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Series solution of ODE near singular points with trig

  1. Nov 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the differential equation
    [itex] (\sin x)y'' + xy' + (x - \frac{1}{2})y = 0 [/itex]

    a) Determine all the regular singular points of the equation
    b) Determine the indicial equation corresponding to each regular point
    c) Determine the form of the two linearly independent solutions near each of the singular points
    d) Give the behavior of all non-zero solutions as x approaches 0.

    I'm mostly concerned about questions b) and c), as those are the ones that I'm stuck at.

    2. Relevant equations
    Singular points of equations with the form [itex] b(x)y'' + p(x)y' + q(x)y = 0 [/itex] occur when [itex] b(x) = 0 [/itex] and [itex] \frac{(x-x_0)p(x)}{b(x)} [/itex] has a definite limit as x -> x0 and [itex]\frac{(x-x_0)^2q(x)}{b(x)} [/itex] has a definite limit as x -> x0.
    Since we're looking for solutions near singular points, I am assuming we have to use the Frobenius method, as that's the only method I'm aware of for doing such a thing. So, assume [itex] y = x^r \sum_{n=0}^{\infty}a_nx^n [/itex].

    3. The attempt at a solution
    So the first thing I attempted was finding the regular singular points.
    Firstly, [itex] \sin x = 0 [/itex] when [itex] x = \pi n, n \in Z [/itex]. So all regular singular points must be in that set.
    Looking at [itex]\frac{xp(x)}{b(x)}[/itex] we get for x = 0: [itex] \frac{x^2}{\sin(x)} = x * (\frac{\sin(x)}{x})^{-1} \\[/itex] and [itex]\frac{x^2q(x)}{b(x)} = \frac{x^3}{\sin(x)} - \frac{x^2}{2\sin(x)} [/itex]. The limit of each of these is 0 because the limit as x approaches 0 of [itex] \frac{\sin(x)}{x} = 1 [/itex]. So 0 looks like a regular singular point. However, for every other possible singular point ([itex] x = \pi, 2\pi, 3\pi ...[/itex]), I don't believe these limits exist, because instead of [itex] \frac{x}{sin(x)} = \frac{0}{0} [/itex] there would be a real number divided by 0. So I believe the only regular singular point is at x = 0.

    Next I tried to find the indicial equation, but no matter how I try it, I get the indicial equation in terms of [itex] a_0 [/itex] and [itex] a_1 [/itex]. Here's my procedure:
    [itex] y = x^r \sum_{n=0}^{\infty} a_nx^n = \sum_{n=0}^{\infty}a_nx^{n+r} \\ [/itex]
    [itex] y' = \sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1} \\[/itex]
    [itex] y'' = \sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2} \\[/itex]

    So here's the differential equation in series form: [itex]\\[/itex]
    [itex] \sum_{n=0}^{\infty} a_n(n+r)(n+r-1)\sin(x)x^{n+r-2} + \sum_{n=0}^{\infty} a_n(n+r)x^{n+r} + \sum_{n=0}^{\infty} a_nx^{n+r+1} - \sum_{n=0}^{\infty} \frac{a_n}{2}x^{n+r} = 0\\[/itex]

    Next, I shifted the indices so all sums were in terms of [itex] x^{n+r+1} \\[/itex]
    [itex] \sum_{n=-2}^{\infty} a_{n+2}(n+r+2)(n+r+1)\sin(x)x^{n+r+1} + \sum_{n=-1}^{\infty} a_{n+1}(n+r+1)x^{n+r+1} + \sum_{n=0}^{\infty} a_nx^{n+r+1} - \sum_{n=-1}^{\infty} \frac{a_{n+1}}{2}x^{n+r+1} \\ [/itex]
    Now I remove terms from each series until I can write the rest of the series starting from n = 0. These extra terms that I removed make up the indicial equation. So for the first sum, I take the first two terms because we are starting at n = -2, and from the second and last sums I remove the first term because those start at n = -1.
    So, the sum of the terms I removed is: [itex] a_{2-2}(2+r-2)(-2+r+1)\sin(x)x^{r-1} + a_{2-1}(2+r-1)(1+r-1)\sin(x)x^{r} + a_{1-1}(-1+r+1)x^{-1+r+1} - \frac{a_{-1+1}}{2}x^{-1+r+1} \\[/itex]
    Simplifying, and removing the factor of [itex] x^{r} \\ [/itex]
    [itex]r(r-1)a_0\sin(x)x^{-1} + (r+1)(r)a_1\sin(x) + ra_0 - \frac{a_0}{2} \\[/itex]

