Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
1  
    
sweet! I don't think I've seen that limit before. –  john mangual 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.