Kansas State University

Discussion

Theorem: Suppose $y(x) = (x-x_0)^r \displaystyle\sum_{n=0}^{\infty}a_n(x-x_0)^n$ is the series solution to the differential equation $p(x)y'' + q(x)y' + r(x)y = 0$ where $p(x)$, $q(x)$ and $r(x)$ are all polynomials and $x_0$ is a regular singular point. Then $y(x)$ converges in the regions $0 < x-x_0< a$ and $-a < x-x_0 < 0$ where $a$ is the distance from $x_0$ to the next nearest singular point of the equation. NOTE: We must consider complex $x$, even though the equation is real.

This theorem is exactly the same as the corresponding theorem for ordinary points, except that the series need not converge at the point $x_0$ and that we use the nearest singular point to $x_0$ excluding $x_0$ itself. The same points raised in the "More Discussion" part of that section apply here as well.

Paradigm

Step 1: Check that $x_0$ is a regular singular point so the theorem applies.

Since all the coefficients are polynomials and the leading coefficient, $x^3+x^2$ is 0 at $x_0=0$, $x_0$ is a singular point. Next we check the appropriate limits. $$ \eqalign { &\lim_{x\to0}\frac{(x-0)x}{x^3+x^2}=\lim_{x\to0}\frac{1}{x+1}=1 \cr &\lim_{x\to0}\frac{(x-0)^2}{x^3+x^2}=\lim_{x\to0}\frac{1}{x+1}=1 \cr } $$ Since both limits exist, $x_0$ is a regular singular point.

Step 2: Find all singular points, including those in the complex plane.

Since all coefficients are polynomial, the singular points are just the roots of the leading coefficient, $x^3+x^2$. The roots are $-1$ and $0$.

Step 3: Find the distances from $x_0$ to the other singular points.

The only singular point other than $x_0=0$ is $-1$ which is 1 unit away from $x_0$.

Step 4: The series solution converges in the regions $0 < x-x_0 < a$ and $-a < x-x_0 < 0$ where $a$ is the smallest distance found in step 3.

Since the only distance found was 1, the series solution converges in the regions $-1 < x < 0$ and $0 < x < 1$. Note that it may also converge outside the region as well; the theorem states a region where it converges but doesn't say anything about where it diverges.

EXAMPLE: Where does the series solution of Bessel's equation of order 1, $x^2y''+xy'+(x^2-1)y=0$ converge?

Step 1: $0^2=0$ so $x_0=0$ is singular. $$ \eqalign { &\lim_{x\to0}\frac{(x-0)x}{x^2}=1 \cr &\lim_{x\to0}\frac{(x-0)^2(x^2-1)}{x^2}=\lim_{x\to0}x^2-1=-1 \cr } $$ So $x_0=0$ is a regular singular point.

Step 2: All coefficients are polynomials so the singular points are the roots of the leading coefficient, $x^2$, which is just $0$.

Step 3: Since 0 is the only singular point, there are no other singular points to find the distance to.

Step 4: Since there is only the one singular point at $x_0=0$, the series converges in the regions $0 < x < \infty$ and $-\infty < x < 0$.

This last example illustrates why one might want to find a series solution about a singular point rather than an ordinary point. If a series solution for Bessel's equation is found about any ordinary point, we can only guarantee that the series will converge in a bounded interval because the singular point at 0 will be some finite distance away from the ordinary point (see problem 10 in section 3). But the expansion about the singular point 0 itself converges in two unbounded intervals. Actually, one solution to this particular equation will converge everywhere, including at the point 0. This is a special bonus to Bessel's equation and we won't be that lucky in general.