    So I'm stuck with terms with [itex] a_1 [/itex] , [itex] \sin(x) [/itex], and exponentials of x. Is this a legitimate indicial equation? If not, where did I go wrong?
  2. jcsd
  3. Nov 14, 2015 #2


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    Homework Helper

    You are going to have to expand [itex]\sin x[/itex] as a taylor series and multiply it by the infinite series for [itex]y''[/itex]. For the origin you get [tex]
    \sum_{m=0}^\infty \sum_{k=0}^\infty \frac{(-1)^m a_k(k+r)(k+r-1)}{(2m+1)!} x^{2m + k + r - 1} + \sum_{n=0}^\infty a_n(n+r)x^{n+r}
    + (x - \tfrac12)\sum_{n=0}^\infty a_nx^{n+r} = 0.[/tex] Now the lowest power of [itex]x[/itex] you can get on the left hand side is [itex]x^{r-1}[/itex], by taking [itex]m= 0[/itex] and [itex]k = 0[/itex] in the product, to get the indicial equation [tex]a_0r(r-1) = 0.[/tex] For the other singular points I would set [itex]x = l\pi + z[/itex] for integer [itex]l[/itex] and [itex]y = \sum_{n=0}^\infty a_n(x - l\pi)^{n+r} = \sum_{n=0}^\infty a_nz^{n+r}[/itex] and use [itex]\sin x = \sin(z + l\pi) = (-1)^l \sin z[/itex] so that the ODE becomes [tex]
    (-1)^l \sin(z)y'' + (l\pi + z)y' + (z + l\pi - \tfrac12)y = 0.[/tex] From this you see that the [itex]x^{r-1}[/itex] coefficient acquires a contribution from the [itex]l\pi y'[/itex] term.
  4. Nov 15, 2015 #3
    Thanks for the reply, although I'm not sure I understand your procedure. Firstly, for the Frobenius method, aren't you supposed to normalize the y'' term and multiply everything by [itex] x^2 [/itex]? I guess that's not really that important, but can you explain how you found the indicial equation or how you can find the recurrence relation when you have that double sum? Also, why isn't x = 0 the only regular singular point? It looks to me like any multiple of [itex]\pi[/itex] makes [itex]\frac{xp(x)}{a(x)}, \frac{x^2q(x)}{a(x)} [/itex] undefined.
  5. Nov 15, 2015 #4


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    There is no such requirement, and here that actually makes your life vastly more difficult.

    If [itex]x_0[/itex] is a regular singular point then the idea of the Frobenius method is that to look for solutions of the form [itex]\sum_{n=0}^\infty a_n(x -x_0)^{n+r}[/itex], where [itex]r[/itex] is fixed by requiring that [itex]a_0 \neq 0[/itex].

    Whatever manipulations you perform on the left hand side of the ODE are aimed at turning it into a power series in [itex](x - x_0)[/itex] so that you can compare cofficients of powers of [itex](x - x_0)[/itex] in order to find [itex]r[/itex] and the [itex]a_k[/itex]. Thus any functions of [itex]x[/itex] which are not of that form need to be expanded as Taylor series about [itex]x_0[/itex], and you will therefore need to multiply power series (unless you can make a simplifying substitution first).

