boundary condition u = 0 on the lower side of R implies that G_n(0) = 0; that is, G_n(0) = A_n + B_n = 0 or B_n = −A_n. This gives

From this and (15), writing , we obtain as the eigenfunctions of our problem

These solutions satisfy the boundary condition u = 0 on the left, right, and lower sides.

To get a solution also satisfying the boundary condition u(x, b) = f(x) on the upper side, we consider the infinite series

From this and (16) with y = b we obtain

We can write this in the form

This shows that the expressions in the parentheses must be the Fourier coefficients b_n of f(x); that is, by (4) in Sec. 11.3,

From this and (16) we see that the solution of our problem is

where

We have obtained this solution formally, neither considering convergence nor showing that the series for u, u_xx and u_yy have the right sums. This can be proved if one assumes that f and f′ are continuous and f″ is piecewise continuous on the interval 0 x a. The proof is somewhat involved and relies on uniform convergence. It can be found in [C4] listed in App. 1.

Unifying Power of Methods. Electrostatics, Elasticity

The Laplace equation (14) also governs the electrostatic potential of electrical charges in any region that is free of these charges. Thus our steady-state heat problem can also be interpreted as an electrostatic potential problem. Then (17), (18) is the potential in the rectangle R when the upper side of R is at potential f(x) and the other three sides are grounded.

Actually, in the steady-state case, the two-dimensional wave equation (to be considered in Secs. 12.8, 12.9) also reduces to (14). Then (17), (18) is the displacement of a rectangular elastic membrane (rubber sheet, drumhead) that is fixed along its boundary, with three sides lying in the xy-plane and the fourth side given the displacement f(x).

This is another impressive demonstration of the unifying power of mathematics. It illustrates that entirely different physical systems may have the same mathematical model and can thus be treated by the same mathematical methods.

12.7 Heat Equation: Modeling Very Long Bars. Solution by Fourier Integrals and Transforms

Our discussion of the heat equation

in the last section extends to bars of infinite length, which are good models of very long bars or wires (such as a wire of length, say, 300 ft). Then the role of Fourier series in the solution process will be taken by Fourier integrals (Sec. 11.7).

Let us illustrate the method by solving (1) for a bar that extends to infinity on both sides (and is laterally insulated as before). Then we do not have boundary conditions, but only the initial condition

where f(x) is the given initial temperature of the bar.

To solve this problem, we start as in the last section, substituting u(x, t) = F(x)G(t) into (1). This gives the two ODEs

and

Solutions are

respectively, where A and B are any constants. Hence a solution of (1) is

Here we had to choose the separation constant k negative, k = −p², because positive values of k would lead to an increasing exponential function in (5), which has no physical meaning.

Use of Fourier Integrals

Any series of functions (5), found in the usual manner by taking p as multiples of a fixed number, would lead to a function that is periodic in x when t = 0. However, since f(x) in (2) is not assumed to be periodic, it is natural to use Fourier integrals instead of Fourier series. Also, A and B in (5) are arbitrary and we may regard them as functions of p, writing A = A(p) and B = B(P). Now, since the heat equation (1) is linear and homogeneous, the function

is then a solution of (1), provided this integral exists and can be differentiated twice with respect to x and once with respect to t.

Determination of A(p) and B(p) from the Initial Condition. From (6) and (2) we get

This gives A(p) and B(p) in terms of f(x); indeed, from (4) in Sec. 11.7 we have

According to (1*), Sec. 11.9, our Fourier integral (7) with these A(p) and B(p) can be written

Similarly, (6) in this section becomes

Assuming that we may reverse the order of integration, we obtain

Then we can evaluate the inner integral by using the formula

[A derivation of (10) is given in Problem Set 16.4 (Team Project 24).] This takes the form of our inner integral if we choose as a new variable of integration and set

Then 2bs = (x − v)p and , so that (10) becomes

By inserting this result into (9) we obtain the representation

Taking as a variable of integration, we get the alternative form

If f(x) is bounded for all values of x and integrable in every finite interval, it can be shown (see Ref. [C10]) that the function (11) or (12) satisfies (1) and (2). Hence this function is the required solution in the present case.

