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Laplace Cylindrical Coordinates (Separation of variables)

  1. Mar 7, 2008 #1
    Hello,

    The following equation:

    [tex]\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+ \frac{\partial^2 u}{\partial z^2} = 0[/tex]

    is solved by separation of variables assuming a solution of the form:

    [tex]u=R(r)Z(z)[/tex]

    In other cases the assumed solution is of the form:

    [tex]u=\psi_{0}+\psi_{1}z+R(r)Z(z)[/tex]

    Could anyone tell me or give me a hint where do the [tex]\psi[/tex]s come from?

    Best Regards
    Phioder
     
  2. jcsd
  3. Mar 8, 2008 #2
    Hello phioder, it seems you are still working on the fourier series problems. The issue with the [tex]\psi[/tex] numbers is not so difficult to understand. It is related to non-homogeneus boundary conditions.This means boundary conditions which are not equal to 0. To see this, let's assume your problem is defined as:

    [tex]\frac{\partial^2 u}{\partial r^2}+ \frac{1}{r} \cdot \frac{\partial u}{\partial r}+ \frac{\partial^2 u}{\partial z^2} = 0[/tex]

    for a cylinder with radius R0 and height L, thus:

    [tex]0 \leq r \leq R_0[/tex]
    [tex]0 \leq z \leq L[/tex]

    subject to the boundary conditions:

    [tex]u(r,0)=T_1[/tex]
    [tex]u(r,L)=T_2[/tex]
    [tex]u(R_0,z)=f(z)[/tex]

    Now this problem can't be solved in the standard way, but there is a workaround if you consider the following reasoning. Assume that the problem can be split up in two parts, one with he boundary conditions at the bottom and the top and one with the boundary on the convex round area and 0 at the top and bottom. The first one is independent of r as can be seen from the boundary conditions. So it is only a function of z. The other one is a standard problem. Now mathematically this is assuming that you can write the solution as:

    [tex]u(r,z)=v(z)+w(r,z)[/tex]

    With v(z) the solution to the first problem and w(r,z) the solution to the second one. Putting this into the differential equation, you get:

    [tex]\frac{\partial^2 w}{\partial r^2}+ \frac{1}{r} \cdot \frac{\partial w}{\partial r}+
    \frac{\partial^2 v}{\partial z^2}+ \frac{\partial^2 w}{\partial z^2} = 0[/tex]

    Because v and w are independent of each other you have the following two resulting equations:

    [tex]\frac{\partial^2 v}{\partial z^2}= 0[/tex]

    [tex]\frac{\partial^2 w}{\partial r^2}+ \frac{1}{r} \cdot \frac{\partial w}{\partial r}+
    \frac{\partial^2 w}{\partial z^2} = 0[/tex]

    The first one for the function v and the second one for the function w. The one for v can be written as an ordinary differential equation because it is only depending on z, thus:

    [tex]\frac{d^2 v}{d z^2}= 0[/tex]

    The boundary conditions are now transformed to become for the first equation:

    [tex]v(0)=T_1[/tex]
    [tex]v(L)=T_2[/tex]

    and for the second these are:

    [tex]w(r,0)=u(r,0)-T_1=T_1-T_1=0[/tex]
    [tex]w(r,L)=u(r,L)-T_2=T_2-T_2=0[/tex]
    [tex]w(R_0,z)=u(R_0,z)-v(z)=f(z)-v(z)[/tex]

    So the second equation has again homogeneus boundary conditions and a transformed one for the round area. If you solve the first one you will end up with:
     
  4. Mar 8, 2008 #3
    There seems to be a database error for which I wasn't able to post everything, so here's part 2:

    [tex]v(z)=A\cdot z + B[/tex]

    Applying the boundary conditions gives you:

    [tex]v(z)=\frac{T_2-T_1}{L} \cdot z + T_1[/tex]

    The last boundary condition for the second equation is therefore:

    [tex]w(R_0,z)=f(z)-\left(\frac{T_2-T_1}{L} \cdot z + T_1\right)[/tex]

    And this equation can be solved using the standard technique. The meaning of [tex]\psi_0[/tex] and [tex]\psi_1[/tex] is now clear. There are:

    [tex]\psi_1=\frac{T_2-T_1}{L} \cdot z[/tex]
    [tex]\psi_0=T_1[/tex]

    In case you have some spare time on this, the solution of the above problem in case the function f(z)=u0, a constant, you obtain:

    [tex]u(r,z)=\frac{T_2-T_1}{L}+T_1+\sum_{n=1}^{\infty} \left( \frac{4}{\pi(2n-1)}\cdot \left(u_0-\frac{T_2-T_1}{2}\right) \cdot \frac{I_0\left(\frac{(2n-1)\pi r}{L}\right)} {I_0\left(\frac{(2n-1)\pi R_0}{L}\right)} \cdot sin\left(\frac{(2n-1)\pi z}{L}\right)\right)+[/tex]

    [tex]\sum_{n=1}^{\infty} \left( \frac{T_2-T_1}{n\pi}\cdot \frac{I_0\left(\frac{2n\pi r}{L}\right)} {I_0\left(\frac{2n\pi R_0}{L}\right)} \cdot sin\left(\frac{2n\pi z}{L}\right)\right)[/tex]

    In case now the temperatures at the top and bottom are 0, you end up with the solution of problem 6.101 given in "Fourier Analysis" by Murray R. Spiegel.

    Hope this helps.
     
    Last edited: Mar 8, 2008
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