# Solving the differential equation m^2 = 0?

1. Nov 26, 2013

### Nat3

I have a problem that starts with the equation:

$r\frac{d^2u}{dr^2} + \frac{du}{dr} = 0$

The solution I'm looking at says to do a substitution, letting $u = r^m$, which after differentiation and simplification results in:

$m^2 = 0$

Up until this point, I understand (well, actually I don't understand how to know to do this substitution, but I understand how the differentiation and simplification happened), but then the next step says that because $m^2 = 0$:

$u(r) = c_1 + c_2 lnr$

Question 1: How do you know to choose $u = r^m$ as the substitution?

Question 2: Why was that form of a solution chosen?

Question 3: Is there a name for this type of substitution and solution?

Thank you!

2. Nov 26, 2013

### LCKurtz

Actually, that wouldn't have my first suggestion. I would have said let $y = u'$ making the equation $ry' + y = 0$ and notice that $ry'+y=(ry)' = 0$ so $ry = C$ and $y = u' = \frac C r$. Hence $u = C\ln r + D$.

If you multiply the original equation by $r$ so it looks like $r^2u'' + ru' = 0$ it becomes an Euler equation for which a standard method would be to try $r^m$, if you have studied those.

3. Nov 26, 2013

### Simon Bridge

(1) The approach is to guess some sort of solution to the DE. It's not actually a substitution.
A good guess to try is one that makes every term in the DE contain the same function.

This particular guess is suggested by $\sum_n r^{n-1}\frac{d^n u}{dr^n}=0$ and works because $r^{n-1}r^{m-n}=r^{m-1}$ ... i.e. the same function for every term in the DE.
Basically - when you see a DE in that pattern, it is likely that u is a polynomial (but may not be).

[ref Euler equation]

(2) $$r\frac{d^2 u}{dr^2}+\frac{du}{dr}=0\\ u(r)=r^{m}\implies [m(m-1)+m]=0\\ \therefore m^2 = 0$$... which tells you that u=1 (or any constant) is a solution ... if somewhat trivial.

Try to think of something else. i.e. notice that the LHS in the DE is a product-rule expansion:
$$r\frac{d^2 u}{dr^2}+\frac{du}{dr}=\frac{d}{dr}\left( r\frac{du}{dr}\right)\\ \frac{d}{dr}\left( r\frac{du}{dr}\right)=0\implies\cdots$$

(3) it's called "trial and error" ;)

IIRC: In the case where a guess of the form u=rm for the above form of DE fails, the solution must therefore be of form u=a+b.ln|r|.

Last edited: Nov 26, 2013
4. Nov 26, 2013

### Nat3

In your solution, $C$ and $D$ are not constants, like they are in $u(r) = c_1 + c_2 lnr$, correct?

We haven't studied Euler equations as far as I can remember, which is probably why I'm so confused! Is $u(r) = c_1 + c_2 lnr$ a standard solution for a Euler equation?

I'm really confused how to jump from a characteristic equation of $m^2 = 0$ to $u(r) = c_1 + c_2 lnr$...

5. Nov 26, 2013

### Simon Bridge

C and D are constants just like yours - it's common to use caps for the arbitrary constants in DE solutions.
The jump is because it is always the case when m=0 so you just remember it like you jump from seeing six rows of seven objects to knowing you have 42 objects.

note: LaTeX can format many functions better if you put a backslash in front of the function abbreviation ... i.e. \ln|r| gets you $\ln|r|$. Also works for log, exp, sin, arcsin, etc.

6. Nov 26, 2013

### LCKurtz

$C$ and $D$ are constants.

For constant coefficient DE's you may recall that when the characteristic equation has a double root $r = \{c,c\}$ you have solutions of $e^{cx}$ and $xe^{cx}$. When you study Euler equations you will learn that when you have a double root of $r = \{c,c\}$ when you try $x^r$, you get solutions $x^c$ and $x^c\ln x$. Your text will probably show you why when you get to that section.