Fig. 299. Solution u(x, t) in Example 1 for U₀ = 100°C, c² = 1 cm²/sec, and several values of t

Use of Fourier Transforms

The Fourier transform is closely related to the Fourier integral, from which we obtained the transform in Sec. 11.9. And the transition to the Fourier cosine and sine transform in Sec. 11.8 was even simpler. (You may perhaps wish to review this before going on.) Hence it should not surprise you that we can use these transforms for solving our present or similar problems. The Fourier transform applies to problems concerning the entire axis, and the Fourier cosine and sine transforms to problems involving the positive half-axis. Let us explain these transform methods by typical applications that fit our present discussion.

Fig. 300. Solution (20) in Example 4

12.8 Modeling: Membrane, Two-Dimensional Wave Equation

Since the modeling here will be similar to that of Sec. 12.2, you may want to take another look at Sec. 12.2.

The vibrating string in Sec. 12.2 is a basic one-dimensional vibrational problem. Equally important is its two-dimensional analog, namely, the motion of an elastic membrane, such as a drumhead, that is stretched and then fixed along its edge. Indeed, setting up the model will proceed almost as in Sec. 12.2.

Physical Assumptions

1. The mass of the membrane per unit area is constant (“homogeneous membrane”). The membrane is perfectly flexible and offers no resistance to bending.
2. The membrane is stretched and then fixed along its entire boundary in the xy-plane. The tension per unit length T caused by stretching the membrane is the same at all points and in all directions and does not change during the motion.
3. The deflection u(x, y, t) of the membrane during the motion is small compared to the size of the membrane, and all angles of inclination are small.

Although these assumptions cannot be realized exactly, they hold relatively accurately for small transverse vibrations of a thin elastic membrane, so that we shall obtain a good model, for instance, of a drumhead.

Derivation of the PDE of the Model (“Two-Dimensional Wave Equation”) from Forces. As in Sec. 12.2 the model will consist of a PDE and additional conditions. The PDE will be obtained by the same method as in Sec. 12.2, namely, by considering the forces acting on a small portion of the physical system, the membrane in Fig. 301 on the next page, as it is moving up and down.

Since the deflections of the membrane and the angles of inclination are small, the sides of the portion are approximately equal to Δx and Δy. The tension T is the force per unit length. Hence the forces acting on the sides of the portion are approximately TΔx and TΔy. Since the membrane is perfectly flexible, these forces are tangent to the moving membrane at every instant.

Horizontal Components of the Forces. We first consider the horizontal components of the forces. These components are obtained by multiplying the forces by the cosines of the angles of inclination. Since these angles are small, their cosines are close to 1. Hence the horizontal components of the forces at opposite sides are approximately equal. Therefore, the motion of the particles of the membrane in a horizontal direction will be negligibly small. From this we conclude that we may regard the motion of the membrane as transversal; that is, each particle moves vertically.

Vertical Components of the Forces. These components along the right side and the left side are (Fig. 301), respectively,

Here α and β are the values of the angle of inclination (which varies slightly along the edges) in the middle of the edges, and the minus sign appears because the force on the left side is directed downward. Since the angles are small, we may replace their sines by their tangents. Hence the resultant of those two vertical components is

Fig. 301. Vibrating membrane

where subscripts x denote partial derivatives and y₁ and y₂ are values between y and y + Δy. Similarly, the resultant of the vertical components of the forces acting on the other two sides of the portion is

where x₁ and x₂ are values between x and x + Δx.

