When we solve the equation $z^{2}u''+p(z)zu'+q(z)u=0$ using Frobenius method, we first find zeros of the indicial polynomial $r_1>r_2$ to get a solution $u_1$. In the case that the difference of roots $r_1-r_2$ is an integer, we might need to find a second linearly independent solution using a different method. It is given on the Wikipedia page (and many other places) that
$$
u_{2}=Cu_{1}\ln x+\sum _{{k=0}}^{\infty }B_{k}x^{{k+r_{2}}}
$$
gives the desired second solution.
My question is: how do we come up with this general form of solution? I try to derive it in various ways, but find it hard to achieve – I have tried to write $u_2 =vu_1$, which reduces the order of the DE and gives
$$
v'= u_1^{-2}\exp\left(- \int \frac{p(z)}{z} dz\right),
$$
but this looks completely different from the formula above.
I do see some vague intuition about the logarithmic term – we are basically saying: if multiplying by $x^{-r}$ does not make the solution analytic, then something stronger like $\ln x$ will. In other words, we create an essential singularity by adding $\ln x$.
I wish to see a more satisfying explanation of this.
Best Answer
The Frobenius method tells you that you can find a solution of the form $u(z)=z^ra(z)$ where $a(z)$ is a normal power series (with a positive radius of convergence). Inserting gives $$ z^2[z^ra''(z)+2rz^{r-1}a'(z)+r(r-1)z^{r-2}a(z)]+zp(z)[z^ra'(z)+rz^{r-1}a(z)]+q(z)z^ra(z)=0 $$ Now one can remove the common power $z^r$ and evaluate at $z=0$ $$ z^2a''(z)+z[2r+p(z)]a'(z)+[r(r-1)+rp(z)+q(z)]a(z)=0\\ [r(r-1)+rp(0)+q(0)]a(0)=0 $$ to obtain the indicial equation. For its solutions we know $1-p_0=r_1+r_2$ by the Viete formulas.
If you set $p(z)=p_0+z\tilde p(z)$ then the solution formula becomes $$ v'=z^{-2r_1}a(z)^{-2}z^{-p_0}\exp(-\int \tilde p(z) ds) $$ Now $n=p_0+2r_1=1+r_1-r_2$ is an integer according to the assumptions for this case, all the other factors are regular power series, so all in all one gets $$ v'(z)=z^{-n}c(z)=c_0z^{-n}+...+c_{n-1}z^{-1}+c_n+c_{n+1}z+... $$ as $n$ is positive, $r_1$ being the larger root. Integrating this gives a logarithmic term from $c_{n-1}z^{-1}$ and a product $z^{1-n}C(z)$ with a power series $C(z)$ from all the other terms. Now the second solution takes the form $$ u_2=c_{n-1}u_1\ln(z)+z^{1-n+r_1}a(z)C(z) $$ Now recall that $1-n+r_1=r_2$ and name the power series product $B(z)=a(z)C(z)$ to get the claimed form of the second solution.