5.2 Solutions about Singular Points

INTRODUCTION

The two differential equations y″ − xy = 0 and xy″ + y = 0 are similar only in that they are both examples of simple linear second-order DEs with variable coefficients. That is all they have in common. Since x = 0 is an ordinary point of the first equation, we saw in the preceding section that there was no problem in finding two distinct power series solutions centered at that point. In contrast, because x = 0 is a singular point of the second DE, finding two infinite series solutions—notice we did not say “power series solutions”—of the equation about that point becomes a more difficult task.

A Definition

A singular point x = x0 of a linear differential equation

a2(x)y″ + a1(x)y′ + a0(x)y = 0 (1)

is further classified as either regular or irregular. The classification again depends on the functions P and Q in the standard form

y″ + P(x)y′ + Q(x)y = 0. (2)

DEFINITION 5.2.1 Regular/Irregular Singular Points

A singular point x0 is said to be a regular singular point of the differential equation (1) if the functions p(x) = (x − x0)P(x) and q(x) = (x − x0)2Q(x) are both analytic at x0. A singular point that is not regular is said to be an irregular singular point of the equation.

The second sentence in Definition 5.2.1 indicates that if one or both of the functions p(x) = (x − x0)P(x) and q(x) = (x − x0)2Q(x) fails to be analytic at x0, then x0 is an irregular singular point.

Polynomial Coefficients

As in Section 5.1, we are mainly interested in linear equations (1) where the coefficients a2(x), a1(x), and a0(x) are polynomials with no common factors. We have already seen that if a2(x0) = 0, then x = x0 is a singular point of (1) since at least one of the rational functions P(x) = a1(x)/a2(x) and Q(x) = a0(x)/a2(x) in the standard form (2) fails to be analytic at that point. But since a2(x) is a polynomial and x0 is one of its zeros, it follows from the Factor Theorem of algebra that x − x0 is a factor of a2(x). This means that after a1(x)/a2(x) and a0(x)/a2(x) are reduced to lowest terms, the factor x − x0 must remain, to some positive integer power, in one or both denominators. Now suppose that x = x0 is a singular point of (1) but that both the functions defined by the products p(x) = (x − x0)P(x) and q(x) = (x − x0)2Q(x) are analytic at x0. We are led to the conclusion that multiplying P(x) by x − x0 and Q(x) by (x − x0)2 has the effect (through cancellation) that x − x0 no longer appears in either denominator. We can now determine whether x0 is regular by a quick visual check of denominators:

If x − x0 appears at most to the first power in the denominator of P(x) and at most to the second power in the denominator of Q(x), then x = x0 is a regular singular point.

Moreover, observe that if x = x0 is a regular singular point and we multiply (2) by (x − x0)2, then the original DE can be put into the form

(x − x0)2 y″ + (x − x0)p(x)y′ + q(x)y = 0, (3)

where p and q are analytic at x = x0.

EXAMPLE 1 Classification of Singular Points

It should be clear that x = 2 and x = −2 are singular points of

(x2 − 4)2y″ + 3(x − 2)y′ + 5y = 0.

After dividing the equation by (x2 − 4)2 = (x − 2)2(x + 2)2 and reducing the coefficients to lowest terms, we find that

.

We now test P(x) and Q(x) at each singular point.

In order for x = 2 to be a regular singular point, the factor x − 2 can appear at most to the first power in the denominator of P(x) and at most to the second power in the denominator of Q(x). A check of the denominators of P(x) and Q(x) shows that both these conditions are satisfied, so x = 2 is a regular singular point. Alternatively, we are led to the same conclusion by noting that both rational functions

p(x) = (x − 2)P(x) =      and     q(x) = (x − 2)2Q(x) =

are analytic at x = 2.