Newton's Second Law Gives the PDE of the Model. By Newton's second law (see Sec. 2.4) the sum of the forces given by (1) and (2) is equal to the mass ρΔA of that small portion times the acceleration ∂²u/∂²; here ρ is the mass of the undeflected membrane per unit area, and ΔA = Δx Δy is the area of that portion when it is undeflected. Thus

where the derivative on the left is evaluated at some suitable point corresponding to that portion. Division by ρΔxΔy gives

If we let Δ0x and Δy approach zero, we obtain the PDE of the model

This PDE is called the two-dimensional wave equation. The expression in parentheses is the Laplacian Δ²u of u (Sec. 10.8). Hence (3) can be written

Solutions of the wave equation (3) will be obtained and discussed in the next section.

12.9 Rectangular Membrane. Double Fourier Series

Now we develop a solution for the PDE obtained in Sec. 12.8. Details are as follows.

The model of the vibrating membrane for obtaining the displacement u(x, y, t) of a point (x, y) of the membrane from rest (u = 0) at time t is

Here (1) is the two-dimensional wave equation with c² = T/ρ just derived, (2) is the boundary condition (membrane fixed along the boundary in the xy-plane for all times t 0), and (3) are the initial conditions at t = 0, consisting of the given initial displacement (initial shape) f(x, y) and the given initial velocity g(x, y), where u_t = ∂u/∂t. We see that these conditions are quite similar to those for the string in Sec. 12.2.

Fig. 302. Rectangular membrane

Let us consider the rectangular membrane R in Fig. 302. This is our first important model. It is much simpler than the circular drumhead, which will follow later. First we note that the boundary in equation (2) is the rectangle in Fig. 302. We shall solve this problem in three steps:

Step 1. By separating variables, first setting u(x, y, t) = F(x, y)G(t) and later F(x, y) = H(x)Q(y) we obtain from (1) an ODE (4) for G and later from a PDE (5) for F two ODEs (6) and (7) for H and Q.

Step 2. From the solutions of those ODEs we determine solutions (13) of (1) (“eigenfunctions” u_mn) that satisfy the boundary condition (2).

Step 3. We compose the u_mn into a double series (14) solving the whole model (1), (2), (3).

Step 1. Three ODEs From the Wave Equation (1)

To obtain ODEs from (1), we apply two successive separations of variables. In the first separation we set u(x, y, t) = F(x, y)G(t). Substitution into (1) gives

where subscripts denote partial derivatives and dots denote derivatives with respect to t. To separate the variables, we divide both sides by c²FG:

Since the left side depends only on t, whereas the right side is independent of t, both sides must equal a constant. By a simple investigation we see that only negative values of that constant will lead to solutions that satisfy (2) without being identically zero; this is similar to Sec. 12.3. Denoting that negative constant by −v², we have

This gives two equations: for the “time function” G(t) we have the ODE

and for the “amplitude function” F(x, y) a PDE, called the two-dimensional Helmholtz³ equation

Separation of the Helmholtz equation is achieved if we set F(x, y) = H(x)Q(y). By substitution of this into (5) we obtain

To separate the variables, we divide both sides by HQ, finding

Both sides must equal a constant, by the usual argument. This constant must be negative, say, −k², because only negative values will lead to solutions that satisfy (2) without being identically zero. Thus

This yields two ODEs for H and Q, namely,

and

Step 2. Satisfying the Boundary Condition

General solutions of (6) and (7) are

with constant A, B, C, D. From u = FG and (2) it follows that F = HQ must be zero on the boundary, that is, on the edges x = 0, x = a, y = 0, y = b; see Fig. 302. This gives the conditions

Hence H(0) = A = 0 and then H(a) = B sin ka = 0. Here we must take B ≠ 0 since otherwise H(x) ≡ 0 and F(x, y) ≡ 0. Hence sin ka = 0 or ka = mπ, that is,

In precisely the same fashion we conclude that c = 0 and p must be restricted to the values p = nπ/b where n is an integer. We thus obtain the solutions H = H_m, Q = Q_n, where

As in the case of the vibrating string, it is not necessary to consider m, n = −1, −2, … since the corresponding solutions are essentially the same as for positive m and n, expect for a factor −1. Hence the functions

are solutions of the Helmholtz equation (5) that are zero on the boundary of our membrane.

