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A Linear Partial Differential Eq.

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Find a vector field [itex]\vec{E}(s,phi,z)[/itex] (cylindrical coordinates) which satisfies:
    [tex]\nabla \times \vec{E} = 0[/tex]
    [tex]\nabla \cdot \vec{E} = q \delta^3(\vec{r}) + (\hat{z} \cdot \vec{E}) \frac{a}{1+az}[/tex]
    where q,a are constants, and [itex]\delta^3(\vec{r})[/itex] is the dirac delta function.

    2. Relevant equations
    I didn't think it was really worth copying these here, but for completeness sake I might as well.

    http://online.physics.uiuc.edu/courses/phys436/fall08/Del%20in%20cylindrical%20and%20spherical%20coordinates%20-%20Wikipedia,%20the...pdf [Broken]

    In Cylindrical Coordinates
    Where E is a vector field, and V is a scalar function.
    [tex]\nabla \cdot \vec{E} = \frac{1}{s} \frac{\partial}{\partial s}(s E_s) + \frac{1}{s}\frac{\partial}{\partial \phi}E_\phi + \frac{\partial}{\partial z} E_z[/tex]
    [tex]\nabla \times \vec{E} = \hat{s}(\frac{1}{s} \frac{\partial}{\partial \phi}E_z - \frac{\partial}{\partial z}E_\phi) +\hat{\phi}(\frac{\partial}{\partial z}E_s - \frac{\partial}{\partial s}E_z) + \hat{z} (\frac{1}{s}\frac{\partial}{\partial s}(s E_\phi) - \frac{1}{s}\frac{\partial}{\partial \phi} E_s)[/tex]
    [tex]\nabla V = \hat{s}\frac{\partial}{\partial s}V+\hat{\phi}\frac{1}{s}\frac{\partial}{\partial \phi}V + \hat{z}\frac{\partial}{\partial z}V[/tex]
    [tex]\nabla^2 V = \frac{1}{s}\frac{\partial}{\partial s}(s\frac{\partial}{\partial s}V) + \frac{1}{s^2}\frac{\partial^2}{\partial \phi^2}V + \frac{\partial^2}{\partial z^2}V[/tex]

    Also, the fact that:
    [tex]\nabla \times (\nabla V)) = 0[/tex]

    3. The attempt at a solution
    I only need to find a solution, and not neccessarily the most general solution. Thus due to cylindrical symmetry of the problem, I can have the vector field not depend on phi. And since the curl is zero (and again using cylindrical symmetry) there should be a solution with E_phi=0.

    The equations therefore reduce to:
    [tex]\frac{\partial}{\partial z} E_s = \frac{\partial}{\partial s} E_z[/tex]
    [tex]\frac{1}{s}\frac{\partial}{\partial s}(s E_s) + \frac{\partial}{\partial z}E_z = q \delta^3(\vec{r}) + E_z \frac{a}{1+az}[/tex]

    Alternatively, since the curl is zero, there exists a scalar function V such that [itex]\vec{E}=-\nabla V[/itex]. Then we just have:
    [tex]\nabla^2 V = - q \delta^3(\vec{r}) + (\frac{\partial}{\partial z} V) \frac{a}{1+az}[/tex]

    Maybe it is easier to solve this equation.
    These are all just linear partial differential equations, but I'm not sure where to go next.

    In the limit [itex]a \rightarrow 0[/itex], the equation is simple (just the field of a point charge in electromagnetism) so I know that the solution in that limit must become:
    [tex]V = \frac{q}{4\pi \sqrt{z^2 + s^2}}[/tex]
    [tex]E_s = \frac{q s}{4\pi (z^2 + s^2)^{3/2}}[/tex]
    [tex]E_\phi= 0[/tex]
    [tex]E_z = \frac{q z}{4\pi (z^2 + s^2)^{3/2}}[/tex]

    Maybe none of what I did was useful. I'm not sure where to go next.
    I have access to Mathematica. If someone knows how to make mathematica solve this so I can play with it to help me see what the answer should look like, that would be very useful too.
    Last edited by a moderator: May 3, 2017
  2. jcsd
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