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2nd order LODE System

  1. Jun 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the following system for [tex]\mathbf{r}(t)[/tex]:


    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 [tex]\mathbf{r}[/tex] is in the form [tex]e^{\lambda t}\mathbf{u}[/tex], where [tex]\lambda[/tex] is an eigenvalue and [tex]\mathbf{u}[/tex] is an eigenvector. Plugging in, I have

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

    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 [tex]\lambda[/tex] and get
    [tex]\lambda = \pm i \sqrt{k \over m}.[/tex]​
    I know that the answer should look something like

    [tex]\mathbf{r} = \mathbf{u}_1\cos{\sqrt{k \over m}}+\mathbf{u}_2\sin{\sqrt{k \over m}},[/tex]​

    but I'm not sure how to find the eigenvectors (u1 and u2) here. Any ideas?
  2. jcsd
  3. Jun 13, 2008 #2


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    Staff Emeritus
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    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.)
  4. Jun 13, 2008 #3
    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
    [tex]x = a_1 \cos{\sqrt{\frac{k}{m}}} + a_2 \sin{{\sqrt{\frac{k}{m}}}}[/tex]
    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:
    [tex]\left(\frac{d^2}{{dt}^2}+\frac{k}{m}\right)\vec{u}=\lambda \vec{u}[/tex]
    [tex]\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}[/tex]
    (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).
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