Now since the factor x − (−2) = x + 2 appears to the second power in the denominator of P(x), we can conclude immediately that x = −2 is an irregular singular point of the equation. This also follows from the fact that p(x) = (x + 2)P(x) = 3/[(x − 2)(x + 2)] is not analytic at x = −2. ≡

In Example 1, notice that since x = 2 is a regular singular point, the original equation can be written as

As another example, we can see that x = 0 is an irregular singular point of x3y″ − 2xy′ + 8y = 0 by inspection of the denominators of P(x) = −2/x2 and Q(x) = 8/x3. On the other hand, x = 0 is a regular singular point of xy″ − 2xy′ + 8y = 0 since x − 0 and (x − 0)2 do not even appear in the respective denominators of P(x) = −2 and Q(x) = 8/x. For a singular point x = x0, any nonnegative power of x − x0 less than one (namely, zero) and any nonnegative power less than two (namely, zero and one) in the denominators of P(x) and Q(x), respectively, imply that x0 is a regular singular point. A singular point can be a complex number. You should verify that x = 3i and x = −3i are two regular singular points of (x2 + 9)y″ − 3xy′ + (1 − x)y = 0.

Any second-order Cauchy–Euler equation ax2y″ + bxy′ + cy = 0, with a, b, and c real constants, has a regular singular point at x = 0. You should verify that two solutions of the Cauchy–Euler equation x2 y″ − 3xy′ + 4y = 0 on the interval (0, ∞) are y1 = x2 and y2 = x2 ln x. If we attempt to find a power series solution about the regular singular point x = 0, namely, y = cnxn, we would succeed in obtaining only the polynomial solution y1 = x2. The fact that we would not obtain the second solution is not surprising since ln x, and consequently, y2 = x2 ln x, is not analytic at x = 0; that is, y2 does not possess a Taylor series expansion centered at x = 0.

Method of Frobenius

To solve a differential equation (1) about a regular singular point, we employ the following theorem due to the eminent German mathematician Ferdinand Georg Frobenius (1849–1917).

THEOREM 5.2.1 Frobenius’ Theorem

If x = x0 is a regular singular point of the differential equation (1), then there exists at least one nonzero solution of the form

(4)

where the number r is a constant to be determined. The series will converge at least on some interval defined by 0 < x − x0 < R.

Notice the words at least in the first sentence of Theorem 5.2.1. This means that, in contrast to Theorem 5.1.1, we have no assurance that two series solutions of the type indicated in (4) can be found. The method of Frobenius, finding series solutions about a regular singular point x0, is similar to the “method of undetermined series coefficients” of the preceding section in that we substitute y = cn(x − x0)n + r into the given differential equation and determine the unknown coefficients cn by a recursion relation. However, we have an additional task in this procedure; before determining the coefficients, we must first find the unknown exponent r. If r is found to be a number that is not a nonnegative integer, then the corresponding solution y = cn (x − x0)n + r is not a power series.

As we did in the discussion of solutions about ordinary points, we shall always assume, for the sake of simplicity in solving differential equations, that the regular singular point is x = 0.

All solutions will be about the regular singular point 0.

EXAMPLE 2 Two Series Solutions

Because x = 0 is a regular singular point of the differential equation

3xy″ + y′ − y = 0, (5)

we try to find a solution of the form y = cnxn + r. Now

so that

which implies r(3r − 2) c0 = 0

(k + r + 1)(3k + 3r + 1) ck + 1 − ck = 0,     k = 0, 1, 2, … .

Since nothing is gained by taking c0 = 0, we must then have

r(3r − 2) = 0, (6)

and (7)

The two values of r that satisfy the quadratic equation (6), r1 = and r2 = 0, when substituted in (7), give two different recurrence relations:

(8)

(9)

Here we encounter something that did not happen when we obtained solutions about an ordinary point; we have what looks to be two different sets of coefficients, but each set contains the same multiple c0. Omitting this term, the series solutions are

(10)

(11)

By the ratio test it can be demonstrated that both (10) and (11) converge for all finite values of x; that is, |x| < ∞. Also, it should be apparent from the form of these solutions that neither series is a constant multiple of the other and, therefore, y1(x) and y2(x) are linearly independent on the entire x-axis. Hence by the superposition principle y = C1y1(x) + C2y2(x) is another solution of (5). On any interval not containing the origin, such as (0, ∞), this linear combination represents the general solution of the differential equation. ≡

Indicial Equation

Equation (6) is called the indicial equation of the problem, and the values r1 = and r2 = 0 are called the indicial roots, or exponents, of the singularity x = 0. In general, after substituting y = cnxn + r into the given differential equation and simplifying, the indicial equation is a quadratic equation in r that results from equating the total coefficient of the lowest power of x to zero. We solve for the two values of r and substitute these values into a recurrence relation such as (7). Theorem 5.2.1 guarantees that at least one nonzero solution of the assumed series form can be found.

