Frobenius Method

Frobenius Method, we consider the general second order linear differential equation,

$\left(x-x_0\right)y^{\prime \prime }+\left(x-x_0\right)P\left(x\right)y^{\prime }+Q\left(x\right)y=0$ (9.1)

and attempt to solve it about a regular singular point $x=x_0$ . For that, we employ the Frobenius method as illustrated in the following theorem. It was discovered by the German mathematician George Frobenius (1848-1917) whose main research was in group theory and analysis. He worked in Berlin and Zurich and published his method for the series solution of linear ordinary differential equations in 1873.

Frobenius Theorem

If $x=x_0$ is a regular singular point of the differential equation , then there exists at least one series solution of this equation, of the form,

$y=\left(x-x_0\right)^r\sum _{n=0}^{\infty }a_n\left(x-x_0\right)^n=\sum _{n=0}^{\infty }a_n\left(x-x_0\right)^{n+r}$ (9.2)

where the number ‘r‘ is a constant that is to be determined. This series converges on some interval $0<\left|x-x_0\right|<R$ , where R is the radius of convergence of the series.
Although, in general, we consider $x=x_0$ as a regular singular point, however, for convenience and simplify we assume $x_0=0$ . If $x_0\ \ne 0$ , the equation can be transformed into one by letting $\left(x-x_0\right)=t$ , say, so that origin is the regular singular point. Thus Eqn.(9.1) can be written as,
$x^2y^{\prime \prime }+xP\left(x\right)y^{\prime }+Q\left(x\right)y=0$ (9.3)
Accordingly, we seek a series solution of this equation of the form,

$\ y=x^r\sum _{n=0}^{\infty }a_nx^n=\sum _{n=0}^{\infty }a_nx^{n+r}$ (9.4)

Differentiating it we obtain,

$y^{\prime }=\sum _{n=0}^{\infty }\left(n+r\right)a_nx^{n+r-1},$

and

$y^{\prime \prime }=\sum _{n=0}^{\infty }\left(n+r\right)\left(n+r-1\right)a_nx^{n+r-1}.$

As the functions P(x) and Q(x) are assumed to be analytic at the origin, they can be expanded as Maclaurin series,

$P\left(x\right)=P_0+P_1\left(x\right)+P_2x^2+...=\sum _{n=0}^{\infty }P_nx^n$ ,

and

$Q\left(x\right)=Q_0+Q_1x+Qx^2+...=\sum _{n=0}^{\infty }Q_nx^n$

Substituting the values of y, y’ y”, P(x) and Q(x) in Eqn.(9.3) we obtain,

$\ x^2\sum _{n=0}^{\infty }\left(n+r\right)\left(n+r-1\right)a_nx^{n+r-2}+x\left(\sum _{n=0}^{\infty }P_nx^n\right)\sum _{n=0}^{\infty }\left(n+r\right)a_nx^{n+r-1}+\left(\sum _{n=0}^{\infty }Q_nx^n\right)\sum _{n=0}^{\infty }a_nx^{n+r}=0,$

or,

$\sum _{n=0}^{\infty }a_n\left[\left(n+r\right)\left(n+r-1\right)+P_nx^n\left(n+r\right)+Q_nx^n\right]=0\$

$a_0\left\{r\left(r-1\right)+P_0r+Q_0\right\}+\sum _{n=0}^{\infty }a_n\left[.\left(n+r\right)\left(n+r-1\right)+P_n\left(n+r\right)+Q_nx^n\right]x^{n+r}=0$ (9.5)

For this equation to be satisfied identically the coefficient of cach power of x must be
zero. Since $a_0\ne 0$ in this first term, we must have,

$r\left(r-1\right)+P_0r+Q_0=0$ (9.6)

This equation is called indicial equation of the differential equation (9.1).
The indicial equation (9.6) of a second order linear differential equation of the form (9.3) can also be found directly without substituting (9.4) into it and carrying out simplification as given above. We can write (9.3) in the standard form,

