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My class is solving the Cauchy-Euler differential equation $a t^2 y'' + b t y' + c y = 0$. The solutions are powers of $t$, $y = t^r$ and then you solve for $r$ using the characteristic equation $a r^2 + (b-a) r + c = 0$. This has two roots $r = \overline{r} \pm \Delta r$ and the general solution is $y = A t^{\overline{r} + \Delta r} + B t^{\overline{r} - \Delta r }$.

What happens if we get double roots, i.e. $(b-a)^2 = 4ac$ or $\Delta r = 0$? We guess a solution of $y = t^r \ln t$. Can derive this guess taking the limit of two distinct roots solution as $\Delta r$ tends to 0? In some sense $$ A t^{\overline{r} + \Delta r} + B t^{\overline{r} - \Delta r } = t^{\overline{r}} ( A t^{ \Delta r} + B t^{ - \Delta r }) \to t^{\overline{r}} (C + D \ln t )$$ for some constants $A,B,C,D$.

In a way, it is plausible the limit works this way because of the integral formula: $$ \int t^{r - 1}\mathrm dt = \begin{cases} \frac{t^r}{r} & r \neq 0\\ \ln t & r = 0 \end{cases}$$ And I expect this is coming into play in the Euler-Cauchy equation. How to make it rigorous, though?

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up vote 6 down vote accepted

Everything works out as long as you are careful about the additive constants in your integrals! What you should actually have written is

$$\int_1^{x} t^{r-1} \, dt = \begin{cases} \frac{x^r - 1}{r} & r \neq 0 \\\ \ln x & r = 0 \end{cases}$$

and then I leave it as an exercise to verify that indeed $\lim_{r \to 0} \frac{x^r - 1}{r} = \ln x$ for positive $x$ (note that this already follows from the above integral formula since integrals respect uniform limits). A great example of the importance of keeping track of that $+C$. This exact limit came up in another math.SE question which I can't track down.

In the application to the Euler-Lagrange equation I guess you can take $A = -B = \frac{1}{\Delta r}$; then the limit becomes $2 \ln t$. Probably you can get this by using the right initial conditions.

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sweet! I don't think I've seen that limit before. – cactus314 Jan 8 '11 at 1:44
@john: it is a straightforward application of l'Hopital's after dividing by ln x. Alternately you can write x^r = e^{r ln x} = 1 + r ln x + ... – Qiaochu Yuan Jan 8 '11 at 2:21

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