It is possible to obtain the indicial equation in advance of substituting y = cnxn + r into the differential equation. If x = 0 is a regular singular point of (1), then by Definition 5.2.1 both functions p(x) = xP(x) and q(x) = x2Q(x), where P and Q are defined by the standard form (2), are analytic at x = 0; that is, the power series expansions

p(x) = xP(x) = a0 + a1x + a2x2 + …     and     q(x) = x2Q(x) = b0 + b1x + b2x2 + … (12)

are valid on intervals that have a positive radius of convergence. By multiplying (2) by x2, we get the form given in (3):

x2 y″ + x[xP(x)] y′ + [x2Q(x)] y = 0. (13)

After substituting y = cnxn + r and the two series in (12) into (13) and carrying out the multiplication of series, we find the general indicial equation to be

r(r − 1) + a0r + b0 = 0, (14)

where a0 and b0 are defined in (12). See Problems 13 and 14 in Exercises 5.2.

EXAMPLE 3 Two Series Solutions

Solve 2xy″ + (1 + x)y′ + y = 0.

SOLUTION

Substituting y = cnxn + r gives

which implies r(2r − 1) = 0 (15)

(k + r + 1)(2k + 2r + 1)ck + 1 + (k + r + 1)ck = 0,     k = 0, 1, 2, . . . . (16)

From (15) we see that the indicial roots are r1 = and r2 = 0.

For r1 = , we can divide by k + in (16) to obtain

(17)

whereas for r2 = 0, (16) becomes

(18)

Thus for the indicial root r1 = we obtain the solution

where we have again omitted c0. The series converges for x ≥ 0; as given, the series is not defined for negative values of x because of the presence of x1/2. For r2 = 0, a second solution is

On the interval (0, ∞), the general solution is y = C1y1(x) + C2y2(x). ≡

EXAMPLE 4 Only One Series Solution

Solve xy″ + y = 0.

SOLUTION

From xP(x) = 0 and x2Q(x) = x and the fact that 0 and x are their own power series centered at 0, we conclude a0 = 0 and b0 = 0 and so from (14) the indicial equation is r(r − 1) = 0. You should verify that the two recurrence relations corresponding to the indicial roots r1 = 1 and r2 = 0 yield exactly the same set of coefficients. In other words, in this case the method of Frobenius produces only a single series solution

≡

Three Cases

For the sake of discussion, let us again suppose that x = 0 is a regular singular point of equation (1) and that indicial roots r1 and r2 of the singularity are real, with r1 denoting the largest root. When using the method of Frobenius, we distinguish three cases corresponding to the nature of the indicial roots r1 and r2.

Case I: If r1 and r2 are distinct and do not differ by an integer, there exist two linearly independent solutions of the form

This is the case illustrated in Examples 2 and 3.

In the next case, we see that when the difference of indicial roots r1 − r2 is a positive integer, the second solution may contain a logarithm.

Case II: If r1 − r2 = N, where N is a positive integer, then there exist two linearly independent solutions of equation (1) of the form

(19)

(20)

where C is a constant that could be zero.

Finally, in the last case, the case when the indicial roots r1 and r2 are equal, a second solution will always contain a logarithm. The situation is analogous to the solution of a Cauchy–Euler equation when the roots of the auxiliary equation are equal.

