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About an equation with eigenvalues

  1. Mar 16, 2009 #1
    Let's say I have the equation p(t)f''(t)=Kf(t) with p(t) a known periodical function, K an unknown constant and f(t) the unknown function.

    This is an eigenvalues problem that once solved gives a set of K={k1, k2,...} eigenvalues.

    I get these eigenvalues and they coincide with the ones obtained by others so I got them right.

    Question
    What if I try to solve the equation with a K that does not belong to the set of eigenvalues?
    I have the initial conditions: f(0) and f'(0), I choose a K which is not an eigenvalue and I try to solve numerically the equation (using MATLAB):

    What happens with f(t)? It is clear that I will get a solution. What is the difference between this solution with a forbidden K and a solution with an allowed K?
     
  2. jcsd
  3. Mar 17, 2009 #2

    HallsofIvy

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    The problem as given clearly has the "trivial solution", f(x)= 0 for all x, as solution. If k is not an eigenvalue, then that will be the only solution.

    The equation Lv= Kv always has the "trivial solution" v= 0.

    K is an eigenvalue of operator L if and only if there is a non-trivial solution to Lv= Kv. That is the definition of "eigenvalue".

    Notice that this is an "existence and uniqueness" question. There always exists the trivial solution. K is an eigenvalue if that solution is not unique.
     
  4. Mar 17, 2009 #3
    For instance, I have the equation:
    f''(t)+M(6+5*sin(2*pi*t))f(t)=0 which is the same as the equation p(t)f''(t)=Kf(t); [p=1/(6+5*sin(2*pi*t)), M=-K]
    initian conditions
    f(0)=1, f'(0)=0.5.

    With M=1.579 which is not an eigenvalue I get (solving the equation with "odesolve" from Mathcad) a f(t) that is sinusoidal but exponentially growing.

    It looks like f(t) exists despite the fact that I solved an equation for a forbidden value of its eigenvalues, M.
     
    Last edited: Mar 17, 2009
  5. Mar 17, 2009 #4

    HallsofIvy

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    This is not an eigenvalue problem. The "trivial solution", f(x)= 0, does not satisfy f(0)= 1 or f'(0)= 0.5.
     
  6. Mar 17, 2009 #5
    An equation of this form: f''(x)+M(6+5*sin(2*pi*x))f(x)=0 appears in many works.

    What I noticed they do, to deal with it, is they suppose that f(x) can be written as a Fourier series (no initial conditions or other restrictions are imposed). They truncate the Fourier sum to 20-30 coefficients, they transform the equation in a matrix equation, calculate the eigenvalues of the matrix eq. and from here they get a set of M=m1, m2, .... eigenvalues.

    Question: What exactly does it mean? That they are looking only for solutions that have a Fourier representation (with a finite number of coefficients) and the eigenvalues they get correspond to the existence of these solutions?

    After getting the set of eigenvalues they start to consider that f(x) has various initial conditions f(0) and f'(0). For each eigenvalues belonging to the set M=m1, m2, .... they get (using "odesolve" from Mathcad) sinusoidal solutions that do not grow in time, they are oscillating no matter what the initial conditions are (excepting the case when these conditions are zero).

    For M outside the set m1, m2, ... (not an eigenvalue of the initial problem) they just say that this is a forbidden domain and do not approach the case any further.

    The final goal for them is to identify when this equation (which describe a physical system) has as solutions oscillations. They choose an M=m1 or M=m2 etc., and after that they work with their physical system in the oscillatory regime.

    In my case M is a parameter that can be varied by hand and in order to pass it from m1 to m2 I have to cross non eigenvalues which generate exponentially growing f(x) (at least these are the solutions I get in Mathcad with "odesolve") compromising the device as long as f(x) is a physical quantity, that normally can not grow indefinitely.

    So my final Question: If I use the method with the Fourier decomposition (in series) can I be sure that for an M that is not an eigenvalue of the equation, no f(x) that start to -infinitum and stop to +infinitum and in the same time have a Fourier decomposition exists?
     
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