Eigenfunctions and Eigenvalues. Having taken care of (5), we turn to (4). Since p² = v² − k² in (7) and λ = cv in (4), we have

Hence to k = mπ/a and p = nπ/b there corresponds the value

in the ODE (4). A corresponding general solution of (4) is

It follows that the functions u_mn(x, y, t) = F_mn(x, y)G_mn(t), written out

with λ_mn according to (9), are solutions of the wave equation (1) that are zero on the boundary of the rectangular membrane in Fig. 302. These functions are called the eigenfunctions or characteristic functions, and the numbers λ_mn are called the eigenvalues or characteristic values of the vibrating membrane. The frequency of u_mn is λ_mn/2π.

Discussion of Eigenfunctions. It is very interesting that, depending on a and b, several functions F_mn may correspond to the same eigenvalue. Physically this means that there may exists vibrations having the same frequency but entirely different nodal lines (curves of points on the membrane that do not move). Let us illustrate this with the following example.

Fig. 303. Nodal lines of the solutions u₁₁, u₁₂, u₂₁, u₂₂, u₁₃, u₃₁ in the case of the square membrane

Fig. 304. Nodal lines of the solution (12) for some values of B₂₁

Step 3. Solution of the Model (1), (2), (3). Double Fourier Series

So far we have solutions (10) satisfying (1) and (2) only. To obtain the solutions that also satisfies (3), we proceed as in Sec. 12.3. We consider the double series

(without discussing convergence and uniqueness). From (14) and (3a), setting we have

Suppose that f(x, y) can be represented by (15). (Sufficient for this is the continuity of f, ∂f/∂x, ∂f/∂y, ∂²f/∂x ∂y in R.) Then (15) is called the double Fourier series of f(x, y). Its coefficients can be determined as follows. Setting

we can write (15) in the form

For fixed y this is the Fourier sine series of f(x, y) considered as a function of x. From (4) in Sec. 11.3 we see that the coefficients of this expansion are

Furthermore, (16) is the Fourier sine series of K_m(y), and from (4) in Sec. 11.3 it follows that the coefficients are

From this and (17) we obtain the generalized Euler formula

for the Fourier coefficients of f(x, y) in the double Fourier series (15).

The B_mn in (14) are now determined in terms of f(x, y). To determine the , we differentiate (14) termwise with respect to t; using (3b), we obtain

Suppose that g(x, y) can be developed in this double Fourier series. Then, proceeding as before, we find that the coefficients are

Result. If f and g in (3) are such that u can be represented by (14), then (14) with coefficients (18) and (19) is the solution of the model (1), (2), (3).

12.10 Laplacian in Polar Coordinates. Circular Membrane. Fourier–Bessel Series

It is a general principle in boundary value problems for PDEs to choose coordinates that make the formula for the boundary as simple as possible. Here polar coordinates are used for this purpose as follows. Since we want to discuss circular membranes (drumheads), we first transform the Laplacian in the wave equation (1), Sec. 12.9,

(subscripts denoting partial derivatives) into polar coordinates r, θ defined by x = r cos θ, y = r sin θ;, thus

By the chain rule (Sec. 9.6) we obtain

Differentiating once more with respect to x and using the product rule and then again the chain rule gives

Also, by differentiation of r and θ we find

Differentiating these two formulas again, we obtain

We substitute all these expressions into (2). Assuming continuity of the first and second partial derivatives, we have u_rθ = u_θr, and by simplifying,

In a similar fashion it follows that

By adding (3) and (4) we see that the Laplacian of u in polar coordinates is

Circular Membrane

Circular membranes are important parts of drums, pumps, microphones, telephones, and other devices. This accounts for their great importance in engineering. Whenever a circular membrane is plane and its material is elastic, but offers no resistance to bending (this excludes thin metallic membranes!), its vibrations are modeled by the two-dimensional wave equation in polar coordinates obtained from (1) with given by (5), that is,