Case III: If r1 = r2, then there always exist two linearly independent solutions of equation (1) of the form

(21)

(22)

Finding a Second Solution

When the difference r1 − r2 is a positive integer (Case II), we may or may not be able to find two solutions having the form y = cnxn + r. This is something we do not know in advance but is determined after we have found the indicial roots and have carefully examined the recurrence relation that defines the coefficients cn. We just may be lucky enough to find two solutions that involve only powers of x; that is, y1(x) = cnxn + r1 (equation (19)) and y2 = bnxn + r2 (equation (20) with C = 0). See Problem 31 in Exercises 5.2. On the other hand, in Example 4 we see that the difference of indicial roots is a positive integer (r1 − r2 = 1) and the method of Frobenius failed to give a second series solution. In this situation, equation (20), with C ≠ 0, indicates what the second solution looks like. Finally, when the difference r1 − r2 is a zero (Case III), the method of Frobenius fails to give a second series solution; the second solution (22) always contains a logarithm and is actually (20) with C = 1. One way to obtain this second solution with the logarithmic term is to use the fact that

(23)

is also a solution of y″ + P(x)y′ + Q(x)y = 0 whenever y1(x) is the known solution. We will illustrate how to use (23) in the next example.

This is (5) in Section 3.2.

EXAMPLE 5 Example 4 Revisited—Using a CAS

Find the general solution of     xy″ + y = 0.

SOLUTION

From the known solution given in Example 4,

we can construct a second solution y2(x) using formula (23). For those with the time, energy, and patience, the drudgery of squaring a series, long division, and integration of the quotient can be carried out by hand. But all these operations can be done with relative ease with the help of a CAS. We give the results:

Here is a good place to use a computer algebra system.

or y2(x) = y1(x) ln x + .

On the interval (0, ∞), the general solution is y = C1y1(x) + C2y2(x). ≡

REMARKS

(i) The three different forms of a linear second-order differential equation in (1), (2), and (3) were used to discuss various theoretical concepts. But on a practical level, when it comes to actually solving a differential equation using the method of Frobenius, it is advisable to work with the form of the DE given in (1).

(ii) When the difference of indicial roots r1 − r2 is a positive integer (r1 > r2), it sometimes pays to iterate the recurrence relation using the smaller root r2 first. See Problems 31 and 32 in Exercises 5.2.

(iii) Since an indicial root r is a root of a quadratic equation, it could be complex. We shall not, however, investigate this case.

(iv) If x = 0 is an irregular singular point, we may not be able to find any solution of the DE of form y = cnxn + r.

5.2 Exercises Answers to selected odd-numbered problems begin on page ANS-13.

In Problems 1–10, determine the singular points of the given differential equation. Classify each singular point as regular or irregular.

  1. x3y″ + 4x2y′ + 3y = 0
  2. x(x + 3)2y″ − y = 0
  3. (x2 − 9)2y″ + (x + 3)y′ + 2y = 0
  4. y″ − y′ + y = 0
  5. (x3 + 4x)y″ − 2xy′ + 6y = 0
  6. x2(x − 5)2y″ + 4xy′ + (x2 − 25)y = 0
  7. (x2 + x − 6)y″ + (x + 3)y′ + (x − 2)y = 0
  8. x(x2 + 1)2y″ + y = 0
  9. x3(x2 − 25)(x − 2)2y″ + 3x(x − 2)y′ + 7(x + 5)y = 0
  10. (x3 − 2x2 − 3x)2y″ + x(x − 3)2y′ − (x + 1)y = 0

In Problems 11 and 12, put the given differential equation into the form (3) for each regular singular point of the equation. Identify the functions p(x) and q(x).

  1. (x2 − 1)y″ + 5(x + 1)y′ + (x2 − x)y = 0
  2. xy″ + (x + 3)y′ + 7x2y = 0

In Problems 13 and 14, x = 0 is a regular singular point of the given differential equation. Use the general form of the indicial equation in (14) to find the indicial roots of the singularity. Without solving, discuss the number of series solutions you would expect to find using the method of Frobenius.

  1. x2y″ + ( x + x2)y′ − y = 0
  2. xy″ + y′ + 10y = 0

In Problems 15–24, x = 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent series solutions about x = 0. Form the general solution on the interval (0, ∞).

