1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Someone please help me understand this.

  1. Mar 15, 2006 #1
    [tex] \frac{\partial^2 \phi}{\partial x^2} + \lambda \phi = 0 [/tex]

    We have to analyze eigenvalues.

    My prof. gave me this answer for [tex]\lambda>0[/tex] it is
    [tex]c*sin(\sqrt(\lambda))x+d*cos(\sqrt(\lambda))x[/tex]

    Shouldn't it be -[tex]\lambda[/tex] inside the squareroot? If not can someone explain how he got this?
     
    Last edited: Mar 15, 2006
  2. jcsd
  3. Mar 15, 2006 #2
    The problem you'er presenting is homogenic.
    sin and cos are natural solutions for this kind of equations.
    pay attention that if you use diff on sin twice, you get -sin.
    same goes for cos.
    c,d = constants
    you need 2 inputs to get the private solution. otherwisw it's a general solution.
     
  4. Mar 15, 2006 #3
    I know its a general solution but I don't understand how he got the positive [tex]\lambda[/tex] inside the squareroot.
     
  5. Mar 15, 2006 #4
    Looks right to me.. for lambda > 0, the eigenvalue is imaginary.

    Yes, it is [tex]\sqrt(\lambda)[/tex] or [tex]sqrt(/lambda))i[/tex]

    should be square root of neg lambda in the first.. give me a while to get latex right..
     
    Last edited: Mar 15, 2006
  6. Mar 15, 2006 #5

    CarlB

    User Avatar
    Science Advisor
    Homework Helper

    Your prof was right.

    If you put a minus sign in the square root, you would have to change the sin and cosine to exponential functions. I.e. [tex]C \exp(\sqrt{-\lambda}x)+D\exp(\sqrt{-\lambda}x).[/tex]

    Carl
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?