We shall consider a membrane of radius R (Fig. 307) and determine solutions u(r, t) that are radially symmetric. (Solutions also depending on the angle θ will be discussed in the problem set.) Then u_θθ = 0 in (6) and the model of the problem (the analog of (1), (2), (3) in Sec. 12.9) is

Fig. 307. Circular membrane

Here (8) means that the membrane is fixed along the boundary circle r = R. The initial deflection f(r) and the initial velocity g(r) depend only on r, not on θ, so that we can expect radially symmetric solutions u(r, t).

Step 1. Two ODEs From the Wave Equation (7). Bessel's Equation

Using the method of separation of variables, we first determine solutions u(r, t) = W(r)G(t). (We write W, not F because W depends on r, whereas F, used before, depended on x.) Substituting u = WG and its derivatives into (7) and dividing the result by c²WG we get

where dots denote derivatives with respect to t and primes denote derivatives with respect to r. The expressions on both sides must equal a constant. This constant must be negative, say, −k², in order to obtain solutions that satisfy the boundary condition without being identically zero. Thus,

This gives the two linear ODEs

and

We can reduce (11) to Bessel's equation (Sec. 5.4) if we set s = kr. Then 1/r = k/s and, retaining the notation W for simplicity, we obtain by the chain rule

By substituting this into (11) and omitting the common factor k² we have

This is Bessel's equation (1), Sec. 5.4, with parameter v = 0.

Step 2. Satisfying the Boundary Condition (8)

Solutions of (12) are the Bessel functions J₀ and Y₀ of the first and second kind (see Secs. 5.4, 5.5). But Y₀ becomes infinite at 0, so that we cannot use it because the deflection of the membrane must always remain finite. This leaves us with

On the boundary r = R we get W(R) = J₀(kR) = 0 from (8) (because G ≡ 0 would imply u ≡ 0). We can satisfy this condition because J₀ has (infinitely many) positive zeros, s = α₁, α₂, … (see Fig. 308), with numerical values

and so on. (For further values, consult your CAS or Ref. [GenRef1] in App. 1.) These zeros are slightly irregularly spaced, as we see. Equation (13) now implies

Hence the functions

are solutions of (11) that are zero on the boundary circle r = R.

Eigenfunctions and Eigenvalues. For W_m in (15), a corresponding general solution of (10) with λ = λ_m = ck_m = cα_m/R is

Hence the functions

with m = 1, 2, … are solutions of the wave equation (7) satisfying the boundary condition (8). These are the eigenfunctions of our problem. The corresponding eigenvalues are λ_m.

The vibration of the membrane corresponding to u_m is called the mth normal mode; it has the frequency λ_m/2π cycles per unit time. Since the zeros of the Bessel function are not regularly spaced on the axis (in contrast to the zeros of the sine functions appearing in the case of the vibrating string), the sound of a drum is entirely different from that of a violin. The forms of the normal modes can easily be obtained from Fig. 308 and are shown in Fig. 309. For m = 1, all the points of the membrane move up (or down) at the same time. For m = 2, the situation is as follows. The function W₂(r) = J₀(α₂r/R) is zero for α₂r/R = α₁, thus r = α₁R/α₂. The circle r = α₁R/α₂ is, therefore, nodal line, and when at some instant the central part of the membrane moves up, the outer part (r > α₁R/α₂) moves down, and conversely. The solution u_m(r, t) has m − 1 nodal lines, which are circles (Fig. 309).