$y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y=0$

and then evaluate $P_0$ and $Q_0$ as,

On substituting these values into Eqn.(9.6) we can obtain the requisite indicial equation.
The types of the roots $r_1\$ , and $r_2\$ , say, of the indicial equation help us in finding the second linearly independent solution of a differential equation, one being given by Eqn.(9.5). Four different cases may arise as given below,
a. The roots $r_1\$ and $r_2\$ may be real and distinct which do not differ by an integer.
b. The roots $r_1\$ , and $r_2\$ may be real which differ by an integer.
c. The roots $r_1\$ and $r_2\$ may be real and equal.
d. The roots $r_1\$ and $r_2\$ may be complex conjugates.
Accordingly various forms of Frobenius solutions are discussed as under:

Forms of Frobenius Solutions Depending on the Nature of the Roots $r_1\$ , and $r_2\$ of the Indicial Equation

The forms of solution of a differential equation depend on the roots $r_1\$ and $r_2\$ , say, of its indicial equation. These solutions are discussed as under:

Case-I Roots $r_1\$ and $r_2\$ are Real and Distanct and $\left(r_1-r_2\right)\ne Integer$

In this case we obtain two linearly independent solution of the given differential equations as,

$y_1=\sum _{n=0}^{\infty }a_nx^{n+r_1}\ \ \ \ \ \ \ \ y_2=\sum _{n=0}^{\infty }a_nx^{n+r_2}\ \ ,x>0$

Case-II Roots $r_1\$ and $r_2\$ are Real and Distanct and $\left(r_1-r_2\right)=Positive\ integer$

The two linearly independent solutions in this case are,

$y_1=\sum _{n=0}^{\infty }a_nx^{n+r_1}\ \ \ and\ \ \ \ \ y_2=A\ y_1\ln x+\sum _{n=1}^{\infty }b_nx^{n+r_2}\ \ ,\ \ x>0,$

where A is a constant , which may be equal to zero. Accordingly $\ y_2$ may or may not contain logarithm term.

Case-III Roots $r_1\$ and $r_2\$ are Real and Distanct and $\left(r_1-r_2\right)\ i.e,\ r_1=r_2=r$

The two linearly independent solutions in this case are,

$\ y_1=\sum _{n=0}^{\infty }a_nx^{n+r}\ \ \ and\ \ \ \ \ y_2=A\ y_1\ln x+\sum _{n=1}^{\infty }b_nx^{n+r}\ \ ,\ \ x>0,\ \$

or,

$y_1=\sum _{n=0}^{\infty }a_nx^{n+r}\ \ \ and\ \ \ \ \ y_2=\ y_1\ln x+\sum _{n=1}^{\infty }b_nx^{n+r}\ \ ,\$ (Assuming A=1 for simplicity)

Here $\ y_2$ always has a logarithm term.

Case-IV Roots $r_1\$ and $r_2\$ are Complex Conjugates say, $p\pm iq$

If the indicial equation has the roots $r_1=p+iq\ \ and\ \ \ r_2=p-iq$ , say, with $p\ne 0,$ then two linearly independent solutions of the given differential equation are the real and imaginary parts of,

$y=x^{p+iq}\sum _{n=0}^{\infty }a_nx^n\ ,$

where the coefficients $a_n$ are found as in Case-I.

First Method to Find the Second Linearly Independent Solution

The second solution can be found by the method of Reduction of Order. It asserts that if $y_1\left(x\right)$ is a solution of the differential equation, $y^{\prime \prime }+P\left(x\right)y^{\prime }+Q\left(x\right)y=0$ , then the second linearly independent solution is given by the formula,

$\ y_2\left(x\right)=y_1\left(x\right)\int _{ }^{ }\frac{e^{-\int _{ }^{ }Pdx}}{y_1^2\left(x\right)}dx.$

The given equation can be written as,

$y^{\prime \prime }+\left(1-\frac{1}{x}\right)y^{\prime }+\frac{1}{x^2}y=0,$

Where $P\left(x\right)=\left(1-\frac{1}{x}\right)$ and $Q\left(x\right)=\frac{1}{x^2}$ . Thus , $-\int _{ }^{ }Pdx=\int _{ }^{ }\left(-1+\frac{1}{x}\right)dx=-x+\ln x$ . So,

