# No solution to harmonic oscillator

1. Oct 23, 2013

### Dustinsfl

1. The problem statement, all variables and given/known data
Given $(\mathcal{L} + k^2)y = \phi(x)$ with homogeneous boundary conditions $y(0) = y(\ell) = 0$ where
\begin{align}
y(x) & = \frac{2}{\ell}\sum_{n = 1}^{\infty} \frac{\sin(k_nx)}{k^2 - k_n^2},\\
\phi(x) & = \frac{2}{\ell}\sum_{n = 1}^{\infty}\sin(k_nx),\\
u_n(x) &= \sqrt{\frac{2}{\ell}}\sum_{n = 1}^{\infty}\frac{\sin(k_nx)}{k^2 - k_n^2},
\end{align}
$\mathcal{L} = \frac{d^2}{dx^2}$, and $k_n = \frac{n\pi}{\ell}$.
If $k = k_m$, there is no solution unless $\phi(x)$ is orthogonal to $u_m(x)$.

2. Relevant equations

3. The attempt at a solution

Why is this?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 23, 2013
2. Oct 23, 2013

### Simon Bridge

Well crunch the numbers - what happens if the condition is not satisfied?

3. Oct 23, 2013

### Dustinsfl

I don't understand what numbers I need to crunch.

4. Oct 23, 2013

### Dustinsfl

For k_m = k_n,
$$\sum_n\frac{0}{0}\sin(k_nx) = \sum_n\sin(k_nx)$$
otherwise, k_m\neq k_n
$$\sum_n\sin(k_nx) = \sum_n\sin(k_nx)$$
so the DE holds.
We have an indeterminant form on the LHS for the first case and equality for the 2nd. How does orthogonality play a role?

5. Oct 23, 2013

### Simon Bridge

What does "orthogonal" mean?

What does $k=k_m$ mean when you only have $k_n$ in your expressions?

6. Oct 23, 2013

### Dustinsfl

$k=k_m = \frac{m\pi}{\ell}$

$k$ is defined in the DE $(\mathcal{L}+k^2)y = \phi(x)$.

Orthogonal is when the inner product is 0

7. Oct 23, 2013

### Simon Bridge

And $k_n = \frac{n\pi}{l}$ and $k_{\text{carrot}}=\frac{\text{carrot}\pi}{l}$... I saw... but why the index "m" when index "n" is used in all the relations above?
What does that mean.......... use words not math?

Similarly for "orthogonal" - what is the relationship between $u$ and $\phi$ when their inner product is zero?
Why is that important to the solutions?

Also curious - how does $u_n(x)$ depend on $n$ when you have it as a sum over all $n$?

Last edited: Oct 23, 2013
8. Oct 23, 2013

### Dustinsfl

Because it says show that for the case when $k=k_m$ the DE has no solution unless $\phi(x)$ is orthogonal to $u_m(x)$. The $k_m$ replaces all $k$ so there will be both $k_m$ and $k_n$.

If there inner product is zero, one of them needs to be a cosine.

9. Oct 23, 2013

### Simon Bridge

I don't mean just say the math in words.
Try to get your head out of the math - this is a physical situation. What is the physical meaning?
I think you need to review the derivation from it's physical foundation. At the moment it seems to be an exercise in abstract mathematics for you.

What is special about $k_m$ is that it is a particular value of $k$, which is one of the $\{k_n\}$ ... which means that one of the terms in the sums, the one where n=m, has a division-by-zero problem. That gives a non-physical result - therefore: no solution.

But isn't that a problem? Shouldn't there be some solutions to the situation - after all, you can physically set one up and watch it work away without blowing up or anything? Either the premise the equations were built on is just wrong or there's something else like............

........ what if that term where n=m did not exist for some reason? Something like that? Then the problem would never come up.

What roles do $\phi$ and $u_n$ play in the solution?
In what situation would you normally care about their orthogonality?

10. Oct 25, 2013

### vela

Staff Emeritus
Where did your expressions for $u_n(x)$, $\phi(x)$, and $y(x)$ come from? As Simon noted, your expression for $u_n(x)$ doesn't make sense. You probably meant, as you wrote in another thread, that $u_n(x) = \sqrt{2/l} \sin k_n x$. The expression for $\phi(x)$ is clearly not orthogonal to $u_m(x)$, so I don't see how it relates to your question.

Go back to the beginning. You're trying to solve the differential equation $(\mathcal{L}+\lambda)y(x) = \phi(x)$ where $\mathcal{L}$ is some Hermitian operator. You have solutions $u_n(x)$ to the homogeneous equation $(\mathcal{L}+\lambda)y(x)=0$ that satisfy the given boundary conditions. In the particular case of the linear oscillator, where $\mathcal{L} = d^2/dx^2$ and the boundary conditions are $y(0)=y(l)=0$, you find that $u_n(x) = \sqrt{2/l}\sin k_n x$ where $\lambda_n=k_n^2$ and $k_n = n\pi/l$.

One important property of Hermitian operators is that the eigenfunctions $\{u_n\}$ of such an operator form a complete set. This means you can express $y$ as a linear combination of the eigenfunctions, i.e.,
$$y(x) = \sum_n a_n u_n(x).$$ Try doing that for $\phi(x)$ as well and plugging the two expansions into the differential equation with $k=k_m$.