Fig. 308. Bessel function J₀(s)

Fig. 309. Normal modes of the circular membrane in the case of vibrations independent of the angle

Step 3. Solution of the Entire Problem

To obtain a solution u(r, t) that also satisfies the initial conditions (9), we may proceed as in the case of the string. That is, we consider the series

(leaving aside the problems of convergence and uniqueness). Setting t = 0 and using (9a), we obtain

Thus for the series (17) to satisfy the condition (9a), the constants A_m must be the coefficients of the Fourier–Bessel series (18) that represents f(r) in terms of J₀(α_mr/R) that is [see (9) in Sec. 11.6 with n = 0, α_{0, m} = α_m and x = r],

Differentiability of f(r) in the interval 0 r R is sufficient for the existence of the development (18); see Ref. [A13]. The coefficients B_m in (17) can be determined from (9b) in a similar fashion. Numeric values of A_m and B_m may be obtained from a CAS or by a numeric integration method, using tables of J₀ and J₁. However, numeric integration can sometimes be avoided, as the following example shows.

12.11 Laplace's Equation in Cylindrical and Spherical Coordinates. Potential

One of the most important PDEs in physics and engineering applications is Laplace's equation, given by

Here, x, y, z are Cartesian coordinates in space (Fig. 167 in Sec. 9.1), etc. u_xx = ∂²u = ∂x², etc. The expression is called the Laplacian of u. The theory of the solutions of (1) is called potential theory. Solutions of (1) that have continuous second partial derivatives are known as harmonic functions.

Laplace's equation occurs mainly in gravitation, electrostatics (see Theorem 3, Sec. 9.7), steady-state heat flow (Sec. 12.5), and fluid flow (to be discussed in Sec. 18.4).

Recall from Sec. 9.7 that the gravitational potential u(x, y, z) at a point (x, y, z) resulting from a single mass located at a point (X, Y, Z) is

and u satisfies (1). Similarly, if mass is distributed in a region T in space with density ρ(X, Y, Z), its potential at a point (x, y, z) not occupied by mass is

It satisfies (1) because (Sec. 9.7) and ρ is not a function of x, y, z.

Practical problems involving Laplace's equation are boundary value problems in a region T in space with boundary surface S. Such problems can be grouped into three types (see also Sec. 12.6 for the two-dimensional case):

1. First boundary value problem or Dirichlet problem if u is prescribed on S.
2. Second boundary value problem or Neumann problem if the normal derivative u_n = ∂u/∂n is prescribed on S.
3. Third or mixed boundary value problem or Robin problem if u is prescribed on a portion of S and u_n on the remaining portion of S.

In general, when we want to solve a boundary value problem, we have to first select the appropriate coordinates in which the boundary surface S has a simple representation. Here are some examples followed by some applications.

Laplacian in Cylindrical Coordinates

The first step in solving a boundary value problem is generally the introduction of coordinates in which the boundary surface S has a simple representation. Cylindrical symmetry (a cylinder as a region T) calls for cylindrical coordinates r, θ, z related to x, y, z by

Fig. 311. Cylindrical coordinates (r 0, 0 θ 2π)

Fig. 312. Spherical coordinates (r 0, 0 θ 2π, 0 φ π)

For these we get immediately by adding u_zz to (5) in Sec. 12.10; thus,

Laplacian in Spherical Coordinates

Spherical symmetry (a ball as region T bounded by a sphere S) requires spherical coordinates r, θ, φ related to x, y, z by

Using the chain rule (as in Sec. 12.10), we obtain in spherical coordinates

We leave the details as an exercise. It is sometimes practical to write (7) in the form

Remark on Notation. Equation (6) is used in calculus and extends the familiar notation for polar coordinates. Unfortunately, some books use θ and φ interchanged, an extension of the notation x = r cos φ, y = r sin φ for polar coordinates (used in some European countries).

Boundary Value Problem in Spherical Coordinates

We shall solve the following Dirichlet problem in spherical coordinates:

The PDE (8) follows from (7) or (7′) by assuming that the solution u will not depend on because the Dirichlet condition (9) is independent of θ. This may be an electrostatic potential (or a temperature) f(φ) at which the sphere S: r = R is kept. Condition (10) means that the potential at infinity will be zero.

