# Solution of 2nd order ODE

Discussion in 'Math' started by boks, Nov 10, 2008.

1. ### boks Thread Starter Active Member

Oct 10, 2008
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0
Can someone explain to me why the solution of $\frac{d^{2}\Phi (\phi)}{d\phi^{2}} = -m_{l}^{2}$ is $\Phi = e^{im_{l}\phi}$?

2. ### steveb Senior Member

Jul 3, 2008
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469
You probably know that the derivative of exp(c*x) with respect to x is c*exp(c*x).

You also need to know that i=sqrt(-1). You may know this, or perhaps you usually use j=sqrt(-1). They tend to use i in physics, and j in electrical engineering.

Anyway, take the derivative of exp(i*ml*phi) with respect to phi twice and remember that i squared is minus one.

3. ### Ratch New Member

Mar 20, 2007
1,068
4
boks,

Actually the solution is Phi = -(1/12)*m^4 + C1*m + C2 , where C1 and C2 are arbitary constants. How did you get a complex term for a solution?

Ratch

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
Ratch, you misread the question. The derivatives are with respect to $\phi$, not with respect to $m_l$. You may have been thrown off by the typo in the equation.

That must have been a good vacation you took!

Last edited: Nov 17, 2008
5. ### Ratch New Member

Mar 20, 2007
1,068
4
steveb,

So what is ml, a constant? And what is the correct equation? I still don't see a sinusoid in the solution.

Yes, my wife and I went to Hot Springs, Arkansas.

Ratch

6. ### steveb Senior Member

Jul 3, 2008
2,433
469
This equation shows up in the quantum mechanical solution for the hydrogen atom, and $m_l$ is the magnetic quantum number. There is a typo in the way boks wrote it, but it was clear from the context what he meant. One issue is the proper amplitude for the solution. There is a customary way to normalize the solution, but I assumed that was not the sticking point.

Last edited: Nov 18, 2008
7. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Based on the problem that was posted, Ratch's solution was a lot closer than the one the OP provided.
For the given problem, I get
$\Phi = \frac{-m^2\phi ^2}{2} + C\phi + D$

SteveB, you mentioned that boks had a typo in the equation he posted. My guess is that the correct DE was missing a factor of $\Phi$ on the right side, like so:
$\frac{d^2 \Phi}{d\phi ^2 } = -m^2 \Phi$

8. ### steveb Senior Member

Jul 3, 2008
2,433
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Well, I would say your answer is the correct answer based on the exact equation posted. It's a matter of judgement whether ratch or the OP is closer. Technically they are both wrong.

Your corrected equation with the $\Phi$ put in on the right hand side is correct, of course. I guess it's good to point this out for the benefit of anyone that is not familiar with the context of the quantum mechanical equation. When I first answered the question I really didn't even notice. It's like one of those sentences that has has a repeated word, and your mind just doesn't even notice it.

9. ### Ratch New Member

Mar 20, 2007
1,068
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Mark44,

If d2(Phi)/phi = -m^2*Phi , then the solution is Phi = A*cos(m*phi) + (B/m)*sin(m*phi)

Ratch

10. ### Mark44 Well-Known Member

Nov 26, 2007
626
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Nope, it's not. Two can play at this pedantic game.
The solution is
$\Phi = A cos(m\phi) + B sin(m\phi)$
To show that this function is the general solution, take the first and second aking derivatives.
$\frac{d}{d\phi}\Phi = -Am sin(m\phi) + Bm cos(m\phi)$
$\frac{d^2}{d\phi ^2}\Phi = -Am^2 cos(m\phi) - Bm^2 cos(m\phi)$
$= -m^2(A cos(m\phi) + B sin(m\phi)) = -m^2\Phi$

In addition, the linear combination of cosine and sine of the solution can also be written as a linear combination of exponential functions (of $im\phi$), similar to what boks originally posted.

11. ### Ratch New Member

Mar 20, 2007
1,068
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Mark44,

Oh yes it is. Our solutions are equivalent. Your "A" is what I designate as "B/m" and your "B" is what I designate as A. They are all constants, albeit different constants. Each solution's double derivative equates to -m^2*Phi . I will admit that your expression of the solution looks nicer.

Would you be so kind as to give me a quick example. Make A and B any nontrivial value, and show the equivalency of the exponential sums.

Ratch

Jul 3, 2008
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13. ### Mark44 Well-Known Member

Nov 26, 2007
626
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I'll take your word for the equivalence of our solutions. For an example,
here's what I was talking about, which was that sine and cosine of the same whatever could be expressed in terms of exponentials. Acos(x) + B sin(x) would be just A times the first thing on the right + B times the second thing on the right. See the wikipedia article on Euler's formula at http://en.wikipedia.org/wiki/Euler's_formula.
$cos x = \frac{e^{ix} + e^{-ix} }{2}$
$sin x = \frac{e^{ix} - e^{-ix} }{2i}$

Mar 20, 2007
1,068
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Mark44,

OK, thanks.

Ratch