  1. 2xy″ − y′ + 2y = 0
  2. 2xy″ + 5y′ + xy = 0
  3. 4xy″ + y′ + y = 0
  4. 2x2y″ − xy′ + (x2 + 1)y = 0
  5. 3xy″ + (2 − x)y′ − y = 0
  6. x2y″ − (x − )y = 0
  7. 2xy″ − (3 + 2x)y′ + y = 0
  8. x2y″ + xy′ + (x2 − )y = 0
  9. 9x2y″ + 9x2y′ + 2y = 0
  10. 2x2y″ + 3xy′ + (2x − 1)y = 0

In Problems 25–30, x = 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the method of Frobenius to obtain at least one series solution about x = 0. Use (21) where necessary and a CAS, if instructed, to find a second solution. Form the general solution on the interval (0, ∞).

  1. xy″ + 2y′ − xy = 0
  2. x2y″ + xy′ + (x2 − )y = 0
  3. xy″ − xy′ + y = 0
  4. y″ + y′ − 2y = 0
  5. xy″ + (1 − x)y′ − y = 0
  6. xy″ + y′ + y = 0

In Problems 31 and 32, x = 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the recurrence relation found by the method of Frobenius first with the largest root r1. How many solutions did you find? Next use the recurrence relation with the smaller root r2. How many solutions did you find?

  1. xy″ + (x − 6)y′ − 3y = 0
  2. x(x − 1)y″ + 3y′ − 2y = 0
    1. The differential equation x4y″ + λy = 0 has an irregular singular point at x = 0. Show that the substitution t = 1/x yields the differential equation

      which now has a regular singular point at t = 0.

    2. Use the method of this section to find two series solutions of the second equation in part (a) about the singular point t = 0.
    3. Express each series solution of the original equation in terms of elementary functions.

Mathematical Model

  1. Buckling of a Tapered Column In Example 4 of Section 3.9, we saw that when a constant vertical compressive force or load P was applied to a thin column of uniform cross section, the deflection y(x) satisfied the boundary-value problem

    EI + Py = 0,     y(0) = 0,     y(L) = 0.

    The assumption here is that the column is hinged at both ends. The column will buckle or deflect only when the compressive force is a critical load Pn.

    1. In this problem let us assume that the column is of length L, is hinged at both ends, has circular cross sections, and is tapered as shown in FIGURE 5.2.1(a). If the column, a truncated cone, has a linear taper y = cx as shown in cross section in Figure 5.2.1(b), the moment of inertia of a cross section with respect to an axis perpendicular to the xy-plane is I = πr4, where r = y and y = cx. Hence we can write I(x) = I0(x/b)4, where I0 = I(b) = π(cb)4. Substituting I(x) into the differential equation in (24), we see that the deflection in this case is determined from the BVP

      x4 + λy = 0,     y(a) = 0,     y(b) = 0,

      where λ = Pb4/EI0. Use the results of Problem 33 to find the critical loads Pn for the tapered column. Use an appropriate identity to express the buckling modes yn(x) as a single function.

    2. Use a CAS to plot the graph of the first buckling mode y1(x) corresponding to the Euler load P1 when b = 11 and a = 1.
      Two illustrations. The first illustration has a column of length L hinged at both ends. It is tapered and has circular cross sections. The radius at the bottom is greater than the radius at the top. A vertical load P acts downward at the center of the top of the column. The second illustration has the column as a truncated cone indicated by dotted lines. The top of the cone is labeled y and the bottom of the cone is labeled x. The tapered side is labeled: y equals c times x. x equals a at the top and x equals b at the bottom. The length of the column is b minus a equals L.

      FIGURE 5.2.1 Tapered column in Problem 34

Discussion Problems

  1. Discuss how you would define a regular singular point for the linear third-order differential equation

    a3(x)y′″a2(x)y″ + a1(x)y′ + a0(x)y = 0.

  2. Each of the differential equations

    x3y″ + y = 0     and     x2y″ + (3x – 1)y′ + y = 0

    has an irregular singular point at x = 0. Determine whether the method of Frobenius yields a series solution of each differential equation about x = 0. Discuss and explain your findings.

  3. We have seen that x = 0 is a regular singular point of any Cauchy–Euler equation ax2y″ + bxy′ + cy = 0. Are the indicial equation (14) for a Cauchy–Euler equation and its auxiliary equation related? Discuss.
  4. Use the method of Frobenius to show that the general solution of the differential equation on is