\n", "\n", "**Theorem (Fuchs)** If $x = x_0$ is a regular singular point, then the solutions of a differential equation:\n", "1. are analytic on some neighbourhood around $x_0$ \n", "2. or they have a pole or a logarithmic term.\n", " \n", "

\n", "\n", "The solution to the ODE can be expressed using a generalised Frobenious series, meaning that any solution can be written as\n", "\n", "$$ y(x) = \\sum_{m=0}^{\\infty} a_m (x - x_0)^{m+r} $$\n", "\n", "where $a_m \\neq 0$, since if it were zero we can absorb a factor of $(x - x_0)$ into $(x - x_0)^r$. This condition leads to the **indicinal equation** for $r$ (i.e. the equation for the index $r$), which is a quadratic. Usually there are two solutions and hence two series, however, if the roots for $r$ differ by an integer, we have to be careful, for reasons that will be explained below. The best way to demonstrate the Frobenious method is through an example. We will solve **Bessel's equation** which in standard form is\n", "\n", "$$ y'' + \\frac{1}{x} y' +(1 - \\frac{s^2}{x^2}) y = 0 $$\n", "\n", "where $s \\geq 0$, $p(x) = 1/x$ and $q(x) = 1 - \\frac{s^2}{x^2}$. So $p$ and $q$ are not analytic at $x = 0$, and $x = 0$ is a singular point. Also $(x \u2013 x_0)p(x)$ and $(x \u2013 x_0)^2 q(x)$ are analytic at $x_0 = 0$, thus $x_0=0$ is a regular singluar point and so we can use the Frobenious method to solve the ODE. Therefore we substitue\n", "\n", "$$ y(x) = \\sum_{m=0}^{\\infty} a_m x^{m+r}, \\ \\ \\ \\ \\ \\ \\ \\ y'(x) = \\sum_{m=0}^{\\infty} a_m (m+r) x^{m+r-1}, \\ \\ \\ \\ \\ \\ \\ \\ y''(x) = \\sum_{m=0}^{\\infty} a_m (m+r) (m+r-1) x^{m+r-2} $$\n", "\n", "into Bessel's equation, noting that the limits are all $m = 0$ since the $m = 0$ and $m = 1$ terms do not necessarily differentiate to zero. If $r$ is not an integer then the leading term does not vanish upon differentiating. Thus we get\n", "\n", "$$ x^2 \\sum_{m=0}^{\\infty} a_m (m+r) (m+r-1) x^{m+r-2} + x \\sum_{m=0}^{\\infty} a_m (m+r) x^{m+r-1} + (x^2 - s^2) \\sum_{m=0}^{\\infty} a_m x^{m+r} = 0 $$\n", "\n", "Now absorb the $x^1, x^2$ pre-factors onto the sum\n", "\n", "$$ \\sum_{m=0}^{\\infty} a_m (m+r) (m+r-1) x^{m+r} + \\sum_{m=0}^{\\infty} a_m (m+r) x^{m+r} - s^2 \\sum_{m=0}^{\\infty} a_m x^{m+r} + \\sum_{m=0}^{\\infty} a_m x^{m+r+2} = 0 $$\n", "\n", "Letting $n = m + 2$ in the final sum:\n", "\n", "$$ ... \\ + \\ ... \\ + \\ ... \\ + \\sum_{n=2}^{\\infty} a_{n-2} x^{n+r} = 0 $$\n", "\n", "Finally, relabel $n \\rightarrow m$, collect in powers of $x$ and split of the $m = 0$ and $m = 1$ terms of the first three\n", "sums:\n", "\n", "$$ a_0 \\left[r(r-1) + r - s^2 \\right] x^r + a_1 \\left[(r+1)r + (r+1) - s^2 \\right] x^{r+1} + \\sum_{m=2}^{\\infty} \\left[a_m \\left[ (m + r)^2 - s^2 \\right] + a_{m-2} \\right] x^{m+r} = 0 $$\n", "\n", "Similarly to the power series method, we require all coefficients infront of each power of $x$ to vanish. The first two terms yield the indicial equation for determinind $r$. The general term inside the summation gives the recurrence relation that generates the coefficients of the power series solution.