Finding General Solution by Method of Frobenius to Differential Equation

Method for solving ordinary differential equations

Some solutions of a differential equation having a regular singular point with indicial roots ${\displaystyle r={\frac {1}{2}}}$ and ${\displaystyle -1}$ .

In mathematics, the method of Frobenius, named after Ferdinand Georg Frobenius, is a way to find an infinite series solution for a second-order ordinary differential equation of the form

$${\displaystyle z^{2}u''+p(z)zu'+q(z)u=0}$$

with ${\textstyle u'\equiv {\frac {du}{dz}}}$ and ${\textstyle u''\equiv {\frac {d^{2}u}{dz^{2}}}}$ .

in the vicinity of the regular singular point ${\displaystyle z=0}$ .

One can divide by ${\displaystyle z^{2}}$ to obtain a differential equation of the form

$${\displaystyle u''+{\frac {p(z)}{z}}u'+{\frac {q(z)}{z^{2}}}u=0}$$

which will not be solvable with regular power series methods if either p(z)/z or q(z)/z ² are not analytic at z = 0. The Frobenius method enables one to create a power series solution to such a differential equation, provided that p(z) and q(z) are themselves analytic at 0 or, being analytic elsewhere, both their limits at 0 exist (and are finite).

Explanation [edit]

The method of Frobenius is to seek a power series solution of the form

$${\displaystyle u(z)=z^{r}\sum _{k=0}^{\infty }A_{k}z^{k},\qquad (A_{0}\neq 0)}$$

Differentiating:

$${\displaystyle u'(z)=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}}$$

$${\displaystyle u''(z)=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}}$$

Substituting the above differentiation into our original ODE:

$${\displaystyle {\begin{aligned}&z^{2}\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r-2}+zp(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r-1)(k+r)A_{k}z^{k+r}+p(z)\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r}+q(z)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }[(k+r-1)(k+r)A_{k}z^{k+r}+p(z)(k+r)A_{k}z^{k+r}+q(z)A_{k}z^{k+r}]\\&=\sum _{k=0}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}\\&=\left[r(r-1)+p(z)r+q(z)\right]A_{0}z^{r}+\sum _{k=1}^{\infty }\left[(k+r-1)(k+r)+p(z)(k+r)+q(z)\right]A_{k}z^{k+r}=0\end{aligned}}}$$

The expression

$${\displaystyle r\left(r-1\right)+p\left(0\right)r+q\left(0\right)=I(r)}$$

is known as the indicial polynomial, which is quadratic inr. The general definition of the indicial polynomial is the coefficient of the lowest power of z in the infinite series. In this case it happens to be that this is the rth coefficient but, it is possible for the lowest possible exponent to be r − 2, r − 1 or, something else depending on the given differential equation. This detail is important to keep in mind. In the process of synchronizing all the series of the differential equation to start at the same index value (which in the above expression isk = 1), one can end up with complicated expressions. However, in solving for the indicial roots attention is focused only on the coefficient of the lowest power ofz.

Using this, the general expression of the coefficient of z ^{k + r} is

$${\displaystyle I(k+r)A_{k}+\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j},}$$

These coefficients must be zero, since they should be solutions of the differential equation, so

$${\displaystyle {\begin{aligned}I(k+r)A_{k}+\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j}&=0\\\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j}&=-I(k+r)A_{k}\\{1 \over -I(k+r)}\sum _{j=0}^{k-1}{(j+r)p^{(k-j)}(0)+q^{(k-j)}(0) \over (k-j)!}A_{j}&=A_{k}\end{aligned}}}$$

The series solution with A _k above,

$${\displaystyle U_{r}(z)=\sum _{k=0}^{\infty }A_{k}z^{k+r}}$$

satisfies

$${\displaystyle z^{2}U_{r}(z)''+p(z)zU_{r}(z)'+q(z)U_{r}(z)=I(r)z^{r}}$$

If we choose one of the roots to the indicial polynomial for r in U _r (z), we gain a solution to the differential equation. If the difference between the roots is not an integer, we get another, linearly independent solution in the other root.

Example [edit]

Let us solve

$${\displaystyle z^{2}f''-zf'+(1-z)f=0}$$

Divide throughout by z ² to give

$${\displaystyle f''-{1 \over z}f'+{1-z \over z^{2}}f=f''-{1 \over z}f'+\left({1 \over z^{2}}-{1 \over z}\right)f=0}$$

which has the requisite singularity atz = 0.