Separating Variables by substituting u(r, φ) = G(r)H(φ) into (8). Multiplying (8) by r², making the substitution and then dividing by GH, we obtain

By the usual argument both sides must be equal to a constant k. Thus we get the two ODEs

and

The solutions of (11) will take a simple form if we set k = n(n + 1). Then, writing G′ = dG/dr, etc., we obtain

This is an Euler–Cauchy equation. From Sec. 2.5 we know that it has solutions G = r^a. Substituting this and dropping the common factor r^a gives

Hence solutions are

We now solve (12). Setting cos φ = w we have sin²φ = 1 − w² and

Consequently, (12) with k = n(n + 1) takes the form

This is Legendre's equation (see Sec. 5.3), written out

For integer n = 0, 1, … the Legendre polynomials

are solutions of Legendre's equation (15). We thus obtain the following two sequences of solution u = GH of Laplace's equation (8), with constant A_n and B_n, where n = 0, 1, …,

Use of Fourier–Legendre Series

Interior Problem: Potential Within the Sphere S. We consider a series of terms from (16a),

Since S is given by r = R for (17) to satisfy the Dirichlet condition (9) on the sphere S, we must have

that is, (18) must be the Fourier–Legendre series of f(φ). From (7) in Sec. 5.8 we get the coefficients

where denotes f(φ) as a function of w = cosφ. Since dw = −sin φ dφ, and the limits of integration and −1 correspond to φ = π and φ = 0, respectively, we also obtain

If f(φ) and f′(φ) are piecewise continuous on the interval 0 φ π, then the series (17) with coefficients (19) solves our problem for points inside the sphere because it can be shown that under these continuity assumptions the series (17) with coefficients (19) gives the derivatives occurring in (8) by termwise differentiation, thus justifying our derivation.

Exterior Problem: Potential Outside the Sphere S. Outside the sphere we cannot use the functions u_n in (16a) because they do not satisfy (10). But we can use the in (16b), which do satisfy (10) (but could not be used inside S; why?). Proceeding as before leads to the solution of the exterior problem

satisfying (8), (9), (10), with coefficients

The next example illustrates all this for a sphere of radius 1 consisting of two hemispheres that are separated by a small strip of insulating material along the equator, so that these hemispheres can be kept at different potentials (110 V and 0 V).

Fig. 314. Partial sums of the first 4, 6, and 11 nonzero terms of (23) for r = R = 1

12.12 Solution of PDEs by Laplace Transforms

Readers familiar with Chap. 6 may wonder whether Laplace transforms can also be used for solving partial differential equations. The answer is yes, particularly if one of the independent variables ranges over the positive axis. The steps to obtain a solution are similar to those in Chap. 6. For a PDE in two variables they are as follows.

1. Take the Laplace transform with respect to one of the two variables, usually t. This gives an ODE for the transform of the unknown function. This is so since the derivatives of this function with respect to the other variable slip into the transformed equation. The latter also incorporates the given boundary and initial conditions.
2. Solving that ODE, obtain the transform of the unknown function.
3. Taking the inverse transform, obtain the solution of the given problem.

If the coefficients of the given equation do not depend on t, the use of Laplace transforms will simplify the problem.

We explain the method in terms of a typical example.

Fig. 317. Traveling wave in Example 1

We have reached the end of Chapter 12, in which we concentrated on the most important partial differential equations (PDEs) in physics and engineering. We have also reached the end of Part C on Fourier Analysis and PDEs.

Outlook

We have seen that PDEs underlie the modeling process of various important engineering application. Indeed, PDEs are the subject of many ongoing research projects.

Numerics for PDEs follows in Secs. 21.4–21.7, which, by design for greater flexibility in teaching, are independent of the other sections in Part E on numerics.

In the next part, that is, Part D on complex analysis, we turn to an area of a different nature that is also highly important to the engineer. The rich vein of examples and problems will signify this. It is of note that Part D includes another approach to the two-dimensional Laplace equation with applications, as shown in Chap. 18.

