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7.3 Singular points and the method of Frobenius

Note: 1 or 1.5 lectures, §8.4 and §8.5 in [EP], §5.4 – §5.7 in [BD]

While behavior of ODEs at singular points is more complicated, certain singular points are not especially difficult to solve. Let us look at some examples before giving a general method. We may be lucky and obtain a power series solution using the method of the previous section, but in general we may have to try other things.

7.3.1 Examples

Example 7.3.1: Let us first look at a simple first order equation

2xy′ - y = 0.

Note that x = 0 is a singular point. If we only try to plug in

 ∞ ∑ k y = akx , k=0

we obtain

 (∑∞ ) (∑∞ ) 0 = 2xy′ - y = 2x kakxk-1 - akxk k=1 k=0 ∑∞ = a0 + (2kak - ak)xk. k=1

First, a0 = 0 . Next, the only way to solve 0 = 2kak - ak = (2k - 1)ak for k = 1,2, 3,... is for ak = 0 for all k . Therefore we only get the trivial solution y = 0 . We need a nonzero solution to get the general solution.

Let us try  r y = x for some real number r . Consequently our solution—if we can find one—may only make sense for positive x . Then  ′ r-1 y = rx . So

0 = 2xy′ - y = 2xrxr-1 - xr = (2r - 1)xr.

Therefore r = 1∕2 , or in other words y = x1∕2 . Multiplying by a constant, the general solution for positive x is

 1∕2 y = Cx .

If C ⇔ 0 then the derivative of the solution “blows up” at x = 0 (the singular point). There is only one solution that is differentiable at x = 0 and that’s the trivial solution y = 0 .

Not every problem with a singular point has a solution of the form  r y = x , of course. But perhaps we can combine the methods. What we will do is to try a solution of the form

y = xrf (x )

where f (x ) is an analytic function.

Example 7.3.2: Suppose that we have the equation

4x 2y′′ - 4x2y′ + (1 - 2x)y = 0,

and again note that x = 0 is a singular point.

Let us try

 ∞∑ ∞∑ y = xr a xk = a xk+r, k k k=0 k=0

where r is a real number, not necessarily an integer. Again if such a solution exists, it may only exist for positive x . First let us find the derivatives

 ∑∞ y′ = (k + r)a xk+r-1, k=0 k ∞ ′′ ∑ k+r-2 y = (k + r)(k + r - 1)akx . k=0
Plugging into our equation we obtain
0 = 4x2y′′ - 4x 2y′ + (1 - 2x)y ( ∞ ) ( ∞ ) ( ∞ ) 2 ∑ k+r-2 2 ∑ k+r-1 ∑ k+r = 4x (k + r)(k + r - 1)akx - 4x (k + r)akx + (1 - 2x) akx ( k=0 ) ( k=0 ) ( ) (k=0 ) ∑∞ k+r ∑∞ k+r+1 ∑∞ k+r ∑∞ k+r+1 = 4(k + r)(k + r - 1)akx - 4(k + r)akx + akx - 2akx k=0 k=0 k=0 k=0 (∑ ∞ ) (∑∞ ) (∑∞ ) (∑∞ ) = 4(k + r)(k + r - 1)akxk+r - 4(k + r - 1 )ak-1xk+r + akxk+r - 2ak-1xk+r k=0 k=1 k=0 k=1 ∑∞ ( ) = 4r(r - 1)a0xr + a0xr + 4(k + r)(k + r - 1)ak - 4(k + r - 1)ak-1 + ak - 2ak-1xk+r k=1 ∑∞ = (4r(r - 1) + 1)a xr + ((4(k + r)(k + r - 1) + 1)a - (4(k + r - 1) + 2)a )xk+r. 0 k k-1 k=1

To have a solution we must first have (4r (r - 1) + 1)a0 = 0 . Supposing that a 0 ⇔ 0 we obtain

4r (r - 1) + 1 = 0.

This equation is called the indicial equation. This particular indicial equation has a double root at  1 r = ∕2 .

OK, so we know what r has to be. That knowledge we obtained simply by looking at the coefficient of xr . All other coefficients of xk+r also have to be zero so

( ) ( ) 4(k + r)(k + r - 1) + 1 ak - 4 (k + r - 1) + 2 ak-1 = 0.

