(Walter Rudin Student Series in Advanced Mathematics) George Simmons, Steven Krantz - Differential Equations_ Theory, Technique, and Practice-McGraw-Hill Science_Engineering_Math (2006)

Section 4.4 Regular Singular Points 173 Now the key idea in solving a differential equation at a regular singular point is to guess a solution of the form ( )y = y(x) = xm · ao + a1x + azx2 + · · · . (4.13) We see that we have modified the guess used in the last section by adding a factor of xm min front. Here the exponent can be positive or negative or zero and may or may not be an integer. In practice-and this is conceptually important to avoid confusion-we assume that we have factored out the greatest possible power of x; thus the coefficient a0 will always be nonzero. We call a series of the type in Equation (4.31 ) a Frobenius series. We now solve the differential equation at a regular singular point just as we did in the last section, except that now our recursion relations will be more complicated-as they will involve m.both the coefficients aj and also the new exponent The method is best understood by examining an example. EXAMPLE 4.8 (4.14) Use the method of Frobenius series to solve the differential equation 2x2y\" + x(2x + l)y' - y = 0 about the regular singular point 0. Solution W riting the equation in the standard form 11 x(2x + 1) , ___1 _ y+ y y0 2x2 2x2 - ' we readily check that ( ) �and x2 · 2x + 1 1 ( )x(2x + 1) -- x. = =- 2x2 2 2x2 2 are both real analytic at x0 = 0. So x0 = 0 is a regular singular point for this differential equation. We guess a solution of the form and therefore calculate that 00 L)my' = + j)ajxm+j-1 j=O and 00 y\" = �)m + j)(m + j - l)ajxm+j-Z. j=O mNote that the derivative series now begin at j = 0 because might not be an integer.

174 Chapter 4 Power Series Solutions and Special Functions Plugging these calculations into the differential equation, written in the form of Equation (4.14), yields 2 00 + j)(m + j - l)ajxm+j z)m j=O + 2 L:j':0(m + j)ajxm+j+I+ L:j:0(m + j)ajxm+j- L:}:oajxm+j= 0. Notice that all sums for all derivatives of y begin at j = 0because m may be nonintegral. We make the usual adjustments in the indices so that all powers of x are xm+j, and break off the odd terms to put on the right-hand side of the equation. We obtain 2 00 + j)(m +j- l)ajXm+j+ 2 00 +j - l)aj-iXm+j l:)m z)m j=I j=I L L+ 00 (m + j)ajxm+j- 00 ajxm+j= -2m(m - l)aoxm.:.._ maoxm+ aoxm. j=I j=I The result is (4.15) 2(m + j)(m + j - l)aj+ 2(m + j - l)aj-l + (m + j)aj- aj= 0 for j = 1, 2, 3, ... together with [ - 2m(m - 1)- m + l]ao= 0. It is clearly not to our advantage to let ao= 0. Thus -2m(m - 1) - m + 1 = 0. This is the indicial equation. The roots of this quadratic equation are m = -1/2, 1. We put each of these values into Equation (4.15) and solve the resulting recursion relation. Now Equation (4.15) says that (2m2 + 2j2 + 4mj - j - m - l)aj= (-2m - 2j + 2)aj-I. Form = -1/2 this is a·= 3 - 2j J -3j + 2j2a·1-1 so 11 a1 = -a0 , a2= --2 a1 = 2-a0 , etc. For m = 1 we have Gj= -2j 3j + 2j2aj-1 so 2 44 a,= --ao, a2 =--a,= -35 ao etc . 5 14 '

Section 4.4 Regular Singular Points 175 Thus we have found the linearly independent solutions ( )a0x-1;2 · 1-x+ 2l x2-+··· and ( � � )aox · 1- x+ x2-+·· . 3 The general solution of our differential equation is then • ( � ) ( � � )y = Ax-112 • 1-x+ x2-+·· +Bx · 1- x+ x2-+· · . 3 There are some circumstances (such as when the indicial equation has a repeated root) that the method we have presented will not yield two linearly independent solutions. We explore these circumstances in the next section. EXERCISES 1. For each of the following differential equations, locate and classify its singular points on the x-axis. (a) x3(x-l)y\"-2(x-l)y'+3xy = 0 (b) x2(x2-l)y\"-x(l-x)y'+2y=0 (c) x2y\"+(2-x)y' = 0 (d) (3x+l)xy\"-(x+l)y'+2y = 0 2. Determine the nature of the point x=0 (i.e., what type of singular point it is) for each of the following differential equations. (a) y\"+(sinx)y = 0 (d) x3y\"+(sinx)y = O (b) xy\"+(sinx)y = 0 (e) x4y\"+(sinx)y = O (c) x2y\"+(sinx)y = 0 3. Find the indicial equation and its roots (for a series solution in powers of x) for each of the following differential equations. (a) x3y\" +(cos 2x-l)y'+2xy = 0 (b)4x2y\"+(2x4-5x)y'+(3x2+2)y=0 (c) x2y\" + 3xy'+4xy = 0 (d) x3y\"-4x2y'+3xy=0 4. For each of the following differential equations, verify that the origin is a regular singular point and calculate two independent Frobenius series solutions: (a) 4xy\"+3y'+y = 0 (c) 2xy\"+(x+l)y'+3y=0 (b) 2xy\"+(3-x)y'-y=0 (d) 2x2y\"+xy'-(x+l)y=0 5. W hen p = 0, Bessel's equation becomes

