# Thread: Calculating the boundary conditions of a differential equation

1. ## 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.

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

Originally Posted by mathstudent1992
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.

Is the small z in z Z''(z) correct?

3. ## 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 ( $zZ''+Z'+v^2Z=0$)

4. ## 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)

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

Originally Posted by romsek
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.