If we plug in r = 1∕2 and solve for ak we get

 ----4(k +-1∕2 --1) +-2- 1- ak = 4 (k + 1∕2)(k + 1∕2 - 1) + 1ak-1 = kak-1.

Let us set a = 1 0 . Then

a1 = 1a0 = 1, a2 = 1a1 = 1, a3 = 1a2 = -1--, a4 = 1a3 = ---1---, ... 1 2 2 3 3 ⋅ 2 4 4 ⋅ 3 ⋅ 2

Extrapolating, we notice that

ak = ----------1----------= 1-. k(k - 1)(k - 2 )⋅⋅⋅3 ⋅ 2 k!

In other words,

 ∑∞ ∑∞ ∑∞ y = akxk+r = 1-xk+1∕2 = x1∕2 -1xk = x1∕2ex. k=0 k=0 k! k=0k!

That was lucky! In general, we will not be able to write the series in terms of elementary functions.

We have one solution, let us call it  1∕2 x y1 = x e . But what about a second solution? If we want a general solution, we need two linearly independent solutions. Picking a0 to be a different constant only gets us a constant multiple of y1 , and we do not have any other r to try; we only have one solution to the indicial equation. Well, there are powers of x floating around and we are taking derivatives, perhaps the logarithm (the antiderivative of x-1 ) is around as well. It turns out we want to try for another solution of the form

 ∑∞ y = b xk+r + (ln x)y, 2 k 1 k=0

which in our case is

 ∞ ∑ k+1∕2 1∕2 x y2 = bkx + (ln x)x e . k=0

We now differentiate this equation, substitute into the differential equation and solve for bk . A long computation ensues and we obtain some recursion relation for bk . The reader can (and should) try this to obtain for example the first three terms

 2b1 --1 6b2---1 b1 = b 0 - 1, b2 = 4 , b3 = 18 , ...

We then fix b 0 and obtain a solution y 2 . Then we write the general solution as y = Ay + By 1 2 .

7.3.2 The method of Frobenius

Before giving the general method, let us clarify when the method applies. Let

 ′′ ′ p(x)y + q(x)y + r(x)y = 0

be an ODE. As before, if p(x0) = 0 , then x0 is a singular point. If, furthermore, the limits

 q(x) r(x) lim (x - x0)---- and lim (x - x0)2---- x→x 0 p(x) x→x 0 p (x )

both exist and are finite, then we say that x0 is a regular singular point.

Example 7.3.3: Often, and for the rest of this section, x0 = 0 . Consider

 2 ′′ ′ 2 x y + x(1 + x)y + (π + x )y = 0.

Write

 q(x) x(1 + x) lim x ----= lim x --------= lim (1 + x) = 1, x→0 p(x) x→ 0 x2 x→0 2 r(x) 2 (π + x2) 2 lxim→0 x ---- = lxi→m0 x ---2----= lix→m0 (π + x ) = π. p (x ) x
So x = 0 is a regular singular point.

On the other hand if we make the slight change

x2y′′ + (1 + x)y′ + (π + x2)y = 0,

then

 q(x) (1-+-x) 1 +-x lix→m0 xp(x) = lixm→0 x x2 = lxi→m0 x = DNE.

Here DNE stands for does not exist. The point 0 is a singular point, but not a regular singular point.

Let us now discuss the general Method of Frobenius4. Let us only consider the method at the point x = 0 for simplicity. The main idea is the following theorem.

Theorem 7.3.1 (Method of Frobenius). Suppose that

p(x)y′′ + q(x)y′ + r(x)y = 0
(7.3)

has a regular singular point at x = 0 , then there exists at least one solution of the form

 ∞ r∑ k y = x akx . k=0

A solution of this form is called a Frobenius-type solution.

The method usually breaks down like this.

(i)
We seek a Frobenius-type solution of the form
 ∑∞ y = a xk+r. k=0 k

We plug this y into equation (7.3). We collect terms and write everything as a single series.