176 Chapter 4 Power Series Solutions and Special Functions Show that its indicial equation has only one root, and use the method of this section to deduce that f:o�Y(x) = (-l)j x2j 22j(j !)2 is the corresponding Frobenius series solution (see also Exercise 7(b) in Section 4.1). 6. Consider the differential equation y II + 1 - -1 y = 0. -yf x3 x2 ( a ) Show that x = 0 is an irregular singular point. ( b ) Use the fact that y1 = x is a solution to find a second independent solution Y2 by the method discussed earlier in the book. ( c ) Show that the second solution y2 that we found in part (b) cannot be expressed as a Frobenius series. 7. Consider the differential equation y\" + .x!b_! y' + !l_ = 0' Xe where p and q are nonzero real numbers and b, care positive integers. It is clear that, if b > 1 or c> 2, then x = 0 is an irregular singular point. ( a ) If b = 2 and c= 3, then show that there is only one possible value of m (from the indicial equation) for which there might exist a Frobenius series solution. ( b ) Show similarly that m satisfies a quadratic equation-and hence we can hope for two Frobenius series solutions, corresponding to the roots of this equation­ if and only if b = 1 and c:=:: 2. Observe that these are exactly the conditions that characterize x = 0 as a \"weak\" or regular singular point as opposed to a \"strong\" or irregular singular point. 8. The differential equation x2y\" + (3x - l)y' + y = 0 (4.16) has x = 0 as an irregular singular point. If ( )Y. = xm ao + a1x + a2x2 + · · · = aoxm + a1xm+I + a2xm+2 + · · · is inserted into Equation (4.16) then show that m = 0 and the corresponding Frobenius series \"solution\" is the power series L:00 y = 1!xj. j=O which converges only at x = 0. This demonstrates that, even when a Frobenius series formally satisfies such an equation, it is not necessarily a valid solution.

ft- MORE ON REGULAR SINGULAR POINTS Section 4.5 More on Regular Singular Points 177 We now look at the Frobenius series solution of \" + p. ' + q. y= 0 y y at a regular singular point from a theoretical point of view. Assuming that 0 is regular singular, we may suppose that 00 00 PjXj and x2·q(x)= L Lx · p(x)= qjxj, j=O j=O valid on a nontrivial interval ( - R, R). We guess a solution of the form 00 00 L Ly= ajxj+m xm ajxj = j=O j=O and calculate 00 m)ajxHm-1 L' y = U+ j=O and 00 Ly\"= U+ m)(j + m - l)ajxj+m-2. j=O Then ( ) ( )Lp(x). y'= �1 00 00 PjXJ· L aj(m + j)xm+1.-I 1=0 pjxj (t ) (f )1=0 = xm-2 aj(m + j)xj = xm-2 j=Of (t )1=0 1=0 xj, Pj-kak(m + k) k=O where we have used the formula (Theorem 4.1) for the product of two power series. Now, breaking off the summands corresponding to k= j, we find that this last is equal to Similarly,

178 Chapter 4 Power Series Solutions and Special Functions cWanecpeul tththeecsoemrimesoenxfparcetsosrioonfsxfmo-r2y• \"T, hpe · y', and q · y into the differential equation, result is and t [ aj[(m+j)(m+j - 1)+(m+})po+q0] � J+ ak[(m+k)Pj-k+qj-d xj= 0. k=O Now of course each coefficient of xj must be 0, so we obtain the following recursion relations: aj[(m+j)(m+j - 1)+(m+})po+qo] j-i +·�:k>: [Cm+k)Pj-k+qj-k]= 0. k=O (Incidentally, this illustrates a point we made earlier, in Section 4.3: That recursion relations need not be binary.) It is convenient, and helps us to emphasize the indicial equation, to write f(m) = m(m - 1)+mp0+q0. Then the recursion relation for j = 0 is aof(m)= 0. The successive recursion relations are aif(m+ l)+ao(mpi+qi)= 0; a2f(m+2)+ao(mp2+q2)+ai[Cm+ l)pi+qi]= 0; ajf(m+j)+ao(mpj+qj)+· · ·+aj-1[(m+j - l)pi+q1]= 0; etc. The first recursion formula tells us, since a0 =fa 0, that f(m) = m(m - 1)+mpo+qo = 0. This is of course the indicial equation. The roots mi, m2 of this equation are called the exponents of the differential equation at the regular singular point.

Section 4.5 More on Regular Singular Points 179 If the roots m1 and m2 are distinct and do not differ by an integer then our procedures will produce two linearly independent solutions for the differential equation. If m1 and m2 do differ by an integer, say m1 = m2 + j for some integer j 2: l, then the recursion procedure breaks down because the coefficient of aj in the (j + 1)51 recursion relation for m2 will be 0-so that we cannot solve for aj. The case m1 = m2 also leads to difficulties-because then our methods only generate one solution. We conclude this section by (i) enunciating a theorem that summarizes our positive results and (ii) providing an example that illustrates how to handle the degenerate cases just described. Theorem 4.3 Supesop that Xo =0 is a rqular singular point for the differential equation y\"+p·y'+q·y=O. xAssmu e that the power series for x · p(x) and 2 • q(x) have radius of convergence R > 0. Supsope that the indicial equation m(m - 1) + mPo + q0 has rotos m 1, m2 with m2 !:: m1• Then the differential equation has at least one solution of the form Yi=x\"'1 L00 aixi j=-0 on the interval (-R. R). In case m1 - m2 is not 7.ero or a positive integer then the differential equation has a second independne t solution Y2 =x•2 L00 bixi j-0 on the interval (- R. R). Mil. EXAMPLE 4.9 Find two independent Frobenius series solutions of xy\" + 2y' + xy = 0. Solution Notice that the constant term of the coefficient of y' is 2 and the constant term of the coefficient of y is 0. Thus the indicial equation is m(m - 1) + 2m + 0 = 0. The exponents for the regular singular point 0 are then 0, -1. Corresponding to m1 = 0, we guess a solution of the form ()() y = .L: ajXj j=O