Use the series solution

$${\displaystyle {\begin{aligned}f&=\sum _{k=0}^{\infty }A_{k}z^{k+r}\\f'&=\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}\\f''&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}\end{aligned}}}$$

Now, substituting

$${\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }&(k+r)(k+r-1)A_{k}z^{k+r-2}-{\frac {1}{z}}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+\left({\frac {1}{z^{2}}}-{\frac {1}{z}}\right)\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-{\frac {1}{z}}\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-1}+{\frac {1}{z^{2}}}\sum _{k=0}^{\infty }A_{k}z^{k+r}-{\frac {1}{z}}\sum _{k=0}^{\infty }A_{k}z^{k+r}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }A_{k}z^{k+r-1}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k-1=0}^{\infty }A_{k-1}z^{k-1+r-1}\\&=\sum _{k=0}^{\infty }(k+r)(k+r-1)A_{k}z^{k+r-2}-\sum _{k=0}^{\infty }(k+r)A_{k}z^{k+r-2}+\sum _{k=0}^{\infty }A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=\left\{\sum _{k=0}^{\infty }\left((k+r)(k+r-1)-(k+r)+1\right)A_{k}z^{k+r-2}\right\}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=\left\{\left(r(r-1)-r+1\right)A_{0}z^{r-2}+\sum _{k=1}^{\infty }\left((k+r)(k+r-1)-(k+r)+1\right)A_{k}z^{k+r-2}\right\}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\\&=(r-1)^{2}A_{0}z^{r-2}+\left\{\sum _{k=1}^{\infty }(k+r-1)^{2}A_{k}z^{k+r-2}-\sum _{k=1}^{\infty }A_{k-1}z^{k+r-2}\right\}\\&=(r-1)^{2}A_{0}z^{r-2}+\sum _{k=1}^{\infty }\left((k+r-1)^{2}A_{k}-A_{k-1}\right)z^{k+r-2}\end{aligned}}}$$

From (r − 1)² = 0 we get a double root of 1. Using this root, we set the coefficient of z ^{k + r − 2} to be zero (for it to be a solution), which gives us:

$${\displaystyle (k+1-1)^{2}A_{k}-A_{k-1}=k^{2}A_{k}-A_{k-1}=0}$$

hence we have the recurrence relation:

$${\displaystyle A_{k}={\frac {A_{k-1}}{k^{2}}}}$$

Given some initial conditions, we can either solve the recurrence entirely or obtain a solution in power series form.

Since the ratio of coefficients ${\displaystyle A_{k}/A_{k-1}}$ is a rational function, the power series can be written as a generalized hypergeometric series.

Roots separated by an integer [edit]

The previous example involved an indicial polynomial with a repeated root, which gives only one solution to the given differential equation. In general, the Frobenius method gives two independent solutions provided that the indicial equation's roots are not separated by an integer (including zero).

If the root is repeated or the roots differ by an integer, then the second solution can be found using:

$${\displaystyle y_{2}=Cy_{1}\ln x+\sum _{k=0}^{\infty }B_{k}x^{k+r_{2}}}$$

where ${\displaystyle y_{1}(x)}$ is the first solution (based on the larger root in the case of unequal roots), ${\displaystyle r_{2}}$ is the smaller root, and the constant C and the coefficients ${\displaystyle B_{k}}$ are to be determined. Once ${\displaystyle B_{0}}$ is chosen (for example by setting it to 1) then C and the ${\displaystyle B_{k}}$ are determined up to but not including ${\displaystyle B_{r_{1}-r_{2}}}$ , which can be set arbitrarily. This then determines the rest of the ${\displaystyle B_{k}.}$ In some cases the constant C must be zero. For example, consider the following differential equation (Kummer's equation with a = 1 and b = 2):

$${\displaystyle zu''+(2-z)u'-u=0}$$

The roots of the indicial equation are −1 and 0. Two independent solutions are ${\displaystyle 1/z}$ and ${\displaystyle e^{z}/z,}$ so we see that the logarithm does not appear in any solution. The solution ${\displaystyle (e^{z}-1)/z}$ has a power series starting with the power zero. In a power series starting with ${\displaystyle z^{-1}}$ the recurrence relation places no restriction on the coefficient for the term ${\displaystyle z^{0},}$ which can be set arbitrarily. If it is set to zero then with this differential equation all the other coefficients will be zero and we obtain the solution 1/z .

See also [edit]

• Fuchs' theorem
• Regular singular point
• Complex differential equation
• Laurent series

External links [edit]

• Weisstein, Eric W. "Frobenius Method". MathWorld.
• Teschl, Gerald (2012). Ordinary Differential Equations and Dynamical Systems. Providence: American Mathematical Society. ISBN978-0-8218-8328-0. (Draft version available online at https://www.mat.univie.ac.at/~gerald/ftp/book-ode/). Chapter 4 contains the full method including proofs.

Finding General Solution by Method of Frobenius to Differential Equation

Source: https://en.wikipedia.org/wiki/Frobenius_method

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