(ii)
The obtained series must be zero. Setting the first coefficient (usually the coefficient of  r x ) in the series to zero we obtain the indicial equation, which is a quadratic polynomial in r .
(iii)
If the indicial equation has two real roots r1 and r2 such that r1 - r2 is not an integer, then we have two linearly independent Frobenius-type solutions. Using the first root, we plug in
 ∑∞ y1 = xr1 akxk, k=0

and we solve for all ak to obtain the first solution. Then using the second root, we plug in

 ∞ r2∑ k y2 = x bkx , k=0

and solve for all bk to obtain the second solution.

(iv)
If the indicial equation has a doubled root r , then there we find one solution
 ∑∞ y = xr a xk, 1 k k=0

and then we obtain a new solution by plugging

 ∑∞ y2 = xr bkxk + (ln x)y1, k=0

into equation (7.3) and solving for the constants bk .

(v)
If the indicial equation has two real roots such that r1 - r2 is an integer, then one solution is
 ∑∞ y1 = xr1 akxk, k=0

and the second linearly independent solution is of the form

 ∑∞ y = xr2 b xk + C(lnx)y , 2 k 1 k=0

where we plug y2 into (7.3) and solve for the constants bk and C .

(vi)
Finally, if the indicial equation has complex roots, then solving for ak in the solution
 r∑∞ k y = x 1 akx k=0

results in a complex-valued function—all the ak are complex numbers. We obtain our two linearly independent solutions5 by taking the real and imaginary parts of y .

The main idea is to find at least one Frobenius-type solution. If we are lucky and find two, we are done. If we only get one, we either use the ideas above or even a different method such as reduction of order (Exercise 2.1.8) to obtain a second solution.

7.3.3 Bessel functions

An important class of functions that arises commonly in physics are the Bessel functions6. For example, these functions appear when solving the wave equation in two and three dimensions. First we have Bessel’s equation of order p :

 2 ′′ ′ ( 2 2) x y + xy + x - p y = 0.

We allow p to be any number, not just an integer, although integers and multiples of 1∕2 are most important in applications.

When we plug

 ∑∞ k+r y = akx k=0

into Bessel’s equation of order p we obtain the indicial equation

 2 r(r - 1) + r - p = (r - p)(r + p ) = 0.

Therefore we obtain two roots r1 = p and r2 = -p . If p is not an integer following the method of Frobenius and setting a0 = 1 , we obtain linearly independent solutions of the form

 p∑∞ (-1 )kx2k y1 = x -2k-----------------------------------, k=0 2 k!(k + p )(k - 1 + p )⋅⋅⋅(2 + p)(1 + p ) ∑∞ (-1 )kx2k y2 = x -p -2k-----------------------------------. k=0 2 k!(k - p )(k - 1 - p )⋅⋅⋅(2 - p)(1 - p )

Exercise 7.3.1: a) Verify that the indicial equation of Bessel’s equation of order p is (r - p )(r + p) = 0 . b) Suppose that p is not an integer. Carry out the computation to obtain the solutions y 1 and y 2 above.

Bessel functions will be convenient constant multiples of y1 and y2 . First we must define the gamma function

 ∫ ∞ Γ (x) = tx-1e-tdt. 0

Notice that Γ (1) = 1 . The gamma function also has a wonderful property

Γ (x + 1) = xΓ (x).

From this property, one can show that Γ (n) = (n - 1)! when n is an integer, so the gamma function is a continuous version of the factorial. We compute:

Γ (k + p + 1) = (k + p )(k - 1 + p )⋅⋅⋅(2 + p)(1 + p)Γ (1 + p ), Γ (k - p + 1) = (k - p )(k - 1 - p )⋅⋅⋅(2 - p)(1 - p)Γ (1 - p ).

Exercise 7.3.2: Verify the above identities using Γ (x + 1) = x Γ (x) .

We define the Bessel functions of the first kind of order p and - p as

 ∑∞ k ( )2k+p J (x) = ----1-----y = ----(--1)----- x- , p 2pΓ (1 + p) 1 k!Γ (k + p + 1 ) 2 k=0∞ k ( ) -----1----- ∑ ----(--1)----- x-2k-p J-p(x) = 2-pΓ (1 - p)y2 = k!Γ (k - p + 1 ) 2 . k=0
As these are constant multiples of the solutions we found above, these are both solutions to Bessel’s equation of order p . The constants are picked for convenience.