CHAPTER 12 REVIEW QUESTIONS AND PROBLEMS

1. For what kinds of problems will modeling lead to an ODE? To a PDE?
2. Mention some of the basic physical principles or laws that will give a PDE in modeling.
3. State three or four of the most important PDEs and their main applications.
4. What is “separating variables” in a PDE? When did we apply it twice in succession?
5. What is d'Alembert's solution method? To what PDE does it apply?
6. What role did Fourier series play in this chapter? Fourier integrals?
7. When and why did Legendre's equation occur? Bessel's equation?
8. What are the eigenfunctions and their frequencies of the vibrating string? Of the vibrating membrane?
9. What do you remember about types of PDEs? Normal forms? Why is this important?
10. When did we use polar coordinates? Cylindrical coordinates? Spherical coordinates?
11. Explain mathematically (not physically) why we got exponential functions in separating the heat equation, but not for the wave equation.
12. Why and where did the error function occur?
13. How do problems for the wave equation and the heat equation differ regarding additional conditions?
14. Name and explain the three kinds of boundary conditions for Laplace's equation.
15. Explain how the Laplace transform applies to PDEs.

16.18 Solve for u = u(x, y):

• 16. u_xx + 25u = 0
• 17. u_yy + u_y − 6u = 18
• 18. u_xx + u_x = 0, u(0, y) = f(y), u_x(0, y) = g(y)

19.21 NORMAL FORM

Transform to normal form and solve:

• 19. u_xy = u_yy
• 20. u_xx + 6u_xy + 9u_yy = 0
• 21. u_xx − 4u_yy = 0

22.24 VIBRATING STRING

Find and sketch or graph (as in Fig. 288 in Sec. 12.3) the deflection u(x, t) of a vibrating string of length π, extending from to x = 0 to x = π, and c² = T/ρ = 4 starting with velocity zero and deflection:

• 22. sin 4x
• 23. sin³x
• 24.

25.27 HEAT

Find the temperature distribution in a laterally insulated thin copper bar c² = K/(σρ = 1.158 cm²/sec) of length 100 cm and constant cross section with endpoints at x = 0 and 100 kept at 0°C and initial temperature:

• 25. sin 0.01 πx
• 26. 50 − |50 − x|
• 27. sin³ 0.1πx

28.30 ADIABATIC CONDITIONS

Find the temperature distribution in a laterally insulated bar of length π with c² = 1 for the adiabatic boundary condition (see Problem Set 12.6) and initial temperature:

• 28. 3x²
• 29. 100 cos 2x
• 30.

31.32 TEMPERATURE IN A PLATE

• 31. Let f(x, y) = u(x, y, 0) be the initial temperature in a thin square plate of side π with edges kept 0°C at and faces perfectly insulated. Separating variables, obtain from the solution

  

  where

  

• 32. Find the temperature in Prob. 31 if

  

33.37 MEMBRANES

Show that the following membranes of area 1 with c² = 1 have the frequencies of the fundamental mode as given (4-decimal values). Compare.

• 33. Circle:
• 34. Square:
• 35. Rectangle with sides
• 36. Semicircle:
• 37. Quadrant of circle: of circle: (α₂₁ = 5.13562 = first positive zero of J₂)

38.40 ELECTROSTATIC POTENTIAL

Find the potential in the following charge-free regions.

• 38. Between two concentric spheres of radii r₀ and r₁ kept at potentials u₀ and u₁, respectively.
• 39. Between two coaxial circular cylinders of radii r₀ and r₁ kept at the potentials and u₀ and u₁, respectively. Compare with Prob. 38.
• 40. In the interior of a sphere of radius 1 kept at the potential f(φ) = cos 3φ + 3 cos φ (referred to our usual spherical coordinates).

SUMMARY OF CHAPTER 12 Partial Differential Equations (PDEs)