$e^{-\int _{ }^{ }Pdx}=e^{-x+\ln x}=e^{-x}e^{\ln x}=xe^{-x}$

Therefore, the above formula for $y_2\left(x\right)$ gives,

$y_2\left(x\right)=y_1\left(x\right)\int _{ }^{ }\frac{xe^{-x}}{x^2e^{-2x}}dx=y_1\left(x\right)\int _{ }^{ }\frac{e^x}{x}dx$

$y_2\left(x\right)=y_1\left(x\right)\int _{ }^{ }\frac{1}{x}\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)dx$

$y_2\left(x\right)=y_1\left(x\right)\int _{ }^{ }\left(\frac{1}{x}+1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+...\right)dx$

$y_2\left(x\right)=a_0xe^{-x}\left[\ln x+x+\frac{x^2}{4}+\frac{x^3}{18}+\frac{x^4}{96}+...\right]$

$\ y_2\left(x\right)=a_0xe^{-x}\ln x+a_0x\left[1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...\right]\left[x+\frac{x^2}{4}+\frac{x^3}{18}+\frac{x^4}{96}+...\right]$

$\ y_2\left(x\right)=a_0xe^{-x}\ln x+a_0\left[x^2-\frac{3}{4}x^3+\frac{11}{36}x^4-\frac{25}{288}x^5+...\right$ . (9.7)

Thus, Eqns. (9.7) give the general series-type solution of the given differential equation as, $y=c_1y_1\left(x\right)+c_2y_{2,}\left(x\right)$ , i.e.,

$\ y=c_1xe^{-x}+c_2\left[xe^{-x}\ln x+a_0\left(x^2-\frac{3}{4}x^3+\frac{11}{36}x^4-\frac{25}{288}x^5+...\right)\right]$

where the series inside the brackets is convergent.

Second Method to Find the Second Linearly Independent Solution

The second linearly independent solution can also be found by the formula;

$y_2\left(x\right)=y_1\left(x\right)\ln x+\sum _{n=1}^{\infty }B_nx^{n+r}.$

In the present case we have $\ r_1=r_2=1\left(=r\right)$ . Thus the above formula gives,

$\ y_2\left(x\right)=y_1\left(x\right)\ln x+\sum _{n=1}^{\infty }B_nx^{n+1}.$

$\ y_2\left(x\right)=y_1\left(x\right)\ln x+\sum _{n=0}^{\infty }b_nx^{n+2}.$ (9.8)

where $B_{n+1}=b_{n\ \ \ \ \ }$ . We find $y_2^{\prime }\$ and $y_2^{^{\prime \prime }}$ and then substitute these values in the given differential equation, $x^2y^{^{\prime \prime }}+\left(x^2-x\right)y^{^{\prime }}+y=0$ to get,

$x^2\left[y_1^{^{\prime \prime }}\ln x+\frac{2}{x}y_1^{^{\prime }}-\frac{y_1}{x}+\sum _{n=0}^{\infty }b_n\left(n+1\right)\left(n+2\right)x^n\right]\ +\left(x^2\ -x\right)\left[y_1^{^{\prime }}\ln x+\frac{y_1}{x}+\sum _{n=0}^{\infty }b_n\left(n+2\right)x^{n+1}\right]+y_1\ln x+\sum _{n=0}^{\infty }b_nx^{n+2}=0,$

$\left[x^2y_1^{^{\prime \prime }}+\left(x^2-x\right)y^{^{\prime }}+y\right]\ln x+\left(2y_1^{^{\prime }}x+y_1x-2y_1\right)+\sum _{n=0}^{\infty }b_n\left(n+1\right)\left(n+2\right)x^{n+2}+\sum _{n=0}^{\infty }b_n\left(n+2\right)x^{n+3}-\sum _{n=0}^{\infty }b_n\left(n+2\right)x^{n+2}+\sum _{n=0}^{\infty }b_nx^{n+2}=0$

