# 2nd order LODE System

1. Jun 13, 2008

### foxjwill

1. The problem statement, all variables and given/known data
Solve the following system for $$\mathbf{r}(t)$$:

$$\frac{d^2\mathbf{r}}{dt^2}=-\frac{k}{m}\mathbf{r}.$$​

2. Relevant equations

3. The attempt at a solution
Now, I know how to solve for the magnitude of r (in fact, since it's the equation for the simple harmonic motion of a spring obeying hooke's law, I have it memorized), but I'd like to be able to solve for the component form. Here's what I've tried so far:

I start by guessing that $$\mathbf{r}$$ is in the form $$e^{\lambda t}\mathbf{u}$$, where $$\lambda$$ is an eigenvalue and $$\mathbf{u}$$ is an eigenvector. Plugging in, I have

$$\left(\lambda^2 + \frac{k}{m}\right) e^{\lambda t}\mathbf{u} = \matbf{0}.$$​

Since exp can never be 0, and it would be meaningless (I think) at this point to have u be 0, we can solve what's left for $$\lambda$$ and get
$$\lambda = \pm i \sqrt{k \over m}.$$​
I know that the answer should look something like

$$\mathbf{r} = \mathbf{u}_1\cos{\sqrt{k \over m}}+\mathbf{u}_2\sin{\sqrt{k \over m}},$$​

but I'm not sure how to find the eigenvectors (u1 and u2) here. Any ideas?

2. Jun 13, 2008

### HallsofIvy

Staff Emeritus
I don't know what "eigenvectors" you are talking about. In fact, I see no reason to solve this as a vector problem at all. The components of r must satisfy exactly the same equation: x'= -(k/m)x, y'= -(k/m)y, z'= -(k/m)z (assuming r is in R3.)

3. Jun 13, 2008

### fantispug

As HallsofIvy stated (and in fact what you essentially did) is solve the exact same equation in the three components: your solutions are precisely of the form
$$x = a_1 \cos{\sqrt{\frac{k}{m}}} + a_2 \sin{{\sqrt{\frac{k}{m}}}}$$
and similarly for y and z (with different constants).

Now you should note that u, as you used it, is NOT actually a eigenvector of the system like you stated and so lambda can not be its eigenvector; for if it were:
$$\left(\frac{d^2}{{dt}^2}+\frac{k}{m}\right)\vec{u}=\lambda \vec{u}$$
so
$$\left(\frac{d^2}{{dt}^2}+\frac{k}{m}\right) e^{\lambda t} \vec{u} = \lambda^2 e^{\lambda t} \vec{u} + 2 \lambda e^{\lambda t} \frac{d\vec{u}}{dt}+ e^{\lambda t} \vec {u}$$
(ick, I suggest you work through it explicitly).

The concept you were really looking for here is constant. u and lambda are constants. Then everything you said makes sense, and you can see that u1 and u2 are determined by boundary conditions (just like in the one variable case).