    In the case of [tex](\sin x)y'' + xy' + (x - \tfrac12)y = 0[/tex] you are going to be multiplying power series: either once as it stands, or twice if you rearrange it as [tex]
    x^2y'' + \frac{x^3}{\sin x} y' + \frac{x^2(x- \frac12)}{\sin x} y = 0.[/tex] Which would you rather? (Also, can you just write down the power series for [itex]x^3/\sin x[/itex] and [itex]x^2(x - \frac12)/\sin x[/itex] from standard references, as you can with the series for [itex]\sin x[/itex] itself, or are you going to have to calculate them yourself?)

    The indicial equation is found by obtaining the coefficient of the lowest power of [itex](x - x_0)[/itex] which occurs after you have turned the ODE into a power series.

    We can recover the coefficients indcutively, starting from the arbitrary non-zero [itex]a_0[/itex]. To get terms of order [itex]x^{n+r}[/itex] from the double sum we need to take all pairs of positive integers [itex](m,k)[/itex] such that [itex]2m + k = n + 1[/itex]. As [itex]n[/itex] increases there will be more and more of these, but one of them will be a multiple of [itex]a_{n+1}[/itex] and the others will be multiples of coefficients that we've already recovered. We can write this as a recurrence of the form [itex]a_{n+1} = F(a_0, \dots, a_n, n)[/itex] (note the express dependence on [itex]n[/itex]) but this is maybe more trouble than it's worth.

    At the origin, for example, comparing coefficients of order [itex]x^{r-1}[/itex], we have to take [itex]m = k = 0[/itex] in the double sum and the other sums don't give us anything, so [tex]
    a_0r(r-1) = 0.
    [/tex] This is the indicial equation. (The roots differ by an integer, so we expect only one linearly independent solution from this; the other will involve a logarithm.)

    At order [itex]x^r[/itex] we have to take the [itex](m,k) = (0,1)[/itex] term from the double sum, and we pick up contributions from the [itex]xy'[/itex] and [itex]-\frac12y[/itex] terms ([itex]n = 0[/itex] in both cases) to find [tex]
    a_1 r(r+1) + a_0r - \tfrac12a_0 = 0.
    [/tex] Now if [itex]r = 0[/itex] then we get [itex]-\frac12 a_0 = 0[/itex] which is a contradiction, so we must take [itex]r = 1[/itex] and [tex]
    2a_1 = -\tfrac12 a_0.[/tex]

    At order [itex]x^{r+1}[/itex] we must take both the [itex](m,k) = (1,0)[/itex] and [itex](m,k) = (0,2)[/itex] terms from the double sum and contributions from the other three sums to get [tex]
    \frac{(-1)a_0r(r-1)}{3!} + a_2(1+r)(2+r) + a_1(1+r) + a_0 - \frac12 a_1 = 0
    [/tex] and actually the first term vanishes, as in this case all terms with [itex]k = 0[/itex] will.

    So far we have [tex]y(x) = a_0\left(x - \frac14 x^2 - \frac{13}{48}x^3 + \dots\right).[/tex] We can, if desired, continue further.

    [itex]y'' + p(x)y' + q(x)y = 0[/itex] has a regular singular point at [itex]x_0[/itex] if and only if [itex]p(x)[/itex] has a pole of at most order 1 at [itex]x_0[/itex] and [itex]q(x)[/itex] has a pole of at most order 2 at [itex]x_0[/itex]. Here, [itex]p(x) = x/\sin x[/itex] has a removable discontinuity at [itex]x = 0[/itex] and a pole of order 1 at [itex]x = n\pi[/itex] for [itex]n \neq 0[/itex], while [itex]q(x) = x/\sin x - 1/(2 \sin x)[/itex] has poles of order 1 at all integer multiples of [itex]\pi[/itex]. Thus all integer multiples of [itex]\pi[/itex] are regular singular points.
    Last edited: Nov 15, 2015
  6. Nov 15, 2015 #5
    Thank you very much, this has been very helpful. However, how am I supposed to find the form of the linearly independent solutions near every regular singular point if there are infinitely many of them? I guess there must be a pattern. Trying that now.
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