When p is not an integer, Jp and J-p are linearly independent. When n is an integer we obtain

 ∑∞ k ( )2k+n J (x) = --(-1)--- x- . n k=0 k!(k + n)! 2

In this case it turns out that

 n Jn(x) = (- 1)J -n(x),

and so we do not obtain a second linearly independent solution. The other solution is the so-called Bessel function of second kind. These make sense only for integer orders n and are defined as limits of linear combinations of J (x) p and J (x) -p as p approaches n in the following way:

 cos(pπ)Jp(x) - J-p(x) Yn(x ) = lpim→n--------------------. sin(pπ)

As each linear combination of Jp(x) and J-p(x ) is a solution to Bessel’s equation of order p , then as we take the limit as p goes to n , Yn(x) is a solution to Bessel’s equation of order n . It also turns out that Yn(x) and Jn(x) are linearly independent. Therefore when n is an integer, we have the general solution to Bessel’s equation of order n

y = AJn(x) + BYn (x ),

for arbitrary constants A and B . Note that Yn(x) goes to negative infinity at x = 0 . Many mathematical software packages have these functions J (x) n and Y (x) n defined, so they can be used just like say sin(x) and cos(x) . In fact, they have some similar properties. For example, - J1(x ) is a derivative of J0(x) , and in general the derivative of Jn(x) can be written as a linear combination of Jn-1(x) and Jn+1(x) . Furthermore, these functions oscillate, although they are not periodic. See Figure 7.4 for graphs of Bessel functions.


PICPIC

Figure 7.4: Plot of the J0(x) and J1(x) in the first graph and Y0(x) and Y1(x) in the second graph.


Example 7.3.4: Other equations can sometimes be solved in terms of the Bessel functions. For example, given a positive constant λ ,

xy′′ + y′ + λ2xy = 0,

can be changed to  2 ′′ ′ 2 2 x y + xy + λ x y = 0 . Then changing variables t = λx we obtain via chain rule the equation in y and t :

 2 ′′ ′ 2 t y + ty + t y = 0,

which can be recognized as Bessel’s equation of order 0. Therefore the general solution is y(t) = AJ (t) + BY (t) 0 0 , or in terms of x :

y = AJ0(λx) + BY 0(λx).

This equation comes up for example when finding fundamental modes of vibration of a circular drum, but we digress.

7.3.4 Exercises

Exercise 7.3.3: Find a particular (Frobenius-type) solution of x2y′′ + xy ′ + (1 + x)y = 0 .

Exercise 7.3.4: Find a particular (Frobenius-type) solution of xy′′ - y = 0 .

Exercise 7.3.5: Find a particular (Frobenius-type) solution of y′′ + 1y′ - xy = 0 x .

Exercise 7.3.6: Find the general solution of 2xy ′′ + y′ - x2y = 0 .

Exercise 7.3.7: Find the general solution of x2y′′ - xy′ - y = 0 .

Exercise 7.3.8: In the following equations classify the point x = 0 as ordinary, regular singular, or singular but not regular singular.

a)
 2 2 ′′ x (1 + x )y + xy = 0
b)
 2 ′′ ′ x y + y + y = 0
c)
xy′′ + x3y′ + y = 0
d)
xy′′ + xy′ - exy = 0
e)
 2 ′′ 2 ′ 2 x y + x y + x y = 0

Exercise 7.3.101: In the following equations classify the point x = 0 as ordinary, regular singular, or singular but not regular singular.

a)
y′′ + y = 0
b)
x3y′′ + (1 + x)y = 0
c)
xy′′ + x5y′ + y = 0
d)
 ′′ sin(x)y - y = 0
e)
cos(x )y′′ - sin(x )y = 0

Exercise 7.3.102: Find the general solution of x2y′′ - y = 0 .

Exercise 7.3.103: Find a particular solution of x2y′′ + (x - 3∕4)y = 0 .

Exercise 7.3.104 (Tricky): Find the general solution of  2 ′′ ′ x y - xy + y = 0 .

4Named after the German mathematician Ferdinand Georg Frobenius (1849 – 1917).

5See Joseph L. Neuringera, The Frobenius method for complex roots of the indicial equation, International Journal of Mathematical Education in Science and Technology, Volume 9, Issue 1, 1978, 71–77.

6Named after the German astronomer and mathematician Friedrich Wilhelm Bessel (1784 – 1846).