The steady state temperature distribution of a rod given by: \begin{equation} \frac{\textrm{d}p(x)y'}{\textrm{d}x} - y = 0,\; 0 \leq x \leq 1,\; \text{and} \;y(0) = 0, \end{equation}
where $y(x)$ is the steady state temperature distribution, and $p(x) = x^s$ is the spatially dependent conductivity of the rod, for some $s \in [0, 1)$. For $0 \leq s < 1$ what is the form of the solution close to $x = 0$ (i.e. the first the term of the series solution)? Does a solution exist for $s = 1$? $s > 1$?
First, simplify the equation into a recognizable form: \begin{align*} &\quad \frac{\textrm{d}p(x)y'}{\textrm{d}x} - y = 0 \\ &\equiv p(x)y'' + p'(x)y' - y = 0 \\ &\equiv y'' + \frac{p'(x)}{p(x)}y' - \frac{1}{p(x)} y = 0 \end{align*}
Since $p(x) = x^s$, thus $p'(x) = sx^{s-1}$ and $p''(x) = s(s-1)x^{s-2}$. Substituting: \begin{align*} &\quad y'' + \frac{p'(x)}{p(x)}y' - \frac{1}{p(x)} y = 0 \\ &\equiv y'' + \frac{s}{x}y - \frac{s^2 - s}{x^2}y = 0 \end{align*}
We can immediately recognize that $x = 0$ is a regular singular point since: \begin{align*} &\lim_{x\rightarrow 0} \frac{sx}{x} = s &\lim_{x\rightarrow 0} -\frac{(s^2 - s)x^2}{x^2} = s - s^2 \end{align*} It is worth noting that these limits exist for all $s$.
So, we look for solutions of the form: \begin{align*} y = \sum_{n=0}^{\infty} a_nx^{n+r} \end{align*} where $r(r-1) + sr + (s - s^2) = 0 = r^2 + (s-1)r + (s - s^2)$ is the corresponding indical equation. Solving for $r$, we obtain: \begin{align} \nonumber &\quad r = \frac{-(s-1) \pm \sqrt{(s-1)^2 - 4(s - s^2)}}{2} \\ \nonumber &\equiv r = \frac{-(s-1) \pm \sqrt{s^2 - 2s + 1 - 4s + 4s^2}}{2} \\ &\equiv r = \frac{-(s-1) \pm \sqrt{5s^2 - 6s + 1}}{2} \end{align}
For what values of $s$ do we get imaginary $r$? That would be when the discriminant of the quadratic formula is less than zero: \begin{align*} 5s^2 - 6s + 1 < 0 \end{align*} Consider when $s = 0$: \begin{align*} &\quad 5s^2 - 6s + 1 = 0 \\ &\equiv s = \frac{6 \pm \sqrt{36 - 20}}{10} \\ &\equiv s = \frac{6 \pm \sqrt{16}}{10} \\ &\equiv s = \frac{6 \pm 4}{10} \\ &\equiv s = 1 \vee s = \frac{1}{5}\\ \end{align*} Since for $(s=0 \wedge d = 5s^2 - 6s + 1) \implies d = 1$, then we know that the discriminant is positive for all values of $s \leq \frac{1}{5}$ and $s \geq 1$, and we have: \begin{align*} r_1 &= \frac{-(s-1) + \sqrt{(s - 1)(s - \frac{1}{5})}}{2} \\ r_2 &= \frac{-(s-1) - \sqrt{(s + 1)(s - \frac{1}{5})}}{2} \end{align*}
The general solution is of the form: \begin{equation*} y = c_1\sum_{n = 0}^{\infty} a_nx^{n + r_1} + c_2\sum_{n = 0}^{\infty} a_nx^{n + r_2} \end{equation*} where $a_n$ might be complex if $\frac{1}{5} s < 1$
We know that $y(0) = 0$, therefore: \begin{align*} &\quad y(0) = 0 = c_1\sum_{n = 0}^{\infty} a_n0^{n + r_1} + c_2\sum_{n = 0}^{\infty} a_n0^{n + r_2} \\ &\equiv 0 = c_1\sum_{n = 0}^{\infty} a_n + c_2\sum_{n = 0}^{\infty} a_n \\ &\equiv 0 = (c_1 + c_2) \\ &\equiv c1 = -c_2 \end{align*}
The first term of this series is: \begin{align*} \nonumber &\quad c_1\left(a_0x^{r_1} - a_0x^{r_2}\right) \\ \nonumber &\equiv c_1a_0\left(x^{\frac{-(s-1) + \sqrt{(s - 1)(s - \frac{1}{5})}}{2}} - x^{\frac{-(s-1) - \sqrt{(s - 1)(s - \frac{1}{5})}}{2}}\right) \\ \end{align*} Let $\frac{1 - s}{2} = \alpha$ and $\frac{\sqrt{(s - 1)(s - \frac{1}{5})}}{2} = \beta$ If $0 \leq s \leq \frac{1}{5}$: \begin{equation*} y \approx c_1a_0\left(x^{\alpha + \beta} - x^{\alpha - \beta}\right) \end{equation*}
If $\frac{1}{5} < s < 1$: \begin{equation*} y \approx c_1a_0\left(x^{\alpha + i\beta} - x^{\alpha - i\beta}\right) \end{equation*} Recall that: \begin{align*} &\quad x^r = e^{rln(x)} \\ &\implies x^{\lambda + i\mu} = e^{\lambda\ln(x)}e^{i\mu\ln(x)} \\ &\equiv x^{\lambda + i\mu} = e^{\lambda\ln(x)}(\cos(\mu\ln(x)) + i\sin(\mu\ln(x))) \end{align*} So we have oscillatory solutions. Thus: \begin{equation*} y \approx c_1a_0x^{\alpha}\left(2i\sin(\beta\ln(x))\right) \end{equation*}
Since the limits determining if $x = 0$ is a regular point exist for all $s$, we can write a (non-trivial) Frobenius series solution for all values of $s$.
Questions:
1) I have not used information about $0 \leq s < 1$, which is bothersome. Did I miss an opportunity to use it to simplify the form of the first term?
2) Is my reasoning for why solutions exist for $s = 1$ and $s > 1$ sound?
