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Help needed in Boundary V.P

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data

    solve the BVP: y'' -2y' + 2y = 0 ... (1)
    reduction
    y(0) = 1
    y(pi) = 1

    2. Relevant equations

    y= emx

    3. The attempt at a solution

    substitution y=emx & its derivatives in (1) we get:

    m2 -2m +2 = 0 = (1+i)(1-i)

    then y = e(c1 sinx + c2 cosx)

    using the conditions given we have:

    y(0) = 1 = c2 cosx implies: c2 = 1/e
    &
    y(pi) = 1 = -ec2 implies: c2 = -1/e

    we thus have that c2 = 1/e = -1/e which is impossible & thus the BVP has apparently no solution.

    Is my solution correct?
    Note that im taking a course in Ordinary D.E.'s

    thx in advance
     
  2. jcsd
  3. Dec 12, 2008 #2


    I assume you mean that this factors as m={(1+i),(1-i)}, and not that 0=(1+i)(1-i), which doesn't make sense.

    now you have solutions for m, which you can plug back into your assumption that [tex]y=e^{mx}[/tex], i believe what you're missing is that you are factoring that solution for m incorrectly in order to obtain the y's

    ~Lyuokdea
     
  4. Dec 13, 2008 #3
    actually yes. what i meant is that m1 = 1+i & m2 = 1-i

    moreover, the general solution when the roots are complex is of the form:

    y = e[tex]\alpha[/tex] { c1 sin ([tex]\beta[/tex] x) + c2 cos([tex]\beta[/tex] x) }

    with [tex]\alpha[/tex] = 1 & [tex]\beta[/tex] = 1 in this exercise.

    so the solution becomes as previously written.
     
  5. Dec 13, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that is exactly what happens. This is an important difference between "initial value problems" and "boundary value problems". We have an "existance and uniqueness theorem" that says that as long as the differential equation is nice there exist a unique solution no matter what the initial conditions are.

    With boundary value problems it is not so simple. For very simple differential equations you can have boundary conditions for which there is no solution or boundary value conditions for which there are an infinite number of solutions.

    A simpler example is y"+ y= 0 with y(0)= 0, y(pi)= 1 which has no solution or y"+ y= 0 with y(0)= 0, y(pi)= 0 which has an infinite number of solutions.

    Think of it as computing the trajectory of a bullet fired from a gun (which would give a fairly simple differential equation). Setting the gun at a specific point and aiming it in a given direction is an example of an "initial value" problem. The bullet will follow a single trajectory whether we can calculate it or not. Setting the gun at a specific point and requiring that the bullet hit a specific point on the target is an example of a "boundary value" problem. It may happen that the bullet can't hit that point (the target is beyond its range) or that there may be two different trajectories that will hit it (one above 45 degrees and the other below).
     
  6. Dec 13, 2008 #5
    so this BVP i have with those conditions has no solution.
     
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