We simplify the following equation as follows:

a. Since $y_1$ as a solution of the given differential equation, so,

$^2y_1^{^{\prime \prime }}+\left(x^2-x\right)y_1^{^{\prime }}+y_1=0$

b. We write ,

$\sum _{n=0}^{\infty }b_n\left(n+2\right)x^{n+3}=\sum _{n=1}^{\infty }b_{n-1}\left(n+1\right)x^{n+2}$

c. We separate the terms corresponding to n=0 in the remaining terms with summation sign to get,

$b_02\ x^2-b_02\ x^2+b_0x^2=b_0x^2$

Accordingly the above equation can be written as,

$\left[2xy_1^{^{\prime }}+xy_1-2y_1+b_0x^2\right]+\sum _{n=1}^{\infty }\left[\left(n+1\right)\left(n+2\right)b_n+\left(n+1\right)b_{n-1}-\left(n+2\right)b_n+b_n\right]x^{n+2}=0,$

or,

$\left[2xy_1^{^{\prime }}+xy_1-2y_1+b_0x^2\right]+\sum _{n=1}^{\infty }\left[\left(n+1\right)\left(n+1\right)b_n+b_{n-1}\right]x^{n+2}=0,\$ (9.9)

Now,

$y_1\left(x\right)=xe^{-x}=x-x^2+\frac{x^3}{2}-\frac{x^4}{6}+\frac{x^5}{24}-...,$

so,

$y_1^{^{\prime }}\left(x\right)=1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-,...$

Substituting the values of $y_1$ and $y_1^{^{\prime }}$ in the above equation (9.9), the first term inside the square brackets becomes,

$2x\left[1-2x+\frac{3x^2}{2}-\frac{2x^3}{3}+\frac{5x^4}{24}-,...\right]+x\left[x-x^2+\frac{x^3}{2}-\frac{x^4}{6}+\frac{x^5}{24}-...\right]-2\left[x-x^2+\frac{x^3}{2}-\frac{x^4}{6}+\frac{x^5}{24}-...\right]+b_0x^2$ .

On collecting the terms with the same power of x it becomes

$\left(-x^2+x^3-\frac{x^4}{2}+\frac{x^5}{6}-\frac{x^6}{24}+...\right)+b_0x^2$ .

Therefore, Equation (9.9) becomes,

$\left(-x^2+x^3-\frac{1}{2}x^4+\frac{1}{6}x^5-\frac{1}{24}x^6+...\right)+b_0x^2+2\left(2b_1+b_0\right)x^3+3\left(3b_2+b_1\right)x^4+4\left(4b_3+b_2\right)x^5+...=0$ .

or,

$\left(-1+b_0\right)x^2+\left(1+4b_1+2b_0\right)x^3+\left(-\frac{1}{2}+9b_2+3b_1\right)x^4+\left(\frac{1}{6}+16b_3+4b_2\right)x^5+,...=0$

Thus, equating the coefficients of various powers x to zero (since, RHS=0), we obtain,

coefficients of $x^2$ equated to zero gives: $-1+b_0=0$ it gives $b_0=1$ ,

coefficients of $x^3$ equated to zero gives: $1+4b_1+2b_0=0$ it gives $b_1=-\frac{3}{4}$ ,

coefficients of $x^4$ equated to zero gives: $\ -\frac{1}{2}+9b_2+3b_1=0$ it gives $b_2=\frac{11}{36}$ ,

coefficients of $x^5$ equated to zero gives: $\frac{1}{6}+16b_3+4b_2=0$ it gives $\ b_3=-\frac{25}{288}$ ,

and so on. Thus substituting the values of these coefficients in Equation (9.8) we obtain the second linearly independent solution of the given differential equation as,

$y=xe^{-x}\ln x+\left(x^2-\frac{3}{4}x^3+\frac{11}{36}x^4-\frac{25}{288}x^5+...\right)$ (9.10)

which is same as obtained in first method.