\n", "\n", "For $m = 0$, after simplification the indicial equation becomes:\n", "\n", "$$ r^2 - s^2 = 0 \\ \\ \\ \\ \\Rightarrow \\ \\ \\ \\ r = \\pm s $$\n", "\n", "where $s$ is positive. Remember that $a_0$ **cannot be zero**.\n", "\n", "For $m = 1$:\n", "\n", "$$ a_1 \\left[ (r+1)^2 - s^2 \\right] = 0 $$\n", "\n", "thus either\n", "\n", "$$ a_1 = 0 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{or} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (r+1)^2 - s^2 = 0 $$\n", "\n", "For now we will only consider the $a_1 = 0$ possibility.\n", "\n", "for $m \\geq 2$:\n", "\n", "$$ a_m = \\frac{-a_{m-2}}{(r+m)^2 - s^2} = \\frac{-a_{m-2}}{m^2 +2rm} $$\n", "\n", "where the last equality comes from the indicial equation, $r^2 = s^2$. Using the recursion relation to evaluate the $a_m$ coefficients and inserting in the generalised power series yields the answer to the ODE\n", "\n", "$$ y(x) = x^r \\left[ 1 - \\frac{x^2}{2(2+r} + \\frac{x^4}{2 \\times 4(2 + 2r)(4 + 2r)} - \\ ... \\right]$$\n", "\n", "Remembering that $a_1 = 0$, the recurrence relation implies that all $a_{odd} = 0$ and thus there is no second series. However, the two roots $r = \\pm s$ will usually yield the two independent solutions.\n", "\n", "Now lets go back to the second indicial equation (from $m = 1$) and consider $\\left[ (r+1)^2 - s^2 \\right] = 0$. Since $r^2 = s^2$ (from the first indicial equation), this requires $2r + 1 = 0$ or $r = -1/2$ and is thus the special case for $s = 1/2$. Therefore we will look for solutions with $s = 1/2$.\n", "\n", "For the $r = 1/2$ solution, the recurrence relation becomes\n", "\n", "$$ a_m = \\frac{- a_{m-2}}{m(m+1)} $$\n", "\n", "and thus\n", "\n", "$$ y(x) = a_0 x^{1/2} \\left[ 1 - \\frac{x^2}{3!} + \\frac{x^4}{5!} - \\ ... \\right] $$\n", "\n", "which can be written as \n", "\n", "$$ y(x) = \\frac{a_0} {x^{1/2}} \\left[ x - \\frac{x^3}{3!} + \\frac{x^3}{5!} - \\ ... \\right] = \\frac{a_0 sinx} {x^{1/2}} $$\n", "\n", "There is no second series, since $a_1 = 0$ for $r = 1/2$.\n", "\n", "Now lets consider the $r = - 1/2$ solution. The recurrence relation becomes\n", "\n", "$$ a_m = \\frac{- a_{m-2}}{m(m-1)} $$\n", "\n", "In this cae, $a_1$ is undetermined and thus the solution is given by two series:\n", "\n", "$$ y(x) = a_0 x^{- 1/2} \\left[ 1 - \\frac{x^2}{2!} + \\frac{x^4}{4!} - \\ ... \\right] + a_1 x^{- 1/2} \\left[ x - \\frac{x^3}{3!} + \\frac{x^5}{5!} - \\ ... \\right] $$\n", "\n", "which can be written as\n", "\n", "$$ y(x) = a_0 \\frac{cosx} {x^{1/2}} + a_1 \\frac{sinx} {x^{1/2}} = y_{GS}(x) $$\n", "\n", "The second term just duplicates the solution we found for $r = 1/2$, so that solution is already present here, which is why we identify this solution as the general solution, $y_{GS}(x)$. This occurs when the roots differ by an integer, as they do here ($r = - 1/2, + 1/2$), which is why we must be careful when roots for $r$ do differ by an integer. 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