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Math Help - Calculating the boundary conditions of a differential equation

  1. #1
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    Calculating the boundary conditions of a differential equation

    Hi, I am doing a problem and need help... I am really confused :/

    Well, the problem says

    "Show that the boundary conditions for zZ''(z)+Z'(z)+v^2Z(z)=0 are |Z(0)|<\infty and Z(L)=0
    (we know that v>0)."

    Z(z) comes from y(z,t)=Z(z)T(t). This problem is about the dynamic behaviour of a hanging cable, so L is the length of the cable and when z=L it means that z is at the highest point where the cable is hanging from, and z=0 is the end point, so |Z(0)| \le L < \infty. However, that's something that the problem tells us, so I cannot just say it is like that, I need to prove it.

    So well, my first thought was to solve the differential equation, but I got to this point

    \lambda=\frac{-1 \pm \sqrt{1-4zv^2}}{2z}

    I know that z cannot be zero, so Z(0) would not be a solution, but that does not prove why |Z(0)|<\infty. So I guess I shouldn't be solving the differential equation.

    Thanks a lot in advance for any help you can give me!

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  2. #2
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    Re: Calculating the boundary conditions of a differential equation

    Quote Originally Posted by mathstudent1992 View Post
    Hi, I am doing a problem and need help... I am really confused :/

    Well, the problem says

    "Show that the boundary conditions for zZ''(z)+Z'(z)+v^2Z(z)=0 are |Z(0)|<\infty and Z(L)=0
    (we know that v>0)."

    Z(z) comes from y(z,t)=Z(z)T(t). This problem is about the dynamic behaviour of a hanging cable, so L is the length of the cable and when z=L it means that z is at the highest point where the cable is hanging from, and z=0 is the end point, so |Z(0)| \le L < \infty. However, that's something that the problem tells us, so I cannot just say it is like that, I need to prove it.

    So well, my first thought was to solve the differential equation, but I got to this point

    \lambda=\frac{-1 \pm \sqrt{1-4zv^2}}{2z}

    I know that z cannot be zero, so Z(0) would not be a solution, but that does not prove why |Z(0)|<\infty. So I guess I shouldn't be solving the differential equation.

    Thanks a lot in advance for any help you can give me!

    Is the small z in z Z''(z) correct?
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  3. #3
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    Re: Calculating the boundary conditions of a differential equation

    Thanks for your reply! Yup, it is z. It is problem 1.B in this PDF file https://drive.google.com/file/d/0B-O...dwc2lsLXF0TDg/

    I know it makes sense, but I don't have a clue how to actually prove it using euqation 1.A.ii ( zZ''+Z'+v^2Z=0)
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    Re: Calculating the boundary conditions of a differential equation

    You don't have to actually prove it, as it constitutes a physical restraint
    (mentioned in the book, just above eq. 10)
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    Re: Calculating the boundary conditions of a differential equation

    Quote Originally Posted by romsek View Post
    Is the small z in z Z''(z) correct?
    Ok, the boundary condition on Z(0) comes directly from the boundary condition on y(0,t).

    |y(0,t)| < infinity because any displacement at the end of the cable cannot exceed the length of the cable.

    y(z,t) = Z(z)T(t), so |y(0,t)| = Z(0)T(t) < infinity and thus Z(0) < infinity.

    You cannot use the approach to solving the diff eq that you have taken. z is a variable, not a constant, so you can't just solve the characteristic equation. This is a non-linear diff eq. If you look at your problem set you'll see it's solved by a combination of Bessel functions so clearly this isn't a 2nd degree linear diff eq with constant coefficients. Those have complex exponentials as solutions.

    I don't think you are really expected to solve this diff eq. Just restrict the bessel functions solution based on the boundary and initial conditions.
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