I want to solve the differential equation $2ty'' + (1 - 2t)y' - y =0$ using Frobenius's method. I understand the method, and I've looked up several examples; however, I can't manage to solve this problem. Here's my attempt.

Let $y = \sum_{n=0}^{\infty} a_{n}t^{n}$ so that $y' = \sum_{n=0}^{\infty} na_{n}t^{n - 1}$ and $y'' = \sum_{n = 0}^{\infty} n(n-1) a_{n}t^{n -2}$.

Plugging these values in, we find

$$2t \sum_{n=0}^{\infty} n(n - 1)a_{n}t^{n - 2} + (1 - 2t)\sum_{n=0}^{\infty}na_{n}t^{n - 1} - \sum_{n=0}^{\infty} a_{n}t^{n} = 0.$$

I am really not sure about how to proceed here. I tried to mess around with the indices in order to make it in a nicer form. I also tried comparing coefficients, but I got nowhere. Note that we can change the index of the second sum to $n = 1$ without losing any terms. Also, we can change the index of the first sum to $n = 2$ without losing any terms. These observations didn't get me anywhere, though.

  • $\begingroup$ is not $t$ meant to be the same as $x$? $\endgroup$ – user121049 Mar 11 at 10:00
  • $\begingroup$ Oops. They should be the same. $\endgroup$ – user381493 Mar 11 at 10:16
  • 1
    $\begingroup$ The method isn't a straightforward power series. You need to start with $y = x^m\sum_{n=0}c_nx^n$, and find out what the lowest power $m$ is. $\endgroup$ – Dylan Mar 11 at 10:42
  • $\begingroup$ Using the factorization $(2tD-1)(D-1)y=0$ one finds for $u=y'-y$ the equation $2tu'-u=0$ with solution $u=C\sqrt{t}$ and thus $$y=C_1e^t+C_2\int e^{t-s}\sqrt{s}\,ds.$$ This you can now compare to the power series solutions. $\endgroup$ – LutzL Mar 11 at 11:28

Plugging these values in, we find $$2t\sum_{n=0}^{\infty}n(n-1)a_nt^{n-2}+(1-2t)\sum_{n=0}^{\infty}na_nt^{n-1}-\sum_{n=0}^{\infty}a_nt^n = 0$$

The next step is to sort things so we have matching powers of $t$. For some particular $k$, what's the coefficient of $t^k$?

To do this, we first bring distribute those polynomials in $t$ across the sums: $$\sum_{n=0}^{\infty}2n(n-1)a_nt^{n-1}+\sum_{n=0}^{\infty}na_nt^{n-1}-\sum_{n=0}^{\infty}2na_nt^{n}-\sum_{n=0}^{\infty}a_nt^n = 0$$ Next, we shift indices so that all of the sums match up. Let $k=n-1$ in the first and second sums, and $k=n$ in the third and fourth: $$\sum_{k=-1}^{\infty}2(k+1)ka_{k+1}t^k+\sum_{k=-1}^{\infty}(k+1)a_{k+1}t^k-\sum_{k=0}^{\infty}2ka_kt^k-\sum_{k=0}^{\infty}a_kt^k = 0$$ And now that they're all lined up, we recombine the sums. They all have the same factor $x^k$, so $$\sum_{k=0}^{\infty}\left[2(k+1)ka_{k+1}+(k+1)a_{k+1}-2ka_k-a_k\right]t^k = 0$$ $$\sum_{k=0}^{\infty}\left[(k+1)(2k+1)a_{k+1}-(2k+1)a_k\right]t^k = 0$$ Since the $k=-1$ terms were both zero, we dropped them from the sum. Now, the coefficient $(2k+1)[(k+1)a_{k+1}-a_k]$ of $a_k$ is zero for each $k\ge 0$, so we must have $a_k=(k+1)a_{k+1}$. That leads to the familiar exponential series $a_n=\frac{a_0}{n!}$.

Now, there's a nagging problem - we only got a one-dimensional solution space. This is a second order equation - where's the other solution? What happened is that the differential equation is singular - that leading $2t$ coefficient of $y''$ is zero at $t=0$. There's another solution, but it doesn't have a power series at zero because it isn't differentiable there.

Can we find that other solution with this method? Well, look at what happens if we don't require the exponents $k$ to be integers. We can solve the system "backwards" to get ever-increasing values of $a_k$ as $k\to-\infty$, but that just leads to a divergent series. We only get a convergent series if the process stops somewhere.

When does it stop? If we're working with integer $k$, it stops when we hit $k=0$. The next step $k=1$ multiplies the coefficient by zero. This is, of course, the solution we've already found.
Our other chance is that $2k+1$ coefficient. Whatever $a_{1/2}$ is, $a_{-1/2}$ can be anything. If we set $a_{-1/2}=0$, then $a_{-3/2}=0$ and so on.

So, that's the alternative: the other solution is of the form $$y=a_{\frac12}t^{\frac12}+a_{\frac32}t^{\frac12}+\cdots+a_{n+\frac12}t^{n+\frac12}+\cdots$$ Can you find that series, too?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy